MatheMatics-ii
B.Tech. I-Year II-Sem (Common to All Branches)
JNTU - Anantapur
Contents
Introduction to the Subject
Syllabus as per R15 Curriculum
List of Important Formulae
L.1
Latest exam question paper with soLutions
May-18 (R15)
QP.1
previous exam question papers with soLutions
May/June-17 (R15)
QP.1
May/June-16 (R15)
QP.1
miD-wise objective type & essay type questions with Key
MID-I (Units-1& 2)
M.1
MID-II (Units- 3, 4 & 5)
M.9
Model Question Papers with Solutions (As per the New External Exam Pattern)
Model Paper-I to III
MP.1
-
L.4
- QP.16
- QP.16
- QP.20
-
M.8
- M.17
- MP.6
unit-wise short & essay questions with soLutions
unit no.
topic no.
UNIt - I
Part-A
Part-B
1.1
1.2
1.3
1.4
1.5
unit name
topic name
LAPLACE tRANSFoRM
question nos.
Q1 -
page nos.
Q60
1.1 - 1.56
ShoRt QUEStIoNS wIth SoLUtIoNS
ESSAy QUEStIoNS wIth SoLUtIoNS
Laplace Transform of Standard Functions
Inverse Transform
First Shifting Theorem
Transforms of Derivatives and Integrals
Q1
Q13
Q13
Q19
Q25
Q29
-
Q12
Q60
Q18
Q24
Q28
Q32
1.1 - 1.4
1.5 - 1.56
1.5
1.9
1.20
1.23
Unit Step Function
Q33 -
Q34
1.26
1.6
Second Shifting Theorem
1.7
Dirac’s Delta Function
1.8
Convolution Theorem
1.9
Q39
1.27
Q40
1.29
Q41 -
Q44
1.30
Laplace Transform of Periodic Function
Q45 -
Q47
1.34
1.10
Differentiation and Integration of Transform
Q48 -
Q53
1.37
1.11
Application of Laplace Transforms to Ordinary Differential
Equations of First and Second Order
Q54 -
Q60
1.41
FoURIER SERIES
Q1 -
Q39
2.1 - 2.54
Part-A
ShoRt QUEStIoNS wIth SoLUtIoNS
Q1 -
Q10
2.1 - 2.4
Part-B
ESSAy QUEStIoNS wIth SoLUtIoNS
Q11 -
Q39
2.5 - 2.54
2.1
Determination of Fourier Coefficients
Q11 -
Q16
2.20
2.2
Fourier Series, Even and Odd Functions
Q17 -
Q22
2.26
2.3
Fourier Series in an Arbitrary Interval
Q23 -
Q27
2.36
2.4
Even and Odd Periodic Continuation Half-range
Fourier sine and cosine Expansions
Q28 -
Q36
2.50
Parseval’s Formula-Complex Form of Fourier Series
Q37 -
Q39
2.54
FoURIER tRANSFoRM
Q1 -
Q40
3.1 - 3.32
Part-A
ShoRt QUEStIoNS wIth SoLUtIoNS
Q1 -
Q17
3.1 - 3.5
Part-B
ESSAy QUEStIoNS wIth SoLUtIoNS
Q18 -
Q40
3.6 - 3.32
sine and cosine integrals
Q18 -
Q20
3.6
Fourier Transform – Fourier Sine and Cosine
Transforms – properties
Q21 -
Q31
3.8
3.3
Inverse Transforms
Q32 -
Q34
3.20
3.4
Finite Fourier Transforms
Q35 -
Q40
3.23
Q1 -
Q52
4.1 - 4.52
UNIt - II
2.5
UNIt - III
3.1
3.2
Q35 -
Fourier Integral Theorem (Only Statement) – Fourier
UNIt - IV PARtIAL DIFFERENtIAL EQUAtIoNS
Part-A
ShoRt QUEStIoNS wIth SoLUtIoNS
Q1 -
Q10
4.1 - 4.4
Part-B
ESSAy QUEStIoNS wIth SoLUtIoNS
Q11 -
Q52
4.5 - 4.52
Arbitrary Constants and Arbitrary Functions
Q11 -
Q26
4.5
4.2
Method of Separation of Variables
Q27 -
Q33
4.13
4.3
Solutions of One-Dimensional Wave Equation,
Q34 -
Q47
4.18
Q48 -
Q52
4.46
4.1
Formation of Partial Differential Equations by Elimination of
Heat Equation
4.4
Two Dimensional Laplace Equation Under Initial
and Boundary Conditions
UNIt - V
Z-tRANSFoRM
Q1 -
Q48
5.1 - 5.40
Part-A
ShoRt QUEStIoNS wIth SoLUtIoNS
Q1 -
Q10
5.1 - 5.7
Part-B
ESSAy QUEStIoNS wIth SoLUtIoNS
Q11 -
Q48
5.8 - 5.40
5.1
Introduction
Q11 -
Q14
5.8
5.2
Inverse Z-transform
Q15 -
Q22
5.11
5.3
Properties – Damping Rule – Shifting Rule
Q23 -
Q29
5.16
5.4
Initial and Final Value Theorems
Q30 -
Q33
5.21
5.5
Convolution Theorem
Q34 -
Q38
5.26
5.6
Solution of Difference Equation by Z-transforms
Q39 -
Q29
5.40
Syllabus
UNIT-I
Laplace transform of standard functions – Inverse transform – First shifting theorem, Transforms of derivatives
and integrals – Unit step function – Second shifting theorem – Dirac’s delta function – Convolution theorem –
Laplace transform of periodic function.
Differentiation and integration of transform – Application of Laplace transforms to ordinary differential equations
of first and second order.
UNIT-II
Fourier Series: Determination of Fourier coefficients – Fourier series – Even and odd functions – Fourier series
in an arbitrary interval – Even and odd periodic continuation – Half-range Fourier sine and cosine expansions –
Parseval’s formula – Complex form of Fourier series.
UNIT-III
Fourier integral theorem (only statement) – Fourier sine and cosine integrals. Fourier transform – Fourier sine
and cosine transforms – Properties – Inverse transforms – Finite Fourier transforms.
UNIT-IV
Formation of partial differential equations by elimination of arbitrary constants and arbitrary functions – Method
of separation of variables – Solutions of one dimensional wave equation, Heat equation and two - dimensional
Laplace’s equation under initial and boundary conditions.
UNIT-V
Z-transform – Inverse Z-transform – Properties – Damping rule – Shifting rule – Initial and final value theorems.
Convolution theorem – Solution of difference equations by Z-transforms.
INTRODUCTION TO THE SUBJECT
Mathematics is a language in which every symbol and every combination has a specific meaning which can be known
by applying logical rules. This is the most challenging and important subject at all the levels of student education.
The mathematical skills are not only required in various work environments ranging from services to manufacturing
but they are also a must if you want to continue your education.
At the basic level, mathematical education begins with learning skills such as identifying different shapes, measuring
space, counting things, using operations such as addition, subtraction, multiplication and division. It also introduces
time which measures duration of a complete day from sunrise to sunset and sunset to sunrise.
Based on this ground knowledge of mathematics, more advanced concepts and operations such as sets and real
numbers, polynomials, coordinate geometry, progressions, mensuration, trigonometry and statistics are developed.
However, you come across all these concepts in various levels in schooling (till 10th class). Indeed, these mathematical
courses are the required disciplines for entering college and university programs in science, biology, engineering,
economics and business.
The tools and techniques explained in the subject Mathematics-II are used in various engineering disciplines for
easy analysis and simplification of concepts.
The table below gives the complete idea about the number of questions that can be asked from each unit in external
examination along with their weightage. This will be helpful for the students to plan and score good marks in their
exams.
Unit No.
1
Unit Name
Laplace Transform
No. of
Weightage
Questions
of marks
Short
Essay
2
1
14
Description
This unit presents a brief study on Laplace
transform of standard functions, inverse
transform, transforms of derivatives and
integrals laplace transform of periodic
function. It also includes application of
Laplace transforms to ordinary differential
equations of first and second order.
2
Fourier Series
2
1
14
This unit deals with the determination of
Fourier coefficients, even and odd functions,
Fourier series in an arbitrary interval, even
and odd periodic continuation, half-range
Fourier sine and cosine expansions,
Parseval’s formula and complex form of
Fourier series.
3
Fourier Transform
2
1
14
This unit emphasizes the concepts of
Fourier integral theorem and Fourier sine and
cosine integrals. It also includes Fourier
transform, Fourier sine and cosine transform
properties, inverse transforms and finite
Fourier transforms.
4
Partial Differential
2
1
14
Equations
This unit introduces Formation of partial
differential equations by elimination of
arbitrary constants and arbitrary functions.
It also includes solutions of one dimensional
wave equation, heat equation and two
dimensional Laplace’s equation under
initial and boundary conditions.
5
Z-Transform
2
1
14
This unit motivates the study of Z-transform
and inverse Z-transform. It also includes
initial and final value theorems and
Convolution theorem.
M.1
MID-I: (Units - 1 & 2)
MID - I
(UNIT S-1 & 2)
OBJECTIVE TYPE &
ESSAY QUESTIONS WITH KEY
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M.2
MATHEMATICS-II [JNTU-ANANTAPUR]
MID-I (Objective Type & Essay Questions with Key)
Objective Type
I.
II.
Fill in the Blanks
1.
The Laplace transform of a unit step function is _____ .
2.
If I(s) =
3.
The final value theorem is
4.
If δ(t) is a unit impulse function, Laplace transform of
5.
The Fourier series expansion of an even function in (– c, c) has only _____ terms.
6.
The Fourier expansion of an odd function has only _____ terms.
7.
In the Fourier series expansion of f(x) = |sinx| in (– π, π), the value of bn = _____.
8.
If f(x) = x4 in (– 1, 1) then the Fourier coefficent, bn = _____.
9.
If f(x) is a periodic function with period 2T, then bn = _____.
10.
The condition for expansion of a function in series are known as _____ conditions.
6(s + 200)
, initial value of i(t) is, _____.
s(s+ 100)
Lt f (t ) = _____ .
t →∞
d2
δ(t) is _____.
d2
Multiple Choice
1.
2.
The inverse Laplace transform of
1
in the region δ < a is _________ .
s−a
(a) –e–atu(t)
(b)
eatu(t)
(c) eatu(–t)
(d)
–eatu(–t)
If X(s) =
2(s + 1)
s 2 2s + 5
, then x (0+) and x( ∞ ) are given by _________ and _________ respectively.
(a) 0, 0
(b)
2, 0
(c) 0, 2
(d)
2, 2
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M.3
MID-I: (Units - 1 & 2)
3.
4.
5.
6.
7.
The inverse Laplace transform of
1
(1–e–as) is _________.
s
(a) u(t) – u(t – a)
(b)
u(t) + u(t – a)
(c) u(t – a) – u(t)
(d)
u(t) – u(t + a)
The final value theorem in Laplace transform determines the _________value of the system output.
(a) Marginal state
(b)
Transient state
(c) Steady state
(d)
None of the above
Laplace transform of e–2t sin h 2t is _________.
(a)
2
(s + 4)
(b)
2
s(s+ 4)
(c)
2
(s − 4)
(d)
2
s(s− 4)
⎧⎪eβt , for t < 0
, its Laplace transform will exist if and only if _________ .
If f(t) = ⎨ αt
⎪⎩e for t > 0
(a) α > Re(s) < β
(b)
α < Re(s) < β
(c) α > Re(s) > β
(d)
α < Re(s) > β
The convolution of x(t) and y(t) is given by x(t)*y(t) = _________.
∞
(a)
∫
(b)
]
[
]
[
]
[
]
[
]
∫ x(t ) y(t – τ) dτ
t
x(t − τ) y(t) dτ
(d)
0
8.
[
0
t
∫
]
t
x (t ) y(t – τ) dτ
0
(c)
[
The advantage of Laplace transform is _________ .
(a)
The total solution is more systematic
(b)
The solution is in frequency domain only
(c)
Initial conditions are incorporated in very first step
(d)
All the above
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M.4
MATHEMATICS-II [JNTU-ANANTAPUR]
9.
10.
11.
A function f(x) defined for 0 < x < 1 can be extended to an odd periodic function in _________.
(a)
(1, 1) such that f(x) = f(–x)
(b)
(–1, 1) such that f(x) = f(–x)
(c)
(1, 1) such that f(x) = 0
(d)
None
In the Fourier series expansion of a function, the Fourier coefficient a0 represents _________
value of the function.
(a)
Average
(b)
Maximum
(c)
Minimum
(d)
Mean
The formulae for finding half range cosine series for the function f(x) in (0, 1) are _________.
(a)
(b)
a0 =
a0 =
2
l
l
∫
f ( x)dx, an =
0
l
1
2l
∫
0
f ( x) dx, an =
2
l
12.
0
l
0
l
∫
1
a0 =
l
l
∫
0
1
f ( x) dx , a n =
l
nπ x
dx
l
l
nπx
dx
l
∫ f ( x) cos
0
The half range sine series for 1 in (0, T) is _________.
(a)
1⎧
sin 3 x
⎫
+ sin 5 x + ...⎬
⎨sinx +
π⎩
3
⎭
(b)
sin 3 x
⎫
⎧
4 π ⎨sinx +
+ sin 5 x + ...⎬
3
⎭
⎩
(c)
4⎧
sin 3 x
⎫
+ sin 5 x + ...⎬
⎨sinx +
π⎩
3
⎭
(d)
sin 3 x
1 ⎧
⎫
+ sin 5 x + ...⎬
⎨sinx +
3
4π ⎩
⎭
[
]
[
]
[
]
nπx
.dx
l
∫
0
]
nπ x
dx
l
∫ f ( x) cos
a0 = 2l f ( x) dx, a n = 2l f ( x) cos
0
(d)
∫ f ( x) cos
1
2l
l
(c)
l
[
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M.5
MID-I: (Units - 1 & 2)
13.
14.
15.
If f(x) = x cos x in (–π, π), then b1 = _________.
(a) –2
(b)
–1
(c) 0
(d)
1
x<0
⎧− x,
If f(x) = ⎨
then f(x) is _________ function.
⎩ x, 0 < x < π
(a)
Odd
(b)
Even and odd
(c)
Even
(d)
None
The half range sine series for f(x) = ex in (0, π) is _________.
(a)
(b)
(c)
(d)
III.
9
2
2
9
2
9
9
2
∞
∑n
n =1
∞
∑n
n =1
∞
n
2
+1
n
2
+1
[1 + ( −1) n +1 e x ] sinx
[1 + ( −1) n +1 e x ] sinx
∑
n2 +1
[1 + ( −1) n +1 e x ]
n
∞
n2 +1
[1 + ( −1) n +1 e x ]
n
n =1
∑
n =1
Match the Following
(i)
1.
Fourier expansion of odd function has
[ ]
(a)
⎛ a 2 + p2 ⎞
2 log ⎜⎜ 2 2 ⎟⎟. 0 < b < a
⎝b +p ⎠
2.
Fourier expansion of even function has
[ ]
(b)
–d
Fc (S)
ds
3.
⎡ e – bt – e – at ⎤
Fc ⎢
⎥=
t
⎢⎣
⎥⎦
[ ]
(c)
– s2[F(U)]
4.
Fs[x f(x)]
[ ]
(d)
sine terms
5.
Fourier expansion of derivative U(x, t)
[ ]
(e)
cosine terms
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M.6
MATHEMATICS-II [JNTU-ANANTAPUR]
(ii)
1.
L[sin (at)]
[ ]
(a)
2.
L[eat sinh (bt)]
[ ]
(b)
3.
L[t cos(at)]
[ ]
(c)
4.
L[cos h (at)]
[ ]
(d)
5.
L[sin (at + b)]
[ ]
(e)
s
2
s – a2
a
2
s + a2
ssin(b) + acos(b)
s2 + a 2
s2 – a 2
(s 2 + a 2 ) 2
b
(s – a) 2 – b 2
KEY
I.
Fill in the Blanks
1.
1
s
2.
6A
3.
Lt s F(s)
s→ 0
4.
s2
5.
cosine
6.
sine
7.
0
8.
0
9.
1
T
10.
α+2T
∫
f(x) sin
α
nππ
. dx
T
Dirichlet’s
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MID-I: (Units - 1 & 2)
II.
M.7
Multiple Choice
1.
(d)
2.
(b)
3.
(a)
4.
(c)
5.
(b)
6.
(b)
7.
(b)
8.
(d)
9.
(b)
10. (d)
11. (a)
12. (c)
13. (d)
14. (a)
15. (b)
III.
Match the Following
(i)
1.
(d)
2.
(e)
3.
(a)
4.
(b)
5.
(c)
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MATHEMATICS-II [JNTU-ANANTAPUR]
(b)
1.
(b)
2.
(e)
3.
(d)
4.
(a)
5.
(c)
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M.9
MID-I: (Units - 1 & 2)
ESSAY QUESTIONS WITH KEY
Q1.
Prove that,
(a)
L{sin at} =
(b)
L{cos at} =
a
2
s + a2
s
2
s + a2
,s>a
(Refer Unit-I / Q15)
, s > a.
Q2.
Find L–1[(2s + 3)/(s3 – 6s2 + 11s – 6)].
Q3.
Prove that,
(i)
L[eat sinh bt] =
b
(s – a)2 – b2
(ii)
L[eat cosh bt] =
s–a
.
(s – a)2 – b2
(Refer Unit-I / Q20
(Refer Unit-I / Q26)
Q4.
Show that L{fn(t)} = sn f(s) – sn – 1f(0) – sn–2f' (0)..... fn– 1(0), where L{f (t)} = f (s).
Q5.
Find the Laplace transform of periodic function f(t) with period T,
where f(t) =
(Refer Unit-I / Q29)
4Et
T
4Et T
– E, 0 ≤ t ≤ = 3E –
, ≤ t ≤ T.
T
2
T 2
Q6.
Using Laplace transform, solve (D2 + 4D + 5)y = 5, given that y(0) = 0, y"(0 ) = 0.
Q7.
Obtain the Fourier series for the function, f(x) = x + x in [–π, π], deduce the value of
(Refer Unit-I / Q47)
(Refer Unit-I / Q54)
∞
2
1
∑n
2
.
(Refer Unit-II / Q12)
n=1
Q8.
⎧ -1
⎪⎪ 2 ( π + x) for - π ≤ x ≤ 0
.
Find the Fourier series in [–π, π] for the function f(x) = ⎨
⎪ 1 ( π - x) for 0 ≤ x ≤ π
⎪⎩ 2
(Refer Unit-II / Q16)
Q9.
Obtain the Fourier series for the function f(x) = |cos x| in (–π, π).
(Refer Unit-II / Q19)
Q10. Find the half-range sine series of f(x) = (x – 1)2 in the interval (0, 1).
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M.10
MATHEMATICS-II [JNTU-ANANTAPUR]
⎧1
⎪⎪ 4 - x ,
Q11. Find the half-range sine series for f(x) = ⎨
⎪x - 3 ,
⎪⎩
4
1
2
1
< x <1
2
.
(Refer Unit-II / Q29)
∞
⎡1
⎤
[f(x)]2 dx = l ⎢ a 02 +
(anna + bbn )⎥ provided the Fourier series for
⎢⎣ 2
⎥⎦
n=1
–l
l
Q12. Prove that
0<x<
∫
∑
f(x) converges uniformly in (– l, l).
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M.11
MID-II: (Units - 3, 4 & 5)
MID - II
(UNIT S-3, 4 & 5)
OBJECTIVE TYPE &
ESSAY QUESTIONS WITH KEY
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M.12
MATHEMATICS-II [JNTU-ANANTAPUR]
MID-II (Objective Type & Essay Questions with Key)
Objective Type
I.
Fill in the Blanks
1.
The Fourier sine integral representation of a function f(x) is _____ .
2.
Fourier transform is a _____ operation.
3.
The Fourier sine transform of
4.
Fourier cosine transform of f(t) is _____.
5.
If F{f(x)} = F(s), then F{f(x – a)} = _____ .
6.
A partial differential equation is said to be of _____ order if number of arbitrary constants equals to the number of
1
is _____.
x
independent variables.
7.
Elimination of two arbitrary functions gives a partial differential equation of order _____.
8.
A partial differential equation is said to be _____ if the dependent variable Z and its derivatives are of degree 1.
9.
The general form of a second-order P.D.E in the function ‘u’ of the two independent variables x, y is _____.
10.
The general solution of two-dimensional heat equation, u(x, y, t) = _____.
11.
∂ 2u
Two-dimensional wave equation is of the form, 2 = _____.
∂t
12.
In a two-dimensional wave equation, C2 is a ____ constant.
13.
In a vibrating rectangular membrane, change in time changes the _____ periodically.
14.
The function Unm(x, y, t) in a rectangular membrane is known as _____ harmonic of the rectangular drum.
15.
Z-transform is defined as _____.
16.
The Z-transform of δ(n) is, _____.
17.
Initial value theorem states that _____ .
18.
If x(n) ⇔ x(z) the z-transform of x(n – n0) is, _____.
19.
The ROC cannot contain any _____.
20.
The region of convergence of the z-transform of a unit step function is, _____.
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M.13
MID-II: (Units - 3, 4 & 5)
II.
Multiple Choice
1.
If Fourier transform of f(x) is F(s) then the inversion formula is _________.
∞
(a)
f(x) = 2 π
∫ F(s) e
− isx
. dx
(b)
f(x) =
−∞
∞
(c)
f(x) = π
∫ F(s) e
− isx
dx
(d)
f(x) =
−∞
2.
(a)
∫
− nπ l f(x) cos
0
nπ x
dx
l
(c)
− nπ
nπ x
. dx
f(x) cos
l
l
∫
∫ F(s) e
− isx
0
(d)
[
]
[
]
[
]
∫
0
⎛s⎞
2 F⎜ ⎟
⎝2⎠
(c)
1 ⎛s⎞
− F⎜ ⎟
2 ⎝2⎠
(d)
1 ⎛s⎞
F⎜ ⎟
2 ⎝2⎠
The Fourier cosine transform of e–x is _________.
s +1
]
nπ x
dx
l
nπ
nπ x
f(x) cos
dx
l
l
(b)
1
[
dx
−∞
∫
⎛s⎞
− 2 F⎜ ⎟
⎝2⎠
2
]
−∞
∞
nπ l f(x) cos
(a)
(c)
[
dx
If Fourier transform of f(x) = f(s) then Fourier cosine transform of e–x is _________.
(a) s2 + 1
6.
−isx
l
0
5.
∫ F(s) e
l
(b)
l
4.
1
π
]
∞
The Fourier sine transform of f(x) in the interval (0, 1) is.
l
3.
1
2π
[
(b)
s2 – 1
(d)
s −1
1
2
The period of a constant function is _____.
(a)
Defined
(b)
Not defined
(c)
Both
(d)
None
If F(λ) is the Fourier transform of f(x), then the Fourier transform of F(ax) is _________ .
(a)
⎡λ ⎤
aF ⎢ ⎥
⎣a ⎦
(b)
1 ⎡λ⎤
F
a ⎢⎣ a ⎥⎦
(c)
⎡λ⎤
F⎢ ⎥
⎣a ⎦
(d)
F[λ]
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
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M.14
7.
MATHEMATICS-II [JNTU-ANANTAPUR]
The inverse finite Fourier sine transform of Fs(n) is F(x) = _________ .
(a)
(c)
8.
∞
2
π
∑
π
2
∑ F (n) sin
Fs (n) cos
n =1
∞
s
n =1
nπ x
n
(b)
nπ x
n
(d)
∑ F (n) sin
π
2
∑ F (n) cos
s
n =1
∞
s
n =1
⎛
∂z ∂z ∂ 2 z ⎞
F⎜⎜ x , y, z, , , 2 ...⎟⎟ = 1
∂x ∂y ∂x ⎠
⎝
(c) F(x, y, z) = 1
]
[
]
[
]
[
]
[
]
[
]
nπ x
n
nπ x
n
Which among the following denotes a partial differential equation?
(a)
9.
∞
2
π
[
(b)
⎛ ∂z ∂z ∂ 2 z ⎞
F⎜⎜ , , 2 ...⎟⎟ = 0
⎝ ∂x ∂y ∂x ⎠
(d)
⎛
∂z ∂z ∂ 2 z ⎞
F⎜⎜ x, y, z, , , 2 ...⎟⎟ = 0
∂x ∂y ∂x ⎠
⎝
A partial differential equation is said to be of higher order if,
(a) Number of arbitrary constants = Number of independent variables
(b) Number of arbitrary constants > Number of independent variables
(c) Number of arbitrary constants < Number of independent variables
(d) Number of arbitrary constants = 0
10.
11.
12.
Obtain partial differential equation by eliminating arbitrary constants from z = ax2 + by2.
(a) z = x zx + y zy
(b)
z=x+y
(c) 2z = x zx + y zy
(d)
2z = x + y
What is the order of the P.D.E for z = a(x + y) + b(x – y) + abt + c?
(a) 1st order
(b)
2nd order
(c) 3rd order
(d)
4th order
A P.D.E is said to be _____, if each term contains the dependent variable or its derivative.
(a) Linear
(b)
Quasi-linear
(c) Non-linear
(d)
Homogeneous
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M.15
MID-II: (Units - 3, 4 & 5)
13.
14.
15.
16.
17.
18.
A P.D.E is said to be parabolic if,
(a) B2 – 4AC < 0
(b)
B2 – 4AC = 0
(c) B2 – 4AC > 0
(d)
B2 – AC = 0
∂ 2u ∂ 2u
+
= 0 is known as __________,
∂x 2 ∂y 2
(a) Two-dimensional Laplace equation
(b)
One-dimensional Laplace equation
(c) Laplace equation
(d)
Flowing heat equation
In a two-dimensional wave equation, C2 = _____.
(a) Tρ
(b)
T
ρ
(c) T + ρ
(d)
T–ρ
The region of convergence is required for evaluating _________ from X(z).
(a)
Inverse Z transform
(b)
Z transform
(c)
Both
(d)
None
If X1(n) ⇔ x1(z) and x2(n) ⇔ x2(z) then z-transform of x1(n) * x2(n) is _________.
(a)
x1(z) – x2(z)
(b)
x1 (z)
x 2 (z)
(c)
x1(z) + x2(z)
(d)
x1(z) x2(z)
The Z-tranform of a system is X(z) =
z
, if the ROC is | z | < 0.2 then x(n) is _________.
z − 0.2
(a)
(0.2)–n u(–n –1)
(b)
–(0.2)–n u(–n –1)
(c)
(0.2)n u(–n –1)
(d)
–(0.2)n u(–n –1)
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]
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[
]
[
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SIA GROUP
M.16
19.
MATHEMATICS-II [JNTU-ANANTAPUR]
The Z-transform of the function F(nt) = ant is _________.
(a) z – at
z
(c)
z− a t
20.
(b)
z− a
z
(d)
None
The ROC of nan.u(n) is _________ for | z | > | a |.
z
III.
(a)
(z − a) 2
(b)
(c)
az
z+ a
(d)
az
z− a
az
(z − a) 2
Match the Following
(i)
∞
1.
∫
xcosx– sinx
∫
xcosx– sinx
0
∞
2.
x
x
0
∞
3.
π –a
e
2
[ ]
(b)
π – ax
e
2
[ ]
(c)
–3π
16
sin s x ds
[ ]
(d)
–π
4
dx
[ ]
(e)
π
8
[ ]
(a)
Z = F1(y) + x F2(y) + x2F3(y)
3
dx
2
∫
∫a
0
∞
5.
(a)
cos
⎛ sinx ⎞
⎜
⎟ dx
⎝ 2x ⎠
0
∞
4.
x
dx
2
[ ]
3
s
2
+s
2
xsinax
∫ 1+ x
2
0
(ii)
1.
The solution of Uxy = 0 is
2.
The solution of
∂ 3z
= 0 is
∂x 3
[ ]
(b)
xp + yq = z
3.
The solution of
∂ 2z
∂ 2z
=
∂x 2
∂y 2
[ ]
(c)
u=
∫ f ( y)dy + φ(x)
4.
The partial differential equation
[ ]
(d)
u=
x3y
+ xf ( y) + φ( y)
6
[ ]
(e)
z = F1(y + x) + F2(y – x)
from z = zx + by + c is,
5.
Solution of Uxx = xy
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t
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M.17
MID-II: (Units - 3, 4 & 5)
(iii)
1.
(an)
[ ]
(a)
e1/z
2.
⎛ 1 ⎞
⎜
⎟
⎝ n +1 ⎠
[ ]
(b)
z
(z– a)
3.
⎛1⎞
⎜ ⎟
⎝ n! ⎠
[ ]
(c)
⎛ z– 1 ⎞
– z log⎜
⎟
⎝ z ⎠
4.
(cos x θ)
[ ]
(d)
5.
(sin x θ)
[ ]
(e)
z sinθ
2
z – 2 zcosθ + 1
z(z– cosθ)
z – 2 zcosθ + 1
2
KEY
I.
Fill in the Blanks
∞
1.
∞
∫
∫
2 π sinλi f(x) sinλ t dt .dλ
0
2.
Linear
3.
s2
2
0
α
4.
∫
Fc(s) = f(t) cosst dt
0
5.
eisa F(s)
6.
First
7.
8.
9.
2
Linear
∂ 2u
∂ 2u
⎛
∂ 2u
∂u ∂u ⎞
A(x, y). 2 + B(x, y).
+ C(x, y). 2 + f ⎜⎜ x, y , u, , ⎟⎟ = 0
∂
y
x
y
∂
∂
∂x ∂y ⎠
∂x
⎝
∞
10.
∞
∑∑ A
m =1 n =1
mn
⎛ mπ x
sin⎜
⎝ a
⎞ ⎛ nπ y ⎞
⎟. sin ⎜
⎟ .exp(–λ2 C2t)
mn
⎠ ⎝ b ⎠
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
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M.18
MATHEMATICS-II [JNTU-ANANTAPUR]
⎛ ∂ 2u
11.
C2 ⎜⎜
12.
Positive
13.
Amplitude
14.
(n, m)th
15.
X(z) =
⎝ ∂x
2
+
∂ 2u ⎞
⎟
∂y 2 ⎟⎠
α
∑ X(n) z
–n
n =0
II.
16.
1
17.
Lim X(z) = x(0)
18.
Z–n0. x(z)
19.
Poles
20.
|z|>1
z– ∞
Multiple Choice
1.
(b)
2.
(c)
3.
(d)
4.
(c)
5.
(b)
6.
(b)
7.
(b)
8.
(d)
9.
(b)
10. (c)
11. (a)
12. (d)
13. (b)
14. (a)
15. (b)
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MID-II: (Units - 3, 4 & 5)
M.19
16. (a)
17. (d)
18. (d)
19. (c)
20. (d)
III.
Match the Following
(i)
1.
(c)
2.
(b)
3.
(e)
4.
(d)
5.
(a)
1.
(c)
2.
(a)
3.
(e)
4.
(b)
5.
(d)
1.
(b)
2.
(c)
3.
(a)
4.
(d)
5.
(e)
(ii)
(iii)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
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M.20
MATHEMATICS-II [JNTU-ANANTAPUR]
ESSAY QUESTIONS WITH KEY
∞
Q1.
∫
Solve the integral equation f(x) cosαx dx= e–αx.
(Refer Unit-III / Q20)
0
∞
Q2.
Find the Fourier transforms of f(x) = e–|x| and deduce that
∫
0
Hence show that F(xe–|x|) = i
cos xt
1+ t 2
dt =
π –|x|
e .
2
4s
.
(1+ s 2 )2
(Refer Unit-III / Q23)
2
Q3.
–x
Find the Fourier cosine transform of e .
Q4.
Find the fourier sine transform of
Q5.
Find the finite cosine transform of, f(x) =
Q6.
Find the finite Fourier sine and cosine transform of f(x) = x (π – x) in 0 < x < π.
Q7.
Obtain the partial differential equation by eliminating the arbitrary constants
(a)
z = (x – a)2 + (y – b)2 + 1
(b)
z = ae – b
2
t
(Refer Unit-III / Q25)
1
.
x(x + a 2 )
(Refer Unit-III / Q26)
2
π
x2
– x+
, 0 < x < π.
3
2π
(Refer Unit-III / Q35)
(Refer Unit-III / Q39)
(Refer Unit-IV / Q12)
cos bx.
Q8.
Form the PDE by eliminating ‘f’ from f(x + y + z, x2 + y2 + z2) = 0.
(Refer Unit-IV / Q24)
Q9.
Solve by separation of variables 3ux + 2uy = 0 with u(x, 0) = 4e– x.
(Refer Unit-IV / Q29)
Q10. A tightly stretched string with fixed end points x = 0 and x = 1 is initially at rest in
its equilibrium position. If itisset on vibrating by giving each of its points a velocity
λx(1–x), find the displacement of the string at any distance x from end at any time ‘t’.
(Refer Unit-IV / Q35)
Q11. Find the deflection u(x, y, t) of the square membrane with a = b = 1 and c = 1,
if the initial velocity is zero and the initial deflection is f(x, y) = A sin πx sin 2πy.
(Refer Unit-IV / Q37)
Q12. A bar of length L is laterally insulated with its ends A and B kept at 0° and 100°
respectively until steady state condition is reached. The temperature at A is raised to 30°
and that at B is reduced to 80° simultaneously. Find the temperature in the rod at a later time.
(Refer Unit-IV / Q46)
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M.21
MID-II: (Units - 3, 4 & 5)
Q13. Find the inverse Z-transform of
3z 2 + z
.
(5z – 1)(5z – 2)
⎡
1 ⎤
when | z | > 5. Determine the region of convergence.
Q14. Find Z–1 ⎢
3⎥
⎣ (z – 5) ⎦
(Refer Unit-V / Q16)
(Refer Unit-V / Q17)
Q15. Find the linearity property of Z-transforms,
(i)
⎛ nπ π ⎞
+ ⎟
cos ⎜
⎝ 2
4⎠
(ii)
⎛ nπ ⎞
+ θ⎟ .
cosh ⎜
⎝ 2
⎠
Q16. If Z(un) =
2 z 2 + 4 z+12
(z-1)4
(Refer Unit-V / Q24)
find u2.
(Refer Unit-V / Q30)
z2
⎪⎧
⎪⎫
⎬ using convolution theorem.
Q17. Evaluate Z–1 ⎨
⎪⎩ (z – 1)(z – 3) ⎪⎭
(Refer Unit-V / Q34)
Q18. Solve the difference equation, using Z-transforms un +2 – un = 2n where u0 = 0, u1 = 1.
(Refer Unit-V / Q43)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
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QP.1
May/June-2017 (R15) Question Paper with Solutions
Code No. : 15A54201/R15
I B.Tech II Semester Regular Examinations
May / June - 2017
R15
Solutions
MATHEMATICS-ii
(Common to All)
Time: 3 Hours
Max. Marks: 70
PART-A
(Compulsory Questions)
1.
Answer the following: (10 × 02 = 20) Marks
(a)
Find L[t2.et.cos4t]. (Unit-I)
(b)
Define unit step function Laplace transform. (Unit-I)
(c)
If f(x) = x4 in (–1, 1) then find the Fourier coefficient of bn. (Unit-II)
(d)
What is Fourier even function (– π, π)? (Unit-II)
(e)
Write the Fourier sine transform of f(t). (Unit-III)
(f)
Find the value of Z(an cosnt). (Unit-V)
(g)
Find the general solution of uxx = xy. (Unit-IV)
(h)
Find the Z-transform of the sequence {x(n)} where x(n) is n.2n. (Unit-V)
(i)
(j)
JK 1 NO
OO . (Unit-V)
Find z–1 KK
L z – 3P
What do you mean by steady state and transient state? (Unit-IV)
PART-B
(Answer all five units, 5 × 10 = 50 Marks)
Unit-I
2.
(a)
(b)
JK
NO
1
OO . (Unit-I, Topic No. 1.8)
Apply convolution theorem for L–1 KKK 3 2
s ^s + 1h O
L
P
t
JK
N
KK –1 sint OOO
dtO . (Unit-I, Topic No. 1.10)
Evaluate L Ke
t
KK
OO
0
L
P
OR
#
JK π ON
d2 x
K O
2 + 9x = cos 2t, if x(0) = 1, x K 2 O = – 1. (Unit-I, Topic No. 1.11)
dt
L P
Unit-II
3.
Solve
4.
Find the Fourier series expansion of f(x) = 2x – x2 in (0, 3) and hence deduce that
(Unit-II, Topic No. 2.3)
5.
π
1
1
1
1
2 – 2 + 2 – 2 + ...... 3 = 12 .
1
2
3
4
OR
Z]
l
]] kx,
0# x #
1
1
1
2
]]
. Deduce the sum of the series 2 – 2 + 2 + ..... 3 .
Obtain half range cosine series for f(x) = []
l
]]
1
3
5
]] k ]l – xg , # x # l
2
\
(Unit-II, Topic No. 2.4)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
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QP.2
MATHEMATICS-II [JNTU-anantapur]
Unit-III
6.
(a)
Find the Fourier sine transform of e–|x|. (Unit-III, Topic No. 3.2)
(b)
Write the conditions of Parseval's identity for Fourier transforms. (Unit-III, Topic No. 3.2)
OR
7.
2
Verify convolution theorem for f(x) = g(x) = e –x . (Unit-III, Topic No. 3.2)
Unit-IV
8.
(a)
(b)
JK xy NO
Form the partial differential equation z = f KK OO by eliminating the arbitrary function. (Unit-IV, Topic No. 4.1)
L 2 P
2u
2u
=4
where u(x, 0) = 8e–3y. (Unit-IV, Topic No. 4.2)
Use the method of separation of variables, solve
2x
2y
OR
9.
Solve the Laplace equation
(Unit-IV, Topic No. 4.4)
JK n π x NO
2 u 2 u
K
O
2 +
2 = 0 subject to the conditions u(0, y) = u(i, y) = u(x, 0) = 0 and u(x, a) = sin K l O
2x
2y
P
L
2
2
Unit-V
10.
(a)
Find the z-transformation of sinnθ. (Unit-V, Topic No. 5.1)
(b)
If U(z) =
2z 2 + 5z + 14
, evaluate U2, U3 using initial value theorem. (Unit-V, Topic No. 5.4)
]z – 1g4
OR
11.
Solve the differential equation un + 2 – 2un + 1 + un = 3n + 5 using z-transforms. (Unit-V, Topic No. 5.6)
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May/June-2017 (R15) Question Paper with Solutions
QP.3
Solutions to May/June-2017, (R15), QP
PART-A
(10 × 2 = 20 Marks)
Q1.
(a)
Find L[t2.et.cos4t].
May/June-17, R(15), Q1(a)
Answer :
Given function is,
t2 et cos 4t
d2
7L 6et cos 4t@A
ds 2
s – 1 WVW
d 2 SR
W
1. 2 SSS
ds ]s – 1g2 + 4 2 W
T
XV
WW
s –1
d 2 SRS
W
2
2 S
S
ds s + 1 – 2s + 16 W
X
T
R
s – 1 NOVWW
d SS d JKK
OW
S
ds SS ds K s 2 – 2s + 17 OWW
L
PX
RTS
VW
d
d 2
S
2
d SS ^ s – 2s + 17h
^ s – 2s + 17h WWW
]s – 1g – ]s – 1g
ds
ds
WW
ds SSS
W
^ s 2 – 2s + 17h2
X
RTS 2
VW
d S ^ s – 2s + 17h ]1g – ]s – 1g]2s – 2g W
SS
W
WW
ds S
^ s 2 – 2s + 17h2
X
RTS 2
V
d S s – 2s + 17 – ^2s 2 – 2s – 2s + 2h WW
SS
W
WW
ds S
^ s 2 – 2s + 17h2
RT
VX
d SSS s 2 – 2s + 17 – 2s 2 + 2s + 2s – 2 WWW
W
ds SS
W
^ s 2 – 2s + 17h2
X
RST 2
VW
d SS – s + 2s + 15 WW
ds SS ^ s 2 – 2s + 17h2 WW
T
X
d
2 d
2
^ s – 2s + 17h . ^ – s 2 + 2s 15h – ^ – s 2 + 2s + 15h . ^ s 2 – 2s + 17h2
ds
ds
2
_^ s 2 – 2s + 17h2i
L[t2 et cos 4t] = ]– 1g2 .
=
=
=
=
=
=
=
=
=
=
=
^ s 2 – 2s + 17h2 . ]– 2s + 2g – ^ – s 2 + 2s + 15h 2 ^ s 2 – 2s + 17h ]2s – 2g
^ s 2 – 2s + 17h4
2
^ s 2 – 2s + 17h 2 ]– s + 1g – 2 ^ s 2 – 2s + 17h ^ – s 2 + 2s + 15h ]2s – 2g
^ s 2 – 2s + 17h4
2 ^ s 2 – 2s + 17h ^^ s 2 – 2s + 17h ]– s + 1g – ^ – s 2 + 2s + 15h ]2s – 2gh
=
^ s 2 – 2s + 17h4
2 ^^ s 2 – 2s + 17h ]– s + 1g – ^ – s 2 + 2s + 15h ]2s – 2gh
=
^ s 2 – 2s + 17h3
2 ^^ – s3 + 3s 2 – 19s + 17h – ^ – 2s3 + 6s 2 + 26s – 30hh
=
^ s 2 – 2s + 17h3
2 ^ – s3 + 3s 2 – 19s + 17 + 2s3 – 6s 2 – 26s + 30h
=
^ s 2 – 2s + 17h3
2 ^ s3 – 3s 2 – 45s + 47h
=
^ s 2 – 2s + 17h3
2 ^ s3 – 3s 2 – 45s + 47h
\
L[t2 et cos 4t] =
.
^ s 2 – 2s + 17h3
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
SIA Group
QP.4
MATHEMATICS-II [JNTU-anantapur]
(b) Define unit step function Laplace transform.
May/June-17, R(15), Q1(b)
Answer :
For answer refer Unit-I, Q33.
(c)
If f(x) = x in (–1, 1)then find the Fourier
4
coefficient of bn.
May/June-17, R(15), Q1(c)
=
Given function is,
f (x) = x4
As f (x) is even function, it doesn't contain sine terms.
1
–1
\
(d) What is Fourier even function (– π, π)?
May/June-17, R(15), Q1(d)
If a function f (x) = f (– x), then f (x) is an even function.
The Fourier series of even function in (– π, π) is
expressed as,
3
a
f (x) = 0 +
a cos nx
2 n=1 n
/
Where
1
a0 =
π
1
an =
π
(e)
π
# f ] xgdx
π
# f ] xg cos nx dx .
–π
Write the Fourier sine transform of f(t).
May/June-17, R(15), Q1(e)
For answer refer Unit-III, Q7, Topic: Fourier sine
Transform.
(f)
Find the value of Z(an cosnt).
May/June-17, R(15), Q1(f)
Answer :
Given that,
x(n) = an cosnt
The Z-transform of the signal x(n) is given as,
Z[x(n)] = X(Z) =
3
/ x]ng.z
–n
n = –3
Z[ancosnt] =
=
3
/ a cos nt.z
n
n=0
3 R
S int
/ SSS e
n=0T
=
=
=
\ Z[ancosnt] =
it
–1hn
n=0L
nN
+ ^e –it a z –1h
2
OO
O
P
RS 3
VW
3
SS
it
–1hn
–it
–1hnW
^
^
+
e az
e a z WW
SS
W
n=0
n=0
T
X
VW
1
1
1 RSS
W
+
S
W
2 S1 – eit a z –1 1 – e –it a z –1 W
T
X
R
WVW
z
z
1 SS
+
S
W
2 SS ^ z – aeit h ^ z – ae – it h WW
T
X
R
V
1 SS z ^ z – ae –it h + z ^ z – aeit h WW
SS
W
2 S ^ z – eit h ^ z – ae –it h WW
T
X
R
V
1 SS z 2 – zae –it + z 2 – zaeit WW
S
W
2 SS z 2 – aze –it – azeit + 1 WW
T
X
R
V
1 SS 2z 2 – az ^e –it + eit h WW
S
W
2 SS z 2 – az ^eit + e –it h + 1WW
T
X
R
V
1 SS 2z 2 – a2z cos t WW
S
W
2 SS z 2 – 2az cos t + 1WW
T
X
1 JKK z 2 – az cos t NOO
.2
2 KK z 2 – 2az cos t + 1 OO
L
P
/
/
z 2 – az cos t
z 2 – 2az cos t + 1
z 2 – az cos t
.
z – 2az cos t + 1
2
(g) Find the general solution of uxx = xy.
–π
Answer :
=
=
Coefficient of bn = 0.
Answer :
=
# f ] xg .sin nπx dx = 0
i.e.,
/ KKK ^e a z
1
=
2
=
Answer :
J
3
=
–n
V
+ e – int WW n –n
WW a z
2
X
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May/June-17, R(15), Q1(g)
Answer :
Given equation is,
uxx = xy
22 u
= xy
2x 2
Integrating with respect to x (keeping y constant) on
Þ
both sides,
2
# 22xu dx = # xy dx
2
Þ
2u
= y
2x
Þ
x2
2u
+ f ^ yh
= y.
2x
2
# xdx
Integrating with respect to x (keeping y constant) on
both sides.
# 22ux dx = 2y # x dx + # f ^ yhdx
2
Þ
u=
y x3
. + f ^ yh
2 3
# 1dx
o n t h e title cover before you buy
QP.5
May/June-2017 (R15) Question Paper with Solutions
Þ
u=
x3 y
+ f ^ yh] xg + φ ^ yh
6
Þ
u=
x3 y
+ x f ^ yh + φ ^ yh
6
\
The general solution is, u =
x3 y
+ x f ^ yh + φ ^ yh
6
where f (y) and f(y) are arbitrary functions.
(h) Find the Z-transform of the sequence {x(n)} where x(n) is n.2n.
Answer :
May/June-17, R(15), Q1(h)
For answer refer Unit-V, Q2(i).
JK 1 NO
OO .
(i) Find Z–1 KK
Lz – 3P
Answer :
May/June-17, R(15), Q1(i)
Given that,
SR 1 WVW
Z –1 SSS
W
z – 3W
T
X
RS 1 VW
WW = an – 1u(n – 1)
As, Z –1 SSS
z – aW
T
X
Here, a = 3
RS 1 VW
WW = 3n – 1u(n – 1)
Z –1 SSS
z – 3W
T
X
= 3n.3– 1u(n – 1)
=
\
(j)
3n
u ]n – 1g
3
n
RS 1 VW
WW = 3 u ]n – 1g .
Z –1 SSS
3
z – 3W
T
X
What do you mean by steady state and transient state?
May/June-17, R(15), Q1(j)
Answer :
Steady State
If the temperature is independent of time it is known as Steady State. In this state, heat flow is constant.
Transient State
If the temperature is dependent of time it is known as Transient State. In this state, heat flow varies with the time.
PART-B
(5 × 10 = 50 Marks)
JK
NO
1
OO .
Q2. (a) Apply convolution theorem for L–1 KKK 3 2
s ^s + 1h O
L
P
Answer :
May/June-17, R(15), Q2(a)
Given function is,
]Z]
b_b
1
`
L–1 ][ 3 2
] s ^ s + 1h bb
\
a
1
t2
1
Let, f (s) = 3 ; f (t) = L–1 ( 3 2 =
2!
s
s
Þ
f (t) =
t2
2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
SIA Group
QP.6
MATHEMATICS-II [JNTU-anantapur]
Let, g (s) =
Þ
1
1
2 = sint
; g (t) = L–1 ( 2
s +1
s +1
2
g(t) = sint
By convolution theorem,
]Z]
b_b
1
`=
L–1 ][ 3 2
] s ^ s + 1h bb
\
a
t
#
f ]ug g ]t – ug du
0
t
=
u2
. sin ]t – ug du
2
#
0
1
=
2
=
=
=
=
=
=
=
=
=
t
# u . sin ]t – ugdu
2
0
VWt
1 RSS 2
d 2
^u h . sin ]t – ug du duWW
sin ]t – ug du –
SSu
W0
2
du
T
X
JK – cos ]t – ug NO WVWt
1 SRS 2 JK – cos ]t – ug NO
OO – 2u KK
OO duW
Su K
W
2 S KL
–1
–1
P
L
P X0
T
VWt
1 RSS 2
u cos ]t – ug – 2 u cos ]t – ug duWWW
S
S
2
0
T
X
R
R
VWVWt
S
1 SS 2
d
]ug . cos ]t – ug du duWWWWWW
SSu cos ]t – ug – 2 SSSu cos ]t – ug du –
2
du
T
XX0
T
JK sin ]t – ug NO
J sin ]t – ug NO VWWt
1 RSS 2
OO – 1. KKK
O du 2W
SSu cos ]t – ug – 2 (u KK
2
–1 P
– 1 OP W0
L
L
T
X
RS
VWVWt
1 RSS 2
u cos ]t – ug – 2 SSS– u sin ]t – ug + sin ]t – ug duWWWWWW
2 SS
0
T
T
XX
R
V
t
R
V
S
1 SS 2
KJ – cos ]t – ug ONOWWWW
Su cos ]t – ug – 2 SSS– u sin ]t – ug + KK
OWWWW
2S
–1
L
PXX0
T
T
1 2
t
7u cos ]t – ug – 2 6– u sin ]t – ug + cos ]t – ug@A0
2
#
#
#
#
#
#
#
#
#
#
1 2
6u cos ]t – ug + 2u sin ]t – ug – 2 cos ]t – ug@t0
2
1
= 76t 2 cos ]t – t g + 2t sin ]t – t g – 2 cos ]t – t g@ – 60 2 cos ]t – 0g + 2 ]0g sin ]t – 0g – 2 cos ]t – 0g@A
2
=
1 2
76t cos ]0g + 2t sin ]0g – 2 cos ]0g@ – 50 + 0 – 2 cos t?A
2
=
1 2
6t ]1g + 2t ]0g – 2 ]1g – 0 – 0 + 2 cos t@
2
=
1 2
6t + 0 – 2 + 2 cos t@
2
=
1 2
6t – 2 + 2 cos t@
2
=
t2
– 1 + cos t
2
\
Z]
_b
2
1
]
b t
`b =
L–1 [] 3 2
– 1 + cos t .
] s ^ s + 1h b
2
\
a
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o n t h e title cover before you buy
QP.7
May/June-2017 (R15) Question Paper with Solutions
t
JK
N
KK –1 sint OOO
dtO .
(b) Evaluate L KKe
OO
t
K 0
L
P
Answer :
May/June-17, R(15), Q2(b)
#
L{x"} + 9L{x} =
s
s2 + 4
Substituting x(0) = 1, x'(0) = k(say) in above equation,
s2.L{x} – s.x(0) – x'(0) + 9L{x} =
Given function is,
t
e
–1
#
L >e –1
s2.L{x} – s(1) – k + 9L{x} =
sin t
t dt
0
t
L{x} (s2 + 9) – s – k =
sin t
–1
t dtH = e .L
#
0
t
# sint t dt
#
... (1)
t
#
Then,
Þ
Þ
Þ
#
s
1
2
s +1
f ] s g = tan–1(∞) – tan–1(s)
π
f ] s g = – tan –1s
2
f ] s g = cot–1s
t
#
0
\
L >e –1
#
0
L{x} =
s3 + ks 2 + 5s + 4k
^ s 2 + 4h^ s 2 + 9h
... (1)
s3 + ks 2 + 5s + 4k = ] A + C g s3 + ]B + Dg s 2 + s ]9A + 4C g + 9B + 4D
Comparing corresponding coefficients on both sides.
... (2)
1
cot –1s
se
1
sin t
–1
t dtH = se cot s .
OR
A + C = 1
... (3)
B + D = k
... (4)
9A + 4C = 5
... (5)
9B + 4D = 4k
... (6)
Solving equations (3) and (5),
1
4
,C=
5
5
Solving equations (4) and (6),
d x
KKJ π NOO
Q3. Solve
2 + 9x = cos 2t, if x(0) = 1, x K 2 O = – 1.
dt
L P
Answer :
May/June-17, R(15), Q3
A=
B = 0, D = k
2
Given differential equation is,
d2 x
2 + 9x = cos2t
dt
Þ x" + 9x = cos2t
... (2)
s3 + ks 2 + 5s + 4k = As3 + 9As + Bs 2 + 9B + Cs3 + 4Cs + Ds 2 + 4D
J
N
sin t
–1 K 1
–1 O
t dtH = e KK s cot sOO
L
P
=
t
s + s3 + 4s + ks 2 + 4k
^ s 2 + 4h^ s 2 + 9h
] As + Bg ^ s 2 + 9h + ]Cs + Dg^ s 2 + 4h
s3 + ks 2 + 5s + 4k
=
2
2
^ s + 4h^ s + 9h
^ s 2 + 4h^ s 2 + 9h
3
–1
L{x} =
s3 + ks 2 + 5s + 4k
As + B Cs + D
+ 2
2
= 2
2
^ s + 4h^ s + 9h
s +4
s +9
f ] s g = 6tan –1 ] s g@s
L >e
s + s ^ s 2 + 4h + k ^ s 2 + 4h
^ s 2 + 4h
Applying partial fractions to equation (1),
Substituting equation (2) in equation (1)
L{x} (s2 + 9) =
1
s +1
3
2
s
+s+k
s2 + 4
2
sin t
2 =
L (
t
Þ
VW
WVW
WW
1
W
f ]ug duWW = f ] s gWW
WW
WW s
X
X
s
s +4
s
s +4
2
L{x} (s2 + 9) =
0
t
SRS
WV
JK 1
NO
SSe –1 sin t dtWWW
= e –1 KK fr ] s gOO
LS
W
t
s
SS
WW
L
P
0
T
X
RS
SRS
SS
SSa L SSS
SS
SS
0
T
T
Consider sin t,
L{sin t} =
s
s + 22
2
Substituting the values of A, B, C and D in equation (2),
1
4
s+k
s+0
s3 + 5s 2 + 5s + 4k
5
5
+
2
=
^ s + 4h^ s 2 + 9h
s2 + 4
s2 + 9
s
4s
k
s3 + 5s 2 + 5s + 4k
+
+
=
^ s 2 + 4h^ s 2 + 9h
5 ^ s 2 + 4h 5 ^ s 2 + 9h s 2 + 9
Applying laplace transform on both sides,
L{x" + 9x} = L{cos2t}
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
Substituting equation (7) in equation (1),
s
4s
k
+
+ 2
Þ L{x} =
2
2
5 ^ s + 4h 5 ^ s + 9h s + 9
SIA Group
... (7)
QP.8
MATHEMATICS-II [JNTU-anantapur]
Applying inverse laplace transform on both sides,
L–1(L{x}) = L–1 )
Þ
s
4s
5
3 + L–1 )
3 + L–1 ( 2
2
2
5 ^ s + 4h
5 ^ s + 9h
s +9
2
1
s
s
1 –1
4
2+ ( 2
2 + kL–1 ( 2
2
L ( 2
5
5
s +4
s +9
s +9
x(t) =
k
1
4
cos 2t + cos 3t. + sin 3t 5
5
9
π
Substituting t =
in equation (8),
2
JK π NO
JK 3π NO k
JK 3 π NO
1
KJ 2 π NO 4
x KK OO = cos KK OO + cos KK OO + sin KK OO
5
L2P
L 2 P 5
L 2 P 9
L 2 P
Þ
x(t) =
Þ
–1=
1
4
k
]– 1g + ]0g + ]– 1g
5
5
9
Þ
–1=
–1 k
–
5
9
Þ
k –1
=
+1
9
5
Þ
k 4
=
9 5
... (8)
4
36
×9 =
5
5
36
Substituting k =
in equation (8),
5
1
4
36
x(t) = cos 2t + cos 3t +
sin 3t
5
5
5×9
Þ
k=
1
4
4
cos 2t + cos 3t + sin 3t
5
5
5
1
= 5cos 2t + 4 cos 3t + 4 sin 3t?
5
1
\
x(t) = 5cos 2t + 4 cos 3t + 4 sin 3t? .
5
=
Q4.
Find the Fourier series expansion of f(x) = 2x – x2 in (0, 3) and hence deduce that
π
1
1
1
1
2 – 2 + 2 – 2 + ...... 3 = 12
1
2
3
4
May/June-17, R(15), Q4
Answer :
Note
In equation
Fourier Series
π2
1
1
1
1
π
1
1
1
1
2 – 2 + 2 – 2 + ..... 3 = 12 is misprinted as 2 – 2 + 2 – 2 + ...... 3 = 12 .
1
2
3
4
1
2
3
4
For answer refer Unit-II, Q27 (up to equation (6))
\
The Fourier series of f (x) = 2x – x2 is given as,
2x – x2 =
Deduction
–9
π2
3
J
N
3
J
N
L
P
n=1
L
P
/ n1 cos KKK 2 n3πx OOO + π3 / 1n . sin KKK 2n3πx OOO
n=1
2
3
in above equation,
2
2
–9 3 1
KKJ 2n π
KJ 3 ON KJ 3 ON
2 KK OO – KK OO = 2
2 cos K 3 ×
2
2
π n=1 n
L P L P
L
Substituting x =
/
3 ON 3
O+
2 OP π
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3
J
N
n=1
L
P
/ 1n . sin KKK 23n π × 23 OOO
o n t h e title cover before you buy
QP.9
May/June-2017 (R15) Question Paper with Solutions
3 –
–9
9
= 2
4
π
9
12 – 9
= 2
4
π
3 π2
×
=
4 9
3
3
/ n1 cos ]n πg + π3 / 1n sin ]nπg
n=1
2
n=1
3
3
/ –n 1 ]– 1g + π3 / 1n ]0g
n=1
n
2
n=1
n
3
/ – ]n– 1g
n=1
+0
2
π2
– ]– 1g1 JKK ]– 1g2 ONO JKK – ]– 1g3 ONO KJK – ]– 1g4 ONO
+ KK –
=
O+K
O+K
O + ....
12
12
2 2 OP KL 3 2 OP KL 4 2 OP
L
π2
1
1
1
1
= 2 – 2 + 2 – 2 + .....
12
1
2
3
4
π2
1
1
1
1
2 – 2 + 2 – 2 + ...... 3 =
.
12
1
2
3
4
OR
Z]
l
]]kx,
0# x #
]]
2
. Deduce the sum of the series
Q5.Obtain half range cosine series for f(x) = []
]]
l
]]k ]l – xg, # x # l
2
1
1
1
\
2 – 2 + 2 + ..... 3 .
1
3
5
Answer :
May/June-17, R(15), Q5
Given function is,
Z]
l
]] kx,
0# x #
]]
2
f(x) = ][
]]
l
]] k ]l – xg, # x # l
2
\
Half-range cosine series is given as,
3
a
npx
f(x) = 20 +
an cos b l l /
... (1)
n=1
2
Where, a0 = l
=
=
=
=
l
#
f (x) dx
0
RS l
VW
SS 2
WW
l
WW
2 SS
SS f ] xg dx +
f ] xg dxWW
lS
WW
SS0
l
WW
S
2
TR
XV
SS l
WW
l
2
S
W
2 SS kxdx + k (l – x) dxWW
S
WW
lS
SS0
WW
l
S
W
2
XV
RST l
WW
SS 2
l
WW
S
2S
SSk xdx + k ]l – xg dxWWW
lS
WW
SS 0
l
WW
S
2
TR l
VW X
SS
WW
l
S 2
W
2k SS
SS xdx + ]l – xg dxWWW
l S
WW
SS0
l
WW
S
2
T
X
#
#
#
#
#
#
#
#
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
SIA Group
QP.10
MATHEMATICS-II [JNTU-anantapur]
l
2
2 2
2k
= l >; x E – < (l – x) F l H
2 0
2
2
l
R
V
2k 1 SS 2 2l
l W
W
. SS^ x h0 – ^]l – xg2h l WW
l 2S
W
2
T
X
R
V
J
k SS l 2
l N2W
= SS – 0 – ]l – l g2 + KKKl – OOO WWW
l S4
2P W
L
T
X
R
V
k SS l 2
l 2 WW
= SS – 0 – 0 + WW
l 4
4
T
X
kl
k 2l 2
= l. 4 = 2
kl
\
a0 = 2
=
2
an = l
=
=
=
=
=
=
=
=
=
l
#
0
npx
f (x) cos b l l dx
RS l
VW
SS 2
W
l
JK nπx NO
JK nπ x NO WWW
2 SS
O dxW
OO dx +
S f ] xg cos KK
f ] xg cos KK
l OP WW
l SS
L l P
L
SS0
WW
l
S
W
2
T
X
RS l
VW
SS 2
WW
l
2 SS kx cos npx dx + k (l – x) cos npx dxWW
WW
l
l
l SS
SS0
WW
l
S
W
2
T
X
RS l
VW
SS 2
WW
l
JK nπ x NO WW
JK n πx NO
2 SS
O dx + k ]l – xg cos KK
O dxW
Sk x cos KK
l OP
l OP WW
l SS
L
L
SS 0
WW
l
S
W
2
T
X
RS l
VW
SS 2
WW
l
J nπ x ON WW
J n πx ON
2k SS
K
K
OO dx + ]l – xg cos KK
OO dxWW
S x cos KK
l SS
L l P
L l P WW
SS0
l
WW
S
2
T
X
RS
V
VW 2l SR
VWl WWW
JK n π xNO
JK nπ x NO
J
N
J
N
2k SSRSS
π
d
d
π
n
x
m
x
OO
OO
W
S
K
O
OO dx dxWW + SS]l – xg cos KKK
] xg KK
]l – xg cos KKK
O dx –
O dx dxWW l WWW
W0 S
l SSSSx cos K l O dx –
dx
dx
l
l
l
ST
L
P
L
P
L
P
L
P
X
T
X 2 WX
T
RS
WV
VW 2l RS
JK JKn π x NO l NO
J JK n π xNO l ON VWl WW
JK nπx NO l
2k SSRSS JK JK n π x NONO l
K
W
S
W
S K K
K
O
O . O – ]– 1gKKsin KK
OO
O . O dxW W
]l – xgKKsin KK
l SSSSx Ksin K l OO . nπ – 1. sin K l O n π dxWW0 + SS
l OP n π OP
l OP nπ OP W l WWW
ST L L
L
P
PP
L
L
L
L
X
T
X 2X
T
RS
V
l
W
J
JK nπx NO VWW 2 RSS l ]l – xg
N
JK nπx NO l
JK n π xNO WWVl WW
2k SSRSS xl
SSS sin KKK nπx OOO – l
sin KK
sin KK
OO dxWW + SS
OO + nπ sin KK l OO dxWW l WWW
l SS n π
l
ST
L l P nπ
L l P X0 T nπ
L
L
P
P X 2W
T
X
RS
VW
l
JK
J
JK n π x NO
JK nπ x NO l NOVWW 2 RSS l ]l – xg JK nπx NO
JK nπx NO l NOWWVl WW
2k SSRSS xl
l
l
K
SS
K
OW
K
OW W
O
K
K
O
O
K
K
O
S
l SSS n π sin K l O – nπ K – cos K l O . nπ OW0 + S nπ sin K l O + nπ K – cos K l O . nπ OW l WW
ST
P
L
L
P
P
L
L
P
L
PX
L
PX 2 W
T
T
X
RS
V
l
WW
V
R
V
R
l
S
JK n πx NO
JKnπ x NOWW 2 SS l ]l – xg JK nπx NO
JK nπx NOWW WW
2k SSSS xl
l2
l2
S
S
K
K
K
K
O
OW
O
OW
l SSSS n π sin K l O + n 2 π 2 cos K l OWW0 + SS n π sin K l O – n 2π 2 cos K l OWW l WW
L
L
L
L
ST
P
PX
P
PX 2 W
T
T
X
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
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#
#
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May/June-2017 (R15) Question Paper with Solutions
QP.11
RSSR 2
V
JK nπl ON
J nπ l ON 0.l
2k SS l
l2
l2
KJ nπ ]0g OON
KJ n π]0g OONWWW
SSSS
O + 2 2 cos KKK
O–
=
sin KK
sin KK
– 2 2 cos KK
O
O
O
O
W
π
l SS 2n π
L 2l P n π
L 2l P n π
L
P n π
L l PWX
TT
RS
VWVW
JK
l NO
SS
WW
K
O
l
–
2
2
l
JK n πl NOWWWW
l
l
2 OP KKJ n π l OON
KKJ n πl OON KL
+ SS l ]l – l g KKJ nπl NOO
W
SS nπ sin K
O – 2 2 cos K l O – n π sin K 2l O + 2 2 cos KK 2l OOWWWWW
l
π
π
n
n
L
P
L
P
L
P
L
PXX
T
R 2
2
2
2
2
S
JK n π NOWVW
]n πg
]n πg
2k S l
l
l
l ]0 g
l
l
l2
KJ nπ ON
SS
=
+ 2 2 cos
sin
sin ]n πg – 2 2 cos ]n πg –
sin KK OO + 2 2 cos KK OOWW
– 0 – 2 2 cos ]0g +
l S 2n π
nπ
2
2
2 nπ L 2 P n π
n π
n π
n π
L 2 PWX
TR 2
V
2
2
W
J n πN
2k SS 2l
l
–l
SS 2 2 cos KKK OOO – 2 2 ]1g + 0 2 2 cos n πWWW
=
l Sn π
W
nπ
L 2 P n π
TR 2
X
V
2
S
W
JK nπ ON
2k S 2l
l
W
S
W
K
O
=
cos K O – 2 2 ]1 – cos n πgW
l SS n 2π 2
W
2
n π
T 2 R L P
X
V
JK n π NO
WW
2k l SS
n
=
.
S2 cos KK OO – 1 – ]– 1g WW
l n 2π 2 S
L 2 P
T
X
R
VW
JK n π ON
2kl SS
= 2 2 SS2 cos KK OO – 1 – ]– 1gnWWW
nπ
L 2 P
T
X
VW
2kl RSS
KJK n π ONO
\ an = 2 2 SS2 cos K O – 1 – ]– 1gnWWW
nπ
L 2 P
T
X
Substituting the values of a0 and an in equation (1),
kl
3
N J
N
J N
2kl KJ
KK2 cos KKK n πOOO – 1 – ]– 1gnOOO cos KKKnπ x OOO
f (x) = 2 +
2
2
2 n=1 n π
L 2 P
L
P L l P
N J
N
J N
kl 2kl 3 1 JK
KK2 cos KKK nπ OOO – 1 – ]– 1gnOOO cos KKK n π xOOO
+ 2
=
2
4
π n=1 n L
L 2 P
P L l P
/
/
\
f (x) =
kl 2kl
+ 2
4
π
3
/ n1
n=1
2
Deduction
N JK nπ x NO
nπ
KKJ
nO
K2 cos 2 – 1 – ]– 1g OO cos KK l OO L
P L
P
... (2)
Substituting x = 0 in equation (2)
f (0) =
Þ
kl 2kl
+ 2
4
π
kl 2kl
+ 2
0 =
4
π
2kl
– kl
= 2
4
π
J
3
J
N
N
J
N
/ n1 KKK2 cos KKK n2πOOO – 1 – ]– 1g OOO cos KKK n πl]0g OOO
n
2
L P
L
P L
NO
JK n π NO
1 JK
K
K O
] gnO ] g
2 K2 cos K 2 O – 1 – – 1 O cos 0
L P
n=1 n L
P
n=1
3
P
/
J
N
/ n1 KKK2 cos JKKK n2πNOOO – 1 – ]– 1g OOO
3
2
n
L P
P
J
N
J
N
JK nπ NO
SRS
WV
1K
– kl π
KK2 cos KKK n π OOO – 1– ]– 1gnOOO Þ
=
×
SSa 2 cos KK OO – 1 – ]– 1gn = 0 if n is oddWWW
2
4
2kl
L 2 P
L 2 P
n = 2, 4, 6 n L
P
T
X
2
R
V
R
V
R
V
J
N
J
N
J
N
W 1S
W 1S
W
1S
–π
K 2πO
K 4π O
K 6π O
Þ
= 2 SSS2 cos KK OO – 1 – ]– 1g2WWW + 2 SSS2 cos KK OO – 1 – ]– 1g4WWW + 2 SSS2 cos KK OO – 1 – ]– 1g6WWW
8
2
2
2
2
4
6
L P
L P
L P
T
X
T
X
T
X
R
V
R
V
J
N
W
S
W
1 SS
1
π
10
KKJ 8 π ONO
K
O
O – 1 – ]– 1g10WW + .....
+ 2 SS2 cos K O – 1 – ]– 1g8WWW + 2 SSS2 cos KK
W
2P
2 OP
8
10
L
L
T
X
T
X
– π2
1
1
1
1
Þ
= 2 62 cos ]πg – 1 – 1@ + 2 62 cos ]2 πg – 1 – 1@ + 2 62 cos ]3π g – 1 – 1@ + 2 62 cos ] 4 πg – 1 – 1@
8
2
4
6
8
2
+
n=1
L
3
/
1
62 cos ]5πg – 1 – 1@ + ...
10 2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
SIA Group
QP.12
MATHEMATICS-II [JNTU-anantapur]
Þ
– π2
1
1
1
1
1
= 2 62 ]– 1g – 1 – 1@ + 2 62 ]1g – 1 – 1@ + 2 62 ]– 1g – 1 – 1@ + 2 62 ]1g – 1 – 1@ + 2 62 ]– 1g – 1 – 1@ + .....
8
2
4
6
8
10
Þ
– π2
1
1
1
1
1
= 2 5– 4? + 2 50? + 2 5– 4? + 2 50? + 2 5– 4? + .....
8
2
4
6
8
10
Þ
– π2
1
1
+ .....
= –1+ 0 – + 0 –
8
9
25
RS 1
VW
1
1
– π2
= – SS 2 + 2 + 2 + .....WWW
S1
8
3
5
T
X
2
π
1
1
1
= 2 + 2 + 2 + .....
8
1
3
5
Þ
Þ
Q6.
\
1
1
1
π2
=
+
+
+
.
.....
8
12 32 52
(a)
Find the Fourier sine transform of e–|x|.
May/June-17, R(15), Q6(a)
Answer :
Given function is,
f (x) = e–|x|
The fourier sine transform of f (x) is given as,
3
Fs[f(x)] =
#
0
=
f (x) . sin sx dx
3
#
e –x sin sx dx [ a |x| = x, x ≥ 0]
0
RS – x
VW3
S e
]– sin sx – s cos sxgWWW = SS 2
Ss +1
W0
T
X
SRS
WVW
e– 0
^ – sin s ]0g – s cos s ]0ghWW
= SS0 – 2
S
W
s +1
T
X
RS – 1
VW
^ – 0 – s ]1ghWW
= SS 2
Ss + 1
W
T
X
–1
]– sg
= 2
s +1
=
\
RS
SS
SSa
T
#
e –ax sin bx dx =
VW
e –ax
WW
]
g
–
a
sin
bx
–
b
cos
bx
WW
a2 + b2
X
s
s +1
2
Fourier sine transform of e–|x| is
s
.
s2 + 1
(b) Write the conditions of Parseval's identity for Fourier transforms.
May/June-17, R(15), Q6(b)
Answer :
The conditions of Parseval's Identity for Fourier Transforms are given as,
(i)
(ii)
1
2p
1
2p
3
#
3
F (s) . G (s) ds =
–3
3
#
–3
#
f (x) .g (x) dx
–3
3
| F (s) | 2 ds =
#
| f (x) | 2 dx
–3
Where F(s) and G(s) are Fourier transform of f(x) and g(x) respectively,
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QP.13
May/June-2017 (R15) Question Paper with Solutions
Parseval's identity for Fourier sine and cosine transforms
2
p
(i)
2
p
(ii)
2
p
(iii)
2
p
(iv)
3
#
3
#
Fc (s) .Gc (s) ds =
0
f (x) .g (x) dx
0
3
#
3
#
Fs (s) .Gs (s) ds =
0
f (x) g (x) dx
0
3
#
3
#
| Fc (s) | 2 ds =
0
| f (x) | 2 dx
0
3
#
3
#
| Fs (s) | 2 ds =
0
| f (x) | 2 dx
0
OR
Q7.Verify convolution theorem for f(x) = g(x) = e
–x 2
.
May/June-17, R(15), Q7
Answer :
Given functions are,
f (x) = g(x) = e –x
2
From the definition of convolution theorem,
F[f (x)*g(x)] = F(s).G(s)
Consider,
F[f(x)*g(x)] =
3
3
# #
f ]ug .g ] x – ug du eisx dx
–3 –3
3
=
3
# #
2
2
2
2
2
2
e –u .e –]x – ug du eisx dx
–3 –3
3
3
=
# #
e –u .e –] x – ug .eisu e –isu eisx dx du
–3 –3
3
3
=
# #
e –u .e –]x – ug du.eisu .eis]x – ug dx
–3 –3
3
=
#
3
#
2
e –u .eisu du
–3
2
e –]x – ug .eis]x – ug dx
–3
Let, x – u = t
dx = dt
U.L : x = ∞, t = ∞
L.L : x = – ∞, t = – ∞
3
=
#
3
e
–u 2
.e du.
–3
#
–3
2
e –t .eist dt
–3
3
=
#
isu
3
2
e –t eist dt
#
2
e –t eist dt
(Replacing u by t)
–3
RS 3
VW2
SS
W
–t2 ist W
e e dtWW
= SS
SS
WW
–3
T
X
#
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QP.14
MATHEMATICS-II [JNTU-anantapur]
=
=
=
=
=
=
si
Let t –
=y
2
dt = dy
RS 3
VW2
SS
W
–_t 2 – ist i W
SS
e
dtWW
SS
WW
–3
T
X
RS 3 KJK 2
2
J is ON2 KJ is ON2ONO V
K
–KKt – ist + KK OO – KK OO OO W
SS
WW
2
2
L P L P P dtW
SS
e L
WW
SS
W
–3
T
X
RS 3 JK is NO2 JK si NO2 VW2
SS
W
–KKt – OO + KK OO
SS
e L 2 P L 2 P dtWWW
SS
WW
–3
T
X
SRS 3
WV2
2
KJ is ON J si N2 W
SS
WW
–KKt – OO KKK OOO
SS
e L 2 P .eL 2 P dtWW
SS
WW
–3
T
X
RS J N2 3 J
2
2 V
W
N
is
is
K
K
O
O
SS KK OO
WW
–KKt – OO
SSeL 2 P
e L 2 P dtWW
SS
WW
–3
T
X
RS 2
VW2
2
3
KJ is ON
SS – s
WW
–KKt – OO
2
SSe 4 .
P dtW
e L
WW
SS
W
–3
T
X
#
#
#
=
N
is NO2 JK is NO2OO
O –K O O
2 OP KL 2 OP O
P
dx
is NO2 JK is NO2
O + KK OO
2 OP
L2P
e
–KK x –
L
dx
–3
JK
3
#
#
=
e
is NO2
O
2 OP
–KK x –
L
.e
JK is NO2
KK OO
L 2 P dx
–3
=
is
Let, x – = q
2
#
KJK is ONO2
K O
eL 2 P
3
#
JK
is NO 2
– KK x – 2 OO
P dx
e L
–3
dx = dq
#
#
U.L : x = ∞, q = ∞
L.L : x = – ∞, q = – ∞
= e
–
s2
4
3
#
2
e – q dq
–3
= e
= πe
–
s2
2
\ F[f (x) * g(x)] = πe
–
s2
2
#
e
–
s2
4
πe
–
s2
4
F(s) =
Similarly,
π .e
–
s2
4
F(s).G(s) = π .e
–
s2
4 .
R 2V
G(s) = G SSe –x WW =
T X
Consider,
... (1)
= πe
= πe
isx
.e dx
2 – isx i
\
dx
#
e
N
KJ
KJ is ON2 KJ is ON2O
–KK x 2 – isx + KK OO – KK OO OO
K
2
2 O
L
L P
F(s).G(s) =
–
π .e
–
s2
4
2s 2
4
– s2
2
– s2
πe 2
From equations (1) and (2)
–3
3
2^ πh
2
πe
\
3
e –_ x
s2
4 .
=
–3
#
#
= e
3
–x2
3
s2
2
4 .2
e – q dq
–
–
0
#
=
LL
JK
3
#
=
e
–KKKK x –
K
–3
L.L : t = – ∞, y = – ∞
RS s2 3
VW2
SS –
W
2
–
y
e dyWWW
= SSe 4
SS
WW
–3
T
X
RS s2
VW2
3
SS –
W
– y2 W
4
= SSe .2 e dyWW
SS
WW
0
T
X
SRS – s2 KJ π NOWVW2
OOWW
= SS2e 4 . KK
S
W
2
L
PX
T
2
s2 ^
πh
–2
= 4.e 4 .
4
R 2V
F(s) = F SSe –x WW =
T X
#
=
U.L : t = ∞, y = ∞
Consider,
KJJK
3
L P P dx
–3
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F[f (x)*g(x)] = F(s).G(s) =
– s2
πe 2
Hence, convolution theorem is verified.
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... (2)
QP.15
May/June-2017 (R15) Question Paper with Solutions
JK xy NO
Form the partial differential equation z = f KK OO by eliminating the arbitrary function.
L 2 P
Answer : May/June-17, R(15), Q8(a)
Q8.
(a)
Given equation is,
xy
z = f c m 2
... (1)
Where f is an arbitrary function.
Differentiating equation (1) partially with respect to 'x' on both sides,
xy y
2z
2x = f ' c m . 2
2
xy y
i.e., p = f ' c m . 2
2
... (2)
Again differntiating equation (1) partially with rspect to 'y' on both sides,
xy x
2z
2y = f ' c m . 2
2
i.e., q = f ' c
xy x
m.
2 2
... (3)
Dividing equation (2) and equation (3),
JK xy NO x
K O
p f 'K 2 O. 2
= JL NP
q
K xy O y
f ' KK OO .
L2P 2
p y
Þ
q = x
Þ
px = qy
\
px = qy is the required partial differential equation.
(b) Use the method of separation of variables, solve
Answer :
2u
2u
where u(x, 0) = 8e–3y.
=4
2x
2y
May/June-17, R(15), Q8(b)
For answer refer Unit-IV, Q31.
OR
22 u 22 u
Q9. Solve the Laplace equation
+
= 0 subject to the conditions u(0, y) = u(i, y) = u(x, 0) = 0 and
2x 2 2y 2
JK nπx NO
OO .
u(x, a) = sin KK
L l P
Answer :
May/June-17, R(15), Q9
Note: In question u(l, y) is misprinted as u(i, y).
For answer refer Unit-IV, Q48.
Q10. (a)
Find the z-transformation of sinnθ.
Answer :
May/June-17, R(15), Q10(a)
Given that,
U(n) = sinnθ
Z{U(n)} =
3
/U z
n
–n
n=0
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING sTUDENTs
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QP.16
MATHEMATICS-II [JNTU-anantapur]
Z{sin nθ} =
3
/ sin nθz
–n
n=0
3
=
/ (e
iθn
– e –iθn –n
2z
2i
n=0
iθn
3
=
/ e2i z
–n
3
–
n=0
=
1
2i
3
/ ^e
–iθn
/ e2i
z –n
n=0
iθ –1hn
z
–
n=0
1
2i
3
/ ^e
n
– iθ –1h
z
n=0
VW
1
1
1 RSS
1
WW
(
2
S
=
2i 1 – eθi z –1
2i S1 – e –θi z –1 W
T
X
1
1
1
(
2
–
=
2i 1 – eθi z –1 1 – e –θi z –1
Z
_
1 ]] 1 – e –θi z –1 – 1 + eθi z –1 bb
[]
`
=
2i ] ^1 – eθi z –1h ^1 – e –θi z –1h bb
\
a
Z
_b
^eθi – e –θih z –1
1 ]]
b
[
`
=
2i ]] 1 – ^eθi + e –θih z –1 + z –2 bb
R\
V a
1 SS z 2 ^eiθ – e –iθh z –1 WW
S
W
=
2i SS z 2 – z ^eiθ + e –iθh + 1WW
T
X
^eiθ – e –iθh
z
2i
=
JK eiθ + e –iθ NO
2
OO + 1
z – 2z KK
2
L
P
z sin θ
= 2
z – 2z cos θ + 1
z sin θ
\ Z[sin nθ] = 2
.
z – 2z cos θ + 1
=
(b) If U(z) =
2z 2 + 5z + 14
, evaluate U2, U3 using initial value theorem.
]z – 1g4
Answer :
May/June-17, R(15), Q10(b)
For answer refer Unit-V, Q31.
OR
Q11. Solve the differential equation un + 2 – 2un + 1 + un = 3n + 5 using z-transforms.
Answer :
May/June-17, R(15), Q11
For answer refer Unit-V, Q45.
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May/June-16 (R15) Question Paper with Solutions
QP.1
Code No.: 15A54201/R15
R15
Solutions
B.Tech I Year II Semester Regular Examinations
May/June - 2016
MATHEMATICS - II
( Common to All )
Time: 3 Hours
Max. Marks: 70
PART-A
( Compulsory Question)
*****
1.
Answer the following: (10 × 02 = 20 Marks)
(a)
Find L[t2.et.cos4t] (Unit- I)
(b)
Find the Laplace Transform of
(c)
What are Dirichlet’s conditions? (Unit- II)
(d)
Express f(x) = x as a Fourier series from – p to p . (Unit- II)
(e)
Write the formula of the Fourier cosine integral of f(x). (Unit- III)
(f)
Write the formula for the inverse Fourier transform of F(s) in ( – 3 , 3 ) (Unit- III)
(g)
Find the value of Z (an cos nt). (Unit- IV)
(h)
Find the Z-transform of the sequence {x(n)} where x(n) is n.2n (Unit- IV)
(i)
Derive a partial differential equation by eliminating the arbitrary function f from the relation:
sin 2t
t . (Unit- I)
f(x2 + y2, x2 – z2) = 0 (Unit- V)
(j)
Form the PDE from the relation z = f(x +it) + g(x – it). (Unit- V)
PART-B
( Answer all five units 5 × 10 = 50 Marks)
UNIT-I
2.
s
Find the inverse Laplace Transform of 2 2 2 by using Convolution theorem. (Unit-I, Topic No. 1.8)
(s + a )
OR
3.
Solve (D2 – D – 2)y = 20 sin2t where y(0) = 1, y'(0) = 2.(Unit-I, Topic No. 1.11)
UNIT-II
4.
1
1
1
1
p2
Find a Fourier Series to represent x – x2 from x = – p to x = p and deduce that 12 = 2 – 2 + 2 – 2 + ...
1
2
3
4
(Unit-II, Topic No. 2.3)
OR
5
p
p
If f(x) = 3 , 0 # x # 3
p
2p
= 0, 3 # x # 3
p 2p
= – 3, 3 #x#p
Then f(x) =
2 <
1
1
cos x – 5 cos 5x + 7 cos 7x + ... F (Unit-II, Topic No. 2.4)
3
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.2
UNIT-III
3
6.
Show that
#
sin pl sin lx
dl = p sin x, for 0 # x # p
2
1 – l2
0
= 0 for x > p (Unit-III, Topic No. 3.1)
OR
7.
3
Find Fourier transform of f(x) = 1 – x for x # 1 = 0 for x $ 1 and hence find
(Unit-III, Topic No. 3.2)
2
#
0
x
x cos x – sin x
cos 2 dx
x3
UNIT-IV
8.
Find the partial differential equation of all spheres whose centre lie on Z - axis and given by equation
x2 + y2 + (z – a)2 = b2, and b being constants (Unit-IV, Topic No. 4.1)
OR
9.
px
A string is stretched and fastened to two points l apart. Motion is started by displacing the string in the form y = a sin l
from which it is released at a time t = 0. Show that the displacement of any point at a distance x from one end at time t is
px
pct
given by y (x,t) = a sin d l n cos d l n . (Unit-IV, Topic No. 4.3)
UNIT-V
10.
Solve the difference equation, using Z-transorm un+2 – un = 2n, Where u0 = 0 and u1 = 1 (Unit-V, Topic No. 5.6)
OR
2
11.
If f(z) =
2z + 3z + 4
, ; z ; >3, then find the values of f(1), f(2), f(3). (Unit-V, Topic No. 5.4)
(z – 3) 3
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May/June-16 (R15) Question Paper with Solutions
QP.3
SOLUTIONS TO MAY/JUNE-2016, R15, QP
PART-A
Q1.
(a)
Find L[t2.et.cos4t]
May/June-16, (R15), Q1(a)
Answer :
The given function is,
t2etcos4t
Consider, et cos4t
< a L [e at cos bt] =
s –1
( s – 1) 2 + ( 4) 2
s –1
=
(s – 1) 2 + 16
L[et cos4t] =
s–a
F
(s – a) 2 + b 2
... (1)
Consider, [t2et cos4t]
d2 < 8 t
L e cos 4t BF
ds 2
Substituting equation (1) in equation (2)
n
< a L [t n f (t)] = (–1) n d n [ f (s)] F
ds
L [t2et cos4t] = (–1)2
L [t2et cos4t] = (–1)2
... (2)
d2
s –1
<
F
ds 2 (s – 1) 2 + 16
RS
V
SS ` (s – 1) 2 + 16 j d (s – 1) – (s – 1) d 8 (s – 1) 2 + 16 B WWW
d
WW
ds
ds
= ds SSS
2
WW
2
SS
8 (s – 1) + 16 B
W
TR
X
V
SS
WW
2
d S ` (s – 1) + 16 j (1) – (s – 1) 2 (s – 1) (1) WW
= ds SS
WW
2
SS
8 (s – 1) 2 + 16 B
WW
S
T
X
2
2
d (s – 1) + 16 – 2 (s – 1)
= ds >
H
2
8 (s – 1) 2 + 16 B
16 – (s – 1)
d
= ds >
2H
8 (s – 1) 2 + 16 B
2
<` (s – 1) 2 + 16 j (–2) (s–1) – 816 – (s – 1) 2 B 2 8 (s – 1) 2 + 16 B 2 (s – 1) F
2
=
2
d ` (s – 1) 2 + 16 j n
2
=
– 2 (s – 1) ` (s – 1) 2 + 16 j< (s – 1) 2 + 16 + 2 ` 16 – (s – 1) 2 jF
=
– 2 (s – 1) 8 (s – 1) 2 + 16 + 32 – 2 (s – 1) 2 B
=
– 2 (s – 1) 8 48 – (s – 1) 2 B
=
2 (s – 1) 8 (s – 1) 2 – 48 B
\ L {t2et cos4t} =
` (s – 1) 2 + 16 j
4
3
` (s – 1) 2 + 16 j
3
` (s – 1) 2 + 16 j
3
` (s – 1) 2 + 16 j
2 (s – 1) 8 (s – 1) 2 – 48 B
8 (s – 1) 2 + 16 B
3
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.4
(b) Find the Laplace Transform of
sin 2t
t .
May/June-16, (R15), Q1(b)
Answer :
The given function is,
sin 2t
t
Let, f(t) = sin2t
SRS
WVW
3
f ( t)
SS
W
f (s) ds WW
SS a L d t n =
WW
SS
W
s
T
X
3
sin 2t
L< t F =
#
#
f (s) ds
s
3
#
=
L [sin 2t] ds
s
3
#
=
< a L [sin at] =
2
ds
s2 + 4
s
a
F
s2 + a2
3
=2
#
s
1
ds
s2 + 4
1
s
= 2 < 2 tan –1 2 F
s
3
–1 s
= < tan 2 F
s
3
s
= tan –1 2 – tan –1 2
s
π
= tan –1 d tan 2 n – tan –1 2
π
s
= 2 – tan –1 2
sin 2t
π
s
\ L < t F = 2 – tan –1 2
(c)
3
< a tan π = 3 and tan –1 (tan θ) = θ F
2
What are Dirichlet’s conditions?
May/June-16, (R15), Q1(c)
Answer :
For answer refer Unit-II, Q3.
(d) Express f(x) = x as a Fourier series from – p to p
May/June-16, (R15), Q1(d)
Answer :
For answer refer Unit-II, Q11.
(e)
Write the formula of the Fourier cosine integral of f(x).
May/June-16, (R15), Q1(e)
Answer :
For answer refer Unit-III, Q18 excluding case (i).
(f)
Write the formula for the inverse Fourier transform of F(s) in ( – 3 , 3 )
May/June-16, (R15), Q1(f)
Answer :
For answer refer Unit-III, Q13.
(g) Find the value of Z (an cos nt).
May/June-16, (R15), Q1(g)
Answer :
n
Given function is a cosnt
Consider, [cosnt]
Z[cosnt] =
z (z – cos nt)
(z – 2z cos nt + 1)
...(1)
2
>a Z [cos θ] =
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z (z – cos θ)
H
(z 2 – 2z cos θ + 1)
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May/June-16 (R15) Question Paper with Solutions
QP.5
From time scaling property of Z-Transform,
z
an x(n) « X ` a j
z
Z[an cosnt] = X ` a j
z` z
– cos nt j
\ Z[an cosnt] = z a2 a z
` j – 2 cos nt + 1
a
a
z
(
z
–
a cos nt)
2
= 2a
(z – 2za cos nt + a 2)
a2
z (z – a cos nt)
= 2
z – 2az cos nt + a 2
z (z – a cos nt)
\ Z[an cosnt] = 2
(z – 2az cos nt + a 2)
(h) Find the Z-transform of the sequence {x(n)} where x(n) is n.2n
May/June-16, (R15), Q1(h)
Answer :
For answer refer Unit-V, Q2 (i).
(i)
Derive a partial differential equation by eliminating the arbitrary function f from the relation:
f(x2 + y2, x2 – z2) = 0.
May/June-16, (R15), Q1(i)
Answer :
Given function is,
f(x2 + y2, x2 – z2) = 0
...(1)
Let, x2 + y2 = u and x2 – z2 = v
... (2)
\ From equations (1) and (2)
f(u,v) = 0
... (3)
Differentiating equation (3) with respect to ‘x’ by chain rule,
2f 2u 2f 2u 2z 2f 2v 2f 2v 2z
2u 2x + 2u 2z 2x + 2v 2x + 2v 2z 2x = 0
2f
2f
2f
2f
Þ
2u (2x) + 2u (0) (p) + 2v (2x) – (2z) 2v p
2z
Where, p = 2x
2f
2f
Þ
2u (2x) + 2v (2x – 2pz) = 0
2f
2f
Þ 2 < x 2u + (x – pz) 2v F = 0
2f
2f
Þ
x 2u + (x – pz) 2v = 0
Differentiating equation (3) with respect to ‘y’ by using chain rule,
2f 2u 2f 2u 2z 2f 2v 2f 2v 2z
2u 2y + 2u 2z 2y + 2v 2y + 2v 2z 2y = 0
2f
2f
2f
2f
Þ
2u (2y) + 2u (0) (q) + 2v (0) + 2v (–2z) q = 0
2z
Where, q = 2y
Þ
... (4)
2f
2f
2 < 2u y + (–qz) 2v F = 0
2f
2f
2u y – qz 2v = 0
2f
2f
Eliminating 2u and 2v from equations (4) and (5),
Þ
...(5)
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.6
x x – pz
=0
y – qz
Þ
– xqz – yx + pyz = 0
Þ
(py – xq)z = xy
This is the required partial differential equation.
(j)
Form the PDE from the relation z = f(x +it) + g(x – it).
May/June-16, (R15), Q1(j)
Answer :
The given function is,
z = f(x + it) + g(x – it)
Let, x + it = u
and x – it = v
z = f(u) + g(v)
...(1)
Differentiating equation (1), partially with respect to ‘x’,
2z
2u
l
l 2v
2x = f (u) 2x + g (v) 2x
< a 2u = 1 and 2v = 1 F
= f l (u) (1) + g l (v) (1)
2x
2x
2z
l
l
2x = f (u) + g (v)
Differentiating equation (2), partially with respect to ‘x’,
...(2)
2u
2v
22 z
= f m (u) 2x + g m (v) 2x
2x 2
= f m (u) + g m (v)
Differentiating equation (1) partially with respect to ‘t’,
2z
2u
l
l 2v
2t = f (u) 2t + g (v) 2t
< a 2u = i, 2v = – i F
= f l (u) (i) + g l (v) (– i)
2t
2t
2z
l
l
2t = if (u) – ig (v)
Differentiating equation (4) partially with respect to ‘t’,
...(3)
... (4)
22 z
2u
2v
= if m (u) 2t – ig m (v) 2t
2t 2
= i 2 f m (u) + i 2 g m (v)
22 z
= – f m (u) – g m (v)
2t 2
Adding equations (3) and (5),
... (5)
22 z
22 z
= f m (u) + g m (v) – f m (u) – g m (v) = 0
2 +
2x
2t 2
22 z
22 z
\ 2 +
=0
2x
2t 2
This is the required partial differential equation.
PART-B
UNIT-I
s
by using Convolution theorem.
Q2. Find the inverse Laplace Transform of 2
(s + a 2) 2
Answer :
The given function is,
s
(s 2 + a 2) 2
s
Let, f (s) = 2
s + a2
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Then, f(t) = L–1 <
s
F
(s 2 + a 2)
QP.7
>a L–1 <
s
F = cos at H
( s 2 + a 2)
>a L–1 <
1
1
F = a sin at H
(s 2 + a 2)
= cosat
1
Let, g (s) = 2
s + a2
Then, g(t) = L–1 <
1
F
(s 2 + a 2)
1
= a sin at
From convolution theorem,
s
L < 2
F=
(s + a 2) 2
–1
RS
SS
SS a L–1 [ f (s) g (s) = f (t) ) g (t) =
SS
S
0
T
t
#
t
#
f (u) g (t – u) du
0
1
= (cosat) ) a (sinat)
VW
WW
f (u) g (t – u) du WW
WW
W
X
t
1
= a
# cos au sin a (t – u) du
0
t
1
= a
# [sin (au + a (t – u) – 2sin (au – a (t – u)] du
0
< a cos A sin B =
sin (A + B) – sin (A – B)
F
2
t
1
= 2a
# [sin (au + at – au) – sin (au – at + au)] du
0
t
1
= 2a
# [sin at – sin (2au – at)] du
0
SRS t
WVW
t
W
1 SS
= 2a S sin at du – sin (2au – at) du WW
SS
WW
S0
W
0
T
X
t
cos (2au – at)
t
1 *
F 4
= 2a sin at 8 (u) B0 + <
2a
0
#
#
cos (2at – at) cos (0 – at)
1
F4
= 2a * 8 t sin at – 0 B + <
–
2a
2a
1
cos at cos at
= 2a < t sin at + 2a – 2a F
1
= 2a [t sin at + 0]
1
= 2a t sin at
\ L–1 <
1
s
F = 2a t sin at
(s 2 + a 2) 2
OR
Q3.
2
Solve (D – D – 2)y = 20 sin2t where y (0) = 1, y'(0) = 2.
May/June-16, (R15), Q3
Answer :
Given differential equation is,
(D 2 – D – 2) y = 20 sin 2t
Þ
Þ
y (0) = 1, yl (0) = 2
d 2 y dy
–
– 2y = 20 sin 2t
dx 2 dx
y m – yl – 2y = 20 sin 2t
<a D = d F
dx
... (1)
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.8
Applying Laplace Transform on both sides of equation (1),
L 8 y m – yl – 2y B = L 8 20 sin 2t B
Þ
Þ
L 8 y m B – L 8 ylB – 2L 8 y B = 20 L 8sin 2t B
8 s 2 L (y) – sy (0) – yl (0) B – 8 s L [y] – y (0) B – 2 L 8 y B = 20
2
s 2 + (2) 2
< a L 8 y n B = s n L (y) – s n – 1 y (0) – s n – 2 yl (0), L 8 sin at B =
Þ
s 2 L 8 y B – s ( 1 ) – ( 2) – < s L 8 y B – 1 F – 2 L 8 y B =
Þ
s 2 L (y) – s – 2 – s L [y] + 1 – 2 L [y] =
Þ
Þ
Þ
Þ
Þ
Þ
40
s2 + 4
40
s2 + 4
40
(s2 – s – 2) L [y] – (s + 1) = 2
s +4
40
+ (s + 1)
(s2 – s – 2) L [y] = 2
s +4
( s + 1)
40
+
L [y] = 2
(s + 4) (s 2 – s – 2) (s 2 – s – 2)
(s + 1)
40
+
L [y] =
(s + 1) (s – 2) (s 2 + 4) (s + 1) (s – 2)
(s2 – s – 2) L [y] – s – 2 + 1 =
L [y] =
3
2
3
... (3)
... (4)
... (5)
2
40 = A (s + s + 4s + 1) + B (s – 2s + 4s – 8)
+ Cs3 – Cs3 – Cs 2 – 2Cs + Ds 2 – Ds – 2D
... (6)
Substituting s = 2 in equation (5),
40 = A (2 + 1) (2 2 + 4) + B (0) + (Cs + D) (0)
Þ
40 = A (3) (8) + 0
40
Þ A = 24
5
` A= 3
Substituting s = – 1 in equation (5),
40 = A (0) + B (– 1 – 2) (( – 1) 2 + 4) + 0 + 0
Þ
a
F
s2 + a2
40
s2 + 4
40
1
+
(s – 2) (s + 1) (s 2 + 4) (s – 2)
40
Consider,
(s – 2) (s + 1) (s 2 + 4)
Applying partial fractions,
40
A
B
Cs + D
= (s – 2) + (s + 1) + 2
(s – 2) (s + 1) (s 2 + 4)
( s + 4)
Þ 40 = A (s + 1) (s 2 + 4) + B (s – 2) (s 2 + 4) + [Cs + D) (s – 2) (s + 1)]
Þ
... (2)
40 = – 3B (5)
40
B = – 15
–8
` B= 3
Substituting s = 0 in equation (5),
Þ
40 = A (1) (4) + B ( – 2) (4) + C (0) + D ( – 2) (1)
Þ
40 = 4A – 8B – 2D
Þ
5
–8
40 = 4 d 3 n – 8 d 3 n – 2D
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Þ
20 + 64
– 2D
3
84
– 2D = 40 – 3
– 2D = 40 – 28
Þ
– 2D = 12
Þ
D =–6
`
D=–6
Þ
Þ
QP.9
40 =
Comparing the co-efficients of s3 on both sides of equation (6),
Þ
Þ
`
0= A+B+C
5 8+
3 – 3 C=0
3
C= 3 =1
C=1
Substituting the corresponding values in equation (4),
5
d –8 n
3
1 ( s) – 6
40
3
= (s – 2) + (s + 1) + 2
2
(s – 2) (s + 4)
s +4
(s – 6)
5
8
= 3 (s – 2) – 3 (s + 1) + 2
(s + 4)
Substituting the corresponding values in equation (3),
(s – 6)
5
8
+ 1
L [y] = 3 (s – 2) – 3 (s + 1) + 2
s +4 s– 2
8
8
s
6
= 3 ( s – 2) – 3 ( s + 1 ) + 2
–
s + 4 s2 + 4
6
s
8
1
1
L [y] = 3 < (s – 2) – (s + 1) F + 2
– 2
s +4
s +4
Applying inverse Laplace transform on both sides,
( 3 × 2)
s
8
1
1
FH
L – 1 > 3 < (s – 2) – (s + 1 ) + < 2
– 2
s +4
s +4
L -1 [L (y)] =
y=
8 8 2t
–t
B + cos 2t – 3 sin 2t
3 e –e
>a L – 1 < s 1– a H = eat
L -1 <
s
F = cosat
s2 + a2
L -1 <
a
F = sin at B
s2 + a2
8
` y = 3 8 e 2t – e –t B + cos 2t – 3 sin 2t
UNIT-II
1
1
1
1
p2
Find a Fourier Series to represent x – x2 from x = – p to x = p and deduce that 12 = 2 – 2 + 2 – 2 + ...
1
2
3
4
Answer :
May/June-16, (R15), Q4
Q4.
The given function is,
f (x) = x – x2 in (–π, π)
From the definition of trigonometric Fourier series,
f(x) =
a0
+
2
1
Where, a0 =
π
π
∫
−π
∞
∑
∞
an cos nx +
n =1
∑b
n
sin nx
π
f ( x)dx = 1
π
... (1)
n =1
# (x – x ) dx =
2
–π
π
1 < x 2 x3 F
–
π 2
3 –π
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.10
(–p) 2 (–p) 3
1 p 2 p3 p 2 p3
1 p 2 p3
= p > 2 – 3 – d 2 + 3 nH = p < 2 – 3 – 2 – 3 F
–1 2p3
– 2p 2
= p < 3 F= 3
– 2p 2
3
\ a0 =
1
an = π
p
#
–p
p
sin nx
1
(x – x 2) cos nx.dx = π >< (x – x 2) n F –
–p
# (1 – 2x) sinnnx dx H
p
–p
V
SRS
J
NOWW
p
p K
SS
WW
–p
K
O
(– cos nx)
1
sin nx
– cos nx
= π SS< (x – x 2) n F – KKK > (1 – 2x) d
dx H OOOWW
n – (– 2)
2
2
SS
n
n
–p K
OOWW
–p
K
–p
SS
L
PWW
T
X
#
>a
# uvdx = u # vdx – # d dxd u # vdx ndx H
V
SRS
J
NWW
p K
SS
W
p
p O
K
O
sin nx
sin nx
1
– cos nx
OOWW
= π SS< (x – x 2) n F – KKK > (1 – 2x) d
.
–
2
n
d
n
H
SS
n2
n3 –p OOOWWW
–p K
–p
K
SS
L
PWW
T
X
p
p
p
sin nx
cos nx
sin nx
1
= π >< (x – x 2) n F + < (1 – 2x)
F + < 2. 3 F H
n 2 –p
n
–p
–p
sin n (–π)
cos n (–π)
sin n (–π)
1
sin nπ
cos nπ + sin nπ
FH
+ (1 – 2 (–π))
+ 2.
= π >< (π – π) 2 n + (1 – 2π)
2. 3 F – < (– π – (– π) 2)
n
n2
n
n2
n3
(–1)
cos nπ
1
= π < 0 + (1 – 2π) 2 + 0 – 0 – (1 + 2π)
– 0F
n
n2
n
(–1)
(–1)
1
= π < (1 – 2π) 2 – (1 + 2π) 2 F
n
n
n
n
8 a sin nπ = 0 B
8 a cos nπ = (–1) n B
n
2π (–1) n (–1) n 2π (–1) n
1 (–1)
F
= π< 2 –
–
–
n
n2
n2
n2
(–1) n
–1
= π < 4π 2 F
n
n
– 4 (–1)
=
n2
– 4 (–1) n
` an =
n2
1
bn = π
p
# (x – x2) . sin nx.dx
–p
RS
VW
p
p
SS
W
1
– cos nx n H
– cos nx WW
= π SS> (x – x 2) d
(
)
.
x
–
1
–
2
dx
WW
n
n
SS
WW
–p –p
S
TR
X
VW
SS
p
p
WW
p
S
sin nx
sin nx
1
– cos nx n H +
WW
= π SS> (x – x 2) d
(
–
)
–
(
)
x
1
2
–
2
dx
d
n
<
F
>
H
2
2
n
SS
WW
n
n
–
p
–p
–p
S
W
T
X
p
p
p
cos
sin
cos
nx
1>
–
nx
–
nx
n H + < ( 1 – 2 x) 2 F + > 2 d
= π ( x – x 2) d
n H
n
n
n3
–p
–p
#
#
–p
p
sin nx 2 cos nx
1
– cos nx n +
= π > (x – x 2 ) d
(1 – 2x) 2 –
H
n
n
n3
–p
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QP.11
RS
VW
SS
sin n (–π) 2 cos n (–π) WWW
– cos nπ n +
1
sin nπ
cos nπ
2 d – cos n (–π) n
+
+
= π SS (π – π 2) d
>
HWW
π
π
π
π
(
–
)
–
2
–
(–
–
(–
)
)
(
1
2
)
–
1
2
n
n
SS
n2
n3
n2
n3
WW
S
T
X
( – 1) n
(– 1) n
(– 1) n
(– 1) n
1<
2
2
= π (π – π ) (– 1) n + (1 – 2π) .0 – 2. 3 + (π + π) (–1) n – (1 + 2π) .0 + 2. 3 F
n
n
( – 1)
( – 1)
1
= π < – (π – π 2) n + (–π 2 – π) n F
n
n
(– 1) n
– 2 ( – 1) n
(– 1) n
(– 1) n
(– 1 ) n
(– 1 ) n
1
1
= π < –π n + π 2 n – π 2 n – π n F = π < –2π n F =
n
n
– 2 (– 1)
` bn =
n
Substituting the corresponding values in equation (1),
3
3
(–1) n
(–1) n
– 2p 2
f(x) = 3 (2) +
( – 2) n sin nx
( – 4) 2 cos nx +
n
n=1
n=1
/
=
– p2
3 –4
/
3
/ (–n1)
2
3
n
cos nx – 2
n=1
/ ( –n1)
n
sin nx
n=1
– p2
– cos x cos 2x cos 3x
< – sin x + sin 2x – sin 3x + .... F
3 – 4 < 1 2 + 2 2 – 3 2 + ... F – 2
1
2
3
2
–p
cos x cos 2x cos 3x
+ ... F
f(x) = x – x2 = 3 + 4 < 2 –
–
1
22
32
=
sin x sin x sin x
+ 2 < 1 – 22 + 33 – ... F
... (2)
Substituting x = 0 in equation (2),
1
1
1
1
p2
0 0 0
0 = – 3 + 4 < 2 – 2 + 2 – 2 + ... F + 2 < 1 – 2 + 3 ... F
1
2
3
4
>
a cos 0o = 1
H
sin 0o = 0
– p2
1
1
1
1
3 + 4 d 1 2 – 2 2 + 3 2 – 4 2 + ... n
1
1
1
1
p2
4 d 2 – 2 + 2 – 2 + ... n = 3
1
2
3
4
1
1
1
1
p2
2 –
2 + 2 –
2 + ... = 3 × 4
1
2
3
4
1
1
1
1
p2
2 –
2 + 2 –
2 + ... = 12
1
2
3
4
p2
1
1 + 1
1
12 = 1 2 – 2 2 3 2 – 4 2 + ...
Hence proved.
0=
Þ
Þ
Þ
`
π
3,
π
0#x# 3
Q5
If f(x) =
π
2π
= 0,
3 #x# 3
2π
=– π
3,
3 #x#π
1
2 <
1
Then f(x) =
cos x – 5 cos 5x + 7 cos 7x + ... F
3
Answer :
OR
May/June-16, (R15), Q5
Given that,
Z] π
]]
,
]] 3
]]
f (x) = ][ 0,
]]
]] π
]] – ,
3
\
π _b
0 # x # 3 bb
bb
π
2π bb`
3 # x # 3 bb
bb
2π
bb
x
#
#
π
b
3
a
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.12
Here, the interval is 0 to 2 p . half-range cosine series is to be determined
Half - range cosine series is expressed as,
3
f(x) = a0 +
/ a cos nx
... (1)
n
n=1
Where,
p
2
a0 = p
#
f (x) dx
0
And
p
2
an = p
#
f (x) cos nx dx
0
RS p/3
VW
2p/3
p
S
WW
2 SS
a0 = p S
f (x) dx +
f (x) dx +
f (x) dx WW
SS
WW
2p/3
p/3
S0
W
T
X V
SRS p/3
WW
2p/3
p
S
W
2S
p
p
= pS d
0 dx +
dx +
–
dx WW
n
<
F
3
3
SS
WW
2p/3
p/3
S0
W
T
X
RS p/3
V
WW
p
SS
W
2 p
–p
= p SS 3
dx + 0 3
dx WW
SS
WW
2p/3
S 0
W
T
X
2 p <8 Bp/3 8 Bp F
= p × 3 x 0 – x 2p/3
#
#
#
#
#
#
#
#
2 p
2p
= 3 > 3 – 0 – < p – 3 FH
2 p p
= 3 <3 – 3F
=0
` a0 = 0
Similarly,
an =
=
=
=
SRS p/3
WVW
2p/3
p
W
2 SS
f (x) cos nx dx +
f (x) cos nx dx +
f (x) cos nx dx WW
p SS
WW
SS 0
2p/3
p/3
W
RST p/3
VW X
p
2p/3
S
WW
2 SS
p
d p n cos nx dx WW
+
+
0
cos
nx
dx
cos
nx
dx
S
pS
3
3
WW
SS 0
p/3
2p/3
W
TR
X
V
SS p/3
WW
p
W
2 SS p
p
cos nx dx + 0 + d – 3 n
cos nx dx WW
p SS 3
WW
SS 0
2p/3
W
T
X
p/3
p
2 p >< sin nx F
sin
nx
<
F H
n 0 –
n
p 3
2p/3
#
#
#
#
#
#
#
#
p
1
p
2 1
= 3 > n <sin n 3 – sin (0) F H – n <sin np – sin 2n 3 F
p
p
2
= 3n <sin n 3 – 0 + sin 2n 3 F
>
a sin np = 0
H
sin (0) = 0
2
p
3np – np n H
= 3n >sin n 3 + sin d
3
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May/June-16 (R15) Question Paper with Solutions
QP.13
2
p
np
= 3n >sin n 3 + sin d np – 3 n H
2
np
p
= 3n >sin n 3 + ( – 1) n sin d – 3 n H
p
p
2
= 3n <sin n 3 – ( – 1) n sin 3 ) F
p
2
= 3n sin n 3 81 – ( – 1) n B
p
2
` an = 3n sin n 3 81 – ( – 1 ) n B
For n = 1
2
p
Þ a1 = 3 (1) sin (1) 3 81 – ( – 1) 1 B
p
2
= 3 sin 3 81 + 1 B
4
p
= 3 sin 3
For n = 2
2
p
Þ a2 = 3 (2) sin 2 3 81 – 1 B
8 a sin (np + q) = ( – 1) n sin q B
8 a sin ( – q ) = – sin q B
=0
` an = 0 for even values of n
4
p
= 3n sin n 3 for odd values of n.
Substituting a0 and an values in equation (1),
3
p
4
f(x) = 0 +
3n sin n 3 cos nx
/
4
= 3
n=1
3
/ < 1n sin n p3 cos nx F
n=1
4
1
1
p
p
1
p
= 3 <sin 3 cos x + 3 cos 3x (0) + 5 sin 5 3 cos 5x + 7 sin 7 3 cos 7x + ... F
31
31
4 3
= 3 < 2 cos x – 2 5 cos 5x + 2 7 cos 7x + ... F
>a sin 5 p3 = sin (2p – p3 ) = sin p3 = d 23 n H
3
4
1
1
= 3 × 2 < cos x – 5 cos 5x + 7 cos 7x + ... F
cos 5x cos 7x
2 <
cos x – 5 + 7 + ... F
3
=
cos 5x cos 7x
2 <
cos x – 5 + 7 + ... F
3
Hence proved.
` f(x) =
UNIT-III
3
Q6.
Show that
#
0
sin pl sinlx
p
dl = 2 sin x, for 0 # x # p
1 – l2
=0
for x > p
May/June-16, (R15), Q6
Answer :
Given integral is,
3
#
0
sin pl sin lx
π
dl = 2 sin x for 0 £ x £ p
1 – l2
=0
for x > p
Consider the function
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.14
p
f(x) = 2 sin x for 0 < x < p
=0
for x > p
... (1)
Here, the interval is 0 to p
Fourier sine integral is expressed as,
3
3
# sin lx #
2
f(x) = p
0
f (t) sin lt dt dl
... (2)
0
Substituting the equation (1) in equation (2),
p
3
# sin lx #
2
f(x) = p
0
0
2 p
= p×2
# sin lx # sin t sin lt dt dl
0
0
3
3
# sin lx #
0
0
3
=
3
# sin lx #
0
0
< ` sin A sin B = 1 cos (A – B) – cos (A + B) F
2
1< `
2 cos lt – t j – cos (lt + t) F dt dl
18
2 cos t (l – 1) – cos t (l + 1) B dt dl
3
1
= 2
a f (x) = f (t)
H
=p
2 sin t
p
3
=
>
p
2 sin t sin lt dt dl
# sin lx< (l 1– 1) sin (l – 1) t – (l 1+ 1) sin (l + 1) t F
p
dl
0
0
3
1
= 2
# sin lx< (l 1– 1) 8sin (l – 1) p – sin (l – 1) (0) BF – (l 1+ 1) 8sin (l + 1) p – sin (l + 1) 0 Bdl
0
# sin lx * < (l1– 1 8sin (pl – p) – 0 F – sin((lpl+ 1+)p) –0 B 4 dl
3
1
= 2
0
# sin lx * < – sin (lp––1pl) F – < sin((lpl+ 1+)p) F 4 dl
3
1
= 2
0
F 4 dl
# sin lx * < sin(1(p– –l)pl F – < –(lsin+ pl
1)
3
1
= 2
0
>
a sin (p – q) = sin q
H
sin (p + q) = – sin q
3
1
= 2
pl sin pl F
# sin lx< sin
1 – l + 1 + l dl
0
3
1
= 2
# sin lx sin pl< 1 –1 l + 1 +1 l Fdl
0
3
1
= 2
# sin lx sin pl< (11 +– ll)+(11 +– ll) Fdl
0
3
1
= 2
# sin lx sin pl< 1 –2l
0
2
F dl
3
1
= 22
#
0
sin lx sin pl
dl
1 – l2
3
#
Þ
0
3
`
#
0
sin lx sin pl
dl = f(x)
1 – l2
sin lx sin pl
p
dl = 2 sin x
1 – l2
Hence proved.
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May/June-16 (R15) Question Paper with Solutions
QP.15
OR
Q7.
3
Find Fourier transform of f(x) = 1 – x for x # 1 = 0 for x 2 1 and hence find
2
#
0
Answer :
x cos x – sin x
x
cos 2 dx .
x3
May/June-16, (R15), Q7
Given function is,
2
f(x) = * 1 – x for – 1 # x # 1
4
0 for x 1 – 1 and x > 1
By definition, the Fourier transform of f (x) is,
1
∞
2π ∫
F(s) =
1
1
(1 − x ) . e
2π ∫
f ( x) . eisx dx =
−∞
2
isx
dx
−1
1
=
isx
1 
2 e
(1 − x ) 
2π 
 is
 isx 

 isx 
 − (−2 x) e  + (−2) e 
 i2s2 

 i 3s3 




 −1
=
isx
1 
2 e
(1 − x ) 
2π 
 is

 isx
 − (2 x) e

 s2


=
 e is
1 
0 − (2) 2
s
2π 

 2 is  
 e −is
+

e
0
2
−
+


 is 3  
 s2


 
=
1  2 is
2 is
−is
−is 
− (e + e ) + is 3 (e − e )
2π  s 2

1
 2 isx 
+
 is 3 e 
 −1

 2 −is 
+
e 
 is 3


1  2
2

− (2 cos s ) + is 3 (2i sin s )
2π  s 2

=
RS
iq
–iq
SS a cos θ = e + e
2
SS
iq
SS
e
e –iq
–
SS
sin θ =
2i
T
2 2 1
. ( s cos s − sin s )
π s3
F(s) = −
VW
WW
WW
WW
WW
X
... (1)
Applying the inverse fourier transform to equation (1),
3
1
2π
#
1
−
–3
∞
2π
∫
−∞
F (s) .e –isx ds = f (x)
2 2 1
π s3
(s cos s − sin s )e −isx ds
1 − x 2 ,
1
–isx

=
(
s
cos
s
–
sin
s
)
e
ds
s3
 0,
–3
1
Substituting x = 2 in equation (2),
3
Þ
2
π
#
∞
2
π
1
∫s
−∞
1
∫s
−∞
3
| x |< 1
otherwise
... (2)
( s cos s − sin s ) e −isx ds = 1–
−∞
∞
∞
1
∫s
= f (x)
3
3
s
s

( s cos s − sin s ) cos − i sin  ds =
2
2

×
π
2
∞
s
( s cos s − sin s ) cos  ds − i ( s cos s − sin s ) sin s  ds
=
2
s3
2
−∞
∫
... (3)
Comparing the real and imaginary parts both sides of equation (3),
∞
1
∫s
−∞
3
s
( s cos s − sin s ) cos  ds =
2
... (4)
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.16
3
#
And
–3
(s cos s – sin s)
s
sin d 2 n ds = 0
s3
... (5)
The integrand of equation (5), indeed, is an odd function of s and hence the integral vanishes. i.e. 0
The integrand on the L.H.S of equation (4) is an even function of s. Hence, equation (4) becomes,
∞
2
∫
( s cos s − sin s )
s3
0
s
cos  ds
2
3π
=
8




∞
∫
−∞
∞

f ( x)dx = 2 f ( x)dx

0

∫
... (6)
Replacing s with x and simplifying, we get from equation (6),
3
2
#
0
3
\
#
0
x
x cos x – sin x
3π
cos d 2 n dx = 8
x3
3π
x
x cos x – sin x
cos d 2 n dx =
x3
16
UNIT-IV
Q8.
Find the partial differential equation of all spheres whose centre lie on Z - axis and given by equation
x2 + y2 + (z – a)2 = b2, a and b being constants.
May/June-16, (R15), Q8
Answer :
Given equation of all spheres whose center lie on Z-axis is,
(x – 0)2 + (y – 0)2 + (z – a)2 = b2
x2 + y2 + (z – a)2 = b2 ...(1)
Where a, b are constants
Differentiating equation (1) partially with respect to ‘x’,
2z
2x + 0 + 2(z – a) 2x = 0
< a 2z = p F
Þ x + (z – a) (p) = 0
2x
Þ
(z – a)p = – x
–x
Þ (z – a) = p
–x
\ (z – a) = p
... (2)
Differentiating equation (1) partially with respect to ‘y’,
2z
0 + 2y + (z – a) 2x = 0
< a q = 2z F
Þ 2y + 2(z – a)q = 0
2y
Þ
(z – a)q = – y
–y
Þ (z – a) = q
Substituting equation (3) in equation (2),
–y
–x
p = q
y
x
Þ
p = q
Þ
xq = py
Þ
xq – py = 0
... (3)
\ xq – py = 0
This is the required partial differential equation.
OR
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May/June-16 (R15) Question Paper with Solutions
QP.17
Q9.
A string is stretched and fastened to two points l apart. Motion is started by displacing the string in
px
the form y = a sin l from which it is released at a time t=0. Show that the displacement of any point
px
pct
at a distance x from one end at time t is given by y ( x ,t) = a sin d l n cos d l n .
Answer :
May/June-16, (R15), Q9
y
y (x, t)
o
l
x
Wave equation is,
22 y
22 y
=
2
2t 2
2x 2
Boundary conditions are,
(i)
... (1)
y(0, t) = 0
(ii)
y(l, t) = 0
2y
(iii) 2t (x, 0) = 0
2y
πx
(iv) 2t (x, 0) = a sin l
y(x, t) = (c1 cos px + c2 sin px) (c3cos pct + c4sin pct) which satisfies the boundary conditions.
... (2)
Applying condition (i) in equation (2),
y(0, t) = c1(c3 cos pct + c4 sin pct)
Where, c3 cos pct + c4 sin pct ¹ 0
` c1 = 0
Substituting in equation (2),
y(x, t) = (c2 sin px)(c3 cos pct + c4 sin pct)
Applying condition (ii) in equation (3),
y(l, t) = c2 sin pl(c3 cos pct + c4 sin pct)
0 = c2 sin pl(c3 cos pct + c4 sin pct)
When c2 ¹ 0 and c3 cos pct + c4 sin pct ¹ 0
\
sin pl = n π
pl = n π
nπ
p= 6
Substituting p value in equation (3),
nπx
nπct
nπct
y(x, t) = c2 sin l ` c3 cos l + c4 sin l
Þ
j
... (4)
Differentiating equation (4) with respect to ‘t’,
2y
nπx d
nπc
nπct
nπc
nπct n
2t (x, t) = c2 sin l – c3 l sin l + c4 l cos l
Applying condition (iii),
2y
nπx d nπc n
2t (x, 0) = c2 sin l c4 l
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.18
nπx
0 = c2 sin l (c4 nπc)
< a c2 ! 0 and sin nπx ! 0 F
6
` c4 = 0
Substituting above value in equation (4),
nπx
nπct
y(x, t) = c2 sin l c3 cos l
nπx
nπct
y(x, t) = bn sin l cos l
[Put c2, c3 = bn]
\ General solution is,
3
y(x, t) =
/ bn sin nπl x cos nπlct
n=1
Applying boundary conditions (iv) in equation (5),
px
y(x, 0) = a sin l
3
px
=
bn sin n l
n=1
px
px
px
Þ
a sin l = b1 sin l + b2 sin 2 l + ...
px
Equating the co-efficient of sin l ,
/
b1 = a and b2, b3...... bn = 0
For n = 1,
p (1) x
p (1) ct
l cos l
px
pct
= a sin l cos l
px
p ct
` y (x, t) = a sin d l n cos d l n
y (x, t) = a sin
Hence proved.
UNIT-IV
Q10. Solve the difference equation, using Z-transorm un+2 – un = 2n, Where u0 = 0 and u1 = 1
May/June-16, (R15), Q10
Answer :
For answer refer Unit-V, Page No. 5.31, Q43.
OR
2z2 + 3z + 4
Q11. If f(z) =
, z > 3 , then find the values of f(1), f(2), f(3).
( z – 3) 3
Answer :
May/June-16, (R15), Q11
Given that,
f(z) =
2z 2 + 3z + 4
( z – 3) 3
... (1)
f(1) = ?
f(2) = ?
f(3) = ?
Þ
3 4
z2 < 2 + z + 2 F
z
f(z) =
3
3
z3 <1 – z F
RS
V
SS
3 4 WWW
S < 2 + z + z 2 F WW
1S
WW
f(z) = z SS
3
SS
W
3
SS d 1 – z n WWW
S
W
T
X
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... (2)
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May/June-16 (R15) Question Paper with Solutions
QP.19
From initial value theorem,
f(0) = Lim f (z)
... (3)
z"3
Substituting equation (2) in equation (3),
RS
V
SS 2 + 3 + 4 WWW
2
z
1S
z WW
f(0) = Lim z SS
3 W
z"3 S
3
SS <1 – F WWW
z
S
W
T
X
1
= 3
=0
f(0) = 0
f(1) is expressed as,
f(1) = Lim z 8 f (z) – f (0) B
... (4)
z"0
Substituting the corresponding values in equation (4),
RS R
VW
V
SS SS
WW
3 4 WW
SS 1 SS 2 + z + 2 WW
WW
z
W
W
f(1) = Lim z SS z SS
3 W – 0W
z"3 S S
WW
SS SS d 1 – 3 n WWW
WW
z
W
SS S
W
T
X
X
RS T
VW
SS 2 + 3 + 4 WW
z z2 W
S
W
= Lim SS
3 W
z"3 S
SS d 1 – 3 n WWW
z
S
W
T
X
3
4
2+ 3 + 3
=
3
d1 – 3 n
3
2+0+0
= 1– 0
=2
\
f(1) = 2
f(2) is expressed as
f(2) = Lim z 2 [f (z) – f (0) – f (1) z –1]
... (5)
z"3
Substituting the corresponding values in equation (5),
RS R
VW
V
SS SS
WW
3 4 WW
SS 1 SS 2 + z + 2 WW
W
z
–1 W
W
W
f(2) = Lim z 2 SS z SS
3 W – 0 – 2z W
WW
z"3
SS SS
WW
3
SS SS d 1 – z n WW
WW
S T
W
X
TR
X
V
SS
W
W
3
4
SS < 2 + + 2 F
WW
z z
S1
2 WW
= Lim z 2 SS z
–
3
z WW
SS
z"3
WW
d1 – 3 n
SS
z
WW
S
TR
X V
SS
3W
SS 2 + 3 + 4 – 2 d 1 – 3 n WWW
z z2
z WW
z2 S
= Lim z SS
WW
3
S
z"3
SS
WW
d1 – 3 n
z
SS
WW
T
X
SRS
WV
3
4
3
3
SS 2 + + 2 – 2 <1 – + 2 – 13 F WWW
z z
z z
z WW
S
= Lim z SS
WW
3
z"3 S
3
SS
WW
d1 – n
z
SS
WW
T
X
8 a (a – b) 3 = a3 + b3 – 3ab + 3ab 2 B
SIA GROUP
MATHEMATICS-II [JNTU-ANANTAPUR]
QP.20
RS
V
SS 2 + 3 + 4 – 2 + 6 + 6 – 2 WWW
z z2
z z 2 z3 W
S
WW
= Lim z SS
3
WW
z"3 S
3
SS
d1 – n
WW
z
S
TRS
X
VW
SS 9 – 2 + 2 WW
S z z 2 z3 WW
= Lim z SS
3 W
z"3 S
SS d 1 – 3 n WWW
z
S
W
RS T
VW X
2
2
SS 9 – +
W
z z 2 WW
S
W
= Lim SS
3 W
z"3 S
SS d 1 – 3 n WWW
z
S
W
X
9 – 0T + 0
= 1– 0
=9
\ f(2) = 9
f(3) is expressed as
f(3) = Lim z3 [f (z) – f (0) – f (1) z –1 – f (2) z –2]
... (6)
z"3
Substituting the corresponding values in equation (6),
RS R
VW
V
SS SS
WW
3 4 WW
+
+
2
S 1 SS
WW
z z 2 WW
2
9
3S
W
W
f(3) = Lim z SS z SS
3 W–0– z – 2W
z WW
z"3
SS SS
3 n WW
d
WW
SS SS 1 – z
WW
S T
W
X
X
RST
VW
N
J
SSK 2z 2 + 3z + 4 O
WW
K
O
= Lim z SSK
– 2z – 9 WW
3 O
O
z " 3 SK
WW
SSKK z d 1 – 3 n OO
WW
O
z
SSKL
W
P
X
RTS
VW
SS 2
3 j3 F – 9z ` 1 – 3 j3 WW
<
+
+
2
z
3
z
4
2
z
z
1
–
–
`
SS
z
z WW
WW
= Lim z SS
3
WW
z"3 S
z`1 – 3 z j
SS
W
X
RS T
VW
SS 2z 2 + 3z + 4 – 2z 2 1 – 3 + 3 – 1
3 + 3 – 1 n WW
–
9
1
z
–
d
n
d
z
z
SS
z2
z3 WW
z2
z3
WW
= Lim SS
3
WW
z"3 S
`1 – 3 z j
SS
W
X
RTS
VW
SS
WW
2
27
9
2
2
SS d 2z + 3z + 4 – 2z + 6z – 6 + z n – 9z + 27 – z + z 2 WW
WW
= Lim SS
3
WW
z"3 S
`1 – 3 z j
SS
W
TRS
X
VW
SS
2 27 9 WW
SS < 9z – 9z – 2 + 27– z – z + z 2 F WW
WW
= Lim SS
3
WW
z"3 S
`1 – 3 z j
SS
W
X
RTS
VW
25
9
SS 0 + 25 –
+ 2 WW
z
S
z WW
= Lim SS
3
WW
3
z"3 S
1
–
`
j
S
W
z
RTS
VW X
25
9
SS 25 –
W
z + z 2 WW
S
W
= Lim SS
3 W
z"3 S
S ` 1 – 3 z j WW
T
X
25 – 0 + 0
=
1– 0
= 25
\ f(3) = 25
\ f(1) = 2, f(2) = 9 and f(3) = 25
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May-2018 (R15) Question Paper with Solutions
QP.1
Code No.: 15A54201
Jawaharlal Nehru Technological University Anantapur
R15
B.Tech. I Year II Semester Regular & Supplementary Examinations
Solutions
May - 2018
MATheMATics-ii
( Common to all )
Time: 3 Hours
Max. Marks: 70
Part-a
(Compulsory Question)
1.
Answer the following: (10 × 2 = 20 Marks)
(a)
Find the Laplace transform of cos2t. (Unit-I)
(c)
s
. (Unit-I)
s 2 + s + 13
Find the half-range sine series for the function f(t) = t – t2 0 < t < 1. (Unit-II)
(d)
Obtain the Fourier series for f(x) = x in the interval (0,2). (Unit-II)
(e)
State convolution theorem of the inverse transforms. (Unit-III)
(f)
Find the Fourier transformation of e–x . (Unit-III)
(g)
Solve py3 + qx2 = 0 by method of separation of variables. (Unit-IV)
(h)
Form the partial differential equation by eliminating the arbitrary constants 2z =
(i)
Find the Z-transformation of (nan). (Unit-V)
(j)
1
Find the inverse Z-transformation of z – 2 . (Unit-V)
(b)
Find the inverse Laplace transform of
2
2
x2 + y
2
2 . (Unit-IV)
a
b
Part-B
(Answer all five units, 5 × 10 = 50 Marks)
Unit-I
2.
(a)
(b)
Z]
]]Esin wt, 0 < t < p/w
Find the Laplace transform of the function f(t) = []
having period 2p/w. (Unit-I, Topic No. 1.9)
p
]]0,
w < t < 2p/w
\
1 at bt
Find the Laplace transform of t (e – e ) (Unit-I, Topic No. 1.4)
OR
3.
Solve by Laplace transform method y'' – 3y' + 2y = 4, where y(0) = 2, y'(0) = 3. (Unit-I, Topic No.1.11)
Unit-II
4.
(a)
Find the Fourier series to represent (x – x2) from x = –p to x = p. (Unit-II, Topic No. 2.2)
(b)
Find the half range cosine series for the function f(x) = x2 in the range 0 # x # p. (Unit-II, Topic No. 2.4)
OR
5.
(a)
(b)
Find the complex form of the Fourier series of f(x) = e–x in – 1 ≤ x ≤ 1 (Unit-II, Topic No. 2.5)
Z]
]] 1 – x if 0 < x < 1
]4
2
as the Fourier series of sine terms. (Unit-II, Topic No. 2.4)
Expand f(x) = ][
]] x – 3 if 1 < x < 1
]
4
2
\
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QP.2
Unit-III
6.
7.
1, |x |< 1
Find the Fourier transform of f(x) = (
Hence deduce that
0, |x |> 1
OR
Find the Fourier transformation of e –a
2 x2
3
#
sinx
x dx (Unit-III, Topic No. 3.2)
0
, a > 0. (Unit-III, Topic No. 3.2)
Unit-IV
8.
(a)
Form the partial differential equation by eliminating the arbitrary functions f(x2 + y2, z–xy) = 0.
(Unit-IV, Topic No. 4.1)
(b)
The ends A and B of a rod 20 cm long have the temperature at 30°C and 80°C until steady state prevails. The
temperature of the ends is changed to 40°C and 60°C respectably. Find the temperature distribution in the rod at
time t. (Unit-IV, Topic No. 4.3)
OR
9.
(a)
Find the partial differential equation by eliminating the arbitrary constants from:
(x – a)2 + (y – b)2 + z2 = c2. (Unit-IV, Topic No. 4.1)
(b)
A tightly stretched string of length 1 with fixed ends is initially in equilibrium position. It is set vibrating by giving
px
each point a velocity v0sin3 ` 1 j . (Unit-IV, Topic No. 4.3)
Unit-V
10.
(a)
z2
Evaluate Z-transformation of (z – 1) (z – 3) using convolution theorem. (Unit-V, Topic No. 5.5)
(b)
1
1
Using Z-transforms solve yn + 4 yn – 1 = un + 3 un–1 where un is a unit step sequence. (Unit-V, Topic No. 5.6)
OR
11.
z2
using convolution theorem. (Unit-V, Topic No. 5.5)
(z – 1) 3
(a)
Evaluate Z-transformation of
(b)
Using Z-transforms solve Un+2 – 2Un+1 + Un = 3n + 5. (Unit-V, Topic No. 5.6)
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May-2018 (R15) Question Paper with Solutions
QP.3
solUTioNs To MAy-2018, R15, QP
Part-a
Q1.
Find the laplace transform of cos2t.
(a)
May-18, (R15), Q1(a)
Answer :
Given function is,
cos2t
L{cos2t} = L &
1 + cos 2t 0
2
1
= 2 [L{1 + cos2t}]
1
= 2 [L{1} + L{cos2t}]
s 0
1 1
= 2 &s + 2
s +4
1 RS s 2 + 4 + s 2 WVW
= 2 SSS
s (s 2 + 4) WW
T
X
1 RSS 2s 2 + 4 VWW
= 2 SS
s (s 2 + 4) WW
T
X
2
+
2 (s 2)
=
2s ( s 2 + 4 )
\
=
s2 + 2
s ( s 2 + 4)
L{cos2t} =
s2 + 2
s ( s 2 + 4)
(b) Find the inverse laplace transform of
s
.
s 2 + s + 13
May-18, (R15), Q1(b)
Answer :
Given function is,
s
s 2 + s + 13
Z]
_
]] s + 1 – 1 bbb
s
2
2
–1
`
. = L [] J
L $ 2
2
s + s + 13
]] KK s + 1 OON + 51 bbb
2P
4 a
\ L
Z]
_b
1
1
]]
bb
s+ 2
2
= L–1 [] J 1 N2 51 – J 1 N2 51 `b
KK s + OO +
]] KK s + OO +
b
2P
4
4 ba
2P
L
\L
]Z]
b_b
]Z]
b_b
1
1
s+ 2
]
b
]
b
2
–1
–1
= L [] J 1 N2 51 `b – L [] J 1 N2 51 `b
]] KK s + OO +
b
]] KK s + OO +
b
2P
4 ba
2P
4 ba
\L
\L
Z]
_b
]]
bb
+ 12
s
b_b
1
b 1 L–1 ]Z]]
= L–1 ][]
2b
2` –
J
N
J
N
b
2
2
2
[
J
J
N
N
51
51
O b
OO `b
]] K 1 O KK
]] KK s + 1 OO + KKK
O bb
]
] K s + 2 O + K 2 OO bb
2
2
P L
P L
P a
P a
\L
\L
RS
VW
s+a
–at
SSa L –1 &
cos b t WW
2
2 0 = e
J
N
–1
–
1
+
+
51 O
51
(s a ) b
SS
WW
1 K 1
= e 2 t cos 2 t – 2 KK
e 2 t sin 2 t OO
SS –1
W
1
1
–at
KK 51
OO
SS L ' (s + a) 2 + b 2 1 = b e sin btWWW
2
L
P
T
X
–1
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QP.4
51
2 t –
–t
= e 2 cos
–1
1
e 2 t sin
51
51
2 t
R
V
–t S
51
51 W
1
= e 2 SSScos 2 t –
sin 2 t WWW
S
W
51
T
X
RS
VW
–
t
51
51
s
. = e 2 SSScos 2 t – 1 sin 2 t WWW .
\
L–1 $ 2
s + s + 13
S
W
51
T
X
(c) Find the half-range sine series for the function f(t) = t – t2, 0 < t < 1.
May-18, (R15), Q1(c)
Answer :
For answer refer Unit-II, Q31.
(d) obtain the Fourier series for f(x) = x in the interval (0, 2).
May-18, (R15), Q1(d)
Answer :
Given function is,
f(x) = x, (0, 2)
The Fourier series of f(x) in the interval (0, 2L) is given as,
3
3
a
n px
n px
f(x) = 20 +
an cos ` L j +
b n sin ` L j
n=1
n =1
/
/
Here, L = 1
2L
#
1
a0 = L
f (x) dx
0
1
= 1
2
# x dx
0
J x2 N 2
= KK 2 OO
L P0
1
= 2 (22 – 02)
=2
2L
#
1
an = L
f (x) cos n px d x
0
2
# x cos npx dx
1
= 1
0
2
2
=x
# cos npx dx
#
–
d
dx (x)
# cos npx dx dx
0
0
RS J sin npx NVW 2
= SSx KK np OOWW –
PX 0
T L
2
# (1)KJK sinnnppx ONO dx
0
L
P
1
1
= np [2 sin np(2) – 0 sin np(0)] – np
2
# sin npx dx
0
1 – cos npx 2
1
= np [2(0) – 0] – np ` np j 0
1
1
= np (0) + 2 2 6cos npx@ 02
np
=0+
=
1
[cos 2np – cos np(0)]
n2 p2
1
(1 – 1) = 0
n2 p2
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May-2018 (R15) Question Paper with Solutions
QP.5
2L
#
1
bn = L
f (x) sin npx dx
0
2
# x sin npx dx
1
= 1
0
2
2
# sin npx dx
=x
#
–
d
d x (x )
# sin npx dx dx
0
0
R – cos npx WV 2
jW 0 –
= SSx `
np
T
X
2
# (1)` – cosnpnpx j dx
0
–1
1
= np [2cos 2np – 0 cos np(0)] + np
2
# cos npx dx
0
1
1
= – np [2(1) – 0] + np
JK sin npx NO 2
K
O
L np P 0
1
1
= – np (2) + 2 2 (sin np(2) – sin np(0))
np
1
2
= – np + 2 2 (0 – 0)
np
2
= – np
⇒
⇒
Substituting the values of a0, an and bn in f(x),
3
3
JK –2 NO
npx
npx
2
f(x) = 2 +
(0) cos ` 1 j +
K np O sin ` 1 j
n=1
n=1 L
P
/
/
/ sin npx .
3
2
f(x) = 1 – np
(e)
n=1
State convolution theorem of the inverse transforms.
May-18, (R15), Q1(e)
Answer :
Let the Fourier transform of f(t) and g(t) be F(w) and G(w) respectively.
i.e., F[f(t)] = F(W)
F[g(t)] = G(W)
Then, by convolution theorem,
f * g = F–1{F(W). G(W)}
2
Find the Fourier transformation of e–x .
(f)
May-18, (R15), Q1(f)
Answer :
Given function is,
f(x) = e–x2
From the definition of Fourier transform,
3
#
F{f(x)}=
f (x) .eisx dx
–3
3
=
#
2
e –x .e isx dx
–3
3
=
#
e –x
2
+isx
dx
–3
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QP.6
3
=
#
e – (x
–3
3
=
#
2
–isx)
e –;(x)
2
dx
is
is 2 is 2
– (2)(x)b 2 l + b 2 l –b 2 l E
dx
–3
3
=
#
e
2
is 2
–;(x) 2 – (2)(x)b i s l + b i s l E + b 2 l
2
2
dx
–3
3
=
#
is
–3
3
=
2
e –b x – 2 l + i
#
2b
2
e – b x – 2 l .e i
is
s l2
2 dx
s 2
2b 2 l
dx
–3
= e
3
s 2
–b 2 l
#
2
e –b x – 2 l dx
is
–3
is
Let, x – 2 = t
⇒
dx – 0 = dt
⇒
dx = dt
is
U.L : If x = + ¥, ( ¥) – 2 = t Þ t = ¥
is
L.L : If x = – ¥, (– ¥) – 2 = t Þ t = – ¥
Substituting the corresponding values in above integral,
F{f(x)} = e
s 2
–b 2 l
3
#
2
e –t .1dt
–3
2
= e –b 2 l
s
3
#
2
e –t dt
–3
\
=e
s 2
–b 2 l
=
s2
p .e – 4
^ ph
>a
3
#
–3
e –x dx = p H
2
s2
F{e–x } = p .e – 4
2
(g) Solve py3 + qx2 = 0 by method of separation of variables.
May-18, (R15), Q1(g)
Answer :
Given equation is,
py3 + qx2 = 0
... (1)
Let the solution of equation (1) be,
z = XY
... (2)
Partially differentiating equation (2) with respect to x and y.
2z
2x = X'Y
⇒
p = X'Y
2z
2y = XY'
⇒
q = XY'
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May-2018 (R15) Question Paper with Solutions
QP.7
Substituting p and q values in equation (1)
(X'Y)y3 + (XY')x2 = 0
⇒
X'Yy3 = – XY'x2
⇒
X' 1
– Y' 1
X . x2 = Y . y 3
Each side is equated to constant k.
– Y' 1
X' 1
& X . 2 = k Y . y3 = k
x
X'
& X = k x2 Y ' = – k y 3
Y
Integrating on both sides,
J x3 N
log X = k KK 3 OO + log C 1
L P
kx3
& log X – log C 1 = 3
kx3
& log (X/C1) = 3
kx3
X
&C =e3
1
& X = C1 e
kx3
3
4
KJ y ON
log Y = – k KK 4 OO + log C2
L P
– ky 4
log Y – log C2 = 4
– ky 4
log (Y/C2) = 4
– ky 4
Y
4
C =e
2
& Y = C2 e
– ky 4
4
Substituting the values of X and Y in equation (2)
z = C1 e
kx3
3
C2 e
–ky 4
4
JK 3
N
y 4 OO
KK x
– 4 OO
P
= A e kKL 3
\
[a C1 C2 = A]
The solution is,
KJK x3 y 4 OON
K
– 4 OO
P
z = A e kKL 3
2
(h) Form the partial differential equation by eliminating the arbitrary constans 2z =
Answer :
x2 y
.
+
a2 b2
May-18, (R15), Q1(h)
For answer refer Unit-IV, Q11(b).
(i)
Find the Z-transformation of (nan).
May-18, (R15), Q1(i)
Answer :
For answer refer Unit-V, Q6.
(j)
1
Find the inverse Z-transformation of z – 2 .
May-18, (R15), Q1(j)
Answer :
Given function is,
1
z–2
RS 1 VW
z–1 SS z – 2 WW = 2n – 1 u(n – 1)
T
X
2n
= 2 u(n – 1)
RS 1 VW 2 n
\ z–1 SS z – 2 WW = 2 u(n – 1)
T
X
RS –1 RS 1 VW
V
Sa z S z – a W = a n –1 u (n – 1)WW
T
T
X
X
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QP.8
Part-B
Unit-I
Z]
]]Esin wt, 0 < t < p/w
Q2. (a) Find the laplace transform of the function f(t) = ][
having period 2p/w.
p
]]0,
w < t < 2 p /w
\
Answer :
May-18, (R15), Q2(a)
Given function is,
Z]
]] E sin wt, 0 < t < p
]
w
f(t) = ][
2p
p
]]0,
]
w <t< w
\
2p
Period, T = w
The Laplace transform for periodic functions is given as,
1
L{f(t)} =
1 – e –sT
=
T
#
f (t) e –st dt
0
2p/w
#
1
1 – e –s(2p/w)
=
1
–2ps
1– e w
=
1
–2ps
1– e w
=
1
–2ps
1– e w
=
E
–2ps
1– e w
f (t) e –st dt
0
2p/w
RS p/w
VW
SS
WW
–st
–st
+
f
(
t
)
e
dt
f
(
t
)
e
dt
SS
WW
p/w
0
T
X
2p/w
RS p/w
VW
SS
W
–st
–st
0.e dtWW
SS E sin wt e dt +
W
0
p/w
T
X
RS p/w
VW
SS
W
–st
SS e sin wt dt + 0WWW
0
T
X
#
#
#
#
#
p/w
#
e –st sin wt dt
0
R
SSSa
T
=
E
–2ps
1– e w
RS e –st
VW p/w
SS 2
2 (– s sin wt – w cos wt)W
W0
s +w
T
X
=
E
–2ps
1– e w
RSRS e –s(p/w)
VW RS e –s(0)
VWVW
SSSS 2
– SS 2
2 (– s sin w (p/w) – w cos w (p/w))W
2 (– s sin w (0) – w cos w (0))W
W
WWW
s +w
s +w
TT
X T
XX
=
E
–2ps
1– e w
RSRS e –sp/w
VW RS 1
VWVW
SSSS 2
(– s sin p – w cos p)WW – SS 2
(– s sin (0) – w cos (0))WWWW
s + w2
s + w2
TT
X T
XX
=
E
–2ps
1– e w
RSRS e –sp/w
VW RS 1
VWVW
SSSS 2
– SS 2
2 (– s (0) – w (–1))W
2 (– s (0) – w (1))W
W
WWW
s +w
s +w
TT
X T
XX
=
E
–2ps
1– e w
RS e –sp/w
VW
1
SS 2
2 (w ) –
2
2 (–w)W
W
s +w
s +w
T
X
=
E
w
j 6e –sp/w + 1@
–2ps ` 2
s + w2
1– e w
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#e
–ax
sin bx dx =
VW
e –ax
2 (– a sin bx – b cos bx)W
W
a +b
X
2
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May-2018 (R15) Question Paper with Solutions
QP.9
–sp/w
\
=
Ew (1 + e )
s + w (1 – e –sp/w) (1 + e –sp/w)
=
Ew
(s 2 + w 2) (1 – e –sp/w)
L{f(t)} =
Ew
(s 2 + w 2) (1 – e –sp/w)
2
2
[a(1 – e–2a) = (1 + e–a) (1 – e–a)]
1
(b) Find the laplace transform of t (eat – ebt)
May-18, (R15), Q2(b)
Answer :
Given function is,
1 at
bt
t (e – e )
Let f(t) = eat – ebt
1
L $ t (e at – e bt) . =
RS
f ( t)
SS
SSa L & t 0 =
s
T
3
#
s
3
=
#
3
#
fr(s) ds
VW
W
fr(s) dsWW
W
X
L {e at – e bt} ds
s
3
=
# [L {e } – L {e }] ds
at
bt
s
3
=
# JKK s 1– a – s 1– b NOO ds
L
s
P
3
=
#
3
1
s – a ds –
s
#
1
s – b ds
s
3
s
= log (s – a) – log (s – b) 3s
\
s–a 3
= log ` s – b j s
s–a
= 0 – log ` s – b j
s–a
= – log ` s – b j
Js – bN
= log KK s – a OO
L
P
JK s – b NO
1
at
bt
L $ t (e – e ) . = log K s – a O
L
P
oR
Q3.
Solve by laplace transform method y'' – 3y' + 2y = 4, where y(0) = 2, y'(0) = 3.
May-18, (R15), Q3
Answer :
Given differential equation is,
y'' – 3y' + 2y = 4.
Applying Laplace transform on both sides,
L{y'' – 3y' + 2y] = L[4]
4
L{y''} – 3L{y'} + 2L{y} = s
4
[s2 L{y} – sy(0) – y'(0)] – 3[s L{y} – y(0)] + 2L{y} = s
⇒
⇒
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QP.10
4
[s2 L{y} – s(2) – 3] – 3[s L{y} – 2] + 2L{y} = s
⇒
⇒
4
s2 L{y} – 2s – 3 – 3s L{y} + 6 + 2L{y} = s
⇒
4
s2 L{y} – 3s L{y} + 2L{y} = s + 2s + 3 – 6
⇒
4
L{y} (s2 – 3s + 2) = s + 2s – 3
⇒
L{y} (s2 – 3s + 2) =
4 + 2s 2 – 3s
s
2s 2 – 3s + 4
s (s 2 – 3s + 2)
⇒
L{y} =
⇒
2s 2 – 3s + 4
L{y} = s (s – 1) (s – 2)
... (1)
Applying partial fractions,
2s 2 – 3s + 4
A
B
C
s (s – 1) (s – 2) = s + s – 1 + s – 2
⇒
... (2)
A (s – 1) (s – 2) + B s (s – 2) + C s (s – 1)
2s 2 – 3s + 4
s (s – 1) (s – 2) =
s ( s – 1 ) (s – 2 )
⇒
2s2 – 3s + 4 = A(s – 1) (s – 2) + Bs(s – 2) + Cs(s – 1)
Substituting s = 0 in equation (3),
2(0)2 – 3(0) + 4 = A(0 – 1) (0 – 2) + B(0) (0 – 2) + C(0) (0 – 1)
⇒ 4 = 2A + 0 + 0
⇒ A = 2.
Substituting s = 1 in equation (3),
2(1)2 – 3(1) + 4 = A(1 – 1) (1 – 2) + B(1) (1 – 2) + C(1) (1 – 1)
⇒ 2 – 3 + 4= 0 – B + 0
⇒ B=–3
Substituting s = 2 in equation (3),
2(2)2 – 3(2) + 4 = A(2 – 1) (2 – 2) + B(2) (2 – 2) + C(2) (2 – 1)
⇒ 8 – 6 + 4 = 0 – 0 + 2C
⇒ C=3
Substituting the values of A, B and C in equation (2),
2s 2 – 3s + 4
2
3
3
s (s – 1) (s – 2) = s – s – 1 + s – 2
... (4)
Substituting equation (4) in equation (1),
2
3
3
L{y} = s – s – 1 + s – 2
Applying inverse Laplace transform on both sides,
2
3
3
L–1[L{y}] = L–1 & s – s – 1 + s – 2 0
3
3
2
⇒
y = L–1 $ s . – L–1 & s – 1 0 + L–1 & s – 2 0
1
1
1
= 2L–1 $ s . – 3 L–1 & s – 1 0 + 3L–1 & s – 2 0
= 2(1) – 3(et) + 3(e2t)
= 2 – 3et + 3e2t
\ y = 2 – 3et + 3e2t
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... (3)
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May-2018 (R15) Question Paper with Solutions
QP.11
Unit-II
Q4.
Find the Fourier series to represent (x – x2) from x = –p to x = p.
(a)
May-18, (R15), Q4(a)
Answer :
For answer refer May/June-16, Q4, (Excluding Deduction).
(b) Find the half range cosine series for the function f(x) = x2 in the range 0 ≤ x ≤ p.
May-18, (R15), Q4(b)
Answer :
Given function is,
f (x) = x2 in [0, p]
The half-range cosine series of a function is given as,
3
a0
nπ x
+ Σ an cos
f(x) =
2 n=1
l
Here,
... (1)
l=p
a0
f(x) = x = 2 +
2
∞
∑a
n
cos n x
... (2)
n =1
Where,
2
a0 =
π
π
∫ f ( x)dx
0
π
π
2 2
2  x3 
x dx =  
a0 =
π
π  3 
0
0
∫
2  π3 − 0 
2π 3
= 
 =
π  3 
3π
π
π
2
2 2
f ( x) cos nx dx ===
x cos nx dx
an =
π0
π0
∫
∫
π
=
2 2
d 2
 
 x cos nxdx −  x cos nxdx dx 
π
 0
 dx
=
2  2 sin nx
sin nx 
x
dx  =
− 2x

π
n
n
0
=
2  2 sin nx
cos nx
sin nx 
x
+ 2x 2 − 2 3 

π
n
n
n 0
=
2   2 sin nπ
cos nπ
sin nπ  
2sin n(0)  
+ 2π
− 2 3  − 0 + 0 −
 π

2
π  
n
n
n  
n3

=
2  cos n π  4
2π
= cos n π
π 
n 2  n 2
∫
∫
∫
∫
π
π
∴ an =
4
n2
( −1) n
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QP.12
Substituting a0 and an in equation (2),
∞
(− 1) cos n x
π2
+
4
x =
2
3
n =1 n
n
∑
2
oR
Q5.
Find the complex form of the Fourier series of f(x) = e–x in – 1 ≤ x ≤ 1
(a)
May-18, (R15), Q5(a)
Answer :
For answer refer Unit-II, Q39.
]Z] 1
1
]] 4 – x if 0 < x < 2
[
(b) Expand f(x) = ]
as the Fourier series of sine terms.
]] x – 3 if 1 < x < 1
]
4
2
Answer :
\
May-18, (R15), Q5(b)
For answer refer Unit-II, Q29.
Unit-III
1, | x | < 1
Find the Fourier transform of f(x) = (
Hence deduce that
0, | x | > 1
Answer :
Q6.
3
#
sinx
x dx
0
May-18, (R15), Q6
Given function is,
1, | x | < 1
f (x) = (
0, | x | > 1
The Fourier transform of the function f(x) is,
3
F[f(x)] =
#e
isx
f (x) dx
–3
1
–1
⇒
F(s) =
#
isx
e f (x) dx +
#e
–1
–3
3
isx
#e
f (x) dx +
isx
f (x) dx
1
1
⇒
#e
F(s) = 0 +
isx
(1) dx + 0
–1
1
⇒
F(s) =
#e
isx
dx
–1
⇒
RS eisx VW1
F(s) = SS is WW
–1
T X
⇒
F(s) =
\
eis – e –is
is
2
= is sinh(is)
SRSa sinh x = e x – e –x WVW
2 X
T
2
= is (i sin s)
6a sinh (ix) = i sin x@
=
2 sin s
s
F(s) =
2 sin s
s
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May-2018 (R15) Question Paper with Solutions
QP.13
Deduction
By inversion formula,
3
#
1
f(x) = 2p
F (s) .e –isx ds
–3
3
#
1
= 2p
2 sin s
s (cos sx – i sin sx)ds
–3
3
#
1
= 2p
3
i
2 sin s
s cos sx ds – 2p
–3
#
2 sin s
sin sx ds
s
–3
3
#
1
= 2p
sin s
s cos sx ds – 0
[a sin sx is an odd function]
–3
3
#
1
= p
sin s
s cos sx ds
–3
3
#
⇒ p f(x) =
sin s
s cos sx ds
–3
3
1, | x | < |
⇒ p(
=
0, | x | > |
(
⇒
#
sin s
s cos sx ds
–3
3
#
p, | x | < |
=
0, | x | > |
sin s
s cos sx ds
–3
3
#
At x = 0, p =
sin s
s cos s(0) ds
–3
3
⇒ p=2
#
sin s
s (1) ds
0
3
⇒
#
p
2 =
sin s
s ds
0
3
⇒
p
2 =
#
sin x
x dx
0
3
\
#
0
sin x
p
x dx = 2
oR
Q7.
2 2
Find the Fourier transformation of e–a x , a > 0.
May-18, (R15), Q7
Answer :
Given function is,
f(x) = e–a2x2
From the definition of Fourier transform,
3
F{f(x)}=
#
f (x) .eisx dx
–3
3
=
#
e –a
2 2
x
.eisx dx
–3
3
=
#
e –a
2 2
x + isx
dx
–3
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QP.14
3
#
=
e – (a
2 2
x –isx)
(b) The ends A and B of a rod 20 cm long have
the temperature at 30°C and 80°C until
steady state prevails. The temperature
of the ends is changed to 40°C and
60°C respectably. Find the temperature
distribution in the rod at time t.
dx
–3
3
#
=
e –;(ax)
2
is
is 2
is 2
–(2)(ax)b 2a l + b 2a l –b 2a l E
dx
–3
3
#
=
e –;(ax)
2
is
is 2
is 2
–(2)(ax)b 2a l + b 2a l + b 2a l E
For answer refer Unit-IV, Q40.
–3
3
#
=
2
e –b ax – 2a l + i
is
2b
s l2
2a dx
oR
–3
Q9.
3
#
=
May-18, (R15), Q8(b)
Answer :
dx
2
e –b ax – 2a l .ei
is
s 2
2b 2a l
(a)
Find the partial differential equation by
eliminating the arbitrary constants from:
dx
(x – a)2 + (y – b)2 + z2 = c2.
–3
May-18, (R15), Q9(a)
Answer :
= e
s 2
–b 2a l
3
#
2
e –b ax – 2a l dx
is
Given equation is,
–3
(x – a)2 + (y – b)2 + z2 = c2
is
Let, ax – 2a = t
Differentiating equation (1) partially with respect to x,
adx – 0 = dt
2z
2(x – a) + 0 + 2z 2x = 0
adx = dt
1
dx = a dt
is
U.L : If x = + ¥, a( ¥) – 2a = t Þ t = ¥
is
L.L : If x = – ¥, a( ¥) – 2a = t Þ t = – ¥
#
s
2
3
#
2
e –t dt
–3
2
e –b 2a l ^ p h
=
a
s
>a
3
#
–3
e
–x 2
F{e–
dx = p H
(a)
p –s 2
} = a 4a 2
e
a2x2
(x – a) + zp = 0
x – a = – zp
... (2)
⇒
2(y – b) + 2zq = 0
⇒
2(y – b + zq) = 0
⇒
(y – b) + zq = 0
JK
2z N
Ka q = 2y OO
L
P
y – b = – zq
... (3)
(– zp)2 + (– zq)2 + z2 = c2
\
⇒
z2 p2 + z2 q2 + z2 = c2
⇒
z2 (p2 + q2 + 1) = c2
The partial differential equation is,
z2 (p2 + q2 + 1) = c2 .
Form the partial differential equation by
eliminating the arbitrary functions f(x2 +
y2, z – xy) = 0.
May-18, (R15), Q8(a)
Answer :
⇒
Substituting equations (2) and (3) in equation (1).
Unit-IV
Q8.
2(x – a + zp) = 0
⇒
s2
p
= a .e – 4a2
\
⇒
JK
2z N
Ka p = 2x OO
L
P
2z
0 + 2(y – b) + 2z 2y = 0
1
e . a dt
–t 2
–3
e –b 2a l
= a
2(x – a) + 2zp = 0
Differentiating equation (1) partially with respect to y,
3
s 2
–b 2a l
⇒
⇒
Substituting the corresponding values in above integral,
F{f(x)} = e
... (1)
For answer refer Unit-IV, Q23.
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(b) A tightly stretched string of length 1
with fixed ends is initially in equilibrium
position. It is set vibrating by giving each
px
point a velocity v0sin3 ` 1 j .
Answer :
May-18, (R15), Q9(b)
For answer refer Unit-IV, Q35.
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May-2018 (R15) Question Paper with Solutions
QP.15
Unit-V
Q10. (a)
Evaluate Z-transformation of
z2
using convolution theorem.
(z – 1) (z – 3)
May-18, (R15), Q10(a)
Answer :
For answer refer Unit-V, Q34.
1
1
(b) Using Z-transforms solve yn + 4 yn – 1 = un + 3 un–1 where un is a unit step sequence.
May-18, (R15), Q10(b)
Answer :
Given equation is,
1
1
yn + 4 yn – 1 = un + 3 un – 1
Applying Z-transform on both sides
1
1
Z & yn + 4 yn –1 0 = Z &u n + 3 u n –1 0
1
1
Z{yn} + 4 Z{yn –1} = Z{un} + 3 Z{un –1}
⇒
⇒
z
1 Y
1 J 1 N
Y + 4 ` z j = z – 1 + 3 KK z – 1 OO
L
P
⇒
J
1 N
1
z
Y KK1 + 4z OO = z – 1 + 3 (z – 1)
L
P
⇒
J 4z + 1 N
3z + 1
Y KK 4z OO = 3 (z – 1)
L
P
(3z + 1) 4z
Y = 3 (z – 1) (4z + 1)
⇒
(3z + 1) 4z
J
1N
3 (z – 1) 4 KK z + 4 OO
L
P
z (3z + 1)
⇒
Y=
J
1N
3 (z – 1) KK z + 4 OO
L
P
Applying partial fractions,
⇒
Y=
... (1)
A
B
3z + 1
N = z –1 +
JK
1
1
O
z+ 4
(z – 1) K z + 4 O
L
P
J
1N
⇒ (3z + 1) = A KK z + 4 OO + B(z – 1)
L
P
–1
Substituting z = 4 in equation (3),
⇒
... (2)
... (3)
J –1 N
J –1 1 N
J –1
N
3 KK 4 OO + 1 = A KK 4 + 4 OO + B KK 4 – 1OO
L P
L
P
L
P
5B
–3
4 +1=0– 4
⇒
– 5B
1
4 = 4
⇒
–1
B= 5
Substituting z = 1 in equation (3)
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QP.16
⇒
J 1N
3(1) + 1 = A KK1 + 4 OO + B(1 – 1)
L
P
5
3 + 1 = A `4j + 0
⇒
= [1n.(n + 1)]1n
/1
n
5A
4= 4
=
16
A= 5
= 1n
m
(m + 1) 1 n–m
m=0
/ (m + 1)
n
m=0
= 1n (1 + 2 + 3 + - - - (n + 1))
Substituting the values of A and B in equation (2)
1
16
3z + 1
J
1 N = 5 (z – 1) – JK + 1 ON
5K z 4 O
(z – 1) KK z + 4 OO
L
P
L
P
Substituting the above value in equation (1),
RS 16
VW
1
SS
W
–
J
N
SS 5 (z – 1) 5 KK z + 1 OO WWW
4P
L
T
X
R
V
1
16z
1 WW
⇒ Y = 15 SSS (z – 1) –
1W
SS
z + 4 WW
T
X
Applying inverse Z-transform on both sides,
z
Y= 3
R
1 S JK 16z NOO –1 JKK z
Z–1(y) = 15 SSSZ –1 KK
O– Z K
1
KK
SS L z – 1 P
z+
S
4
L
T
J –1 NnVW
1 RS
\
yn = 15 SS16 (1) – KK 4 OO WW .
L PX
T
oR
Q11. (a)
JK z 2
z NO
KK
2 .
z – 1 OO
(
z
–
1
)
L
P
z 3
Z–1 ` z – 1 j = Z–1
=
\
(n + 1) (n + 2)
2
(n + 1) (n + 2)
z 3
Z–1 ` z – 1 j =
2
(b) Using Z-transforms solve Un+2 – 2Un+1 + Un
= 3n + 5.
Answer :
May-18, (R15), Q11(b)
For answer refer May/June-17, Q11.
NOWVW
OOW
OOWW
WW
PX
z3
Evaluate Z-transformation of
(z – 1) 3
using convolution theorem.
May-18, (R15), Q11(a)
Answer :
Given function is,
z3
(z – 1) 3
z
Z–1 ` z – 1 j = 1n
Then,
z 2
z
z
Z–1 ` z – 1 j = Z–1 ` z – 1 . z – 1 j
= 1n.1n
/1
n
=
m
.1 n–m
m=0
[a From convolution theorem]
/1
n
= 1n .
m
.1 –m
m=0
= 1n (1 + 1 + 1 + . . . . .+1)
= 1n (n + 1)
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MP.1
Model Question Papers with Solutions
R15
Jawaharlal Nehru Technological University Anantapur
B.Tech. I Year II Semester Examination
MODEL
PAPER
1
MATHEMATICS-II
(Common to All Branches)
Time: 3 Hours
Max. Marks: 70
PART - A
(20 MARKS)
1.
Solutions
(a)
Define the Laplace transform and state its domain and Kernel.
(Unit-I / Q1)
(b)
Find the Laplace transform of eat.
(Unit-I / Q3)
(c)
Give the Euler’s formulae for Fourier series.
(Unit-II / Q2)
(d)
Give the Fourier series expansion for odd periodic functions.
(Unit-II / Q7)
(e)
State Fourier integral theorem.
(Unit-III / Q1)
(f)
Find Fourier sine transform series of xe–ax.
(Unit-III / Q9)
(g)
Define order and degree with reference to partial differential equation.
(Unit-IV / Q1)
(h)
Form partial differential equation by eliminating the arbitrary
1 2 2y
a e + b.
2
(Unit-IV / Q4)
(i)
State and prove the linearity property of Z-transform.
(Unit-V / Q4)
(j)
State and prove convolution theorem.
constants z = axey +
(Unit-V / Q10)
PART - B
(50 MARKS)
UNIT - I
2.
Prove that,
(a)
L{sin at} =
(b)
L{cos at} =
a
2
s + a2
s
s2 + a 2
,s>a
(Unit-I / Q15)
, s > a.
OR
3.
Using Laplace transform, solve (D2 + 4D + 5)y = 5, given that y(0) = 0, y"(0 ) = 0.
(Unit-I / Q54)
UNIT - II
Q4.
 -1
 2 ( π + x) for - π ≤ x ≤ 0
Find the Fourier series in [–π, π] for the function f(x) = 
.
 1 ( π - x) for 0 ≤ x ≤ π
 2
(Unit-II / Q16)
OR
5.
Find the half-range sine series of f(x) = (x – 1)2 in the interval (0, 1).
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UNIT - III
∞
6.
Find the Fourier transforms of f(x) = e
–|x|
and deduce that
cos xt
∫ 1+ t
2
dt =
0
Hence show that F(xe–|x|) = i
π –|x|
e .
2
4s
.
(1+ s 2 )2
(Unit-III / Q23)
OR
7.
Find the finite cosine transform of, f(x) =
π
x2
– x+
, 0 < x < π.
3
2π
(Unit-III / Q35)
UNIT - IV
8.
Form the PDE by eliminating ‘f’ from f(x + y + z, x2 + y2 + z2) = 0.
(Unit-IV / Q24)
OR
9.
Find the deflection u(x, y, t) of the square membrane with a = b = 1 and c = 1,
if the initial velocity is zero and the initial deflection is f(x, y) = A sin πx sin 2πy.
(Unit-IV / Q37)
UNIT - V
10.

1 
Find Z–1 
when | z | > 5. Determine the region of convergence.
3
 (z – 5) 
(Unit-V / Q17)
OR
11.
Find the linearity property of Z-transforms,
(i)
 nπ π 
+ 
cos 
 2
4
(ii)
 nπ 
+ θ .
cosh 
 2

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MP.3
Model Question Papers with Solutions
R15
Jawaharlal Nehru Technological University Anantapur
B.Tech. I Year II Semester Examination
MODEL
PAPER
2
MATHEMATICS-II
(Common to All Branches)
Time: 3 Hours
Max. Marks: 70
PART - A
Solutions
(20 MARKS)
1.
(Unit-I / Q8)
(a)
Define second shifting theorem.
(b)
 3(s2 – 2)2 
Find the inverse Laplace transformation of 
.
5
 2s

(Unit-I / Q12)
(c)
Give the Dirichlet conditions for Fourier expansion.
(Unit-II / Q3)
(d)
Give the Fourier series expansion for even periodic functions.
(Unit-II / Q6)
(e)
State the properties of Fourier transform.
(Unit-III / Q5)
(f)
Find Fourier sine transform of
(g)
Write the three possible solutions of
(h)
Form partial differential equation by eliminating the arbitrary constants
1
.
x
(Unit-III / Q10)
∂ u ∂2 u
.
=
∂ t ∂ x2
z = ax + by + a2 + b2.
(i)
Show that, Z(nan) =
(j)
 z2 + 2 
Find Z–1 
2.
 (z – 1) 
(Unit-IV / Q9)
(Unit-IV / Q2)
az
(z - a)2
(Unit-V / Q6)
.
(Unit-V / Q9)
PART - B
(50 MARKS)
UNIT - I
2.
Find L–1[(2s + 3)/(s3 – 6s2 + 11s – 6)].
(Unit-I / Q20)
OR
3.
Prove that,
(i)
L[eat sinh bt] =
(ii)
L[eat cosh bt] =
b
(s – a)2 – b 2
s–a
(s – a)2 – b 2
.
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UNIT - II
4.
Obtain the Fourier series for the function f(x) = |cos x| in (–π, π).
(Unit-II / Q19)
OR
1 2 ∞

2
[f(x)]
dx
=
l
(anan + bbn ) provided the Fourier series for
 a0 +
Prove that
 2

n=1
–l
l
5.
∑
∫
(Unit-II / Q37)
f(x) converges uniformly in (– l, l).
UNIT - III
6.
2
–x
Find the Fourier cosine transform of e .
(Unit-III / Q25)
OR
7.
Find the finite Fourier sine and cosine transform of f(x) = x (π – x) in 0 < x < π.
(Unit-III / Q39)
UNIT - IV
8.
Solve by separation of variables 3ux + 2uy = 0 with u(x, 0) = 4e– x.
(Unit-IV / Q29)
OR
9.
A tightly stretched string with fixed end points x = 0 and x = 1 is initially at rest in
its equilibrium position. If itisset on vibrating by giving each of its points a velocity
λx(1–x), find the displacement of the string at any distance x from end at any time ‘t’.
(Unit-IV / Q35)
UNIT - V
10.
If Z(un) =
2 z 2 + 4 z+12
(z-1) 4
(Unit-V / Q30)
find u2.
OR
11.
z2


 using convolution theorem.
Evaluate Z–1 
 (z – 1)(z – 3) 
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MP.5
Model Question Papers with Solutions
R15
MODEL
PAPER
Jawaharlal Nehru Technological University Anantapur
B.Tech. I Year II Semester Examination
3
MATHEMATICS-II
(Common to All Branches)
Time: 3 Hours
Max. Marks: 70
Solutions
PART - A
1.
(a)
(20 MARKS)
Write about the existence of Laplace transform.
(b)
Using Laplace transform, evaluate
(Unit-I / Q2)
∞
∫ te
–t
sin t dt.
(Unit-I / Q6)
(c)
Define even and odd function.
(d)
Define half-range Fourier cosine series of f(x).
(Unit-II / Q5)
(Unit-II / Q10
(e)
Using Fourier integral show that,
(f)
Define inverse finite Fourier sine and cosine transforms.
(g)
Form the partial differential equation by eliminating the arbitrary
0
∞
∫
0
π
1– cos πλ
sin(xλ)dλ =
, 0 < x < π.
2
λ
(Unit-III / Q3)
(Unit-III / Q16)
function from z = f(x2– y2).
(h)
(Unit-IV / Q5)
Write the boundary conditions for the following problem. A rectangular plate
is bounded by the line x = 0,y = 0 and y = b. Its surfaces are insulated.
The temperature along x = 0 and y = 0 are kept at 0ºC and the others are kept at 100ºC.
(i)
Write the properties of multiplication by n and division by n of Z-transforms.
(j)
Find the Z-transform of,
(i) 2n + 3
(ii) cos 3n.
(Unit-IV / Q10)
(Unit-V / Q7)
(Unit-V / Q3)
PART - B
(50 MARKS)
UNIT - I
n
n
n–1
n–2
f(0) – s f' (0)..... fn– 1(0), where L{f (t)} = f (s).
OR
2.
Show that L{f (t)} = s f(s) – s
3.
Find the Laplace transform of periodic function f(t) with period T,
where f(t) =
(Unit-I / Q29)
4Et
T
4Et T
– E, 0 ≤ t ≤ = 3E –
, ≤ t ≤ T.
T
2
T 2
(Unit-I / Q47)
UNIT - II
∞
4.
Obtain the Fourier series for the function, f(x) = x + x2 in [–π, π], deduce the value of
∑n
1
2
.
(Unit-II / Q12)
n=1
OR
5.
1
1
 4 - x , 0 < x < 2
Find the half-range sine series for f(x) = 
.
x - 3 , 1 < x < 1
 4
2
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UNIT - III
∞
6.
∫
Solve the integral equation f(x) cosαx dx= e–αx.
(Unit-III / Q20)
0
OR
7.
Find the fourier sine transform of
1
.
x(x 2 + a 2 )
(Unit-III / Q26)
UNIT - IV
8.
Obtain the partial differential equation by eliminating the arbitrary constants
(a)
z = (x – a)2 + (y – b)2 + 1
(b)
z = ae – b
2
t
(Unit-IV / Q12)
cos bx.
OR
9.
A bar of length L is laterally insulated with its ends A and B kept at 0° and 100°
respectively until steady state condition is reached. The temperature at A is raised to 30°
and that at B is reduced to 80° simultaneously. Find the temperature in the rod at a later time.
(Unit-IV / Q46)
UNIT - V
10.
Find the inverse Z-transform of
3z 2 + z
.
(5z – 1)(5z – 2)
(Unit-V / Q16)
OR
11.
Solve the difference equation, using Z-transforms un +2 – un = 2n where u0 = 0, u1 = 1.
Look for the SIA GROU P LOGO
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List of Important Formulae
L.1
LIST OF IMPORTANT FORMULAE
UNIT - I
1.
Laplace Transform: Laplace transform is a mathematical tool that transforms a time-domain function to a frequency domain
function and vice-versa.
2.
Shifting Theorem: Let f(t) be a function defined for all the positive values of ‘t’. Then, the first shifting theorem states that,
For L[f(t)]= f (s )
L[eat f(t)]= f ( s − a), where ( s − a) > 0
3.
Unit Step Function: The unit step function or Heaviside unit function is defined as,
⎧ 0, t < a
u(t – a) or H(t – a) = ⎨
⎩1, t > a
Where, a > 0
4.
Second Shifting Theorem: The second shifting theorem states that,
For L[f(t)] = f (s) and
⎧ f (t − a) ; t > a
g(t) = ⎨
; t<a
⎩0
Laplace transform of g(t) is given by,
L[g(t)] = e–as f (s )
5.
Dirac’s Delta Function: The Dirac’s delta function or unit impulse function can be defined as the limiting form of,
⎧ 0, t < a
⎪1
⎪
fε(t – a) = ⎨ , a ≤ t ≤ a + ε
⎪ε
⎪⎩ 0 , t > a
6.
Laplace Transform of a Periodic Function: The Laplace transform of a periodic function is given by,
T
∫e
L{f(t)} =
0
− st
f (t ) dt
1 − e − st
Where, T – Period of f(t).
7.
Linearity Property of Laplace Transform
If L[f(t)] = f (s) and
L[g(t)] = g (s ) , then
L [c1 f(t) + c2 g(t)] = c1 L[f(t)] + c2 L[g(t)]
= c1 f (s) + c2 g (s)
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L.2
MATHEMATICS-II [JNTU-ANANTAPUR]
UNIT - II
1.
Periodic Function: A function f (x) defined for all real numbers is said to be periodic, if there is a real number T < 0 such that,
f (x + T) = f (x) for all x
2.
Euler’s Formulae for Fourier Series: The Fourier series expansion of function f(x) in the interval k ≤ x ≤ k + 2π is given by,
a0 ∞
+
(an cos nx + bn sin nx)
2 n =1
∑
f(x) =
3.
Dirichlet’s Conditions
1.
f (x) is uniformly bounded on the fundamental interval (–π, π) of period 2π, that is, | f (x)| < M for all –π < x < π where M
is a constant.
2.
f (x) has no more than a finite number of finite discontinuities in (–π, π).
3.
f (x) has a finite number of strict maxima or minima in (–π, π).
4.
Even Function: A function f(x) is said to be an even function if f(–x) = f(x)
5.
Odd Function: A function f(x) is said to be an odd function if f(–x) = –f(x).
6.
Half-Range Fourier Sine Series: The half-range Fourier sine series of f(x) defined in the interval (0, π) is represented
as,
∞
f(x) =
7.
∑ b sin nx
n
n =1
Half-Range Fourier Cosine Series: The half-range Fourier cosine series of f(x) defined in the interval (0, π) is
represented as,
f(x) =
a0
+
2
∞
∑a
n cos nx
n=1
UNIT - III
1.
Fourier integral theorem: The Fourier integral theorem of f(x) is defined as,
1
f(x) =
π
2.
0 −∞
∞
∞
0
0
2
sin λx f (t ) sin λt dt dλ
π
∫
∫
Fourier Cosine Integral: The Fourier cosine integral of f(x) is defined as,
f(x) =
4.
∫ ∫ f (t ) cos λ(t − x)dt.dλ
Fourier Sine Integral: The Fourier sine integral of f(x) is defined as,
f(x) =
3.
∞ ∞
∞
∞
0
0
2
cos λx f (t ) cos λ t dt dλ
π
∫
∫
Fourier Transform of Function f(x): The Fourier transform of a function f(x) is defined as,
∞
F(s) = F.T{f(x)} =
∫ f ( x) e
isx
dx
−∞
5.
Linear Property: If f(x) and g(x) are two functions whose Fourier transform are given as F(s) and G(s) respectively, then,
F[a f(x) + b g(x)] = a. F(s) + b G(s)
Where, a and b are arbitrary constants.
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List of Important Formulae
6.
L.3
Scaling Property: If f(x) is a function whose complex Fourier transform is given by F(s),
Then,
F{f(ax)} =
1 ⎛s⎞
F⎜ ⎟
a ⎝a⎠
Where, a ≠ 0.
7.
Shifting Property: If a function f(x) has a complex Fourier transform F(s),
Then,
F{f(x–a)} = eisa.F(s)
8.
Modulation Property: If a function f (x) has a complex Fourier transform F(s), then,
F{f(x) cos ax} =
9.
1
[F(s + a) + F(s – a)]
2
Fourier Sine Transform: The Fourier sine transform of the function f(x) is given as,
∞
Fs(S) = FST{f(x)} =
∫ f ( x) sin sx dx
0
10.
Fourier Cosine Transform: The Fourier cosine transform of the function f(x) is defined as,
∞
Fc(S) = FCT {f(x)} =
∫ f ( x) cos sx dx
0
11.
Inverse Fourier Transform: The inverse Fourier transform of f(S) is defined as,
1
f(x) = I.F.T {F(S)} =
2π
∞
∫ F ( S )e
−isx
ds
−∞
UNIT - IV
1.
Order: The order of a partial differential equation is equal to the order of highest partial derivative for a particular equation.
2.
Degree: The degree of a partial differential equation is equal to the power of highest partial derivative.
3.
Wave Equation: The expression for one dimensional wave equation is given as,
2
∂2 y
2 ∂ y
c
=
∂t 2
∂x 2
4.
Heat Equation: The general solution of the heat equation is given as,
2
∂u
2 ∂ u
=C
∂t
∂x 2
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MATHEMATICS-II [JNTU-ANANTAPUR]
UNIT - V
1.
Linearity Property Transform: Let x, y, z are said to be any constants and un, vn, wn be any distinct functions then.
Z(xun + yvn – zwn) = xZ(un) + yZ(vn) – Z(wn)
2.
Damping Rule (or) Change of Scale of Property: The two change of scale properties or damping rule is given as,
1.
If Z(un) = u (z ) then Z(a–n un) = u (az)
2.
⎛z⎞
If Z(un) = u (z ) , then z(an un) = u ⎜ ⎟
⎝a⎠
Lt f (z) = f (0)
3.
Initial Value Theorem of Z-transform: If z(f (n)) = f (z), then
4.
f (n) =
Final Value Theorem of Z-Transform: If zf (n) = f (z), then nLt
→∞
5.
Convolution Theorem Statement: If Z–1[f(z)] = f (α) and Z–1[g(z)] = g(α)
z →∞
Lt (z – 1)f (z)
z →1
α
Then, Z–1[f(z).g(z)] = f (α)* g(α) = Σ f(β) g(α – β)
β=0
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1.1
UNIT-1 (Laplace Transform)
UNIT
LAPLACE TRANSFORM
1
PART-A
SHORT QUESTIONS WITH SOLUTIONS
Q1. Define the Laplace transform and state its domain and Kernel.
Model Paper-I, Q1(a)
Ans:
Laplace Transform: Laplace transform is a mathematical tool that transforms a time-domain function to a frequency domain
function and vice-versa.
Definition: Laplace transform of a function f(t) is a linear integral transform which is defined as,
f (s) = L{f(t)}
∞
=
∫ f (t ).e
− st
.dt
0
Where,
f(t) – Time-domain function
e–st – Kernel of the transform [k(s, t)]
L – Laplace transform operator
s – Complex variable = σ + jω
f (s) – Complex variable function.
Laplace transform helps in better understanding of both time-domain and frequency domain functions and their properties.
Q2. Write about the existence of Laplace transform.
Model Paper-III, Q1(a)
Ans:
Existence of Laplace Transform: Laplace transform of any function f(t) exists, if it satisfies the following two conditions,
Condition 1: f(t) is Piece-wise Continuous: A function is said to be piece-wise continuous in an interval [a, b], if it is continuous
and has finite limits at the extreme points, in any sub-interval range of [a, b].
Condition 2: f(t) is of Exponential Order ‘b’: A function, f(t) is said to be of exponential order ‘b’, if there exists ‘M’ and ‘b’ such
that,
|f(t)| < Mebt
Q3. Find the Laplace transform of eat .
Ans: Given function is eat
Model Paper-I, Q1(b)
Let, f(t) = eat
From the definition of Laplace transform,
∞
L{f(t)} =
∫e
− st
. f (t ) dt
0
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MATHEMATICS-II [JNTU-ANANTAPUR]
∞
∴
∫e
L{eat} =
− st
∫e
.e .dt =
0
[
−( s − a ) t
.dt
]
− (s − a)
L{tn} =
1
=
s−a
1
s−a
∴ L{e at } =
Q4.
Prove that L{tn} =
s n+ 1
, n ≥ 0.
n ( n − 1)( n − 2 )( n − 3).... 2 . 1
L{1}
s × s × s × s....
=
n!
n! 1
. = n +1
n
s
s s
k
⎛
⎜Q L{k } = ⇒ L{1} =
s
⎝
n!
s
n +1
Q5. State first shifting theorem.
Ans: Let f(t) be a function defined for all the positive values of
‘t’. Then, the first shifting theorem states that,
n!
s n +1
For L[f(t)] = f (s )
Consider L.H.S,
L[eat f(t)] = f ( s − a), where ( s − a) > 0 .
L{tn}
∞
From the definition of Laplace transform,
Q6.
∞
−t
Model Paper-III, Q1(b)
∞
∫ te
∞
∴ L{tn} = e
− st n
−t
sin t dt
.t dt
Consider, sin t
L[sin t] =
∞
∞⎡
∞
⎤
⎡ ∞
d n − st ⎤
− st
n
⎥
⎢
⎢
−
t
e
dt
.
t
e .dt ⎥ dt
=
dt
⎥
⎢ 0
⎥
⎢
0
⎦0 0 ⎣
⎣
⎦
∫
∫
∫
Then, L[t sint] =
(Q Integration by Parts)
[ ]
∞
∞
⎡ e − st ∞ ⎤
⎛
e −st
n
0 ⎥
− ⎜ n.t n −1.
= ⎢t .
⎜
−s ⎥
−s
⎢
⎦0 0 ⎝
⎣
∫
⎞
⎟dt
⎟
⎠
1
s 2 +1
a ⎞
⎛
⎜Q L[sin at ] = 2
⎟
s + a2 ⎠
⎝
−d
L[sin t]
ds
n
⎛
⎞
⎜Q L[t n f (t )] = (−1) n d f (s ) ⎟
n
⎜
⎟
ds
⎝
⎠
−d ⎡ 1 ⎤
ds ⎢⎣ s 2 + 1 ⎥⎦
⎡ (s 2 + 1)(0) − (1)(2s ) ⎤
= −⎢
⎥
( s 2 + 1) 2
⎦⎥
⎣⎢
=
∞
= (0 ) +
... (1)
0
0
n −st n −1
e .t dt
s
∫
⎛ ⎛ u ⎞ ' vu ′ − u v ′ ⎞
⎜Q ⎜ ⎟ =
⎟
⎜ ⎝v⎠
⎟
v2
⎝
⎠
0
∞
n − st n −1
e .t dt
L{t } =
s
∫
=
0
2s
( s + 1) 2
2
∞
⇒
n
n −1
L{t } = L{t }
s
n
∫
L[f(t)] = e −st f (t ) dt
... (1)
0
On substituting, n = n – 1, in equation (1), we get,
n −1
L{t n − 2 }
L{t } =
s
n–1
sin t dt.
Given function is,
0
⇒
∫ te
0
∫
∫
Using Laplace transform, evaluate
Ans:
− st
L{f (t)} = e f (t ) dt
n
1⎞
⎟
s⎠
Hence proved.
Ans: Given that,
L{tn} =
n n −1 n − 2 n − 3 2 1
.
.
.
.... − L{t n − n }
s s
s
s
s s
=
∴ L{t n } =
n!
... (3)
∴ On generalizing equation (1), we get,
0 −1
− (s − a)
=
n−2
L{t n − 3 }
s
L{tn–2} =
0
∞
e −( s − a ) t 0
=
On substituting, n = n – 2, in equation (1), we get,
∞
at
∞
... (2)
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L[tsint] =
∫e
0
− st
(t sin t ) dt =
2s
( s 2 + 1) 2
on the TITLE COVER before you buy
1.3
UNIT-1 (Laplace Transform)
On substituting s = 1, we get,
∞
∫e
−t
t sin t dt =
0
∞
∫
∴ e−t t sint dt =
0
2 (1)
(1 + 1) 2
=
2
1
=
4
2
1
2
Q7. Define unit step function.
Ans: The unit step function or Heaviside unit function is defined as,
⎧ 0, t < a
u(t – a) or H(t – a) = ⎨
⎩1, t > a
Where, a > 0
The graphical representation of unit step function is as illustrated in figure below.
u(t – a)
1
0
a
t
Figure: Unit Step Function
Q8. Define second shifting theorem.
Ans:
Model Paper-II, Q1(a)
Statement: Let f(t) be a function defined for all the positive values of ‘t’.
Then, the second shifting theorem states that,
For L[f(t)] = f (s) and
⎧ f (t − a) ; t > a
g(t) = ⎨
; t<a
⎩0
Laplace transform of g(t) is given by,
L[g(t)] = e–as f (s )
Another Form
⎧1 t > 0
For L{F(t)} = f (s) and H(t) = ⎨
⎩0 t < 0
Then, L{F(t – a) H(t – a)} = e–as f (s)
Q9. Define Dirac’s delta function.
Ans: The Dirac’s delta function or unit impulse function can be defined as the limiting form of,
⎧ 0, t < a
⎪1
⎪
fε(t – a) = ⎨ , a ≤ t ≤ a + ε
⎪ε
⎪⎩ 0 , t > a
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SIA GROUP
1.4
MATHEMATICS-II [JNTU-ANANTAPUR]
This function is graphically represented in figure below.
fε(t – a)
1
ε
a
a +ε
t
Figure: Unit Impulse Function
In general, Dirac delta function is given by,
{ f ε (t − a)}
δ(t – a) = εLt
→0
Q10. Define the Laplace transform of a periodic function.
Ans: A function f(t) is said to be periodic, if it satisfies f(t + T) = f(t).
The Laplace transform of a periodic function is given by,
T
∫e
L{f(t)} =
0
− st
f (t ) dt
1 − e − st
Where, T – Period of f(t).
Q11. Define inverse Laplace transform.
Ans: If f (s ) is considered as the Laplace transform of f(t), then f(t) is called as inverse Laplace transform of f (s ) .
i.e., f(t) = L−1{ f (s )}
Where, L–1 – Inverse Laplace transform operator.
⎡ 3(s 2 − 2) 2 ⎤
Q12. Find the inverse Laplace transformation of ⎢
⎥.
5
⎣⎢ 2s
⎦⎥
Model Paper-II, Q1(b)
Ans: The given function is,
–1
⎡ 3( s 2 − 2) 2 ⎤
⎥
5
⎣ 2s
⎦
L ⎢
⇒
⎡ 3( s 2 − 2) 2 ⎤ 3 −1 ⎡ s 4 + 4 − 4 s 2 ⎤
L ⎢
⎥
⎥ = L ⎢
5
s5
⎦⎥
⎣⎢
⎣ 2s
⎦ 2
–1
=
=
3 ⎡ −1 ⎡ s 4 ⎤ −1 ⎡ 4 ⎤ −1 ⎡ 4s 2 ⎤ ⎤ 3 ⎡ −1 ⎡ 1 ⎤
−1 ⎡ 1 ⎤
−1 ⎡ 1 ⎤ ⎤
⎢L ⎢ 5 ⎥ + L ⎢ 5 ⎥ − L ⎢ 5 ⎥⎥ = ⎢ L ⎢ ⎥ + 4 L ⎢ 5 ⎥ − 4 L ⎢ 3 ⎥ ⎥
s
2
2 ⎢⎣ ⎣⎢ s ⎦⎥
⎣s ⎦
⎣ s ⎦⎦
⎣ ⎣ ⎦
⎣s ⎦
⎣⎢ s ⎦⎥ ⎥⎦
3 ⎡ 4×t 4
t2 ⎤
− 4. ⎥
⎢1 +
2 ⎣⎢
4!
2! ⎦⎥
⎛
⎞
⎡1 ⎤
⎜Q L−1 ⎢ ⎥ = 1 and ⎟
⎣s⎦
⎜
⎟
⎜
n ⎟
⎜ L−1 ⎛⎜ 1 ⎞⎟ = t ⎟
⎜
⎟
⎝ s n +1 ⎠ n! ⎠
⎝
⎤ 3 t4
3 ⎡ t4
3⎡
t4
t2 ⎤
− 2t 2 ⎥ = + − 3t 2
⎢1 + 4. − 4. ⎥ = ⎢1 +
2 ⎢⎣
24
2 ⎥⎦
2 ⎢⎣ 6
⎦⎥ 2 4
2
2
4
⎡
⎤
−
3
(
2
)
3
s
t
2
∴ L−1 ⎢
⎥ = + − 3t
5
⎥⎦ 2 4
⎣⎢ 2 s
=
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
1.5
UNIT-1 (Laplace Transform)
PART-B
ESSAY QUESTIONS WITH SOLUTIONS
1.1 LAPLACE TRANSFORM OF STANDARD FUNCTIONS
Q13. Define Laplace transform, domain of the function and Kernel for the Laplace transform. Also write
about the existence of Laplace transform.
Ans:
For answer refer Unit-I, Q1 and Q2.
Q14. List out the various properties of Laplace transform.
Ans:
Properties of Laplace Transform
(i)
Linearity Property
If L[f(t)]= f (s) and
L[g(t)] = g (s ) , then
L [c1 f(t) + c2 g(t)] = c1 L[f(t)] + c2 L[g(t)]
= c1 f (s) + c2 g (s)
Where, c1 and c2 are constants
k
(s > 0) and k is a constant
s
(ii)
L [k] =
(iii)
L [t] =
(iv)
L [tn] =
(v)
L [eat] =
(vi)
L [sin at] =
(vii)
L [cos at] =
(viii)
L [ sin h at] =
s2 − a 2
(ix)
L [cos h at] =
s − a2
(x)
L [eat.tn] =
(xi)
L [eat sin bt] =
1
s2
n!
, where n is a positive integer
s n +1
1
, where (s – a) > 0
s−a
a
s2 + a2
(if s > 0)
s
2
s + a2
(if s > 0)
a
s
2
(if s > |a|)
(if s > |a|)
n!
( s − a) n +1
b
( s − a) 2 + b 2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.6
MATHEMATICS-II [JNTU-ANANTAPUR]
s−a
(xii)
L [eat cos bt] =
(xiii)
L [eat sin h bt] =
(xiv)
L [eat cos h bt] =
(xv)
1 ⎛s⎞
L[f(at)] = f ⎜ ⎟
a ⎝a⎠
( s − a) 2 + b 2
( s − a) 2 − b 2
=
1 ⎡ 2 ja ⎤
⎥
⎢
2 j ⎢⎣ s 2 − ( ja ) 2 ⎥⎦
s−a
(Q (a – b) (a + b) = a2 – b2)
( s − a) 2 − b 2
=
∴ L{sin at} =
(a)
L{sin at} =
(b)
L{cos at} =
a
s2 + a 2
s
s2 + a 2
,s>a
(b)
, s > a.
Model Paper-I, Q2
a
s2 + a2
L{cosat} =
s
s + a2
2
,s > a
Consider L.H.S,
L{cos at}
a
,s > a
By the definition of Laplace transform,
s2 + a2
Consider L.H.S,
L{sin at}
By the definition of Laplace transform,
∞
∫
− st
L{f(t)} = e . f (t ) dt
0
∞
∞
∫
∫
− st
L{f (t)}= e . f (t )dt
− st
∴ L{cos at}= e . cos at dt
0
0
∞
∴
1 ⎛ 2 ja ⎞
a
⎜
⎟ = 2
2 j ⎝ s2 + a2 ⎠
s + a2
Hence proved.
Ans:
L{sinat} =
1 ⎛ s + ja − s + ja ⎞
⎜
⎟
2 j ⎜⎝ (s − ja )(s + ja ) ⎟⎠
b
Q15. Prove that,
(a)
=
∞
e jat + e − jat
− st ⎛
⎜
e
.
=
⎜
2
⎝
0
∫
− st
L{sin at}= e .sin at dt
∫
0
∞
jat
− e − jat
− st ⎛ e
= e .⎜⎜
2j
⎝
0
∫
⎞
⎟.dt
⎟
⎠
∞
∞
⎤
1 ⎡ − st jat
⎢
+
e
e
dt
e − st .e − jat .dt ⎥
.
.
=
2⎢
⎥
0
⎣0
⎦
∫
=
∞
∞
⎤
1 ⎡ −st jat
⎢ e .e dt − e −st .e − jat dt ⎥
2j ⎢
⎥
0
⎣0
⎦
=
∞
∞
⎤
1 ⎡ −( s − ja ) t
⎢ e
.dt − e −( s + ja )t .dt ⎥
2j ⎢
⎥
0
⎣0
⎦
∫
∫
∫
∫
[
] [
⎞
⎟ dt
⎟
⎠
]
∞
∞
e − ( s + ja ) 0 ⎤
1 ⎡⎢ e − ( s − ja ) t 0
⎥
−
=
− ( s + ja ) ⎥
2 j ⎢ − ( s − ja )
⎣
⎦
=
∫
∞
∞
⎤
1 ⎡ −( s− ja ) t
⎢ e
.dt + e −( s + ja )t .dt ⎥
2⎢
⎥
0
⎣0
⎦
∫
∫
[
] [
∞
=
1 ⎡ 0 −1
0 −1 ⎤
+
2 ⎢⎣ − ( s − ja) − ( s + ja ) ⎥⎦
=
1 ⎡ 0 −1
0 −1 ⎤
−
⎢
⎥
2 j ⎣ − ( s − ja ) − ( s + ja ) ⎦
=
1⎡ 1
1 ⎤
+
⎢
⎥
2 ⎣ s − ja s + ja ⎦
=
1 ⎡ 1
1 ⎤
−
⎢
2 j ⎣ s − ja s + ja ⎥⎦
=
1 ⎡ s + ja + s − ja ⎤
2 ⎢⎣ ( s − ja )(s + ja ) ⎥⎦
Look for the SIA GROU P LOGO
]
− −
e −( s + ja )t 0
1 ⎡ e ( s ja ) t 0
+
= ⎢
− (s + ja )
2 ⎢ − (s − ja)
⎣
∞
on the TITLE COVER before you buy
⎤
⎥
⎥
⎦
1.7
UNIT-1 (Laplace Transform)
=
⎤
1⎡
2s
⎢ 2
2⎥
2 ⎣⎢ s − ( ja ) ⎦⎥
[Q (a – b) (a + b) = a2 – b2]
=
1 ⎡ 2s ⎤
2 ⎢⎣ s 2 + a 2 ⎥⎦
=
s
s2 + a 2
∴ L{cos at} =
3
3
L{cos 7t} + L{cos t}
2
2
=
s
3⎛
⎜
2 ⎝ s2 + 72
=
3⎡ s
s ⎤
+ 2
⎢
2
2 ⎣ s + 49 s + 1 ⎥⎦
⎞ 3⎛ s ⎞
⎟
⎟+ ⎜ 2
⎠ 2 ⎝ s +1⎠
s ⎤
⎡
⎢Q L{cos at} = 2
⎥
s + a2 ⎦
⎣
3⎡ s
s ⎤
+ 2 ⎥
⎢
2
2 ⎣ s + 49 s + 1 ⎦
Q17. Find the Laplace Transform of the following
functions,
(i) f(t) = kt, k is a real constant >1
δ '(t).
(ii) f(t) = tδ
Ans:
(i)
f(t) = kt, k is a Real Constant >1
Given that,
f(t) = kt
Where,
K is real constant > 1
From the definition of Laplace transform,
∴ L{3 cos 3t cos 4t} =
s
s2 + a2
Hence proved.
Q16. Find L[3 cos 3t cos 4t].
Ans: Given Laplace function is, L[3 cos 3t cos 4t]
Consider (cos 3t ⋅ cos 4t)
On multiplying and dividing by 2, we get,
cos 3t.cos 4t =
=
1
( 2 cos 3t . cos 4t )
2
1
[cos(3t + 4t) + cos(3t – 4t)]
2
∞
f (s ) = L[f(t)]=
[Q 2 cos A cos B = cos( A + B ) + cos( A − B )]
=
=
∴
=
cos 3t cos 4t =
1
1
cos 7t +
cos t
2
2
e − st dt
... (1)
0
⇒
∞
⎡ t e − st ⎤
e − st t
k . log k dt
−
f (s ) = ⎢ k
⎥
⎣ − s ⎦0 0 − s
∫
1
[cos 7t + cos t ]
2
[Q cos (– t) = cost]
1
1
cos 7t +
cos t
2
2
t
∞
1
[cos 7t + cost (–t)]
2
=
∫k
⎡ d x
⎤
x
⎢Q dx (a ) = a log a⎥
⎦
⎣
∞
... (1)
⇒
1 log k − st t
e k dt
f (s ) = +
s
s 0
⇒
f (s ) =
∫
1 log k
. f (s)
+
s
s
Taking Laplace transform of 3 cos 3t cos 4t, we get,
L{3 cos 3t cos 4t} = L{3(cos 3t . cos 4t )}
⎧ ⎛1
1
⎞⎫
= L ⎨3⎜ cos 7t + cos t ⎟ ⎬
2
⎠⎭
⎩ ⎝2
[Q From equation (1)]
3
⎫
⎧3
= L ⎨ cos 7t + cos t ⎬
2
2
⎭
⎩
⎫
⎧3
⎫
⎧3
= L ⎨ cos 7t ⎬ + L⎨ cos t ⎬
⎭
⎩2
⎭
⎩2
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[Q From equation (1)]
⇒
1
⎡ log k ⎤
f ( s ) ⎢1 −
⎥ = s
s
⎣
⎦
⇒
⎡ s − log k ⎤ 1
f ( s) ⎢
⎥=
s
⎣
⎦ s
⇒
f (s ) =
1
( s − log k )
∴ L[ k t ] =
1
( s − log k )
SIA GROUP
1.8
(ii)
MATHEMATICS-II [JNTU-ANANTAPUR]
δ'(t)
f(t) = tδ
On applying Laplace transform on both sides,
Given that,
L[f(t)] = L[sin(2ωt) cos(2ωt0) – cos(2ωt) sin(2ωt0)]
f(t) = t δ'(t)
= L[sin(2ωt) cos(2ωt0)] – L[cos(2ωt) sin(2ωt0)]
Considering R.H.S and integrating,
∫ t.δ' (t)dt = t.∫ δ' (t ).dt − ∫ (t ' ∫ δ' (t ).dt ) dt
∫
= t .δ(t ) − 1.δ(t ).dt
= cos(2ωt0) × L[sin 2ωt)] – sin(2ωt0) × L[cos(2ωt)]
= cos( 2ωt0 ) ×
... (2)
2ω
s
− sin 2ωt0 × 2
2
s + (2ω) 2
s + (2ω)
2
Here, δ(t) = 1 for t = 0. But, at the same instant t.δ(t) = 0.
a ⎤
⎡
⎢Q L (sin at ) = s 2 + a 2 ⎥
⎢
⎥
s
⎢ L (cos at ) =
⎥
2
2 ⎥
⎢⎣
s +a ⎦
On substituting above value in equation (2), we get,
⇒
∫ t.δ' (t).dt = t.(0) – ∫ δ(t).dt
⇒
∫ t.δ' (t).dt = 0 – ∫ δ(t).dt
⇒
∫ t.δ' (t).dt = – ∫ δ(t).dt
∴ t.δ' (t ) = − δ(t )
=
... (3)
By definition of Laplace transform,
∴ L[ f (t )] =
2ω cos(2ωt 0 )
s 2 + 4ω2
−
s. sin(2ωt 0 )
s 2 + 4ω2
2ω cos( 2ωt 0 ) − s. sin( 2ωt 0 )
s 2 + 4ω 2
F(s) = L [f(t)]
ωt + θ)
t cos(ω
(ii)
= L [t.δ'(t)]
= L [– δ(t)]
Let, f(t) = t cos(ωt + θ)
[Q From equation (3)]
∞
⇒
F(s) =
∫ − δ(t ).e
− st
cos(A + B) = cosA cosB – sinA sinB
∴
dt
f(t) = t cos(ωt + θ) = t[cosωt cosθ – sinωt sinθ]
0
= t cosωt cosθ – t sinωt sinθ
And also, for an unit impulse function or signal,
⎧1, t = 0
δ(t) = ⎨
⎩0, else where
∴
f (s) = – (1) e
− st
|t =0 = – e–s(0) = – 1
∴ L[t.δ' (t )] = L[−δ(t )] = −1
Q18. Find the Laplace transform for the following
functions,
(i)
ω (t – t0)
sin 2ω
(ii)
ω t + θ ).
t cos(ω
On applying Laplace transform on both sides, we get,
⇒
L[f(t)] = L[t cosωt cosθ – t sinωt sinθ]
= L[t cosωt cosθ] – L[t sinωt sinθ]
= cosθ L[t cosωt] – sinθ L[t sinωt]
⎡ s 2 − ω2 ⎤
2
2 2 ⎥ – sinθ ×
⎣ (s + ω ) ⎦
= cosθ × ⎢
⎡ 2ωs ⎤
⎢ 2
2 2⎥
⎣ (s + ω ) ⎦
⎡
s2 − a2
2 as ⎤
and L(t sin at ) = 2
⎢Q L(t cos at ) = 2
⎥
2 2
(s + a )
(s + a 2 ) 2 ⎦⎥
⎣⎢
Ans:
(i)
ω(t – t0)
sin 2ω
Let, f(t) = sin 2ω(t – t0) = sin(2ωt – 2ωt0)
Since,
( s 2 − ω2 ) cos θ 2ωs sin θ
− 2
=
( s 2 + ω2 ) 2
( s + ω2 ) 2
sin(A – B) = sinA cosB – cosA sinB
f(t) = sin(2ωt – 2ωt0)
= sin(2ωt) cos(2ωt0) – cos(2ωt) sin(2ωt0)
Look for the SIA GROU P LOGO
... (1)
∴ L[ f (t )] =
( s 2 − ω2 ) cos θ − 2ωs. sin θ
(s 2 + ω2 ) 2
on the TITLE COVER before you buy
1.9
UNIT-1 (Laplace Transform)
1.2 INVERSE TRANSFORM
s+3
⎡
⎤
⎥.
2
⎣ s − 10s + 29 ⎦
Q19. Find L–1 ⎢
Ans: The given function is,
s+3
⎤
⎡
L–1 ⎢ 2
⎥
s
s
−
10
+
29
⎦
⎣
The numerator and denominator can be written as,
s + 3 = (s – 5) + 8 and
⇒
⇒
s2 – 10s
+ 29 = (s – 5)2 + 22
⎡ ( s − 5) + 8 ⎤
s+3
⎡
⎤
–1
L–1 ⎢ 2
⎥
⎥ =L ⎢
⎣ s − 10 s + 29 ⎦
⎢⎣ ( s − 5) 2 + 2 2 ⎥⎦
⎤
⎡
⎡ ( s − 5) ⎤
8
= L–1 ⎢
+ L–1 ⎢
2
2 ⎥
2
2⎥
⎢⎣ ( s − 5) + 2 ⎥⎦
⎣⎢ ( s − 5) + 2 ⎦⎥
⎤
⎡ −1 ⎡
⎤
s−a
= e at cos bt ⎥
⎢Q L ⎢
2
2⎥
⎢⎣ ( s − a ) + b ⎥⎦
⎥⎦
⎢⎣
⎡
⎤
4× 2
= e5t.cos 2t + L–1 ⎢
2
2⎥
⎢⎣ (s − 5) + 2 ⎥⎦
⎤
2
2
2⎥
⎣ ( s − 5) + 2 ⎦
⎡
= e5t.cos 2t + 4L–1 ⎢
⎤
⎡
⎤
b
at
−1 ⎡
sin
=
e
bt
⎥
⎢Q L ⎢
⎥
2
2
⎢⎣ ( s − a ) + b ⎦⎥
⎥⎦
⎢⎣
= e5t cos 2t + 4e5t sin 2t
= e5t [cos 2t + 4 sin 2t ]
s+3
⎤
⎡
5t
∴ L−1 ⎢ 2
⎥ = e (cos 2t + 4 sin 2t )
⎣ s − 10s + 29 ⎦
Q20. Find L–1[(2s + 3)/(s3 – 6s2 + 11s – 6)].
Model Paper-II, Q2
Ans: The given function is,
L–1[ (2s + 3)/(s3 – 6s2 + 11s – 6)]
Consider,
⇒
2s + 3
s − 6s 2 + 11s − 6
3
2s + 3
⎡
⎤
⎡
⎤
2s + 3
⎢ 3
⎥ = ⎢
⎥
2
⎣ s − 6 s + 11s − 6 ⎦
⎣ ( s − 1)( s − 2)( s − 3) ⎦
On applying partial fractions, we get,
∴
2s + 3
A
B
C
=
+
+
( s − 1)( s − 2)( s − 3)
s −1
s−2
s −3
⇒
2s + 3
A( s − 2)( s − 3) + B ( s − 1)( s − 3) + C ( s − 1)( s − 2)
=
( s − 1)( s − 2)( s − 3)
( s − 1)( s − 2)( s − 3)
⇒
... (1)
2s + 3 = A(s – 2) (s – 3) + B(s – 1)(s – 3) + C(s – 1)(s – 2)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (2)
SIA GROUP
1.10
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting s = 1 in equation (2), we get,
2(1) + 3 = A(1 – 2)(1 – 3) + B(1 – 1)(1 – 3) + C(1 – 1)(1 – 2)
⇒
2 + 3 = A(–1)(–2) + B(0) (–2) + C(0) (–1)
⇒
5 = A(2) + 0 + 0
⇒
5 = 2A
∴A=
5
2
... (3)
On substituting s = 2 in equation (2), we get,
2(2) + 3 = A(2 – 2)(2 – 3) + B(2 – 1)(2 – 3) + C(2 – 1)(2 – 2)
⇒
4 + 3 = A(0)(–1) + B(1)(–1) + C(1)(0)
⇒
7 = 0 + B(–1) + 0
⇒
7 = –B
∴ B = −7
... (4)
On substituting s = 3 in equation (2), we get,
2(3) + 3 = A(3 – 2)(3 – 3) + B(3 – 1)(3 – 3) + C(3 – 1)(3 – 2)
⇒
⇒
6 + 3 = A(1)(0) + B(2)(0) + C(2)(1)
9 = 2C
∴C =
9
2
... (5)
On substituting equations (3), (4) and (5) in equation (1), we get,
9 2
5 2
( −7 )
2s + 3
+
+
=
s
− 3)
(
s
−
s
−
(
1
)
(
2
)
( s − 1)( s − 2)( s − 3)
=
5
9
7
–
+
2( s − 1)
s−2
2( s − 3)
On applying inverse Laplace transform on both sides, we get,
9 ⎫
2s + 3
⎫ L−1 ⎧ 5 − 7 +
⎧
⎬
⎨
L−1 ⎨ 3
⎬=
2
2
(
1
)
2
2
(
s
s
s
−
−
− 3) ⎭
⎩
⎩ s − 6 s + 11s − 6 ⎭
9 ⎫
2s + 3
⎫ L−1 ⎧ 5 ⎫
⎧
−1 ⎧
−1 ⎧ 7 ⎫
L−1 ⎨ 3
⎬ – L ⎨
⎨
⎬
⎬ + L ⎨
⎬=
2
⎩s − 2⎭
⎩ 2( s − 1) ⎭
⎩ s − 6 s + 11s − 6 ⎭
⎩ 2( s − 3) ⎭
=
5 −1 ⎧ 1 ⎫
9 −1 ⎧ 1 ⎫
−1 ⎧ 1 ⎫
L ⎨
⎬ –7 L ⎨
⎬
⎬ + L ⎨
2
s
−
2
⎩ s − 1⎭
⎩ s − 3⎭
⎭ 2
⎩
=
5 t
9
.e – 7e2t + e3t
2
2
⎛
⎞
⎛ 1 ⎞
⎜⎜Q L−1 ⎜
⎟ = e at ⎟⎟
⎝s−a⎠
⎝
⎠
2s + 3
9e 3t
⎫ 5 t
⎧
2t
∴ L−1 ⎨ 3
=
−
+
e
e
7
⎬
2
⎩ s − 6 s 2 + 11s − 6 ⎭ 2
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
1.11
UNIT-1 (Laplace Transform)
–1
3
3
Q21. Find L [1/(s + 1)s ].
Ans: The given function is,
⎧⎪
⎫⎪
1
L−1 ⎨ 3
3⎬
⎪⎩ ( s + 1) s ⎪⎭
Adding and subtracting s3 in numerator, we get,
3
3
⎧⎪
⎫⎪
1
⎪1 + s − s ⎫⎪
−1 ⎧
L−1 ⎨ 3
L
=
⎬
⎬
⎨
⎪⎩ ( s + 1) s 3 ⎪⎭
⎪⎩ ( s 3 + 1) s 3 ⎪⎭
3
⎫⎪
s3
⎪ (1 + s )
−1 ⎧
−
= L ⎨ 3
⎬
⎪⎩ ( s + 1) s 3 ( s 3 + 1) s 3 ⎪⎭
1 ⎫
−1 ⎧ 1
= L ⎨ 3− 3
⎬
+ 1⎭
s
s
⎩
⎧ 1 ⎫ −1 ⎧ 1 ⎫
⎬− L ⎨ 3 ⎬
⎩ s3 ⎭
⎩ s + 1⎭
−1
= L ⎨
t2
⎧ 1 ⎫
− L−1 ⎨ 3
=
⎬
2!
⎩ s + 1⎭
Resolving
=
t2
⎧ 1 ⎫
− L−1 ⎨ 3
⎬
2
⎩ s + 1⎭
=
⎧⎪
⎫⎪
t2
1
− L−1 ⎨
⎬
2
2
⎪⎩ ( s + 1)( s − s + 1) ⎪⎭
1
( s + 1)( s 2 − s + 1)
1
( s + 1)( s 2 − s + 1)
⇒
=
⎡
tn ⎤
−1 ⎧ 1 ⎫
⎢Q L ⎨ n +1 ⎬ = ⎥
⎩ s ⎭ n! ⎦⎥
⎣⎢
[Q ( a 3 + b 3 ) = ( a + b )( a 2 − ab + b 2 )]
... (1)
into partial fraction, we get,
A
Bs + C
+
s +1 s2 − s +1
... (2)
1 = A(s2 – s + 1) + (Bs + C) (s + 1)
... (a)
On substituting s = –1, we get,
1 = A((–1)2 – (–1) + 1) + (B(–1) + C) (–1 + 1)
⇒
1 = A(1 + 1 + 1) + (– B + C)0
⇒
1 = 3A + 0
∴ A=
1
3
On substituting A =
⇒
1
in equation (a), we get,
3
1=
( s 2 − s + 1)
+ ( Bs + C )( s + 1)
3
1=
s2 s 1
− + + Bs 2 + Bs + Cs + C
3 3 3
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (b)
SIA GROUP
1.12
MATHEMATICS-II [JNTU-ANANTAPUR]
On comparing the constant terms on L.H.S and R.H.S in equation (b), we get,
1 =
1
+C
3
⇒
C =1–
⇒
C =
∴C =
1
3
3 −1 2
=
3
3
2
3
On substituting C =
1 =
2
in equation (b), we get,
3
s2 s 1
2
2
− + + Bs 2 + Bs + s +
3 3 3
3
3
... (c)
On comparing the coefficient of ‘s’ on L.H.S and R.H.S in equation (c), we get,
0=
−1
2
+B+
3
3
⇒
B=
1 2
−
3 3
⇒
B=
1− 2
3
∴B =
−1
3
On substituting the values of A, B and C in equation (2), we get,
∴
−1
2
1
s+
3
3 + 3
=
s + 1 ( s 2 − s + 1)
( s + 1)( s 2 − s + 1)
1
=
1 ⎛ 1 ⎞ 1 ⎛⎜ − s + 2 ⎞
⎟
⎜
⎟+
3 ⎜⎝ ( s + 1) ⎟⎠ 3 ⎜⎝ ( s 2 − s + 1) ⎟⎠
=
−s+2 ⎞
1⎛ 1
+
⎟
⎜
3 ⎝ s +1 s2 − s +1⎠
On substituting the above partial fractions in equation (1), we get,
⎧1 ⎛ 1
⎧⎪
⎫⎪ t 2
− s + 2 ⎞⎫
1
− L−1 ⎨ ⎜
+ 2
⎟⎬
=
L–1 ⎨ 3
3⎬
2
⎪⎩ ( s + 1) s ⎪⎭
⎩ 3 ⎝ s + 1 s − s + 1 ⎠⎭
=
t 2 1 −1 ⎧ 1 ⎫ 1 −1 ⎧ − s + 2 ⎫
− L ⎨
⎬
⎬− L ⎨
2 3
⎩ s + 1 ⎭ 3 ⎩ s 2 − s + 1⎭
=
t 2 1 −t 1 −1 ⎧ s − 2 ⎫
− e + L ⎨ 2
⎬
2 3
3
⎩ s − s + 1⎭
Look for the SIA GROU P LOGO
⎡
−1 ⎧ 1 ⎫
− at ⎤
⎬=e ⎥
⎢Q L ⎨
⎩s+ a⎭
⎦
⎣
on the TITLE COVER before you buy
... (3)
1.13
UNIT-1 (Laplace Transform)
Consider,
s−2
2
s − s +1
2
⎛1⎞
Adding and subtracting ⎜ ⎟ in denominator.
⎝2⎠
⇒
s−2
2
s − s +1
=
s−2
2
⎛1⎞ ⎛1⎞
s − s +1 − ⎜ ⎟ + ⎜ ⎟
⎝2⎠ ⎝2⎠
2
=
s−2
2
⎛ 1⎞ 3
⎜s − ⎟ +
2⎠ 4
⎝
=
2
=
s−2
2
⎛1⎞
⎛1⎞
s2 − s + ⎜ ⎟ +1 − ⎜ ⎟
2
⎝ ⎠
⎝2⎠
s−2
2
1 ⎞ ⎛ 3 ⎞⎟
⎛
⎜ s − ⎟ + ⎜⎜
2 ⎠ ⎝ 2 ⎟⎠
⎝
2
=
2
s
2
⎛ 1 ⎞ ⎛⎜ 3 ⎞⎟
⎜s − ⎟ + ⎜
2 ⎠ ⎝ 2 ⎟⎠
⎝
2
−
2
2
1 ⎞ ⎛ 3 ⎞⎟
⎛
⎜ s − ⎟ + ⎜⎜
2 ⎠ ⎝ 2 ⎟⎠
⎝
2
1 1
+
2
2 2
−
=
2
2
2
2
1 ⎞ ⎛⎜ 3 ⎟⎞
1 ⎞ ⎛⎜ 3 ⎞⎟
⎛
⎛
+
+
s
−
s
−
⎟
⎟
⎜
⎜
2 ⎠ ⎝⎜ 2 ⎟⎠
2 ⎠ ⎝⎜ 2 ⎠⎟
⎝
⎝
s−
1
1
2
2
2
−
−
=
2
2
2
2
2
2
1 ⎞ ⎜⎛ 3 ⎞⎟
1 ⎞ ⎜⎛ 3 ⎞⎟
1 ⎞ ⎜⎛ 3 ⎞⎟
⎛
⎛
⎛
⎜s − ⎟ + ⎜
⎜s − ⎟ + ⎜
⎜s − ⎟ + ⎜
2 ⎠ ⎝ 2 ⎠⎟
2 ⎠ ⎝ 2 ⎠⎟
2 ⎠ ⎝ 2 ⎠⎟
⎝
⎝
⎝
s−
1
3
1
1
−2
s−
2
2
2
2
+
−
=
2
2 =
2
2
2
2
2
2
1 ⎞ ⎜⎛ 3 ⎞⎟
1 ⎞ ⎛⎜ 3 ⎞⎟
1 ⎞ ⎛⎜ 3 ⎞⎟
1 ⎞ ⎜⎛ 3 ⎞⎟
⎛
⎛
⎛
⎛
⎜s − ⎟ + ⎜
⎜s− ⎟ +⎜
⎜s − ⎟ + ⎜
⎜s − ⎟ + ⎜
2 ⎠ ⎝ 2 ⎠⎟
2 ⎠ ⎝ 2 ⎠⎟
2 ⎠ ⎝ 2 ⎠⎟
2 ⎠ ⎝ 2 ⎠⎟
⎝
⎝
⎝
⎝
s−
3
3
3
1
1
×
s−
3
2
2
2
3
2
.
−
−
=
2
2
2
2 =
2
2
2
2
3
⎞
⎛
1 ⎞ ⎛⎜ 3 ⎞⎟
1⎞ ⎜ 3⎟
⎛
⎛
⎛ 1 ⎞ ⎛⎜ 3 ⎟⎞
⎛ 1 ⎞ ⎛⎜ 3 ⎟⎞
+
−
+
s
−
s
⎟
⎜
⎜
⎟
⎜s − ⎟ + ⎜
⎜s − ⎟ + ⎜
2 ⎠ ⎜⎝ 2 ⎟⎠
2 ⎠ ⎜⎝ 2 ⎟⎠
⎝
⎝
2 ⎠ ⎝ 2 ⎟⎠
2 ⎠ ⎝ 2 ⎟⎠
⎝
⎝
s−
3
1
s−
s−2
2
2
− 3.
=
⇒ 2
2
2
2
2
s − s +1 ⎛
⎞
⎛
1 ⎞ ⎛ 3 ⎞⎟
1⎞
3⎟
⎛
s− ⎟ +⎜
⎜
⎜ s − ⎟ + ⎜⎜
2 ⎠ ⎜⎝ 2 ⎟⎠
2 ⎠ ⎝ 2 ⎟⎠
⎝
⎝
On applying inverse Laplace transform on both sides, we get,
⎧
⎫
⎪
⎪
3
1
s−
⎪
⎪⎪
−1 ⎪
2
2
−1 ⎧ s − 2 ⎫
− 3.
⎬
L ⎨ 2
⎬= L ⎨
2
2
2
2
⎪⎛
⎩ s − s + 1⎭
1 ⎞ ⎛⎜ 3 ⎞⎟ ⎪
1 ⎞ ⎛⎜ 3 ⎞⎟
⎛
⎜s − ⎟ +⎜
⎪⎜ s − ⎟ + ⎜
⎪
2 ⎠ ⎝ 2 ⎟⎠ ⎪⎭
2 ⎠ ⎝ 2 ⎟⎠
⎝
⎪⎩ ⎝
⎧
⎫
⎧
⎫
⎪
⎪
⎪
⎪
3
1
s−
⎪
⎪⎪
⎪
⎪⎪
−1 ⎪
−1 ⎪
2
2
− 3L ⎨
= L ⎨
⎬
⎬
2
2
2
⎪⎛
⎪⎛
1 ⎞ ⎛⎜ 3 ⎞⎟ ⎪
1 ⎞ ⎛⎜ 3 ⎞⎟ ⎪
⎪⎜ s − ⎟ + ⎜
⎪
⎪⎜ s − ⎟ + ⎜
⎪
2 ⎠ ⎝ 2 ⎟⎠ ⎪⎭
2 ⎠ ⎝ 2 ⎟⎠ ⎪⎭
⎪⎩ ⎝
⎪⎩ ⎝
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.14
MATHEMATICS-II [JNTU-ANANTAPUR]
⎛ 3⎞
⎛ 3⎞
t/2
t/2
⎟t
= e cos⎜⎜ ⎟⎟ t − 3e sin⎜⎜
⎟
2
2
⎝ ⎠
⎝ ⎠
⇒
⎛
⎞
⎫⎪
⎧
s−a
at
⎜Q L−1 ⎪⎨
⎬ = e cos bt ⎟
2
2
⎜
⎟
⎪⎩ (s − a) + b ⎪⎭
⎜
⎟
⎜
⎟
⎫⎪
b
⎪
−1 ⎧
at
= e sin bt ⎟
L ⎨
⎜⎜
2
2⎬
⎟
⎪⎩ ( s − a) + b ⎪⎭
⎝
⎠
⎤
⎡ ⎛ 3⎞
⎧ s−2 ⎫
t/2
⎟ t − 3 sin 3 t ⎥
L−1 ⎨ 2
⎬ = e ⎢cos⎜⎜
2 ⎥
2 ⎟⎠
⎩ s − s + 1⎭
⎦
⎣⎢ ⎝
... (4)
On substituting equation (4) in equation (3), we get,
⇒
⎛ 3⎞ ⎤
t 2 1 −t 1 t / 2 ⎡ ⎛ 3 ⎞
⎟⎟ t ⎥
⎟⎟ t − 3 sin ⎜⎜
− e + e ⎢cos ⎜⎜
2 3
3
⎝ 2 ⎠ ⎥⎦
⎣⎢ ⎝ 2 ⎠
⎞ ⎤
⎛
⎧⎪
⎫⎪ t 2 1 −t 1 t / 2 ⎡ ⎛ 3 ⎞
1
⎟ t − 3 sin⎜ 3 ⎟ t ⎥
∴ L−1 ⎨ 3
= − e + e ⎢cos⎜
3⎬
⎜ 2 ⎟ ⎥
3
⎪⎩ (s + 1)s ⎪⎭ 2 3
⎢⎣ ⎜⎝ 2 ⎟⎠
⎠ ⎦
⎝
Q22. Find inverse Laplace transforms of
5s − 2
.
s 2 (s + 2)(s − 1)
⎧⎪
⎫⎪
5s − 2
Ans: The given function is, ⎨ 2
⎬
⎪⎩ s ( s + 2)(s − 1) ⎪⎭
On taking partial fractions, we get,
5s − 2
s ( s + 2)(s − 1)
2
=
D
B
A
C
+ 2 +
+
s −1
s
s+2
s
⇒ 5 s – 2 = A s(s + 2) (s – 1) + B (s+2) (s – 1) + C (s – 1) s2 + D (s + 2) s2
On substituting s = 0 in equation (1), we get,
– 2 = 0 + B(0 + 2) (0 – 1) + 0 + 0
⇒
– 2 = B(– 2)
∴ B =1
On substituting s = 1 in equation (1), we get,
5 – 2 = A(1) (1 + 2) (1 – 1) + B(1 + 2) (1 – 1) + C(1 – 1) (1)2 + D(1 + 2)(1)2
⇒
3 = 3D
∴ D =1
On substituting s = – 2 in equation (1), we get,
5 (– 2) – 2 = A(– 2) (– 2 + 2) (– 2 – 1) + B(– 2 + 2) (– 2 – 1) + C(– 2 – 1) (– 2)2 + D(– 2 + 2)(–2)2
⇒ – 10 – 2 = C (– 2 – 1) (– 2)2
⇒
– 12 = C (– 3) (4)
∴ C =1
On equating s3 terms on both sides of equation (1), we get,
0=A+C+D
⇒ A=–C–D
=–1–1
∴ A = −2
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
... (1)
1.15
UNIT-1 (Laplace Transform)
On substituting the corresponding values of A, B, C and D in equation (1), we get,
5s − 2
∴
2
s ( s + 2)( s − 1)
=–
1
1
1
2
+
+ 2 +
s −1
s
s+2
s
On applying inverse Laplace transform on both sides, we get,
⎧⎪
⎫⎪
5s − 2
1
1
1 ⎤
−1 ⎡ − 2
+ 2+
+
L–1 ⎨ 2
⎬= L ⎢
⎪⎩ s ( s + 2)(s − 1) ⎪⎭
s + 2 s − 1 ⎥⎦
s
⎣ s
⎧ 1 ⎫
⎧1 ⎫
⎧1 ⎫
⎧ 1 ⎫
= – 2 L–1 ⎨ ⎬ + L–1 ⎨ 2 ⎬ + L–1 ⎨
⎬ + L–1 ⎨
⎬
s
+
2
s
⎭
⎩
⎩ ⎭
⎩s ⎭
⎩ s −1 ⎭
= – 2(1) + t + e–2t + et
n ⎞
⎛
⎜Q L−1 ⎛⎜ 1 ⎞⎟ = t ⎟
⎜
⎝ s n +1 ⎠ n! ⎟
⎜
⎟
at ⎟
−1 ⎛ 1 ⎞
⎜
=
L
e
⎟
⎜
⎜
⎟
⎝s−a⎠
⎝
⎠
= – 2 + t + et + e–2t
⎫⎪
⎧⎪
5s − 2
t
−2 t
∴ L−1 ⎨ 2
⎬ = −2 + t + e + e
⎪⎩ s ( s + 2)( s − 1) ⎪⎭
⎤
⎡
s
Q23. Find L–1 ⎢ 2
⎥.
2
2
⎣ (s + 1) (s + 9) (s + 25) ⎦
Ans: The given function is,
⎤
⎡
s
L−1 ⎢ 2
⎥
2
2
⎣ ( s + 1) ( s + 9) ( s + 25) ⎦
Consider,
1
( s + 1) ( s 2 + 9)
2
Taking partial fractions, we get,
As + B Cs + D
1
+ 2
= 2
2
s +1
s +9
( s + 1) ( s + 9)
2
⇒
... (A)
1
( As + B) ( s 2 + 9) + (Cs + D ) (s 2 + 1)
=
( s 2 + 1) ( s 2 + 9)
( s 2 + 1) ( s 2 + 9)
⇒
1= (As + B) (s2 + 9) + (Cs + D) (s2 + 1)
1= As3 + 9As + Bs2 + 9B + Cs3 + Cs + Ds2 + D
⇒
1= (A + C)s3 + (B + D)s2 + (9A + C)s + (9B + D)
On comparing the coefficients of s3, s2, s and constants in equation (1), we get,
0 =A+C
... (1)
0 =B+D
0 = 9A + C
1= 9B + D
From equation (2), A = – C
... (3)
... (4)
... (5)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (2)
SIA GROUP
1.16
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting A = – C in equation (4), we get,
⇒
–9C + C = 0
⇒
–8C = 0
∴C = 0
∴A= 0
⇒
From equation (3), B = –D
On substituting B = –D in equation (5), we get,
⇒
1= –9D + D
1= – 8D
∴ D=
−1
8
∴ B=
1
8
On substituting the corresponding values in equation (A), we get,
1
1
0.s −
8
8 +
= 2
2
2
2
( s + 1) s + 9
( s + 1) ( s + 9)
0. s +
1
∴
=
1⎡ 1
1 ⎤
− 2
⎢
2
8 ⎣ s + 1 s + 9 ⎥⎦
Consider,
⎡
⎤
s
1
=⎢ 2
⎥. 2
2
( s + 1) ( s + 9) (s + 25)
⎢⎣ (s + 1)(s + 9) ⎥⎦ s + 25
s
2
2
Consider,
2
s
2
2
( s + 1) ( s + 25)
s
( s 2 + 1) ( s 2 + 25)
=
1⎡ 1
1 ⎤
s
.
−
8 ⎢⎣ s 2 + 1 s 2 + 9 ⎥⎦ s 2 + 25
=
⎤
s
s
1⎡
−
⎥
⎢
8 ⎣⎢ ( s 2 + 1) ( s 2 + 25) ( s 2 + 9) ( s 2 + 25) ⎦⎥
=
=
As + B
s +1
2
+
... (B)
Cs + D
... (C)
s 2 + 25
( As + B ) ( s 2 + 25) + (Cs + D ) ( s 2 + 1)
( s 2 + 1) ( s 2 + 25)
⇒
s = (As + B) (s2 + 25) + (Cs + D) (s2 + 1)
⇒
s = As3 + 25 As + Bs2 + 25B + Cs3 + Cs + Ds2 + D
s = (A + C) s3 + (B + D)s2 + (25 A + C) s + (25 B + D)
Comparing the coefficients of
s3, s2,
... (6)
s and constants in equation (6), we get,
0 =A+C
... (7)
0 =B+D
... (8)
1 = 25A + C
... (9)
0 = 25B + D
... (10)
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1.17
UNIT-1 (Laplace Transform)
From equation (7), A = –C
On substituting A = –C in equation (9), we get,
⇒ 1= 25(–C) + C
⇒ 1= –24C
∴C=
−1
24
∴ A=
1
24
From equation (8), B = –D
On substituting B = –D in equation (10), we get,
⇒
0 = 25(–D) + D
⇒
0 = –24D
∴ D=0
∴B =0
On substituting corresponding values in equation (C), we get,
∴
⎛ −1⎞
1
s + 0 ⎜ ⎟s + 0
24
s
24
+⎝ 2 ⎠
=
2
2
2
s +1
s + 25
( s + 1) ( s + 25)
=
1 ⎡ s
s ⎤
− 2
⎢
2
24 ⎣ s + 1 s + 25 ⎥⎦
... (11)
Consider,
s
2
2
=
2
=
( s + 9) ( s + 25)
⇒
s
2
( s + 9) ( s + 25)
As + B
2
s +9
+
Cs + D
s 2 + 25
( As + B ) ( s 2 + 25) + (Cs + D ) ( s 2 + 9)
( s 2 + 9) ( s 2 + 25)
s = (As + B) (s2 + 25) + (Cs + D) (s2 + 9)
⇒
s = As3 + 25 As + Bs2 + 25B + Cs3 + 9Cs + Ds2 + 9D
⇒
s = (A + C)s3 + (B + D)s2 + (25 A + 9C)s + (25 B + 9D)
On comparing the coefficients of s3, s2, s and constants in equation (12), we get,
0 =A+C
0 =B+D
1 = 25A + 9C
0 = 25B + 9D
From equation (13), we get, A = – C
On substituting A = – C in equation (15), we get,
1= 25(– C) + 9C
⇒
1 = – 16C
∴C =
−1
16
∴A =
1
16
... (D)
... (12)
... (13)
... (14)
... (15)
... (16)
From equation (14), we get, B = – D
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.18
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting B = –D in equation (16), we get,
0 = 25B + 9D
⇒
0 = 25(–D) + 9D
⇒
0 = –16D
∴D = 0
∴B = 0
On substituting corresponding values in equation (D), we get,
⎛ −1⎞
1
s + 0 ⎜ ⎟s + 0
s
16
+⎝ 2 ⎠
= 162
2
2
( s + 9) ( s + 25)
s +9
s + 25
s
2
2
( s + 9) ( s + 25)
=
s ⎤
1 ⎡ s
−
16 ⎢⎣ s 2 + 9 s 2 + 25 ⎥⎦
... (17)
On substituting equations (11) and (17) in equation (B), we get,
s
2
2
2
( s + 1)( s + 9)( s + 25)
∴
s
2
2
2
( s + 1) ( s + 9 ) ( s + 25)
=
1⎡ 1 ⎛ s
s ⎞ 1 ⎛ s
s ⎞⎤
− 2
− 2
⎟− ⎜ 2
⎟⎥
⎢ ⎜ 2
8 ⎣ 24 ⎝ s + 1 s + 25 ⎠ 16 ⎝ s + 9 s + 25 ⎠⎦
=
s ⎞ 1⎛ s
s ⎞⎤
1 ⎡1 ⎛ s
−
−
⎟
⎟− ⎜
⎜
8 × 8 ⎢⎣ 3 ⎝ s 2 + 1 s 2 + 25 ⎠ 2 ⎝ s 2 + 9 s 2 + 25 ⎠ ⎥⎦
=
⎤
1 ⎡
s
s
s
s
−
−
+
⎢
⎥
64 ⎣⎢ 3( s 2 + 1) 3( s 2 + 25 ) 2 ( s 2 + 9 ) 2( s 2 + 25 ) ⎦⎥
=
s
s
− 2s + 3s ⎤
1 ⎡
−
+
⎢ 2
⎥
2
64 ⎢⎣ 3( s + 1) 2( s + 9) 6( s 2 + 25) ⎥⎦
=
1
64
⎤
⎡
s
s
s
−
+
⎥
⎢
2
2
2
⎢⎣ 3( s + 1) 2( s + 9 ) 6 ( s + 25 ) ⎥⎦
=
1
64
⎤
⎡
s
s
s
−
+
⎥
⎢
2
2
2
⎢⎣ 3( s + 1) 2( s + 9 ) 6 ( s + 25 ) ⎥⎦
Taking inverse Laplace transform on both sides, we get,
⎧
⎞ ⎫⎪
⎫⎪
s
s
s
s
⎪
−1 ⎪ 1 ⎛
−1 ⎧
⎟⎬
−
+
∴ L ⎨ 2
⎬ = L ⎨ ⎜⎜
2
2
2
2
2
⎪⎩ ( s + 1) ( s + 9) ( s + 25) ⎪⎭
⎪⎩ 64 ⎝ 3( s + 1) 2 ( s + 9 ) 6 ( s + 25 ) ⎟⎠ ⎪⎭
=
⎫⎪⎤
⎫⎪ 1 −1 ⎧⎪
s
s
1 ⎡ 1 −1 ⎧⎪ s ⎫⎪ 1 −1 ⎧⎪
⎢ L ⎨ 2
⎬⎥
⎬+ L ⎨ 2
⎬− L ⎨ 2
2
2
64 ⎣⎢ 3
⎪⎩ ( s + 5 ) ⎪⎭⎦⎥
⎪⎩ ( s + 3 ) ⎪⎭ 6
⎪⎩ ( s + 1) ⎪⎭ 2
=
1
64
=
1 ⎡ 2 cos t − 3 cos 3t + cos 5t ⎤
⎥
64 ⎢⎣
6
⎦
=
1
[ 2 cos t − 3 cos 3t + cos 5t ]
384
1
1
⎡1
⎤
⎢ 3 cos t − 2 cos 3t + 6 cos 5t ⎥
⎣
⎦
⎡ −1 ⎧ s ⎫
⎤
= cos at ⎥
⎢Q L ⎨ 2
2⎬
⎩s + a ⎭
⎣
⎦
⎫⎪
⎧⎪
s
1
[ 2 cos t − 3 cos 3t + cos 5t ]
∴ L−1 ⎨ 2
⎬=
2
2
⎪⎩ ( s + 1)( s + 9)( s + 25) ⎪⎭ 384
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1.19
UNIT-1 (Laplace Transform)
Q24. Find the inverse Laplace transform of,
(2s2 − 6s + 5)
.
(s3 − 6s 2 + 11s − 6)
Ans: The given function is,
(2s 2 − 6s + 5)
( s 3 − 6s 2 + 11s − 6)
Resolving (s3 – 6s2 + 11s – 6) into factors, we get,
1 1 –6 11 –6
0 1 –5 6
1 –5
6
0
i.e., s2 – 5s + 6
⇒
s3 – 6s2 + 11s – 6 = (s – 1) (s2 – 5s + 6)
=(s – 1) (s2 – 3s – 2s + 6)
= (s – 1) (s (s – 3) – 2(s – 3))
= (s – 1) (s – 2) (s – 3)
⇒
2s 2 − 6s + 5
2s 2 − 6s + 5
=
s 3 − 6s 2 + 11s − 6 ( s − 1)(s − 2)( s − 3)
... (1)
On applying partial fraction to equation (1), we get,
⇒
⇒
⇒
A
B
C
(2s 2 − 6s + 5)
+
+
=
(
s
−
1
)
(
s
−
2
)
(
s
− 3)
( s − 1)(s − 2)(s − 3)
(2 s 2 − 6s + 5)
( s − 1)( s − 2)( s − 3)
=
... (2)
A( s − 2)( s − 3) + B( s − 1)( s − 3) + c(s − 1)( s − 2)
( s − 1)( s − 2)( s − 3)
A(s – 2) (s – 3) + B(s – 1) (s – 3) + C(s – 1) (s – 2) = 2s2 – 6s + 5
... (3)
On substituting s = 1 in equation (3) we get,
A(1 – 2) (1 – 3) + B(1 – 1) (1 – 3) + C(1 – 1)(1 – 2) = 2(1)2 – 6(1) + 5
⇒
2A = 2 – 6 + 5
⇒
2A = 1
∴ A=
1
2
On substituting s = 2 in equation (3), we get,
A(2 – 2) (2 – 3) + B(2 – 1) (2 – 3) + C(2 – 1) (2 – 2) = 2(2)2 – 6(2) + 5
⇒
–B = 8 – 12 + 5
⇒
–B = 1
∴ B = −1
On substituting s = 3 in equation (3), we get,
A(3 – 2) (3 – 3) + B(3 – 1) (3 – 3) + C(3 – 1) (3 – 2) = 2(3)2 – 6(3) + 5
⇒
2C = 18 – 18 + 5
⇒
2C = 5
∴ C = 5/ 2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.20
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting A, B, C values in equation (2), we get,
∴
1
1
5
(2s 2 − 6s + 5)
−
+
=
s
−
s
−
s
− 3)
2
(
1
)
(
2
)
2
(
( s − 1)(s − 2)(s − 3)
On applying inverse Laplace transform on both sides, we get,
⎡ (2 s 2 − 6s + 5) ⎤
1
1
5 ⎤
−1 ⎡
L−1 ⎢
−
+
⎥ = L ⎢
⎥
2
(
1
)
(
2
)
2
(
s
s
s
−
−
− 3) ⎦
s
s
s
(
1
)(
2
)(
3
)
−
−
−
⎣
⎦⎥
⎣⎢
1 ⎞ −1 ⎛ 1 ⎞ −1 ⎛ 5 ⎞
−1 ⎛
⎟⎟ − L ⎜⎜
⎟⎟ + L ⎜⎜
⎟⎟
= L ⎜⎜
⎝ 2(s − 1) ⎠
⎝ ( s − 2) ⎠
⎝ 2( s − 3) ⎠
=
⎤
⎡
1 −1 ⎛ 1 ⎞ −1 ⎛ 1 ⎞ 5 −1 ⎛ 1 ⎞ 1 t
5
−1 ⎛ 1 ⎞
L ⎜
⎟ = e at ⎥
⎟−L ⎜
⎟+ L ⎜
⎟ = e − e 2t + e 3t ⎢Q L ⎜
2
2
⎝ s−a⎠
⎝ s −1⎠
⎝ s−2⎠ 2
⎝ s−3⎠ 2
⎦
⎣
⎡ ( 2 s 2 − 6 s + 5) ⎤ 1 t
5 3t
2t
∴ L−1 ⎢
⎥ = e −e + e
2
⎢⎣ ( s − 1)( s − 2)( s − 3) ⎥⎦ 2
1.3 FIRST SHIFTING THEOREM
Q25. State and prove first shifting theorem.
Ans:
Statement
For answer refer Unit-I, Q5.
Proof: Given that,
... (1)
L{f(t)} = f (s )
∞
∫
− st
∴ L{f(t)} = e f (t ) dt
0
∞
∴
∫
− st at
L{e f(t)}= e [ e f (t )]dt
at
0
∞
=
∫e
−( s − a )t
f (t ) dt
... (2)
0
On comparing equations (1) and (2), we get,
L{e at f (t )} = f ( s − a)
Hence, the statement is proved.
Q26. Prove that,
b
(s − a)2 − b 2
(i)
L[eat sinh bt] =
(ii)
L[eat cosh bt] =
s−a
.
(s − a)2 − b 2
Ans:
(i)
Model Paper-II, Q3
b
(s − a)2 − b 2
Consider, L[sinh bt]
L[eat sinh bt] =
∞
∫
− st
Since, L[f(t)] = e . f (t ) dt
0
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1.21
UNIT-1 (Laplace Transform)
∞
⇒
L[sinh bt] =
∫
(ii)
e −st . sinh bt dt
0
∞
=
∫e
ebt − e −bt
⎜
2
⎝
− st ⎛
⎜
0
∞
⎞
⎟dt
⎟
⎠
at
− at
⎛
⎜Q sinh(at ) = e − e
⎜
2
⎝
s−a
(s − a)2 − b 2
Consider, L[cosh bt]
L(eat cosh bt) =
∫
− st
Since, L[f(t)] = e f (t ) dt
0
⎞
⎟
⎟
⎠
∞
∫
− st
⇒ L[cosh bt] = e cosh bt dt
0
∞
⎛ ebt + e −bt
= e − st ⎜⎜
2
⎝
0
∞
∫
1
(e − st .e bt − e − st e −bt ) dt
=
2
∫
0
at
− at
⎛
⎜Q cosh at = e + e
⎜
2
⎝
∞
∞
⎤
1 ⎡ − ( s −b ) t
dt − e −( s +b )t dt ⎥
= ⎢ e
2⎢
⎥
0
⎣0
⎦
∫
∫
∞
∫
0
∞
∞
⎤
1 ⎡ − ( s −b ) t
⎢
e
dt + e −( s +b ) t dt ⎥
=
2⎢
⎥
0
⎣0
⎦
∫
1 ⎡⎛⎜ e −∞ − e 0 ⎞⎟ ⎛⎜ e −∞ − e 0 ⎞⎟⎤
−
⎥
= ⎢⎜
2 ⎣⎢⎝ − ( s − b) ⎟⎠ ⎜⎝ − ( s + b ) ⎟⎠⎦⎥
=
=
=
⇒
L[sin h(bt)]=
1⎡ 1
1 ⎤
−
2 ⎢⎣ s − b s + b ⎥⎦
1 ⎡ s + b − ( s − b) ⎤
2 ⎢⎣ ( s + b)( s − b) ⎥⎦
1 ⎡ 2b ⎤
2 ⎢⎣ s 2 − b 2 ⎥⎦
=
1 ⎡ e −∞ − e 0 e −∞ − e 0 ⎤
+
⎥
⎢
2 ⎢⎣ − ( s − b) − (s + b) ⎥⎦
=
1⎡ 1
1 ⎤
+
2 ⎢⎣ s − b s + b ⎥⎦
=
1 ⎡ s + b + s −b ⎤
2 ⎢⎣ ( s − b)( s + b) ⎥⎦
=
1 ⎛ 2s ⎞
⎜
⎟
2 ⎝ s 2 − b2 ⎠
b
s − b2
Now, L[eat sinh bt] =
⇒
b
( s − a) 2 − b 2
L[cosh bt] =
s
s − b2
2
L[eat cosh bt] =
( s − a)
( s − a) 2 − b 2
(Q L[eat f(t)] = f (s – a))
∴ L[e at sinh bt ] =
∫
∞
∞⎤
⎡
1 ⎢⎛⎜ e −( s−b ) t ⎞⎟ ⎛⎜ e −( s+b )t ⎞⎟ ⎥
+
=
2 ⎢⎜⎝ − ( s − b) ⎟⎠ ⎜⎝ − (s + b) ⎟⎠ ⎥
0
0⎦
⎣
1 ⎡⎛ 0 − 1 ⎞ ⎛ 0 − 1 ⎞⎤
⎟−⎜
⎟⎥
⎢⎜
2 ⎣⎜⎝ − ( s − b) ⎟⎠ ⎜⎝ − ( s + b) ⎟⎠⎦
2
⎞
⎟
⎟
⎠
1
(e −st .ebt + e −st .e −bt ) dt
=
2
∞
∞⎤
⎡
1 ⎢⎛⎜ e −( s −b ) t ⎞⎟ ⎛⎜ e −( s +b) t ⎞⎟ ⎥
−
=
2 ⎢⎜⎝ − (s − b) ⎟⎠ ⎜⎝ − (s + b) ⎟⎠ ⎥
0
0 ⎦
⎣
=
⎞
⎟ dt
⎟
⎠
b
(s − a)2 − b 2
Hence proved.
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
[Q L[eat f(t)] = f (s – a)]
∴ L[e at cosh bt ] =
(s − a)
(s − a) 2 − b 2
Hence proved.
SIA GROUP
1.22
MATHEMATICS-II [JNTU-ANANTAPUR]
⎧ ⎛
1
⎞⎫
Q27. Evaluate L ⎨ e t ⎜ cos2t + sinh2t ⎟ ⎬ .
2
⎠⎭
⎩ ⎝
Ans: The given function is,
⎧ ⎛
1
⎞⎫
L ⎨e t ⎜ cos 2t + sinh 2t ⎟ ⎬
⎠⎭
2
⎩ ⎝
... (1)
On simplifying equation (1), we get,
⎧ ⎛
1
1
⎫
⎧
⎞⎫
L ⎨e t ⎜ cos 2t + sinh 2t ⎟ ⎬ = L ⎨e t cos 2t + et sinh 2t ⎬
2
2
⎠⎭
⎩
⎭
⎩ ⎝
t
= L{e cos 2t} +
1
L{et sinh 2t}
2
... (2)
Consider L{et cos 2t},
Here, a = 1, b = 2
⇒
L{et cos 2t}=
=
⇒
L{et cos 2t}=
⎡
(s − a) ⎤
at
⎢Q Le cos bt =
⎥
(s − a) 2 + b 2 ⎦⎥
⎣⎢
( s − 1)
2
( s − 1) + 2
2
s −1
2
s − 2s + 1 + 4
s −1
s − 2s + 5
... (3)
2
Consider L{et sinh 2t},
⇒
L{et sinh 2t} =
=
⇒
L{et sinh 2t] =
⎡
⎤
b
t
⎢Q L[e sinh bt ] =
2
2⎥
( s − a) − b ⎦⎥
⎣⎢
2
( s − 1) 2 − 2 2
2
s − 2s + 1 − 4
2
2
... (4)
2
s − 2s − 3
On substituting equations (3) and (4) in equation (2), we get,
1
2
s −1
⎧ ⎛
1
⎞⎫
+
L ⎨e t ⎜ cos 2t + sinh 2t ⎟ ⎬ = 2
2
2
s − 2s − 3
2
⎠⎭ s − 2 s + 5
⎩ ⎝
=
s −1
2
s − 2s + 5
+
1
2
s − 2s − 3
⎧ ⎛
1
1
s −1
⎞⎫
+ 2
∴ L⎨e t ⎜ cos 2t + sinh 2t ⎟⎬ = 2
2
−
+
−
2
5
2s − 3
s
s
s
⎠⎭
⎩ ⎝
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1.23
UNIT-1 (Laplace Transform)
–3t
Q28. Find the Laplace transform of the function e (2 cos 5t – 3 sin 5t).
Ans: The given function is,
e–3t (2 cos 5t – 3 sin 5t)
The Laplace transform of above expression can be obtained as,
L[e–3t (2 cos 5t – 3 sin 5t) = L[e–3t (2 cos 5t)] – L[e–3t. (3 sin 5t)]
= 2L[e–3t.cos 5t] – 3 L[e–3t. sin 5t]
s+a
⎛
− at
⎜Q L (e cos bt ) =
( s + a) 2 + b 2
⎜
⎜
b
⎜ L ( e − at sin bt ) =
⎜
( s + a) 2 + b 2
⎝
⎡
⎤ ⎡
⎤
s+3
5
⎥ − 3⎢
⎥
2
2
⎣ ( s + 3) + 25 ⎦ ⎣ ( s + 3) + 25 ⎦
=2⎢
=
=
∴ L[e −3t (2 cos 5t − 3 sin 5t )] =
⎞
⎟
⎟
⎟
⎟
⎟
⎠
2( s + 3)
15
−
2
( s + 3) + 25 ( s + 3)2 + 25
2 s + 6 − 15
2
( s + 3) + 25
=
2s − 9
( s + 3) 2 + 25
2s − 9
( s + 3) 2 + 25
1.4 TRANSFORMS OF DERIVATIVES AND INTEGRALS
Q29. Show that L{fn(t)} = sn f(s) – sn – 1f(0) – sn–2f' (0)..... fn– 1(0), where L{f (t)} = f (s).
OR
Model Paper-III, Q2
Prove the theorem of derivatives on Laplace transforms.
Ans: Given that,
L[f(t)]= f (s)
And
L[f n(t)] = snf(s) – sn – 1 f(0) – sn – 2 f' (0) ... f n–1(0)
Consider,
∞
L[ f ′(t )] =
∫e
–st
f' (t) dt
0
∞
= [f
(t )e − st ]∞
0
–
∫ f (t) (– s)e
–st dt
0
= [e
− st
∞
f (t )] 0 –
⎡Q
⎢⎣
∫ f ' (t ).dt = f (t )⎤⎥⎦
∞
∫e
− st
( − s ). f (t ) dt
0
∞
∫
= – f(0) + s e–st f (t) dt = –f (0) + s L[f (t)]
0
= s f (s) – f(0)
Similarly,
L[f'' (t)] = s2 f (s) – sf (0) – f' (0)
∴
L[fn (t)] = sn f (s) – sn–1f (0) – sn–2 f' (0).... fn–1(0)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.24
MATHEMATICS-II [JNTU-ANANTAPUR]
Q30. Find the Laplace transform of the following
functions using the theorem on transforms of
derivatives.
(a)
cos at
(b)
t sin at.
Ans:
(a)
⇒
⇒
⇒
⇒
L{2 a cos at – a2t sin at} = s2L{t sin at} – s(0.sin a
0) – (sin a.0 + a(0) cos a(0))
2
L{2 a cos at} – L{a t sin at}= s2L{t sin at} – 0 – 0
2aL{cos at} – a2 L{t sin at}= s2L{t sin at}
L{t sin at} (s2 + a2) = 2a L{cos at}
⇒
L{t sin at} =
cos at
2a
s +a
2
2
cos at
=
Let,
f(t) = cos at
∴ L{t sin at} =
Then,
f ''(t)= – a2 cos at
According to the theorem of derivatives, we have,
L{f n(t)} = sn L{f(t)} – sn – 1 f(0) – sn – 2 f '(0)....
L{– a2 cos at} = s2 L{cos at} – s.cos a(0) – (– a sin a(0))
⇒
2
2
– a L{cos at } = s L{cos at} – s.1 – 0
⇒
2
2
L{cost at} (s + a ) = s
⇒
L{cos at} =
s
s2 + a2
∴ L{cos at} =
(b)
s
2 as
2
(s + a 2 ) 2
2as
(s 2 + a 2 ) 2
∫
Ans: Given that,
L[f(t)] = f (s )
⎡t
⎤ f (s)
L ⎢ f (u ) du ⎥ =
s
⎢0
⎥
⎣
⎦
∫
Consider L.H.S,
⎡t
⎤
L ⎢ f (u ) du ⎥ =
⎢
⎥
⎣0
⎦
∫
⎤
⎡t
e − st ⎢ f (u ) du ⎥ dt
⎥
⎢
0
⎦
⎣0
∞
∫
∫
∞
⎛
⎞
⎜Q L[ f (t )] = e −st f (t )dt ⎟
⎜
⎟
0
⎝
⎠
s + a2
∫
2
tsin at
⎡ e − st
=⎢
⎢ −s
⎣
The given function is,
t sin at
t
∫
0
∫
Then,
f '(t) = t(a cos at) + sin at(1)
=0+
f '(t) = sin at + at cos at
1
L[f (t)]
s
And f ''(t) = [sin at + at cos at]1
⎛
⎜Q
⎜
⎝
= a cos at + [at (– a sin at) + cos at.a]
= a cos at – a2t sin at + a cos at
L{f n(t)} = sn L{f(t)} – sn–1 f(0) – sn – 2 f'(0) ......
∴
t
∫
∴ L f (u ) du =
0
L{f''(t)} = s2 L{f(t)} – sf(0) – f '(0)
Look for the SIA GROU P LOGO
0
⎞
0
⎠
∫ f (u ) du = 0 ⎟⎟
1
=
f (s)
s
f''(t) = 2 a cos at – a2t sin at
According to the theorem of derivatives,
∫
⎤
⎡ d ⎛t
⎞
⎢Q ⎜ f (u ).du ⎟ = f (t )⎥
⎟
⎥
⎢ dt ⎜
⎝0
⎠
⎦
⎣
f(t) = t sin at
⇒
∞
∞
⎤
1
⎥
f ( u ) du +
e − st f ( t ) dt
s
⎥
0
⎦0
Let,
⇒
s + a2
⎧t
⎫ 1
Q31. Prove that L ⎪⎨ f(u) du ⎪⎬ = f(s), where L{f(t )} = f (s).
⎪⎩ 0
⎪⎭ s
∴ L{f''(t)} = s2 L{f(t)} – sf(0) – f '(0)
⇒
s
2
s
⎛
⎞
⎜Q L{cos at } 2
⎟
s + a2 ⎠
⎝
The given function is,
f '(t)= – a sin at
.
f (s )
s
Hence proved.
on the TITLE COVER before you buy
1.25
UNIT-1 (Laplace Transform)
⎡ t e− t sint ⎤
Q32. Find L ⎢
dt ⎥ .
t
⎢0
⎥
⎣
⎦
∫
Ans: The given function is,
⎡ t e −t sin t ⎤
dt ⎥
⎢
t
⎥⎦
⎢⎣ 0
∫
Consider, sin t
⇒
L[sin t] =
⇒ L[e–t sin t] =
1
s +1
a ⎞
⎛
⎜Q L[sin at ] = 2
⎟
s + a2 ⎠
⎝
1
( s + 1) 2 + 1
⎞
⎛
b
⎟
⎜Q L[e −at sin bt ] =
2
2⎟
⎜
(s + a) + b ⎠
⎝
2
Then,
⎡ e −t sin t ⎤
L⎢
⎥ =
⎦⎥
⎣⎢ t
∫
∫
L[e −t sin t ]ds
s
∞
=
∞
⎛
⎞
⎜Q L ⎡ f (t ) ⎤ = f ( s)ds ⎟
⎥
⎢
⎜
⎟
⎣ t ⎦ s
⎝
⎠
∞
ds
∫ ( s + 1)
s
2
+1
∞
⎡ −1 ⎛ s + 1 ⎞⎤
⎟⎥
= ⎢tan ⎜
⎝ 1 ⎠⎦ s
⎣
1
1
⎡
⎤
dx = tan −1 ( x / a ) ⎥
⎢Q 2
a
x + a2
⎣
⎦
∫
= tan–1 (∞) – tan–1 (s + 1)
= π/2 – tan–1 (s +1)
⇒
⎡ e −t sin t ⎤
–1
L⎢
⎥ = cot (s +1)
t
⎦⎥
⎣⎢
π⎞
⎛
⎜Q tan −1 ∞ = ⎟
2⎠
⎝
π
⎛
−1
−1 ⎞
⎜Q cot θ = − tan θ ⎟
2
⎝
⎠
⎤ 1
⎡t
Since, L ⎢ g (t ) dt ⎥ = g ( s ) , we get,
⎥ s
⎢0
⎦
⎣
∫
⎡ t e −t sin t ⎤
1
⎥ dt = . cot −1 (s + 1)
L⎢
t
s
⎢
⎥
⎣0
⎦
∫
⎡ t e −t sin t ⎤ 1
dt ⎥ = ⋅ cot −1 ( s + 1)
∴ L⎢
t
⎥ s
⎢0
⎦
⎣
∫
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.26
MATHEMATICS-II [JNTU-ANANTAPUR]
1.5 UNIT STEP FUNCTION
Q33. Define unit step function and find the Laplace
transform of unit step function.
Ans:
Unit Step Function
For answer refer Unit-I, Q7.
Laplace Transform of Unit Step Function: From the definition
of Laplace transform,
∫e
− st
f (t ) dt
[e − st ]∞2
−s
=
0 − e −2 s
−s
⇒ L{u(t – 2)} =
∞
L{f(t)}=
=
e −2 s
s
... (2)
From equation (1),
0
∞
L{u(t – a)}=
∫e
− st
L{t2u(t – 2)} = (–1)2
u (t − a ) at
d2
[L{u(t – 2)}]
ds 2
0
=
∫e
− st
∫
.0dt + e − st .1dt
0
= 0+
a
[ ]
∞
e − st a
=
d 2 ⎡ e −2 s ⎤
⎢
⎥
ds 2 ⎢⎣ s ⎥⎦
=
d ⎡ s.( −2e −2 s ) − e −2 s .1 ⎤
⎥
⎢
ds ⎣⎢
s2
⎦⎥
=
d ⎡ − 2se−2 s − e −2 s ⎤
⎢
⎥
ds ⎢⎣
s2
⎥⎦
−s
=
− as
e −∞ − e − as − e
=
−s
−s
=
e − as
s
∴ L{u (t − a)} =
[Q L{tn f(t)} = (−1)n
∞
a
e − as
s
Q34. Evaluate L{t2 u(t – 2)}.
Ans: The given Laplace function is,
L{t2u (t – 2)}
Consider, L{u(t – 2)}
Here,
= ( −1)
... (1)
=
∞
∫
− st
∴ L{u(t – 2)}= e u (t − 2)dt
−1
s4
0
∞
∫
[Q L{f (t)} = e −st . f (t ) dt ]
=–
0
∞
∫
= e
− st
∫
u (t − 2 ) + e
0
[Q From equation (2)]
⎡ e −2 s (2 s + 1) ⎤
⎥
⎢
s2
⎦⎥
⎣⎢
⎡ s 2 .{e −2 s .2 + (2s + 1)(−2e −2 s )} − e −2s (2s + 1)( 2s) ⎤
= (−1) ⎢
⎥
s4
⎣⎢
⎦⎥
⎧ 0, t < a
u(t – 2) = ⎨
⎩ 1, t > 2
2
d
ds
dn
f (s) ]
ds n
− st
u (t − 2) dt
[ 2 s 2e − 2 s − 4 s 3e − 2s − 2 s 2 e − 2s − 4 s 2 e − 2s − 2 se −2 s ]
e −2 s
[ −4 s 3 − 4 s 2 − 2 s ]
s4
=
e −2 s
(4 s 3 + 4 s 2 + 2s )
s4
=
2 s.e −2 s
( 2 s 2 + 2 s + 1)
4
s
2
∞
2
∫
∫
= e − st .0 + e − st .1dt
0
2
∞
∫
= 0+ e
− st
.dt
∴ L{t 2 u (t − 2)} =
2e −2 s
s3
(1 + 2s + 2s 2 )
2
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
1.27
UNIT-1 (Laplace Transform)
Upper Limit
When t = ∞; x = ∞
On substituting the corresponding values in
equation (4), we get,
1.6 SECOND SHIFTING THEOREM
Q35. State and prove second shifting theorem.
Ans:
∞
∫
Statement
−s ( x+ a )
f ( x ) dx
L[g(t)] = e
For answer refer Unit-I, Q8.
0
∞
Proof
∫
− sx − sa
= e .e f ( x ) dx
Given that,
0
∞
... (1)
L[f(t)] = f (s)
=e
And
... (2)
According to the definition of Laplace transform,
L[f(t)] =
f ( x) dx
∫
... (3)
L[g(t)] =
∫e
.g (t ) dt
∫
a
∞
a
∫
e −st .0 dt + e −st . f (t − a ) dt
0
⎧⎪
⎫⎪
1
∴ L–1 ⎨
= e–2tsint
2
2⎬
⎪⎩ ( s + 2) + 1 ⎪⎭
a
∞
⎞
⎛
b
⎟
⎜Q L[e −at sin bt ] =
2
2⎟
⎜
(s + a) + b ⎠
⎝
∫
− st
= 0 + e f (t − a ) dt
a
According to Second shifting theorem,
∞
L{g(t)] =
∫e
... (2)
⎧⎪
⎫⎪
1
Consider, L–1 ⎨
2
2⎬
⎪⎩ ( s + 2) + 1 ⎪⎭
[Q From equation (2)]
∫
... (1)
⎧⎪ e −2 s ⎫⎪
⎫⎪
e −2 s
⎪
−1 ⎧
L–1 ⎨ 2
⎬= L ⎨
2
2⎬
⎪⎩ s + 4 s + 5 ⎪⎭
⎪⎩ (s + 2) + 1 ⎪⎭
∞
e − st .g (t ) dt + e − st .g (t ) dt
0
=
[Q From equation (1)]
Equation (1) can also be written as,
∞
∫
= e − sa . f ( s )
⎧⎪ e −2 s
⎫⎪
⎬
L–1 ⎨ 2
⎪⎩ s + 4 s + 5 ⎪⎭
0
=
[Q From equation (3)]
Q36. Find L–1 [e–2s/(s2 + 4s + 5)].
Ans: The given function is,
∞
− st
= e − sa L[ f (t )]
Hence, the statement is proved.
e −st . f (t ) dt
0
− st
f (t − a ) dt
⎧⎪ e −2 s
⎫⎪
L–1 ⎨
⎬ = f(t – 2) H(t – 2)
2
⎪⎩ ( s + 2) + 1 ⎪⎭
... (4)
a
Let,
[Q L−1[e − as f (s ) = f (t − a) H (t − a) ]
t – a=x ⇒ t=x +a
= [e–2(t – 2).sin(t – 2)]H(t – 2)
[Q f(t) = e–2t sin t]
dt – 0 = dx
⇒
− sx
∴ L[ g (t )] = e − sa f ( s )
∞
∴
∫e
0
⎧ f (t − a ) ; t > a
g(t) = ⎨
;t < a
⎩0
∴
− sa
= e 4−2t . sin(t − 2) H (t − 2)
dt = dx
Lower Limit
When t = a ; x = 0
⎫⎪
⎧⎪ e −2 s
4− 2t
. sin(t − 2).H (t − 2)
∴ L−1 ⎨
⎬=e
2
⎪⎩ ( s + 2) + 1 ⎪⎭
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.28
MATHEMATICS-II [JNTU-ANANTAPUR]
⎧
2π
⎛ 2π ⎞
⎪⎪cos⎜ t − 3 ⎟, t > 3
⎠
⎝
Q37. Find L[g(t)] where, g(t) = ⎨
2π .
⎪
0,
t<
3
⎩⎪
On substituting corresponding values in equation (2),
we get,
L[g(t)] =
∞ −s⎛ u + 2 π ⎞
⎟
⎜
3 ⎠
⎝
∫e
cosu.du
0
Ans: The given function is,
∞
⎧ ⎛ 2π ⎞
2π
⎪⎪cos ⎜⎝ t − ⎟⎠ , t >
3
3
g(t) = ⎨
2π
⎪ 0,
t<
⎪⎩
3
=
∫e
− su
.e
−
L[g(t)] =
∫e
− st
... (1)
−2 π ∞
s
3 . e − su . cos u. du
∫
=e
∫e
=
∫
2π 3
∫
∫e
− st
e − st (0)dt +
0
∞
⎛ 2π ⎞
e −st cos ⎜ t −
⎟dt
3 ⎠
⎝
2π 3
∫
∴L[ g (t )] = e
∫
∫e
− st
2π 3
⎛ 2π ⎞
cos⎜ t −
⎟dt
3 ⎠
⎝
Q38. If L[f(t)] =
... (2)
2π
=u
3
⇒
t=u+
⇒
dt = du
2π
3
For Lower Limits
For t =
−2 π
.s ⎛
3
s ⎞
⎜ 2
⎟
⎝ s +1⎠
9s 2 − 12s + 5
(s − 1) 3
. Find L[f(3t)] using
change of scale property.
Ans: The Laplace transform of f(3t) is obtained as,
Let,
t–
⎛ s ⎞
⎟
⎜ 2
⎝ s +1 ⎠
s ⎞
⎛
⎜Q L{cos at} = 2
⎟
s
a2 ⎠
+
⎝
⎛ 2π ⎞
e − st cos⎜ t − ⎟dt
3 ⎠
⎝
2π 3
∞
=
2π
.s
3
g (t )dt
∞
=0+
−
2π 3
0
=
=e
∞
g (t )dt +
−2 π
s
3 .L[cos u ]
⎛ ∞
⎞
⎜Q e − su cos t dt = L[cos t ]⎟
⎜
⎟
⎝ 0
⎠
g (t ) dt
− st
=e
0
0
2π 3
cosu.du
0
On applying Laplace transform, we get,
∞
2π
s
3
2π
2π 2π
−
,u=
=0
3
3
3
For Upper Limits
2π
t = ∞, u = ∞ –
= ∞.
3
Look for the SIA GROU P LOGO
L[f (3t)] =
1 ⎛s⎞
f⎜ ⎟
3 ⎝3⎠
⎛
1 ⎛ s ⎞⎞
⎜⎜Q L[ f ( at )] = f ⎜ ⎟ ⎟⎟
a ⎝ a ⎠⎠
⎝
⎡ ⎛ s ⎞2
⎤
⎛s⎞
⎢ 9⎜ ⎟ − 12 ⎜ ⎟ + 5 ⎥
1 ⎢ ⎝3⎠
⎝3⎠
⎥
= ⎢
3
⎥
3
⎛s
⎞
⎥
⎢
⎜ − 1⎟
⎥⎦
⎢⎣
⎝3 ⎠
⎡ 9 × s2
⎤
s
− 12. + 5 ⎥
1⎢ 9
3
⎥
= ⎢
3⎢
( s − 3) 3
⎥
⎢⎣
⎥⎦
27
1 ⎡ s 2 − 4s + 5 ⎤
= ⎢
⎥ × 27
3 ⎣⎢ ( s − 3) 3 ⎦⎥
=
∴ L[ f (3t )] =
9( s 2 − 4 s + 5)
( s − 3) 3
9( s 2 − 4 s + 5)
( s − 3) 3
on the TITLE COVER before you buy
1.29
UNIT-1 (Laplace Transform)
L[e–3t
Q39. Find
property.
sinh 3t] using change of scale
Ans: The given function is,
L[e–3t sinh 3t]
Let, f(t) = e–t sinht
L[sinh t] =
=
a ⎤
⎡
⎢Q L[sinh at ] = s 2 − a 2 ⎥
⎣
⎦
1
s − 12
2
1
1.7 DIRAC’S DELTA FUNCTION
Q40. Define Dirac delta function or the unit impulse
function. Also obtain its Laplace transform.
Ans:
Dirac Delta Function
For answer refer Unit-I, Q9.
Laplace Transform of Dirac Delta Function: The Laplace
transform of Dirac delta function is given by,
L{δ(t – a)} = L ⎡⎢ Lt f ε (t − a ) ⎤⎥
⎣ε→0
⎦
= Lt L{ f ε (t − a)}
2
s −1
ε →0
t
⎞
⎛
b
⎟
⎜Q L[e −a sinh bt =
2
2⎟
⎜
(s + a) − b ⎠
⎝
1
s + 2s + 1 − 1
=
1
s + 2s
∫
⎡a
ε →0 ⎢
⎣0
From the change of scale property,
⎡
⎤
⎢
⎥
1
1
⎢
⎥
∴ L[e–3t sinh 3t] =
3 ⎢ ⎛ s ⎞2
⎛ s ⎞⎥
⎢ ⎜ ⎟ + 2⎜ ⎟ ⎥
⎝ 3 ⎠ ⎦⎥
⎣⎢ ⎝ 3 ⎠
⎛
1 ⎛ s ⎞⎞
⎜⎜Q L[ f (at )] = f ⎜ ⎟ ⎟⎟
a
⎝ a ⎠⎠
⎝
⎡
⎤
⎡
⎤
⎢
⎢
⎥
1 ⎥
1
1
1
⎢
⎥ =
⎢ 2
⎥
=
2
3 ⎢s
3 ⎢ s + 6s ⎥
2s ⎥
⎢⎣ 9 + 3 ⎥⎦
⎢⎣ 9 ⎥⎦
∴ L[ e −3t sinh 3t ] =
⎤
a
a +ε
⎥
⎦
1 ⎡ 9 ⎤
3 ⎢⎣ s 2 + 6 s ⎥⎦
3
s 2 + 6s
3
2
s + 6s
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎞
⎟
, a ≤ t ≤ a + ε⎟
⎟
⎟
, t>a
⎠
, t<a
∞
a+ ε
⎤
⎡a
− st
− st 1
− st
= Lt ⎢ e .0dt + e . dt + e .0dt ⎥
ε →0 ⎢
ε
⎥
a
a+ ε
⎦
⎣0
∫
2
=
∞
⎛
⎧0
⎜
⎜Q f ε (t − a ) = ⎪⎨ 1
⎜
⎪ε
⎜
⎪⎩ 0
⎝
2
=
a+ε
= Lt ⎢ ∫ e − st f ε (t − a ) dt + ∫ e − st f ε (t − a )dt + ∫ e − st f ε (t − a) dt ⎥
1
L[e– t sinh t] = 2
2
s + 1 + 2s − 1
=
⎤
⎡∞
⎢ e −st f ε (t − a )dt ⎥
Lt
=
ε →0 ⎢
⎥
⎦
⎣0
∞
⎛
⎞
⎜Q L{ f (t )} = e −st . f (t )dt ⎟
⎜
⎟
0
⎝
⎠
∫
1
L[e–t sinh t] =
( s + 1) 2 − 1
∫
∫
⎤
⎡ 1 a+ε
⎢
Lt
e − st dt ⎥
=
ε →0 ⎢ ε
⎥
⎦
⎣ 0
∫
[ ]
⎡ 1 e −st a +ε ⎤
a ⎥
= Lt ⎢
ε →0 ⎢ ε
−s ⎥
⎦
⎣
⎡ e − s ( a +ε ) − e − sa ⎤
= Lt ⎢
⎥
ε →0 ⎢
− εs
⎦⎥
⎣
(
)
(
)
⎡ e − sa e − sε − 1 ⎤
= Lt ⎢
⎥
ε →0 ⎢
− εs
⎥⎦
⎣
⎡ 1 − e − sε ⎤
− sa
⎥
⎢
= e εLt
→0 ⎢
⎣ εs ⎥⎦
On applying L.Hospital’s rule, we get,
⎡ − (−ε)e − sε ⎤
− sa
e
Lt
.
⎥
⎢
L{δ(t – a)} =
ε →0 ⎢
ε
⎥⎦
⎣
− sa
e − sε
= e . εLt
→0
= e − sa .1
= e–sa
∴ L{δ(t − a )} = e − sa
SIA GROUP
1.30
MATHEMATICS-II [JNTU-ANANTAPUR]
1.8 CONVOLUTION THEOREM
Q41. State and prove convolution theorem.
Limits
For t = u ⇒ v = 0
For t = ∞ ⇒ v = ∞
Ans:
⎧⎪∞
⎫⎪
e − su f (u )⎨ e − sv g (v )dv ⎬du
⎪⎩ 0
⎪⎭
0
∞
Statement: Consider two functions f(t) and g(t) for t > 0. If
f (s ) and g (s ) are Laplace transforms of these functions then,
L{y(t)} =
∫
∫
∞
L{ f (t)*g(t)} = f ( s)· g ( s)
=
∫e
− su
f (u ){ g ( s )}du
0
Proof
∞
Let,
∫
− su
= g ( s ) e f (u ) du = g ( s )· f ( s )
y(t) = f(t) * g(t)
o
t
Then, y(t) =
∫ f (u ) g (t − u ) du then,
0
∴ L{ y(t )} = f ( s).g ( s)
Q42. Using Laplace transform solve,
Taking Laplace transform, we get,
t
∞
⎧t
⎫⎪
⎪
⎨ f (u ) g (t − u )du ⎬dt
⎪⎩ 0
⎪⎭
∫e ∫
− st
L{y(t)}=
0
y(t) = 1 – e–t +
∫ y(t − u) sin u du.
0
Ans: Given function is,
∞ t
∫∫e
=
− st
f (u ) g (t − u ) du dt
t
∫
−t
y(t) = 1 − e + y (t − u ) sin u du
0 0
On changing the order of integration from figure, we get,
u
... (1)
0
From the definition of convolution,
For y(t) = f(t) × g(t),
t
u=t
t=u
y(t) =
t=∞
t
u=0
... (2)
0
R
O
∫ f (u ) g (t − u ) du
Figure
On comparing equation (2) with equation (1), we get,
y(t) = 1 – e–t + y(t) × sin t
Taking Laplace transform, we get,
L{y(t)} = L{1} – L{e–t} + L{y(t) × sint}
Where,
R – Region of integration
=
1
1
−
+ L{ y (t )}.L{sin t}
s s +1
Lt t → u to ∞
1
⎫
⎧
⎪⎪
⎪⎪Q L(1) = s
⎬
⎨
⎪ L[e − at ] = 1 ⎪
⎪⎩
s + a ⎪⎭
Lt u → 0 to ∞
∞∞
⇒
L{y(t)} =
∫∫e
− st
f (u ) g (t − u ) dt du
0 u
∞∞
=
∫∫e
− st + su − su
f (u ) g (t − u ) du dt
⇒ L{y(t)} =
1
1
+ L{ y (t )}. 2
s ( s + 1)
s +1
0 u
⎧⎪∞
⎫⎪
− su
− s (t −u )
e
f
u
g (t − u )dt ⎬ du
(
)
⎨ e
=
⎪⎩ 4
⎪⎭
0
∞
∫
∫
⎡ ⎡1
1 ⎤
⎤
⎢Q L ⎢ sin at ⎥ = 2
⎥
⎦ s + a2 ⎦
⎣ ⎣a
⇒ L{ y (t )} − L{ y (t )}.
1
2
s +1
=
1
s (s + 1)
Let,
t–u=v
⇒
dt = dv
Look for the SIA GROU P LOGO
⎡ s 2 + 1 − 1⎤
1
⇒ L{ y (t )}⎢ 2
⎥ =
s (s + 1)
⎢⎣ s + 1 ⎥⎦
on the TITLE COVER before you buy
1.31
UNIT-1 (Laplace Transform)
⇒ L{ y (t )}.
s2
1
=
2
s (s + 1)
s +1
⇒ L{y(t)} =
⇒ L{y(t)} =
∴
s2 +1
1
×
s (s + 1)
s2
s2 +1
s 3 ( s + 1)
⎧ 2
⎫
−1 ⎪ s + 1 ⎪
y(t) = L ⎨ 3
⎬
⎪⎩ s (s + 1) ⎪⎭
1 ⎫⎪
⎪ 1
−1 ⎧
+ 3
= L ⎨
⎬
⎪⎩ s( s + 1) s ( s + 1) ⎪⎭
⎫⎪
1 ⎫ −1 ⎧⎪ 1
⎧1
= L−1 ⎨ −
⎬+ L ⎨ 3
⎬
⎪⎩ s ( s + 1) ⎪⎭
⎩ s s + 1⎭
⎫⎪
⎧1 ⎫
⎧ 1 ⎫ −1 ⎧⎪ 1
= L−1 ⎨ ⎬ − L−1 ⎨
⎬+ L ⎨ 3
⎬
⎪⎩ s (s + 1) ⎪⎭
⎩s ⎭
⎩ s + 1⎭
⎧⎪ 1 ⎫⎪
⎧1⎫
⎧ 1 ⎫
= 1 – et + L–1 ⎨ 3
⎬ = 1 − e t + L−1 ⎨ 3 ⎬.L−1 ⎨
⎬
⎪⎩ s (s + 1) ⎪⎭
⎩s ⎭
⎩ s + 1⎭
⎛ t2 ⎞
= 1 − e t + ⎜ ⎟(e −t )
⎜2⎟
⎝ ⎠
⎛ t2
∴ y (t ) = 1 − e t + ⎜
⎜2
⎝
⎡ −1 ⎡ 1 ⎤ t n ⎤
⎢Q L ⎢ n +1 ⎥ =
⎥
⎣ s ⎦ n! ⎥
⎢
⎢ − ⎡ 1 ⎤
⎥
− at
=
e
⎢ L 1⎢
⎥
⎥
⎣s + a⎦
⎣⎢
⎦⎥
⎞ −t
⎟( e )
⎟
⎠
Q43. Using convolution theorem find, L–1[1/(s2 – 1) (s2 + 25)].
Ans: The given function is,
⎡
⎤
1
L−1 ⎢ 2
⎥
2
⎣⎢ ( s − 1)( s + 25) ⎦⎥
Let, f (s ) =
1
2
s −1
⎧ 1 ⎫
then, f (t) = L−1 ⎨ 2
⎬ = sin ht
⎩ s − 1⎭
Let, g (s ) =
1
2
s + 25
⎧ 1 ⎫ sin 5t
then, g(t) = L−1 ⎨ 2
⎬=
5
⎩ s + 25 ⎭
⇒
f ( s) g ( s) =
1
( s 2 − 1)(s 2 + 25)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.32
MATHEMATICS-II [JNTU-ANANTAPUR]
On applying Laplace transform on both sides, we get,
⎫⎪
⎫⎪
1
⎪
1
⎪ 1
−1 ⎧
−1 ⎧
−1
⇒ L−1{ f ( s) g ( s )} = L ⎨ 2
⎬ = L ⎨ 2
⎬ (Q L { f ( s) g (s )} = f (t ) * g (t ))
2
2
⎪⎩ ( s − 1)( s + 25 ) ⎪⎭
⎪⎩ ( s − 1) ( s + 25 ) ⎪⎭
= f (t) * g(t)
t
=
∫
0
sin 5(t − u )
sin ht
du
5
⎛
⎜Q f (t ) * g (t ) =
⎜
⎝
t
∫
0
⎞
f (t ) g (t − u)du ⎟
⎟
⎠
On multiplying and dividing by 2, we get,
t
⎪
−1 ⎧
⎫⎪
2
1
⇒ L ⎨
⎬ = 5 × 2 sin ht sin 5( t − u ).du
2
2
⎪⎩ ( s − 1)( s + 25) ⎪⎭
0
∫
t
1
2 sin ht sin 5(t − u ).du
=
10
∫
[Q 2 sin A sin B = cos(A – B) – cos(A + B)]
0
t
=
1
cos( ht − (5t − 5u )) − cos( ht + (5t − 5u )) du
10
∫
0
=
1
10
t
∫ cos( ht − 5t + 5u ) − cos( ht + 5t − 5u ) du
0
t
1 ⎡ sin(ht − 5t + 5u ) sin(ht + 5t − 5u ) ⎤
+
=
⎥
10 ⎢⎣
5
5
⎦0
=
1 ⎡ sin( ht − 5t + 5t ) sin( ht + 5t − 5t ) sin( ht − 5t + 0) sin( ht + 5t − 0) ⎤
+
−
−
⎥
10 ⎢⎣
5
5
5
5
⎦
=
1 ⎧1
⎫
⎨ (sin ht + sin ht − sin( ht − 5 t ) − sin( ht + 5t )) ⎬
10 ⎩ 5
⎭
=
1
{2 sin ht − [sin ( ht − 5t ) + sin( ht + 5t )]}
50
=
1
{2 sin ht − ( 2 sin ht cos 5t )} (Q 2 sin A cos B = sin(x + y) + sin(x – y))
50
=
2 sin ht
{1 – cos 5t}
50
=
sin ht
{1 – cos 5t}
25
⎫⎪ sin ht
⎧⎪
1
{1 − cos 5t}
∴ L−1 ⎨ 2
⎬=
2
25
⎪⎩ ( s − 1)( s + 25) ⎪⎭
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
1.33
UNIT-1 (Laplace Transform)
⎤
.
2
2⎥
⎣ (s + 6s + 13) ⎦
⎡
s+3
Q44. Find the inverse Laplace transformation of ⎢
Ans: The given function is,
s+3
2
( s + 6 s + 13) 2
⎡
⎤
⎡
⎤
s+3
s+3
–1
–1
L ⎢ 2
⎢
⎥
2 ⎥ =L
2
2
⎣⎢ ( s + 6 s + 13) ⎦⎥
⎣⎢ [( s + 3) + 4][( s + 3) + 4] ⎦⎥
Let, f (s) =
⇒
( s + 3) 2 + 2 2
⎡
⎤ –3t
s+3
cos 2t
f(t) = L−1 ⎢
2
2⎥=e
⎣⎢ ( s + 3) + 2 ⎦⎥
And g(s) =
⇒
s+3
⎡
s+a
−1 ⎛
⎢Q L ⎜⎜
(
s
a)2 + b 2
+
⎝
⎣⎢
⎤
⎞
⎟ = ( e − at cos bt ) ⎥
⎟
⎠
⎦⎥
⎡
1
−1 ⎛
⎢Q L ⎜⎜
2
(
)
s
a
+
+ b2
⎝
⎣⎢
⎞
bt ⎤
⎟ = e − at sin ⎥
⎟
b ⎦⎥
⎠
1
( s + 3) 2 + 2 2
⎛
1
⎜
⎜ ( s + 3) 2 + 2 2
⎝
–1
g(t) = L
⎞
⎟ = e −3t sin 2t
⎟
2
⎠
From convolution theorem, we get,
⎡
⎤
s+3
L ⎢ 2
2⎥ =
⎣⎢ ( s + 6 s + 13) ⎦⎥
–1
t
t
∫
f (u ) g (t − u ) du
(Q L [ f (s ) g (s )] = f(t) * g(t) =
∫ f (u ) g (t − u )du )
0
0
∞
=
–1
∫e
− 3u
cos 2u.e −3(t −u )
0
sin 2(t − u )
du
2
t
1
e − 3u cos 2u e − 3t e 3u sin( 2t − 2u ) du
=
2
∫
[Q e
–3u
3u
0
.e = e = 1]
0
t
=
1 − 3t
e
cos 2u sin( 2t − 2u ) du
2
∫
0
On multiply and divide with 2, we get,
1 e −3 t
=
2 2
=
t
∫ 2 cos 2u sin( 2t − 2u ) du
0
1 − 3t t
e
[sin( 2u + 2t − 2u ) − sin( 2u − 2 t + 2u )]du [Q 2 cos A sin B = sin (A + B) – sin (A – B)]
4
∫
0
=
t
t
t
⎤
e −3t
e − 3t ⎡
[sin 2t − sin( 4u − 2t )]du =
⎢ sin 2t du − sin( 4u − 2t ) du ⎥
4
4 ⎢
⎥
0
0
⎦
⎣0
∫
∫
∫
t
t
e −3t ⎡
⎡ − cos( 4t − 2t ) ⎤ ⎤ e −3t ⎡
⎡ cos( 4t − 2t ) cos(0 − 2t ) ⎤ ⎤
t
⎢
−
t
du
sin
2
−
⎢
⎥ ⎥=
⎢sin 2t[t ]o + ⎢
= 4 ⎢
⎥⎥
4
⎥
4
4
4
⎣
⎦
⎣
⎦⎦
0⎦
⎣
0
⎣
∫
e −3t
=
4
⎡
⎡ cos 2t cos 2t ⎤ ⎤ e −3t
−
[t sin 2t ]
⎢t sin 2t + ⎢
⎥=
4 ⎥⎦ ⎦
⎣ 4
4
⎣
⎡
⎤ e −3t
s+3
∴ L−1 ⎢ 2
t sin 2t
=
2⎥
4
⎣⎢ ( s + 6 s + 13) ⎦⎥
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.34
MATHEMATICS-II [JNTU-ANANTAPUR]
1.9 LAPLACE TRANSFORM OF PERIODIC FUNCTION
Q45. If f(t) is a periodic function with period T, prove that L{f(t)} =
1
1− e
Ans: According to the definition of Laplace transform,
− sT
T
∫e
− st
f(t) dt .
0
∞
∫e
L[f(t)] =
− st
f ( t ) dt
0
2T
T
∫
=
e − st f ( t ) dt +
0
∫
3T
e − st f ( t ) dt +
∫e
− st
f ( t ) dt + …
... (1)
2T
T
As f(t) is a periodic function with period T, substituting t= u, t = u + T, t = u + 2T, … respectively, in equation (1), we get,
T
∫
L{f(t)} =
T
∫
0
∴
T
∫
e − su f (u ) du + e − s ( u + T ) f ( u + T ) du + e − s ( u + 2T ) f ( u + 2T ) du + …
0
0
f(u) = f(u + T) = f(u + 2T) = …
T
∫e
L{f(t)} =
T
− su
f (u ) du + e
− sT
0
∫e
T
− su
f (u ) du + e
0
− 2 sT
∫e
− su
f (u ) du + …
0
T
∫
− sT
+ e − 2 sT + …) e − su f (u ) du
= (1 + e
0
∴ L{ f (t )} =
T
1
=
∫e
1 − e − sT
− st
1 ⎞
⎛
−x
−2 x
⎜Q t = u and 1 + e + e ..... =
⎟
1 − e−x ⎠
⎝
f (t ) dt
0
T
1
∫e
1 − e − sT
− st
f (t )dt
0
Hence proved.
Q46. Find L[f(t)] where f(t) is given by,
f(t) = t, 0 < t < b
= 2b – t, b < t < 2b, where 2b is the period of f(t)
Ans: Given that,
f(t) = t, 0 < t < b
= 2b – t, b < t < 2b
Where, 2b is the period of f(t)
Laplace transform of a periodic function f(t) with a period ‘T’ is given by,
T
L{ f (t )} =
∴
L{f(t)} =
=
∫e
− st
f (t ).dt
0
1 − e − sT
1
1− e
− 2 bs
1
1 − e − 2 bs
2b
∫e
− st
f ( t ) dt
(Q T = 2b)
0
2b
⎡b
⎤
⎢ e − st f (t )dt + e − st f (t ) dt ⎥
⎢0
⎥
b
⎣
⎦
∫
∫
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
... (1)
1.35
UNIT-1 (Laplace Transform)
1
=
1 − e −2 bs
=
=
=
=
1
1 − e − 2 bs
1
1 − e −2bs
1
1 − e −2 bs
1
1− e
− 2bs
2b
⎤
⎡ b − st
⎢ e tdt + e − st (2b − t )dt ⎥
⎥⎦
⎢⎣ 0
b
∫
∫
[Q From equation (1)]
2b
2b
b
⎫
⎧⎡ b
⎞ ⎤⎪
⎛ d
⎞ ⎤ ⎡⎢
⎛d
⎪⎢
− st
− st
− st
− st
⎨ t e dt − ⎜ (t ) e dt ⎟dt ⎥ + ( 2b − t ) e .dt − ⎜ ( 2b − t ) e dt ⎟ dt ⎥ ⎬
⎠ ⎥⎦ ⎪
⎝ dt
⎠ ⎥⎦ ⎢⎣
⎝ dt
⎪⎩ ⎢⎣ 0
b
b
0
⎭
∫
∫
∫
∫
⎧⎡ ⎛ − st
⎪⎢ ⎜ e
⎨ t⎜
⎪⎩⎢⎣ ⎝ − s
b
b
⎛ e −st
⎞ 1 −st ⎤ ⎡
⎟ +
e dt ⎥ + ⎢ ( 2b − t )⎜
⎜ −s
⎟
⎥ ⎢
⎝
⎠0 s 0
⎦ ⎣
⎧⎛ ⎛ −st
⎪⎜ ⎜ e
⎨⎜ t ⎜
⎪⎩⎜⎝ ⎝ − s
⎞ 1 ⎛ e −st
⎟ + ⎜
⎜
⎟
⎠0 s ⎝ − s
⎧⎡ − st
⎪ e
⎨⎢t
⎪⎩⎢⎣ − s
∫
b
b
b
b
⎛ − st
⎞ ⎞⎟ ⎡
⎟ + ⎢( 2 b − t ) ⎜ e
⎟
⎜ −s
⎟ ⎟ ⎢
⎝
⎠0 ⎠ ⎣
2b
∫
∫
2b 2b
⎤⎫
− st
⎞
⎟ − ( −1) ( e ) dt ⎥ ⎪⎬
⎟
⎥⎪
−s
⎠b
b
⎦⎭
∫
2b
⎛ − st
⎞
⎟ −1⎜e
⎟
s ⎜⎝ − s
⎠b
2b ⎤ ⎫
⎞
⎟
⎟
⎠b
⎥ ⎪⎬
⎥⎪
⎦⎭
2b ⎫
2b
⎡ e − st ⎤
⎤ ⎡ e − st ⎤ ⎡ e − st ⎤
⎡ e − st ⎤
⎥ +⎢ 2 ⎥
⎥ + ⎢t
⎥ + ⎢ 2 ⎥ + ⎢2b
⎥⎦ 0 ⎢⎣ − s ⎥⎦ 0 ⎢⎣ − s ⎥⎦ b ⎢⎣ s ⎥⎦ b ⎢⎣ s ⎥⎦ b
⎪
⎬
⎪⎭
⎧⎪⎡ b.e − sb
⎤ ⎡ e − sb
e −bs ⎤ ⎡ 2be −2bs be − sb ⎤ ⎡ e − s 2 b e −bs ⎤ ⎫⎪
e 0 ⎤ ⎡ e −2bs
− 0⎥ + ⎢ 2 −
b
b
2
2
+
−
−
− 2 ⎥⎬
⎥
⎢
⎥+⎢
⎥+⎢
⎨⎢
s ⎦⎥ ⎣⎢ s 2
−s
− s ⎦⎥ ⎣⎢ s
− s 2 ⎦⎥ ⎣⎢
s ⎦⎥ ⎪⎭
⎪⎩⎣⎢ − s
⎥⎦ ⎣⎢ − s
=
1
1 − e −2bs
=
e −bs 1 e −2bs
e −bs
1
2be −2bs e −bs
e −2 bs e −bs ⎫⎪
⎪⎧ − e −bs
b
b
−
+
−
+
+
−
+
− 2 ⎬
b
2
2
b
⎨
s
s
s
s
s2
s2
1 − e −2 bs ⎪⎩ s
s2
s ⎪⎭
=
1
1− e
− 2bs
1
⎡1
⎤
− bs
− 2bs
− e −bs ) + ( −be −bs − 2be − 2bs + 2be −bs + 2be −2bs − be −bs ) ⎥
⎢ 2 (− e + 1 + e
s
⎣s
⎦
⎤
⎡ − 2e −bs + 1 + e −2bs
1
+ 0⎥ = 2
(1 − 2e −bs + e −2bs )
⎢
2
s
⎥⎦ s (1 − e −2bs )
⎢⎣
=
1
1 − e −2bs
=
1
(1 − e −bs ) 2
s (1 − e −2bs )
[Q a2 – 2ab + b2 = (a – b)2]
=
(1 − e −bs )(1 − e −bs )
s 2 (1 + e −bs )(1 − e −bs )
[Q (a – b)2 = (a – b) (a – b); (a2 – b2) = (a + b) (a – b)]
=
2
1 − e − bs
s 2 (1 + e −bs )
=
1 ⎡ 1 − e − bs ⎤
⎢
⎥
s 2 ⎢⎣1 + e −bs ⎥⎦
=
1 ⎡ e −bs / 2 (e bs / 2 − e − bs / 2 ) ⎤
⎥
⎢
s 2 ⎣⎢ e −bs / 2 (e bs / 2 + e −bs / 2 ) ⎦⎥
=
1
s2
=
⎛ bs ⎞
tan h⎜ ⎟
s2
⎝ 2⎠
∴ L[ f (t )] =
⎛ bs ⎞
tan h⎜ ⎟
s
⎝ 2⎠
⎛ 1 + e − x e − x / 2 (e x / 2 − e − x / 2 ) ⎞
⎜Q
⎟
⎜ 1 + e − x = e − x / 2 (e x / 2 + e x / 2 ) ⎟
⎝
⎠
⎡ e bs / 2 − e − bs / 2 ⎤
⎥
⎢ bs / 2
+ e −bs / 2 ⎥⎦
⎢⎣ e
1
x
−x
⎛
⎜Q tan hx = e − e
⎜
e x + e−x
⎝
⎞
⎟
⎟
⎠
1
2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.36
MATHEMATICS-II [JNTU-ANANTAPUR]
Q47. Find the Laplace transform of periodic function f(t) with period T,
4Et
T
4Et T
− E, 0 ≤ t ≤ = 3E −
, ≤ t ≤ T.
T
2
T 2
where f(t) =
Model Paper-III, Q3
Ans: Given that,
f(t) is a periodic function of period T and is given by,
T
⎧ 4 Et
⎪ T −E ; 0≤t ≤ 2
f(t) = ⎨
4 Et T
⎪3E −
; ≤t ≤T
2
T
⎩
Laplace transform of a periodic function f(t) with period ‘T’ is given by,
L{f(t)} =
=
1
1 − e −sT
1
1− e
− sT
T
∫e
− st
f (t ) dt
0
T /2
∫
T
e − st f (t ) dt +
0
∫e
− st
f (t ) dt
T /2
T
⎡T / 2
4 Et ⎞ ⎤
⎤
− st ⎡ 4 Et
− st ⎛
⎢
−
e
E
dt
e
E
+
−
3
⎟dt ⎥
⎜
=
⎥
⎢ T
T ⎠ ⎥
(1 − e − sT ) ⎢⎣ 0
⎦
⎣
⎝
T/2
⎦
1
=
1
1 − e − sT
∫
∫
T
T /2
⎡⎧ E T / 2
⎫⎪
⎫⎪⎤ ⎧⎪ T
4E
⎢⎪⎨ 4
t.e − st dt − E e − st dt ⎬⎥ + ⎨3E e − st dt −
te − st dt ⎬
⎢⎪ T
T
⎪⎭
⎪⎭⎥⎦ ⎪⎩ T / 2
T /2
0
0
⎣⎩
∫
∫
⎡
− st
e − st
⎢ 4 E ⎛⎜ − te
− 2
=
− sT ⎢ T ⎜
s
(1 − e ) ⎢
⎝ s
⎣
1
∫
T /2
⎞
⎟
⎟
⎠0
⎛ e − st
+ E⎜
⎜ s
⎝
T /2
⎞
⎟
⎟
⎠0
∫
⎛ 3Ee − st
+⎜
⎜ −s
⎝
T
− st
− st
⎛
⎞
⎟ − 4 E ⎜ − te − e
⎟T
T ⎜⎝
s
s2
⎠
2
T ⎤
⎞ ⎥
⎟
⎟T ⎥
⎠ ⎥
2⎦
=
⎡
⎢ 4E
⎢
(1 − e − sT ) ⎢ T
⎢⎣
=
1
⎡ 2 E − sT / 2 4 E − sT / 2 4 E E − sT / 2 E 3E − sT 3E − sT / 2 4 E − sT 4 E − sT 2 E − sT / 2 4 E − sT / 2 ⎤
− 2e
+ 2+ e
− −
+
+
+ 2e
−
− 2e
e
e
e
e
e
⎢−
⎥
s
s
s
s
s
s
Ts
Ts
1 − e − sT ⎣ s
Ts
Ts
⎦
=
4E
1
⎤
⎡E
− 2e − sT / 2 + e − sT / 2 − 1 + 3e − sT / 2 − 3e − sT + 4e − sT − 2e − sT / 2 + 2 − e − sT / 2 + 1 + e − sT − e − sT / 2 ⎥
− sT ⎢ s
Ts
(1 − e ) ⎣
⎦
=
4E
1
⎤
⎡ E − sT
− 1) + 2 (e − sT + 1 − 2e − sT / 2 )⎥
(e
− sT ⎢ s
Ts
(1 − e ) ⎣
⎦
=
4E
1 ⎡− E
⎤
(1 − e − sT ) + 2 (1 − e − sT / 2 ) 2 ⎥
− sT ⎢
Ts
1− e
⎣ s
⎦
1
⎡⎛ − T − sT / 2
⎞
⎢⎜ 2 e
1
e − sT / 2 ⎟⎟ ⎛
⎜
−
−⎜0 − 2
⎢
s
s2 ⎟ ⎝
s
⎢⎜⎜
⎟
⎠
⎣⎢⎝
⎤
⎡
⎛ e − sT / 2 1 ⎞ 3E −sT
4 E ⎢⎛⎜ Te − sT e −sT
⎞⎥
− sT / 2
⎜
⎟
+
−
−
−
−
− 2
E
e
e
⎢ −
⎟⎥
⎜ s
s ⎟⎠ s
T ⎢⎜⎝
s
s
⎠⎥
⎝
⎢⎣
⎦⎥
(
(
− E ⎛⎜ 1 − e −sT
=
s ⎜⎝ 1 − e −sT
)
)
(
⎞ 4 E 1 − e −sT / 2
⎟+
⎟ Ts 2 1 − e −sT
⎠
(
⎛ − T −sT / 2
⎞ ⎜ 2 e
e −sT / 2
⎟ −⎜
−
⎟ ⎜
s
s2
⎠ ⎜
⎝
⎞⎤
⎟⎥
⎟⎥
⎟⎥
⎟
⎠ ⎦⎥
)
[Q a2 – 2ab + b2 = (a – b)2]
)
2
−E
4E
(1 − e− sT /2 )2
×
+
1
=
s
Ts2 (1 − e−sT /2 )(1 + e−sT /2 )
[Q a2 – b2 = (a – b) (a + b)]
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
1.37
UNIT-1 (Laplace Transform)
=
− E 4 E ⎛⎜ 1 − e − sT / 2 ⎞⎟ − E 4E ⎛ 1 − e −sT / 4 .e − sT / 4 ⎞
⎟
+ 2
+ 2⎜
=
s
s
Ts ⎜⎝ 1 + e −sT / 2 ⎟⎠
Ts ⎜⎝ 1 + e − sT / 4 .e −sT / 4 ⎟⎠
− sT / 4
⎛
⎜1− e
− E 4E ⎜
e sT / 4
+ 2⎜
=
s
Ts
e − sT / 4
⎜⎜ 1 + sT / 4
e
⎝
=
⎞
⎟
sT / 4
− e − sT / 4 ⎞⎟
⎟ − E + 4 E ⎛⎜ e
⎟ = s Ts 2 ⎜ e sT / 4 + e −sT / 4 ⎟
⎝
⎠
⎟⎟
⎠
⎛ e x − e− x
⎞
⎜Q
⎟
=
⎜ e x + e − x tanh x ⎟
⎝
⎠
− E 4E
⎛ sT ⎞
+ 2 tanh⎜ ⎟
s
Ts
⎝ 4 ⎠
∴ L{ f (t )} = −
E 4E
⎛ sT ⎞
+
tanh⎜ ⎟
s Ts 2
⎝ 4 ⎠
1.10 DIFFERENTIATION AND INTEGRATION OF TRANSFORM
Q48. Show that L{tn(f(t))} = (–1)n
dn
{ f (s)} where, n = 1, 2, 3,.....
dsn
OR
Prove that L{tn f(t)} = (–1)n
dn
{ f (s)} where, n = 1, 2, 3,.....
dsn
Ans: Given that,
dn
{ f (s)}
ds n
L[tn f (t)] = (–1)n
... (1)
Where, n = 1, 2, 3,.....
From the definition of Laplace transform,
∞
f (s)=
∫e
–st
f(t) dt
0
= L[ f(t)]
On differentiating on both sides with respect to s, we get,
⇒
d
d
f (s ) =
ds
ds
∞
=
∞
∫e
–st
f(t)dt
0
d –st
e f(t)dt =
ds
∫
0
∞
∫ – te
–st
f(t)dt
0
∞
∫
= – e–st [t f(t)]dt = – L[t f(t)]
0
⇒ L[t f (t)] = −
d
f (s)
ds
... (2)
By mathematical induction, equation (2) can be written as,
L[tn f(t)] = (–1)n
dn f
(s)
ds n
Hence proved.
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.38
MATHEMATICS-II [JNTU-ANANTAPUR]
Consider,
Q49. Find L[t sin 3t cos 2t].
Ans: Consider, t sin 3t cos 2t
L[sin t] =
On multiplying and dividing by 2, we get,
L[t sin t] = (–1)
t
t sin3t cos2t = [2 sin 3t cos 2t]
2
t
[sin(3t + 2t) + sin(3t – 2t)]
2
[Q sin(A + B) + sin(A – B) = 2 sin A cos B]
t
[sin 5t + sint]
2
d ⎡ 1 ⎤
ds ⎢⎣ s 2 +1 ⎥⎦
=
2s
( s 2 + 1) 2
∴
L[t sin 3t cos 2t] =
=
⎡1
⎤
L[t sin 3t cos 2t] = L ⎢ [t sin 5t + t sin t ⎥
⎣2
⎦
1
[L(t sin 5t) + L(t sin t)]
2
∴ L[t sin 3t cos 2t ] =
L[sin 5t] =
[Q L(sin at) =
s 2 + 52
∴ L[t sin 5t] = (–1)
... (1)
a
s2 + a2
Ans: The given function is,
te–t sin 2t
The Laplace transform of te–t sin 2t is given by,
]
= (–1)
d
(f(s)), for n = 1, 2, 3, .... ]
ds n
d ⎛ 1 ⎞
⎜
⎟
ds ⎝ s 2 + 5 2 ⎠
{
10s
( s + 52 ) 2
d ⎡
2
⎤
⎢
2
ds ⎣ s + 1 + 2 s + 4 ⎥⎦
d ⎡
′
⎞
⎛ u ⎞ vu′ − uv′ ⎟
⎜ ⎟ =
v2 ⎟
⎝v⎠
⎠
⎡ − 2s ⎤
= − 5⎢ 2
2 2⎥
⎣ (s + 5 ) ⎦
2
}
2
⎤
= (–1) ds ⎢ 2
⎥
⎣ s + 2s + 5 ⎦
= (–1)
=
=
2
d
d s ( s + 1) 2 + 2 2
⎤
⎡
b
at
⎢QL e sin bt =
2
2⎥
(s − a) + b ⎦
⎣
= (–1)
⎡ ( s 2 + 52 )(0) − (1)(2s) ⎤
= − 5⎢
⎥
(s 2 + 52 )2
⎦⎥
⎣⎢
⎛
⎜Q
⎜
⎝
d
L{e −t sin 2t}
ds
n
⎤
⎡
n
n d
[ f (s )]⎥
⎢Q L{t f (t )} = (−1)
n
ds
⎥⎦
⎢⎣
n
= ( −1)(5)
5s
s
+ 2
2
( s + 25)
(s + 1) 2
2
5s
s
+
( s 2 + 252 ) ( s 2 + 1) 2
L{te–t sin 2t} = (–1)
d ⎛ 5 ⎞
⎜
⎟
ds ⎝ s 2 + 5 2 ⎠
[Q L[tn f(t)] = (– 1)n
1 ⎡ 10s
2s ⎤
+
⎥
⎢
2 ⎣ ( s 2 + 52 ) 2 ( s 2 + 1) 2 ⎦
Q50. Find the Laplace transform of te–t sin 2t.
Consider,
5
... (3)
On substituting equations (2) and (3), in equation (1),
Applying Laplace transform on both sides, we get,
=
1
s2 +1
we get,
1
[t sin 5t + t sin t]
2
⇒ t sin3t cos2t =
=
⎡ ( s 2 + 1)(0) − 2s ⎤ (−1)(−2 s)
= (−1) ⎢
⎥=
2
2
2
2
⎥⎦ ( s + 1)
⎢⎣ ( s + 1)
=
=
1
s 2 + 12
=
... (2)
Look for the SIA GROU P LOGO
∴L{te −t sin 2t} =
0 − 2( 2 s + 2)
( s 2 + 2 s + 5) 2
′
⎛
⎞
⎜Q ⎛⎜ u ⎞⎟ = vu ′ − uv′ ⎟
2
⎜ ⎝v⎠
⎟
v
⎝
⎠
2( 2 s + 2 )
2
( s + 2 s + 5) 2
4(s + 1)
( s 2 + 2s + 5) 2
4( s + 1)
( s + 2s + 5) 2
2
on the TITLE COVER before you buy
1.39
UNIT-1 (Laplace Transform)
⎡[f(t)]⎤
⎥ =
⎣ t ⎦
Q51. If L[f(t)] = f (s), then prove that L⎢
∞
∫
∞
f(s)ds
∫
Q52. Using Laplace transform evaluate, e – at sin 2 t/t dt .
s
provided Lim f(t) exists.
t→0 t
0
Ans: The given function is,
∞
∫
Ans: Given that,
0
L{f (t)} = f (s)
⎧ f (t ) ⎫
And L ⎨
⎬ =
⎩ t ⎭
∞
∫ f ( s)ds for
Lt
t →0
s
f(t)
exists
t
∞
∫e
L{f (t)} = f (s ) =
Given integral is exactly the Laplace of
− st
f (t ) dt
Integrating on both sides with respect to ‘s’, we get,
∞ ⎡∞
∫
f ( s ) ds =
∫ ⎢⎢∫ e
s
s
∫
− st
⎣0
⎤
f (t )dt ⎥ ds
⎥
⎦
∫∫
e − st f (t ) ds dt
... (1)
Consider, sin2 t
⇒
⎧1 − cos 2t ⎫
⎬
L[sin2t] = L ⎨
2
⎩
⎭
1 − cos 2θ ⎞
⎛
2
⎟
⎜Q sin θ =
2
⎝
⎠
∞∞
=
sin 2 t
with s = a.
t
⎧ sin 2 t ⎫
e − at . sin 2 t
dt
L⎨
=
⎬
t
⎩ t ⎭ s=a 0
s=a
0
∴
sin 2 t
dt
t
∞
From the definition of Laplace transform,
∞
e − at .
=
1
1
L{}
1 − L{cos 2t}
2
2
=
s
1 1 1
. − .
2 s 2 s2 + 4
0 s
[Changing the order of integration]
∞
=
∫
0
⎤
⎡∞
f (t ) ⎢ e − st ds ⎥ dt
⎥
⎢s
⎦
⎣
∫
s ⎞
1
⎛
⎜Q L{1} = , L{cos 2t} = 2
⎟
s
s + a2 ⎠
⎝
[Q f (t) is independent of s]
∞
=
⎡e
− st
⎤
∫ f (t )⎢⎢⎣ − t ⎥⎥⎦
0
∞
=
⎡ 1
∫ f (t ) ⎢⎣− t (e
=
∞
dt
s
−∞
0
∞
=
∫
⎤
⎡ 1
f (t )⎢− (0 − e − st )⎥ dt
⎦
⎣ t
∞
⎡ e − st ⎤
f (t ) ⎢
⎥ dt
⎢⎣ t ⎥⎦
0
=
∫
0
∞
=
∫
0
∴
⎧ f (t ) ⎫
L⎨
⎬=
⎩ t ⎭
⎤
− e − st ) ⎥ dt
⎦
e − st
⎧ f (t ) ⎫
f (t )
dt = L ⎨
⎬
t
⎩ t ⎭
∞
∫ f (s)ds
s
Hence proved.
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎧⎪ sin 2 t ⎫⎪
L⎨
⎬=
⎪⎩ t ⎪⎭
s ⎤
1 ⎡1
−
2 ⎢⎣ s s 2 + 4 ⎥⎦
∞
1⎛1
∫ 2 ⎜⎝ s − s
2
s
s ⎞
⎟ ds
+4⎠
∞
⎛
⎞
⎜Q L ⎡ f (t ) ⎤ = f (s )ds ⎟
⎢
⎥
⎜
⎟
⎣ t ⎦ s
⎝
⎠
∫
∞
⎧⎪ sin 2 t ⎫⎪ 1 ⎡ 1
s
⎤
L⎨
ds − 2
ds
⎬=
⎢
⎪⎩ t ⎪⎭ 2 ⎣ s
s + 4 ⎥⎦
s
∫
∞
⎤
1 ⎡1
2s
ds ⎥
=
⎢ ds −
2
2 ⎣s
2( s + 4) ⎦
s
∫
∞
∞
⎤
1⎡ 1
1
2s
⎢
ds
ds ⎥
−
=
2
2⎢ s
2 s +4 ⎥
s
⎣s
⎦
∫
⎛
⎜Q
⎝
∫
∫s
2
2s
⎞
ds = log(s 2 + a 2 ) ⎟
2
+a
⎠
SIA GROUP
1.40
MATHEMATICS-II [JNTU-ANANTAPUR]
∞
∞
1
⎡1
⎤
2
= ⎢ log( s) − log(s + 4)⎥
4
⎣2
⎦s
[
= log( s)1/ 2 − log( s 2 + 4)1 4
1 ⎧ 2s
2s ⎫
−
=
⎬ ds
⎨
2 ⎩ s2 + a2 s2 + b2 ⎭
s
∫
]
∞
s
∞
∞
⎫⎪
1 ⎧⎪
2s
2s
ds
ds ⎬
−
= ⎨ 2
2
2
2
2⎪ s +a
s +b
⎪⎭
0
⎩s
∫
∞
⎡
⎤
s
= log ⎢ 2
14⎥
⎢⎣ (s + 4) ⎥⎦ s
⎡
⎤
s
= –log ⎢ 2
14⎥
⎣⎢ (s + 4) ⎦⎥
=
⎡ s2 + 4 ⎤
⎡ ( s 2 + 4)1 4 ⎤
1
log
log
⎢ 2 ⎥
=
⎢
2 14 ⎥ = 4
⎢⎣ s ⎥⎦
⎣⎢ (s )
⎦⎥
⎧⎪ sin 2 t ⎫⎪
L⎨
⎬ =
⎪⎩ t ⎪⎭
⇒
∞
∫
e −at .
0
∫
[
1
log(s 2 + a 2 ) − log( s 2 + b 2 )
2
1 ⎡ ⎪⎧ s 2 + a 2 ⎪⎫⎤
= ⎢log ⎨ 2
⎬⎥
2 ⎢⎣ ⎪⎩ s + b 2 ⎪⎭⎥⎦
⎡ s2 + 4 ⎤
1
sin 2 t
dt = log ⎢ 2 ⎥
t
4
⎣⎢ s ⎦⎥
=
]
∞
s
∞
s
⎛ s2 + a2
1 ⎧⎪ ⎛ ∞ ⎞
⎨log ⎜ ⎟ − log ⎜⎜ 2
2
2 ⎪⎩ ⎝ ∞ ⎠
⎝ s +b
⎞ ⎫⎪
⎟⎬
⎟⎪
⎠⎭
For s = a,
∞
∫
∴ e
− at
0
⎛ s 2 + a 2 ⎞ ⎫⎪
1 ⎧⎪
⎟
⎜
−
log(
1
)
log
= ⎨
⎜ s 2 + b 2 ⎟ ⎬⎪
2 ⎪⎩
⎠⎭
⎝
sin 2 t
1
⎪⎧ a 2 + 4 ⎫⎪
dt = log ⎨ 2 ⎬
t
4 ⎪⎩ a ⎪⎭
Q53. Using Laplace transform, evaluate
∞
∫
0
1 ⎛ s2 + a2
= − log⎜ 2
2 ⎜⎝ s + b 2
(cosat − cosbt)
. dt.
t
Ans: The given function is,
∞
∫
0
s
s +a
2
2
–
s +b
2
s ⎞
⎛
⎜Q L{cos at} = 2
⎟
s
a2 ⎠
+
⎝
∞
⎧
∫ ⎨⎩ s
s
s
2
+a
2
−
∫
∞
⇒
s
2
1
⎪⎧ s 2 + b 2 ⎫⎪
⎧ cos at − cos bt ⎫
e − st ⎨
⎬
⎬dt = log ⎨ 2
2
t
⎪⎩ s + a 2 ⎪⎭
⎩
⎭
0
Let, s = 0,
Consider, cos at – cos bt
∴ L{cos at – cos bt} = L{cos at} – L{cos bt}
⎧ cos at − cos bt ⎫
L⎨
⎬ =
t
⎭
⎩
(Q log(1) = 0)
∞
⇒
(cosat-cosbt)
. dt
t
=
⎞
⎟
⎟
⎠
⎧⎪ 0 + b 2 ⎫⎪
1
⎧ cos at − cos bt ⎫
log
e0 ⎨
dt
=
⎨
⎬
⎬
2
t
⎪⎩ 0 + a 2 ⎪⎭
⎭
⎩
0
∫
∞
1 ⎧⎪ ⎛ b ⎞
⎧ cos at − cos bt ⎫
⎬dt = ⎨log⎜ ⎟
⎨
2 ⎪⎩ ⎝ a ⎠
t
⎭
⎩
0
∫
⎫
⎬ ds
s +b ⎭
⎪
⎬
⎪⎭
s
2
2
=
∞
⎛
⎞
⎜Q L ⎡ f (t ) ⎤ = f (s )ds ⎟
⎢ t ⎥
⎜
⎟
⎦ s
⎣
⎝
⎠
∫
∞
s ⎫
2 ⎧ s
− 2
⎬ ds
⎨ 2
2
2 ⎩s + a
s + b2 ⎭
s
∫
⎛b⎞
⎝a⎠
∞
∴
∫
0
Look for the SIA GROU P LOGO
1⎧
⎛ b ⎞⎫
⎨2 log ⎜ ⎟ ⎬
2⎩
⎝ a ⎠⎭
= log ⎜ ⎟
On multiplying and dividing by 2, we get,
=
2⎫
cos at − cos bt
⎛b⎞
dt = log⎜ ⎟
t
⎝a⎠
on the TITLE COVER before you buy
1.41
UNIT-1 (Laplace Transform)
1.11 APPLICATION OF LAPLACE TRANSFORMS TO ORDINARY DIFFERENTIAL EQUATIONS OF
FIRST AND SECOND ORDER
Q54. Using Laplace transform, solve (D2 + 4D + 5)y = 5, given that y(0) = 0, y"(0 ) = 0.
Model Paper-I, Q3
Ans: The given differential equation is,
(D2 + 4D + 5)y = 5
y(0) = 0, y"(0) = 0
⇒
... (1)
D2y + 4Dy + 5y = 5
dy ⎞
⎛
⎜Q D =
⎟
dx ⎠
⎝
d 2 y 4dy
+
+ 5y = 5
dx
dx 2
⇒
y'' + 4y' + 5y = 5
Applying Laplace transform on both sides, we get,
L{y'' + 4y' + 5y} = L{5}
L{y''} + 4L{y'} + 5L{y} = L{5}
s 2 L{ y} − sy(0) − y ' (0) + 4{sL{ y} − y (0)} + 5L{ y} =
(s2 + 4s + 5)L{y} – (s + 4)y(0) – y'(0) =
5
s
[Q L{yn} = sn L{y} – sn–1 y(0) – sn – 2 y'(0)....]
5
s
Since y(0) = 0, y'(0) = 0, the above equation becomes,
(s2 + 4s + 5)L{y} – (s + 4)0 – 0 =
L{y} =
The roots of
s2 +
5
s
5
2
s ( s + 4 s + 5)
4s + 5 are,
s=
− 4 ± 16 − 4 × 5 − 4 ± − 4 −4 ± 2i 2(−2 ± i )
=
=
=
2
2
2
2
= –2 ± i
= –2 + i, – 2 – i
⇒
L{y} =
A
B
C
5
+
= +
s s + (2 − i ) s + (2 + i )
s ( s + ( 2 − i ))( s + ( 2 + i ))
... (1)
A( s + ( 2 − i ))( s + ( 2 + i )) + Bs ( s + (2 + i )) + C ( s )( s + (2 − i ))
5
=
s( s + (2 − i ))(s + (2 + i ))
s ( s + ( 2 − i ))( s + (2 + i ))
5 = A(s + (2 – i))(s + (2 + i)) + Bs(s + (2 + i)) + C(s) (s + (2 – i))
... (2)
On substituting s= 0 in equation (2), we get,
⇒
5 = A(2 – i) (2 + i)
⇒
5 = A[4 – (i2)]
⇒
5 = A[5]
∴ A =1
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.42
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting s = –(2 – i) in equation (2), we get,
5 = B(– (2 – i)) (– 2 – i + 2 + i)
⇒
5 = B(–2 + i) (2i)
⇒
5 = B(– 4i + 2i2)
⇒
5 = B(– 4i – 2)
∴B =
−5
4i + 2
On substituting s= – (2 + i) in equation (2), we get,
5 = C(– 2 – i)(– 2 – i + 2 – i)
⇒
5 = C(– 2 – i)(– 2i)
⇒
5 = C(2 + i)(2i)
⇒
5 = C(4i + 2i2)
⇒
5 = C [4i – 2]
∴C =
5
4i − 2
On substituting the corresponding values of A, B, C in equation (1), we get,
∴ L–1{L(y)} =
⎤
⎤
⎡
1
5
1
5 ⎡
1
−
⎥
⎢
⎥+
⎢
s ( 4i + 2 ) ⎣ ( s + ( 2 − i )) ⎦ 4i − 2 ⎣ ( s + ( 2 + i )) ⎦
On applying inverse Laplace transform on both sides, we get,
⎡1
L–1{L(y)} = L–1 ⎢ −
⎢⎣ s
y = 1−
∴ y = 1−
⎞⎤
⎛
⎞
⎛
5
1
5
1
⎟⎥
⎜⎜
⎟⎟ +
⎜⎜
( 4i + 2) ⎝ ( s + ( 2 − i )) ⎠ ( 4i − 2) ⎝ s + ( 2 + i ) ⎟⎠ ⎥⎦
⎛
⎧ 1 ⎫
− at ⎞
⎜⎜Q L−1 ⎨
⎬ = e ⎟⎟
+
s
a
⎩
⎭
⎝
⎠
5
5
e − ( 2 −i )t +
e − (2 +i )t
4i + 2
4−2
5 − ( 2 − i )t
5 − ( 2 + i )t
+
e
e
4i + 2
4i − 2
Q55. Using Laplace transform, solve (D2 + 2D – 3)y = sin x, y(0) = y'(0) = 0.
Ans: Given that,
(D2 + 2D – 3)y = sin x
⇒
d2y
dx 2
+2
dy
− 3 y = sin x
dx
dy ⎞
⎛
⎜Q D = ⎟
dx ⎠
⎝
y'' + 2y' – 3y = sin x
... (1)
On applying Laplace transform on both sides of equation (1), we get,
L{y'' + 2y' – 3y} = L{sin x}
⇒
L{y''} + 2L{y'} – 3L{y} = L{sin x}
⇒
{s2 L(y) – s(y(0)) – y'(0)} + 2{s L(y) – y(0)} – 3 L{y} = L{sin x} {Q L{yn} = sn L{y} – sn – 1 y{0} – sn – 2 y'(0) ...
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... (2)
1.43
UNIT-1 (Laplace Transform)
Since, y(0) = 0 and y'(0) = 0, equation (2) becomes,
⇒
⇒
[s2 L{y} – s(0) – 0] + 2[s L(y) – 0] – 3 L(y) = L{sin x}
[s2 L(y)] + 2[s L(y)] – 3 L(y) = L{sin x}
⇒
L(y)[s2 + 2s – 3] =
⇒
L(y) =
⇒
L(y) =
⎧⎪
1 ⎫⎪
⎨Q L (sin x ) =
⎬
⎪⎩
1 + s 2 ⎪⎭
1
1 + s2
=
L(y) =
1
2
(1 + s )( s 2 + 2 s − 3)
1
2
2
(1 + s )( s + 3s − s − 3)
1
(1 + s )(s ( s + 3) − 1( s + 3))
2
1
... (3)
2
(1 + s )( s + 3)( s − 1)
On applying partial fractions to equation (3), we get,
1
(1 + s 2 )( s + 3)( s − 1)
=
As + B
( s 2 + 1)
+
C
D
+
( s + 3) ( s − 1)
... (4)
1 = (As + B)(s + 3)(s – 1) + C(s2 + 1)(s – 1) + D(s2 + 1)(s + 3)
= (As + B)(s2 + 3s – s – 3) + C(s3 – s2 + s – 1) + D(s3 + s + 3s2 + 3)
= (As + B)(s2 + 2s – 3) + C(s3 – s2 + s – 1) + D(s3 + 3s2 + s + 3)
= As3 + Bs2 + 2As2 + 2Bs – 3As – 3B + Cs3 – Cs2 + Cs – C + Ds3 + 3Ds2 + Ds + 3D
1 = s3(A + C + D) + s2(B + 2A – C + 3D) + s(2B – 3A + C + D) + (– 3B – C + 3D)
⇒
On comparing the coefficients of s3, s2, s and constant terms on both sides of equation (4), we get,
A + C + D =0
2A + B – C + 3D = 0
– 3A + 2B + C + D = 0
– 3B – C + 3D = 1
Solving equations (5) and (8), we get,
A+C+D=0
– 3B – C + 3D = 1
A − 3B + 4 D = 1
... (5)
... (6)
... (7)
... (8)
... (9)
... (10)
Solving equations (6) and (7), we get,
2A + B – C + 3D = 0
– 3A + 2B + C + D = 0
− A + 3B + 4D = 0
... (11)
On solving equations (9) and (10), we get,
A – 3B + 4D = 1
– A + 3B + 4D = 0
8D = 1
∴D =
1
8
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (12)
SIA GROUP
1.44
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting equation (11) in equation (10), we get,
⎛1⎞
– A + 3B + 4 ⎜ ⎟ = 0
⎝8⎠
−1
⇒
– A + 3B =
2
−1
⇒
– (A – 3B) =
2
⇒
A − 3B =
1
2
2 A − 6B = 1
⇒
From equation (8), we get,
C = – 3B + 3D – 1
On substituting equation (14) in equation (6), we get,
2A + B – (– 3B + 3D – 1) + 3D = 0
⇒ 2A + B + 3B – 3D + 1 + 3D = 0
2A + 4B = –1
Solving equations (13) and (15), we get,
... (13)
... (14)
... (15)
2A – 6B = 1
2A + 4B = –1
–
–
+
–10B = 2
B =
∴ B=
−1
−2
=
5
10
−1
5
... (16)
On substituting the value of B in equation (12), we get,
⎛ −1⎞ 1
A–3⎜ ⎟ =
⎝ 5 ⎠ 2
⇒
A=
=
∴A=
3
1
–
2 5
5−6
10
−1
10
... (17)
On substituting the value of ‘A’ and ‘D’in equation (5), we get,
1
−1
+ C + =0
10
8
C=
=
∴C =
1 1
−
10 8
4−5
40
−1
40
Look for the SIA GROU P LOGO
... (18)
on the TITLE COVER before you buy
1.45
UNIT-1 (Laplace Transform)
On substituting the values of A, B, C and D in the equation (4), we get,
⎛ −1 ⎞ 1
−1
1
⎜ ⎟s −
⎝ 10 ⎠ 5 + 40 + 8
L(y) =
s + 3 s −1
s2 +1
=
− 1 ⎧⎪ s ⎫⎪ ⎛ − 1 ⎞⎧⎪ 1 ⎫⎪ ⎛ − 1 ⎞⎧ 1 ⎫ 1 ⎧ 1 ⎫
⎬
⎬+ ⎨
⎬ + ⎜ ⎟⎨
⎬ + ⎜ ⎟⎨
⎨
10 ⎪⎩ s 2 + 1 ⎪⎭ ⎝ 5 ⎠⎪⎩ s 2 + 1 ⎪⎭ ⎝ 40 ⎠⎩ s + 3 ⎭ 8 ⎩ s − 1 ⎭
... (19)
By applying inverse Laplace transform on both sides of equation (19), we get,
L–1{L(y)} =
y=
⎧⎛ − 1 ⎞⎛ s ⎞ ⎛ 1 ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎛ 1 ⎞⎫
L−1 ⎨⎜ ⎟⎜ 2 ⎟ − ⎜ ⎟⎜ 2 ⎟ − ⎜
⎟+ ⎜
⎟⎬
⎩⎝ 10 ⎠⎝ s + 1 ⎠ ⎝ 5 ⎠⎝ s + 1 ⎠ 40 ⎝ s + 3 ⎠ 8 ⎝ s − 1 ⎠⎭
− 1 −1 ⎧⎪ s ⎫⎪ 1 −1 ⎧⎪ 1 ⎫⎪ 1 −1 ⎧ 1 ⎫ 1 −1 ⎧ 1 ⎫
L ⎨
L ⎨ 2
⎬−
⎬− L ⎨ 2
⎬+ L ⎨
⎬
10
⎪⎩ s + 1 ⎪⎭ 40
⎪⎩ s + 1 ⎪⎭ 5
⎩ s + 3⎭ 8
⎩ s − 1⎭
Q
y=
−1
1
1 −3t 1 t
e + e
{cos t} − {sin t} −
10
5
40
8
... (20)
L(cos t) =
L(sin t) =
L(e–3t) =
L(et) =
Q56. Solve the differential equation
s
s2 + 1
1
2
s +1
1
and
s+3
1
s −1
d2 x
+ 9x = sin t using Laplace transforms given that x(0) = 1, x'(0) = 0.
dx 2
Ans: The given differential equation is,
d 2x
+ 9 x = sin t
dt 2
⇒
⎛ dx
⎞
= x′ ⎟
⎜Q
dt
⎝
⎠
x'' + 9x = sin t
Taking Laplace transform on both sides, we get,
L[x'' + 9x] = L[sin t]
L[x''] + 9L[x] =
a ⎞
⎛
⎜Q L(sin at ) = 2
⎟
s +1⎠
⎝
1
2
s +1
s2L[x] – s.x(0) – x'(0) + 9L[x] =
1
2
s +1
Since, x(0) = 1, x'(0) = 0, equation (1) becomes,
s2 L[x] – s(1) – 0 + 9 L[x] =
⇒
s2L[x] – s + 9 L[x] =
⇒
s2L[x] + 9 L[x] =
⇒
L[x] (s2 + 9) =
L[x] =
(Q L[ y n ] = s n L[ y ] − s n−1 y (0) − s n −2 y ′(0)....)
... (1)
1
2
s +1
1
2
s +1
1
+s
s2 +1
1
2
s +1
+s
s
1
+ 2
2
( s + 1)( s + 9) (s + 9)
... (2)
2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.46
MATHEMATICS-II [JNTU-ANANTAPUR]
Consider,
1
( s 2 + 1)( s 2 + 9 )
By using partial fractions, we get,
1
As + B Cs + D
+ 2
= 2
( s + 1)( s 2 + 9)
s +1
s +9
2
⇒
1 = (As + B) (s2 + 9) + (Cs + D) (s2 + 1)
= As3 + 9As + Bs2 + 9B + Cs3 + Cs + Ds2 + D
1 = (A + C) s3 + (B + D)s2 + (9A + C) s + (9B + D)
Comparing s3, s2, s and constant terms on both sides, we get,
0=A +C
0=B+D
0 = 9A + C
1 = 9B + D
From equation (3), we get,
A=–C
On substituting equation (7) in equation (5), we get,
0 = 9(– C) + C
⇒
0 = – 9C + C
⇒
0 = – 8C
... (3)
... (4)
... (5)
... (6)
... (7)
∴C = 0
∴A= 0
From equation (4), we get,
B =–D
On substituting equation (8) in equation (6), we get,
1= 9(– D) + D
⇒
1= – 9D + D
⇒
1= – 8 D
∴D = −
∴B =
... (8)
1
8
1
8
On substituting the corresponding values in equation (2), we get,
L[x] =
s
1
+ 2
2
( s + 1)( s + 9) s + 9
2
1
1⎤
⎡
⎢ 0s + 8 0s − 8 ⎥
s
+ 2
⎥+ 2
=⎢ 2
⎢ s +1 s + 9 ⎥ s + 9
⎣
⎦
=
s
1 1
1 1
− .
+
.
8 s2 + 1 8 s2 + 9 s2 + 9
=
1 ⎤
s
1⎡ 1
−
.
+
8 ⎢⎣ s 2 + 1 s 2 + 32 ⎥⎦ s 2 + 32
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1.47
UNIT-1 (Laplace Transform)
Applying inverse Laplace transform, we get,
s ⎫
1 ⎤
−1 ⎧ 1 ⎡ 1
+ 2
− 2
L–1 {L[x]} = L ⎨ ⎢ 2
2⎥
2⎬
⎩8 ⎣ s +1 s + 3 ⎦ s + 3 ⎭
x=
=
1 ⎡ −1 ⎧ 1 ⎫ −1 ⎧ 1 ⎫ −1 ⎧ s ⎫⎤
L ⎨
+L ⎨ 2
⎬−L ⎨ 2
2⎬
2 ⎬⎥
8 ⎢⎣ ⎩ s 2 + 1 ⎭
⎩s + 3 ⎭
⎩ s + 3 ⎭⎦
1
1⎛
⎞
⎜ sin t − sin 3t ⎟ + cos 3t
3
8⎝
⎠
⎞
⎛
1 ⎫ 1
−1 ⎧
⎜Q L ⎨ 2 2 ⎬ = sin at⎟
⎩s + a ⎭ a
⎟
⎜
⎟
⎜
s
⎧
⎫
⎜ L−1 ⎨ 2
= cos at ⎟
2⎬
⎠
⎝
⎩s + a ⎭
1
1⎛
⎞
∴ x = ⎜ sin t − sin 3t ⎟ + cos 3t
3
8⎝
⎠
Q57. Solve the following differential equation using the Laplace transforms,
d2 y 2dy
+
+ 2y = 5 sint and y(0) = y'(0) = 0.
dt
dt 2
Ans: Given that,
d 2 y 2dy
+
+ 2 y = 5 sin t and y(0) = y'(0) = 0
dt
dt 2
⇒
⎛ dy
⎞
= y′ ⎟
⎜Q
dt
⎝
⎠
y'' + 2y' + 2y = 5 sin t
Applying Laplace transform on both sides, we get,
L{y'' + 2y' + 2y} = L{5 sin t}
L{y''} + 2L{y'} + 2L{y} = 5L{sin t}
[s2 y(s) – sy(0) – y'(0)] + 2[sy(s) – y(0)] + 2y(s) = 5L{sin t}
[ Q L{yn} = snL{y} – sn–1.y{0} – sn–2.y1{0} .....]
... (1)
Since, y(0) = 0 and y'(0) = 0, equation (1) becomes,
[s2y(s) – 0 – 0] + 2[sy(s) – 0] + 2y(s) = 5L{sin t}
⇒
s2 y(s) + 2sy(s) + 2y(s) = 5 L{sin t}
⇒
y(s) (s2 + 2s + 2) = 5 ×
⇒
y(s) =
a ⎤
⎡
⎢Q L{sin at} = s 2 + a 2 ⎥
⎦
⎣
1
2
s +1
5
( s + 1)(s 2 + 2s + 2)
2
By using partial fractions, we get,
y(s) =
5
( s + 1)(s + 2s + 2)
2
2
=
As + B
s +1
2
+
Cs + D
... (2)
s + 2s + 2
2
⇒
5 = (As + B) (s2 + 2s + 2) + (Cs + D) (s2 + 1)
⇒
5 = As3 + 2As2 + 2As + Bs2 + 2Bs + 2B + Cs3 + Cs + Ds2 + D
⇒
5 = s3(A + C) + S2(2A + B + D) + s(2A + 2B + C) + (2B + D)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (3)
SIA GROUP
1.48
MATHEMATICS-II [JNTU-ANANTAPUR]
3
2
Thus, comparing the coefficients of s , s , s and constant from equation (3), we get,
A+C=0
... (4)
2A + B + D = 0
... (5)
2A + 2B + C = 0
... (6)
2B + D = 5
... (7)
From equation (4),
A=–C
... (8)
On substituting equation (8) in equations (5) and (6), we get,
2(– C) + B + D = 0
⇒
– 2C + B + D = 0
... (9)
And,
2(– C) + 2B + C = 0
⇒
– 2C + 2B + C = 0
⇒
– C + 2B = 0
⇒
C = 2B
... (10)
On substituting equation (10) in equation (9), we get,
– 2(2B) + B + D = 0
⇒
– 4B + B + D = 0
⇒
– 3B + D = 0
... (11)
On solving equations (7) and (11), we get,
2B+D=5
–3B+D=0
+ –
–
5B=5
∴B = 1
On substituting equation (12) in equation (7), we get,
2(1) + D = 5
⇒
D=5–2
∴D = 3
Thus, from equation (10), we get,
C = 2B = 2(1) = 2
∴C = 2
From equation (8), we get,
A =–C=–2
∴ A = −2
On substituting corresponding values in equation (2), we get,
y(s) =
− 2s + 1
s2 +1
+
2s + 3
s 2 + 2s + 2
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1.49
UNIT-1 (Laplace Transform)
On applying inverse Laplace on both sides, we get,
⇒
2s + 3 ⎤
−1 ⎡ − 2 s + 1
+ 2
L–1[y(s)] = L ⎢ 2
⎥
⎣ s + 1 s + 2s + 2 ⎦
⇒
2s + 3 ⎤
−1 ⎡ − 2 s + 1 ⎤
−1 ⎡
y(t) = L ⎢ 2
⎥
⎥+L ⎢ 2
⎣ s + 2s + 2 ⎦
⎣ s +1 ⎦
⇒
2s
3
⎤
⎤
−1 ⎡ − 2 s ⎤
−1 ⎡ 1 ⎤
−1 ⎡
−1 ⎡
y(t) = L ⎢ 2
⎥
⎥+ L ⎢ 2
⎥+ L ⎢ 2
⎥+L ⎢ 2
⎣ s + 2s + 2 ⎦
⎣ s + 2s + 2 ⎦
⎣ s + 1⎦
⎣ s + 1⎦
⎤
⎤
1
1 ⎤
s
s ⎤
−1 ⎡
−1 ⎡
−1 ⎡
−1 ⎡
= − 2 L ⎢ 2 2 ⎥ + L ⎢ 2 2 ⎥ + 2L ⎢
⎥ + 3L ⎢
2
2
2⎥
⎣ s +1 ⎦
⎣s +1 ⎦
⎣⎢ ( s + 1) + 1 ⎦⎥
⎣⎢ ( s + 1) + 1 ⎦⎥
⎡
⎡ s + 1−1 ⎤
⎤
1
s ⎤ −1 ⎡ 1 ⎤
−1 ⎡
−1
−1
+
3
L
= − 2L ⎢ 2 2 ⎥ + L ⎢ 2 2 ⎥ + 2L ⎢
⎢
⎥
2
2
2
2⎥
⎣ s +1 ⎦
⎣ s +1 ⎦
⎣⎢ (s + 1) + 1 ⎦⎥
⎣⎢ (s + 1) + 1 ⎦⎥
⎡ −1 ⎡
⎡
⎤⎤
⎤
⎤
1 ⎤
s +1
1
1
⎤
−1 ⎡
−1 ⎡
− L−1 ⎢
⎥ + 3L ⎢
⎥ + L ⎢ 2 2 ⎥ + 2⎢ L ⎢
2
2⎥
2
2⎥
2
2⎥
⎣ s +1 ⎦
⎣ s +1 ⎦
⎢⎣ ( s + 1) + 1 ⎥⎦ ⎦⎥
⎢⎣ ( s + 1) + 1 ⎥⎦
⎣⎢ ⎢⎣ ( s + 1) + 1 ⎥⎦
⎡
−1
= − 2L ⎢
⇒
s
2
2
y(t) = –2 cos t + sin t + 2 [e–t cos t – e–t sin t] + 3 e–t sin t
⎡ −1 ⎡ s ⎤
⎤ 1 − at
⎡
1
⎡ a ⎤
= e . sin bt ,
= cos at , L−1 ⎢ 2
= sin at , L−1 ⎢
⎢Q L ⎢ 2
2⎥
2⎥
2
2⎥
⎣s + a ⎦
⎣s + a ⎦
⎢
⎣⎢ ( s + c ) + b ⎦⎥ b
⎢
⎤
⎡
s+a
⎢
= e − at cos bt
L−1 ⎢
2
2⎥
⎢
+
+
(
)
s
a
b
⎥
⎢
⎦
⎣
⎣
⇒
⇒
⎤
⎥
⎥
⎥
⎥
⎥
⎦
y(t) = – 2cos t + sin t + 2e–t cos t – 2e–t sin t + 3e–t sin t
y(t) = – 2 cos t + sin t + 2e–t cos t + e–t sin t
∴ y (t ) = 2 cos t (e −t − 1) + sin t ( e −t + 1)
Q58. Using Laplace transform solve (D3 – D2 + 4D – 4)y = 68 exsin2x, y = 1, Dy = –19, D2y = –37 at x = 0.
Ans: Given differential equation is,
(D3 – D2 + 4D – 4)y = 68 ex sin 2x
d3y
dx
⇒
⇒
⇒
3
−
d2y
dx
2
+4
dy
dy ⎞
⎛
− 4 y = 68e x sin 2 x ⎜ Q D = ⎟
⎝
dx
dx ⎠
y''' – y'' + 4y' – 4y = 68 ex sin 2x
Applying Laplace transform on both sides, we get,
L{y''' – y'' + 4y' – 4y} = L{68 ex sin 2x}
L{y'''} – L{y''} + 4{y'} – 4{y}= 68 L {ex sin 2x}
[s3 L{y} – s2y(0) – sy'(0) – y''(0)] – [s2 L{y} – sy(0) – y'(0)] + 4[sL{y} – y(0)] – 4L{y} = 68 L{ex sin 2x}
(Q L{ y n } = s n L{ y} − s n −1 y (0) − s n−2 y ′(0)...)
⇒
s3 L{y} – s2 y(0) – sy'(0) – y''(0) – s2 L{y} + sy(0) + y'(0) + 4sL{y} – 4y(0) – 4L{y} =
136
( s − 1) 2 + 22
{
}
⎤
⎡
b
at
⎥
⎢Q L e sin bt =
( s − a) 2 + b 2 ⎦⎥
⎣⎢
⇒
L{y}(s3 – s2 + 4s – 4) –y''(0) – y'(0)(s – 1) + y(0)(– s2 + s – 4) =
136
( s − 1) 2 + 4
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.50
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting the given conditions, we get,
L{y}(s3 – s2 + 4s – 4) – (– 37) – (– 19) (s –1) + (–s2 + s – 4) =
136
( s − 1) 2 + 4
136
⇒
y(s) (s3 – s2 + 4s – 4) + 37 + 19 s – 19 – s2 + s – 4 =
⇒
y(s) (s3 – s2 + 4s – 4) – s2 + 20s + 14 =
⇒
y(s) (s3 – s2 + 4s – 4) =
136
+ s2 – 20 s – 14
s 2 − 2s + 5
⇒
y(s) (s3 – s2 + 4s – 4) =
136 + ( s 2 − 20s − 14)( s 2 − 2s + 5)
s 2 − 2s + 5
⇒
y(s) (s3 – s2 + 4s – 4) =
136 + s 4 − 2s 3 + 5s 2 − 20s 3 + 40s 2 − 100s − 14s 2 + 28s − 70
s 2 − 2s + 5
⇒
y(s) =
2
s − 2s + 1 + 4
136
2
s − 2s + 5
s 4 − 22 s 3 + 31s 2 − 72s + 66
( s 2 − 2s + 5)(s 3 − s 2 + 4 s − 4)
Consider,
s3 – s2 + 4s – 4
1
∴
∴
1
−1
4
−4
0
1
0
4
1
0
4
0
s3 – s2 + 4s – 4 = (s – 1) (s2 + 4)
y(s) =
s 4 − 22 s 3 + 31s 2 − 72 s + 66
( s − 1)( s 2 + 4)( s 2 − 2 s + 5)
Taking partial fractions of above equation, we get,
y(s) =
s 4 − 22 s 3 + 31s 2 − 72 s + 66
A
Bs + C
Ds + E
+ 2
+ 2
=
2
2
s − 1 s + 4 s − 2s + 5
( s − 1)( s + 4)( s − 2 s + 5)
⇒
s4 – 22s3 + 31s2 – 72s + 66 = A (s2 + 4)(s2 – 2s + 5) + (Bs + C)(s – 1)(s2 – 2s + 5) + (Ds + E) (s – 1)(s2 + 4)
⇒
s4 – 22s3 + 31s2 – 72s + 66 = A (s4 – 2s3 + 9s2 – 8s + 20) + (Bs + C)(s3 – 3s2 + 7s – 5) + (Ds + E) (s3 – s2 + 4s – 4)
On substituting, s = 1 in equation (2), we get,
1 – 22 + 31 – 72 + 66 = A(1 – 2 + 9 – 8 + 20)
⇒
4 = A(20)
∴A =
1
5
On equating the coefficients of s4, s 3, s2 and s, we get,
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... (1)
... (2)
1.51
UNIT-1 (Laplace Transform)
4
On substituting equation (3) in equation (5), we get,
s -terms
⇒
1= A + B + D
B+D=1–A
⇒
B+D=1–
⇒
⇒
B+D=
B=
⎛4
⎞
146
⇒ 7⎜ − D ⎟ –3C + 4D – E =
5
⎝5
⎠
1
5
⇒
4
5
4
–D
5
⇒
–3C – 3D – E =
146 28
–
5
5
⇒
– 3C – 3D – E =
118
5
⇒
3C + 3D + E =
... (3)
3
s -terms
⇒
28
146
–7D – 3C + 4D – E =
5
5
– 22 = –2A – 3B + C – D + E
⎛1⎞
– 22 = –2 ⎜ ⎟ (– 3B + C) + (– D + E)
⎝5⎠
– 22 =
⇒
– 3B + C – D + E = –
... (8)
On substituting equation (3) in equation (6), we get,
−2
– 3B + C – D + E
5
⇒
−118
5
⎛4
⎞
−352
–5 ⎜ − D ⎟ + 7C – 4D + 4E =
5
⎝5
⎠
108
5
... (4)
⇒
– 4 + 5D + 7C – 4D + 4E =
−352
5
s2-terms
31 = 9A + (7B – 3C) + (4D – E)
⇒
⇒
⎛1⎞
31 = 9 ⎜ ⎟ + 7B – 3C + 4D – E
⎝5⎠
9
31 – = 7B – 3C + 4D – E
5
146
7B – 3C + 4D – E =
5
⇒
7C + D + 4E = –
⇒
7C + D + 4E =
... (5)
C + 2D + E =
– 72 = – 8A + (– 5B + 7C) + (– 4D + 4E)
– 72 =
– 2C – D =
12C + 12 D + 4 E = –
−12
−108
+ 3D + C − D + E =
5
5
C + 2D + E =
−108 12
+
5
5
⇒
C + 2D + E =
−96
5
... (10)
Multiplying equation (8) with ‘4’ and subtracting with
equation (9), we get,
⎛4
⎞
−108
− 3⎜ − D ⎟ + C − D + E =
5
⎝5
⎠
⇒
22
5
... (6)
On substituting equation (3) in equation (4), we get,
⇒
−118
5
(–) (–) (–)
1⎞
⎛
⎜Q A = ⎟
5⎠
⎝
⇒
... (9)
−96
5
3C + 3D + E =
−8
– 5B + 7C – 4D + 4E
5
−352
– 5B + 7C – 4D + 4E =
5
−332
5
Solving equation (7) and equation (8), we get,
s-terms
⇒
352
+4
5
7C + D + 4 E =
472
5
−332
5
(–) (–) (–)
... (7)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
5C + 11 D = –
140
5
... (11)
SIA GROUP
1.52
MATHEMATICS-II [JNTU-ANANTAPUR]
Multiplying equation (10) with equation (11) and adding with equation (11), we get,
– 22C – 11D =
242
5
5C + 11 D = –
– 17 C =
∴C =
140
5
102
5
−6
5
... (12)
On substituting equation (12) in equation (10), we get,
22
⎛−6⎞
– 2⎜
⎟− D =
5
⎝ 5 ⎠
⇒
D=
12 22
−
5 5
∴D = −2
... (13)
On substituting equations (12) and (13) in equation (7), we get,
−6
−96
+ 2(−2) + E =
5
5
E=
⇒
−96 6
+ +4
5
5
∴ E = −14
On substituting ‘D’ value in equation (3), we get,
B=
4
– (–2)
5
B=
14
5
On substituting A, B, C, D and E values in equation (1), we get,
14
⎛ 6⎞
s +⎜− ⎟
1
5
⎝ 5 ⎠ + (−2) s + (−14)
+
y(s) =
2
5(s − 1)
s 2 − 2s + 5
5(s + 4)
=
1
14 s − 6
( 2 s + 14)
+
−
5( s − 1) 5( s 2 + 4) s 2 − 2 s + 5
Look for the SIA GROU P LOGO
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1.53
UNIT-1 (Laplace Transform)
Taking Laplace inverse on both sides, we get,
⎧ 1
14s − 6
2s + 14 ⎫
− 2
_
L–1 {y(s)} = L−1 ⎨
⎬
2
⎩ 5( s − 1) 5( s + 4) s − 2s + 5 ⎭
y(t) =
=
1
s
⎫
1 −1 ⎧ 1 ⎫ 14 −1 ⎧ s ⎫ 6 −1 ⎧ 1 ⎫
⎫
−1 ⎧
−1 ⎧
L ⎨
⎬
⎬ − 14 L ⎨ 2
⎬ − 2L ⎨ 2
⎬− L ⎨ 2
⎬+ L ⎨ 2
5
⎩ s − 2s + 5 ⎭
⎩ s − 2s + 5 ⎭
⎩s + 4⎭ 5 ⎩s + 4⎭
⎩ s − 1⎭ 5
⎧ s −1+ 1 ⎫
⎫
1 x 14
6 1
1
−1 ⎧
e + cos 2 x − × sin 2 x − 2 L−1 ⎨
⎬ − 14 L ⎨
⎬
2
2
5
5
5 2
⎩ ( s − 1) + 4 ⎭
⎩ ( s − 1) + 4 ⎭
⎛
⎞
⎧ 1 ⎫ at −1 ⎧ s ⎫
⎧ a ⎫
⎜⎜Q L−1 ⎨
= cos at , L−1 ⎨ 2
= sin at ⎟⎟
⎬ = e ,L ⎨ 2
2⎬
2⎬
−
s
a
⎩
⎭
⎩s + a ⎭
⎩s + a ⎭
⎝
⎠
=
⎧
⎫
⎧
⎫
⎧
⎫
1 x 14
s −1
3
1
1
e + cos 2 x − sin 2 x − 2 L−1 ⎨
− 2 L−1 ⎨
− 14L−1 ⎨
2
2⎬
2
2⎬
2
2⎬
5
5
5
⎩ ( s − 1) + 2 ⎭
⎩ ( s − 1) + 2 ⎭
⎩ ( s − 1) + 2 ⎭
=
1 x 14
3
1
1
e + cos 2 x − sin 2 x − 2e x cos 2 x − 2 × e x sin 2 x − 14 × e x sin 2 x
5
5
5
2
2
⎞
⎛
⎧
⎫ at
⎧
⎫ 1 at
s−a
1
⎜Q L−1 ⎨
= e cos bt , L−1 ⎨
= e sin bt ⎟
2
2⎬
2
2⎬
⎟
⎜
⎩ ( s − a) + b ⎭
⎩ ( s − a) + b ⎭ b
⎠
⎝
=
3
e x 14
+ cos 2 x − sin 2 x − 2e x cos 2 x − e x sin 2 x − 7e x sin 2 x
5 5
5
=
3
e x 14
+ cos 2 x − sin 2 x − 2e x cos 2 x − 8e x sin 2 x
5
5
5
ex
⎛ 14
⎞
⎛3
⎞
+ cos 2 x⎜ − 2e x ⎟ − sin 2 x⎜ + 8e x ⎟
=
5
⎝ 5
⎠
⎝5
⎠
∴ y (t ) =
ex
⎛ 14
⎛3
⎞
⎞
+ cos 2 x⎜ − 2e x ⎟ − sin 2 x ⎜ + 8e x ⎟
⎝5
⎝ 5
⎠
⎠
5
Q59. Solve, y'' – 3y' + 2y = 4t + e3t when, y(0) = 1, y'(0) = – 1.
Ans: Given that,
y''– 3y' + 2y = 4t + e3t
On applying Laplace transform the above equation becomes,
L[y'' – 3y' + 2y] = L[4t + e3t]
⇒
⇒
L[y"] – 3L[y'] + 2L[y] = 4L[t] + L[e3t]
[s2L[y] – sy(0) – y'(0)] – 3[sL[y] – y(0)] + 2L[y] =
4
s
2
+
1
s −3
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎡Q L{ y n } = s n L{ y} − s n −1 y (0) − s n − 2 y ' (0).....⎤
⎥
⎢
⎥
⎢Q L{e at } = 1
⎥⎦
⎢⎣
s−a
SIA GROUP
1.54
MATHEMATICS-II [JNTU-ANANTAPUR]
Since, given y(0) = 1 and y'(0) = –1
[s2L[y] – s(1) – (–1) – 3[sL[y] – 1] + 2L[y] =
s2L[y] – s + 1 – 3sL[y] + 3 + 2L[y] =
⇒ (s2 – 3s + 2)L[y] – s + 1 + 3 =
⇒
(s2 – 3s + 2)L[y] =
⇒
(s – 1)(s – 2)L[y] =
⇒
4
s2
+
4
s
2
+
1
s−3
1
s−3
1
4
+
2
s
−3
s
1
4
+s–4
2 +
s −3
s
4( s − 3) + 1( s 2 ) + ( s − 4)( s 2 )( s − 3)
s 2 ( s − 3)
L[y] =
s 4 − 7 s 3 + 13s 2 + 4s − 12
s 2 ( s − 1)(s − 2)( s − 3)
By using partial fractions, we get,
L[y] =
s 4 − 7s 3 + 13s 2 + 4s − 12
2
s (s − 1)(s − 2)(s − 3)
=
A B
C
D
E
+ 2+
+
+
s s
s −1 s − 2 s − 3
⇒ s4 – 7s3 + 13s2 + 4s – 12 = A s(s – 1) (s – 2) (s – 3) + B(s – 1) (s – 2) (s – 3) + C s2 (s – 2) (s – 3) +
D s2(s – 1) (s – 3) + E s2(s – 1) (s – 2)
On substituting s = 0 in equation (2), we get,
0 – 0 + 0 + 0 – 12 = 0 + B (0 – 1) (0 – 2) (0 – 3) + 0 + 0 + 0
– 12 = B(– 6)
∴B = 2
On substituting s = 1 in equation (2), we get,
14 – 7(1)3 + 13(1)2 + 4(1) – 12= 0 + 0 + C(1)2 (1 – 2) (1 – 3) + 0 + 0
1 – 7 + 13 + 4 – 12 = C(– 1) (– 2)
– 1 = 2C
∴C =
−1
2
On substituting s = 2 in equation (2), we get,
24 – 7(2)3 + 13(2)2 + 4(2) – 12 = 0 + 0 + 0 D(2)2 (2 – 1)(2 – 3) + 0
8 = D(4)(1)(–1)
D=
−8
4
D = –2
On substituting s = 3 in equation (2), we get,
34 – 7(3)3 + 13(3)2 + 4(3) – 12 = 0 + 0 + 0+ 0 + E(3)2 (3 – 1) (3 – 2)
9 = E(9) (2) (1)
∴E =
1
2
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... (1)
... (2)
1.55
UNIT-1 (Laplace Transform)
Comparing ‘s4’ terms on both sides of equation (2), we get,
1=A+C+D+E
1
⎛ −1 ⎞
1 = A + ⎜ ⎟ + (−2) +
2
⎝ 2 ⎠
1=A–2
∴A= 3
⇒
On substituting corresponding values in equation (1), we get,
−1
1
3 2
−2
2
+
+ 2
L[y] = + 2 +
s s
s −1 s − 2 s − 3
Applying inverse Laplace transform on both sides we get,
2
1 1 ⎫
⎧3 2 1 1
−
+
L−1{L[ y ]} = L–1 ⎨ + 2 −
⎬
−
−
s
2
s
1
s
2
2 s − 3⎭
s
⎩
1 –1 ⎡ 1 ⎤
⎡1 ⎤
⎡ 1 ⎤ 1 –1 ⎡ 1 ⎤
⎡ 1 ⎤
+
y = 3L−1 ⎢ ⎥ + 2L−1 ⎢ 2 ⎥ –
L ⎢
L ⎢
– 2 L–1 ⎢
⎥
⎥
⎥
2
⎣s⎦
⎣s ⎦ 2
⎣s − 2⎦
⎣ s − 3⎦
⎣ s −1 ⎦
= 3(1) + 2(t) –
1 t
1
e − 2e 2t + e 3t
2
2
⎛
⎡ 1 ⎤
at ⎞
⎜⎜Q L−1 ⎢
⎥ = e ⎟⎟
−
s
a
⎦
⎣
⎝
⎠
1
1
∴ y = 3 + 2t − e t − 2e 2t + e 3t
2
2
Q60. Solve the following differential equation using the Laplace transforms
dx
d2 x
2t
2 – 2 dt + x = e with
dt
dx
= –1 at t = 0.
dt
Ans: Given differential equation is,
x(0) = 2,
d 2x
dx
− 2 + x = e2t
2
dt
dt
⎛ dx
⎞
= x′ ⎟
⎜Q
⎝ dt
⎠
⇒ x'' – 2x' + x = e2t
Applying Laplace transform on both sides, we get,
L[x'' – 2x' + x] = L[e2t]
⇒
L[x''] – 2L[x'] + L[x] =
1
s−2
1 ⎞
⎛
at
⎜Q L[ e ] =
⎟
s
a⎠
−
⎝
1
s−2
[Q L[yn] = snL{y} – sn – 1. y(0) – sn–2. y'(0)...]
⇒ [s2L[x] – s x(0) – x'(0)] – 2(s L[x] – x(0)) + L[x] =
Since, x(0) = 2 and x' =
dx
= – 1 equation (1) becomes,
dt
[s2 L[x] – s(2) – (–1)] – 2(sL[x] – 2) + L[x] =
⇒
... (1)
s2L[x] – 2s + 1 – 2sL[x] + 4 + L[x] =
1
s−2
1
s−2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
1.56
MATHEMATICS-II [JNTU-ANANTAPUR]
⇒
(s2 – 2s + 1)L[x] – 2s + 1 + 4 =
⇒
(s2 – 2s + 1)L[x] =
=
⇒
(s – 1)2 L[x] =
L[x] =
1
s−2
1 + (2s − 5)(s − 2)
1
+ 2s – 5 =
( s − 2)
( s − 2)
1 + (2s 2 − 4s − 5s + 10)
( s − 2)
1 + 2s 2 − 9s + 10
( s − 2)
2s 2 − 9s + 11
( s − 1) 2 (s − 2)
On applying partial fractions, we get,
L[x] =
A
B
C
2s 2 − 9s + 11
+
+
=
2
2
( s − 2)
( s − 1) (s − 2) ( s − 1) ( s − 1)
... (2)
⇒ A(s – 1) (s – 2) + B(s – 2) + C(s – 1)2 = 2s2 – 9s + 11
On substituting s = 1 in equation (3), we get,
A(1 – 1) (1 – 2) + B(1 – 2) + C(1 – 1)2 = 2(1)2 – 9(1) + 11
⇒
– B = 2 – 9 + 11
⇒
– B =4
... (3)
∴ B = −4
On substituting s = 2 in equation (3), we get,
A(2 – 1) (2 – 2) + B(2 – 2) + C(2 – 1)2 = 2(2)2 – 9 (2) + 11
C = 2(4) – 18 + 11
= 8 – 18 + 11
=1
∴ C =1
On comparing coefficient of s2 in equation (3), we get,
A + C =2
A + 1 =2
A =2–1=1
∴ A =1
On substituting A, B, C in equation (2), we get,
L(x) =
1
4
1
−
+
2
( s − 1) ( s − 1)
( s − 2)
On applying inverse Laplace transform on both sides, we get,
⇒
⎡ 1
4
1 ⎤
+
⎥
L–1 L{x} = L–1 ⎢ ( s − 1) −
2
(s − 2) ⎥⎦
( s − 1)
⎢⎣
⎛ 1 ⎞
⎟ – 4 L–1
⎝ s −1 ⎠
x = L–1 ⎜
⎛ 1 ⎞ −1 ⎛ 1 ⎞
⎟
⎜
⎜ ( s − 1) 2 ⎟ + L ⎜⎝ s − 2 ⎟⎠
⎠
⎝
x = et – 4 t.et + e2t
⎛ 1 ⎞
⎟ = eat; L–1
[Q L–1 ⎜
⎝ s−a⎠
⎛ 1 ⎞
⎟
⎜
⎜ (s − a) 2 ⎟ = t eat]
⎠
⎝
∴ x = e t − 4t e t + e 2 t
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2.1
UNIT-2 (Fourier Series)
UNIT
FOURIER SERIES
2
PART-A
SHORT QUESTIONS WITH SOLUTIONS
Q1.
Define a periodic function.
Ans: A function f (x) defined for all real numbers is said to be periodic if there is a real number T < 0 such that,
f (x + T) = f (x) for all x
... (1)
And the number T is called its period.
∴ From equation (1),
f(x ± kT) = f (x) for all x, and
k = 1, 2, 3, ... i.e., 2T, 3T, 4T etc., are also periods of f (x).
Q2.
Give the Euler’s formulae for Fourier series.
Model Paper-I, Q1(c)
Ans: The Fourier series expansion of function f(x) in the interval k ≤ x ≤ k + 2π is given by,
f(x) =
a0 ∞
(an cos nx + bn sin nx)
+
2 n =1
∑
Where,
1
a0 =
π
1
an=
π
bn=
1
π
k +2 π
∫ f ( x)dx
k
k +2 π
∫ f ( x) cos nx dx
k
k + 2π
∫ f ( x) sin nx dx
k
The above Fourier coefficients a0, an and bn are referred to as Euler’s formulae.
Q3.
Give the Dirichlet conditions for Fourier expansion.
Model Paper-II, Q1(c)
Ans:
Dirichlet’s Conditions
1.
f (x) is uniformly bounded on the fundamental interval (–π, π) of period 2π, that is, | f (x)| < M for all –π < x < π where M is
a constant.
2.
f (x) has no more than a finite number of finite discontinuities in (–π, π).
3.
f (x) has a finite number of strict maxima or minima in (–π, π).
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Q4.
MATHEMATICS-II [JNTU-ANANTAPUR]
π as a Fourier series.
Expand f(x) = x , 0 < x < 2π
2
Ans: Given that,
f(x) = x2
The Fourier series is given as,
∞
a0
(an cos nx + bn sin nx)
+
f(x) =
2 n =1
∑
a0 =
1
π
1
a0 =
π
⇒
an =
1
π
1
an =
π
⇒
=
... (1)
2π
∫ f ( x)dx
0
2π
∫
x 2 dx =
0
1 ⎡ x3 ⎤
⎢ ⎥
π ⎣⎢ 3 ⎦⎥
2π
0
1
1
8π 2
3
3
3
[(
2
)
0
]
π
−
=
π
(
8
)
=
=
3π
3π
3
2π
∫ f ( x) cos nxdx
0
2π
2π
1 ⎡ 2 sin nx
⎛ − sin nx ⎞⎤
⎛ − cos nx ⎞
− 2 x⎜
⎟⎥
⎟ + 2⎜
x cos nx.dx = ⎢ x
2
π⎣
n
⎝ n 3 ⎠⎦ 0
⎠
⎝ n
0
∫
2
⎤
1 ⎡⎛
4π
⎞
⎢⎜ 0 + 2 cos 2nπ − 0 ⎟ − (0 + 0 − 0) ⎥
π ⎣⎝
n
⎠
⎦
an = 4/n2
1
bn =
π
2π
∫
2π
x 2 sin nx.dx =
0
(sin nx) 2 cos nx ⎤
1 ⎡ 2 ( − cos nx )
+ 2x
+
x
⎥
π ⎢⎣
n
n2
n3 ⎦ 0
=
1 ⎡⎛⎜ − 4π 2 cos 2nπ
2 cos 2nπ ⎞⎟ ⎛
2
+0+
− ⎜0 + 0 + 3
⎢⎜
3
⎟
n
π ⎣⎢⎝
n
n
⎠ ⎝
=
⎤ −4π
1 ⎡⎛⎜ − 4 π 2
2 ⎞
+ 0 + 3 ⎟ − (0 + 0 + 2 n 3 )⎥ =
⎢⎜
⎟
π ⎣⎢⎝ n
n
n ⎠
⎥⎦
⎞⎤
⎟⎥
⎠⎦⎥
On substituting a0, an, bn values in equation (1), we get,
∴ x2 =
Q5.
4 2
π +
3
∞
4
∑n
n =1
2
cos nx −
∞
4π
sin nx
n
n =1
∑
Define even and odd function.
Model Paper-III, Q1(c)
Ans:
Even Function: A function f(x) is an even function if f(–x) = f(x)
Examples: cos x, sec x, x2
Odd Function: A function f(x) is an odd function if f(–x) = –f(x).
Examples: sin x, tan x, x3
Q6.
Give the Fourier series expansion for even periodic functions.
Model Paper-II, Q1(d)
Ans: Generally, a periodic function f(x) defined in the interval (–π, π) is represented by,
f(x) =
a0
+
2
∞
∑
n =1
an cos nx +
∞
∑b
n
sin nx
... (1)
n =1
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2.3
UNIT-2 (Fourier Series)
Ans: A periodic function f(x) defined in the interval (–π, π) is
represented by,
Where,
1
a0 =
π
an =
1
π
1
bn =
π
π
∫ f ( x ) dx
∞
... (2)
f(x) =
−π
π
∫ f ( x) cos nx dx
1
a0 =
π
π
... (4)
−π
1
an =
π
When f(x) is an even function,
From equation (2), we get,
a0 =
⇒
a0 =
1
π
2
π
f ( x) dx
−π
⇒
an =
2
π
∫ f ( x) dx
1
a0 =
π
0
π
∫ f ( x) cos nx dx
an =
π
∫ f ( x) cos nx dx
0
∫
a0
+
2
∞
∑a
n
⇒
2
an =
π
Q7.
∫ f ( x ) dx
... (2)
−π
π
∫ f ( x) cos nx dx
... (3)
−π
π
∫ f ( x) sin nx dx
... (4)
−π
2
bn =
π
cos nx
∴ f(x) =
∫ f ( x)dx
Give the Fourier series expansion for odd
periodic functions.
Model Paper-I, Q1(d)
π
∫ f ( x) cos nx dx = 0
−π
π
∫ f ( x) sin nx dx
−π
π
∫ f ( x) sin nx dx
0
∑b
n
sin nx
n =1
2
bn =
π
0
0
(Q f(x) is odd)
−π
Where,
π
∫ f ( x) cos nx . dx
∫ f ( x)dx = 0
∞
n=1
π
π
Therefore, it can be concluded that if a function f(x) is
odd in the interval (–π, π), its Fourier series expansion consists
of only sine terms.
Where,
2
a0 =
π
1
π
1
bn =
π
1
f ( x ) sin x dx = 0
π −π
Therefore, it can be concluded that if a function f(x) is
even in the interval (–π, π), its Fourier series expansion
consists of only cosine terms.
f(x) =
π
sin nx is an odd function and f(x) sin nx is an even
function.
∴ From equation (4),
π
∴
... (1)
As ‘cos nx’ is an even function, here f(x) cos nx becomes
an odd function
∴ From equation (3),
−π
Since, sin nx is an odd function and f(x) sin nx also
beomes an odd function
∴ From equation (4),
bn =
∑
When f(x) is an odd function,
From equation (2), we get,
π
As ‘cos nx’ is an even function, f(x) cos nx is also an
even function.
∴ From equation (3), we get,
1
an =
π
1
bn =
π
π
∫
∑
Where,
... (3)
−π
∫ f ( x) sin nx dx
∞
a0
+ an cosnx + bn sin nx
2 n=1
n=1
π
∫ f ( x) sin nx dx
0
Q8.
Define the Fourier series for functions having
period ‘2l’.
Ans: Consider a function f(x), which is periodic in the interval
(k, k + 2l) and period ‘2l’. Then, the fourier series expansion of
f(x) is represented as,
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MATHEMATICS-II [JNTU-ANANTAPUR]
f(x) =
a0
+
2
∞
∑
an cos
n =1
nπx
+
l
∞
∑b sin
n
n=1
nπx
l
Where,
1
a0 =
l
an =
1
l
1
bn =
l
Q9.
k + 2l
∫ f ( x)dx
k
k + 2l
∫
f ( x ) cos
nπx
dx
l
f ( x ) sin
nπx
dx
l
k
k + 2l
∫
k
Define half range Fourier sine series of f(x).
Ans: The half-range Fourier sine series of f(x) is defiend in the interval (0, π) is represented as,
∞
f(x) =
∑ b sin nx
n
n =1
2
Where, bn =
π
π
∫ f ( x) sin nx dx
0
Q10. Define half-range Fourier cosine series of f(x).
Model Paper-III, Q1(d)
Ans:
The half-range Fourier cosine series of f(x) defined in the interval (0, π) is represented as,
f(x)=
a0
+
2
∞
∑a
n cos nx
n=1
Where,
2
a0 =
π
2
an =
π
π
∫ f ( x)dx
and
0
π
∫ f ( x) cos nx dx
0
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2.5
UNIT-2 (Fourier Series)
PART-B
ESSAY QUESTIONS WITH SOLUTIONS
2.1 DETERMINATION OF FOURIER COEFFICIENTS
π , π ).
Q11. Express f(x) = x as a Fourier series in (–π
Ans:
Given expression is,
f(x) = x in (–π,π)
The Fourier series representation of f(x) in (–π, π) is
f(x) =
a0
+
2
1
Where, a0 =
π
=
∞
∞
∑a
n cos nx +
n =1
π
∫
−π
given by,
∑b
n =1
n
sin nx
... (1)
π
π
1 ⎡ x2 ⎤
x.dx =
⎢ ⎥
π ⎣⎢ 2 ⎦⎥
−π
−π
1
f(x) dx =
π
∫
1
1 2
[(π)2 – (–π)2] =
[π – π2]
2π
2π
∴ a0 = 0
1
an =
π
=
π
∫ x. cos nx dx
−π
[∫
) ]
∫ (∫
π
1
x cos nx dx – 1 cos nx.dx dx
–π
π
π
1 ⎡ x. sin nx
sin nx ⎤
−
dx ⎥
= ⎢
n
π ⎣ n
⎦ −π
∫
π
1 ⎡ x. sin nx ( − cos nx ) ⎤
−
=
⎥
π ⎢⎣ n
n2
⎦ −π
π
1 ⎡ x. sin nx cos nx ⎤
+
= ⎢
⎥
π ⎣ n
n 2 ⎦ −π
=
1 ⎧ π sin nπ cos nπ ⎡ − π sin n(− π) cos n(− π) ⎤ ⎫
+
−⎢
+
⎨
⎥⎬
n
n2
n2
π ⎩ n
⎣
⎦⎭
=
1 ⎧ π sin nπ cos nπ ⎡ π sin nπ cos nπ ⎤ ⎫
+
−⎢
+
⎬
⎨
π ⎩ n
n2
n 2 ⎥⎦ ⎭
⎣ n
=
1 ⎧ π sin nπ cos nπ π sin nπ cos nπ ⎫
+
−
−
⎬
⎨
n
n2
n2 ⎭
π ⎩ n
=
1
{0} = 0
π
∴ an = 0
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MATHEMATICS-II [JNTU-ANANTAPUR]
1
bn =
π
π
∫
−π
1
f(x). sin nx dx =
π
π
∫ x.sin nx dx
−π
π
⎤
⎧⎪ ⎛
1 ⎡
⎞ ⎫⎪
= ⎢ x sin nx dx − ⎨ 1⎜ sin nx.dx ⎟dx ⎬⎥
π ⎢
⎪⎩ ⎝
⎠ ⎪⎭⎥
⎦ −π
⎣
∫
∫ ∫
π
1 ⎡ x.(− cos nx)
( − cos nx ) ⎤
−
dx ⎥
π ⎢⎣
n
n
⎦ −π
∫
=
π
1 ⎡ − x. cos nx sin nx ⎤
+ 2 ⎥
=
n
π ⎢⎣
n ⎦ −π
=
1 ⎧ − π cos nπ sin nπ ⎡ − (− π) cos n (− π) sin n( − π) ⎤ ⎫
+
−⎢
+
⎨
⎥⎬
n
n
n2
n2
⎦⎭
⎣
π ⎩
=
1 ⎧ − π cos nπ sin nπ ⎡ π cos nπ sin nπ ⎤ ⎫
+ 2 −⎢
− 2 ⎥⎬
⎨
π ⎩
n
n
n ⎦⎭
⎣ n
=
1 ⎧ − π cos nπ sin nπ π cos nπ sin nπ ⎫
+
−
+ 2 ⎬
⎨
n
n
π ⎩
n2
n ⎭
=
1 ⎧ − 2π cos nπ 2 sin nπ ⎫
+
⎬
⎨
π ⎩
n
n2 ⎭
=
− 2 cos nπ
2 sin nπ
+
n
πn 2
(Here, sin nπ = 0)
=
−2
−2
cos nπ =
(–1)n
n
n
For n = 1, 2, 3...
(Q cos nπ = (–1)n for n = 1, 2, 3...)
−2
(−1) n
n
On substituting the values of a0, an and bn in equation (1), we get,
∴ bn =
∞
f(x) =
−2
∑ ⎡⎢⎣ n
n =1
∞
⎤
(−1) n sin nx ⎥ = –2
⎦
n =1
⎡ (−1) n
∑ ⎢⎢⎣
n
⎤
sin nx ⎥
⎥⎦
1
1
⎡
⎤
∴ f ( x) = −2⎢− sin x + sin 2 x − sin 3x + ...⎥
2
3
⎣
⎦
∞
π , π], deduce the value of
Q12. Obtain the Fourier series for the function, f(x) = x + x in [–π
2
1
∑n
n =1
2
.
Model Paper-III, Q4
Ans: The given function is,
f (x) = x + x2 in (–π, π)
From the definition of trigonometric Fourier series,
a0
+
f(x) =
2
∞
∑
n =1
∞
an cos nx +
∑b
n
sin nx
... (1)
n =1
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2.7
UNIT-2 (Fourier Series)
1
Where, a0 =
π
π
∫
f ( x ) dx =
−π
1
π
π
π
∫
( x + x 2 ) dx =
−π
1 ⎡ x2 x3 ⎤
⎢ + ⎥
π ⎣⎢ 2
3 ⎥⎦
−π
1 ⎡ π 2 π3 π 2 π 3 ⎤
1 ⎡ π 2 π 3 ⎛ ( −π ) 2 ( −π )3 ⎞ ⎤
−
+ ⎥
+
−
+
⎢
⎥
=
= ⎢ +
π ⎢⎣ 2
3
2
3 ⎥⎦
π ⎣⎢ 2
3 ⎝⎜ 2
3 ⎠⎟ ⎦⎥
=
∴ a0 =
1
an =
π
=
1 ⎡ 2π3 ⎤ 2π2
⎥ =
⎢
π ⎣⎢ 3 ⎦⎥
3
2π 2
3
π
∫
( x + x 2 ) cos nx.dx =
−π
π
π
1 ⎡⎡
sin nx ⎤
sin nx ⎤
⎢ ⎢ ( x + x 2 ).
−
(1 + 2 x).
.dx ⎥
⎥
π ⎢⎣
n ⎦ −π
n
⎥
−π
⎣
⎦
∫
π
⎡
⎤
π
−π
⎛⎡
1 ⎢⎡
(− cos nx) ⎤ ⎞⎟⎥
⎛ − cos nx ⎞
⎜⎢ +
2 sin nx ⎤
⎥
+
−
−
dx
(
x
x
).
(
1
2
x
)
(
2
)
.
⎜
⎟
π ⎢⎢⎣
n ⎥⎦ −π ⎜⎜ ⎢
⎥ ⎟⎟⎥
n2
⎝ n 2 ⎠ −π − π
⎦ ⎠⎦
⎝⎣
⎣
∫
⎡
⎛ d
⎞ ⎤
⎢Q uvdx = u vdx − ⎜ u vdx ⎟dx ⎥
dx
⎝
⎠ ⎦
⎣
∫
∫
∫
∫
π
π
π ⎤
⎡
⎛⎡
1 ⎢⎡
⎛ − cos nx ⎞ ⎤
⎛ sin nx ⎞ ⎥
2 sin nx ⎤
(
x
x
).
–
(1
2
x
)
2.
+
⎜⎢ +
⎟
= π ⎢ ⎣⎢ +
⎝⎜ n 2 ⎠⎟ ⎥⎦ −π ⎝⎜
n ⎦⎥ −π ⎜⎝ ⎣
n3 ⎠ – π ⎥
⎣
⎦
=
π
π
π
1 ⎡⎡
sin nx ⎤
cos nx ⎤
⎡ sin nx ⎤ ⎤
⎡
⎢ ⎢( x + x 2 ).
+ ⎢(1 + 2 x). 2 ⎥ − ⎢2. 3 ⎥ ⎥
⎥
n ⎦ −π ⎣
π ⎣⎢ ⎣
n ⎦ −π ⎣
n ⎦ − π ⎦⎥
=
1⎡
sin nπ
cos nπ
sin nπ
sin n(−π) −(1 + 2( −π)). cos n(−π) + 2 sin n( −π ) ⎤
+ (1 + 2π). 2 − 2. 3 − (−π + (−π) 2 ).
(π + π) 2 .
⎥
⎢
n2
n3
⎦
π⎣
n
n
n
n
=
1⎡
(−1) n
cos nπ ⎤
⎢0 + (1 + 2π) 2 + 0 − 0 − (1 − 2π). 2 − 0⎥
π ⎢⎣
n
n
⎥⎦
[Q sin nπ = 0]
=
1⎡
(−1) n
(−1) n ⎤
⎢(1 + 2π) 2 − (1 − 2π). 2 ⎥
π ⎣⎢
n
n ⎦⎥
[Q cos nπ = (–1)n]
=
1 ⎡ (−1) n 2π.(−1) n (−1) n 2π(−1) n ⎤ 1 ⎡ (−1)n ⎤
+
− 2 +
⎥ = ⎢4π 2 ⎥
⎢
π ⎢⎣ n 2
n2
n
n 2 ⎥⎦ π ⎢⎣
n ⎥⎦
=
4(−1) n
n2
∴ an =
4(−1) n
n2
∴ a1 =
−4
(1) 2
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a2 =
a3 =
a4 =
4
(2) 2
−4
(3) 2
4
(4) 2
And bn =
=
=
1
π
and so on.
π
∫ (x + x
2
). sin nx.dx
−π
π
π
1 ⎡⎡
− cos nx ⎤
⎛ − cos nx ⎞⎤
⎢ ⎢ ( x + x 2 )⎜
.dx ⎥
⎟⎥ − (1 + 2 x ).
π ⎢⎣
n ⎠⎦ − π
n
⎥
⎝
−π
⎦
⎣
∫
π
π
π
⎡⎡
1 ⎢⎡ ⎡
sin nx ⎤
sin nx ⎤ ⎤
⎛ − cos nx ⎞ ⎤
(x + x2 ) ⎜
+ ⎢ ⎢ (1 + 2 x ). 2 ⎥ − (2). 2 .dx ⎥ ⎥
⎢
⎟
⎥
⎝
n ⎠ ⎦ −π ⎢ ⎣
π ⎢⎣
⎥⎥
n ⎦ −π – π
n
⎣
⎦⎦
⎣
∫
π
π
⎡ ⎛ − cos nx ⎞ π ⎤
1⎡
sin nx ⎤
⎡
2 ⎛ − cos nx ⎞ ⎤
⎟ ⎥ + ⎢(1 + 2 x ). 2 ⎥ − ⎢ 2⎜
⎟ ⎥
= ⎢( x + x ) ⎜
π⎣
n
n ⎦ − π ⎢⎣ ⎝ n 3 ⎠ − π ⎥⎦
⎝
⎠⎦ − π ⎣
π
1⎡
sin nx 2 cos nx ⎤
2 ⎛ − cos nx ⎞
⎟ + (1 + 2 x ). 2 +
= ⎢( x + x )⎜
⎥
π⎣
n
n
n 3 ⎦ −π
⎠
⎝
⎡
⎤
sin nπ
cos nπ
2 ⎛ − cos n π ⎞
+2
( −π + ( −π ) 2 ) ⎥
⎢ ( π + π ) ⎜⎝
⎟⎠ + (1 + 2 π ).
2
3
n
1
n
n
⎥
= ⎢
⎢
π
sin n ( −π ) 2 cos n ( −π ) ⎥
⎛ − cos n ( −π ) ⎞
−
⎢
⎥
⎜⎝
⎟⎠ − (1 + 2( −π )).
n
n2
n3
⎣
⎦
=
1⎡
(−1)n
( −1) n
(−1)n
(−1)n ⎤
2
+ (1 + 2π ).0 + 2. 3 − (π 2 − π)( −1)
− (1 − 2π).0 − 2. 3 ⎥
⎢(π + π )( −1)
n
n
π ⎣⎢
n
n ⎥⎦
n
1⎡
2(−1)n
(−1)n
2(−1)n ⎤
2 ( −1)
2
−
π
+
π
+
+
+
π
−
π
(
)
(0)
(
)
–
(0)
–
⎥
= π⎢
n
n
n3
n3 ⎦⎥
⎣⎢
=
∴ bn =
∴
(−1) n ⎤
1 ⎡ ( −1)n
( −1)n
(−1)n
( −1)n ⎤ 1 ⎡
− 2(−1) n
− π2
+ π2
−π
⎥ =
⎢ −π
⎥ = ⎢− 2π
n ⎥⎦
n
n
n
n ⎥⎦ π ⎢⎣
π ⎢⎣
n
− 2(−1) n
n
b1 =
2
−2
; b2 =
1
2
b3 =
2
−2
and so on.
; b4 =
3
4
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.9
UNIT-2 (Fourier Series)
On substituting the corresponding values in equation (1), we get,
f (x) =
f (x)=
∞
2π 2
+
3(2)
4(−1) n
∑
n2
n =1
∞
. cos nx +
− 2(−1) n
. sin nx
n
n =1
∑
⎡ ( −1)
⎤
1
( −1)
(1)
π2
.cos 2 x +
.cos(3x ) +
.cos 4 x + ...⎥ +
+ 4 ⎢ 2 cos( x) +
2
2
2
3
(2)
(3)
(4)
⎣ (1)
⎦
2
2
2
⎡ (2)
⎤
⎢⎣ 1 .sin x − 2 .sin 2 x + 3 sin 3 x − 4 .sin 4 x + ....⎥⎦
π2
cos 2 x cos 3 x cos 4 x
sin 2 x sin 3 x sin 4 x
⎡
⎤
⎡
⎤
− 4 ⎢ cos x −
+
−
+ .....⎥ + 2 ⎢ sin x −
+
−
+ ....⎥
x+ x =
2
2
2
3
2
3
4
2
3
4
⎣
⎦
⎣
⎦
2
... (2)
On substituting x = π in equation (2), we get,
⇒
⇒
⇒
⇒
π + π2 =
π2 ⎡
cos2π cos3π cos4π
sin 2π
⎤ ⎡
⎤
− 4 ⎢cos π − 2 + 2 − 2 + ....⎥ + 2 ⎢sin π −
+ ....⎥
3
2
3
4
2
⎣
⎦ ⎣
⎦
π + π2 =
π2
1
1
1
⎡
⎤
− 4 ⎢ −1 − 2 − 2 − 2 + ...⎥ + 2[0 − 0 + 0....]
3
2
3
4
⎣
⎦
π + π2 –
1
1
1
⎛
⎞
= − 4⎜ − 1 − 2 − 2 − 2 + .... ⎟
2
3
4
⎝
⎠
π2
3
⎛ 1⎞
1
1
1
⎛
⎞
π + π2 ⎜ 1 − ⎟ = 4⎜1 + 2 + 2 + 2 + .... ⎟
3
⎝
⎠
3
4
⎝ 2
⎠
1
1
1
π 2π 2
+
= 1 + 2 + 2 + 2 + .....
4 4×3
2
3
4
∞
∑n
π π2
+
=
4 6
⇒
∞
∴
∑n
n =1
1
2
=
n =1
1
2
π π2
+
4 6
Q13. Expand f(x) = ex, – π < x < π as a Fourier series. Derive a series for
π
.
sinh π
Ans: Given function is, f(x) = ex, – π < x < π
The Fourier series expansion of f(x) is given as,
f(x) =
a0
+
2
∞
∑a
n
n =1
cos nx +
∞
∑b
n
sin nx
... (1)
n =1
π
1
f ( x) dx
Where, a0 =
π −π
∫
π
=
1 x π
1
1
e x dx = [e ]−π = [e π − e − π ]
π
π −π
π
∫
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.10
MATHEMATICS-II [JNTU-ANANTAPUR]
On multiplying and dividing R.H.S by 2, we get,
a0 =
2 ⎡ e π − e −π ⎤
1 2 π
. [e − e − π ] = ⎢
⎥
2
π ⎢⎣
π 2
⎥⎦
⎡
eπ − e−π ⎤
⎢Q sinh π =
⎥
2 ⎦
⎣
2
= . sinh π
π
∴ a0 =
2
sinh π
π
1
an =
π
π
∫ f ( x) cos nxdx
−π
π
1
e x cos nxdx
=
π −π
∫
Consider,
∫e
x
cos nxdx = In
∫
∫
⎡d
⎤
∫
x
x
In = cos nx e dx − ⎢ (cos nx ). e dx ⎥ dx
dx
⎣
⎦
∫
⇒
x
x
In = cos nx.e − − n sin nx.e dx
⇒
x
x
In = e cos nx + n e sin nxdx
⇒
In = e x cos nx + n ⎢sin nx e x dx − ⎜
⇒
In = e x cos nx + n[sin nx.e x − n cos nxe x dx ]
⇒
In = e cos nx + n[e sin nx − n e cos nxdx]
⇒
x
x
x
2
In = e cos nx + n.e sin nx − n e cos nxdx
⇒
In = ex(cos nx + nsin nx) – n2In
⇒
In + n2In = ex(cos nx + nsin nx)
⇒
In(1 + n2) = ex(cos nx + nsin nx)
∫
⎡
⎣
⎞
⎤
∫ ⎝ dx (sin nx )∫ e dx ⎟⎠ dx ⎥⎦
x
∫
x
∫
x
x
∫
⇒
⇒
⎛d
∫
In =
∫
ex
(cos nx + nsin nx)
1 + n2
e x cos nx dx =
ex
1+ n2
(conx + nsin nx)
... (2)
π
∴
1
e x cos nxdx
an =
π −π
∫
π
⎤
1 ⎡ ex
(cos nx + n sin nx )⎥
= ⎢
2
π ⎢⎣1 + n
⎥⎦ − π
Look for the SIA GROU P LOGO
[Q From equation (2)]
on the TITLE COVER before you buy
2.11
UNIT-2 (Fourier Series)
=
=
=
=
1
π (1 + n 2 )
1
π (1 + n 2 )
1
π(1 + n )
2
[( e π (cos nπ + n sin nπ )) − ( e − π (cos( − n π) + n sin( − nπ ))]
[ e π (cos n π + 0) − e − π (cos( − nπ ) + 0 )]
[e π cos nπ − e − π cos nπ]
[Q cos(– θ) = cos θ]
cos nπ(e π − e − π )
... (3)
π(1 + n 2 )
On multiplying and dividing the equation (3) by 2 on R.H.S, we get,
an = 2.
=
∴ an =
∴
bn =
=
Consider,
(e π − e − π )
2
π(1 + n 2 )
cos nπ
2(−1) n
π(1 + n 2 )
.
sinh π
2(−1) n sinh π
π(1 + n 2 )
1
π
1
π
∫e
x
π
∫ f ( x) sin nxdx
−π
π
∫e
x
sin nxdx
−π
sin nxdx = In
∫
∫
⎡d
∫
⎤
x
x
In = sin nx e dx − ⎢ (sin nx ). e dx ⎥ dx
dx
⎣
⎦
∫
⇒
x
x
In = sin nx.e − n cos nx.e dx
⇒
x
x
In = e . sin nx − n e cos nxdx
⇒
⎡
⎡d
⎤ ⎤
In = e x sin nx − n ⎢cos nx e x dx − ⎢ (cos nx ) e x dx ⎥ dx ⎥
dx
⎣
⎦ ⎦
⎣
⇒
In = e x sin nx − n[cos nx .e x − − n sin nx.e x dx ]
⇒
x
x
x
In = e sin nx − n[e cos nx + n e sin nxdx]
⇒
2
x
x
x
In = e sin nx − ne cos nx − n e sin nxdx
⇒
In = e x (sin nx − n cos nx) − n 2 I n
⇒
⇒
⇒
∫
∫
∫
∫
∫
∫
∫
In + n2In = ex(sin nx – ncos nx)
In(1 + n2) = ex(sin nx – ncos nx)
In =
ex
(sin nx − n cos nx )
1 + n2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.12
MATHEMATICS-II [JNTU-ANANTAPUR]
∫
e x sin nxdx =
ex
(sin nx – ncosnx)
1 + n2
... (4)
π
∴
1
e x sin nxdx
bn =
π −π
∫
π
⎤
1 ⎡ ex
(sin nx − n cos nx )⎥
= ⎢
2
π ⎢⎣1 + n
⎥⎦ − π
=
=
=
=
1
π(1 + n 2 )
1
π(1 + n 2 )
1
π(1 + n 2 )
[Q From equation (3)]
[e π (sin nπ − n cos nπ) − e − π (sin( − nπ) − n cos( − nπ))]
[eπ(0 – ncosnπ) – e– π(0 – ncosnπ)]
[Q cos(–θ) = cos θ]
[ − ne π cos nπ + e − π n cos nπ]
n cos nπ
[− e π + e − π ]
π(1 + n 2 )
... (5)
On multiplying and dividing equation (5) by 2 on R.H.S, we get,
bn =
∴ bn =
⎡
eπ − e−π ⎤
⎥
⎢Q sinh π =
2
⎢
⎥
⎢
e−π − eπ ⎥
⎢⇒ − sinh π =
⎥
2 ⎦
⎣
2n( −1) n ( −1)
n cos nπ [e − π − e π ]
sinh π
×
2
=
2
π(1 + n 2 )
π(1 + n 2 )
2n( −1) n+1 sinh π
π(1 + n 2 )
On substituting the values of a0, an and bn in equation (1), we get,
2
sinh π
2 sinh π
π
+
f(x) =
2
π
ex =
∞
( −1) n
∑ 1+ n
n =1
2
2 sinh π
π
cos nx +
∞
( −1) n
sinh π 2 sinh π ⎡
+
cos nx +
⎢
2
π
π
⎣⎢ n =1 1 + n
∑
∞
∑
n =1
∞
∑
n(−1) n +1
n =1
n (−1) n +1
1+ n2
1+ n 2
sin nx
⎤
sin nx ⎥
⎦⎥
∴ The required Fourier series is,
ex =
Deriving the Series for
⇒
2 sinh π
π
⎡1
⎢ +
⎣⎢ 2
∞
( −1) n
∑ 1+ n
n =1
2
∞
cos nx +
∑
n =1
n( −1) n +1
1+ n
2
⎤
sin nx ⎥
⎦⎥
π
: On substituting x = 0 in equation (6), we get,
sinhπ
e0 =
2 sinh π ⎡ 1
⎢ +
π ⎢⎣ 2
∑
1=
2 sinh π ⎡ 1
⎢ +
π ⎣⎢ 2
∑1+ n
∞
(−1) n
cos 0 +
2
n =1 1 + n
∞
n =1
(−1) n
2
.1 +
⎤
n( −1) n +1
sin 0⎥
2
⎥⎦
n =1 1 + n
∞
∑
n( −1) n +1 ⎤
.0⎥
2
n =1 1 + n
⎦⎥
∞
∑
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
... (6)
2.13
UNIT-2 (Fourier Series)
⇒
⎡1
π
= 2⎢ 2 +
⎢⎣
sinh π
⇒
⎡ 1 (−1)1 (−1) 2 (−1) 3 (−1) 4
⎤
π
+ .....⎥
+
+
+
= 2⎢ +
2
2
2
2
sinh π
1+ 3
1+ 4
⎣⎢ 2 1 + 1 1 + 2
⎦⎥
⇒
π
(−1)
1
1
⎤
⎡ 1 (−1)
+
+ ...⎥
+
+
= 2⎢ +
2
2
sinh π
1+ 3
1+ 4 2
⎦
⎣ 2 1+1 1+ 2
⇒
π
1
1
1
⎡1 1
⎤
−
+
+ .....⎥
= 2⎢ − +
2
2
2
sinh π
⎣2 2 1+ 2 1+ 3 1+ 4
⎦
∴
∞
( − 1) n
∑ 1+ n
n =1
2
⎤
+ 0⎥
⎥⎦
π
1
1
⎡ 1
⎤
= 2⎢
–
+
– ......⎥
2
2
2
sinhπ
⎣1+ 2 1+ 3 1+ 4
⎦
π , π ).
Q14. If f(x) = cosh ax, expand f(x) as a Fourier Series in (–π
Ans: Given function is,
f (x) = cosh ax in (–π,π)
The Fourier series representation of f(x) is given by,
Where,
f(x) =
∞
a0 ∞
an cos nx +
bn sin nx
+
2 n =1
n =1
a0 =
1
π
=
∑
1
π
π
∫
f ( x) dx =
−π
π
⎛ e ax + e − ax
⎜
⎜
2
− π⎝
∫
∑
1
π
... (1)
π
∫ cosh ax dx
−π
ax
− ax
⎛
⎜Q cosh ax = e + e
⎜
2
⎝
⎞
⎟ dx
⎟
⎠
⎞
⎟
⎟
⎠
π
π
⎤
1 ⎡ ax
⎢ e dx + e − ax dx ⎥
2π ⎢
⎥
−π
−π
⎦
⎣ −π
aπ
− aπ
⎡⎛ ax ⎞ π ⎛ − ax ⎞ π ⎤
e ⎤ ⎡ a −aπ eaπ ⎤⎤
e ⎟ ⎥
1 ⎢⎜ e ⎟
1 ⎡⎡ e
⎜
+
−
+
−
=
=
⎢
⎥
2π ⎢⎜⎝ a ⎟⎠ −π ⎜⎝ − a ⎟⎠ −π ⎥ 2π ⎣ ⎢⎣ a
a ⎥⎦ ⎢⎣ −a −a ⎥⎦⎦
⎣
⎦
π
=
1
2π
=
aπ
e− aπ ⎤ ⎡ −e− aπ eaπ ⎤⎫ 1 ⎡ e − e – e + e ⎤
1 ⎧⎡ e
−
+
+ ⎥⎬ =
⎢
2π ⎨⎢⎣ a
a
a
a ⎥⎦
a ⎥⎦ ⎢⎣ a
a ⎦ ⎭ 2π ⎣ a
⎩
=
1
2π
=
1
2
[ e aπ – e – aπ ] =
sinh aπ
aπ
aπ
∫
(e ax + e −ax ) dx =
∫
∫
aπ
∴ a0 =
− aπ
− aπ
aπ
⎡ 2e aπ 2e – aπ ⎤
–
⎢
⎥
a ⎥⎦
⎢⎣ a
2
sinh aπ
aπ
1
∴ an =
π
π
∫ f ( x) cos nx dx
−π
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.14
MATHEMATICS-II [JNTU-ANANTAPUR]
1
=
π
π
1
cosh ax cos nx dx =
π
∫
−π
π
⎛ e ax + e −ax
⎜
⎜
2
− π⎝
∫
⎞
⎟ cosnx dx
⎟
⎠
ax
− ax
⎛
⎜Q cosh ax = e + e
⎜
2
⎝
⎞
⎟
⎟
⎠
π
π
⎤
1 ⎡⎢ ax
e cos nx dx + e − ax cos nx dx ⎥
=
2π ⎢
⎥
−π
⎦
⎣ −π
∫
π
π
⎧⎡ ax
⎤ ⎫⎪
⎤
⎡ e − ax
⎪ e
a
nx
n
nx
(
cos
sin
)
(
a
cos
nx
n
sin
nx
)
+
−
+
+
⎥ ⎬
⎥
⎢ 2
⎨⎢ 2
2
2
⎪⎩⎣⎢ a + n
⎦⎥ −π ⎪⎭
⎦⎥ − π ⎣⎢ a + n
1
=
2π
=
∫
⎧⎪⎡ e aπ
⎤ ⎡ e − aπ
⎤
a
n
n
n
π
+
π
(
cos
sin
)
( a cos n( − π) + n sin n ( − π) ⎥ +
⎥−⎢ 2
⎨⎢ 2
2
2
⎪⎩⎢⎣ a + n
⎥⎦ ⎢⎣ a + n
⎥⎦
1
2π
⎡ e − aπ
⎤ ⎡ e aπ
⎤ ⎫⎪
−
π
+
π
(
cos
sin
)
[ − a cos n( − π) + n sin n ( − π)]⎥ ⎬
a
n
n
n
⎢ 2
⎥−⎢ 2
2
2
⎢⎣ a + n
⎥⎦ ⎢⎣ a + n
⎥⎦ ⎪⎭
=
⎧⎪⎡ ae aπ cos nπ ne aπ sin nπ ae − aπ cos nπ ne − aπ sin nπ ⎤
+
−
+
⎥ +
⎨⎢ 2
2
a2 + n2
a2 + n2
a 2 + n 2 ⎥⎦
⎪⎩⎢⎣ a + n
1
2π
⎡ − ae − aπ cos nπ ne − aπ sin nπ ae aπ cos nπ ne aπ sin nπ ⎤ ⎫⎪
+
−
+
⎢
⎥⎬
2
2
a2 + n2
a2 + n2
a 2 + n 2 ⎥⎦ ⎪⎭
⎢⎣ a + n
=
1
2π
⎧⎪ ae aπ cos nπ ae − aπ cos nπ ae − aπ cos nπ ae aπ cos nπ ⎫⎪
+
−
−
⎨ 2
⎬
⎪⎩ a + n 2
a2 + n2
a2 + n2
a 2 + n 2 ⎪⎭
=
1
2π
⎧⎪ 2ae aπ cos nπ 2ae − aπ cos nπ ⎫⎪
−
⎨
⎬
⎪⎩ a 2 + n 2
a 2 + n 2 ⎪⎭
=
1
2π
⎧⎪ 2ae aπ cos nπ − 2ae −aπ cos nπ ⎫⎪
2 a cos nπ
⎨
⎬=
2
2
⎪⎩
⎪⎭ π( a 2 + n 2 )
a +n
=
=
∴ an =
bn =
2 a cos n π
π(a 2 + n 2 )
[Q sin nπ = 0]
[Q sin nπ = 0]
⎧⎪ e aπ − e − aπ ⎫⎪
⎨
⎬
2
⎪⎩
⎪⎭
aπ
− aπ ⎞
⎛
⎜Q sinh aπ = e − e
⎟
⎜
⎟
2
⎝
⎠
× sinh aπ
2a sinh aπ cos nπ
π (a 2 + n 2 )
2a sinh aπ cos nπ
π(a 2 + n 2 )
1
π
1
=
π
π
∫
f ( x ) sin nx dx =
−π
π
⎛ e ax + e −ax
⎜
⎜
2
− π⎝
∫
1
π
π
∫ cosh ax sin nx dx
−π
⎞
⎟ sin nx dx
⎟
⎠
π
π
⎤
1 ⎡⎢ ax
+
e
sin
nx
dx
e − ax sin nx dx ⎥
=
2π ⎢
⎥
−π
⎣− π
⎦
∫
∫
Look for the SIA GROU P LOGO
ax
− ax
⎛
⎜Q cosh ax = e + e
⎜
2
⎝
⎞
⎟
⎟
⎠
⎤
⎡
e ax
ax
Q
=
sin
(a sin nx − n cos nx)⎥
e
nxdx
⎢
2
2
a +b
⎦⎥
⎣⎢
∫
on the TITLE COVER before you buy
2.15
UNIT-2 (Fourier Series)
π
π
⎧⎡ e ax
⎤ ⎫⎪
⎡ e −ax
⎤
⎪
(
sin
cos
)
(
a
sin
nx
n
cos
nx
)
a
nx
n
nx
+
−
−
−
⎥ ⎬
⎢ 2
⎥
⎨⎢ 2
2
2
⎪⎩⎣⎢ a + b
⎦⎥ − π ⎪⎭
⎦⎥ −π ⎣⎢ a + b
1
=
2π
=
⎧⎪⎡ e aπ
⎤
e − aπ
(
sin
cos
)
( a sin n ( − π) − n cos n( − π) ⎥ +
a
n
n
n
π
−
π
−
⎨⎢ 2
2
2
2
a +b
⎪⎩⎣⎢ a + b
⎦⎥
1
2π
⎡ e − aπ
⎤ ⎫⎪
e aπ
( − a sin nπ − n cos nπ) − 2
[ − a sin n ( − π) − n cos n ( − π)]⎥ ⎬
⎢ 2
2
2
a +b
⎣⎢ a + b
⎦⎥ ⎪⎭
⎤ ⎫⎪
⎧⎪⎡ e aπ
e − aπ
π
+
π
−
π
−
π
(
a
sin
n
n
cos
n
a
sin
n
n
cos
n
)
⎥⎬
π
−
π
−
π
+
π
+
(
a
sin
n
cos
n
a
sin
n
n
cos
n
)
⎨⎢ 2
2
a 2 + b2
⎦⎥ ⎪⎭
⎪⎩⎣⎢ a + b
1
=
2π
=
1
{0} = 0
2π
∴ bn = 0
On substituting a0, an and bn values in equation (1), we get,
2
∞
∞
2a sinh aπ cos nπ
aπ sinh aπ
( 0)
f(x) =
cosnx
+
+
2
π(a 2 + n 2 )
n =1
n =1
∑
∑
∴ f ( x) =
sinh aπ
+
aπ
∞
∑
2a cos(nπ ) sinh aπ cos nx
π (a 2 + n 2 )
n =1
⎧⎪ 0 for
2
⎪⎩ x for
Q15. Find the Fourier series to represent the function f(x) given, f(x) = ⎨
−π≤x≤0
0≤x≤π
.
Ans: Given equation is,
⎧⎪ 0 for − π ≤ x ≤ 0
f ( x) = ⎨ 2
⎪⎩ x for 0 ≤ x ≤ π
The Fourier series representation of f (x) in the interval [–π, π] is given by,
f (x) =
a0
+
2
1
Where, a0 =
π
∞
∑a
∞
n
cos nx +
n =1
n
sin nx
... (1)
n =1
π
∫
∑b
f ( x) dx =
−π
0
π
⎤
1⎡
⎢ f ( x ) dx + f ( x ) dx ⎥
π⎢
⎥
0
⎣−π
⎦
∫
∫
0
π
⎤ 1 ⎡ ⎛ x3 ⎞π ⎤
1⎡
1 3
π3
2
[ π − 0] =
= ⎢ (0) dx + x dx ⎥ = ⎢0 + ⎜⎜ ⎟⎟ ⎥ =
π⎢
⎥ π ⎢ ⎝ 3 ⎠ 0 ⎥ 3π
3π
0
⎣− π
⎦
⎣
⎦
∫
∴ a0 =
∫
π2
3
Similarly,
1
an =
π
π
∫ f ( x) cos nx dx
−π
0
π
⎤
1⎡
= ⎢ f ( x ) cos nx dx + f ( x ) cos nx dx ⎥
π⎢
⎥
0
⎦
⎣−π
∫
∫
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.16
MATHEMATICS-II [JNTU-ANANTAPUR]
0
π
⎤
1⎡
⎢
+
(
0
)
cos
nx
dx
x 2 cos nx dx ⎥
=
π⎢
⎥
0
⎦
⎣− π
∫
=
∫
⎡ 0 + x 2 cosnx dx −
⎢⎣
1
π
∫
∫( ∫
)
π
2 x . cosnx dx dx ⎤
⎥⎦ 0
π
1 ⎡ 2 sin nx
sin nx ⎤
−2 x
= ⎢x
π⎣
n
n ⎥⎦ 0
∫
π
⎤
1 ⎡ x 2 sin nx 2
−
x ⋅ sin nx dx ⎥
= ⎢
π ⎣⎢
n
n
⎥⎦ 0
∫
[∫
]
π
⎤
1 ⎡ x 2 sin nx 2
− x sin nx dx − (1 sin nx dx) dx ⎥
= ⎢
π ⎣⎢
n
n
⎦⎥ 0
∫ ∫
⎡
⎡d
⎤ ⎤
⎢Q uvdx = u vdx − ⎢ u vdx ⎥ dx ⎥
⎣ dx
⎦ ⎦
⎣
∫
∫
∫
∫
π
1 ⎧⎪ x 2 sin nx 2 ⎡ (− cos nx )
cos nx ⎤ ⎫⎪
− ⎢x
+
dx ⎥ ⎬
= ⎨
π ⎪⎩
n
n⎣
n
n
⎦ ⎪⎭0
∫
1 ⎧⎪ x 2 sin nx 2 ⎡ − x cos nx sin nx ⎤ ⎫⎪
− ⎢
+ 2 ⎥⎬
= ⎨
π ⎪⎩
n
n⎣
n
n ⎦ ⎪⎭
1 ⎧⎪ x 2 sin nx 2 x cos nx 2 sin nx ⎫⎪
+
−
= ⎨
⎬
π ⎪⎩
n
n2
n 3 ⎪⎭
1
π
=
π sin nπ 2 cos nπ 2 sin nπ
+
−
n
n2
πn 3
2
= 0+
∴ an =
2
n
2
2
n
0
π
0
⎡ ⎧⎪ π 2 sin n π 2 πcos n π 2sin n π ⎫⎪
⎤
+
−
−
0
⎢⎨
⎥
{
}
⎬
n
n2
n 3 ⎭⎪
⎢⎣ ⎩⎪
⎥⎦
=
=
π
2
n2
( −1) n − 0
[Q sin nπ = 0 and cos nπ = (–1)n]
( −1) n
( −1) n
Similarly,
1
bn =
π
π
∫ f ( x) sin nx dx
−π
0
π
⎤
1⎡
⎢
⎥
+
f
(
x
)
sin
nx
dx
f
(
x
)
sin
nx
dx
=
π⎢
⎥
−
π
0
⎣
⎦
∫
∫
0
π
⎤
1⎡
⎢
(0) sin nx dx + x 2 sin nx dx ⎥
=
π⎢
⎥
0
⎣− π
⎦
∫
∫
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.17
UNIT-2 (Fourier Series)
π
π
⎤
⎤
1⎡
1⎡ 2
2
= ⎢0 + x sin nx dx ⎥ = ⎢ x sin nx dx ⎥
π⎢
⎥ π⎢
⎥
⎣ 0
⎦
⎦
⎣0
∫
[
∫
]
=
π
1 2
x ⋅ sin nx dx − ( 2 x sin nx dx ) dx
0
π
=
1⎡ 2
(− cos nx) ⎤
x ⋅ sin nx dx − 2 x
dx ⎥
⎢
n
π⎣
⎦0
∫
∫
∫
∫
π
∫
π
1 ⎡ 2 (− cos nx) 2
⎤
+
x ⋅ cos nx dx ⎥
= ⎢x
π⎣
n
n
⎦0
∫
⎫⎪
1 ⎪⎧ x 2 cos nx 2 ⎡
x ⋅ cos nx dx − (1 ⋅ cos nx dx ) dx ⎤ ⎬
+
= π ⎨−
⎥⎦ ⎪
n
n ⎢⎣
⎪⎩
⎭
∫
∫ ∫
π
0
π
1 ⎪⎧ − x 2 cos nx 2 ⎡ sin nx
sin nx ⎤ ⎪⎫
+ ⎢x ⋅
−
dx ⎥ ⎬
= ⎨
π ⎩⎪
n
n⎣
n
n
⎦ ⎭⎪ 0
∫
1 ⎧⎪ − x 2 cos nx 2 ⎡ x sin nx (− cos nx) ⎤ ⎫⎪
+ ⎢
−
= ⎨
⎥ ⎬⎪
π ⎪⎩
n
n⎣ n
n2
⎦⎭
1 ⎧⎪ x 2 cos nx 2 ⎡ x sin nx cos nx ⎤ ⎫⎪
+ ⎢
+
= ⎨−
⎥⎬
π ⎪⎩
n
n⎣ n
n 2 ⎦ ⎪⎭
1 ⎧⎪ x 2 cos nx 2 x sin nx 2 cos nx ⎫⎪
+
+
= ⎨−
⎬
π ⎪⎩
n
n2
n 3 ⎪⎭
π
0
π
0
π
0
1 ⎧⎪ − π 2 cos nπ 2π sin nπ 2 cos nπ 2 cos 0 ⎫⎪
+
+
−
⎨
⎬
π ⎪⎩
n
n2
n3
n3 ⎪⎭
=
⎧⎪ − π cos nπ 2 sin nπ 2 cos nπ
2 ⎫⎪
− 3⎬
+
+
=⎨
2
3
n
⎪⎩
πn
πn ⎪⎭
n
− π(−1) n
2
2
+ 0 + 3 (−1) n − 3
n
πn
πn
=
∴ bn =
2
−π
( −1) n + 3 [(−1) n − 1]
n
πn
On substituting a0, an and bn in equation (1), we get,
∞
∴ f ( x) =
π2
( −1)n
+2
cos nx +
6
n2
n =1
∑
∞
⎡ −π
∑ ⎢⎣ n
n =1
( −1) n +
2
πn 3
⎤
[( −1) n − 1⎥ sin nx
⎦
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.18
MATHEMATICS-II [JNTU-ANANTAPUR]
⎧−1
⎪ ( π + x) for − π ≤ x ≤ 0
π , π ] for the function f(x) = ⎨ 12
Q16. Find the Fourier series in [–π
.
⎪ (π − x) for 0 ≤ x ≤ π
⎩2
Model Paper-I, Q4
Ans: Given equation is,
⎧ –1
⎪⎪ 2 ( π + x ) for – π ≤ x ≤ 0
f(x) = ⎨
⎪ 1 ( π – x) for 0 ≤ x ≤ π
⎪⎩ 2
The Fourier series expansion of f (x) in the interval [–π, π] is given by,
f (x) =
a0
+
2
∞
∑
∞
an cos nx +
n =1
∑b
n
sin nx
... (1)
n =1
Where,
1
a0 =
π
=
π
∫ f ( x) dx
−π
π
0
π
⎤
⎤ 1 ⎡ 0 −1
1
1⎡
⎢ f ( x ) dx + f ( x ) dx ⎥ = ⎢
(π + x ) dx +
(π − x) dx ⎥
2
π⎢
⎥
⎥ π ⎢−π 2
0
⎦
⎣
0
⎣−π
⎦
∫
∫
∫
∫
π
0
0
π
⎤ 1 ⎡ −1 ⎛
2 ⎞
2 ⎞ ⎤
⎛
1 ⎡ −1
1
x
1
x
⎢
⎥
⎢
⎜ πx + ⎟ + ⎜ πx − ⎟ ⎥
(π + x ) dx +
(π − x) dx =
=
π⎢ 2
2
⎢ 2 ⎜⎝
⎥
π
2 ⎟⎠ −π 2 ⎜⎝
2 ⎟⎠ 0 ⎥
0
⎦
⎣ −π
⎣
⎦
∫
∫
=
1 ⎡ −1 ⎡ ⎛
( −π ) 2 ⎞ ⎤ 1
2
⎢ ⎢0 − ⎜ ( −π ) +
⎥+
π ⎢⎣ 2 ⎣ ⎝
2 ⎟⎠ ⎦ 2
⎡⎛ 2 π 2 ⎞
⎤⎤
⎢ ⎜ π − ⎟ − 0⎥ ⎥
2
⎝
⎠
⎣
⎦ ⎥⎦
=
1 ⎡ − 1 ⎡ 2π 2 − π 2 ⎤ π 2 ⎤
1 ⎡ − 1 ⎛⎜ 2 π 2 ⎞⎟ 1 ⎪⎧ π 2 ⎫⎪⎤
+ ⎨ ⎬⎥ = ⎢ ⎢
⎢ ⎜π −
⎥+ ⎥
π ⎢⎣ 2 ⎢⎣
2
π ⎢⎣ 2 ⎝
2 ⎟⎠ 2 ⎪⎩ 2 ⎪⎭⎥⎦
⎥⎦ 4 ⎥⎦
=
1 ⎡ − π 2 ⎛⎜ π 2 ⎞⎟ ⎤
+
⎢
⎜ 4 ⎟ ⎥⎥
π ⎣⎢ 4
⎠⎦
⎝
∴ a0 = 0
Similarly,
1
an =
π
π
∫ f ( x) cos nx dx
−π
0
π
π
⎤ 1⎡0 1
⎤
1
1⎡
⎢
⎢
⎥
⋅
+
⋅
−
π
+
+
x
nx
dx
f
(
x
)
cos
nx
dx
f
(
x
)
cos
nx
dx
(
)
cos
(π − x) cos nx dx ⎥
=
=
π⎢
2
⎥ π ⎢− π 2
⎥
0
0
⎣− π
⎣
⎦
⎦
∫
=
∫
∫
0
π
⎤
1
1 ⎡ −1
⎢
(π − x ) cos nx dx ⎥
(π + x) cos nx dx +
π⎢ 2
2
⎥
0
⎣ −π
⎦
∫
∫
Look for the SIA GROU P LOGO
∫
⎡
⎡d
⎤ ⎤
⎢Q uvdx = u vdx − ⎢ u vdx ⎥ dx ⎥
⎣ dx
⎦ ⎦
⎣
∫
∫
∫
∫
on the TITLE COVER before you buy
2.19
UNIT-2 (Fourier Series)
[
]
[
]
=
π⎫
0
1
1 ⎧ −1
( π + x ) cos nx − (1 cos nx dx ) dx + ( π − x) cos nx − (−1 cos nx dx ) dx ⎬
⎨
−π
0⎭
2
π ⎩2
=
1
π
π
0
⎧⎪ − 1 ⎡
sin nx
sin nx ⎤
1⎡
sin nx
sin nx ⎤ ⎫⎪
π
+
−
+
π
−
+
dx
x
x
(
).
(
)
⎨ ⎢
⎥ ⎬
n
n
n
n ⎥⎦ −π 2 ⎢⎣
⎦ 0 ⎪⎭
⎪⎩ 2 ⎣
=
1
π
π
0
⎧⎪ − 1 ⎡
sin nx ( − cos nx ) ⎤ ⎫⎪
sin nx ( − cos nx ) ⎤
1⎡
x
(
).
−
+
π
−
+
⎨ ⎢( π + x).
⎥
⎢
⎥ ⎬
n
n
n2
n2
⎦ −π 2 ⎣
⎦ 0 ⎪⎭
⎪⎩ 2 ⎣
=
1
π
=
π
1 ⎪⎧ − 1 ⎡ n( π + x) sin nx + cos nx ⎤ 0
1 ⎡ n( π − x ) sin nx − cos nx ⎤ ⎪⎫
+
⎨ ⎢
⎥
⎢
⎥ ⎬
π ⎪2 ⎣
n2
n2
⎦ −π 2 ⎣
⎦ 0 ⎪⎭
⎩
=
1 ⎧ −1
1
0
π⎫
⎨ 2 [n( π + x ).sin nx + cos nx ]−π + 2 [n(π − x ) sin nx − cos nx ]0 ⎬
π ⎩ 2n
2n
⎭
=
1 ⎧ −1
⎨
π ⎩ 2n2
∫
∫ ∫
∫ ∫
∫
∫
π
⎧⎪ − 1 ⎡ ( π + x ) sin nx cos nx ⎤ 0
1 ⎡ ( π − x) sin nx cos nx ⎤ ⎫⎪
+
+
−
⎨ ⎢
⎬
n
n
n 2 ⎥⎦ − π 2 ⎢⎣
n 2 ⎥⎦ 0 ⎪⎭
⎪⎩ 2 ⎣
{[n(π − 0)sin n(0) + cos n(0) ] − [( n(π − π) sin n( −π) + cos n( −π)]}
+
=
∫
1
2n
2
{[n(π − π) sin nπ – cos nπ] − [(n(π − 0)sin n(0) – cos n(0)]}⎫⎬⎭
1
1
{–1 + cos nπ – cosnπ + 1} = 0
2 {–[1 – cos(– nπ)] + [– cos nπ – (–1)]} =
2n π
2n 2 π
∴ an = 0
Similarly,
bn =
1
π
π
∫ f ( x).sin nx.dx
−π
=
π
π
⎤ 1 ⎡ 0 −1
1 ⎡0
⎤
1
+
+
π
(
x
).sin
nx
.
dx
(π − x) sin nx.dx ⎥
=
⎢ f ( x ).sin nx.dx + f ( x ) sin nx.dx ⎥
⎢∫
∫
π ⎢
2
0
⎦
⎥⎦ π ⎣ −π 2
0
⎣− π
=
1 ⎧ −1 ⎡ π + x sinnx dx − 1 sinnx dx dx⎤ 0 + 1 ⎡ π − x sinnx dx − −1 sinnx dx dx⎤ π ⎫
(
).
(
) ⎥
(
).
(
) ⎥ ⎬
⎨
⎦ −π 2 ⎣⎢
⎦0 ⎭
π ⎩ 2 ⎣⎢
=
1
π
∫
∫
∫
1
=
π
∫ ∫
∫
∫ ∫
π
0
⎧⎪ − 1 ⎡
( − cos nx)
( − cos nx) ⎤
1 ⎡ − ( π − x) cos nx
( − cos nx) ⎤ ⎫⎪
−
+
+
dx
⎨ ⎢( π + x )
⎥
⎢
⎥ ⎬
n
n
n
n
⎦ −π 2 ⎣
⎦ 0 ⎪⎭
⎪⎩ 2 ⎣
∫
∫
π
⎧⎪ − 1 ⎡ − (π + x) cos nx sin nx ⎤ 0
1 ⎡ − ( π − x ) cos nx sin nx ⎤ ⎫⎪
+
−
+
⎨ ⎢
⎢
⎥ ⎬
⎥
n
n
n 2 ⎦ −π 2 ⎣
n 2 ⎦ 0 ⎪⎭
⎪⎩ 2 ⎣
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.20
MATHEMATICS-II [JNTU-ANANTAPUR]
π
⎧⎪ − 1 ⎡ − n( π + x) cos nx + sin nx ⎤ 0
1 ⎡ − n ( π − x ) cos nx − sin nx ⎤ ⎫⎪
+
⎨ ⎢
⎥
⎢
⎥ ⎬
n2
n2
⎦ −π 2 ⎣
⎦ 0 ⎪⎭
⎪⎩ 2 ⎣
1
=
π
=
1 ⎧ −1
1
⎫
[ − n( π + x) cos nx + sin nx ]0−π + 2 [ − n ( π − x) cos nx − sin nx ]0π ⎬
⎨
π ⎩ 2n 2
2n
⎭
=
1 ⎧ –1
⎨
π ⎩ 2n 2
{[ –n(π + 0) cos n(0) + sin n(0)] − [ – n(π − π)cos n(–π) + sin n(–π)]}
+
=
=
∴ bn =
1
⎫
– n(π − π) cos nπ – sin nπ ] − [ – n(π − 0) cos n(0) – sin n(0)]}⎬
2 {[
2n
⎭
1
1
1 ⎧ −1
1
⎫
[nπ – sin nπ – sin nπ + nπ] =
(2nπ – 2sin nπ)
⎨ 2 [ − nπ + sin nπ] + 2 [ − sin nπ + nπ]⎬ =
2
2n 2 π
π ⎩ 2n
2n
⎭ 2n π
2 nπ
=
2
2n π
1
n
[Q sin nπ = 0]
1
n
On substituting the values of a0, an and bn in equation (1), we get,
∞
f (x) = 0 +
∑[(0)cos nx + b sin nx]
n
n =1
∞
f(x) =
∑
n =1
∞
bn sin nx =
⎛1⎞
∑ ⎜⎝ n ⎟⎠ sin nx
n =1
1
1
⎡
⎤
∴ f ( x ) = ⎢sin x + sin2 x + sin3 x + ...⎥
2
3
⎣
⎦
2.2 FOURIER SERIES EVEN, AND ODD FUNCTIONS
Q17. Give the Fourier series expansion for even periodic functions.
Ans: Generally, a periodic function f(x) defined in the interval (–π, π) is represented by,
f(x) =
a0
+
2
∞
∑
an cos nx +
n =1
∞
∑b
n
sin nx
... (1)
n =1
Where,
1
a0 =
π
an =
1
π
1
bn =
π
π
∫ f ( x )dx
... (2)
−π
π
∫ f ( x) cos nx dx
... (3)
−π
π
∫ f ( x) sin nx dx
... (4)
−π
When f(x) is an Even Function: From equation (2), we get,
1
a0 =
π
π
∫ f ( x) dx
−π
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2.21
UNIT-2 (Fourier Series)
2
a0 =
π
⇒
When f (x) is an Odd Function: From equation (2), we get,
π
∫ f ( x) dx
0
As ‘cos nx’ is an even function, f(x) cos nx is also an
even function.
∴ From equation (3), we get,
an =
1
π
2
an =
π
⇒
π
a0 =
an =
π
∫ f ( x) cos nx dx
Since, sin nx is an odd function and f(x) sin nx also
becomes an odd function
∴ From equation (4), we get,
1
π
1
bn =
π
1
bn = ∫ f (x)sin nx dx = 0
π −π
Therefore, it can be concluded that if a function f(x) is
even in the interval (–π, π), its Fourier series expansion consists
of only cosine terms.
⇒
bn =
2
π
cos nx
n =1
∴ f ( x) =
an =
2
π
π
0
Where, bn =
π
∫ f ( x) sin nx dx
−π
π
∫ f ( x) sin nx dx
0
∞
∑b
n
sin nx
2
π
π
∫ f ( x) sin nx dx
0
f ( x ) cos nx . dx
0
Q18. Give the Fourier series expansion for odd
periodic functions.
Ans: Generally, a periodic function f(x) defined in the interval
(–π, π) is represented by,
f(x) =
π
n =1
∫ f ( x)dx
∫
∫ f ( x) cos nx dx = 0
−π
Therefore, it can be concluded that if a function f(x) is
odd in the interval (–π, π), its Fourier series expansion consists
of only sine terms.
∞
Where,
2
π
π
∴ From equation (4), we get,
π
a0 =
(Q f(x) is odd)
Sin nx is an odd function, f (x) sin nx becomes an even
function.
0
n
∫ f ( x)dx = 0
−π
As ‘cos nx’ is an even function, f(x) cos nx becomes an
odd function
−π
∑a
π
∴ From equation (3), we get,
∫ f ( x) cos nx dx
a
∴ f ( x) = 0 +
2
1
π
a0
+
2
∞
∑a
∞
n cos nx +
n=1
∑ b sin nx
Q19. Obtain the Fourier series for the function
π , π ).
Model Paper-II, Q4
f(x) = |cos x| in (–π
Ans: Given equation is,
f(x)= | cos x |
Since, f(x) = | cos x | is an even function, its Fourier
series consists of cosine terms only
n
n =1
... (1)
Where,
a0 =
1
π
1
an =
π
bn =
1
π
∴
π
∫ f ( x )dx
f(x) =
a0
+
2
∞
∑a
n
cos nx
... (1)
n =1
... (2)
−π
Where,
π
∫ f ( x) cos nx dx
... (3)
−π
π
∫ f ( x) sin nx dx
... (4)
−π
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
1
a0 =
π
π
∫
−π
f ( x ) dx =
2
π
π
∫ f ( x)dx
0
π
π
⎡
⎤
⎢Q f ( x)is even function i.e., f ( x ) dx = 2 f ( x) dx ⎥
⎢
⎥
−π
0
⎣
⎦
∫
∫
SIA GROUP
2.22
MATHEMATICS-II [JNTU-ANANTAPUR]
π
π
⎤
⎤ 2 ⎡π 2
2⎡
⎢
⎥
⎢
⎥
|
cos
x
|
dx
cos
–
cos
.
x
dx
x
dx
=
=
⎥
π⎢
π⎢
⎥
π
⎣0
⎦
2
⎣0
⎦
∫
=
∴ a0 =
∫
[
∫
]
π
4
2
2
(sin x ) 02 – (sin x ) ππ = [1 + 1] =
2
π
π
π
4
π
1
an =
π
π
∫ f ( x) cos nx dx
−π
π
⎛ π
⎞
⎜Q f ( x )dx = 2 f ( x)dx ⎟
⎜
⎟
0
⎝ −π
⎠
π
=
2
| cos x | cos nx dx
π
∫
∫
0
∫
π
⎤
⎡π 2
2⎢
⎥
+
−
cos
x
cos
nx
dx
(
cos
x
)
cos
nx
dx
=
⎥
π⎢
π2
⎦
⎣0
∫
∫
⎡
⎢ Q cos x is positive in the interval
⎣
⎛ π ⎞⎤
⎛ π⎞
⎜ 0, ⎟ and negative in the interval ⎜⎝ 2 , π⎟⎠ ⎥
⎦
⎝ 2⎠
π
⎤
⎡π 2
1⎢
⎥
−
2
cos
nx
cos
x
dx
2
cos
nx
cos
x
dx
= π⎢
⎥
π2
⎦
⎣0
∫
∫
π
⎤
⎡π 2
1⎢
{cos( nx + x ) + cos( nx − x)}dx − {cos( nx + x ) + cos( nx − x)}dx⎥
=
⎥
π⎢
π2
⎦
⎣0
∫
∫
[Q 2 cos A cos B = cos (A + B) + cos (A – B)]
π
⎡π 2
⎤
1⎢
{cos( n + 1) x + cos( n − 1) x}dx − {cos( n + 1) x + cos( n − 1) x}dx⎥
=
⎥
π⎢
π2
⎣0
⎦
∫
∫
π2
π
⎡ π2
⎫⎤
⎫ ⎧π
1 ⎢⎧⎪
⎪
⎪ ⎪
= ⎢⎨ cos( n + 1) x dx + cos( n − 1) x dx ⎬ − ⎨ cos( n + 1) x dx + cos( n − 1) xdx ⎬⎥⎥
π ⎪
⎪⎭⎥⎦
π2
0
⎭⎪ ⎪⎩π 2
⎢⎣⎩ 0
∫
=
∫
∫
∫
π2
π
1 ⎡⎧ sin(n + 1) x sin(n − 1) x ⎫
⎧ sin(n + 1) x sin(n − 1) x ⎫ ⎤
+
+
⎢⎨
⎬ −⎨
⎬ ⎥
π ⎢⎩ n + 1
n − 1 ⎭0
n − 1 ⎭ π 2 ⎥⎦
⎩ n +1
⎣
⎡ ⎧⎛
⎫
π
π⎞
sin(n + 1)
sin( n − 1)
⎪
+
−
1 ⎢ ⎪⎪⎜
sin(
1).(0)
sin(
1)(0)
n
n
⎞⎪
2 +
2⎟ −⎛
+
= ⎢ ⎨⎜
⎜
⎟⎬
π ⎢⎪
n +1
n −1 ⎟ ⎝
n +1
n −1 ⎠ ⎪
⎜
⎟
⎠
⎢⎣ ⎩⎪⎝
⎭⎪
⎧
π
π ⎞ ⎫⎤
⎛
⎪⎪⎛ sin(n + 1)π sin(n − 1)π ⎞ ⎜ sin(n + 1) 2 sin(n − 1) 2 ⎟⎪⎪⎥
⎟ ⎬⎥
− ⎨⎜
+
+
⎟ −⎜
n −1 ⎠ ⎜
n +1
n − 1 ⎟ ⎪⎥
⎪⎝ n + 1
⎟ ⎪⎥
⎜
⎪⎩
⎠ ⎭⎦
⎝
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2.23
UNIT-2 (Fourier Series)
⎡ ⎧⎛
⎫ ⎧
π
π⎞
π
π ⎞ ⎫⎤
⎛
sin(n + 1).
sin( n − 1)
sin(n + 1)
sin(n − 1)
⎥
⎪⎪ ⎪⎪
1 ⎢ ⎪⎪⎜
⎟
⎜
2
2
2
2 ⎟ ⎪⎪ ⎥
⎢ ⎨⎜
+
− (0 + 0)⎬ − ⎨(0 + 0) − ⎜
+
⎬
⎟
⎟
= π ⎢⎪
n +1
n −1
n +1
n −1
⎪ ⎪
⎜
⎟
⎜
⎟ ⎪⎥
⎠
⎝
⎠ ⎪⎭ ⎦⎥
⎢⎣ ⎪⎩⎝
⎪⎭ ⎪⎩
[Q sinnπ = 0]
π
π
π
π⎤
⎡
sin( n − 1) ⎥
sin( n + 1)
sin( n − 1)
sin( n + 1)
1⎢
2
2 +
2 +
2 +
⎥ where n ≠ 1.
= ⎢
π⎢
n +1
n −1
n +1
n −1 ⎥
⎦⎥
⎣⎢
=
1⎡ 1 ⎧
π
π⎫
1 ⎧
π
π ⎫⎤
⎨sin(n + 1) + sin(n + 1) ⎬ +
⎨sin( n − 1) + sin(n − 1) ⎬ ⎥
⎢
2
2 ⎭ n –1 ⎩
2
2 ⎭⎦
π ⎣n +1 ⎩
=
1⎡ 1 ⎧
π⎫ 1 ⎧
π ⎫⎤
⎨2sin(n + 1) ⎬ +
⎨2sin(n − 1) ⎬⎥
⎢
2 ⎭ n −1 ⎩
2 ⎭⎦
π ⎣n +1 ⎩
=
2 ⎡ 1 ⎧ ⎛ nπ π ⎞ ⎫ 1 ⎧ ⎛ nπ π ⎞ ⎫⎤
⎨sin ⎜ + ⎟ ⎬ +
⎨sin ⎜ − ⎟ ⎬⎥
⎢
π ⎣ n + 1 ⎩ ⎝ 2 2 ⎠ ⎭ n − 1 ⎩ ⎝ 2 2 ⎠ ⎭⎦
=
2 ⎡ 1 ⎧ nπ
π
nπ π ⎫ 1 ⎧ nπ
π
nπ π ⎫⎤
⎨sin cos + cos sin ⎬ +
⎨sin cos − cos sin ⎬⎥
⎢
2
2
2 ⎭ n −1 ⎩ 2
2
2
2 ⎭⎦
π ⎣ n +1 ⎩ 2
=
2 ⎡ 1 ⎧sin nπ
⎛ nπ ⎞ ⎫ 1 ⎧ ⎛ nπ ⎞
⎛ nπ ⎞ ⎫⎤
(0) + cos ⎜ ⎟ (1)⎬ +
(0) − cos ⎜ ⎟ (1)⎬⎥
⎨
⎨sin
⎢
⎝ 2 ⎠ ⎭ n −1 ⎩ ⎜⎝ 2 ⎟⎠
⎝ 2 ⎠ ⎭⎦
π ⎣ n +1 ⎩ 2
=
2⎡ 1
⎛ nπ ⎞ 1 ⎧
⎛ nπ ⎞ ⎫⎤
cos ⎜ ⎟ +
⎨− cos ⎜⎝ ⎟⎠ ⎬⎥
2 ⎭⎦
π ⎣⎢ n +1 ⎝ 2 ⎠ n −1 ⎩
[Q sin(A + B) = sinA cosB + cosA sinB]
nπ
nπ ⎤
⎡
cos ⎥
cos
2⎢
2
2 −
= ⎢
⎥
π ⎢ n +1
n −1 ⎥
⎢⎣
⎥⎦
=
nπ 2 ⎡ n − 1 − (n + 1) ⎤
2⎡ 1
1 ⎤
nπ
cos
−
cos
= ⎢
⎥
⎥
⎢
2
π ⎣ n + 1 n −1⎦
π ⎣ (n + 1)(n − 1) ⎦
2
=
2 ⎡ −2 ⎤
nπ
nπ
2 ⎡ n − 1 − n − 1⎤
cos
= ⎢ 2 ⎥ cos
⎢
⎥
2
2
π
2
π ⎣ n −1 ⎦
⎣ n − 1⎦
∴ an =
−4
π (n − 1)
2
cos
nπ
if n ≠ 1
2
π
⎤
⎡π 2
2⎢
2
2
⎥
a1 = π ⎢ cos x dx – cos dx ⎥ = 0
π
2
⎦
⎣0
∫
∫
∴ a1 = 0
On substituting the values of a0, an in equation (1), we get,
∴ f ( x) =
2 4
−
π π
∞
∑n
n=2
1
2
−1
cos
nπ
cos nx
2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.24
MATHEMATICS-II [JNTU-ANANTAPUR]
Q20. Show that for – π < x < π , sinax =
2sinaπ ⎡ sinx
2sin2x 3sin3x
⎤
–
+
– ....⎥ (a is not an integer).
⎢ 2
π
⎦
⎣1 – a 2 22 – a 2 32 – a 2
Ans: Given equation is,
2 sin aπ ⎡ sin x 2 sin 2 x 3 sin 3 x
⎤
– 2 2 + 2 2 –....⎥ , for –π < x < π
⎢
2
2
π
2 –a
3 –a
⎣1 –a
⎦
since, sinax is an odd function, its Fourier series consists of sine terms only.
sinax =
∞
∴
sinax =
∑b
n
sin nx
... (1)
n =1
Where,
2
bn =
π
π
∫
π
f ( x) sin nx dx =
0
2
sin ax sin nx dx
π
∫
0
π
1
[cos( a – n ) x – cos( a + n) x ] dx
=
π
∫
[Q 2sinA sinB = cos(A – B) – cos(A + B)]
0
=
π
π
⎤
1⎡
⎢ [cos( a – n) x – [cos( a + n) x ⎥ dx
π⎢
⎥
0
⎦
⎣0
∫
∫
π
1 ⎡ sin(a – n) x sin(a + n) x ⎤ 1 ⎡ sin( a – n) π sin( a + n ) π ⎤
–
–
– 0⎥
= ⎢
=
π⎣ a–n
a + n ⎥⎦ 0 π ⎢⎣ a – n
a+n
⎦
=
=
1 ⎡ sin aπ cos nπ – cos aπ sin nπ sin aπ cos nπ + cos aπ sin nπ ⎤
–
⎥
π ⎢⎣
a–n
a+n
⎦
[Q sin(A + B) = sinA cosB + cosA sinB ; sin(A – B) = sinA cosB – cosA sinB]
1 ⎡ sin aπ cos nπ sin aπ cos nπ ⎤
–
⎥
π ⎢⎣
a–n
a+n
⎦
[Q sinnπ = 0]
1
1 ⎞ 1
⎛ 1
n⎛a+n–a+n⎞
–
⎟ = sin aπ (–1) ⎜
⎟
= sin aπ cos nπ ⎜
π
⎝a–n a+n⎠ π
⎝ a2 – n2 ⎠
=
∴ bn =
[Q cosnπ = (–1)n]
(–1) n 2n
sin aπ
π(a 2 – n 2 )
(–1) n 2n
π(a 2 – n 2 )
sin aπ
On substituting the value of ‘bn’ in equation (1), we get,
2 sin aπ
sinax =
π
∞
n(–1) n
sin nπ
∑
2
2
n =1 a – n
∞
⎤ 2 sin aπ ⎛ sin x
2 sin aπ ⎡ (–1) n +1 .n
2 sin 2 x 3 sin 3 x
⎞
sin nπ⎥ =
– 2
– ... ⎟
+ 2
⎢
⎜ 2
=
2
2
2
2
2
π ⎣⎢ n =1 n – a
π
2 –a
3 –a
⎝1 – a
⎠
⎦⎥
∑
2 sin 2 x 3 sin 3x
2 sin aπ ⎛ sin x
⎞
– ...⎟
– 2
+ 2
⎜ 2
2
2
2
π ⎝1 – a
2 –a
3 –a
⎠
Hence proved.
∴ sin ax =
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2.25
UNIT-2 (Fourier Series)
Q21. Find the Fourier series to represent the function
π < x < π.
f(x) = | sin x |, –π
=
1 ⎫ ⎧ 1
1 ⎫⎤
−1 ⎡
⎧ 1
(−1) n+1 ⎨
+
+
⎬−⎨
⎬
n
1
1
n
n
1
1
n ⎭⎥⎦
π ⎢⎣
+
−
+
−
⎩
⎭ ⎩
=
2
2 ⎤
−1 ⎡
( −1) n +1
−
⎢
2
π ⎣
1− n
1 − n 2 ⎥⎦
Ans: Since, f(x) = | sin x| is an even function, its Fourier series
consists of cosine terms only.
a0
+
f (x) =
2
∴
∞
∑a
cos nx
n
... (1)
n =1
−2
=
2
− ( n − 1) π
π
Where, a0 =
=
2
π
1
f ( x) dx
π −π
∫
=−
π
∫
f ( x ) dx
2[1 + (−1) n ]
,n>2
π(n 2 − 1)
an = 0, when n is odd,
[Q f (x) is even function]
And an = −
0
π
⎡ π
⎤
⎢Q f ( x ) dx = 2 f ( x) dx ⎥
⎢ −π
⎥
0
⎣
⎦
∫
[( −1) n ( −1)1 − 1]
∫
π
1
| sin x | cos x dx
If n = 1, a1 =
π −π
∫
π
=
2
sin x dx
π0
=
2
[− cos x] π0 = 2 [−(−1) + 1] = 4
π
π
π
4
, when n is even, n > 2
π( n 2 − 1)
∫
π
2
sin x cos x dx
=
π0
∫
π
1
1 cos 2 x
sin 2 x dx = −
2 0
π
π0
=
1
(1 − cos 2 π)= 0
2π
π
an =
1
f ( x ) cos nx dx
π −π
∫
π
π
=
∫
Hence the Fourier series can be written as,
2
f ( x ) cos nx dx
=
π0
∫
⇒
2 4
f (x) = −
π π
∴
2 4
f (x) = −
π π
π
2
sin x cos nx dx
=
π0
∫
=
1
π
cos 2 nx
2
−1
∑ (2n)
n =1
∞
cos 2n x
∑ (2n − 1)(2n + 1)
n =1
π
∫ [sin (n + 1) x − sin( n − 1) x]dx
0
π
=
∞
1 ⎡ − cos(n + 1) x cos(n − 1) x ⎤
+
n +1
n − 1 ⎥⎦ 0
π ⎢⎣
=
1 ⎡ − cos (n + 1)π cos (n − 1)π
1
1 ⎤
+
+
−
⎢
π⎣
n +1
n −1
n + 1 n − 1 ⎥⎦
=
− 1 ⎡ ( −1) − 1 (−1) − 1⎤
+
⎢
⎥
π ⎣ n +1
1− n ⎦
Q22. Obtain a Fourier expansion for
the interval − π < x < π .
Ans: f (x) =
1− cosx in
1− cos x in (–π, π)
Since f (x) =
1 − cos( − x ) = 1− cos x
∴
f (x) is even in (–π, π)
∴
bn = 0 for all n.
Fourier series of f (x) is given by,
n +1
n +1
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
a0
+
f (x) =
2
a
∑a
n
cos nx
n =1
SIA GROUP
2.26
MATHEMATICS-II [JNTU-ANANTAPUR]
Where,
1
a0 =
π
π
π
1
f (x) dx =
π
−π
∫
∫
−π
π
2
1− cos x dx =
π
∫
0
π
2 ⎡
x⎤
⎢
2 sin ⎥
1− cos x dx =
π ⎢ 0 2⎥
∫
⎣
⎦
x⎤
⎡
⎢Q 1 − cos x = 2 sin 2 ⎥
⎦
⎣
π
−4 2 ⎡
x⎤
cos ⎥
=
π ⎢⎣
2 ⎦0
a0 =
4 2
π
1
an =
π
2
an =
π
π
2
f (x)cos nx dx =
π
−π
∫
π
∫
0
π
∫
1− cos x cos nx dx
0
x
2 2
2 sin cos nx dx =
2
π
π
π
x
2
sin cos nx dx =
2
π
0
∫
⎡
⎤
∫ ⎢⎣sin ⎛⎜⎝ n + 2 ⎞⎟⎠ x − sin ⎛⎜⎝ n − 2 ⎞⎟⎠ x⎥⎦ dx
1
1
0
π
2 ⎡ − cos(n + 1 / 2) x + cos(n − 1 / 2) x ⎤
=
⎢
⎥
( n − 1 / 2) ⎦ 0
π ⎣ ( n + 1 / 2)
=
an =
2 ⎡ sin nπ + sin nπ + 1 − 1 ⎤
2
⎢ n + 1/ 2 n − 1/ 2 n + 1/ 2 n − 1/ 2 ⎥ =
⎦
π ⎣
π
n − 1/ 2 − n − 1/ 2 ⎤
⎡
⎢0 + 0 +
⎥
n 2 − 1/ 4
⎣
⎦
− 2
π( n 2 − 1 / 4 )
∴ Fourier series is given by,
f (x) =
2 2
+
π
∞
∑
n =1
2
2 2
=
–
π
π
− 2
cos nx
π( n 2 − 1 / 4)
∞
∑
n =1
1
cos nx
( n − 1 / 4)
2
2.3 FOURIER SERIES IN AN ARBITRARY INTERVAL
0 ≤ x ≤1
⎧ πx,
.
Q23. Obtain the Fourier series for the function f(x) = ⎨
π(2
−
x),
1
≤x≤2
⎩
Ans: Given that,
f(x) = πx when 0 ≤ x ≤ 1 and
f(x) = π(2 – x) when 1 ≤ x ≤ 2
Fourier series is,
f(x) =
∞
a0
nπx
an cos
+
+
2 n =1
l
∑
∞
∑b
n
n =1
sin
nπx
l
... (1)
∴2l = 2 ⇒ l = 1
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.27
UNIT-2 (Fourier Series)
Then equation (1) becomes,
f(x) =
a0
+
2
∞
∞
∑a
n cos nπx +
n =1
∑b
n =1
n
sin nπx
... (2)
Here,
1
a0 =
l
2l
∫
2
f ( x ) dx =
0
0
2
1
=
∫
∫ f ( x) dx
∫
πx dx + π( 2 − x) dx
1
0
1
2
⎡ ( 2 − x) 2 ⎤
⎡ x2 ⎤
= π ⎢ ⎥ + π⎢
⎥
⎢⎣ − 2 ⎥⎦1
⎢⎣ 2 ⎥⎦ 0
⎛ 1 0 ⎞ ⎡⎛ (2 − 2) 2
= π ⎜ − ⎟ + π⎢⎜⎜
⎝ 2 2 ⎠ ⎣⎢⎝ − 2
=
⎞ ⎛ (2 − 1) 2 ⎞⎤
⎟−⎜
⎟
⎟ ⎜ − 2 ⎟⎥
⎠ ⎝
⎠⎦⎥
1⎞ π
⎛
π
π
+ π ⎜0 + ⎟ =
+ =π
2⎠ 2
⎝
2
2
And,
2l
1
n πx
dx
f ( x) cos
an =
l
l 0
∫
2
=
∫ f ( x) cos
0
nπx
dx
1
[Q l = 1]
2
=
∫ f (x) cos nπx dx
0
1
=
∫
2
∫
f (x) cos nπx dx + f (x ) cos nπx dx
0
=
1
1
2
0
1
∫ πx cos nπx dx + ∫ π (2 − x). cos nπx dx
1
2
⎡
⎡ ⎛ sin nπx ⎞ ⎛ cos nπx ⎞ ⎤
⎛ sin nπx
⎛ cos nπx ⎞ ⎞⎤
⎟ − 1⎜ − 2 2 ⎟ ⎥ + π ⎢(2 − x)⎜⎜
= π ⎢ x⎜
− ( −1) ⎜ − 2 2 ⎟ ⎟⎟⎥
⎝ n π ⎠ ⎠ ⎦1
⎣ ⎝ nπ ⎠ ⎝ n π ⎠ ⎦ 0
⎝ nπ
⎣
2
1
⎡ (2 − x ) sin nπx cos nπx ⎤
⎡ x sin nπx cos nπx ⎤
=π⎢
− 2 2 ⎥
+ 2 2 ⎥ +π⎢
n
π
nπ
n π ⎦0
n π ⎦1
⎣
⎣
⎡⎛ (1) sin nπ(1) cos nπ(1) ⎞ ⎛ (0) sin nπ(0) cos nπ(0) ⎞⎤
+
+
= π ⎢⎜
⎟−⎜
⎟⎥
nπ
nπ
n2π2 ⎠ ⎝
n 2 π 2 ⎠⎦
⎣⎝
⎡⎛ (2 − 2) sin nπ(2) cos nπ(2) ⎞ ⎛ (2 − 1) sin nπ(1) cos nπ(1) ⎞⎤
−
−
⎟−⎜
⎟⎥
+ π ⎢⎜
nπ
nπ
n2 π2 ⎠ ⎝
n 2 π 2 ⎠⎦
⎣⎝
⎡⎛
cos nπ ⎞ ⎛
1
= π ⎢⎜ 0 + 2 2 ⎟ − ⎜ 0 + 2 2
π
π
n
n
⎝
⎠
⎝
⎣
⎞⎤
⎟⎥ + π
⎠⎦
⎡⎛
cos 2nπ ⎞ ⎛
cos nπ ⎞⎤
⎢⎜ 0 − 2 2 ⎟ − ⎜ 0 − 2 2 ⎟⎥ [Q cos 2nπ = (–1)2n = 1]
n π ⎠ ⎝
n π ⎠⎦
⎣⎝
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.28
MATHEMATICS-II [JNTU-ANANTAPUR]
1
cos 2nπ cos nπ ⎤
⎡ cos nπ
2× π
= π ⎢ 2 2 − 2 2 − 2 2 + 2 2 ⎥ = 2 2 [cos nπ – 1]
n π
n π
n π ⎦ n π
⎣n π
2π
=
∴
n 2 π2
[Q cos nπ = (–1)n]
[(–1)n – 1]
n is even
⎧0,
⎪
a n = ⎨ − 4π
⎪ 2 2 , n is odd
⎩n π
1
l
bn =
2l
∫
f ( x ) sin
0
2
1
nπx
dx =
l
1
∫ f ( x) sin
0
nπx
dx
1
[Q l = 1]
2
=
∫ f ( x) sin nπx dx
0
2
1
=
∫
∫
πx. sin nπx dx + π( 2 − x) sin nπx dx
1
0
2
1
⎡
⎡ ⎛ − cos nπx ⎞ ⎛ − sin nπx ⎞⎤
⎛ − cos nπx ⎞
⎛ − sin nπx ⎞⎤
⎟ − (−1)⎜
⎟⎥
= π ⎢ x⎜
⎟⎥ + π ⎢(2 − x )⎜
⎟ − 1⎜
2 2
nπ ⎠
nπ ⎠ ⎝ n π ⎠ ⎦ 0
⎝
⎝ n 2 π 2 ⎠⎦1
⎣
⎣ ⎝
1
2
⎡ cos nπx cos nπx ⎤
cos nπx sin nπx ⎤
⎡ x cos nπx sin nπx ⎤
⎡
+
= π ⎢−
+ 2 2 ⎥ + π⎢(2 − x )
− 2 2 ⎥ = π ⎢−
nπ
nπ ⎥⎦
⎣
nπ
nπ
n π ⎦1
n π ⎦0
⎣
⎣
∴ bn = 0
On substituting the above values in equation (2), we get,
π
4π
f(x) =
– 2
2
π
=
∞
cos nπx
n2
n =1,3,5,7....
∑
1
1
π 4π ⎡ 1
⎤
− 2 ⎢ 2 cos πx + 2 cos 3πx + 2 cos 5πx + ...⎥
2 π ⎣1
5
3
⎦
Q24. Find the Fourier series corresponding to the following f(x) defined in (–2, 2) as follows,
⎧2, − 2 ≤ x ≤ 0
of f(x) in (–2, 2).
f(x) = ⎨
0 <x<2
⎩ x,
Ans: Given function is,
⎧2, − 2 ≤ x ≤ 0
f(x) = ⎨
⎩ x, 0 < x < 2
a0
+
Let, f(x)=
2
∞
∑
a n cos
n =1
nπx
+
l
∞
∑b
n =1
n
sin
n πx
in [–l, l]
l
Here,
l =2
∴
1
a0 =
l
l
∫ f ( x) dx
−l
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
... (A)
2.29
UNIT-2 (Fourier Series)
1
=
2
=
2
∫
1⎡
⎢ f ( x ) dx +
2⎢
⎣−2
0
f ( x ) dx =
−2
∫
2
∫
0
2
2
⎤ 1⎡
⎤ 1⎡0
⎡ x2 ⎤ ⎤
0
⎢
⎢
⎥
⎥
+
2
dx
x
dx
[ 2 x] + ⎢ ⎥ ⎥
f ( x ) dx =
=
−2
2⎢
⎥ 2⎢
⎥
⎢⎣ 2 ⎥⎦ 0 ⎥
0
⎣− 2
⎦
⎦
⎦
⎣
∫
∫
4⎤ 6
1⎡
2( 2 ) + ⎥ = = 3
⎢
2⎦ 2
2⎣
... (1)
∴ a0 = 3
∴
1
an =
2
=
2
∫ f ( x) cos
−2
nπx
dx
2
0
2
0
2
1⎡
nπx
nπx ⎤ 1 ⎡
nπx
nπx ⎤
⎢ f ( x) cos
dx + f ( x) cos
dx ⎥ = ⎢2 cos
dx + x cos
dx ⎥
2⎢
2
2
⎥ 2⎢
2
2
⎥
0
⎣− 2
⎦
−
2
0
⎦
⎣
∫
∫
⎡
nπx
2 sin
1 ⎢⎢
2
=
2 ⎢ nπ
⎢
2
⎣
0
nπx
2
nπ
2
∫
2
x sin
+
−2
2
−
∫
0
0
∫
⎡
nπx ⎤
nπx
0
− cos
⎥ 1⎢ 4
n
x
2
2
π
2
2 dx ⎥ = ⎢ sin
( 2 sin nπ − 0) −
+
nπ
2 −2 nπ
nπ
nπ
⎥ 2 ⎢ nπ
⎢
2
⎥
⎣
2
⎦
sin
2⎤
⎥
⎥
⎥
⎥
0⎦
⎡ 4
[sin 0 − sin(nπ)]+ 2 (0 − 0)⎤⎥ 1 4
1 ⎢ nπ
⎡
⎤
nπ
⎥ = ⎢ 2 2 (cos nπ − 1)⎥
= ⎢
4
2⎢
⎥
2 ⎣n π
⎦
+
(cos nπ − cos 0)
⎢⎣ n 2 π 2
⎥⎦
∴
an =
2
2
(cos nπ − 1) = 2 2 ((−1) n − 1)
2
n π
n π
2
⎧ 0
⎪
an = ⎨ − 4
⎪⎩ n 2 π 2
∴
1
bn =
2
2
when n is even
∫ f ( x) sin
−2
nπx
dx
2
1⎡
nπx
⎢ f ( x ) sin
dx +
2⎢
2
⎣− 2
0
=
... (2)
when n is odd
∫
2
∫ f ( x) sin
0
nπx ⎤
dx ⎥
2
⎥⎦
0
2
nπx
nπx ⎤
1⎡
dx + x sin
dx ⎥
= ⎢ 2 sin
2 ⎢−2
2
2
⎥⎦
0
⎣
∫
⎡ ⎛
n πx ⎞
⎢ 2⎜ − cos
⎟
1⎢ ⎝
2 ⎠
= ⎢
nπ
2
⎢
2
⎢⎣
∫
0
2
⎤
nπx ⎞
⎛
nπx ⎥
x⎜ − cos
2 − cos
⎟
2 ⎠
2 dx ⎥
+ ⎝
−
⎥
nπ
nπ
0
⎥
2
2
⎥⎦
0
∫
−2
⎡
nπx
sin
⎢
1 ⎢− 4
2
2
2
[cos 0 − cos nπ] −
2 cos nπ +
=
nπ
nπ nπ
2 ⎢ nπ
⎢
2
⎣
2⎤
⎥ 1 −4
4
⎤
⎥ = ⎡
[1 − cos nπ] −
cos nπ + 0⎥
⎢
⎥ 2 ⎣ nπ
nπ
⎦
⎥
0⎦
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.30
MATHEMATICS-II [JNTU-ANANTAPUR]
=
∴ bn =
1 ⎡− 4 4
4
⎤
cos nπ −
cos nπ ⎥
+
nπ
2 ⎢⎣ nπ nπ
⎦
−2
nπ
... (3)
On substituting equations (1), (2) and (3) in equation (A), we get,
3
f (x) =
–
2
( 2 n −1)
∞
−2
nπx
nπx
3
sin
cos
+
=
–
2 2
2
nπ
2
2
n =1
n =1,3,5 n π
∑
∑
4
( 2 n −1)
nπx
4
−
cos
2 2
2
n =1,3,5 n π
∑
∞
∑ nπ sin
n =1
2
nπx
2
⎧0, if − 2 < x < −1
⎪
Q25. Find the Fourier series for the functions f(x) = ⎨ K, if − 1 < x < 1 .
⎪0, if1 < x < 2
⎩
Ans: The Fourier series is,
f(x) =
a0
+
l
∞
⎛ nπx ⎞
a n cos⎜
⎟+
⎝ l ⎠
n =1
∑
∞
∑b
n
n =1
⎛ n πx ⎞
sin ⎜
⎟
⎝ l ⎠
... (1)
l
1
f ( x) dx
Where, a0 =
l −l
∫
Here period 2l = 4
⇒
l =2
a0 =
2
1
2
⎤ 1 ⎡ −1
⎤
1⎡
⎢ f ( x) dx ⎥ = ⎢ ∫ f ( x )dx + ∫ f ( x)dx + ∫ f ( x )dx⎥
2 ⎣⎢− 2
−1
1
⎦⎥
⎦⎥ 2 ⎢⎣− 2
∫
=
−1
1
2
⎤
1⎡
⎢ 0.dx + Kdx + 0 dx ⎥
2 ⎣⎢ − 2
−1
1
⎦⎥
=
1
1
[ Kx]
−1
2
=
K
[1 − (−1)]
2
∫
∫
∫
∴ a0 = K
l
an =
1
⎛ nπx ⎞
f ( x) cos⎜
⎟ dx
l −l
⎝ l ⎠
=
1
⎛ nπx ⎞
f ( x ) cos⎜
⎟dx
2 −2
⎝ 2 ⎠
∫
2
∫
−1
1
2
1⎡
⎛ nπx ⎞
⎛ nπx ⎞
⎛ nπx ⎞ ⎤
⎟dx + f ( x) cos⎜
⎟dx + f ( x) cos⎜
⎟dx ⎥
= ⎢ f ( x) cos⎜
2 ⎢−2
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠ ⎥⎦
−1
+1
⎣
∫
∫
Look for the SIA GROU P LOGO
∫
on the TITLE COVER before you buy
2.31
UNIT-2 (Fourier Series)
=
−1
1
2
1⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
⎛ nπx ⎞
⎟ dx ⎥
⎟dx + 0. cos⎜
⎟ dx + K . cos⎜
⎢ 0. cos⎜
2 ⎣⎢ − 2
⎝ 2 ⎠ ⎦⎥
⎝ 2 ⎠
⎝ 2 ⎠
−1
1
∫
∫
∫
1
⎤
⎡
⎛ ⎛ nπx ⎞ ⎞
⎥
⎢
⎜ sin ⎜
⎟⎟
1⎢
2 ⎠⎟
⎝
⎜
0+ K
=
+ 0⎥
⎥
⎟
⎜
nπ
2⎢
⎥
⎢
⎟
⎜
2
⎠ −1
⎝
⎦⎥
⎣⎢
=
1 2K
.
2 nπ
⎡
⎛ nπ ⎞⎤ 2K ⎛ nπ ⎞
sin⎜ ⎟
⎢ 2 sin ⎜ 2 ⎟⎥ =
nπ ⎝ 2 ⎠
⎝ ⎠⎦
⎣
l
bn =
1
⎛ nπx ⎞
f ( x ) sin ⎜
⎟dx
l −l
⎝ l ⎠
∫
1
=
1
⎛ nπx ⎞
K . sin⎜
⎟ dx
2 −1
⎝ 2 ⎠
∫
1
⎡ ⎛ nπx ⎞ ⎤
cos⎜
⎟
K ⎢ ⎝ 2 ⎠⎥
⎥
=− ⎢
2 ⎢ ⎛ nπ ⎞ ⎥
⎢ ⎜⎝ 2 ⎟⎠ ⎥
⎣
⎦ −1
=−
2K
[ 0] = 0
2 nπ
On substituting the corresponding values in equation (1), we get,
f(x) =
K
+
2
∞
2K
⎛ nπ ⎞
⎛ nπx ⎞
⎟
2 ⎠
∑ nπ sin ⎜⎝ 2 ⎟⎠ cos⎜⎝
n =1
This is the required fourier series.
⎧ 2
⎪⎪ x
Q26. Find the Fourier series of the function f(x) = ⎨ 2
⎪π
⎪⎩ 4
π
π
<x<
2
2
.
π
3π
if < x <
2
2
if −
Ans: Consider the Fourier series,
f(x) =
nπx
a0 ∞
nπx ∞
+ Σ an cos
+ Σ bn sin
n =1
l
l
2 n =1
But the length of the interval is 2π,
⇒
l = π/2
∴ The Fourier series reduces to,
f(x) =
∞
a0 ∞
+ Σ an cos 2nx + Σ bn sin 2 nx
n=1
2 n =1
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.32
MATHEMATICS-II [JNTU-ANANTAPUR]
Where,
3π / 2
1
a0 =
f ( x ) dx
l −π / 2
∫
3π
⎡ π2
⎤
2
2⎢
π2 ⎥
2
x dx +
dx
a0 =
4 ⎥
π⎢ π
π
2
⎦
⎣− 2
∫
2 ⎡ x3 ⎤
= ⎢ ⎥
π ⎣⎢ 3 ⎦⎥
=
a0 =
⎛
⎜⎝Q l =
∫
π
π⎞
⎟
2⎠
2 ⎡ π π ⎤ π ⎡ 3π π ⎤
2 π2 3π
+ . [x ]π 2 =
⎢ + ⎥+ ⎢ − ⎥
2
4
π
3π ⎣ 8
8 ⎦ 2 ⎣ 2 2⎦
−π 2
3
2
3
2 2π 3 π 2 π π 2 π 2
+ .
+
=
6
2
3π 8
2 2
4π 2
6
⇒ a0 =
2π2
3
⇒
a0 π 2
=
2
3
3π
1 2
nπx
dx
f ( x ) cos
an =
l
l −π
∫
2
π
3π
2
2
⎫⎪
2 ⎧⎪ 2 2
π
cos 2nx dx ⎬
= ⎨ x cos 2nx dx +
4
π ⎪− π
⎪⎭
−π 2
⎩ 2
∫
∫
π⎞
⎛
⎜Q l = ⎟
2⎠
⎝
π
3π
2
2 ⎡ 2 ⎛ sin 2nx ⎞
π ⎡ sin 2nx ⎤ 2
⎛ − cos 2nx ⎞ ⎛ − sin 2nx ⎞⎤
2
+
+
⎟ − 2 x⎜
⎟
⎜
⎟
= ⎢x ⎜
⎥
2
2
⎢
⎥
π ⎣ ⎝ 2n ⎠
⎝ 4n
⎠ ⎝ 8n
⎠ ⎦ − π 2 ⎣ 2n ⎦ π 2
2
=
2 ⎡ π2
π π2
1
⎛ − π ⎞⎤
n
sin
2
− sin 2n⎜
⎟⎥ + 2
⎢
π ⎣ 8n
2 8n
⎝ 2 ⎠⎦ πn
–
⎡π
π π
⎛ − π ⎞⎤
⎢ 2 cos 2n 2 + 2 cos 2n⎜ 2 ⎟⎥
⎝
⎠⎦
⎣
⎡
π
π⎤
3π
⎛ − π ⎞⎤ π ⎡
⎢sin 2 n 2 − sin 2n⎜ 2 ⎟ ⎥ + 4n ⎢sin 2n 2 − sin 2n 2 ⎥
2πn ⎣
⎦
⎠⎦
⎝
⎣
1
2
=
⎤
2 ⎡ 2π2
1 ⎡ 2π
1
π
⎤
sin nπ⎥ + 2 ⎢ cos nπ⎥ − 2 [2 sin nπ] + [sin 3nπ − sin nπ]
⎢
4n
π ⎢⎣ 8n
⎦ 2n π
⎥⎦ πn ⎣ 2
=
π
π
π
1
1
sin nπ + 2 cos nπ − 2 sin nπ − sin nπ + sin 3nπ
2n
4n
4n
πn
n
π
1 ⎞
1
⎛ π
− 2 ⎟ sin nπ + 2 cos nπ + sin 3nπ
=⎜
4n
n
⎝ 4n πn ⎠
∴ sin n π = 0
=
1
cos nπ
n2
Look for the SIA GROU P LOGO
sin 3n π = 0
cos n π = ( −1) n
on the TITLE COVER before you buy
2.33
UNIT-2 (Fourier Series)
an =
1
( −1) n
2
n
3π
1
nπx
f ( x) sin
dx
Finally, bn =
l −π
l
2
∫
2
3π
2
2
f ( x) sin 2nx dx
=
π −π
∫
(Q l = π / 2)
2
π
3π
2
2 π2
sin 2nx dx
x 2 sin 2nx dx +
=
π −π
π +π 4
2
∫
2
∫
2
2
π
3π
2 ⎡ 2 ⎛ − cos 2 nx ⎞
π ⎡ − cos 2nx ⎤ 2
⎛ − sin 2nx ⎞ ⎛ cos 2nx ⎞ ⎤
+
+
2
⎟ − 2 x⎜
⎟
⎜
⎟
= ⎢x ⎜
⎥
2
3
⎢
⎥
π⎣ ⎝
2n
⎠
⎝ 4n
⎠ ⎝ 8n ⎠ ⎦ −π 2 ⎣ 2 n ⎦ π 2
2
2
=
2 ⎡ − π2
1
π
π π2
1
⎛−π⎞
⎛ − π ⎞⎤
cos 2n + 0 + 3 cos 2n + cos 2n⎜
⎟ + 0 − 3 cos 2n⎜
⎟⎥
⎢
2
2 8n
π ⎣⎢ 8n
4n
4n
⎝ 2 ⎠
⎝ 2 ⎠⎦⎥
−
π ⎡
π⎤
3π
cos 2n − cos 2n ⎥
⎢
2
2⎦
4n ⎣
=
⎤ π
2 ⎡ − π2
1
1
π2
cos nπ + 3 cos nπ + cos nπ − 3 cos nπ⎥ − [cos 3nπ − cos nπ]
⎢
8n
π ⎢⎣ 8n
4n
4n
⎥⎦ 4n
=
2
π
[0 + 0] − [cos 3nπ − cos nπ]
π
4n
[
−π
(−1) 3n − ( −1) n
bn =
4n
]
⎡Q cos 3n π = ( −1) 3n ⎤
⎢
⎥
⎢⎣ cos nπ = ( − 1) n ⎥⎦
bn = 0, if n is even
On substituting a0, an and bn in equation (1), we get,
f(x) =
∴
∞
π 2 ∞ ( −1) n
cos 2nx + Σ 0. sin 2nx
+Σ
2
n =1
3 n =1 n
f(x) =
π 2 ∞ (−1) n
+Σ
cos 2nx
3 n =1 n 2
f(x) =
π 2 cos 2 x cos 4 x cos 6 x cos 8 x
− 2 +
− 2 +
+ ...
3
1
22
3
42
Q27. Find Fourier series of period 3 to represent f(x) = 2x – x2 in (0, 3). Also find Fourier series of period 2 to
represent f(x) = 2x – x2 in (0, 2).
Ans: The first step in finding the Fourier series of period 2l is to represent f(x) = 2x – x2 in (0, 2l) and deduce the required results.
f(x) =
a0 ∞
nπx ∞
nπx
+ Σ an cos
+ Σ bn sin
n =1
2 n =1
l
l
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (1)
SIA GROUP
2.34
MATHEMATICS-II [JNTU-ANANTAPUR]
2l
1
f ( x) dx
Here, a0 =
l0
∫
2l
1
(2 x − x 2 )dx
=
l0
∫
2l
1 ⎡ 2 x 2 x3 ⎤
= ⎢
− ⎥
l⎣ 2
3 ⎦0
a0 =
1 ⎡ 2 8l 3 ⎤
⎥
⎢4l −
l⎣
3 ⎦
⎛ 2l 2 ⎞
a0
⎟
= 2l ⎜⎜ l −
2
3 ⎟⎠
⎝
... (2)
2l
And
1
nπx
f ( x) cos
dx
an =
l0
l
∫
2l
1
nπx
( 2 x − x 2 ) cos
dx
=
l0
l
∫
2l
⎤
nπx
nπx ⎤
1⎡
⎡
2
x
x
dx
x
dx
dx ⎥
−
−
−
⎢
(
2
)
cos
(
2
2
)
cos
=
⎢
⎥
l⎢
l
l
⎣
⎦ 0 ⎥⎦
⎣
∫
∫
∫
2l
⎡
nπx
nπx ⎤ ⎤
⎡
sin
sin
⎢
1⎢
2
l − (2 − 2 x )
l ⎥dx ⎥⎥
= ⎢(2 x − x ) n
⎢
nπ ⎥ ⎥
π
l⎢
⎢
⎥
l
l
⎣
⎦ ⎦⎥ 0
⎣⎢
∫
⎡
⎡
⎛
⎞
nπx ⎢
⎜ − cos nπx ⎟
⎢
sin
⎜
1⎢
2
l − ⎢(2 − 2 x)
l ⎟−
= ⎢(2 x − x )
⎜
⎟
2
⎢
nπ
l
⎜ ⎛⎜ nπ ⎞⎟ ⎟
⎢
⎢
⎜
⎟
l
⎢
⎢⎣
⎝ ⎝ l ⎠ ⎠
⎣
2l
∫
⎛
⎞ ⎤⎤
⎜ cos nπx ⎟ ⎥ ⎥
⎜
⎟ ⎥⎥
l
dx
⎜2
2 ⎟ ⎥⎥
⎜ ⎛⎜ nπ ⎞⎟ ⎟ ⎥ ⎥
⎜
⎟
⎝ ⎝ l ⎠ ⎠ ⎥⎦ ⎥⎦ 0
2l
⎡
⎡
⎤⎤
⎛
⎞
nπx ⎢
nπx ⎥ ⎥
⎜ − cos nπx ⎟
⎢
sin
sin
1⎢
2
l − ⎢(2 − 2 x )⎜
l ⎟−2
l ⎥⎥
= ⎢(2 x − x )
⎜
⎟
2
3 ⎥⎥
⎢
nπ
l
⎛ nπ ⎞ ⎥ ⎥
⎜ ⎛⎜ nπ ⎞⎟ ⎟
⎢
⎢
⎜
⎟
⎜
⎟
l
⎢
⎢⎣
⎝ l ⎠ ⎥⎦ ⎥⎦ 0
⎝ ⎝ l ⎠ ⎠
⎣
2l
⎡
⎡
⎛
⎞
⎛
⎞⎤ ⎤
nπx ⎢
⎜ − cos nπx ⎟
⎜ sin nπx ⎟⎥ ⎥
⎢
sin
1⎢
2
l − ⎢(2 − 2 x)⎜
l ⎟ + (−2)⎜
l ⎟⎥ ⎥
= ⎢(2 x − x )
⎜
⎜
2 ⎟
3 ⎟⎥ ⎥
⎢
nπ
l
⎜ ⎛⎜ nπ ⎞⎟ ⎟
⎜ ⎛⎜ nπ ⎞⎟ ⎟⎥ ⎥
⎢
⎢
⎜
⎟
⎜ l
⎟
l
⎢
⎢⎣
⎝ ⎝ l ⎠ ⎠
⎝ ⎝ ⎠ ⎠⎥⎦ ⎥⎦ 0
⎣
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.35
UNIT-2 (Fourier Series)
⎡
⎛
⎞⎤
⎜
⎟⎥
⎢
⎤
⎜ − cos 2nπ
l2
1
2nπ ⎟ ⎥ ⎡
2 sin 2nπ ⎢
2
sin
0
(
0
2
)
−
−
−
−
− 0⎥
= ( 4l − 4l ) nπ − ⎢(2 − 4l )⎜
⎢
⎟
2
3 ⎥
2 2
l
n π
⎥⎦
⎛ nπ ⎞ ⎟⎥ ⎣⎢
⎜ ⎛ nπ ⎞
⎢
⎜ ⎟ ⎟
⎜ ⎜ l ⎟
l
⎥
⎢⎣
l
⎝ ⎠ ⎠⎦
⎝ ⎝ ⎠
=
1⎡
l2
2l 2 ⎤
⎢0 + 2(1 − 2l ) 2 2 − 0 − 2 2 ⎥
l ⎣⎢
n π
n π ⎦⎥
=
2l
[1 − 2l − 1]
n 2π2
an = –
4l 2
n2 π2
2l
Finally, bn =
sin 2 n π = 0 sin 0 = 0
cos 2 n π = 1 cos 0 = 1
... (3)
2l
1
nπx
1
nπx
( 2 x − x 2 ) sin
dx
f ( x) sin
dx =
l0
l
l0
l
∫
∫
⎡
nπx ⎞
⎛
⎜ − cos
⎟
1⎢
2 ⎜
l ⎟−
= ⎢(2 x − x )
nπ
l⎢
⎜
⎟
⎜
⎟
⎢⎣
l
⎝
⎠
2l
∫
⎡
nπx ⎞⎤ ⎤
⎛
⎜ − cos
⎟ ⎥
⎢
l ⎟⎥⎥ dx ⎥
⎢ (2 − 2 x ) ⎜
nπ
⎜
⎟⎥ ⎥
⎢
⎜
⎟ ⎥
l
⎝
⎠⎦⎥ ⎦ 0
⎣⎢
⎡
⎡
⎞
⎛
nπx ⎞ ⎢
⎛
⎜ − sin nπx ⎟
⎢
−
cos
⎜
⎟
1⎢
2
l ⎟ − ⎢(2 − 2 x )⎜
l ⎟−
= ⎢(2 x − x )⎜
⎜
2 ⎟
nπ
l⎢
⎟ ⎢
⎜
⎜ ⎛⎜ nπ ⎞⎟ ⎟
⎟ ⎢
⎜
⎟
⎜
l
⎝
⎠ ⎢
⎢
⎝ ⎝ l ⎠ ⎠
⎣
⎣
2l
∫
⎡ ⎛
⎞⎤ ⎤ ⎤
⎢ ⎜ sin nπx ⎟⎥ ⎥ ⎥
⎟⎥ ⎥ ⎥
⎢2⎜
l
⎢ ⎜ π 2 ⎟⎥ ⎥ dx ⎥
⎢ ⎜ ⎛⎜ n ⎞⎟ ⎟⎥ ⎥ ⎥
⎢⎣ ⎜⎝ ⎝ l ⎠ ⎟⎠⎥⎦ ⎥⎦ ⎥
⎦0
2l
⎡
⎡
⎛
⎞
⎛
⎞⎤ ⎤
nπ x ⎞ ⎢
⎛
⎜ − sin nπx ⎟
⎜ cos nπx ⎟⎥ ⎥
⎢
−
cos
⎟
⎜
1⎢
2
l ⎟ − ⎢(2 − 2 x)⎜
l ⎟ + (−2)⎜
l ⎟⎥ ⎥
⎜
⎜
= ⎢(2 x − x )⎜
⎢
2 ⎟
3 ⎟⎥ ⎥
nπ
l
⎟
⎜
⎜ ⎛ nπ ⎞ ⎟
⎜ ⎛ nπ ⎞ ⎟⎥ ⎥
⎢
⎟ ⎢
⎜
⎜ ⎜ l ⎟ ⎟
⎜ ⎜ l ⎟ ⎟⎥ ⎥
l
⎠ ⎢
⎝
⎢
⎝ ⎝ ⎠ ⎠
⎝ ⎝ ⎠ ⎠⎦ ⎦ 0
⎣
⎣
⎡
⎛
⎛ − cos nπx ⎞
⎜
⎢
⎜
⎟
⎜
1⎢
2 ⎜
l
⎟ − ( 2 − 2 x )⎜
= ⎢(2 x − x )
nπ
l
⎜
⎟
⎜
⎜
⎟
⎢
⎜
l
⎝
⎠
⎢⎣
⎝
=
bn =
⎛
⎞
− sin nπx ⎟
⎜
⎜
⎟
l
+ (−2)⎜
2 ⎟
⎛ nπ ⎞ ⎟
⎜
⎜ ⎟ ⎟
⎜
⎝ l ⎠ ⎠
⎝
2l
⎤
− cos nπx ⎞⎟⎥
⎟⎥
l
3 ⎟⎥
⎛ nπ ⎞ ⎟ ⎥
⎜ ⎟ ⎟
⎝ l ⎠ ⎠⎥⎦ 0
⎤ 1⎡− l
1⎡ l
2l 2
⎤
2
4l (1 − l )⎥
⎢− (4l − 4l ) cos 2nπ − 0 − 3 3 (cos 2nπ − cos( 0))⎥ = ⎢
l ⎢⎣ nπ
nπ
⎥⎦ l ⎣ nπ
⎦
− 4l(1− l)
nπ
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (4)
SIA GROUP
2.36
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting equations (2), (3) and (4) in equation (1), we get,
⎛ 2l ⎞
f(x) = 2l ⎜1 − ⎟ +
3⎠
⎝
∞
nπx
− 4l 2
+
cos
2 2
l
n =1 n π
∑
∞
nπx
− 4l (1 − l )
sin
nπ
l
n =1
∑
2 ∞
⎛ 2l ⎞ 4l
2x – x2 = 2l ⎜1 − ⎟ − 2
3⎠ π
⎝
∞
nπx 4l (1 − l )
nπx
1
1
−
cos
sin
2
l
l
π n=1 n
n =1 n
∑
∑
... (5)
To represent the Fourier series with period 3 in (0, 3), substitute 2l = 3 or l = 3/2 value in equation (5), we get,
⎛
6⎜1 −
n
x
π
3
9
1
2
⎞
⎛
− ⎝
cos
2x – x2 = 3⎜1 − ⎟ − 2
2
π
3
⎝ 3 ⎠ π n =1 n
∞
∑
⇒
2x – x2 =
9
π2
∞
3⎞
⎟
2⎠
∞
1
∑ n sin
n =1
2nπx
3
1
2nπx 3 ∞ 1
2nπx
cos
sin
+
2
3
π n =1 n
3
n =1 n
∑
∑
... (6)
Equation (6), is the required Fourier series of period 3.
To represent the Fourier series with period 2 in (0, 2)
Let us substitute 2l = 2
⇒
l = 1 in equation (5), we get,
⇒
2x – x2 = 2⎜1 −
⇒
⎛
⎝
2x –
x2
2⎞ 4
⎟−
3 ⎠ π2
2 4
= − 2
3 π
∞
1
∑n
n =1
2
∞
1
4(1 − 1) ∞ 1
cos
n
x
sin nπx
π
−
2
π n =1 n
n =1 n
∑
∑
cos nπx
This is the required result.
2.4
EVEN AND ODD PERIODIC CONTINUATION HALF-RANGE FOURIER SINE AND COSINE
EXPANSIONS
Model Paper-I, Q5
Q28. Find the half-range sine series of f(x) = (x – 1)2 in the interval (0, 1).
Ans: Given that,
f (x) = (x – 1)2 interval (0, 1)
The half-range fourier sine series is given by,
∞
⎛ nπx ⎞
f (x) = (x –1)2 = ∑ bn sin ⎜
⎟
n =1
⎝ l ⎠
∞
= ∑ bn sin (nπx)
n =1
2
Where, bn =
l
2
=
1
l
[Q l = 1]
⎛ nπx ⎞
⎟ dx
l ⎠
∫ f ( x).sin⎜⎝
0
1
∫ f ( x).sin(nπx)dx
[Q l = 1]
0
1
=
2
( x − 1) 2 . sin (nπx )dx
1
∫
0
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.37
UNIT-2 (Fourier Series)
1
2
= 2 ( x − 2 x + 1). sin (nπx )dx
∫
0
1
1
⎡1
⎤
2
⎢
π
−
π
+
2
x
sin(
n
x
)
dx
2
x
sin(
n
x
)
dx
sin(nπx)dx ⎥
=
⎢
⎥
0
0
⎣0
⎦
∫
∫
∫
Let, bn = 2 [I1 + I2 +I3]
1
Consider, I1 =
∫x
2
sin( nπx) dx
0
1
⎡ ⎡ − cos( nπx) ⎤
⎡ − cos( nπx ) ⎤ ⎤
= ⎢x2 ⎢
⎥ dx ⎥
⎥ − ( 2 x) ⎢
π
n
nπ
⎦ ⎦0
⎣
⎦
⎣ ⎣
∫
1
⎡ − x2
⎤
2
cos( nπx ) +
x cos( nπx )dx ⎥
=⎢
nπ
⎣⎢ nπ
⎦⎥ 0
∫
1
⎡− x2
sin( nπx) ⎤ ⎤
2 ⎡ ⎡ sin( nπx) ⎤
dx ⎥ ⎥
cos( nπx ) +
− 1×
⎢x⎢
=⎢
⎥
nπ ⎣ ⎣ nπ ⎦
nπ
⎦ ⎥⎦ 0
⎣⎢ nπ
∫
1
⎡ − x2
2 ⎡ x
1
⎤⎤
cos( nπx) +
sin(nπx ) −
sin(nπx)dx ⎥ ⎥
=⎢
⎢
nπ ⎣ nπ
nπ
⎦ ⎦⎥ 0
⎣⎢ nπ
∫
1
⎡ − x2
2 ⎡ x
1 ⎡ − cos( nπx) ⎤ ⎤ ⎤
cos( nπx) +
=⎢
⎢ sin(nπx) −
⎥ ⎥⎥
nπ ⎣ nπ
nπ ⎢⎣
nπ
⎦ ⎦ ⎦⎥ 0
⎣⎢ nπ
1
⎡ − x2
2 ⎡ x
1
⎤⎤
cos( nπx ) +
sin(nπx ) + 2 2 cos( nπx )⎥ ⎥
=⎢
⎢
nπ ⎣ nπ
n π
⎦ ⎥⎦ 0
⎢⎣ nπ
1
⎤
⎡ − x2
2x
2
cos( nπx) + 2 2 sin(nπx) + 3 3 cos( nπx )⎥
=⎢
n π
n π
⎥⎦ 0
⎢⎣ nπ
2
2
2
⎡ −1
⎤ ⎡
⎤
= ⎢ cos nπ + 2 2 sin( nπ) + 3 3 cos( nπ) ⎥ − ⎢0 + 0 + 3 3 cos 0 ⎥
n π
n π
n π
⎣ nπ
⎦ ⎣
⎦
=
−1
2
2
( −1) n + 0 + 3 3 ( −1) n − 3 3
nπ
n π
n π
=
(−1) n ⎡ 2
2
⎤
− 1⎥ − 3 3
⎢
2 2
nπ ⎣ n π
⎦ n π
[Q cos (nπ) = (–1)n, sin (nπ) = 0]
1
∫
Consider, I2 = − 2 x sin( nπx ) dx
0
1
⎡ ⎡ − cos( nπx ) ⎤
− cos( nπx) ⎤
= − 2⎢ x ⎢
dx ⎥
⎥ − 1×
π
n
nπ
⎦
⎦0
⎣ ⎣
∫
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.38
MATHEMATICS-II [JNTU-ANANTAPUR]
1
1
⎡− x
⎤
cos( nπx) +
cos( nπx )dx ⎥
= − 2⎢
nπ
⎦0
⎣ nπ
∫
1
⎡− x
1 ⎡ sin(nπx ) ⎤ ⎤
cos( nπx ) +
= − 2⎢
⎥
nπ ⎢⎣ nπ ⎥⎦ ⎦ 0
⎣ nπ
1
1
⎡− x
⎤
cos( nπx ) + 2 2 sin(nπx)⎥
= − 2⎢
n π
⎦0
⎣ nπ
⎡⎧ − 1
1
1
⎫⎤
⎫ ⎧
= − 2⎢⎨ cos nπ + 2 2 sin nπ⎬ − ⎨0 + 2 2 sin 0⎬⎥
n π
⎭⎦
⎭ ⎩ n π
⎣ ⎩ nπ
⎡ −1
⎤
n
= − 2 ⎢ ( −1) + 0 − 0 − 0 ⎥
⎣ nπ
⎦
=
2
( −1) n
nπ
1
∫
I3 = sin(nπx)dx
0
1
⎡ − cos nπx ⎤
=⎢
⎥
⎣ nπ ⎦ 0
=
−1
[cos nπ − cos 0]
nπ
=
−1
[(−1) n − 1]
nπ
=
1
[1 − ( −1) n ]
nπ
⎡ ⎧⎪ ( −1) n ⎡ 2
2 ⎪⎫ ⎧ 2
⎫ ⎧1
⎤
1 − ( −1) n
− 1⎥ − 3 3 ⎬ + ⎨ (−1) n ⎬ + ⎨
Hence, bn = 2 ⎢⎨
⎢
2 2
⎦ n π ⎪⎭ ⎩ nπ
⎭ ⎩ nπ
⎢⎣ ⎪⎩ nπ ⎣ n π
{
}⎫⎬⎤⎥
⎭⎥⎦
⎡ (−1) n ⎡ 2
2
1 ⎤
⎤
− 1 + 2 − 1⎥ − 3 3 + ⎥
= 2⎢
⎢
2 2
nπ ⎥⎦
⎦ n π
⎢⎣ nπ ⎣ n π
⎡ (−1) n ⎡ 2 ⎤
2
1 ⎤
− 3 3+ ⎥
= 2⎢
⎢
2 2⎥
nπ ⎥⎦
⎢⎣ nπ ⎣ n π ⎦ n π
=
=
∴
4
3 3
n π
4
n π
3 3
(−1) n −
4
3 3
n π
+
2
nπ
[(−1) − 1]+ n2π
n
∞
f (x)= ∑ bn sin( nπx )
n =1
∞ ⎡ 4
2 ⎤
n
= ∑ ⎢ 3 3 [( −1) − 1] + ⎥ sin( nπx )
nπ ⎦
n =1 ⎣ n π
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2.39
UNIT-2 (Fourier Series)
1
⎧1
⎪4 − x , 0 < x < 2
.
Q29. Find the half-range sine series for f(x) = ⎨
3
1
⎪x −
<x <1
,
4
2
⎩
Model Paper-III, Q5
Ans: Given that,
1
⎧1
⎪4 − x , 0 < x < 2
f(x) = ⎨
3
1
⎪x −
, < x <1
2
⎩ 4
The half-range Fourier sine series is given by,
∞
f(x) =
∑b
n
n =1
⎛ nπx ⎞
sin⎜
⎟
⎝ l ⎠
∞
=
∑b
n
sin(nπx )
[Q l = 1]
n =1
Where,
bn =
2
l
2
=
1
l
⎛ nπx ⎞
⎟ dx
l ⎠
∫ f ( x).sin⎜⎝
0
l
∫ f ( x).sin( nπx) dx
0
l
⎤
⎡1/ 2
= 2⎢ f1 ( x ) sin(nπx ) dx + f 2 ( x) sin(nπx ) dx ⎥
⎥
⎢0
1/ 2
⎦
⎣
∫
∫
l
⎤
⎡1/ 2 ⎛ 1
3⎞
⎞
⎛
= 2⎢ ⎜ − x ⎟ sin nπx dx + ⎜ x − ⎟ sin(nπx ) dx ⎥
4⎠
⎥
⎢ 0 ⎝4
⎠
⎝
1/ 2
⎦
⎣
∫
∫
= 2[I1 + I2]
Consider,
1/ 2
I1 =
⎛1
⎞
∫ ⎜⎝ 4 − x ⎟⎠ sin( nπx) dx
0
1/ 2
⎡⎡ 1
⎤ ⎡ − cos( nπx ) ⎤
⎡ − cos( nπx) ⎤ ⎤
= ⎢ ⎢ − x⎥ ⎢
− (−1) ⎢
⎥ dx ⎥
⎥
nπ
nπ
⎦ ⎦0
⎣
⎦
⎣⎣ 4 ⎦ ⎣
∫
1/ 2
⎤
⎡ ⎛ 1− 4x ⎞
1
cos( nπx )dx ⎥
⎟ cos( nπx) −
= ⎢− ⎜
nπ
⎦0
⎣ ⎝ 4nπ ⎠
∫
1/ 2
⎤
⎡⎛ 4 x − 1 ⎞
1
⎟ cos( nπx ) − 2 2 sin(nπx)⎥
= ⎢⎜
n π
⎦0
⎣⎝ 4nπ ⎠
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SIA GROUP
2.40
MATHEMATICS-II [JNTU-ANANTAPUR]
⎤
⎡⎧⎛ ⎛ 1 ⎞ ⎞
⎫
⎥
⎢⎪⎜ 4⎜ ⎟ − 1 ⎟
⎪
⎢⎪⎜ ⎝ 2 ⎠ ⎟ cos⎛⎜ nπ ⎞⎟ − 1 sin⎛⎜ nπ ⎞⎟⎪ − ⎧⎛⎜ 0 − 1 ⎞⎟ cos 0 − 1 sin 0⎫⎥
⎬⎥
⎬ ⎨
= ⎢⎨⎜ 4nπ ⎟
2 2
n 2π2
⎝ 2 ⎠⎪ ⎩⎝ 4nπ ⎠
⎝ 2 ⎠ n π
⎭
⎟
⎥
⎢⎪⎪⎜
⎪⎭
⎠
⎥⎦
⎢⎣⎩⎝
=
1
1
1
⎛ nπ ⎞
⎛ nπ ⎞
cos⎜ ⎟ − 2 2 sin ⎜ ⎟ +
4nπ
2
π
2
4
n
π
n
⎝ ⎠
⎝ ⎠
1
I2 =
⎛
3⎞
∫ ⎜⎝ x − 4 ⎟⎠.sin( nπx) dx
1/ 2
1
⎡⎛
3 ⎞ ⎡ − cos nπx ⎤
⎡ − cos nπx ⎤ ⎤
− 1.⎢
= ⎢⎜ x − ⎟.⎢
⎥ dx ⎥
⎥
4 ⎠ ⎣ nπ ⎦
⎣ nπ ⎦ ⎦1 / 2
⎣⎝
∫
1
⎤
⎡ ⎛ 4x − 3 ⎞
1
cos( nπx )dx ⎥
⎟ cos( nπx) +
= ⎢− ⎜
π
π
4
n
n
⎠
⎦ 1/ 2
⎣ ⎝
∫
1
⎤
⎡⎛ 3 − 4 x ⎞
1
⎟ cos( nπx ) + 2 2 sin(nπx)⎥
= ⎢⎜
n π
⎦ 1/ 2
⎣⎝ 4nπ ⎠
⎡⎧ 1
1
1
⎛ nπ ⎞⎫⎤
⎛ nπ ⎞
⎫ ⎧ 1
cos nπ + 2 2 sin nπ⎬ − ⎨
cos ⎜ ⎟ + 2 2 sin⎜ ⎟⎬⎥
= ⎢ ⎨−
n π
⎝ 2 ⎠⎭⎦⎥
⎝ 2 ⎠ n π
⎭ ⎩ 4nπ
⎣⎢⎩ 4nπ
⎡⎧ − 1
1
⎛ nπ ⎞ ⎫ ⎤
⎛ nπ ⎞
⎫ ⎧ 1
(−1) n + 0⎬ − ⎨
cos ⎜ ⎟ + 2 2 sin⎜ ⎟⎬⎥
= ⎢⎨
⎝ 2 ⎠⎭⎦⎥
⎝ 2 ⎠ n π
⎭ ⎩ 4nπ
⎣⎢⎩ 4nπ
=
− (−1) n
1
1
⎛ nπ ⎞
⎛ nπ ⎞
−
cos⎜ ⎟ − 2 2 sin⎜ ⎟
4 nπ
4 nπ ⎝ 2 ⎠ n π
⎝ 2 ⎠
∴ bn = 2[I1 + I2]
⎡ 1
1
(−1) n
1
1
1
⎛ nπ ⎞ ⎤
⎛ nπ ⎞
⎛ nπ ⎞
⎛ nπ ⎞
−
−
cos⎜ ⎟ − 2 2 sin⎜ ⎟⎥
cos⎜ ⎟ − 2 2 sin⎜ ⎟ +
= 2⎢
⎝ 2 ⎠⎦⎥
⎝ 2 ⎠ n π
⎝ 2 ⎠ 4nπ 4nπ 4nπ
⎝ 2 ⎠ n π
⎣⎢ 4nπ
⎡ −2
1
(−1) n ⎤
⎛ nπ ⎞
−
= 2⎢ 2 2 sin⎜ ⎟ +
⎥
⎝ 2 ⎠ 4nπ 4nπ ⎥⎦
⎢⎣ n π
⎡ −4
1
(−1) n ⎤
⎛ nπ ⎞
−
⎥
= ⎢ 2 2 sin⎜ ⎟ +
⎝ 2 ⎠ 2nπ 2nπ ⎦⎥
⎣⎢ n π
=
−4
1
⎛ nπ ⎞
sin ⎜ ⎟ +
[1 − ( −1) n ]
n π
⎝ 2 ⎠ 2 nπ
2 2
∞
∴ f ( x) =
∑ ⎡⎢⎣ 2nπ [1 − (−1)
n =1
1
n
]−
4
n π
2 2
sin
nπ ⎤
sin( nπx )
2 ⎥⎦
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2.41
UNIT-2 (Fourier Series)
Q30. Obtain half-range sine series for ex in 0 < x < 1.
Ans: Given that,
f(x) = ex for 0 < x < 1
The half-range Fourier sine series is given by,
∞
f(x) =
∑b
n
n =1
⎡ nπx ⎤
sin ⎢
⎥
⎣ l ⎦
∞
=
∑b
n
sin(nπx )
n =1
Where,
bn =
2
l
l
⎡ nπx ⎤
dx
l ⎥⎦
∫ f ( x) sin ⎢⎣
0
1
∫
= 2 f ( x) sin( nπx)dx
[Q l = 1]
0
1
∫
x
= 2 e sin( nπx)dx
0
1
⎡
⎤
ex
(sin(nπx ) − nπ cos( nπx) ⎥
= 2⎢ 2
2
⎣⎢1 + ( nπ)
⎦⎥ 0
=
2
n 2π 2 + 1
⎤
⎡
e ax
ax
−
[
a
sin
bx
b
cos
bx
]
⎥
⎢Q e sin bxdx = 2
a + b2
⎦⎥
⎣⎢
∫
[ e x [sin( nπx ) − nπ cos( nπx)]]10
⎡e{(sin( nπ) − nπ cos( nπ )}⎤
⎢
⎥
n π + 1 ⎣ − e 0 .{sin 0 − nπ cos 0} ⎦
2
=
=
2 2
2
2 2
n π +1
2
=
n π +1
=
n π +1
=
n π +1
2 2
2
2 2
2nπ
2 2
[e{0 − nπ( −1) n } − 1.{0 − nπ}]
[Q cos(nπ) = (– 1)n, sin(nπ) = 0]
[ −2.718nπ( −1) n + nπ]
[ nπ(1 − 2.718( −1) n )]
[1 − 2.718( −1) n ]
∞
∴ f(x) =
∑b
n
sin(nπx )
n =1
∞
=
⎤
⎡ 2n π
(1 − 2.718(−1) n ⎥ sin(nπx)
2 2
+
1
⎦
n =1
∑ ⎢⎣ n π
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.42
MATHEMATICS-II [JNTU-ANANTAPUR]
2
Q31. Find the half-range sine series of f(t) = t – t in the interval 0 < t < 1.
Ans: Given that,
f (t) = t – t2
0<t<1
The Fourier half-range sine series of f (t) in the interval (0, l) is given by,
∞
f (t) =
∑b
n
n =1
⎛ nπt ⎞
sin⎜
⎟
⎝ l ⎠
... (1)
Where,
bn =
2
l
l
⎛ nπt ⎞
⎟dt
l ⎠
∫ f (t ) ⋅ sin⎜⎝
0
Here, l = 1
∞
∴ f (t) =
∑b
n
⋅ sin(nπt )
n =1
1
2
⎛ nπt ⎞
(t − t 2 ) sin⎜
⎟dt
bn =
1
⎝ 1 ⎠
∫
0
1
⎡1
⎤
2
⎢
= 2 t sin(nπt )dt − t sin(nπt )dt ⎥
⎢0
⎥
0
⎣
⎦
∫
∫
1
⎡ ⎡ − cos( nπt ) ⎤
⎡ − cos( nπt ) ⎤ ⎤ ⎡ 2 ⎡ − cos( nπt ) ⎤
⎡ − cos( nπt ) ⎤ ⎤
dt ⎥ ⎥
= 2⎢t ⎢
⎥dt ⎥ − ⎢t ⎢
⎥− ⎢
⎥ − (2t ) ⋅ ⎢
π
π
π
n
n
n
nπ
⎦ ⎦ ⎣ ⎣
⎣
⎦
⎦⎦ 0
⎣
⎦
⎣⎣
∫
∫
1
⎡⎡ − 1
1
2
⎤⎤
⎤ ⎡ −1
cos( nπt )dt ⎥ − ⎢ t 2 cos( nπt ) +
t cos( nπt )dt ⎥ ⎥
= 2⎢⎢ t cos( nπt ) +
π
π
π
π
n
n
n
n
⎦⎦ 0
⎦ ⎣
⎣⎣
∫
∫
1
⎡⎧ − 1
1 ⎡ sin(nπt ) ⎤ ⎫ ⎧⎪ − 1 2
2 ⎡ ⎡ sin(nπt ) ⎤
⎡ sin(nπt ) ⎤ ⎤ ⎫⎪⎤
− ⎢
t⎢
= 2⎢⎨ t cos( nπt ) +
⎬ − ⎨ t cos( nπt ) +
⎢
⎥
⎢
⎥
⎥ dt ⎥ ⎬⎥
nπ ⎣ nπ ⎦ ⎭ ⎪⎩ nπ
nπ ⎣ ⎣ nπ ⎦
⎣ nπ ⎦ ⎦ ⎪⎭⎦⎥ 0
⎣⎢⎩ nπ
∫
1
⎡⎧ − t
1
2 ⎡ t
⎫ ⎧⎪ − t 2
⎛ − cos( nπt ) ⎞⎤ ⎫⎪⎤
cos( nπt ) +
sin(nπt ) − ⎜
⎟ ⎥ ⎬⎥
= 2⎢⎨ cos( nπt ) + 2 2 sin(nπt )⎬ − ⎨
⎢
nπ ⎣ nπ
n π
⎭ ⎪⎩ nπ
⎝ n 2 π 2 ⎠⎦ ⎪⎭⎦⎥ 0
⎣⎢⎩ nπ
1
⎡−t
⎤
1
2t
2
t2
cos( nπt ) − 2 2 sin(nπt ) − 3 3 cos( nπt )⎥
= 2⎢ cos( nπt ) + 2 2 sin(nπt ) +
nπ
n π
n π
n π
⎢⎣ nπ
⎥⎦ 0
⎡
⎛ −t t2
2 ⎞
2t
⎛ 1
+
− 3 3 ⎟ + sin( nπt )⎜ 2 2 − 2 2
= 2 ⎢cos( nπt )⎜⎜
⎟
n
n
π
π
π
π
n
n
n
π
⎝
⎠
⎝
⎣⎢
1
⎞⎤
⎟⎥
⎠ ⎦⎥ 0
⎡⎧
2 ⎞
2 ⎞⎫ ⎧
2 ⎞
⎞ ⎫⎤
⎛ 1
⎛
⎛ 1
⎛ −1 1
+
− 3 3 ⎟ + sin(nπ)⎜ 2 2 − 2 2 ⎟⎬ − ⎨cos(0)⎜ 0 + 0 − 3 3 ⎟ + sin(0)⎜ 2 2 − 0 ⎟⎬⎥
= 2⎢⎨cos( nπ)⎜
n π ⎠⎭ ⎩
n π ⎠
⎠⎭⎥⎦
⎝n π
⎝
⎝n π
⎝ nπ nπ n π ⎠
⎢⎣⎩
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2.43
UNIT-2 (Fourier Series)
⎡⎧
⎫ ⎧ −2
⎫⎤
n⎛ −2 ⎞
= 2⎢⎨(−1) ⎜ 3 3 ⎟ + 0⎬ − ⎨ 3 3 + 0⎬⎥
⎭⎥⎦
⎝ n π ⎠ ⎭ ⎩n π
⎢⎣⎩
[Q cos (nπ) = (– 1)n, sin(nπ) = sin0 = 0]
2 ⎡ x2 ⎤
= ⎢ ⎥
π ⎢⎣ 2 ⎥⎦
2 ⎤
⎡ −2
n
= 2 ⎢ 3 3 ( −1) + 3 3 ⎥
n π ⎦
⎣n π
=
1 2 π
[ x ]0
π
=
1 2
[ π − 0] = π
π
⎡ 2
n ⎤
= 2 ⎢ 3 3 [1 − ( −1) ]⎥
⎣n π
⎦
2
an =
π
; if n is even
;
if n is odd
f (t) =
∑n π
8
n =1
∞
8
t – t2 =
π
3 3
3
∑
sin(nπt )
n
0
∫
12
+
1
32
+
1
52
+ ... =
π
=
2 ⎡x
1
⎤
sin nx −
sin nx dx ⎥
⎢
π ⎣n
n
⎦0
=
2 ⎡x
1 ⎡ − cos nx ⎤⎤
⎢ sin nx − ⎢
⎥
π ⎣n
n ⎣ n ⎥⎦⎦ 0
∫
π
Q32. Obtain half-range cosine series f(x) = x in the interval
1
π
2 ⎡ ⎡ sin nx ⎤
sin nx ⎤
− 1×
dx ⎥
= ⎢x⎢
⎥
π⎣ ⎣ n ⎦
n
⎦0
[Q l = 1]
3
0 ≤ x ≤ π. Hence show that
0
∫
8 ⎡
1
1
⎤
⎢sin( πt ) + 3 sin( 2πt ) + 3 sin( 3πt ) + ...⎥
3
2
π3 ⎣
⎦
=
∫ f ( x).cos nx dx
π
sin(nπt )
n =1
π
2
x cos nx dx
=
π
Now substituting the value of ‘bn’ in equation (1), we get,
∞
0
∴ a0 = π
⎡ 4
n ⎤
= ⎢ 3 3 [1 − ( −1) ]⎥
⎣n π
⎦
⎧ 0
⎪
∴ bn = ⎨ 8
⎪⎩ n 3 π3
π
π
π2
.
8
2 ⎡x
1
⎤
= ⎢ sin nx + 2 cos nx ⎥
π ⎣n
n
⎦0
Ans: Given that,
f(x) = x
0≤x≤π
=
2 ⎡⎧ π
1
1
⎫⎤
⎫ ⎧
⎢⎨ sin nπ + 2 cos nπ⎬ − ⎨0 + 2 cos 0⎬⎥
π ⎣⎩ n
n
⎭⎦
⎭ ⎩ n
=
2 ⎡ ⎧⎪
( −1) n ⎫⎪ ⎧ 1 ⎫ ⎤
⎢ ⎨0 + 2 ⎬ − ⎨ 2 ⎬ ⎥
π ⎢⎣ ⎪⎩
n ⎪⎭ ⎩ n ⎭ ⎥⎦
The half range Fourier cosine series is given by,
∞
∑
a0
an cos nx
+
f(x) =
2 n =1
... (1)
[Q cos nπ = (– 1)n, sin nπ = 0]
Where,
a0 =
and an =
2
π
2
π
π
=
∫ f ( x)dx
2 ⎡ (−1) n − 1 ⎤
⎥
⎢
π ⎢⎣ n 2 ⎥⎦
0
π
∫ f ( x).cos nx dx
0
Now,
π
2
x dx
a0 =
π
∫
0
∴ an =
2
n π
2
[( −1) n − 1]
when n = even
⎧0,
⎪
−
4
an = ⎨
, when n = odd
⎪⎩ n 2 π
On substituting the values of a0 and an in equation (1),
we get,
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.44
MATHEMATICS-II [JNTU-ANANTAPUR]
∞
π
2
+
[(−1) n − 1] cos nx
2 n =1 n 2 π
∑
f(x) =
∞
x=
π
⎡ −4 ⎤
+
cos nx
2 n =1,3,5 ⎢⎣ n 2 π ⎥⎦
⇒
x=
π 4
1
−
cos nx
2 π n =1,3,5 n 2
⇒
x=
1
1
π 4⎡1
1
⎤
cos x + 2 cos 3 x + 2 cos 5 x + 2 cos 7 x + ......⎥
−
2 π ⎢⎣12
3
5
7
⎦
⇒
x=
π 4⎡
1
1
1
⎤
− ⎢cos x + cos 3x + cos 5 x +
cos 7 x + ......⎥
2 π⎣
9
25
49
⎦
∑
∞
∑
... (2)
On substituting x = 0 in equation (2), we get,
π 4⎡1
1
1
1
⎤
− ⎢ 2 cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + ......⎥
2 π ⎣1
7
5
3
⎦
⇒
0=
⇒
−4 ⎡ 1
1
1
1
⎤ −π
+ +
+
+ ......⎥ =
π ⎢⎣12 32 52 7 2
2
⎦
⇒
1
12
+
1
32
+
1
52
+
1
72
π2
8
+ ...... =
Hence proved.
π – x) in 0 ≤ x ≤ π .
Q33. Find the half-range cosine series of f(x) = x(π
Ans: Given that,
f(x) = x(π – x) 0 ≤ x ≤ π
i.e., f(x) = x(π – x) x∈[0, π]
The half-range cosine series expansion of f(x) in [0, π] is given by,
a0 ∞
an cos nx
+
f(x) =
2 n =1
∑
2
a0 =
π
... (1)
π
∫ f ( x)dx
0
π
=
2
x ( π − x )dx
π
∫
0
=
π
π
⎤
2⎡
⎢ πx dx − x 2 dx ⎥
π⎢
⎥
0
⎣0
⎦
=
π
π
⎤
2⎡
⎢π x dx − x 2 dx⎥
π⎢
⎥
0
⎣ 0
⎦
∫
∫
∫
∫
π
π⎤
⎡
2 ⎢ ⎡ x 2 ⎤ ⎡ x3 ⎤ ⎥
π⎢ ⎥ − ⎢ ⎥
=
π ⎢ ⎢⎣ 2 ⎥⎦ ⎢⎣ 3 ⎥⎦ ⎥
0
0⎦
⎣
=
2 ⎡π 2
1
⎤
[ π − 0] − [ π 3 − 0]⎥
π ⎢⎣ 2
3
⎦
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.45
UNIT-2 (Fourier Series)
=
2 ⎡ π 3 π3 ⎤
⎢ − ⎥
π ⎢⎣ 2
3 ⎥⎦
=
2 π3
×
π 6
=
π2
3
∴ a0 = π 2 / 3
an =
2
π
π
∫ f ( x) cos nx dx
0
π
2
x( π − x ). cos nx dx
=
π
∫
0
π
π
⎤
2⎡
⎢
π
−
x
cos
nx
dx
x 2 cos nx dx ⎥
=
π⎢
⎥
0
⎣0
⎦
∫
∫
π
π
⎤
2⎡
⎢
π
−
x
cos
nx
dx
x 2 cos nx dx ⎥
=
π⎢
⎥
0
⎣ 0
⎦
∫
∫
2 ⎡ ⎧ ⎡ sin nx ⎤
sin nx dx ⎫ ⎧ 2 ⎡ sin nx ⎤
⎡ sin nx
−
− 2( x ) ⎢
= ⎢ π⎨ x ⎢
⎬ − ⎨x ⎢
⎥
⎥
π ⎢⎣ ⎩ ⎣ n ⎦
n
⎣ n
⎭ ⎩ ⎣ n ⎦
∫
∫
π
⎤ ⎫⎤
⎥ dx ⎬⎥
⎦ ⎭⎥⎦ 0
2
⎫⎪⎤
2 ⎡ ⎧ x sin nx 1
⎫ ⎧⎪ x sin nx 2
sin nx dx ⎬ − ⎨
x sin nx dx ⎬⎥
−
−
= ⎢π⎨
n
n
n
π ⎣⎢ ⎩ n
⎪⎭⎦⎥
⎭ ⎪⎩
∫
∫
π
0
π
2 ⎡ ⎧ x sin nx cos nx ⎫ ⎧⎪ 2 sin nx 2 ⎧ ⎡ cos nx ⎤
⎛ cos nx ⎞ ⎫⎫⎪⎤
− ⎜−
+
− ⎨ x ⎢−
⎟dx ⎬⎬⎥
⎬ − ⎨x
= ⎢π⎨
⎥
2
n
n⎩ ⎣
n ⎦
n ⎠ ⎭⎪⎭⎦⎥
π ⎣⎢ ⎩ n
n ⎭ ⎪⎩
⎝
0
∫
π
2 ⎡ ⎧ x sin nx cos nx ⎫ ⎧⎪ x 2 sin nx 2 ⎧ x cos nx sin nx ⎫⎫⎪⎤
− ⎨−
+
+
= ⎢ π⎨
⎬⎬⎥
⎬−⎨
n
n⎩
n
π ⎢⎣ ⎩ n
n 2 ⎭ ⎪⎩
n 2 ⎭⎪⎭⎥⎦
0
π
⎤
2 ⎡ ⎧ x sin nx cos nx ⎫ x 2 sin nx 2
2
−
− 2 x cos nx + 3 sin nx ⎥
+
= ⎢π⎨
2 ⎬
π ⎢⎣ ⎩ n
n
n
n
n ⎭
⎥⎦ 0
π
⎤
2 ⎡ πx sin nx π cos nx x 2 sin nx 2
2
+
−
− 2 x cos nx + 3 sin nx ⎥
= ⎢
2
π ⎣⎢
n
n
n
n
n
⎦⎥ 0
=
⎫⎪ ⎧ π cos 0
2 ⎡⎧⎪ π 2 sin nπ π cos nπ π 2 sin nπ 2
2
2
⎫⎤
+
−
−
π
π
+
π
n
n
cos
sin
0
0
0
sin
0
−
+
−
−
+
⎢⎨
⎬
⎨
⎬⎥
n
n
π ⎢⎣⎪⎩
⎪⎭ ⎩
n2
n2
n3
n2
n3
⎭⎥⎦
=
2 ⎡ ⎧⎪
( −1) n π
2π( −1) n
⎪⎫ ⎧ π ⎫⎤
0
0
−
−
+
⎢ ⎨0 +
⎬ − ⎨ 2 ⎬⎥
π ⎢⎣ ⎪⎩
n2
n2
⎭⎪ ⎩ n ⎭⎥⎦
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
[Q cos (nπ) = (– 1)n, sin (nπ) = 0]
SIA GROUP
2.46
MATHEMATICS-II [JNTU-ANANTAPUR]
2 ⎡ − π(−1) n π ⎤
− 2⎥
⎢
π ⎢⎣ n 2
n ⎥⎦
2
n
= 2 [ − π{( −1) + 1}]
n π
=
=
−2
n2
[(−1) n + 1]
; if n is odd
⎧0
⎪
∴
an = ⎨ − 4
⎪ 2 ; if n is even
⎩n
On substituting the values of a0 and an in equation (1), we get,
⇒
f(x) = x(π – n) =
π2 / 3 π2
+
when ‘n’ is odd
2
6
f(x) = x(π – x) =
π2 / 3
+
2
∞
∑− n
n =1
4
2
cos nx when ‘n’ is even
∞
=
∴ f (x) =
cos nx
π2
−4
2
6
n =1 n
∑
1
1
π2 ⎧
⎫
− 4⎨cos x + cos2x + cos3x +.....⎬
6
4
9
⎩
⎭
Q34. Find the half-range cosine series of f(x) = x(2 – x) in 0 ≤ x ≤ 2.
Ans: Given that,
f(x) = x(2 – x) 0 ≤ x ≤ 2
The half-range cosine series is given by,
∞
f (x) =
a0
nπx
an cos
+
L
2 n =1
∑
... (1)
Where,
a0 =
2
L
L
∫ f ( x)dx
0
And
2
an =
L
Here,
L
⎛ nπx ⎞
⎟ dx
L ⎠
∫ f ( x) cos⎜⎝
0
L =2
2
a0 =
2
x( 2 − x ) dx =
2
∫
0
2
2
∫
0
2
∫
2 xdx − x 2 dx
0
2
⎡ x 2 ⎤ ⎡ x3 ⎤
1 3
2
= 2⎢ ⎥ − ⎢ ⎥ = [2 − 0] − [2 − 0]
3
2
3
⎣⎢ ⎦⎥ 0 ⎣⎢ ⎦⎥ 0
= 4−
∴ a0 =
8
4
=
3
3
4
3
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.47
UNIT-2 (Fourier Series)
2
2
⎛ nπx ⎞
x ( 2 − x ) cos ⎜
an =
⎟ dx =
⎝ 2 ⎠
2
∫
0
2
=
∫
0
2
∫ (2 x − x
0
2
⎛ nπx ⎞
) cos ⎜
⎟ dx
⎝ 2 ⎠
2
⎛ nπx ⎞
⎛ nπx ⎞
2
2 x cos ⎜
⎟ dx
⎟ dx − x cos ⎜
⎝ 2 ⎠
⎝ 2 ⎠
∫
0
2
2
⎡ ⎡ ⎛ nπx ⎞ ⎤
⎡ ⎡ ⎛ nπx ⎞ ⎤
⎛ nπx ⎞ ⎤
⎛ nπx ⎞ ⎤
sin ⎜
sin ⎜
⎟ ⎥
⎟⎥
⎟ ⎥
⎟⎥
⎢ ⎢ sin ⎜
⎢ ⎢ sin ⎜
⎝ 2 ⎠ ⎥
⎠
⎝
⎠
⎝
⎠
⎝
2 ⎥
2
2 ⎥
= 2⎢ x ⎢
dx
− 2( x )
dx ⎥ − ⎢ x 2 ⎢
−
⎥
⎢ ⎢ nπ / 2 ⎥
⎢ ⎢ nπ / 2 ⎥
⎥
nπ / 2
nπ / 2
⎥
⎢ ⎢
⎢ ⎢
⎥
⎥
⎥
⎥⎦ 0
⎢⎣ ⎣
⎥⎦ 0 ⎢⎣ ⎣
⎦
⎦
∫
∫
2
2
⎡ 2
⎛ nπx ⎞ 2
⎛ nπx ⎞ ⎤ ⎡ 2 2 ⎛ nπx ⎞ 4
⎛ nπx ⎞ ⎤
x sin ⎜
sin⎜
= 2⎢ x sin ⎜
⎟−
⎟ dx ⎥ − ⎢ x sin⎜
⎟−
⎟ dx
⎝ 2 ⎠ nπ
⎝ 2 ⎠ ⎦ 0 ⎣ nπ
⎝ 2 ⎠ nπ
⎝ 2 ⎠ ⎥⎦ 0
⎣ nπ
∫
∫
2 ⎡
2⎤
⎡
⎧ ⎡
⎡
⎛ nπx ⎞ ⎤ ⎤ ⎢
⎛ nπx ⎞ ⎫ ⎥
⎛ nπx ⎞ ⎤
cos
cos
−
cos
−
−
⎜
⎟
⎜
⎟
⎜
⎟
⎢
⎢
⎝ 2 ⎠ ⎥ ⎥⎥ ⎢ 2 2 ⎛ nπx ⎞ 4 ⎪⎪ ⎢
⎝ 2 ⎠ ⎪⎪ ⎥
⎝ 2 ⎠⎥
2
⎛ nπx ⎞ 2 ⎢
⎥ −
⎥−
x sin ⎜
dx⎬ ⎥
x⎢
⎟−
⎟−
= 2⎢ x sin ⎜
⎨
⎢
⎝ 2 ⎠ nπ ⎢ nπ / 2 ⎥ ⎥
⎝ 2 ⎠ nπ ⎪ ⎢ nπ / 2 ⎥
⎢ nπ
nπ
nπ / 2
⎪ ⎥
⎢
⎢
⎥⎥ ⎢
⎢
⎥
⎢
⎪⎭ ⎥
⎪
⎦
⎣
⎦
⎣
⎩
⎣
⎦0 ⎣
0⎦
∫
2
2
⎡ 2
4
⎛ nπx ⎞
⎛ nπ x ⎞ ⎤ ⎡ 2 2 ⎛ nπ x ⎞ 4 ⎧ − 2 x
⎛ nπ x ⎞ 2
⎛ nπ x ⎞ ⎫ ⎤
cos ⎜
cos ⎜
= 2 ⎢ x sin ⎜
⎟ + 2 2 cos ⎜
⎟−
⎟ ⎥ − ⎢ x sin ⎜
⎟+
⎟ dx ⎬ ⎥
⎨
⎝
⎠
⎝
⎝
⎠
⎝
⎠
⎠
⎝ 2 ⎠ ⎭⎦
π
π
n
n
n
n
n
π
π
π
2
2
2
2
n π
⎦0 ⎣
⎣
⎩
0
∫
2
⎡
⎧
⎡ ⎛ nπx ⎞ ⎤⎫ ⎤
sin ⎜
2 ⎢
⎟ ⎪⎥
⎪
⎡2
4
⎛ nπx ⎞
⎛ nπx ⎞ 2 ⎢⎢ ⎝ 2 ⎠ ⎥⎥⎪ ⎥
⎛ nπx ⎞ ⎤ ⎢ 2 2 ⎛ nπx ⎞ 4 ⎪ − 2
= 2 ⎢ x sin ⎜
+
−
x
x
cos
sin
cos
−
+
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎨
⎬
⎝ 2 ⎠ n π ⎢ n π / 2 ⎥⎪ ⎥
⎝ 2 ⎠ nπ ⎪ nπ
⎝ 2 ⎠ ⎥⎦ 0 ⎢ nπ
⎝ 2 ⎠ n2π 2
⎣ nπ
⎢
⎥⎪ ⎥
⎢
⎦⎭ ⎥⎦ 0
⎣
⎢⎣
⎩⎪
2
⎡ 4
16
8
8
⎛ nπx ⎞
⎛ nπx ⎞ 2 2 ⎛ nπx ⎞
⎛ nπx ⎞
⎛ nπx ⎞ ⎤
x sin⎜
x sin ⎜
=⎢
⎟ + 2 2 cos⎜
⎟−
⎟ − 2 2 x cos ⎜
⎟ + 3 3 sin ⎜
⎟
⎝ 2 ⎠ n π
⎝ 2 ⎠ nπ
⎝ 2 ⎠ n π
⎝ 2 ⎠ n π
⎝ 2 ⎠ ⎥⎦ 0
⎣ nπ
⎡⎧ 4
8
2 2
16 ⎫ ⎛ nπx ⎞ ⎧ 8
= ⎢⎨ x −
x + 3 3 ⎬ sin ⎜
⎟ +⎨ 2 2 − 2 2
⎝
⎠
π
π
n
n
2
n π ⎭
n π
⎩n π
⎣⎩
2
⎫ ⎛ nπx ⎞ ⎤
x⎬ cos ⎜
⎟⎥
⎭ ⎝ 2 ⎠ ⎦0
2
⎡⎧ 2 ⎛
8 ⎞ ⎛ nπx ⎞ ⎫ ⎧⎛ 8
⎞
⎛ nπx ⎞ ⎫⎤
= ⎢⎨ ⎜ 2 x − x 2 + 2 2 ⎟ sin ⎜
⎟ ⎬ + ⎨⎜ 2 2 (1 − x)⎟ cos ⎜
⎟ ⎬⎥
⎝ 2 ⎠ ⎭⎥⎦
⎠
n π ⎠ ⎝ 2 ⎠ ⎭ ⎩⎝ n π
⎢⎣⎩ nπ ⎝
0
⎡⎧ 2 ⎧
8 ⎫ ⎛ nπ( 2) ⎞ ⎫ ⎧ ⎧ 8
⎫
⎛ nπ( 2) ⎞ ⎫ ⎤
2
⎟ ⎬⎥
⎟ ⎬ + ⎨ ⎨ 2 2 (1 − 2)⎬ cos ⎜
= ⎢ ⎨ ⎨2(2) − 2 + 2 2 ⎬ sin ⎜
⎝ 2 ⎠ ⎭ ⎦⎥
n π ⎭ ⎝ 2 ⎠ ⎭ ⎩⎩ n π
⎭
⎣⎢ ⎩ nπ ⎩
⎡⎧ 2 ⎛
8 ⎞ ⎛ nπ(0) ⎞ ⎫ ⎧⎛ 8
⎞
⎛ nπ(0) ⎞ ⎫⎤
− ⎢⎨ ⎜ 2(0) − 0 + 2 2 ⎟ sin ⎜
⎟ ⎬⎥
⎟ ⎬ + ⎨⎜ 2 2 (1 − 0)⎟ cos ⎜
⎝ 2 ⎠ ⎭⎦⎥
⎠
n π ⎠ ⎝ 2 ⎠ ⎭ ⎩⎝ n π
⎣⎢⎩ nπ ⎝
⎡⎧ 16
8
8
⎫⎤
⎫ ⎧ 16
= ⎢⎨ 3 3 sin(nπ) − 2 2 cos( nπ)⎬ − ⎨ 3 3 sin(0) + 2 2 cos(0)⎬⎥
n π
n π
⎭⎦
⎭ ⎩n π
⎣⎩ n π
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.48
MATHEMATICS-II [JNTU-ANANTAPUR]
= 0−
=
8
n π
−8
2 2
n π
2
2
(−1) n − 0 −
8
n π2
2
×1
[Q sin nπ = 0, cos nπ = (–1)n]
[1 + ( −1) n ]
⎧0
⎪
∴ an = ⎨ − 16
⎪⎩ n 2 π 2
; if n is odd
; if n is even
On substituting the values of a0 and an is equation (1), we get,
4
3
x(2 – x) = +
2
∞
−8
∑n π
2 2
n =1
⎛ nπx ⎞
[1 + ( −1) n ] cos ⎜
⎟
⎝ 2 ⎠
∞
− 16
4
⎛ nπx ⎞
+
cos ⎜
⎟
2
2
⎝ 2 ⎠
6 n =1 n π
∑
=
2 16
= 3− 2
π
∞
∑n
n =1
1
2
[Considering for even values of n]
⎛ nπx ⎞
cos ⎜
⎟
⎝ 2 ⎠
⎡1
⎤
⎛ 6πx ⎞
⎛ 4πx ⎞ 1
⎛ 2πx ⎞ 1
⎟ + .....⎥
⎢ 2 cos ⎜⎝ 2 ⎟⎠ + 2 cos⎜⎝ 2 ⎟⎠ + 2 cos ⎜⎝
2 ⎠
4
6
⎣2
⎦
=
2 16
−
3 π2
=
2 16 ⎡ 1
1
1
⎤
− 2 ⎢ cos πx + cos 2πx + cos 3πx + .....⎥
3 π ⎣4
16
36
⎦
Q35. Find the half-range sine series of f(x) = x defined in the interval (0, 2).
Ans: Given range is,
Extended rangeis,
0 < x < 2 ⎫ for the function f(x) = x
⎬
−2< x<0⎭
The new function is symmetrical about the origin and therefore, represents on odd function in (–2, 2)
y
A
x1
M
−2 −1
L
0 1 2
x
y'
Figure
Hence, the fourier series for f(x) over the full period (–2, 2) will contain only sine terms given by,
∞
f(x) =
∑b
n
n =1
sin
n πx
2
[Ql = 2]
2
Where, bn =
n πx
2
f (x )sin
dx
20
2
∫
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
2.49
UNIT-2 (Fourier Series)
2
∫
=
x sin
0
nπx
dx
2
⎡Q sin nπ = 0, sin 0 = 0 ⎤
⎢
⎥
n
⎣cos nπ = ( −1) , cos 0 = 1⎦
2
⎡
⎞⎤
⎛
⎢ ⎛⎜ − cos nπx ⎞⎟ ⎜ − sin nπx ⎟⎥
2 ⎟ − 1⎜
2 ⎟⎥
= ⎢ x⎜
⎟
⎜
⎢ ⎜
nπ
⎟ ⎜ ⎛ nπ ⎞ 2 ⎟⎥
⎜
⎟
⎢
⎥
2
⎠ ⎜ ⎜⎝ 2 ⎟⎠ ⎟⎥
⎢⎣ ⎝
⎠⎦ 0
⎝
2
nπx
nπx ⎤
4
⎡ − 2x
+ 2 2 sin
cos
=⎢
2
2 ⎥⎦ 0
n π
⎣ nπ
2
−2⎡
nπx 2
nπx ⎤
x cos
− sin
=
nπ ⎢⎣
2
nπ
2 ⎥⎦ 0
=
−2 ⎡
2nπ 2
2 πn ⎤
−
2 cos
sin
− [0 − 0]
⎢
nπ ⎣
nπ
2
2 ⎥⎦
=
−2
[2 cos nπ − cos 0] + 24 2 [sin nπ − sin 0]
nπ
n π
=
−4
cos nπ
nπ
[Q sin nπ = sin (0) = 0]
4(− 1)
nπ
n
bn = −
If n = 1, 2, 3, ........
4
−4
4
−4
, b2 =
, b3 =
, b4 =
etc.
π
2π
3π
4π
b1 =
∞
∴
∑b
f(x) =
n
n =1
= b1 sin
sin
nπx
2
πx
4πx
3πx
2πx
+ b2 sin
+ b3 sin
+ b4 sin
+…
2
2
2
2
πx 4
4πx
3πx 4
4
2πx 4
−
sin
+ sin
sin
sin
−
+…
π
2
2
4π
2
3π
2 2π
x=
4πx
3πx 1
4 ⎡ πx 1
2πx 1
⎤
− sin
+ sin
− sin
+ …⎥
sin
π ⎢⎣
2
2
4
2
3
2 2
⎦
=
Q36. Obtain sine series for f(x) = π x – x2 in 0 < x < π .
Ans: Given that,
f (x) = πx – x2
∞
The sine series is
∑b
n
sin nx in (0, π)
n =1
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
2.50
MATHEMATICS-II [JNTU-ANANTAPUR]
Where,
2
π
bn =
π
∫
f ( x). sin nx.dx =
0
2
π
π
∫
(πx − x 2 ). sin nx.dx =
0
π
π
⎤
2 ⎡
⎢ πx sin nx.dx − x 2 .sin nx.dx ⎥
π ⎢
⎥⎦
0
⎣0
∫
∫
Applying integration by parts, we get,
2
=
π
π
π
⎧⎪ ⎡ ⎛ − cos nx ⎞
⎛ − cos nx ⎞ ⎤ ⎫⎪
⎛ − cos nx ⎞ ⎤ ⎡ 2 ⎛ − cos nx ⎞
−
x
dx
x
x
1
2
dx
−
−
π
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎨ ⎢
⎥ ⎬
⎥ ⎢
⎝ n ⎠ ⎦ 0 ⎪⎭
⎝ n ⎠ ⎦0 ⎣ ⎝ n ⎠
⎪⎩ ⎣ ⎝ n ⎠
2
=
π
π⎫
π
⎧
cos nx
sin nx ⎤ ⎡ 2 cos nx 2 ⎛ ⎛ − sin nx ⎞
⎛ − sin nx ⎞ ⎞⎤ ⎪
⎪⎡
dx ⎟ ⎟⎟⎥ ⎬
+ π 2 ⎥ − ⎢− x
− ⎜⎜ x⎜
⎟ − 1⎜
⎨⎢ − π x
n
n⎝ ⎝ n ⎠
n
⎠ ⎠⎥⎦ 0 ⎪
⎝ n
n ⎦ 0 ⎢⎣
⎪⎩⎣
⎭
∫
∫
∫
π⎫
π
⎧
⎡ 2 cos nx 2 ⎛
cos nx
sin nx ⎤
sin nx ⎛ cos nx ⎞ ⎞⎤ ⎪
2 ⎪⎡
+ π 2 ⎥ − ⎢− x
− ⎜⎜ − x
− ⎜ 2 ⎟ ⎟⎟⎥ ⎬
= ⎨⎢ − π x
n
n⎝
n
n
π ⎪⎣
n ⎦ 0 ⎢⎣
⎝ n ⎠ ⎠⎥⎦ 0 ⎪
⎭
⎩
2
=
π
π
π
⎧⎪⎡
cos nx
sin nx ⎤
2
⎡ 2 cos nx 2
⎤ ⎫⎪
x
x
x
sin
nx
cos
nx
−
−
+
+
−
π
+
π
⎨⎢
⎢
⎥ ⎬
⎥
n
n
n2 ⎦0 ⎣
n2
n3
⎦ 0 ⎪⎭
⎪⎩⎣
π
2
π
cos nx
sin nx
sin nx 2 cos nx ⎤
⎡
2 cos nx
− 2x
−
⎢− π x n + π 2 + x
⎥
n
n
n2
n3 ⎦0
⎣
2
=
π
⎡
⎛ cos nx ⎞⎤
⎛ − sin nx ⎞
2 ⎛ − cos nx ⎞
⎟ + (−2)⎜ 3 ⎟⎥
⎟ − ( π − 2 x )⎜
⎢(π x − x )⎜
2
n
⎠
⎝
⎝ n ⎠⎦ 0
⎠
⎝ n
⎣
=
π
⎧⎪⎡ 2
cos 0 ⎤ ⎫⎪
⎛ − sin 0 ⎞
⎛ − sin nπ ⎞ ⎛ cos nπ ⎞⎤ ⎡
2 ⎛ − cos nπ ⎞
⎟⎥ – ⎢0 − (π − 2(0))⎜
⎟ − 2 3 ⎥⎬
⎟ − 2⎜
⎟ − (π − 2π)⎜
⎨⎢(π − π )⎜
2
2
3
n
⎪⎩⎣
n ⎦ ⎪⎭
⎠
⎝
⎠
⎝ n
⎠ ⎝ n ⎠⎦ ⎣
⎝ n
=
2
π
=
2 ⎧⎡0 ( )(0) 2 cos nπ ⎤ ⎡0 2 ⎤ ⎫ 2 ⎧− 2 cos nπ + 2 ⎫
−
⎬
⎨⎢ − − π
⎥ − ⎢ − ⎥⎬ = ⎨ 3
n3 ⎭
π ⎩⎣
n 3 ⎦ ⎣ n 3 ⎦⎭ π ⎩ n
=
2 2
4
×
{1 − cos nπ}= 3 [1 – (– 1)n]
π n3
πn
⎧0; if n is even
⎪
∴ bn = ⎨ 8
⎪ 3 ; if n is odd
⎩ πn
Thus, f(x) = πx – x2
∞
∑ πn
f ( x) =
n =1
8
3
. sin nx
or
f ( x) =
8⎛
sin 3x sin 5x
⎞
⎜ sin x + 3 + 3 + ...⎟
π⎝
3
5
⎠
2.5 PARSEVAL’S FORMULA-COMPLEX FORM OF FOURIER SERIES
l
Q37. Prove that
∞
⎤
⎡1
[f(x)]2 dx = l⎢ a 20 +
(anna + b bn )⎥ provided the Fourier series for f(x) converges uniformly in
n =1
⎦⎥
⎣⎢ 2
–l
∫
∑
(– l, l).
OR
Derive the Parseval’s formula.
Model Paper-II, Q5
Ans:
Given that,
The Fourier series of f(x) converges uniformly in (– l, l)
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UNIT-2 (Fourier Series)
The Fourier expansion of f(x) in (– l, l) is given as,
a0
+
2
f(x) =
∞
⎛
∑ ⎝⎜ a
n
cos
n =1
nπx
nπx ⎞
+ bn sin
l
l ⎠⎟
... (1)
Where,
1
a0 =
l
an =
bn =
l
∫ f ( x) dx
... (2)
–l
1
l
1
l
l
⎛ nπ x ⎞
⎟ dx and
l ⎠
∫ f ( x ) cos ⎜⎝
–l
l
... (3)
⎛ nπx ⎞
⎟dx
l ⎠
∫ f ( x) sin⎜⎝
−l
... (4)
On multiplying equation (1) by f(a) and integrating term by term from – l to l, we get,
l
∫
−l
f ( x). f ( x) dx = ao
2
l
∫
−l
l
l
∞ ⎡
⎛ nπx ⎞ ⎤⎥
⎛ nπx ⎞
f ( x)dx + Σ ⎢a n f ( x ) cos⎜
⎟dx
⎟dx + bn f ( x ) sin⎜
n =1⎢
l ⎠
l ⎠ ⎥
⎝
⎝
−l
⎦
⎣ −l
∫
∫
... (5)
On substituting equations (2), (3) and (4) in equation (5), we get,
l
=
∫
[ f ( x )]2 dx =
−l
2
= a0
∞
ao
[ a o L] + Σ [a n ( a n l ) + bn (bn l ) ]
n =1
2
l ∞ 2
+ Σ [an l + bn2 l ]
2 n =1
⎡ a02 ∞ 2
⎤
2
l
= ⎢ + Σ (an + bn )⎥
⎣⎢ 2 n =1
⎦⎥
l
⎡ a2 ∞
⎤
∴ [ f ( x )]2 dx = l ⎢ 0 + Σ ( an2 + bn2 ) ⎥
⎢⎣ 2 n =1
⎥⎦
−l
∫
Hence proved.
Q38. Derive the complex form of Fourier series.
Ans: The Fourier series expansion of a periodic function f(x) of period 2l is given as,
f(x) =
a0 ∞ ⎡
⎛ nπx ⎞
⎛ nπx ⎞⎤
+ Σ ⎢an cos⎜
⎟ + bn sin⎜
⎟⎥
2 n =1⎣
⎝ l ⎠
⎝ l ⎠⎦
Where, a0 =
1
l
an =
1
l
1
bn =
l
l
∫ f ( x)dx
−l
l
⎛ nπx ⎞
⎟ dx and
l ⎠
∫ f ( x) cos⎜⎝
−l
l
⎛ nπx ⎞
⎟ dx
l ⎠
∫ f ( x) sin⎜⎝
−l
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MATHEMATICS-II [JNTU-ANANTAPUR]
jnπx
⎡ ⎛ jnπx
−
a0 ∞ ⎢ ⎜ e l + e l
+ Σ ⎢an ⎜
f(x) =
2 n =1⎢ ⎜
2
⎜
⎣⎢ ⎝
⇒
jnπx
⎞
⎛ jnπx
⎟
⎜ e l − e− l
⎟ + bn ⎜
2i
⎟⎟
⎜⎜
⎠
⎝
⎞⎤
⎟⎥
⎟⎥
⎟⎟⎥
⎠⎦⎥
jθ
− jθ
⎛
⎜Q cos θ = ⎛⎜ e + e
⎜
⎜
2
⎝
⎝
jθ
jθ
⎞
⎟ sin θ = e − e
⎟
2j
⎠1
⎞
⎟
⎟
⎠
jnπ x
jnπ x
jnπ x
jnπ x
a0 ∞ ⎡ an l
an − l
bn l
bn − l ⎤
e
e
e
e
+
Σ
⎢
+
+
−
⎥
f(x) = 2 n=1 2
2
2j
2j
⎣⎢
⎦⎥
jnπx
jnπx
a0 ∞ ⎡ l ⎡ an bn ⎤ − l ⎡ an bn ⎤ ⎤
⎢
+
+
Σ
+
e
e
⎢ − ⎥⎥
⎥
⎢
= 2 n =1
⎢⎣
⎣ 2 2 j ⎦ ⎥⎦
⎣ 2 2j⎦
jnπx
jnπx
a0 ∞ ⎡⎛ an − jbn ⎞ l
⎛ an + jbn ⎞ − l ⎤
⎢
⎥
e
e
+
+
Σ
⎟
⎜
⎟
⎜
f(x) = 2 n =1
2
2
⎢⎣⎝
⎥⎦
⎠
⎝
⎠
a0
1
1
(a – jbn); C–n =
(a + jbn)
, Cn =
2
2 n
2 n
On substituting the corresponding values in above equation, we get,
Let, C0 =
∞
⎡
f(x) = C0 + Σ ⎢Cne
jnπ x
l
n=1 ⎢
⎣
+ C−ne
−
jnπ x ⎤
l ⎥
... (1)
⎥⎦
The value of ‘Cn’ can be calculated as,
1
(an − jbn )
2
On substituting the value of ‘an’ and ‘bn’ in equation (2), we get,
Cn =
... (2)
l
l
1 ⎡1
1
⎛ nπx ⎞ ⎤⎥
⎛ nπx ⎞
⎢
f ( x) cos⎜
f ( x ) sin⎜
⎟dx
⎟dx − j
Cn =
l
2 ⎢l
⎝ l ⎠ ⎥⎦
⎝ l ⎠
−l
⎣ −l
∫
∫
l
1⎡
⎛ nπ x ⎞
⎛ nπ x ⎞ ⎤⎥
⎢
j
−
cos
sin
⎜
⎟
⎜⎝ l ⎟⎠ f ( x)dx
= 2l ⎢
⎝ l ⎠
⎥
⎣ −l
⎦
∫
l
1
= 2l
⇒
Cn =
1
2
∫
e
−j
nπx
l .f
( x ) dx
[Q e–iθ = cosθ – sinθ]
−l
l
∫ f ( x)e
− jnπ x
l .dx
... (3)
−l
On substituting n = – n in equation (3), we get,
Cn =
1
2l
l
∫
f ( x )e
jnπx
l .dx
... (4)
−l
From equation (3) and (4), we get,
1
Cn = 2l
∴
l
∫
f ( x) e
−
jnπx
l .dx
−l
for n = 0, ± 1, ± 2, +3,....
Equation (1) can be written as,
∞
f ( x) = Σ C n e
jnπx
l
n = −∞
This is the required complex from of Fourier series.
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2.53
UNIT-2 (Fourier Series)
Q39. Find the complex form of the Fourier series of f(x) = e–x in − |≤ x ≤| .
Ans: Given function is,
f(x)= e–x
Interval, −| ≤ x ≤ |
Complex form of Fourier series = ?
The complex form of fourier series is given as,
∞
f(x) = Σ C n e
jnπx
l
... (1)
n = −∞
Here,
l = 1 and
1
Cn =
2l
l
∫
f ( x )e
− jnπx
l dx
−l
The value of Cn can be obtained as,
1
Cn =
2l
l
∫e
− x − jnπx
e
dx
−l
l
1
− (1+ jnπ )
dx
= 2 e
∫
−l
1 ⎡ e −(1+ jnπ) ⎤
⎥
= ⎢
2 ⎢⎣ − (1 + jnπ) ⎥⎦
1
−1
1 ⎡ e −(1+ jnπ) + e (1+ jnπ) ⎤
= ⎢−
⎥
2 ⎣⎢
1 + jnπ
⎦⎥
=
e − (1+ jnπ) − e − (1+ jnπ)
2(1 + jnπ)
=
e1.e jnπ − e −1.e − jnπ
2(1 + jnπ)
=
e[cos nπ + j sin nπ] − e −1[cos nπ − j sin nπ]
2(1 + jnπ)
=
e(−1) n − e −1 (−1) n
2(1 + jnπ)
[Q cos nπ = (–1)n, sin nπ = 0]
⎡ e − e −1 ⎤ (−1) n ×1 − jnπ
= ⎢ 2 ⎥ (1 + jnπ)(1 − jnπ)
⎢⎣
⎥⎦
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MATHEMATICS-II [JNTU-ANANTAPUR]
=
∴ Cn =
(−1) n (1 − jnπ) sinh1
1+ n π
2 2
x
−x
⎛
⎜Q sinh x = e − e
⎜
2
⎝
⎞
⎟
⎟
⎠
(−1) n (1 − jnπ) sinh 1
1+ n2π2
On substituting the corresponding values in equation (1), we get,
∞ ⎡ ( −1) n (1 − jnπ) sinh1 ⎤
jnπx
⎥e
e–x = n =Σ−∞ ⎢
, which is required complex fourier series.
2 2
1+ n π
⎣⎢
⎦⎥
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3.1
UNIT-3 (Fourier Transform)
UNIT
FOURIER TRANSFORM
3
PART-A
SHORT QUESTIONS WITH SOLUTIONS
Q1. State Fourier integral theorem.
Ans: The Fourier integral theorem of f(x) is defined as,
f(x) =
Model Paper-I, Q1(e)
∞ ∞
1
π
∫ ∫ f (t ) cos λ(t − x)dt.dλ
0 −∞
Q2. Define Fourier sine integral and cosine integral.
Ans:
Fourier Sine Integral: The Fourier sine integral of f(x) is defined as,
∞
∞
∞
∞
0
0
2
sin λx f (t ) sin λ t dt dλ
f(x) =
π
0
0
Fourier Cosine Integral: The Fourier cosine integral of f(x) is defined as,
∫
f(x) =
∫
2
cos λx f (t ) cos λ t dt dλ
π
∫
∫
∞
Q3.
Using Fourier integral show that,
∫
0
π
1 − cos πλ
λ )dλ
λ =
sin(xλ
, 0 < x < π.
2
λ
Model Paper-III, Q1(e)
π
Ans: Let, f (x) = , 0 < x < π
2
Applying Fourier sine integral to f (x),
2
f (x) =
π
=
=
=
2
π
2
π
2
π
∞
=
∫
0
∴ f ( x) =
∞
∫
∞
sin( λ x ) dλ
∫ f (t ) sin λtdt
0
π
0
∞
∫ sin( λ x) d λ ∫ 1sin(λt ) dt
0
∞
∫
0
− cos(λt ) ⎤
sin( λ x ) d λ ⎡
⎢
⎣
0
∞
∫
0
λ
π
⎥
⎦0
1 − cos( λπ )
sin (λx)dλ
λ
... (1)
π
1 − cos( λπ )
sin (λx) dλ =
λ
2
π
[0 ≤ x ≤ π]
2
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Q4. Define the Fourier transform of function f(x).
Ans: The Fourier transform of a function f(x) is defined as,
F(s) = F.T{f(x)} =
∫ f ( x) e
Define Fourier sine and cosine transforms of
f(x).
Ans:
Fourier Sine Transform: The Fourier sine transform of the
function f(x) is,
∞
isx
Q7.
dx
−∞
∞
Q5. State the properties of Fourier transform.
Ans:
Model Paper-II, Q1(e)
The properties of Fourier transforms are as follows,
(a)
Linear Property: If f(x) and g(x) are two functions whose
Fourier transform is given by, F(s) and G(s) respectively, then,
F[a f(x) + b g(x)] = a. F(s) + b G(s)
Where, a and b are arbitrary constants.
(b)
Scaling Property: If f(x) is a function whose complex
Fourier transform is given by F(s),
Then,
1 ⎛s⎞
F⎜ ⎟
a ⎝a⎠
F{f(ax)} =
Where, a ≠ 0.
(c)
Shifting Property: If a function f(x) has a complex Fourier
transform F(s),
Then,
F{f(x–a)} = eisa.F(s)
(d)
Modulation Property: If a function f (x) has a complex
Fourier transform F(s), then,
Fs(S) = FST{f(x)} =
0
Fourier Cosine Transform: The Fourier cosine transform of
the function f(x) is defined as,
∞
Fc(S) = FCT {f(x)} =
Q8.
Find
Fourier
1
Fc [f (x)] =
2
a<x <b
x < a and x > b
Ans: By the definition,
1
2
=
∫ f ( x) . e
isx
of,
∞
∫ f ( x ) cos sx dx
0
∞
⎤
⎡a
⎢∫ f ( x) cos sx dx + ∫ f ( x ) cos sx dx ⎥
⎥⎦
⎢⎣ 0
a
a
⎤
1 ⎡
⎢∫ cos x cos sx dx + 0 ⎥
2 ⎣⎢ 0
⎦⎥
=
a
⎤
1 ⎡
⎢∫ cos(1 + s) x + cos (1− s ) xdx⎥
2 ⎣⎢ 0
⎦⎥
=
1
2
=
1 ⎡ sin(1 + s) x sin(1 − s) x ⎤
+
2 ⎢⎣ 1 + s
1 − s ⎥⎦ 0
=
1 ⎡ sin(1 + s)a sin(1 − s )a ⎤
+
2 ⎢⎣ 1 + s
1 − s ⎥⎦
F{f(x)} = F(s)
∞
transform
Ans: Given that,
f (x)= cos x, 0 < x < a
= 0, x ≥ a
Fourier cosine transform of f (x) is,
=
⎧⎪e ikx ,
f(x) = ⎨
⎪⎩0,
cosine
⎧cosx, 0 < x < a
f(x) = ⎨
.
x≥a
⎩ 0,
F{f(x) cos ax}=
Q6.
∫ f ( x) cos sx dx
0
=
1
[F(s + a) + F(s – a)]
2
Find the Fourier transform of,
∫ f ( x) sin sx dx
dx
−∞
a
⎤
⎡a
⎢∫ cos(1 + s) x dx + ∫ cos (1 − s) x dx⎥
⎥⎦
⎢⎣ 0
0
a
⎡
a
b
∞
a
b
⎤
= ⎢ ∫ f ( x ) .e isx dx + ∫ f ( x ) e isx dx + ∫ f ( x) . e isx dx⎥
⎢
⎣ −∞
⎥
⎦
⎤
⎡ b
ikx
isx
= ⎢0 + e . e dx + 0⎥
⎥
⎢ a
⎦
⎣
∫
b
=
∫e
a
=
ikx
.e
isx
ei (k + s) x
dx =
i( k + s )
b
Q9. Find Fourier sine transform series of xe–ax.
Model Paper-I, Q1(f)
Ans:
To find the Fourier sine transform of xe–ax
Fourier sine transform is given by,
a
e i ( k + s )b − e i ( k + s ) a
i (k + s )
Fs[x f (x)] =
−d
Fc [ f ( x)]
ds
∞
Fc [ f ( x )] =
∫ f ( x) cos sxdx
0
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3.3
UNIT-3 (Fourier Transform)
∞
Fs
(e–ax)
=
∫e
− ax
cos sxdx
0
Fs (e–ax) =
Fs (xe–ax) =
a
s + a2
2
−d ⎛ a ⎞
⎜
⎟
ds ⎝ s 2 + a 2 ⎠
⎛ − 2s
= − a⎜
⎜ 2
2
⎝ s +a
(
(s
=
2as
2
+ a2
)
2
⎞
⎟
⎟
⎠
)
2
Q10. Find Fourier sine transform of
1
.
x
Model Paper-II, Q1(f)
Ans: Given that,
f (x) =
1
x
Fourier sine transform of f (x) is,
Fs[f(x)] =
⎡1 ⎤
Fs ⎢ ⎥ =
⎣x⎦
2
π
2
π
∞
∫ f ( x) sin sx dx
0
∞
1
∫ x sin sx dx
0
Let, sx = t ⇒ s dx = dt ⇒ dx =
=
=
2
π
2
π
∞
s
∫ t sin t
0
∞
∫
0
=
2 π
π 2
=
π
2
dt
s
dt
s
sin t
dt
t
⎡ ∞ sin t
π⎤
dt = ⎥
⎢Q
2 ⎥⎦
⎢⎣ 0 t
∫
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MATHEMATICS-II [JNTU-ANANTAPUR]
–ax
Q11. Find the Fourier cosine transforms of e
sin ax.
Ans: Fourier cosine transform for f (x) is,
Fc (S) =
∞
∞
0
0
∫ f ( x) cos sx dx = ∫ e
− ax
∞
=
sin ax cos sx dx
∞
1
1 − ax
[e − ax sin(a + s ) x ]dx +
e sin( a − s ) xdx
20
20
∫
∫
[Q 2sinA cosB = sin(A + B) + sin(A – B)]
∞
∞
⎤
⎤
e − ax
1⎡
1⎡
e − ax
[ −a sin( a + s ) x − ( a + s ) cos( a + s ) x]⎥ + ⎢ 2
[−a sin( a − s) x − (a − s) cos( a − s ) x]⎥
= ⎢ 2
2
2
2 ⎣ a + (a + s)
2 ⎣ a + ( a − s)
⎦0
⎦0
⎤
⎡
e − ax
−ax
(− a sin bx − b cos bx) ⎥
⎢Q e sin bx dx = 2
2
a +b
⎦⎥
⎣⎢
∫
=
⎤
1⎡
1
1
[a + s ] + 2
( a − s )⎥
⎢ 2
2
2
2 ⎣ a + (a + s )
a + (a − s)
⎦
Q12. Find the Fourier cosine transform of 5e–2x + 2e–5x.
Ans: Given that,
f (x) = 5e–2x + 2e–5x
The Fourier cosine transform is given as,
∞
Fc ( f (x)) =
∫ f ( x) cos px dx
0
∞
∫
∞
∞
0
0
−2 x
+ 2e −5 x ) cos px dx = 5e − 2 x cos px dx + 2e −5 x cos px dx
= (5e
∫
0
∞
∫
=5 e
−2x
∫
∞
∫
cos px dx + 2 e −5 x cos px dx
0
0
∞
∞
⎤
⎡ e −5 x
⎤
⎡ e−2x
px
p
px
px
p
px
(
2
cos
sin
)
2
(
5
cos
sin
)
+
−
+
−
+
=5⎢ 2
⎥
⎢
⎥
2
2
2
⎦0
⎣5 + p
⎦0
⎣2 + p
⎡ ∞ − ax
⎤
e − ax
( −a cos bx + b sin bx )⎥
⎢Q e cos bx dx = 2
2
a +b
⎢⎣ 0
⎥⎦
∫
⎡
= 5 ⎢0 +
⎣
=
⎡
2 ⎤
5 ⎤
⎥ + 2 ⎢0 + 2
⎥
p + 4⎦
p + 25 ⎦
⎣
2
10
10
+ 2
p + 4 p + 25
2
∴ It is the required solution.
Q13. Define the inverse Fourier transform.
Ans: The inverse Fourier transform of f(S) is defined as,
f(x) = I.F.T {F(S)} =
1
2π
∞
∫ F ( S )e
−isx
ds
−∞
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3.5
UNIT-3 (Fourier Transform)
Q14. Define the inverse Fourier sine and cosine transform.
Ans:
Inverse Fourier Sine Transform: The inverse Fourier transform of f(s) is defined as,
∞
2
Fs ( S ) sin sx ds
f(x) = I.F.T {F(S)} =
π0
∫
Inverse Fourier Cosine Transform: The inverse Fourier cosine transform of Fc(S) is defined as,
∞
f(x) = IFCT {Fc(S)} =
2
Fc (S ) cos sx ds
π
∫
0
Q15. Define finite Fourier sine and cosine transforms.
Ans:
Finite Fourier Sine Transform: The finite Fourier sine transform of f(x) which is defined in 0 < x < k as,
k
nπx
dx , n = 1, 2, 3,.....
k
0
Finite Fourier Cosine Transform: The finite Fourier cosine transform of f(x) which is defined in 0 < x < k as,
Fs(n) =
∫ f ( x) sin
k
Fc(n) =
∫ f ( x) cos
0
nπx
dx , n = 1, 2, 3,...
k
Q16. Define inverse finite Fourier sine and cosine transforms.
Ans:
Inverse Finite Fourier Sine Transform: The inverse finite Fourier sine transform of Fs(n) is defined as,
Model Paper-III, Q1(f)
∞
nπx
2
Fs (n) sin
, 0<x<k
k n =1
k
Inverse Finite Fourier Cosine Transform: The inverse finite Fourier cosine transform of Fc(n) is defined as,
∑
F(x) =
2
F(x) =
k
∞
∑ F (n) cos
c
n =1
nπx
, 0<x<k
k
Q17. Find the finite Fourier cosine transform of f(x) = x2 in (0, l).
Ans: Given that,
f (x) = x2 in (0, 1)
Finite Fourier cosine transform of f (x) is given by,
l
Fc(n) =
∫
0
⎛ nπx ⎞
f ( x ) cos ⎜
⎟ dx =
⎝ l ⎠
2
= x
1
∫x
2
cos nπxdx (Q l =1)
0
1
1
1
− cos nπx ⎤
sin nπx
2 ⎡ cos πx
⎢x −
−
−
dx ⎥
nπ 0 nπ ⎢
nπ 0
nπ
⎥
0
⎣
⎦
∫
1
1
1 2
2
2 sin nπx
[ x sin nπx ] + 2 2 [ x cos nπx ] –
=
2 2
0
0
nπ
nπ
n π
n π
=
2
2
1
(sin nπ) + 2 2 (cos nπ) – 3 3 sin nπ
n π
nπ
nπ
=
sin nπ
nπ
1
0
2 ⎤
2
2
⎡
n
⎢1 − n 2 π 2 ⎥ + n 2 π 2 cos nπ = n 2 π 2 (−1)
⎣
⎦
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SIA GROUP
3.6
MATHEMATICS-II [JNTU-ANANTAPUR]
PART-B
ESSAY QUESTIONS WITH SOLUTIONS
3.1 FOURIER INTEGRAL THEOREM (ONLY STATEMENT) – FOURIER SINE AND COSINE
INTEGRALS
Q18. Obtain an expression for sine and cosine integrals.
Ans: According to Fourier integral theorem,
f(x) =
1
π
∞ ∞
∫ ∫ f (t ) cos(λt − λx)dt.dλ
0 −∞
∞ ∞
1
f (t )[cos λ t cos λ x + sin λ t.sin λx ]dt .dλ
=
π 0 −∞
∫∫
=
1
π
∞ ∞
∫∫
f (t ) cos λt cos λx dtdλ +
0 −∞
∞
1
π
∞
(Q cos(A – B) = cosA cosB + sinA sinB)
∞ ∞
∫ ∫ f (t ) sin λt. sin λx dt.dλ
0 −∞
∞
∞
1
1
cos λ x f (t ). cos λ t dt dλ +
sin λx f (t ) sin λt dt.dλ
=
π0
π0
−∞
−∞
∫
∫
∫
∫
... (1)
Case (i): If f(x) is an odd function and f(t) cosλt is also an odd function and f(t) sinλt is an even function. Then, the first term on the
right side of equation (1) becomes zero.
∞
∞
0
−∞
1
sin λx
f(x) =
π
∫
∫ f (t ) sin λt dt.dλ
∞
∞
0
0
2
sin λ x f (t ).sin λ t dt.d λ
π
∫
f ( x) =
∫
... (2)
Equation (2) represents the Fourier sine integral.
Case (ii): If f(x) is an even function and f(t) sinλt is also an even function and f(t) cosλt is an odd function then the second term on
the right side of equation (1) becomes zero.
∞
1
cos λx
f(x) =
π
∫
0
∴ f(x) =
∞
∫ f (t ) cos λt dt.dλ
−∞
∞
∞
0
0
2
cos λ x f (t ) cos λt dt.dλ
π
∫
∫
... (3)
Equation (3) is known as Fourier cosine integral.
∞
Q19. Using Fourier integral theorems show that
cos ωx
∫ 1+ ω
2
dω =
0
∞
Ans: To prove that
cos ωx
∫ 1+ ω
0
2
dω =
π −x
e .
2
π −x
e using Fourier integral theorem.
2
The Fourier cosine integral is given as,
∞
⎡∞
⎤
2
⎢ f (t ) cos λtdt ⎥ dλ
λ
cos
x
f(x) =
π
⎢0
⎥
0
⎣
⎦
∫
∫
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... (1)
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3.7
UNIT-3 (Fourier Transform)
Let, f (x) = e , f (t) = e and put λ = ω in equation (1), we get,
–x
–t
∞
⎡∞
⎤
2
⎢ e −t cos ωt dt ⎥ dω
ω
cos
x
e =
π
⎢
⎥
0
⎣0
⎦
∫
–x
∫
∞
⇒
–x
e
2 ⎡
⎛ 1 ⎞⎤
dω
=
⎢cos ωx ⎜⎝ 2
2⎟
⎠ ⎥⎦
π ⎣
1
+
ω
0
∫
∞
∞
⎤
⎡ e −ax
a ⎤
− ax
(−a cos px + p sin px)⎥ = 2
[Q L {cos at} = e cos px dx = ⎢ 2
2
2⎥
⎦⎥ 0 a + p ⎦⎥
⎣⎢ a + p
0
∫
∞
∞
∴
⇒
e–x =
⇒
π −x
e =
2
cos ω x
∫ 1+ ω
2
dω =
0
2 cos ωx
dω
π 1 + ω2
0
∫
∞
cos ωx
∫ 1+ ω
2
dω
0
π −x
e
2
∞
∫
αx dx= e–ααx.
Q20. Solve the integral equation f(x) cosα
Model Paper-III, Q6
0
Ans: Given equation is,
∞
∫ f ( x) cos αxdx = e
–αx
... (1)
0
∞
∫ f ( x) cos αxdx = F (α)
... (2)
c
0
∴ From equations (1) and (2), it can be observed that,
Fc(α) = e–αx
By inversion formula, we get,
... (3)
∞
f(x) =
2
[e −αx cos(αx )]dx
π
∫
0
2⎛ α ⎞
⎟
f(x) = ⎜
π ⎝ α + x2 ⎠
∞
⎡
⎢Q L{cos at} = e − as cos psds =
⎢
0
⎣
∫
∞
⎤
⎡ e −as
a ⎤
(−a cos ps + p sin ps)⎥ =
⎢ 2
⎥
2
2
+
a
p
⎥⎦ 0
a + p 2 ⎦⎥
⎣⎢
On substituting equation (4), in equation (2), we get,
f(x) =
2⎛ α ⎞
⎜
⎟ in Fc(α)
π ⎝ α2 + x2 ⎠
∞
=
∫ f ( x) cos αxdx
0
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SIA GROUP
... (4)
3.8
MATHEMATICS-II [JNTU-ANANTAPUR]
We get,
∞
α
2⎛
∫ π ⎜⎝ α
Fc(α) =
2
⎞
+x
0
2⎟
⎠
cos αxdx
∞
⇒
2 ⎛ α ⎞
⎟ cos αxdx
⎜
π ⎝ α2 + x 2 ⎠
0
∫
Fc(α) =
... (5)
From equation (3),
Fc(α) = e–αx
On substituting Fc(α) = e–αx in equation (5), we get,
∞
e
–αx
α
2
cos αxdx
=
π (α 2 + x 2 )
0
∫
... (6)
From equations (1) and (6), we get,
∞
∞
0
α
2
∫ f ( x) cos αxdx = ∫ π (α
2
0
+ x2 )
cos αxdx
= e–αx
Hence proved.
3.2 FOURIER TRANSFORM –
FOURIER SINE AND COSINE TRANSFORMS – PROPERTIES
Q21. State and prove the properties of Fourier transform.
Ans: The salient properties of Fourier transforms are,
(a)
Linear Property: If f(x) and g(x) are two functions whose Fourier transform is given by, F(s) and G(s) respectively, then,
F[a f(x) + b g(x)] = a. F(s) + b G(s)
Where, a and b are arbitrary constants.
Proof: By definition∞of Fourier transform,
∫
F(s) = e isx f ( x) dx
Similarly,
−∞
∞
∫
isx
G(s) = e g ( x ) dx
−∞
∞
∫
F[a f(x) + b g(x)] = e isx [ af ( x) + bg ( x)]dx
−∞
∞
=
∫e
∞
isx
[ af ( x)]dx +
−∞
∫
isx
[bg ( x)]dx
−∞
∞
=a
∫e
∞
∫
e isx f ( x ) dx + b e isx g ( x) dx
−∞
−∞
= a. F(s) + b. G(s).
∴ F [af ( x) + b.g ( x)] = a.F ( s) + b.G(s)
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3.9
UNIT-3 (Fourier Transform)
(b)
Scaling Property: If f(x) is a function whose complex Fourier transform is given by F(s),
Then,
F{f(ax)} =
1 ⎛s⎞
F⎜ ⎟
a ⎝a⎠
Where, a ≠ 0.
Proof : By the definition of Fourier transform we get,
∞
F(s) =
∫e
isx
. f ( x )dx
isx
. f ( ax ) dx
−∞
∞
F{f(ax)}=
∫e
−∞
Let, ax = u
⇒
x=
⇒
dx =
u
1
= (u)
a
a
du
a
∞
∴
f{f(ax)} =
∫e
is
(ua ). f (u ).⎛ du ⎞
⎜ ⎟
⎝ a ⎠
−∞
=
=
1
a
∞
∫e
i ( s a )u
. f (u ).du
−∞
1 ⎛s⎞
.F ⎜ ⎟
a ⎝a⎠
[Q By definition of Fourier transform]
(c)
Shifting Property: If a function f(x) has a complex Fourier transform F(s),
F{f(x – a)} = eisa.F(s).
Proof : By definition of Fourier transform,
∞
F(s) =
∫e
isx
. f ( x )dx
isx
. f ( x − a ) dx
−∞
∞
F{f(x–a)} =
∫e
−∞
Let, x – a = u
⇒
x= u+a
dx= du
∞
∴
∫
F{f(x – a)}= e is ( u + a ) . f (u ).du
−∞
∞
=
∫e
isu + isa
. f (u ).du
−∞
∞
=
∫e
isu isa
e . f (u ).du
−∞
∞
isa
=e
∫e
isu
. f (u ).du = eisa . F(s)
−∞
∴ F{ f ( x − a)} = e isa .F ( s)
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SIA GROUP
3.10
(d)
MATHEMATICS-II [JNTU-ANANTAPUR]
Modulation Property: If a function f (x) has a complex Fourier transform F(s), then,
1
[F(s + a) + F(s – a)]
2
Proof: By the definition of Fourier transform,
F{f(x) cosax} =
∞
∫e
F(s) =
isx
f ( x) dx
isx
.[ f ( x ) cos ax]dx
isx
. cos ax f ( x ) dx
−∞
∞
∫e
F{f(x).cosax} =
−∞
∞
∫e
=
−∞
∞
=
⎡ e iax + e −iax ⎤
e isx .⎢
⎥ f ( x ) dx
2
⎢
⎣
⎦⎥
−∞
∫
1
=
2
1
=
2
=
1
2
[Q cosθ =
(e iθ + e −iθ )
]
2
∞
∫ [e
isx iax
+ e isx .e −iax ] f ( x) dx
ix ( s + a )
+ e ix ( s −a ) ] f ( x )dx
e
−∞
∞
∫ [e
−∞
∞
∫
[ e ix ( s + a ) . f ( x )]dx +
−∞
=
1
1
F ( s + a) + F ( s − a)
2
2
=
1
[ F ( s + a ) + F ( s − a )]
2
1
2
∞
∫ [e
ix ( s − a )
. f ( x )]dx
−∞
1
[ F (s + a) + F ( s − a )
2
Q22. Show that the Fourier sine transform of,
∴ F { f ( x ) cos ax} =
, for 0 < x < 1
⎧x
⎪
2
x
,
for 1< x < 2 is 2 sin s(1− cos s) .
−
f(x)= ⎨
s2
⎪0
, for x > 2
⎩
Ans: Given that,
f (x)= x, when 0 < x < 1 limits lies between 0 and 1.
f (x)= 2 – x, when 1 < x < 2 limits lies between 1 and 2.
f (x) = 0, when x > 2 limits lies between 2 and ∞.
By sine transform,
∞
f (x) =
∫ f ( x) sin sxdx
0
1
fs(s) =
∫
2
∫
0
1
fs(s) =
∫
0
∞
∫
f ( x ) sin sxdx + f ( x ) sin sxdx + f ( x ) sin sxdx
1
2
2
∫
x sin sxdx + ( 2 − x ) sin sxdx + 0
1
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3.11
UNIT-3 (Fourier Transform)
By integration by parts, we get,
1
fs(s) =
1
⎡ cos sx ⎤
⎢− x s ⎥ +
⎣
⎦0
∫
0
2
∫
=
=
=
2
s
2
1
sin s −
sin s
s2
s2
sin 2 s
−
s2
sin s
=
s2
2 sin s
s2
−
sin 2 s
=
s2
2 sin s − sin 2 s
0
cos xt
1+ t
that F(xe–|x|) = i
2
dt =
π –|x|
e . Hence show
2
4s
.
(1 + s 2 )2
=
Model Paper-I, Q6
=
−2
∫
=
f ( x ) e isx dx
−∞
Given that,
f(x) = e–|x|
=
(is ) − 1
⇒
F[f(x)] = ?
The Fourier transform of f(x) is given as,
∞
−2
−2
f ( x ) e dx
e −| x| e isx dx
f(x) =
∞
0
∫
∫
e x e isx dx + e − x e isx dx
−∞
∫e
−∞
−2
− ( s 2 + 1)
= f(s)
1+ s2
=
1
2π
2
2π
0
∞
0
=
2
=
∞
∫e
− isx
f ( s ) ds
−∞
2
1+ s2
in the above equation,
we get,
−∞
=
− s2 −1
1
f(x) =
2π
∞
∫
(Q i2 = –1)
( −1) s 2 − 1
On substituting, f(s) =
isx
−∞
⇒ F[f(x)] =
i s −1
2 2
∴ The inverse Fourier transform of f(s) is f(x).
⎧⎪e − x if x > 0
f(x) = e–|x| = ⎨ x
⎪⎩ e if x < 0
∫
−2
=
2
∞
Fourier transform of f(x) =
F[f(x)] =
∞
=1⎢
Ans: We know that,
⇒
0
⎡ is − 1 − 1 − is ⎤
⎥
⎣ (1 + is )(is − 1) ⎦
s2
Q23. Find the Fourier transforms of f(x) = e–|x| and
∫
∞
e0
e −∞
e −∞
e0
−
+
−
1 + is 1 + is is − 1 is − 1
2 sin s (1 − cos s )
deduce that
0
1 ⎤
0⎡ 1
−
=e ⎢
⎥
⎣ 1 + is is − 1 ⎦
s2
∞
0
⎡ e − x (1+is ) ⎤
⎡ e x (1+is ) ⎤
⎥
⎥ +⎢
=⎢
⎣⎢ 1 + is ⎦⎥ − ∞ ⎣⎢ is − 1 ⎦⎥ 0
(sin 2 s − sin s)
+
∫
⎡ e x (is −1) ⎤
⎡ e x (1+ is ) ⎤
⎥ +⎢
⎥
=⎢
⎢⎣ 1 + is ⎥⎦ − ∞ ⎢⎣ is − 1 ⎥⎦ 0
cos s 1 ⎛ sin s ⎞ cos s 1
− 2 (sin 2 s − sin s )
+ ⎜
⎟+
s
s⎝ s ⎠
s
s
1
∫
e x(1+ is ) dx + e x( is −1) dx
−∞
1
− cos s 1 ⎛ sin sx ⎞ ⎡
cos s ⎤ 1 ⎡ sin sx ⎤
+ ⎜
−
⎟ + 0+
s
s ⎝ s ⎠ 0 ⎢⎣
s ⎥⎦ s ⎢⎣ s ⎥⎦1
=−
=
=
cos sx
cos x ⎤
cos sx
⎡
dx + ⎢− ( 2 − x)
dx
+ −
s
s ⎥⎦1
s
⎣
1
=
∞
0
2
x + isx
∫
dx + e
− x + isx
dx
0
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f(x) =
1
π
∞
∫e
− isx
−∞
∞
∫e
− isx
−∞
∞
∫e
−∞
⎛ 2 ⎞
⎜
⎟ ds
⎝1+ s2 ⎠
⎛ 1 ⎞
⎜
⎟ ds
⎝1+ s2 ⎠
− isx ⎛
1 ⎞
⎜
⎟ ds
⎝1+ s2 ⎠
SIA GROUP
3.12
MATHEMATICS-II [JNTU-ANANTAPUR]
Q24. Find the Fourier transform of,
Given that,
f(x) = e–|x|
⇒
∞
1
π
∫e
for | x | < a
⎧1,
and hence evaluate
f(x) = ⎨
⎩0, for | x | > a > 0
− isx ⎛
1 ⎞
⎜
⎟ ds = e–|x|
⎝1+ s2 ⎠
−∞
∞
∫
0
∞
⇒
⎛ 1 ⎞
e − isx ⎜
⎟ ds = πe–|x|
2
+
s
1
⎝
⎠
−∞
∫
⇒
∫
⎛ cos sx
∫ ⎜⎝ 1 + s
−
2
−∞
⇒
cos sx
∫ 1+ s
∫
a
∫
f (t ) e ist dt = 1 . e ist dt
−∞
−a
a
⎡ e ist ⎤
e isa − e −isa
= ⎢
=
⎥
is
⎣ is ⎦ − a
i sin sx ⎞
⎟ ds = πe–|x|
1+ s2 ⎠
=
On equating real and imaginary parts, we get,
∞
−∞
sin as.cos xs
ds.
s
∞
F(s) =
⎛ 1 ⎞
(cos sx − i sin sx ) ⎜
⎟ ds = πe–|x|
2
s
+
1
⎝
⎠
−∞
∞
∫
Ans: The Fourier transform of the function f (x) is,
∞
⇒
∞
sin x
dx and
x
2i sin sa 2 sin sa
=
is
s
⎡
e isa − e −isa ⎤
=
Q
sin
sa
⎢
⎥
2i
⎣
⎦
ds = πe–|x|
2
−∞
To evaluate the integrals (a) and (b), apply the inversion
formula of F(s),
Let, s = t,
∞
⇒
cos xt
∫ 1+ t
2
∞
1
F ( s ) e −isx ds
f (x) =
2π − ∞
∫
dt = πe–|x|
−∞
∞
⇒
2
cos xt
∫ 1+ t
dt = πe–|x|
2
(a)
On substituting, F(s) =
... (1)
2 sin sa
in equation (1),
s
0
∞
∞
∴
cos xt
∫ 1+ t
2
dt =
0
F(xnf(x)) = – i
=–i
∫
π −| x |
e
2
=
d
f(s)
ds
d ⎛ 2 ⎞
⎟
⎜
ds ⎝ 1 + s 2 ⎠
= –i2
= –i2
= –i2
=i
1
2 sin sa −ixs
e ds
f (x) =
2π − ∞ s
d ⎛ 1 ⎞
⎜
⎟
ds ⎝ 1 + s 2 ⎠
=
4s
2 2
(1 + s )
Look for the S IA GROUP LOGO
∫
−∞
∞
∫
−∞
sin sa
(cos xs − i sin xs ) ds
s
∞
sin sa cos xs
i sin sa sin xs
ds −
ds
s
π
s
−∞
∫
∞
f (x) =
(1 + s 2 ) 2
(1 + s 2 ) 2
1
π
∞
The integrand in the second integral is an odd function
and hence this integral vanishes. Therefore,
(1 + s 2 ) 0 − 1( 2 s )
−2 s
1
π
1 sin sa cos sx
ds
π −∞
s
∫
∞
⇒
=
1 sin sa cos sx ds
π −∞
s
∫
= π f (x)
⎧π for | x |≤ a
=⎨
⎩0 for | x |> a
on the TITLE COVER before you buy
... (2)
3.13
UNIT-3 (Fourier Transform)
(b)
If x = 0, then equation (2) yields,
∞
sin sa
ds = π
s
−∞
∫
∞
⇒
∫
0
π
sin sa
ds =
s
2
2
Q25. Find the Fourier cosine transform of e − x .
Ans: Given that,
2
f (x) = e − x
Fourier cosine transform of f(x) is given as,
Model Paper-II, Q6
∞
∫
Fc (p) = f ( x) cos( px) dx
0
∞
∴ Fc (p) =
∫e
− x2
cos px dx
0
∞
Let, Fc (p) =
∫e
− x2
cos px dx = I
... (1)
0
On differentiating equation (1) on both sides with respect to ‘p’, we get,
dI
=
dp
∞
∫e
− x2
0
d
(cos px) dx
dp
∞
=
∫e
− x2
( − x sin px) dx
0
On multiplying and dividing by 2 on RHS we get,
∞
2
dI
1
( −2 xe − x ) sin px dx
=
2
dp
∫
0
∞
⇒
2
dI 1
sin px ( −2 xe − x ) dx
=
dp 2
∫
0
∞
∞
⎤
2
1⎡
dI
⎤
⎡d
− x2
⎢
sin px (−2 xe )dx − ⎢ sin px ⎥ (−2 xe − x )dx dx ⎥
=
dp 2 ⎢
⎥
⎦
⎣ dx
0
0
⎦
⎣
∫
∫
∫
⎡Q f ( x) g ( x ) dx = f ( x ) g ( x ) dx − ⎡ f ′( x ) g ( x ) dx ⎤ dx ⎤
⎢⎣
⎢⎣
⎥⎦ ⎥⎦
∫
[
1⎡
dI
− x2
= ⎢ sin px.e
dp 2 ⎢
⎣
] − ∫ [( p cos px)e ]dx⎤⎥⎥
∞
∞
− x2
0
0
∞
⎤
2
dI
1 ⎡
⎢
−
−
[
0
0
]
p
e − x cos pxdx⎥
=
dp
2 ⎢
⎥
0
⎦
⎣
∫
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎦
∫
∫
∫
⎡Q − 2 xe − x 2 dx = e − x 2 ⎤
⎥⎦
⎢⎣
∫
[ Q e– ∞ = 0, sin 0 = 0]
SIA GROUP
3.14
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting p = 0 in equation (2), we get,
I = c e–0 = c
∞
dI
− p − x2
e cos px dx
=
dp
2
∫
0
∴
−p
I
=
2
π
2
c=
[ From equation (1)]
π
in equation (2), we get,
2
On substituting c =
dI − p
=
dp
I
2
Integrating on both sides, we get,
⇒
1
∫ I dI = ∫
⇒
⇒
log I =
π −
e
I=
2
−p
dp
2
∴
1 ⎡− p2 ⎤
⎥ + log c
⎢
2 ⎢⎣ 2 ⎥⎦
[ Q loge x = a ⇒ x = ea]
⇒
I = ce
On substituting p = 0 in equation (1), we get,
... (2)
∫e
−x
f(x) =
cos 0 dx
∫e
−x
FS(p) =
.dx
1
x( x + a 2 )
∫ f ( x) sin px dx
0
⇒
1
1
dt = 2x dx ⇒ dx =
dt =
dt
2x
2 t
FS(p) =
sin px
∫ x( x
∫e
−t
0
dt
d
{FS ( p )} =
dp
2 t
1
(t ) −1 / 2 e −t dt
I=
2
∫
∞
∫ x( x
=
d
. (sin px ) dx
+ a 2 ) dp
1
2
0
∞
0
∫ x( x
1
2
0
∞ 1
−1
t 2 e −t dt
∫
∞
=
0
∫ (x
0
⇒
⇒
1 ⎛1⎞
I = Γ⎜ ⎟
2 ⎝2⎠
I=
... (1)
On differentiating equation (1) with respect to ‘p’, we
∞
⇒
dx
+ a2)
2
0
get,
∞
1
2
.
2
∞
I=
1
x(x 2 + a 2 )
∞
2
Let, t = x2
⇒
p2
4
Fourier cosine transform of f(x) is given as,
0
⇒
π −
e
2
− x2
Model Paper-III, Q7
∞
I=
[ From equation (1)]
Given that,
0
I =
p2
4
Ans:
∞
I=
π −
e
2
Fc{p} = I =
Q26. Find the Fourier sine transform of
p2
4
2
p2
4
∴ The fourier cosine transform of e
− p2
+ log c
log I =
4
−
[ From equation (3)]
1
π
2
π
I=
2
∞
⎤
⎡
⎢Q Γ ( p) = e −t t p −1dt ⎥
⎥
⎢
0
⎦
⎣
∫
2
x +a
1
2
+ a2 )
.x. cos px dx
cos px dx
... (2)
∞
1
Q
+ a2 )
=
2
∫ 2 ye
−( x 2 + a 2 ) y 2
dy
0
On multiplying cos px on both sides of above equation,
1
⎡
⎤
⎢Q Γ 2 = π ⎥
⎣
⎦
we get,
cos px
... (3)
Look for the S IA GROUP LOGO
2
x +a
2
∞
=
∫
cos px.2 ye −( x
2
+ a2 ) y2
dy
0
on the TITLE COVER before you buy
3.15
UNIT-3 (Fourier Transform)
Integrating on both sides with respect to x from 0 to ∞.
∞
cos px
∫x
2
0
+a
2
∞∞
dx =
∫ ∫ cos px.2 ye
−( x 2 + a 2 ) y 2
dy dx
0 0
∞∞
=
∫ ∫ cos px.2 ye
−x2y2
.e − a
2 2
y
dy dx
0 0
∞
=
∫
2 ye − a
2 2
y
0
⎤
⎡∞ 2 2
⎢ e − x y cos px dx ⎥ dy
⎥
⎢0
⎦
⎣
∫
... ( 3)
∞
Let,
∫e
− x2 y 2
cos px dx = S
... (4)
0
On differentiating on both sides with respect to p, we get,
dS
=
dp
∞
∫e
− x2 y 2
0
d
(cos px ) dx
dp
∞
=
∫e
− x2 y 2
( − x sin px ) dx
0
=
=
=
1
2y2
2
− x2 y 2
(− x sin px )dx
0
∫ (−2 xy .e
2
2
− x2 y2
) sin px dx
0
∞
1
2y
∫ 2 y .e
∞
1
2y
∞
∫ [sin px.(−2 xy e
2 − x2 y 2
2
)]dx
0
∞
∞
1 ⎡
⎤ ⎤⎥
⎡d
2 − x2 y2
2 − x2 y2
⎢
−
−
−
px
xy
e
px
xy
e
dx
sin
2
sin
(
2
)
=
⎥ dx ⎥
⎢ dx
2 y 2 ⎢⎣
⎦ ⎦
⎣
0
0
∫
∫
(
2 2
1 ⎡
⎢ sin px.e − x y
=
2
2 y ⎢⎣
1
=
2y2
=
∴
⇒
1
2y
) − ∫ [( p cos px)(e
∞
∫
∞
−x2y2
0
0
∞
⎤
⎡
2 2
⎢0 − 0 − p (e − x y cos px )dx ⎥
⎥
⎢
0
⎦
⎣
[ − pS ] =
2
∫
− pS
2y
2
]
⎤
) ⎥ dx
⎥
⎦
⎡Q − 2 xe − x 2 dx = e − x 2 ⎤
⎥⎦
⎢⎣
∫
[Q e–∞ = 0, sin 0 = 0]
∞
⎡
⎤
2 2
⎢Q S = e − x y cos px dx ⎥
⎢
⎥
0
⎣
⎦
∫
− pS
dS
=
dp
2 y2
dS
dp
=−p 2
S
2y
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
3.16
MATHEMATICS-II [JNTU-ANANTAPUR]
Integrating on both sides with respect to p, we get,
1
p
∫ S dS = − ∫ 2 y
2
Therefore, equation (5), becomes,
∞
dp
S =
∫e
− x2 y 2
cos px dx
0
⇒
log S = −
⇒
log S =
p2 1
.
+ log C1
2 2y2
− p2
4y2
+ log C1
−p
S = C1e
⇒
=
2
π − p2 / 4 y2
e
2y
π − p2 / 4 y2
e
in equation (3), we get,
2y
On substituting S =
/4 y2
... (5)
On substituting p = 0 in equation (4), we get,
∞
cos px
∫x
0
2
+ a2
∞
dx =
∫ 2 ye
−a2 y2
.
0
π − p2 / 4 y2
.e
dy
2y
∞
S=
∫e
− x2 y 2
dx
∞
[Q cos 0 = 1]
Let, x2y2 = t
∴
⇒
dx =
xy =
1
S=
p
2y t
1
e −t
=
dt
2y t
⇒
1
t −1/ 2 e −t dt
2y
∫
∞ 1
−1
1
t 2 e −t dt
S=
2y
∫
⇒
S=
1 ⎛1⎞
Γ⎜ ⎟
2y ⎝ 2 ⎠
⇒
S=
1
π
2y
S=
∞
⎤
⎡
⎢Q Γ( p ) = e − t t p −1dt ⎥
⎥
⎢
0
⎦
⎣
∫
2
π − pa
.e
2a
π ⎡ e − ap ⎤
FS(P) = 2a ⎢ − a ⎥
⎦⎥
⎣⎢
=
∴ FS ( P ) =
π
when p = 0 in equation (5),
2y
π − ap
e
+ C1
− 2a
π
− 2a 2
∞
π
= C1 e 0
2y
π
C1 =
2y
Look for the S IA GROUP LOGO
e − ap + C1
Q27. Find the Fourier cosine transform of e–ax and
we get,
⇒
π − 2. 2 a .a
.e
2a
d
π − Pa
e
[ FS ( p)] =
dp
2a
⎡
⎤
⎛1⎞
⎢Q Γ⎜ ⎟ = π ⎥
⎝2⎠
⎣
⎦
π
2y
On substituting S =
π.
d
π − Pa
[ FS ( p)] =
e dp
2a
dp
0
⇒
⎞
⎟
⎟
⎠ dy
On integrating above equation with respect to ‘P’ on
both sides, we get,
0
⇒
∫
=
∞
S=
dy
0
dt
0
⇒
2
⎛
∞ −a 2 ⎜ y 2 + p
2 2
⎜
4
a
y
π e ⎝
=
dt
1
∫
y − p2 / 4 y2
t
2 t
∞
∴
2 2
0
⇒
ydx =
∫
π e−a
=
0
hence evaluate
cospx
∫a
0
2
+ p2
dx .
Ans: Given function is,
Fourier cosine transmission of f(x) is given as,
f(x) = e–ax
on the TITLE COVER before you buy
3.17
UNIT-3 (Fourier Transform)
From definition,
∞
Fc(p) =
∫ f ( x) cos pxdx
0
∞
=
∫e
− ax
cos pxdx =
0
a
... (1)
2
a + p2
∞
⎡
⎤
⎢Q L{cos at} = e − ax cos pxdx
⎥
⎢
⎥
0
⎢
⎥
∞
⎢
⎧ e − ax
⎫ ⎥
=⎨ 2
(− a cos px + p sin px )⎬ ⎥
⎢
2
⎢
⎩a + p
⎭0 ⎥
⎢
⎥
a
⎢
⎥
= 2
⎢
⎥
a + p2
⎣⎢
⎦⎥
∫
By Fourier cosine transform, we get,
∞
f(x) =
2
Fc ( p ) cos pxdp
π
∫
0
∞
⇒
f(x) =
2
a
cos pxdp
2
π a + p2
0
⇒
e–ax =
2a
π
∞
∫
cos px
∫a
0
2
+ p2
∞
cos px
∫a
2
0
+ p2
[Q From equation (1)]
dp
dp = π e − ax
2a
∴ The Fourier cosine transform of e
–ax
is
∞
a
2
a +p
2
and
π − ax
cos px
e
dp =
2
2
2a
+p
∫a
0
⎧1, | x | < a
Q28. Find the Fourier transform of, f(x) = ⎨
and hence find the value of
⎩0, | x | > a
∞
∫
0
sinax
dx .
x
Ans: Given that,
⎧1 | x | < a
f(x) = ⎨
⎩0 | x | > a
The Fourier transform of the function f(x) is,
∞
∫
isx
F[f(x)] = e f ( x ) dx
−∞
−a
=
∫
−∞
a
e isx f ( x ) dx +
∫
−a
∞
∫
e isx f ( x) dx + e isx f ( x )dx
a
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
3.18
MATHEMATICS-II [JNTU-ANANTAPUR]
a
a
⎡ e isx ⎤
⎥
= e (1) dx = ⎢
⎢⎣ is ⎥⎦ −a
−a
∫
=
isx
1 x –x
(e – e )]
[Q sin hx =
2
2
= sin h(isa )
is
2 sin sa
s
[Q sin h(ix) = isinx]
We get,
∫e
isx
f ( x) dx
Let f(x) =
=
1
2π
∴
− isx
1
2π
∫
−∞
−∞
⇒
1
2π
∫
−∞
∞
∫
−∞
1
sin sxdx
x
∞
⎧1 , | x |< a
2 sin sa −isx
.e ds = ⎨
s
⎩0 , | x |> a
−∞
∫
2 sin sa
.(cos sx − i sin sx ) ds
s
2 sin sa
i
cos sxds −
s
2π
∞
∫
−∞
=
∞
∫
−∞
=
= cos x – i sin x]
Let, x = 0
−∞
∞
⇒
∫
0
∞
or
∫
0
1
∫
x
0
∞
cos sx dx + i
∫
=
sin sa
ds = π
s
∫ (cos sx + i sin sx)
∫e
isx
1
1
.
sin sx dx
x
0
∞
2 sin sa
sin sxds
s
⎧π , | x |< a
sin sa
cos sxds = ⎨
s
⎩0 ,| x |> a
∫
∞
0
1
dx
x
dx
x
0
∞
2 sin sa
. cos sx ds
s
∞
cos sxdx
x
∞
=
∫e
isx
( x) −1 / 2 dx
0
∞
[Q Second integral is an odd function]
⇒
1
∫
0
⎧ 1 ⎫
⎧ 1 ⎫
FC ⎨
⎬ + iFS ⎨
⎬=
⎩ x⎭
⎩ x⎭
[Q e
∞
∫
∞
– ix
1
⇒
2π
x
⎧ 1 ⎫
And FC{f(x)} = FC ⎨
⎬
⎩ x⎭
ds
Taking LHS,
∞
is self reciprocal under Fourier sine
1
0
∞
∫ F ( s )e
x
transform,
∞
By inversion formula,
1
2π
1
Ans: To prove
−∞
f(x) =
is self-reciprocal under Fourier
⎧ 1 ⎫
FS{f(x) = FS ⎨ ⎬
⎩ x⎭
∞
F(s) =
x
sine transform.
e isa − e −isa
is
∴ F [f(x)] = F(s) =
1
Q29. Prove that
=
∫
1
−1
e isx ( x ) 2 dx
0
Let, isx = – t
x=−
t
is
⇒ dx = −
dt
is
⇒
1
∞
−1
π
sin as
ds =
2
s
⎧ 1 ⎫
⎧ 1 ⎫
∴ FC ⎨
⎬ + iFS ⎨
⎬=
⎩ x⎭
⎩ x⎭
π
sin ax
dx =
2
x
−1
⎧ 1 ⎫
⎧ 1 ⎫
− t ⎛ − t ⎞ 2 ⎛ − dt ⎞
iF
+
.
e
−
=
⇒ FC ⎨
⎟
⎜
⎟
⎜
⎬
⎬
S⎨
is ⎠ ⎝ is ⎠
⎝
⎩ x⎭
⎩ x⎭
0
Look for the S IA GROUP LOGO
⎛ − t ⎞ 2 ⎛ − dt ⎞
e .⎜ ⎟ ⎜
⎟
⎝ is ⎠ ⎝ is ⎠
0
∫
−t
∞
1
∫
on the TITLE COVER before you buy
3.19
UNIT-3 (Fourier Transform)
1
=
=
=
(−1).(−1) 2
−1 ∞
∫
1
−1+1
(is ) 2
∫
S
1/ 2
1
−1
f(x) =
(e −t t 2 ) dt
1
2
=
s
1
2
1
2
∞
=−
+i
1
2
∫
∞
=−
∫
2
)
x(1 + x 2 )
dx with
x (1 + x 2 )
0
... (2)
dx
(1 + x 2 ) sin px − sin px
dx
∞
⎡ (1 + x 2 ) sin px
sin px ⎤
−
dx
=− ⎢
2
2 ⎥
(
1
)
(
1
)
+
+
x
x
x
x
⎥
⎢
⎦
⎣
0
∫
2
π
∞
∴
∞
dk
sin px
sin px
dx +
dx
=−
dp
x
x (1 + x 2 )
0
0
∫
∫
∞
dk
sin px
π
dx
−
+
2
dp = 2
x
x
+
(
1
)
0
π
∫
⎡ ∞ sin ax
⎤
π
⎢Q
dx = (if a > 0)⎥
2
⎢ 0 x
⎥
⎣
⎦
π
2s
As Fourier sine transform of
2
dx
( x 2 + 1 − 1) sin px
0
∫
π
⎧ 1 ⎫
Fs ⎨
⎬=
2
s
⎩ x⎭
A function f(x) is equal to F(s) then the function f(x) is
said to be self reciprocating.
∴
∫ x(1 + x
∞
=−
1
s
+i
x 2 sin px
0
2s
2s
Equating real and imaginary parts on both sides, we get,
⎧ 1 ⎫
FC ⎨
⎬=
⎩ x⎭
cos px
∫ 1+ x
∫
1
2
⎛ 1
1 ⎞
⎟ π
+i
= ⎜⎜
2 s ⎟⎠
⎝ 2s
=
dx ... (1)
∞
⎤
⎡ 1
= π⎥
⎢Q
2
⎦
⎣
π
2
x sin px
dk
d
dx
{FC [ f ( x)]} = −
=
1+ x2
dp
dp
0
π
π
cos + i sin
4
4 π
=
s
=
0
respect to ‘p’, we get,
π
⎛
⎜ cos + i sin ⎟
2
2⎠
⎝
=
s
2
0
cos px
∫ f ( x) cos px dx = ∫ 1 + x
∞
1
π ⎞2
1
∞
On differentiating, k = F C[f(x)] =
1
=
∞
0
s
(i ) 2
.
1
Let, k = FC[f(x)] =
∫
1−
1+ x 2
1+ x2
To find the Fourier cosine transform of f(x).
∞
⎡
⎤
⎢Q n = e − x ( x) n −1 dx ⎥
⎢
⎥
0
⎣
⎦
(i )
1
Ans: Given that,
0
(i 2 )1 / 2
(is)
Q30. Find the Fourier cosine transform of, f(x) =
0
1
−1+1 ∞
( −1) 2
is
1
−1
−t 2
e t dt
1
is
1
s
x
self reciprocal under Fourier sine transform.
Hence proved.
. Hence,
1
x
Again differentiating with respect to ‘p’, we get,
dp
{FC [ f ( x)]} =
2
d2
is
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
∞
d2
dp 2
⇒
cos px
∫ 1+ x
2
dx
0
{FC [ f ( x)]} = FC[f(x)]
(D2 – 1) FC[f(x)] = 0
SIA GROUP
3.20
MATHEMATICS-II [JNTU-ANANTAPUR]
General solution of the above is,
FC[f(x)] = C1ep + C2e–p
Put p = 0 in equation (1), then
∞
1
∫ 1+ x
FC[f(x)] =
2
... (3)
dx = [tan −1 x]∞
0 =
0
a
d
[ F (α )] = 2
dα
α + a2
⇒
On integrating equation (1) with respect to ‘α’, we get,
π
2
∞
F (α) = Fs [ f ( x )] =
π
d
{FC [ f ( x )]} = − (when p = 0)
And
dp
2
π
[Q From equations (3) and (4)]
2
On solving the above two equations, we get,
F(α) = 0
On substituting α, f (α) values in equation (2), we get,
π −p
∴ FC[f(x)] = 0e + e
2
0 = tan–1(0) + c
1
1+ x
2
is, FC [ f ( x)] =
Q31. Find the Fourier sine transform of
e
⇒
p −P
e
2
c=0
x
, where
Hence the Fourier Sine transform of
Where a > 0
By definition of Fourier sine transform,
∞
3.3 INVERSE TRANSFORMS
Q32. Find the inverse Fourier cosine transform f(x)
f ( x) sin αx dx
⎧⎪ a − p ,
2
w hen p < 2a
⎪⎩ 0
w hen p ≥ 2a
of FC (p) = ⎨
e
− ax
x
sin αx dx = F(α)
∞
d
[ F (α)] =
dα
∫
0
∞
d
[ F ( α)] =
dα
∫
0
e − ax ⎛ d
⎞
sin αx⎟ dx
⎜
⎠
x ⎝ dα
e − ax
x. cos αx dx
x
⎧⎪ a − p , when p < 2 a
2
FC ( p ) = ⎨
, when p ≥ 2 a
⎪⎩ 0
By the definition of inverse Fourier cosine transform, we
have,
∞
2
FC ( p ) cos pxdp
f(x) =
π
∫
0
=
⎡
a ⎤
⎢Q e − ax cos bx dx =
⎥
2
2
⎢
⎥
+
a
b
⎣ 0
⎦
∞
∫
Look for the S IA GROUP LOGO
,
Ans: Given that,
Differentiating on both sides with respect to ‘α’, we get,
⇒
e − ax
−1 ⎛ α ⎞
= tan ⎜ ⎟ ,
⎝a⎠
a
where a > 0
0
0
⎛α⎞
F (α ) = Fs [ f ( x )] = tan −1 ⎜ ⎟
⎝a⎠
∴
e − ax
f(x) =
x
∫
0= 0 + c
–ax
a > 0.
Ans: Given that,
Fs[ f(x)] =
... (2)
For α = 0, we get,
p
∞
dα
−1 ⎛ α ⎞
F (α) = Fs [ f ( x )] = tan ⎜ ⎟ + c
⎝a⎠
∴
π
C1 = 0 and C2 =
2
∫
+ a2
∫
C1 – C 2 = −
Fs[ f(x)] =
a
2
⎡ ∞ a
⎤
−1 ⎛ x ⎞
⎢Q
⎥
=
+
dx
tan
c
⎜
⎟
⎝ a⎠ ⎥
⎢ 0 x2 + a2
⎣
⎦
π
and
2
∴ The Fourier cosine transform of
∫α
0
... (4)
∴ From equation (3), we get,
C1 + C2 =
... (1)
∞
2a
⎤
2⎡
⎢ FC ( p) cos pxdp + FC ( p) cos pxdp⎥
π⎢
⎥
2a
⎦
⎣0
∫
∫
∞
2a
⎤
p⎞
2⎡ ⎛
= ⎢ ⎜ a − ⎟ cos pxdp + (0) cos pxdp ⎥
π⎢ ⎝
2⎠
⎥
2a
⎣0
⎦
∫
∫
on the TITLE COVER before you buy
3.21
UNIT-3 (Fourier Transform)
2⎡
⎢ a cos pxdp −
π⎢
⎣0
2a
=
∫
2a
∫
0
2⎡
1
⎢a cos pxdp −
π⎢
2
⎣ 0
2a
=
∫
⎤
p
cos pxdp⎥
2
⎥
⎦
2a
∫
0
⎤
p cos px.dp ⎥
⎥
⎦
2 ⎡ ⎛ sin px ⎞
1⎡
= π ⎢a ⎜⎝ x ⎟⎠ − 2 ⎢⎣ p cos pxdp −
⎢⎣
0
2a
∫
∫{ ∫
}
2a
⎤
⎤
(1). cos pxdp dp ⎥ ⎥
⎦0 ⎥
⎦
2a
2 ⎡ ⎡ sin 2ax sin x(0) ⎤ 1 ⎡ sin px
sin px ⎤ ⎤
a
p
dp
⎢
−
−
−
= π ⎢ x
⎥ ⎥
x ⎥⎦ 2 ⎢⎣
x
x
⎦ 0 ⎥⎦
⎢⎣ ⎣
∫
2a ⎤
=
2 ⎡⎛ a
⎞ 1 ⎡ p sin px ⎛ − cos px ⎞⎤
⎢⎜ sin 2ax ⎟ − ⎢
−⎜
⎟⎥
π ⎢⎝ x
⎠ 2⎣ x
⎝ x 2 ⎠⎦ 0
⎣
=
2 ⎡a
1 ⎧ p sin px cos px ⎫
⎢ sin 2ax − ⎨
+
⎬
x
2⎩
π ⎢x
x 2 ⎭0
⎣
=
2 ⎡ a sin 2 ax 1 ⎡ ⎧ 2 a sin 2ax cos 2ax ⎫ ⎡ ⎧ 0 sin( 0) x cos x (0) ⎫ ⎤
+
− ⎢⎨
+
⎬⎥
⎢
⎬⎢ − ⎨
x
x
x
2 ⎣⎩
π⎣
x 2 ⎭⎣ ⎩
x 2 ⎭⎦
=
2 ⎡ a sin 2ax a sin 2ax cos 2ax
1 ⎤
−
−
+ 2⎥
π ⎢⎣
x
x
2x 2
2x ⎦
=
2⎡ 1
cos 2ax ⎤
−
⎥
⎢
2
π ⎣ 2x
2x 2 ⎦
=
2 ⎡1 − cos 2ax ⎤
⎥
π ⎢⎣ 2 x 2
⎦
=
2 sin 2 ax
×
π
x2
=
2 sin 2 ax
πx 2
⎥
⎥
⎦
2a ⎤
⎥
⎥⎦
Q33. Find the inverse Fourier sine transform f(x) of FS (p) =
p
1+ p 2
.
Ans: Given that,
FS ( p) =
p
1+ p 2
By the definition of inverse Fourier sine transform,
∞
f (x) =
2
FS ( p ) sin( px) dp
π
∫
0
2
f (x) =
π
∞
⎛
p
∫ ⎜⎜⎝ 1 + p
0
2
⎞
⎟ sin( px ) dp
⎟
⎠
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (1)
SIA GROUP
3.22
MATHEMATICS-II [JNTU-ANANTAPUR]
Again differentiating equation (3) with respect to ‘x’,
Multiply and divide by ‘p’, we get,
2
f (x)=
π
⇒
⇒
2
f (x) =
π
f (x) =
2
π
we get,
∞
⎛ p
⎜
⎜1+ p2
0⎝
∫
∞
∫
0
⎞ p
⎟ × sin px dp
⎟ p
⎠
∞
d
dp
d
−2
( f ′( x )) =
. (cos px )
2
dx
π 1 + p dx
0
∫
p 2 sin( px )
.dp
p(1 + p 2 )
∞
p 2 + 1 −1
∫ p(1 + p
0
2
)
∞
f ′′(x) =
⇒
∫
⇒
∞
⎤
⎞
2 ⎡ ⎛⎜ 1
1
⎟ sin( px)dp ⎥
⎢
−
f (x) =
π ⎢ ⎜⎝ p p(1 + p 2 ) ⎟⎠
⎥
⎦
⎣0
⇒
∞
∞
⎤
2 ⎡ sin( px)
sin( px)
⎥
⎢
.dp −
dp
f (x) =
2
π⎢
p
⎥
(
1
)
+
p
p
0
⎦
⎣0
⇒
∞
⎤
2 ⎡π
sin( px)
⎥
.
dp
f (x) = ⎢ −
π ⎢2
p(1 + p 2 ) ⎥⎦
0
⎣
2
f (x) = 1 −
π
∞
∞
2
0
sin px
∫ p (1 + p
2
0
)
⎛ ∞ sin px π ⎞
⎜Q
= ⎟
⎜
p
2⎟
⎝ 0
⎠
)
... (2)
∫
⇒
2
f ′(x) = 0 −
π
∞
0
... (6)
d
(f(x)) = c1ex + c2e–x(–1)
dx
= c1ex – c2e–x
Let substitute x = 0 in equation (2), we get,
⇒
2
f (x) = 1 −
π
∞
sin p ( 0)
∫ p (1 + p
2
0
)
dp
∞
2
(0) dp
π
∫
0
dp
d
. (sin px )
p (1 + p 2 ) dx
p cos px
∫ p(1 + p
2
0
− 2 cos px
dp
f ′(x) =
π 1+ p2
0
∫
f ′′(x ) – f(x) = 0
= 1−
)
=1
Let x = 0 is equation (3), we get,
∞
df ( x) − 2 cos p(0)
dp
=
dx
π
1+ p2
0
∫
.dp
∞
∞
⇒
... (4)
[From equation (1)]
f(x) = c1ex + c2e–x
⇒
∞
⎤
d
sin px
d ⎡ 2
⎢1 −
( f ( x )) =
dp ⎥
2
dx
dx ⎢ π p (1 + p ) ⎥
0
⎣
⎦
∫
f ′′(x) = f(x)
dp
f(x) = c1 e M1 x + c2 e M 2 x
dp
Differentiating equation (2) with respect to ‘x’, we get,
d
2
f ′(x) = dx (1) − π
1+ p2
M= 1 =±1
∴ M1 = +1 and M2 = –1.
The general solution of the differential equation (5) is
given by,
∫
dp
∞
0
p sin px
On differentiating equation (6) with respect to x, we get,
sin px
∫ p(1 + p
∫
d 2 f ( x)
– f (x) = 0
... (5)
dx 2
The auxillary equation of the above equation is,
M2 – 1 = 0
⇒ M2 = 1
∫
∫
∞
⇒
∫
2 π 2
∴ f (x) = × −
π 2 π
⇒
⇒
∞
⎤
⎞
2 ⎡ ⎛⎜ (1 + p 2 )
1
⎟ sin( px)dp ⎥
⎢
−
f (x) =
π ⎢ ⎜⎝ p(1 + p 2 ) p(1 + p 2 ) ⎟⎠
⎥
⎦
⎣0
∫
∫
2
f ′′(x) =
π
sin px.dp
(Adding and subtracting 1 in the numerator)
⇒
dp
−2
( − sin px × p )
π 1+ p2
0
... (3)
Look for the S IA GROUP LOGO
df ( x) − 2 cos 0
dp
=
dx
π 1+ p2
0
∫
on the TITLE COVER before you buy
... (7)
3.23
UNIT-3 (Fourier Transform)
∞
−2
1
dp (Q cos 0 = 1)
=
π 1+ p2
0
1
=
2π
∫
=
[
−2
tan −1 x
π
]
∞
0
[
=
− 2 ⎡π
⎤
− 0⎥
π ⎢⎣ 2
⎦
=
−2 π
×
π 2
]
df ( x)
and f(x) values in equation (7),
dx
⇒ 1 = c1e0 + c2e–0 and –1 = c1e0 – c2e–0
⇒ c1 + c2 = 1
... (8)
and
c1 – c2 = –1
... (9)
Solving equations (8) and (9), we get,
⇒ c1 = 0 and c2 = 1
On substituting c1 and c2 values in equation (6), we get,
∴ f(x) = (0)ex + 1 . e–x
By the definition of inverse Fourier transform,
1
1
[1 − 0] −
[0 − 1]
2π( y − ix )
2π( y + ix )
ipx
1
1
+
2π( y − ix ) 2π( y + ix )
=
y + ix + y − ix
2π( y − ix )( y + ix )
2y
2
2π( y − i 2 x 2 )
y
2
π( y + x 2 )
∴ F −1[ F ( p)] = f ( x) =
1
2π
FINITE FOURIER TRANSFORMS
f(x) =
π
x2
–x+
, 0<x<π
2π
3
Finite cosine of f(x) is given that,
.F ( p ) dp
ipx −| p| y
e
Fc(n) =
dp
nπx
dx
l
π
∞
0
− ipx py
∫
e dp + e −ipx e − py dp
0
(Q | p |= − p if p ≤ 0
and = + p if p ≥ 0)
0
∞
⎤
1 ⎡
⎢ e p ( y −ix ) dp + e − p ( y +ix ) dp ⎥
=
2π ⎢
⎥
0
⎦
⎣−∞
∫
∫ f ( x) cos
0
−∞
−∞
Model Paper-I, Q7
f (x) =
∞
∫e
y
π ( y 2 + x2 )
x2
π
−x+
, 0 < x < π.
3
2π
Ans: Given function is,
−∞
∫e
(Q i2 = –1)
Q35. Find the finite cosine transform of,
∞
∫e
=
l
=
0
=
=
3.4
Q34. Find the inverse Fourier transform f(x) of
F(p) = e–|p|y.
Ans: Given that,
F(p) = e–|p|y
1
=
2π
∞
1
1
[e 0 − e − ∞ ] −
[e −∞ − e 0 ]
2π( y − ix )
2π( y + ix )
=
f ( x) = e − x
1
2π
0
1
1
⎡ −∞
⎤
⎢Q e = ∞ = ∞ = 0⎥
e
⎣
⎦
we get,
f(x) =
∫
=
= –1
On substituting
−∞
1
e − p ( y +ix ) dp
2π
1 ⎡ e p ( y −ix) ⎤
1 ⎡ e − p ( y +ix ) ⎤
+
=
⎥
⎢
⎥
⎢
2π ⎣⎢ y − ix ⎦⎥ − ∞ 2π ⎣⎢ − ( y + ix ) ⎦⎥
∫
−2
tan −1 ( ∞) − tan −1 (0)
π
∫
e p ( y −ix ) dp +
0
1
⎡
⎤
dx = tan −1 x ⎥
⎢Q
2
x
+
1
⎣
⎦
=
∞
0
∫
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎛π
x 2 ⎞⎟
nπx
= ⎜⎜ − x +
⎟ cos π dx
3
2
π
⎠
0⎝
∫
π
⎛π
x2 ⎞
⎟ cos nx dx
= ⎜⎜ − x +
π ⎟⎠
3
2
⎝
0
∫
⎛π
x 2 ⎞ sin nx
= ⎜⎜ − x + ⎟⎟
n
2π ⎠
⎝3
π
π
−
0
sin nx ⎛
x⎞
⎜ − 1 + ⎟ dx
n ⎝
π⎠
0
∫
SIA GROUP
3.24
MATHEMATICS-II [JNTU-ANANTAPUR]
=0–
=
−1
n
1
n
π
⎛x
⎞
∫ ⎜⎝ π − 1⎟⎠ sin nx dx
[Q sin nx |0π = 0]
0
⎧⎪⎛
x ⎞ − cos nx
⎨⎜ − 1 + ⎟
n
π
⎠
⎪⎩⎝
π
π
−
0
∫
0
− cos nx
n
⎛ 1 ⎞ ⎫⎪
⎜ ⎟ dx ⎬
⎝ π ⎠ ⎪⎭
π
⎤
1
π⎞
n
1 ⎡⎛
= 2 ⎢⎜ − 1 + π ⎟(−1) − (−1)⎥ − n π − cos nx dx
⎠
⎦
n ⎣⎝
0
∫
=
1 ⎛⎜ 1 − sin nx
–
n
n 2 ⎜⎝ nπ
=
1
−0
n2
π⎞
⎟
⎟
0⎠
= 1/n2 ∀ n > 0
When,
n = 0,
π
∫
x2 ⎞
⎛
Fc(n) = ⎜⎜ π 3 − x + ⎟⎟ dx
2π ⎠
0⎝
π
π3
π2
π2
⎡π
x 2 x3 ⎤
+
= ⎢ x−
–
+
⎥ =
3
6π
2
2 6π ⎦ 0
⎣3
=
π2
π2
–
2
2
=0
π.
Q36. Find the finite Fourier sine and cosine transform of f(x) defined by f(x) = 4x where 0 < x < 4π
Ans: Given function is,
f(x) = 4x for 0 < x < 4π
By the definition of finite Fourier sine transform,
l
FS(n) =
⎛ nπx ⎞
⎟dx
l ⎠
∫ f ( x) sin⎜⎝
0
4π
=
⎛ nπx ⎞
∫ 4 x. sin⎜⎝ 4π ⎟⎠dx
0
4π
=
⎛ nx ⎞
∫ 4 x. sin⎜⎝ 4 ⎟⎠dx
0
4π
⎡
⎧ d
⎛ nx ⎞
⎛ nx ⎞ ⎫ ⎤
= ⎢4 x sin⎜ ⎟dx − ⎨
(4 x). sin ⎜ ⎟dx ⎬dx ⎥
⎝ 4 ⎠
⎝ 4 ⎠ ⎭ ⎥⎦ 0
⎢⎣
⎩ dx
∫
∫
∫
Look for the S IA GROUP LOGO
on the TITLE COVER before you buy
3.25
UNIT-3 (Fourier Transform)
4π
⎡ ⎛
⎛
⎛ nx ⎞ ⎞
⎛ nx ⎞ ⎞ ⎤
⎜ − cos⎜ ⎟ ⎟ ⎥
⎢ ⎜ − cos ⎜ ⎟ ⎟
⎝ 4 ⎠ ⎟ − 4.⎜
⎝ 4 ⎠⎟ ⎥
⎜
= ⎢ 4 x⎜
⎜ ⎛ n ⎞ ⎟dx ⎥
⎢
⎛n⎞ ⎟
⎜ ⎜ ⎟ ⎟ ⎥
⎢ ⎜ ⎜ ⎟ ⎟
⎝ ⎝ 4 ⎠ ⎠ ⎦0
⎣ ⎝ ⎝4⎠ ⎠
∫
4π
⎡
⎛ nx ⎞
⎛ nx ⎞ ⎤
4 × 4 cos⎜ ⎟ ⎥
⎢ − 4 x × 4 cos⎜ 4 ⎟
⎝ ⎠+
⎝ 4 ⎠ dx ⎥
=⎢
n
n
⎢
⎥
⎢
⎥
⎣
⎦0
∫
4π
⎡
⎛ nx ⎞ ⎤
⎛ nx ⎞
16 cos⎜ ⎟ ⎥
⎢ − 16 x cos ⎜ 4 ⎟
⎝ 4 ⎠ dx ⎥
⎝ ⎠+
=⎢
n
n
⎢
⎥
⎢
⎥
⎣
⎦0
∫
4π
4π
⎡
⎡
⎤
⎛ nx ⎞
⎛ nx ⎞
⎢ − 16 x cos⎜ ⎟
⎢ − 16 x cos⎜ ⎟
⎥
⎛ nx ⎞
⎝ 4 ⎠ 16
⎝ 4 ⎠ 16
+
+
cos⎜ ⎟dx ⎥ = ⎢
=⎢
⎢
⎢
n
n
n
n
⎝ 4 ⎠ ⎥
⎢
⎢
⎥
⎢⎣
⎣
⎦0
∫
∫
4π
⎛ ⎛ nx ⎞ ⎞ ⎤
⎜ sin ⎜ ⎟ ⎟ ⎥
⎜ ⎝ 4 ⎠ ⎟⎥
⎜ ⎛ n ⎞ ⎟⎥
⎜⎜ ⎜ ⎟ ⎟⎟ ⎥
⎝ ⎝ 4 ⎠ ⎠ ⎦⎥ 0
4π
⎡
⎡
⎤
⎤
⎛ nx ⎞
⎛ nx ⎞
⎢ − 16 x cos⎜ ⎟
⎢ − 16 x cos⎜ ⎟
⎥
⎥
4
4
×
16
4
nx
64
nx
⎛ ⎞
⎛ ⎞
⎝ ⎠
⎝ ⎠
+ 2 sin⎜ ⎟⎥
+
sin⎜ ⎟⎥ = ⎢
=⎢
⎢
⎢
n
n×n
n
⎝ 4 ⎠⎥
⎝ 4 ⎠⎥
n
⎢
⎢
⎥
⎥
⎣
⎣
⎦0
⎦0
⎡⎧
⎫⎤
⎫ ⎧
⎛ n(0) ⎞
⎛ n(4π) ⎞
⎟
⎟
⎢⎪ − 16(4π) cos⎜
⎪⎥
⎪ ⎪ − 16(0) cos ⎜
⎝ 4 ⎠ 64 ⎛ n(0) ⎞⎪⎥
⎝ 4 ⎠ 64 ⎛ n(4π) ⎞⎪ ⎪
⎢⎪
+ 2 sin⎜
+ 2 sin⎜
⎟⎬
⎟⎬ − ⎨
= ⎢⎨
n
n
n
n
⎝ 4 ⎠⎪⎥
⎝ 4 ⎠⎪ ⎪
⎪
⎥
⎢
⎪⎭⎥
⎪
⎪⎭ ⎪⎩
⎦
⎣⎢⎩
⎡⎧
⎫ ⎧
⎫⎤
⎛ 4nπ ⎞
⎛0⎞
⎟
⎢⎪ − 64π cos⎜
⎪ ⎪ 0 cos ⎜ ⎟
⎪⎥
⎪⎥
⎝ 4 ⎠ 64 ⎛ 4nπ ⎞⎪ ⎪
⎝ 4 ⎠ + 64
⎢⎪⎨
+
−
sin
sin
(
0
)
⎟
⎜
⎬ ⎨
⎬⎥
=⎢
2
2
n
n
4
n
⎠⎪ ⎪
⎝
n
⎪⎥
⎢⎪⎪
⎪⎭ ⎪⎩
⎪⎭⎥⎦
⎢⎣⎩
⎡⎧ − 64π cos (nπ) 64
⎤
⎫
+ 2 sin(nπ)⎬ − {0 + 0}⎥
= ⎢⎨
n
n
⎭
⎣⎩
⎦
⎡ − 64π(− 1)n 64 ⎤
+ 2 (0)⎥
=⎢
n
n
⎣⎢
⎦⎥
=
− 64π(−1) n
n
∴ FS(n) =
− 64π(−1) n
n
(Q cos nπ = (−1)
n
(Q sin(0) = 0)
and sin nπ = 0
)
(Q sin nπ = 0)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
3.26
MATHEMATICS-II [JNTU-ANANTAPUR]
The finite Fourier cosine transform is given by,
l
FC(n) =
⎛ nπx ⎞
⎟ dx
l ⎠
∫ f ( x) cos⎜⎝
0
4π
=
⎛ nπx ⎞
4 x cos⎜
⎟ dx =
⎝ 4π ⎠
0
∫
4π
⎛ nx ⎞
∫ 4 x. cos⎜⎝ 4 ⎟⎠dx
0
4π
4π
⎡
⎧ d
⎛ nx ⎞
⎛ nx ⎞ ⎫ ⎤
( 4 x) cos ⎜ ⎟ dx ⎬dx ⎥
= ⎢ 4 x cos⎜ ⎟ dx − ⎨
⎝ 4 ⎠
⎝ 4 ⎠ ⎭ ⎥⎦ 0
⎢⎣
⎩ dx
∫
∫
∫
⎡ ⎛ ⎛ nx ⎞ ⎞
⎛ ⎛ nx ⎞ ⎞ ⎤
⎜ sin⎜ ⎟ ⎟ ⎥
⎢ ⎜ sin⎜ ⎟ ⎟
4
4 ⎠⎟
⎝
⎜
⎢
− 4⎜⎜ ⎝ ⎠ ⎟⎟dx ⎥
= 4 x⎜
⎟
⎢
⎥
⎛n⎞
⎛n⎞
⎜ ⎜ ⎟ ⎟ ⎥
⎢ ⎜ ⎜ ⎟ ⎟
⎝ ⎝ 4 ⎠ ⎠ ⎦0
⎣ ⎝ ⎝4⎠ ⎠
∫
4π
4π
⎡
⎡ ⎛
⎡
⎛ nx ⎞
⎛ nx ⎞ ⎤ ⎤
⎛ nx ⎞ ⎞
⎛ nx ⎞ ⎤
4 sin⎜ ⎟ ⎥
⎢ 16 x sin ⎜ ⎟
⎢ ⎜ 4 sin⎜ ⎟ ⎟
⎢ − cos ⎜ ⎟ ⎥ ⎥
⎝ 4 ⎠ 16 ⎢
⎝ 4 ⎠ ⎥⎥
⎝ 4 ⎠⎟−4
⎝ 4 ⎠ dx ⎥
⎜
⎢
−
=⎢
= ⎢ 4 x⎜
⎟
⎢
⎥
⎢
n
n
n
n
⎛ n ⎞ ⎥⎥
⎟
⎢
⎢ ⎜ ⎟ ⎥⎥
⎥
⎢ ⎜
⎢⎣
⎠
⎣ ⎝ 4 ⎠ ⎦ ⎥⎦ 0
⎣ ⎝
⎦0
∫
4π
4π
⎡
⎡
⎤
⎤
⎛ nx ⎞
⎛ nx ⎞
⎢16 x sin⎜ ⎟
⎢16 x sin ⎜ ⎟
⎥
⎥
4
4
16
4
nx
64
nx
⎞
⎞
⎛
⎛
⎝ ⎠
⎝ ⎠
+ × . cos⎜ ⎟⎥ = ⎢
+ 2 cos ⎜ ⎟⎥
=⎢
⎢
⎢
n
n
n n
⎝ 4 ⎠⎥
⎝ 4 ⎠⎥
n
⎢
⎢
⎥
⎥
⎣
⎣
⎦0
⎦0
⎡⎧
⎫⎤
⎫ ⎧
⎛ n(0) ⎞
⎛ n(4π) ⎞
⎟
⎟
⎢⎪16(4π) sin⎜
⎪⎥
⎪ ⎪16(0) sin⎜
⎛ n(0) ⎞⎪⎥
⎛ n(4π) ⎞⎪ ⎪
⎝ 4 ⎠ 64
⎝ 4 ⎠ 64
⎢⎪
+
−
+
cos
cos
⎟⎬
⎜
⎟⎬ ⎨
⎜
= ⎢⎨
n
n
n2
n2
⎝ 4 ⎠ ⎪⎥
⎝ 4 ⎠⎪ ⎪
⎢⎪
⎥
⎪⎭⎥
⎪
⎪⎭ ⎪⎩
⎣⎢⎩
⎦
⎡⎧
⎤
⎫
⎛ 4nπ ⎞
⎟
⎢⎪ 64π. sin⎜
⎥
⎪
64
⎫⎥
⎛ 4nπ ⎞⎪ ⎧
⎝ 4 ⎠ 64
⎢⎪⎨
cos
0
cos
0
+
−
+
(
)
⎟
⎜
⎬
⎬
⎨
=⎢
n
n2
⎝ 4 ⎠⎪ ⎩ n 2
⎭⎥
⎢⎪⎪
⎥
⎪
⎭
⎣⎢⎩
⎦⎥
⎡⎧ 64π. sin nπ 64
⎫ ⎧ 64 ⎫⎤
+ 2 cos nπ⎬ − ⎨ 2 (1)⎬⎥
= ⎢⎨
n
n
⎭ ⎩n
⎭⎦
⎣⎩
⎡ 64π × (0) 64(−1) n 64 ⎤
+
− 2⎥
=⎢
n
n2
n ⎦⎥
⎣⎢
=
=
∴ FC (n) =
[Q sin nπ = 0 and cos nπ = (–1)n]
64(−1) n 64
− 2
n2
n
64
n2
(
((−1) − 1)
n
)
64
(−1)n − 1
n2
Look for the S IA GROUP LOGO
on the TITLE COVER before you buy
3.27
UNIT-3 (Fourier Transform)
2
x⎞
⎛
Q37. Find the finite Fourier sine transform of f(x) defined by f(x) = ⎜ 1 − ⎟ , where 0 < x < π .
⎝ π⎠
Ans: Given function is,
2
⎛ x⎞
f(x) = ⎜ 1– ⎟ , where 0 < x < π
⎝ π⎠
The Finite Fourier sine transform of f(x) is given by,
l
FS(n) =
⎛ nπx ⎞
⎟dx
l ⎠
∫ f ( x) sin⎜⎝
0
π
2
π
x⎞
⎛ nπx ⎞
⎛
= ⎜1 − π ⎟ sin ⎜ π ⎟ dx =
⎝
⎠
⎠
⎝
0
∫
⎡⎛ x ⎞ 2
= ⎢⎜1 − ⎟ sin( nx ) dx −
⎢⎝ π ⎠
⎣
⎧⎪ d ⎛ x ⎞ 2
⎫⎪ ⎤
⎨ ⎜1 − ⎟ sin( nx ) dx ⎬dx ⎥
⎪⎭ ⎥⎦
⎪⎩ dx ⎝ π ⎠
0
∫
∫
2
x⎞
⎛
⎜1 − ⎟ sin (nx )dx
⎝ π⎠
0
π
∫
∫
π
⎡⎛ x ⎞ 2 ⎛ − cos( nx) ⎞
⎛ x ⎞⎛ − 1 ⎞⎛ − cos( nx) ⎞ ⎤
= ⎢⎜1 − ⎟ ⎜
⎟dx ⎥
⎟ − 2⎜1 − ⎟⎜ ⎟⎜
n
n
⎠ ⎥⎦ 0
⎝ π ⎠⎝ π ⎠⎝
⎠
⎢⎣⎝ π ⎠ ⎝
∫
π
⎤
⎡⎛ x ⎞ 2 ⎛ − cos( nx ) ⎞ 2 ⎛ x ⎞
⎜1 − ⎟(cos( nx ) )dx ⎥
⎟−
= ⎢⎜1 − ⎟ ⎜
n
⎠ nπ ⎝ π ⎠
⎦⎥ 0
⎣⎢⎝ π ⎠ ⎝
∫
π
⎡⎛ x ⎞ 2 ⎛ − cos( nx) ⎞ 2 ⎡⎛ x ⎞
⎫ ⎤⎤
⎧ d ⎛ x⎞
⎟−
= ⎢⎜1 − ⎟ ⎜
⎢⎜1 − ⎟ cos( nx)dx − ⎨ ⎜1 − ⎟ cos( nx)dx ⎬dx ⎥ ⎥
n
⎠ nπ ⎣⎢⎝ π ⎠
⎢⎣⎝ π ⎠ ⎝
⎭ ⎦⎥ ⎥⎦ 0
⎩ dx ⎝ π ⎠
∫
∫
∫
π
⎡⎛ x ⎞ 2 ⎛ − cos( nx) ⎞ 2 ⎡⎛ x ⎞⎛ sin(nx) ⎞
⎛ − 1 ⎞⎛ sin(nx) ⎞ ⎤ ⎤
⎟dx ⎥ ⎥
⎟ − ⎜ ⎟⎜
⎟−
= ⎢⎜1 − ⎟ ⎜
⎢⎜1 − ⎟⎜
n
⎝ π ⎠⎝ n ⎠ ⎦ ⎥⎦ 0
⎠ nπ ⎣⎝ π ⎠⎝ n ⎠
⎢⎣⎝ π ⎠ ⎝
∫
π
⎡⎛ x ⎞ 2 ⎛ − cos( nx) ⎞ 2 ⎡⎛ x ⎞⎛ sin(nx) ⎞ 1 ⎛ sin(nx) ⎞ ⎤ ⎤
⎟−
⎟+ ⎜
⎟dx ⎥ ⎥
= ⎢⎜1 − ⎟ ⎜
⎢⎜1 − ⎟⎜
n
⎠ nπ ⎣⎝ π ⎠⎝ n ⎠ π ⎝ n ⎠ ⎦ ⎥⎦
⎢⎣⎝ π ⎠ ⎝
0
π
⎡⎛ x ⎞ 2 ⎛ − cos( nx) ⎞ 2 ⎡⎛ x ⎞⎛ sin(nx ) ⎞ 1 ⎛ − cos( nx ) ⎞⎤ ⎤
= ⎢⎜1 − ⎟ ⎜
⎟⎥ ⎥
⎜
⎟+
⎟−
⎢⎜1 − ⎟⎜
n
n2
⎠⎦ ⎥⎦ 0
⎠ nπ ⎣⎝ π ⎠⎝ n ⎠ π ⎝
⎢⎣⎝ π ⎠ ⎝
∫
π
⎡⎛ x ⎞ 2 ⎛ − cos( nx) ⎞ 2 ⎛ x ⎞⎛ sin(nx) ⎞
2 ⎛ cos( nx) ⎞⎤
= ⎢⎜1 − ⎟ ⎜
⎟⎥
⎟+ 2 ⎜
⎟ − ⎜1 − ⎟⎜
2
n
⎠⎥⎦ 0
⎠ nπ ⎝ π ⎠⎝ n ⎠ nπ ⎝ n
⎢⎣⎝ π ⎠ ⎝
⎡⎧⎪⎛ π ⎞ 2 ⎛ − cos nπ ⎞ ⎛ 0 ⎞ 2 ⎛ − cos( n × 0) ⎞⎫⎪ ⎧ 2 ⎛ π ⎞⎛ sin(n × π) ⎞
⎟−
⎟ − ⎜1 − ⎟ ⎜
⎟⎬ − ⎨ ⎜1 − ⎟⎜
= ⎢⎢⎨⎜1 − π ⎟ ⎜
π
π
π⎠ ⎝
n
n
n
n
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎪
⎪
⎩
⎭
⎣⎩
2 ⎛ 0 ⎞⎛ sin(n × 0) ⎞⎫ ⎧ 2 ⎛ cos( nπ) ⎞ 2 ⎛ cos( n × 0) ⎞⎫⎤
⎟ ⎬⎥
⎟− 2 ⎜
⎟⎬ + ⎨ 2 ⎜
⎜1 − ⎟⎜
2
nπ ⎝ π ⎠⎝
n
n2
⎠⎭⎦⎥
⎠ nπ ⎝
⎠ ⎭ ⎩ nπ ⎝ n
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
3.28
MATHEMATICS-II [JNTU-ANANTAPUR]
⎡⎧ 1 ⎫ ⎧ 2 ⎛ π ⎞⎛ 0 ⎞ 2
⎫ ⎧⎪ 2
= ⎢⎨0 + n ⎬ − ⎨ nπ ⎜1 − π ⎟⎜ n ⎟ − nπ (1 − 0)(0)⎬ + ⎨ 2
⎢⎣⎩
⎠⎝ ⎠
⎭ ⎩ ⎝
⎭ ⎪⎩ nπ
[
⎫⎤
⎛ (−1) n ⎞
⎜
⎟ − 2 ⎛⎜ 1 ⎞⎟⎪⎬⎥
2
2
2
⎜ n ⎟ nπ ⎝ n ⎠⎪⎥
⎝
⎠
⎭⎦
[Q cos nπ = (–1)n]
]
⎡⎧ 1 ⎫
⎧ 2
⎫⎤
n
= ⎢⎨ n ⎬ − {0}+ ⎨ 3 2 (−1) − 1 ⎬⎥
⎩n π
⎭⎦
⎣⎩ ⎭
=
∴ FS (n) =
1 2(−1) n
2
+ 3 2 − 3 2
n n π
n π
1 2( −1) n
2
+ 3 2 − 3 2
n nπ
nπ
Q38. Find the finite Fourier sine and cosine transforms of f(x) defined by f(x) =
x
, where 0 < x < 4.
π
x
, for 0 < x < 4.
π
The finite Fourier sine transform of the function f(x) is given by,
Ans: Given function is, f(x) =
l
FS(n) =
∫
0
⎛ nπx ⎞
f ( x ) sin ⎜
⎟dx =
⎝ l ⎠
4
⎛ nπx ⎞
⎟ dx
4 ⎠
⎛x⎞
∫ ⎜⎝ π ⎟⎠ sin⎜⎝
0
4
⎡x
⎧d ⎛x⎞
⎛ nπx ⎞
⎛ nπx ⎞ ⎫ ⎤
=⎢
sin ⎜
⎟ dx − ⎨ ⎜ ⎟ sin ⎜
⎟ dx ⎬dx ⎥
⎝ 4 ⎠
⎝ 4 ⎠ ⎭ ⎥⎦ 0
⎢⎣ π
⎩ dx ⎝ π ⎠
∫
∫
∫
4
⎡ ⎡
⎛
⎛ nπx ⎞ ⎤
⎛ nπx ⎞ ⎞ ⎤
⎡ ⎡
⎛ nπx ⎞ ⎤
⎜ − cos⎜
⎟⎥
⎟⎟ ⎥
⎢ ⎢ − cos⎜
⎟⎥
⎢ ⎢ − 4 cos⎜
x
1
4
4
⎝
⎠
⎝
⎠
⎜
⎟
4 ⎠⎥ 1
x
⎥
⎢
⎝
⎢
⎥
⎢
−
dx
⎢
=
=
+
⎢ π ⎢ ⎛ nπ ⎞ ⎥
π ⎜ ⎛ nπ ⎞ ⎟ ⎥
⎢π ⎢
⎥ π
nπ
⎜ ⎜ ⎟ ⎟ ⎥
⎢ ⎢ ⎜ ⎟ ⎥
⎢
⎥
⎝ ⎝ 4 ⎠ ⎠ ⎦ 0 ⎣⎢ ⎢⎣
⎣ ⎣ ⎝ 4 ⎠ ⎦
⎦
∫
4
4
∫
⎡ ⎡
⎡ ⎡
⎤
⎛ nπx ⎞ ⎤
⎛ nπx ⎞ ⎤
⎟⎥
⎟⎥
⎢ ⎢ − 4 cos⎜
⎢ ⎢ − 4 cos⎜
⎥
4
x
⎝ 4 ⎠⎥
⎝ 4 ⎠ ⎥ + 4 cos⎛ nπx ⎞dx ⎥ = ⎢ x ⎢
=⎢ ⎢
+ 2
⎟
⎜
⎢
⎥
⎢
⎢π ⎢
⎥
π
n
π
⎥ nπ 2
nπ
4
n
π
⎠
⎝
⎢ ⎢
⎥
⎢ ⎢
⎥
⎥
⎦
⎦
⎣ ⎣
⎦ 0 ⎢⎣ ⎣
∫
4
⎤
⎡ ⎡
⎛ nπx ⎞ ⎤
⎟⎥
⎥
⎢ ⎢ − 4 cos⎜
x
⎝ 4 ⎠ ⎥ + 4 × 4 sin⎛ nπx ⎞⎥
⎟
⎜
=⎢ ⎢
=
⎢π ⎢
nπ
⎥ n 2 π3
⎝ 4 ⎠⎥
⎥
⎢ ⎢
⎥
⎦
⎦0
⎣ ⎣
⎛ nπx ⎞ ⎤
4 cos⎜
⎟ ⎥
⎝ 4 ⎠ ⎥
dx
⎥
nπ
⎥
⎦⎥ 0
4
⎡ ⎛ nπx ⎞ ⎤ ⎤
⎟ ⎥⎥
⎢ sin ⎜
⎢ ⎝ 4 ⎠ ⎥⎥
⎢ ⎛ nπ ⎞ ⎥ ⎥
⎢ ⎜ ⎟ ⎥⎥
⎣ ⎝ 4 ⎠ ⎦ ⎥⎦ 0
4
⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
⎢ − 4 x cos ⎜ 4 ⎟ 16 sin ⎜ 4 ⎟ ⎥
⎠⎥
⎝
⎠
⎝
⎢
+
2
2 3
⎢
⎥
nπ
n π
⎢
⎥
⎣
⎦0
⎡⎧
⎛ nπ(0) ⎞ ⎫⎤
⎛ nπ(0) ⎞
⎛ nπ(4) ⎞
⎛ nπ(4) ⎞ ⎫ ⎧
⎟ ⎪⎥
⎟ 16 sin⎜
⎟ ⎪ ⎪ − 4(0) cos ⎜
⎟ 16 sin ⎜
⎢⎪ − 4(4) cos⎜
⎝ 4 ⎠ ⎪⎥
⎝ 4 ⎠+
⎝ 4 ⎠⎪− ⎪
⎝ 4 ⎠+
⎢⎪
⎬⎥
⎬ ⎨
= ⎢⎨
nπ 2
n 2π3
nπ 2
n 2 π3
⎪⎥
⎪ ⎪
⎢⎪
⎪⎭⎥
⎪⎭ ⎪⎩
⎢⎣⎪⎩
⎦
⎡⎧ − 16 cos(nπ) 16 sin (nπ)⎫ ⎧ 16 sin (0)⎫⎤
+
= ⎢⎨
⎬ − ⎨0 + 2 3 ⎬⎥
nπ 2
n 2 π3 ⎭ ⎩
n π ⎭⎦
⎣⎩
[Q sin(0) = 0]
⎡⎧ − 16 cos (nπ) 16 sin (nπ)⎫
⎤
+
⎬ − {0 + 0}⎥
= ⎢⎨
2
2 3
nπ
n π
⎭
⎣⎩
⎦
Look for the S IA GROUP LOGO
on the TITLE COVER before you buy
3.29
UNIT-3 (Fourier Transform)
=−
16 cos (nπ ) 16 sin (nπ )
+
nπ 2
n 2 π3
=−
16(− 1)n 16(0)
16(− 1)n
+
−
=
nπ 2
n 2 π3
nπ 2
[Q sin nπ = 0 and cos nπ = (–1)n]
16(− 1)
nπ 2
The finite Fourier cosine transform of f(x) is given by,
n
∴
FS(n) = −
l
FC(n)=
⎛ nπx ⎞
⎟ dx
l ⎠
∫ f ( x) cos⎜⎝
0
4
⎛ x ⎞ ⎛ nπx ⎞
= ⎜ π ⎟ cos⎜ 4 ⎟ dx =
⎝ ⎠ ⎝
⎠
0
∫
4
⎡x
⎧ d ⎛ x⎞
⎛ nπx ⎞
⎛ nπx ⎞ ⎫ ⎤
cos⎜
⎟ dx − ⎨ ⎜ ⎟ cos ⎜
⎟ dx ⎬dx ⎥
⎢
⎝ 4 ⎠
⎝ 4 ⎠ ⎭ ⎥⎦ 0
⎢⎣ π
⎩ dx ⎝ π ⎠
∫
∫
4
⎡ ⎛ ⎛ nπx ⎞ ⎞
⎛
⎛ nπx ⎞ ⎞ ⎤
⎜ − sin⎜
⎟⎟
⎟⎟ ⎥
⎢ ⎜ sin⎜
x ⎜ ⎝ 4 ⎠⎟
1⎜
4 ⎠⎟ ⎥
⎝
⎢
−
dx =
=
⎢ π ⎜ ⎛ nπ ⎞ ⎟
π ⎜ ⎛ nπ ⎞ ⎟ ⎥
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎢
⎥
⎝ ⎝ 4 ⎠ ⎠ ⎦0
⎣ ⎝ ⎝ 4 ⎠ ⎠
∫
4
⎡
⎤
⎛ nπx ⎞
⎢ x 4 sin⎜ 4 ⎟
⎥
π
n
x
4
⎞
⎛
⎞
⎛
⎠−
⎝
⎥ =
dx
sin
⎟
⎜
= ⎢⎜ ⎟
nπ
⎢⎝ π ⎠
nπ 2
⎝ 4 ⎠ ⎥
⎢
⎥
⎣
⎦0
∫
4
⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
⎢ 4 x sin⎜ 4 ⎟ 4 × 4 × cos⎜ 4 ⎟ ⎥
⎠⎥
⎝
⎠+
⎝
=
=⎢
2
2 3
⎢
⎥
nπ
n π
⎢
⎥
⎣
⎦0
∫
4
⎡ ⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞ ⎤
4 sin⎜
⎟⎥
⎟ ⎥
⎢ ⎢ 4 sin⎜
⎝ 4 ⎠⎥ − 1
⎝ 4 ⎠ dx ⎥
⎢x ⎢
⎢π ⎢
⎥
⎥ π
nπ
nπ
⎢ ⎢
⎥
⎥
⎦
⎣ ⎣
⎦0
∫
⎡
⎛ nπx ⎞
4 sin⎜
⎟
⎢
x
⎞
⎛
⎝ 4 ⎠− 4
⎢⎜ ⎟
⎢⎝ π ⎠
nπ
nπ 2
⎢
⎣
4
⎡
⎛ nπx ⎞ ⎤ ⎤
⎢ − cos ⎜ 4 ⎟ ⎥ ⎥
⎝
⎠ ⎥⎥
⎢
n
π
⎢ ⎛ ⎞ ⎥⎥
⎢ ⎜ 4 ⎟ ⎥⎥
⎣ ⎝ ⎠ ⎦⎦ 0
4
⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
⎢ 4 x sin⎜ 4 ⎟ 16 cos⎜ 4 ⎟ ⎥
⎠⎥
⎝
⎠+
⎝
⎢
2
2 3
⎢
⎥
nπ
n π
⎢
⎥
⎣
⎦0
⎡⎧
⎛ nπ( 0) ⎞ ⎫⎤
⎛ n π (0 ) ⎞
⎛ nπ( 4) ⎞ ⎫ ⎧
⎛ nπ( 4) ⎞
⎟ ⎪⎥
⎟ 16 cos⎜
⎟ ⎪ ⎪ 4(0) sin ⎜
⎟ 16 cos⎜
⎢ ⎪ 4( 4) sin ⎜
⎪
⎝ 4 ⎠ ⎪⎥
⎝ 4 ⎠+
⎝ 4 ⎠⎪− ⎪
⎝ 4 ⎠+
⎢
⎬
⎬ ⎨
= ⎨
⎢⎪
nπ 2
n 2 π3
nπ 2
n 2 π3
⎪⎥
⎪ ⎪
⎢⎪
⎪⎭⎥⎦
⎪
⎪
⎭ ⎩
⎣⎩
⎡⎧16 sin nπ 16 cos nπ ⎫ ⎧ 16 cos 0 ⎫⎤
= ⎢⎨
+
⎬ − ⎨0 + 2 3 ⎬⎥
2
n2π3 ⎭ ⎩
n π ⎭⎦
⎣⎩ nπ
16 ⎤
⎡ 16 sin nπ 16 cos nπ
+
=⎢
− 2 3⎥
n 2 π3
n π ⎦
⎣ nπ 2
[Q cos 0 = 1]
⎡16(0) 16(− 1)n
16 ⎤
= ⎢ 2 + 2 3 − 2 3⎥
n π
n π ⎦⎥
⎣⎢ nπ
[Q sin nπ = 0 and cos nπ = (–1)n]
=
∴ FC (n) =
16(− 1)n
16
16
− 2 3 = 2 3 ( −1) n − 1
2 3
n π
n π
n π
[
]
16 ⎡
(−1)n − 1⎤⎦
n 2π 3 ⎣
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
3.30
MATHEMATICS-II [JNTU-ANANTAPUR]
π – x) in 0 < x < π .
Q39. Find the finite Fourier sine and cosine transform of f(x) = x (π
Model Paper-I, Q7
Ans: Given function is, f(x) = x (π – x) in 0 < x < π.
The Finite Fourier sine transform of f(x) is given by,
l
FS(n) =
⎛ nπx ⎞
⎟dx
l ⎠
∫ f ( x) sin⎜⎝
0
π
=
⎛ nπx ⎞
f ( x ) sin ⎜
⎟dx =
⎝ π ⎠
∫
0
π
=
∫
π
∫ f ( x) sin nx dx
0
π
x( π − x) sin nx dx =
0
∫ ( πx − x
2
) sin nx dx
0
[
]
π
π
⎡ ⎧d
⎫ ⎤
= (πx − x 2 ) sin( nx ) dx − ⎢ ⎨ ( πx − x 2 ) sin nxdx ⎬dx ⎥
0
dx
⎭ ⎦0
⎣ ⎩
∫
π
∫
∫
π
⎡
⎡
⎛ − cos nx ⎞ ⎤
2 ⎡ − cos nx ⎤ ⎤
⎟dx ⎥
= ⎢(πx − x )⎢
⎥ ⎥ − ⎢ (π − 2 x )⎜
n
n ⎠ ⎦0
⎝
⎦⎦ 0 ⎣
⎣
⎣
π
∫
π
⎡⎡
⎡
⎡ − cos nx ⎤ ⎤
⎛ − cos nx ⎞
⎛ − cos nx ⎞ ⎤ ⎤
= ⎢( πx − x 2 ) ⎢
⎟ − ( −2) ⎜
⎟dx ⎥ dx ⎥
⎥ − ⎢ ⎢( π − 2 x ⎜
⎥
n
n
⎣ n ⎦ ⎦ 0 ⎢⎣ ⎣
⎝
⎠
⎝
⎠ ⎦ ⎥⎦ 0
⎣
π
∫
∫
∫
π
⎡⎡
⎡
⎛ − sin nx ⎞ ⎤ ⎤
⎛ − cos nx ⎞
2 ⎡ − cos nx ⎤ ⎤
⎟dx ⎥ ⎥
⎟ + (2) ⎜
= ⎢(πx − x )⎢ n ⎥ ⎥ − ⎢⎢(π − 2 x ⎜
2
n
⎠
⎣
⎦ ⎦ 0 ⎢⎣⎣
⎝
⎠ ⎦ ⎥⎦ 0
⎝ n
⎣
∫
π
∫ ∫
π
⎡⎡
⎡
⎛ − sin nx ⎞
⎛ sin nx ⎞ ⎤ ⎤
2 ⎛ − cos nx ⎞ ⎤
⎟ ⎥ − ⎢ ⎢( π − 2 x )⎜
⎟ − 2 ⎜ 2 ⎟ dx ⎥ ⎥
= ⎢ ( πx − x )⎜
2
n
⎝
⎠ ⎦ 0 ⎢⎣ ⎣
⎝ n
⎠
⎝ n ⎠ ⎦ ⎦⎥ 0
⎣
π
∫
π
⎡
⎡
⎛ − cos nx ⎞⎤
⎛ − sin nx ⎞
2 ⎛ − cos nx ⎞⎤
⎟⎥
⎟−2 ⎜
⎟⎥ − ⎢(π − 2 x )⎜
= ⎢(πx − x )⎜
2
n ⎠⎦ 0 ⎣
⎝ n3 ⎠⎦ 0
⎠
⎝ n
⎝
⎣
π
∫
π
⎡
sin nx
cos nx ⎤
⎡
2 ⎛ − cos nx ⎞⎤
⎟⎥ + ⎢(π − 2 x ) 2 − 2 3 ⎥
= ⎢(πx − x )⎜
n ⎠⎦ 0 ⎣
n
n ⎦0
⎝
⎣
⎡⎧
⎛ − cos nπ ⎞
sin nπ
sin n(0) ⎫ ⎧ cos nπ cos n (0) ⎫⎤
⎧
2 ⎛ − cos n(0) ⎞ ⎫
−
⎟ − (π (0 ) − (0 ) ⎜
⎟ ⎬ + ⎨(π − 2 π) 2 − ( π − 2(0))
⎬ − 2⎨
⎬⎥
n
n
⎝
⎠
⎝
n
n 2 ⎭ ⎩ n3
n 3 ⎭⎦
⎠⎭ ⎩
2
= ⎢ ⎨(π × π − π )⎜
⎣⎢ ⎩
⎡ ⎧⎪ ( −1) n 1 ⎫⎪⎤
= ⎢ − 2⎨ 3 − 3 ⎬ ⎥
n ⎪⎭⎦⎥
⎣⎢ ⎪⎩ n
=
∴ Fs (n ) =
−2
n3
2
n3
[(−1) − 1]
n
[1 − (−1) n ]
Look for the S IA GROUP LOGO
on the TITLE COVER before you buy
3.31
UNIT-3 (Fourier Transform)
The finite Fourier cosine transform of f(x) is given by,
l
FC(n) =
⎛ nπx ⎞
⎟dx
l ⎠
∫ f ( x) cos⎜⎝
0
π
=
⎛ nπx ⎞
⎟ dx
π ⎠
∫ x(π − x) cos⎜⎝
0
π
=
∫ x(π − x) cos(nx)dx
0
π
=
∫ (πx − x
2
) cos( nx) dx
0
[
]
π
π
⎡ ⎧d
⎫ ⎤
2
= (πx − x ) cos( nx)dx − ⎢ ⎨ (πx − x ) cos( nx )dx ⎬dx ⎥
0 ⎣ ⎩ dx
⎭ ⎦0
2
∫
π
∫
∫
π
⎡ ⎧
⎡
⎛ sin nx ⎞ ⎫ ⎤
2 ⎛ sin nx ⎞ ⎤
⎟ ⎬dx ⎥
⎟ ⎥ − ⎢ ⎨( π − 2 x )⎜
= ⎢ ( πx − x )⎜
⎝ n ⎠ ⎭ ⎦⎥ 0
⎝ n ⎠ ⎦ 0 ⎢⎣ ⎩
⎣
π
∫
π
⎡⎡
⎡
d
⎛ sin nx ⎞ ⎤ ⎤
⎛ sin nx ⎞
2 ⎛ sin nx ⎞ ⎤
( π − 2 x) ⎜
= ⎢ ( πx − x )⎜
⎟dx ⎥ ⎥
⎟ dx −
⎟ ⎥ − ⎢ ⎢ ( π − 2 x) ⎜
dx
⎝ n ⎠ ⎦ ⎥⎦ 0
⎝ n ⎠
⎝ n ⎠ ⎦ 0 ⎢⎣ ⎣
⎣
∫
π
∫
∫
π
⎡
⎡
⎛ − cos nx ⎞ ⎤
⎛ − cos nx ⎞
2 ⎛ sin nx ⎞ ⎤
⎟dx ⎥
⎟ − (−2)⎜
⎟ ⎥ − ⎢ ( π − 2 x )⎜
= ⎢(πx − x )⎜
2
n
⎝ n2 ⎠ ⎦0
⎠
⎝ n
⎠⎦ 0 ⎣
⎝
⎣
∫
π
π
⎡
⎡
⎛ − cos nx ⎞ ⎛ sin nx ⎞⎤
2 ⎛ sin nx ⎞ ⎤
⎟ − 2⎜
⎟⎥ − ⎢(π − 2 x)⎜
⎟⎥
= ⎢(πx − x )⎜
⎝ n 2 ⎠ ⎝ n 3 ⎠⎦ 0
⎝ n ⎠⎦ 0 ⎣
⎣
π
π
⎡
⎡
⎛ cos nx ⎞ ⎛ sin nx ⎞⎤
2 ⎛ sin nx ⎞ ⎤
⎟⎥ + ⎢(π − 2 x)⎜ 2 ⎟ + 2⎜ 3 ⎟⎥
= ⎢(πx − x )⎜
n
⎠⎦ 0 ⎣
⎝
⎝ n ⎠ ⎝ n ⎠⎦ 0
⎣
⎡⎧
⎧
⎛ cos nπ ⎞
⎛ cos n(0) ⎞⎫ ⎧ sin nπ sin n(0) ⎫⎤
2 ⎛ sin nπ ⎞
2 ⎛ sin n(0) ⎞ ⎫
−
⎟⎬ + 2 ⎨
⎬⎥
= ⎢⎨(π × π − π )⎜ n ⎟ − ( π(0) − (0) ⎜ n ⎟ ⎬ + ⎨( π − 2π)⎜ 2 ⎟ − ( π − 2(0))⎜
n3 ⎭⎥⎦
⎝
⎠⎭ ⎩
⎝
⎠
⎝ n ⎠
⎝ n 2 ⎠⎭ ⎩ n 3
⎢⎣⎩
⎡
⎛ ( −1) n
⎜
= ⎢ 0 + ( − π )⎜ 2
⎢⎣
⎝ n
⎤
⎞
⎟ − ( π)⎛⎜ (1) ⎞⎟ + 2( 0) ⎥
⎟
⎝ n2 ⎠
⎥⎦
⎠
⎡ π(−1) n − π ⎤
− 2⎥
= ⎢−
n2
n ⎥⎦
⎢⎣
=
∴
FC(n) =
−π
n
2
−π
n2
[1 + (−1) ]
n
[1 + (−1) ]
n
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
3.32
MATHEMATICS-II [JNTU-ANANTAPUR]
x2 π
− , where 0 < x < π .
2π 6
Q40. Find the finite cosine transform of f(x) defined by f(x) =
Ans: Given function is,
f(x) =
x2 π
− , where 0 < x < π.
2π 6
The Finite Fourier cosine transform of f(x) is given by,
l
FC(n) =
⎛ nπx ⎞
⎟ dx
l ⎠
∫ f ( x) cos⎜⎝
0
π
⎛ x 2 π ⎞ ⎛ nπx ⎞
⎟
⎜
= ⎜ 2π − 6 ⎟ cos ⎜ π ⎟ dx =
⎠
⎠ ⎝
0⎝
∫
π
⎛ x2 π ⎞
⎜
− ⎟
⎜ 2π 6 ⎟ cos(nx) dx
⎠
⎝
0
∫
π
π
⎡⎛ x 2 π ⎞
⎤ ⎡ ⎧⎪ d ⎛ x 2 π ⎞
⎫ ⎤
⎟ cos nxdx ⎪⎬dx ⎥
= ⎢⎜
− ⎟ cos nxdx ⎥ − ⎢ ⎨ ⎜
−
⎜
⎪⎭ ⎥⎦
⎢⎣⎝ 2π 6 ⎟⎠
⎥⎦ 0 ⎢⎣ ⎪⎩ dx ⎜⎝ 2π 6 ⎟⎠
0
∫
∫
∫
π
π
π
⎡⎛ x 2 π ⎞⎛ sin nx ⎞⎤ ⎡ ⎛ x ⎞⎛ sin nx ⎞ ⎤ π
− ⎟⎜
⎢⎜⎜
⎟dx ⎥
⎟⎥ − ⎢ ⎜ ⎟⎜
⎢⎣⎝ 2π 6 ⎟⎠⎝ n ⎠⎥⎦ 0 ⎣ ⎝ π ⎠⎝ n ⎠ ⎦ 0
⎡⎛ x 2 π ⎞⎛ sin nx ⎞⎤ ⎡ ⎧⎛ 2 x ⎞⎛ sin nx ⎞⎫ ⎤
⎜
⎟
= ⎢⎜ 2π − 6 ⎟⎜ n ⎟⎥ − ⎢ ⎨⎜ 2π ⎟⎜ n ⎟⎬dx ⎥ =
⎠⎭ ⎦⎥ 0
⎠⎥⎦ 0 ⎢⎣ ⎩⎝ ⎠⎝
⎢⎣⎝
⎠⎝
∫
∫
π
⎡⎛ x 2 π ⎞⎛ sin nx ⎞ ⎤
⎡ x sin nx
⎧ d ⎛ x ⎞ sin nx ⎫ ⎤
= ⎢⎜
− ⎟⎜
dx ⎬dx ⎥
dx − ⎨ ⎜ ⎟
⎟⎥ − ⎢
⎟
⎜
n
n
⎩ dx ⎝ π ⎠
⎭ ⎦⎥
⎣⎢⎝ 2π 6 ⎠⎝ n ⎠ ⎦⎥ 0 ⎣⎢ π
∫
∫
∫
π
π
⎡⎛ x 2 π ⎞⎛ sin nx ⎞ ⎤
⎡ x ⎡ − cos nx ⎤
⎛ 1 ⎞⎛ − cos nx ⎞ ⎤
= ⎢⎜
− ⎟⎜
−⎢ ⎢
−
dx
⎥
⎟
⎜
⎟
⎜
⎟
⎥
⎥
2
⎝ π ⎠⎝ n 2 ⎠ ⎦ 0
⎦
⎢⎣⎜⎝ 2π 6 ⎟⎠⎝ n ⎠ ⎥⎦ 0 ⎣ π ⎣ n
∫
π
π
⎡⎛ x 2 π ⎞⎛ sin nx ⎞⎤ ⎡ − x ⎛ cos nx ⎞
⎤
1
⎟
= ⎢⎜
− ⎜
⎜ 2 ⎟ + 2 cos nxdx ⎥
⎟⎥ − ⎢
⎜
⎢⎣⎝ 2π 6 ⎟⎠⎝ n ⎠⎥⎦ 0 ⎣ π ⎝ n ⎠ πn
⎦0
∫
π
π
⎡⎛ x 2 π ⎞⎛ sin nx ⎞ ⎤
⎡ − x ⎛ cos nx ⎞
1 ⎛ sin nx ⎞ ⎤
− ⎟⎜
−⎢
+
.
⎜
⎜
⎥
⎟
⎟
⎟
= ⎢⎜⎜
⎥
2
2
⎟
⎣⎢⎝ 2π 6 ⎠⎝ n ⎠ ⎦⎥ 0 ⎣ π ⎝ n ⎠ πn ⎝ n ⎠ ⎦ 0
π
π
⎡⎛ x 2 π ⎞⎛ sin nx ⎞⎤
⎡ x ⎛ cos nx ⎞ sin nx ⎤
− ⎟⎜
+ ⎢ ⎜ 2 ⎟−
⎥
⎟
= ⎢⎜⎜
3 ⎥
⎢⎣⎝ 2π 6 ⎟⎠⎝ n ⎠⎥⎦ 0 ⎣ π ⎝ n ⎠ πn ⎦ 0
⎡⎧⎪⎛ π 2 π ⎞⎛ sin nπ ⎞ π ⎛ cos nπ ⎞ ⎛ sin nπ ⎞⎫⎪
− ⎟⎜
⎟+ ⎜
⎟−⎜
⎟⎬ −
= ⎢⎨⎜⎜
⎢⎣⎪⎩⎝ 2π 6 ⎟⎠⎝ n ⎠ π ⎝ n 2 ⎠ ⎝ πn3 ⎠⎪⎭
=
=
∴ FC (n) =
⎧⎪⎛ 0 2 π ⎞⎛ sin n(0) ⎞ 0 ⎛ cos n(0) ⎞ sin n(0) ⎫⎪⎤
− ⎟⎜
⎟+ ⎜
⎟−
⎬⎥
⎨⎜⎜
2
πn 3 ⎪⎭⎥⎦
⎠
⎪⎩⎝ 2π 6 ⎟⎠⎝ n ⎠ π ⎝ n
⎡⎧⎪⎛ π π ⎞
⎫⎪ ⎧⎛
⎫⎤
1(−1) n
π⎞
− 0⎬ − ⎨⎜ 0 − ⎟(0) + 0 − 0⎬⎥
⎢⎨⎜ − ⎟(0) +
2
6⎠
⎪⎭ ⎩⎝
n
⎢⎣⎪⎩⎝ 2 6 ⎠
⎭⎥⎦
(−1) n
n2
(−1)n
n2
Look for the S IA GROUP LOGO
on the TITLE COVER before you buy
4.1
UNIT-4 (Partial Differential Equations)
UNIT
4
PARTIAL DIFFERENTIAL
EQUATIONS
PART-A
SHORT QUESTIONS WITH SOLUTIONS
Model Paper-I, Q1(g)
Q1. Define order and degree with reference to partial differential equation.
Ans:
Order: The order of a partial differential equation is equal of the order to highest partial derivative for a particular equation.
Degree: The degree of a partial differential equation is equal to the power of highest partial derivative.
Examples
(i)
x
∂z
∂z
+y
=z
∂x
∂y
IOrder 1 and degree 1.
(ii)
(iii)
∂2N ∂2N
+ 2 = 3t 2 + x
2
∂x
∂t
Order 2 and degree 1.
⎛ ∂2N
⎜
⎜ ∂t 2
⎝
2
⎞
⎟ + ∂N = t − x
⎟
∂x
⎠
Order 2 and degree 2.
Q2. Form partial differential equation by eliminating the arbitrary constants z = ax + by + a2 + b2.
Model Paper-II, Q1(h)
Ans:
Given that,
z = ax + by + a2 + b2
... (1)
On differentiating equation (1) partially with respect to ‘x’ and ‘y’, we get,
∂z
=a
∂x
⇒
zx = a
⇒
a =p
[Q
∂z
= zx = p]
∂x
[Q
∂z
∂y = zy = q]
∂z
= zy = b
∂y
⇒ b= q
On substituting the values of ‘a’ and ‘b’ in equation (1), we get,
⇒ z = px + qy + p2 + q2.
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.2
MATHEMATICS-II [JNTU-ANANTAPUR]
Q3.
Form partial differential equation by eliminating
the arbitrary constants z = (x2 + a2) (y2 + b2).
Ans: Given that,
z = (x2 + a2) (y2 + b2)
... (1)
Differentiating equation (1) partially with respect to ‘x’,
we get,
On differentiating equation (4) partially with respect to
‘y’, we get,
∂z
1
= zy = axey + a2.e2y.2
∂y
2
∂z
= zx = 2x (y2 + b2)
∂x
⇒
2
∂z
= p]
[Q zx =
∂x
2
p = 2x (y + b )
Q5.
p
= y 2 + b2
2x
⇒
q = (aey)x + (aey)2
[Q
⇒
q = p.x + p2
[From equation (2)]
Form the partial differential equation by
eliminating the arbitrary function from z = f(x2– y2).
Ans:
p
2x
On differentiating equation (1) partially with respect to
‘y’, we get,
∴ y2 + b 2 =
Model Paper-III, Q1(g)
Given that,
z = f(x2 – y2)
... (1)
On differentiating equation (1) partially with respect to
‘x’, we get,
∂z
= zx = p = f ′ (x2 – y2).2x
∂x
∂z
= zy = 2y(x2 + a2)
∂y
⇒
q = 2y(x2 + a2)
or
[Q zy =
∂z
=q]
∂y
f ′ (x2 – y2) =
2
On substituting the values of x 2 + a2 , y 2 + b2 in
equation (1), we get,
z=
f ′ (x2 – y2) =
q p
.
2 y 2x
Form partial differential equation by
eliminating the arbitrary constants z = axey +
1 2 2y
a e + b.
2
Ans: Given that,
Model Paper-I, Q1(h)
1 2 2y
a e +b
... (1)
2
On differentiating equation (4) partially with respect to
‘x’, we get,
z = ax ey +
∂z
= zx = aey
∂x
p = aey
... (3)
−q
p
=
2y
2x
∴4 xyz = pq
⇒
−q
2y
From equation (2) and equation (3), we get,
pq
z=
4 xy
Q4.
... (2)
∂z
= z = q = f ′ (x2 – y2) . (–2y)
∂y y
q
∴ x +a =
2y
2
p
2x
On differentiating equation (1) partially with respect to
‘y’, we get,
q
= x 2 + a2
2y
⇒
∂z
= zy = q]
∂y
⇒
[Q
∂z
= zx = p]
∂x
... (2)
Look for the SIA GROU P LOGO
⇒
p −q
=
x
y
⇒
py = – qx
∴ py = qx = 0
Q6.
Form the partial differential equation by
eliminating the arbitrary function from xyz =
f(x2 + y2 + z2).
Ans: Given that,
xyz = f(x2 + y2 + z2)
... (1)
On differentiating equation (1), partially with respect to
x, we get,
⇒
∂z ⎞
∂z ⎞
⎛
⎛
y⎜ z + x ⎟ = f ′( x 2 + y 2 + z 2 ) ⎜ 2 x + 2 z ⎟
∂
x⎠
x
∂
⎝
⎝
⎠
... (2)
y(z + xp) = f ′( x 2 + y 2 + z 2 ) (2 x + 2 zp )
... (3)
on the TITLE COVER before you buy
4.3
UNIT-4 (Partial Differential Equations)
On differentiating equation (1), partially with respect to
y, we get,
⎛
∂z ⎞
x⎜⎜ z + y ⎟⎟ = f ′( x 2 + y 2 + z 2 )
y⎠
∂
⎝
⇒
⎛
∂z ⎞
⎜⎜ 2 y + 2 z ⎟⎟
∂y ⎠
⎝
x(z + yq) = f ′( x + y + z ) (2 y + 2 zq )
2
2
Integrating equation (1) wtih respect to a and b partially,
we get,
2
Consider,
⇒
⇒
... (4)
Dividing equations (3) and (4), we get,
Consider,
y ( z + xp ) (2 x + 2 zp)
=
x( z + yq) (2 y + 2 zq)
∴
Form the partial differential equation by
eliminating the arbitrary function from, z = f
(sinx + cosy).
Ans: Given that,
z = f(sinx + cosy)
... (1)
On differentiating equation (1) with respect to ‘x’, we get,
∂z
= zx = f ′ (sinx + cosy).cosx
∂x
f ′ (sinx + cosy) =
p
cos x
z =0+0
z = 0 (constant)
∴z =k
Q9.
Model Paper-II, Q1(g)
∂u ∂ 2u
=
... (1)
∂t ∂x 2
The three possible solutions of equation(1) are,
... (2)
∂z
q= – siny . f ′ (sinx + cosy) [Q ∂y = zy = q]
−q
sin y
∂ u ∂ 2u
.
=
∂ t ∂x 2
Given partial differentiation equation is,
∂z
= z = f ′ (sinx + cosy) (–siny)
∂y y
f ′ (sinx + cosy) =
Write the three possible solutions of
Ans:
On differentiating equation (1) with respect to ‘y’, we get,
⇒
∂z
∂b
get,
∂z
p = cosx. f ′ (sinx + cosy) [Q
= zx = p]
∂x
⇒
0 = y3 + 0
y3 = 0 ⇒ y = 0
⇒ 0 = 0 + x2
⇒ x2 = 0 ⇒ x = 0
On substituting the values of x and y in equation (1), we
(y + zq) y (z + xp) = (x + zp) x (z + yq)
Q7.
∂z
∂a
... (3)
Equating equations (3) and (2), we get,
2
(i)
u (x, t) = (A1epx + B1e–px) e p
(ii)
u (x, t) = A2 + B2x
(iii)
u (x, t) = (A3cospx + B3sinpx) e – p
t
2
t
Q10. Write the boundary conditions for the following
problem. A rectangular plate is bounded by the
line x = 0, y = 0 and y = b. Its surfaces are
insulated. The temperature along x = 0 and y =
0 are kept at 0ºC and the others are kept at 100ºC.
Ans:
Model Paper-III, Q1(h)
Let, u(x, y) be a function.
According to the given data, the rectangular plate is
shown in figure.
−q
p
=
cos x sin y
y = b, u = 100
∴ p siny = –q cosx
or
x =a
u = 100
x=0
u=0
p siny + q cosx = 0
Q8.
Solve by the method of separation of variable
py3 + qx2 = 0.
Ans: Given that,
py3 + qx2 = 0
Let p = a and q = b
∴ z = ay3 + bx2
... (1)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
y = 0, u = 0
Figure
The boundary conditions are,
(i) u(0, y) = 0
(ii) u(x, 0) = 0
(iii) u(a, y) = 0
(iv) u(x, b) = 0
SIA GROUP
4.4
MATHEMATICS-II [JNTU-ANANTAPUR]
PART-B
ESSAY QUESTIONS WITH SOLUTIONS
4.1 FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS BY ELIMINATION OF ARBITRARY
CONSTANTS AND ARBITRARY FUNCTIONS
Q11. Obtain partial differential equation by eliminating the arbitrary constants for the following equations,
(a)
z = ax2 + by2
(b)
2z =
x2
a2
+
y2
b2
.
Ans:
(a)
z = ax2 + by2
Given that,
z = ax2 + by2
... (1)
On differentiating equation (1) partially with respect to ‘x’ and ‘y’, we get,
⇒
zx =
∂z
= 2ax
∂x
a=
zx
2x
Let, p = zx
zy =
b =
∂z
= 2by
∂y
zy
2y
Let, q = zy
On substituting the values of ‘a’ and ‘b’ in equation (1), we get,
⇒
⎛ zx ⎞ 2 ⎛ z y ⎞ 2
z = ⎜ ⎟ x + ⎜⎜ ⎟⎟ y
⎝ 2x ⎠
⎝ 2y ⎠
⇒
z=
⇒
z=
⇒
x. z x y. z y
+
2
2
x.z x + y.z y
2
2z = x.zx + y.zy
∴ 2 z = x. p = yq
(b)
2z =
x2 y 2
+
a2 b2
Given that,
2z =
x2 y 2
+
a 2 b2
Look for the SIA GROU P LOGO
... (1)
on the TITLE COVER before you buy
4.5
UNIT-4 (Partial Differential Equations)
On differentiating equation (1) partially with respect to
‘x’, we get,
⇒
2.
∂z 2 x
=
∂x a 2
⇒
2.zx =
⇒
zx =
⇒
On differentiating equation (1) partially with respect to
‘y’, we get,
p=
∂z
= zy = q = 2(y – b)
∂y
2x
⇒
a2
⇒
q=
a
2
x
a2
⇒ a2 =
x
p
[Q p = zx =
∂z
]
∂x
⇒
b2 =
⇒
⇒
z =
(b)
z = ae − b
2
⇒
cos bx
... (1)
2
∂z
= – abe −b t sin bx
∂x
... (2)
[
2
∂ ⎡ ∂z ⎤ ∂
− abe −b t . sin(bx )
=
⎢
⎥
∂x ⎣ ∂x ⎦ ∂z
⇒
Q12. Obtain the partial differential equation by
eliminating the arbitrary constants
(a) z = (x – a)2 + (y – b)2 + 1
t
cos bx.
Model Paper-III, Q8
Ans:
(a)
z = (x – a)2 + (y – b)2 + 1
Given that,
z = (x – a)2 + (y – b)2 + 1
... (1)
On differentiating equation (1) partially with respect to
‘x’, we get,
]
∂2z
−b 2 t
−
abe
cos bx (b )
=
∂x 2
∂2z
2
= − ab 2e −b t cos bx
∂x 2
... (3)
On differentiating equation (1) partially with respect to
‘t’, we get,
∂z
= zx = p = 2(x – a)
∂x
p
2
t
On differentiating equation (2) with respect to ‘x’,
∴2 z = px = qy
x–a =
2
we get
px 2 qy 2
+
2z =
x
y
⇒
cos bx
2
∂z
= – ae −b t sin bx (b)
∂x
2
x
y
+
2z =
⎛ x ⎞ ⎛ y⎞
⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟
⎝ p⎠ ⎝q⎠
2
t
On differentiating equation (1) partially with respect to
‘x’, we get,
y
q
z = ae −b
2
−b
z = ae
∂z
]
[Q q = zy =
∂y
y
2
(b)
p2 q2
+
+1
4
4
Given that,
b2
On substituting the values of a2 and b2 in equation (1),
we get,
⇒
2
⎛ p⎞ ⎛q⎞
z = ⎜ ⎟ + ⎜ ⎟ +1
⎝ 2 ⎠ ⎝ 2⎠
∴ 4z = p 2 = q 2 + 4
y
b
... (3)
2
∂z 2 y
=
2.
∂y b 2
zy =
q
2
Substituting equations (2) and (3) in equation (1), we get,
x
On differentiating equation (1) partially with respect to
‘y’, we get,
⇒
y–b=
∂z
− b 2t
cos bx.( −b 2 )
= ae
∂t
⇒
2
∂z
= − ab 2 e −b t cos bx
∂t
... (4)
From equations (3) and equation (4), we get,
... (2)
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∂2z
∂x
2
=
∂z
∂t
SIA GROUP
4.6
MATHEMATICS-II [JNTU-ANANTAPUR]
Q13. Form the partial differential equation by
eliminating the arbitrary constants a, b from
1
1
2z = (x + a) 2 + (y − a) 2 + b .
Ans: Given that,
[
]
2z = x + a + y − a + b
On differentiating with respect to x, we get,
1
2∂ z
+0
=
∂x
2 x+a
4∂z
=
∂x
⇒
4p =
⇒ x+a =
⇒ x+a =
1
⇒
x+a
x+a
(1 − l 2 − m 2 ) .z = a
1
(1 − l 2 − m 2 )
l+
16 p 2
1
–x
... (1)
⇒ l+
∂z
=0
∂x
(1 − l 2 − m 2 ) . p = 0
m+
(1 − l 2 − m 2 ) .
4∂z
=
∂y
m+
(1 − l 2 − m 2 ) .q = 0
1
y−a
y−a
y + x=
1
16 q 2
16q 2
∴16( x + y ) =
1
16q
2
1
16 p 2
1
p2
∂z
= q]
∂y
[Q
+a
+
+
+
... (4)
... (5)
m = − (1 − l 2 − m 2 ) .q
... (2)
On substituting the value of ‘a’ from equation (1) in
equation (2), we get,
y =
∂z
=0
∂y
l = − (1 − l 2 − m 2 ) . p and
1
... (3)
Eliminating l, m from equations (2), (3) and (4).
From equations (3) and (4), we get,
1
1
⇒ y−a =
4q
y =
∂z
= p]
∂y
[Q
On differentiating equation (2) partially with respect to
y, we get,
1
2∂z
+0
=
∂y
2 y−a
⇒ y–a =
... (2)
On differentiating equation (2) partially with respect to x
on both sides, we get,
1
4p
4q =
⇒
(1 − l 2 − m 2 )
lx + my +
16 p 2
On differentiating given equation with respect to y,
we get,
⇒
n =
Therefore, equation (1), becomes,
1
a=
⇒
Q14. Form the differential equations of all planes
which are at a constant distance ‘a’ from the
origin.
Ans: It is required to find the differential equation of all planes
which are at a constant distance ‘a’ from origin.
The equation of the plane in normal form as,
lx + my + nz = a
... (1)
Where l, m, n are the direction cosines of the normal
from the origin to the plane.
Then we have l2 + m2 + n2 = 1
1
16 p 2
Squaring on both sides and then adding the above two
equations, we get,
l2 + m2 = (1 – l2 – m2)p2 + (1 – l2 – m2) q2
⇒
l2 + m2 = (1 – l2 – m2) (p2 + q2)
⇒
l2 + m2 = p2 + q2 – l2p2 – l2q2 – m2p2 – m2q2
⇒
l2 (1 + p2 + q2) + m2 (1 + p2 + q2) = p2 + q2
⇒
(l2 + m2) (1 + p2 + q2) = p2 + q2
−x
1
16q 2
1
q2
Look for the SIA GROU P LOGO
... (6)
⇒
∴
l2 + m2 =
p2 + q 2
(1 + p 2 + q 2 )
(1 – l2 – m2) = 1 – (l2 + m2)
=1–
p2 + q2
2
1+ p + q
2
⎡
⎤
1
=⎢
2
2⎥
⎢⎣1 + p + q ⎥⎦
on the TITLE COVER before you buy
4.7
UNIT-4 (Partial Differential Equations)
On substituting (1 – l2 – m2) =
l = −
m =−
1
1+ p2 + q2
1
(1 + p 2 + q 2 )
1
(1 + p 2 + q 2 )
in equations (5) and (6), we get,
. p and
.q
On substituting the values of l, m, (1 – l2 – m2) in equation (2), we get,
1
−
⇒
(1 + p 2 + q 2 )
−1
⇒
(1 + p 2 + q 2 )
⇒
. px −
1
(1 + p 2 + q 2 )
.qy +
1
(1 + p 2 + q 2 )
.z = a
[ px + qy − z ] = a
px + qy – z = – a
(1 + p 2 + q 2 )
2
2
px + qy + a (1 + p + q ) = z
∴ The required differential equation of all planes which are at a constant distance ‘a’ from origin is given by,
z = px + qy + a (1 + p 2 + q 2 )
Q15. Form the partial differential equation by eliminating constants a, b and c from
x 2 y2 z 2
+
+
=1
a2 b2 c2
Ans: Given equation is,
x2
y2
z2
+
+
=1
... (1)
a2
b2
c2
In equation (1), there are 3 arbitrary constants and 2 independent variables. Thus, differential equation of order greater than
1 is otained. Treating ‘z’ as a function of ‘x’ and ‘y’.
Differentiating equation (1) partially with respect to ‘x’ and ‘y’, we get,
2x
a2
x
2
a
Similarly,
2y
b
2
y
b
2
+
+
+
+
×
∂z
=0
∂x
×
∂z
=0
∂x
2
×
∂z
=0
∂y
2
×
2z
c2
z
c
2
2z
c
z
c
... (2)
∂z
∂y = 0
... (3)
Again differentiating equation (2) partially with respect to ‘x’, we get,
1
a2
+
2
z
1 ⎛ ∂z ⎞
⎜
⎟
+ 2
2
c
c ⎝ ∂x ⎠
⎛ ∂2 z ⎞
⎜
⎟
⎜ ∂x 2 ⎟ = 0
⎝
⎠
2
⎛ ∂2 z ⎞
c 2 ⎛ ∂z ⎞
⎜
⎟
⎟
⎜
+
z
+
⎜ ∂x 2 ⎟ = 0
a 2 ⎝ ∂x ⎠
⎝
⎠
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (4)
SIA GROUP
4.8
MATHEMATICS-II [JNTU-ANANTAPUR]
From equation (2),
x
a
2
+
z
c
2
×
On substituting the value of f ′ (x2 + y2 + z2) from equation
(2) in equation (3), we get,
∂z
=0
∂x
1+ p
2 x + 2 zp
⇒
1 + q= (2y + 2zq) .
⇒
1 + q=
⇒
⇒
⇒
∴
(1 + q) (x + zp) = (y + zq) (1 + p)
x + zp + xq + zpq = y + yp + zq + zpq
x – y = yp + zq – zp – xq
x – y = (y – z) p + (z – x)q
Multiplying with c2 on both sides, we get,
⇒
c
2
a
2
=–
z ⎛ ∂z ⎞
⎜ ⎟
x ⎝ ∂x ⎠
On substituting
c2
value in equation (4), we get,
a2
Q17. Form the partial differential equation by
eliminating the arbitrary function from z = yf(x)
+ x g(y).
Ans: Given that,
z = y f (x) + x g(y)
... (1)
By differentiating equation (1) partially with respect to
x, we get,
2
⎛ ∂2 z ⎞
z ⎛ ∂z ⎞ ⎛ ∂z ⎞
⎟
− ⎜ ⎟ + ⎜ ⎟ +z⎜
⎜ ∂x 2 ⎟ = 0
x ⎝ ∂x ⎠ ⎝ ∂x ⎠
⎝
⎠
∴ The required partial differential equation is,
2
⎛ ∂z ⎞
z ∂z
∂2 z
= ⎜ ⎟ +z 2
x ∂x
⎝ ∂x ⎠
∂x
∂z
= y f '(x) + g(y)
∂x
Q16. Obtain the partial differential equation by
eliminating the arbitrary function from x+y+z =
f(x2+y2+z2).
Ans: Given that,
x + y + z = f(x2 + y2 + z2)
... (1)
Differentiating equation (1) partially with respect to ‘x’,
we get,
1+
⇒
⇒
∂z
∂z
2
2
2
f ′( x 2 + y 2 + z 2 )
= 2 xf ′( x + y + z ) + 2 z.
∂x
∂x
1 + zx = f ′ (x2 + y2+ z2) [2x + 2z zx]
1 + p = (2x + 2z.p) f ′ (x2 + y2 + z2) [Q
f ′ (x2 + y2 + z2) =
( y + zq)(1 + p)
( x + zp)
∂z
= zx = p]
∂x
1+ p
2 x + 2 zp
... (2)
On differentiating equation (1) with respect to ‘y’,
we get,
∴ yf '(x) + g(y) = p
... (2)
By differentiating equation (1), partially with respect to
y, we get,
∂z
= f (x) + xg'(y)
∂y
∴ f (x) + xg'(y) = q
... (3)
Since, equations (1), (2) and (3) are not sufficient to
eliminate f, g, f ' and g', the second order partial derivatives are
determined as,
∂2 z
= r = y f ''(x)
∂x 2
∂2z
= s = f '(x) + g'(y)
∂x∂y
∂2z
= t = x g''(y)
∂y 2
... (3)
⎡ ∂z
⎤
= zy = q ⎥
⎢Q
⎣ ∂y
⎦
Look for the SIA GROU P LOGO
... (5)
... (6)
From equation (2),
y f '(x) = p – g(y)
f ′( x) =
1 + zy = 2y f ′ (x2 + y2 + z2) + 2z zy f ′ (x2 + y2 + z2)
1 + q = (2y + 2zq) f ′ (x2 + y2 + z2)
... (4)
1
[ p − g ( y )]
y
... (7)
From equation (3),
x g'(y) = q – f (x)
g ′( y ) =
1
[q − f ( x )]
x
on the TITLE COVER before you buy
... (8)
4.9
UNIT-4 (Partial Differential Equations)
Q19. Form the partial differential equation by
eliminating the arbitrary function from
z = xf 1 (x+t) + f 2 (x+t).
From equation (5), we get,
s = f '(x) + g'(y)
s=
⇒
⇒
⇒
1
1
[p – g(y)] + [q – f (x)]
y
x
Ans: Given equation is,
z = x f1 (x + t) + f2 (x + t)
x[ p − g ( y )] + y[q − f ( x)]
s=
xy
To form the partial differential equation by eliminating
arbitrary functions.
∴ Differentiate equation (1), with respect to ‘x’ and ‘t’.
sxy = x[p – g(y)] + y[q – f (x)]
xys = xp – x g(y) + yq – y f (x)
xys = px + qy – [y f (x) + x g(y)]
(Q From equation (1))
First, differentiating with respect to ‘x’, we get,
∂z
= f (x + t) + xf1′ + f 2′
∂x 1
xys = px + qy − z
This is the required partial differential equation.
Q18. Form the partial differential equation by
eliminating the arbitrary function from z = f(y)
+ φ (x + y).
Ans: The given function is,
z = f(y) + φ(x + y)
... (1)
By differentiating equation (1), partially with respect to
x, we get,
⇒
∂z
= 0 + φ'(x + y) . 1
∂x
⇒
∂z
= p = φ'(x + y)
∂x
⇒
∂z
= xf1′ + f 2′
∂t
∂2 z
= f1′ + xf1′′+ f1′ + f 2′′
∂x 2
∂2 z
= xf1′′ + f1′ + f 2′′
∂x∂t
... (3)
∂2 z ∂ 2 z
+
= 2 f1′ + xf1′′ + f 2′′ + xf1′′ + f 2′′
∂x 2 ∂t 2
= 2 f1′ + 2 xf1′′ + 2 f 2′′
= 2 ( f1′ + xf1′′ + f 2′′)
⎛ ∂2z ⎞
⎟
= 2 ⎜⎜
⎟
∂
x
∂
t
⎝
⎠
‘x’,
⇒
∂2 z
= s = φ''(x + y)
∂x ∂y
∴
From equations (4) and (5), we get,
r = s or r – s = 0
This is the required partial differential equation.
... (6)
On adding equations (4) and (5), we get,
∂2z
= r = φ'' (x + y)
... (4)
⇒
∂x 2
On differentiating equation (3), partially with respect to
∂2z
= φ''(x + y).1
∂x∂y
... (5)
Again differentiating equation (3) partially with respect
to ‘x’, we get,
∂2z
= φ'' (x + y) .1
∂x 2
⇒
... (4)
∂2 z
= xf1′′ + f 2′′
∂t 2
... (2)
On differentiating equation (2) partially with respect to
‘x’, we get,
⇒
... (3)
Again differentiating equations (2) and (3) partially with
respect to ‘x’ and ‘t’, we have,
∂z
= f '(y)+ φ'(x + y) . 1
∂y
∂z
= q = f '(y) + φ'(x + y)
∂y
... (2)
On differentiating with respect to ‘t’, we get,
= 2 f1′ + xf1′′+ f 2′′
By differentiating equation (1), partially with respect to
y, we get,
⇒
... (1)
[Q From equation (6)]
⎛ ∂2z ⎞
∂2 z ∂ 2 z
⎜
⎟
2
+
=
⎜ ∂x∂t ⎟
∂x 2 ∂t 2
⎝
⎠
... (5)
⇒
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎛ ∂2 z ⎞ ∂ 2 z
⎜
⎟
−
2
⎜ ∂x∂t ⎟ + ∂t 2 = 0
∂x 2
⎝
⎠
∂2z
This is the required partial differential equation.
SIA GROUP
4.10
MATHEMATICS-II [JNTU-ANANTAPUR]
Q20. Form the partial differential equations by
eliminating the arbitrary functions z = f(2x + y) +
g(3x – y).
Ans: Given equation is,
z = f(2x + y) + g(3x – y)
... (1)
On differentiating equation (1) partially, with respect to
‘x’, we get,
∂z
= f'(2x + y).2 + g'(3x – y).3
∂x
⇒
∂z
= p = 2f '(2x + y) + 3g'(3x – y)
∂x
⇒
Q21. Form the partial differential equation by
eliminating the arbitrary function from
⎛ y⎞
z = yf ⎜ ⎟ .
⎝x⎠
Ans: Given that,
⎛
z = y.f ⎜
⎝
⎛
zx = y. f ′⎜
⎝
... (2)
∂z
= f'(2x + y).1 + g'(3x – y).(–1)
∂y
∂z
= q = f '(2x + y) – g'(3x – y)
∂y
⇒
px 2
⎛y⎞
′
−
f
⎜
⎟
⇒
=
y2
⎝x⎠
... (3)
∂2z
= 2f"(2x + y).2 + 3g"(3x – y).3
∂x 2
⇒
p2 =
∂2z
= 4f"(2x + y) + 9g"(3x – y)
∂x 2
... (4)
∂y 2
⎛ y⎞
⎛ y ⎞⎛ 1 ⎞
zy = y. f ′⎜ ⎟⎜ ⎟ + f ⎜ ⎟.1
⎝ x⎠
⎝ x ⎠⎝ x ⎠
⇒
q2 =
∂ z
∂y 2
= f "(2x + y) + g"(3x – y)
⎛ y⎞
⎛y⎞
y
⇒ f ⎜ ⎟ = q – . f ′⎜ ⎟
x
x
⎝ ⎠
⎝x⎠
∂2z
= 2f"(2x + y).1 + 3g"(3x – y)(–1)
⇒
∂x∂y
pq =
∂2z
= 2f"(2x + y) – 3g"(3x – y)
∂xδy
... (6)
Adding equations (4) and (6), we get,
= 4f"(2x + y) + 9g"(3x – y) + 2f"(2x + y) – 3g"(3x – y)
= 6f"(2x + y) + 6g"(3x – y)
= 6[f "(2x + y) + g"(3x – y)]
= 6(q2)
[ Q From equation (5)]
2
2
p + pq = 6q
∴ p2 – 6q2 + pq = 0 is the required differential equation.
p2 + pq
Look for the SIA GROU P LOGO
... (3)
On substituting equation (3) in equation (1), we get,
⇒
z = y [q –
⇒
z =
... (5)
On differentiating equation (2) partially, with respect to
‘y’, we get,
⇒
⎛ y⎞
⎛ y⎞ ⎛ y⎞
q = ⎜ ⎟ f ′⎜ ⎟ + f ⎜ ⎟
⎝x⎠
⎝ x⎠ ⎝x⎠
= f"(2x + y).1 – g"(3x – y)(–1)
2
... (2)
On differentiating equation (1) with respect to ‘y’, we get,
⇒
On differentiating equation (3) partially with respect to
‘y’, we get,
∂2 z
y ⎞⎛ − 1 ⎞
⎟⎜ ⎟
x ⎠⎝ x 2 ⎠
⎛ − y2 ⎞ ⎛ y ⎞
⎜
⎟
p = ⎜ 2 ⎟ f ′⎜ x ⎟
⎝ x ⎠ ⎝ ⎠
On differentiating equation (2) partially, with respect to
‘x’, we get,
⇒
... (1)
On differentiating equation (1) with respect to ‘x’, we get,
On differentiating equation (1) partially, with respect to
‘y’, we get,
⇒
y⎞
⎟
x⎠
y ′⎛ y ⎞
f ⎜ ⎟]
x
⎝x⎠
⎛ y⎞
y
[qx – y f ′⎜ ⎟ ]
x
⎝x⎠
⎛ y⎞
On substituting f ′⎜ ⎟ value from equation (2), we get,
⎝x⎠
⎛ px 2 ⎞
⎜−
⎟
⎜ y2 ⎟ ]
⎝
⎠
⇒
y
z=
[qx – y
x
⇒
px 2
y
z=
(qx +
)
y
x
⇒
y
z=
x
⇒
z=
⇒
z = px + qy
⎛ qxy + px 2 ⎞
⎜
⎟
⎜
⎟
y
⎝
⎠
1
.x(qy + px)
x
on the TITLE COVER before you buy
4.11
UNIT-4 (Partial Differential Equations)
Q22. Form the partial differential equation by eliminating the arbitrary function from, z =
Ans: Given that,
z = eax + by f (ax – by)
On differentiating equation (1) with respect to ‘x’, we get,
eax + by
f (ax – by).
... (1)
∂z
= zx
∂x
= eax+by. f ′ (ax – by) . a + f(ax – by) . eax + by . a
p = a [ eax + by f ′ (ax – by) + eax + by . f(ax – by)]
p = a [eax + by f ′ (ax – by) + z]
[Q From equation (1)]
p
= eax + by . f ′ (ax – by) + z
a
... (2)
On differentiating equation (1) partially with respect to ‘y’, we get,
∂z
=z
∂y y
= eax+by. f ′ (ax – by) . (–b) + f(ax – by) . eax + by . (b)
⇒
q = b [–eax + by . f ′ (ax – by) + f(ax – by) .eax + by]
⇒
q = b [–eax + by. f ′ (ax – by) + z]
[Q From equation (1)]
q
= –eax + by f ′ (ax – by) + z
b
... (3)
Adding equation (2) and equation (3), we get,
p q
+ = eax+by f ′ (ax – by) + z – eax + by f ′ (ax – by) + z
a b
⇒
pb + qa
= 2z
ab
⇒ pb + qa = 2abz.
Q23. Form the partial differential equation by eliminating arbitrary function from f(x2 + y2, z – xy) = 0.
Ans: Given equation is,
f (x2 + y2, z – xy) = 0
... (1)
Let,
x2 + y2 = u and z – xy = v
... (2)
∴ From equations (1) and (2), we get,
f (u, v) = 0
... (3)
On differentiating equation (3) with respect to x, by using chain rule, we get,
∂f ∂u ∂f ∂u ∂z ∂f ∂v ∂f ∂v ∂z
+
+
+
=0
∂u ∂x ∂u ∂z ∂x ∂v ∂x ∂v ∂z ∂x
⇒
∂f
∂f
(2x) +
∂u
∂u
Where, p =
⇒
(0) p +
∂f
∂f
(– y) +
1. p = 0
∂v
∂v
∂z
∂x
∂f
∂f
∂f
(– y) +
(2x) +
(p) = 0
∂v
∂v
∂u
⇒ (2x)
∂f
∂f
=0
+ (p – y)
∂v
∂u
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (4)
SIA GROUP
4.12
MATHEMATICS-II [JNTU-ANANTAPUR]
On differentiating equation (3) with respect to ‘y’ by
using chain rule, we get,
∂f ∂u
∂u ∂y
⇒
On differentiating equation (1) partially with respect to
y, we get,
∂f ∂v ∂f ∂v ∂z
∂f ∂u ∂z
+
+
=0
∂v ∂y ∂v ∂z ∂y
∂u ∂z ∂y
+
∂f ⎡ ∂u ∂u ∂z ⎤ ∂f ⎡ ∂v ∂v ∂z ⎤
⎢ + . ⎥+ ⎢ + . ⎥
∂u ⎣ ∂y ∂z ∂y ⎦ ∂v ⎣ ∂y ∂z ∂y ⎦ = 0
∂f
∂f
∂f
∂f
(0) (q) +
(– x) +
1.q = 0
(2 y) +
∂v
∂v
∂u
∂u
Where,
⇒
∂z
=q
∂y
Let
⇒
∂f
∂f
∂f
(2 y) –
x +
q=0
∂v
∂v
∂u
⇒
(2 y)
Eliminating
∂f
∂f
+ (q – x)
=0
∂v
∂u
⇒
... (5)
∂f
∂f
and
from equations (4) and (5), we get,
∂v
∂u
2 x ( p − y)
=0
2 y (q − x )
∂f ⎡
∂z ⎤ ∂f ⎡
∂z ⎤
1 + 1. ⎥ + ⎢2 y + 2 z. ⎥ = 0
⎢
∂u ⎣
∂y ⎦ ∂v ⎣
∂y ⎦
∂z
=q
∂y
∂f
∂f
(1 + q ) + ( 2 y + 2 zq ) = 0
∂u
∂v
Eliminating
∂f
∂f
between equations (2) and (3),
and
∂v
∂u
we get,
(1 + p)
(1 + q)
2( x + zp )
=0
2( y + zq)
⇒
(1 + p). 2(y + zq) – 2(x + zp).(1 + q) = 0
⇒
2x (q – x) – (p – y) 2y = 0
⇒
(1 + p) (y + zq) – (x + zp).(1 + q) = 0
⇒
2 xq – 2x2 – 2yp + 2y2 = 0
⇒
y + zq + yp + zpq – x – xq – zp – zpq = 0
⇒
(y – z)p + (z – x)q = x – y
⇒
⇒
2
2
xq – x – yp + y = 0
2
2
xq – yp = x – y
∴ The required partial differential equation by eliminating
arbitrary function from f (x2 + y2, z – xy) = 0 is xq – yp = x2 – y2.
Q24. Form the PDE by eliminating ‘f’ from f(x + y + z,
x2 + y2 + z2) = 0.
Model Paper-I, Q8
Ans: Given that,
f(x + y + z, x2 + y2 + z2) = 0
Let, x + y + z = u
And x2 + y2 + z2 = v
∴
f(u, v) = 0
... (1)
On differentiating equation (1) partially with respect to
‘x’ by chain rule, we get,
∴ The partial differential equation by eliminating ‘f ’
from f(x + y + z, x2 + y2 + z2) = 0 is,
(y – z)p + (z – x)q = x – y
Q25. Form the partial differential equation by eliminating
arbitrary function φ from φ(x + y + z, x2 + y2 – z2) =
0.
Ans: Given that,
φ(x + y + z, x2 + y2 – z2) = 0
Let, x + y + z = u and
x2 + y2 – z2 = v
⇒ φ(u, v) = 0
... (1)
Differentiating equation (1) partially with respect to x by
chain rule, we get,
∂φ ⎡ ∂u ∂u ∂z ⎤ ∂φ ⎡ ∂v ∂v ∂z ⎤
+ .
+
+ .
=0
∂u ⎢⎣ ∂x ∂z ∂x ⎥⎦ ∂v ⎢⎣ ∂x ∂z ∂x ⎥⎦
∂f ⎡ ∂u ∂u ∂z ⎤ ∂f ⎡ ∂v ∂v ∂z ⎤
+ .
+
+ .
=0
∂u ⎢⎣ ∂x ∂z ∂x ⎥⎦ ∂v ⎢⎣ ∂x ∂z ∂x ⎥⎦
⇒
∂f
∂u
∂z ⎤ ∂f
⎡
⎢1 + 1. ∂x ⎥ + ∂v
⎣
⎦
⇒
∂z ⎤
⎡
⎢ 2 x + 2 z. ∂x ⎥ = 0
⎣
⎦
∂f
∂f
(1 + p ) + ( 2 x + 2 zp ) = 0
∂u
∂v
∂φ ⎡
∂z ⎤ ∂φ ⎡
∂z ⎤
1 + (1) ⎥ + ⎢2 x + (−2 z ) ⎥ = 0
∂u ⎢⎣
∂x ⎦ ∂v ⎣
∂x ⎦
∂φ
[1 + p ]+ ∂φ [2 x − 2 zp ] = 0
∂u
∂v
Where,
⇒
∂z
Let
=p
∂x
⇒
... (3)
... (2)
Look for the SIA GROU P LOGO
p=
∂z
∂x
on the TITLE COVER before you buy
... (2)
4.13
UNIT-4 (Partial Differential Equations)
Similarly differentiating equation (1) partially with respect
to y,
Similarly differentiating equation (1), partially with
respect to y, we get,
∂φ ⎡ ∂u ∂u ∂z ⎤ ∂φ ⎡ ∂v ∂v ∂z ⎤
⎢ + . ⎥+ ⎢ + . ⎥ =0
∂u ⎣ ∂y ∂z ∂y ⎦ ∂v ⎣ ∂y ∂z ∂y ⎦
⇒
∂z ⎤
∂φ ⎡
∂z ⎤ ∂φ ⎡
⎢1 + (1) ⎥ + ⎢2 y + (−2 z ) ⎥ = 0
∂y ⎦
∂u ⎣
∂y ⎦ ∂v ⎣
∂φ
[1 + q] + ∂φ [2 y − 2 zq] = 0
∂u
∂v
Where,
⇒
q =
Eliminating
... (3)
∂z
∂y
∂φ
∂φ
and
between equations (2) and (3),
∂u
∂v
⇒
∂φ ⎡ ∂u ∂u ∂z ⎤ ∂φ ⎡ ∂v ∂v ∂z ⎤
⎢ + ⋅ ⎥ =0
⎢ + ⋅ ⎥+
∂u ⎣ ∂y ∂z ∂y ⎦ ∂v ⎣ ∂y ∂z ∂y ⎦
⇒
∂φ ⎡ 1
∂z ⎤ ∂φ ⎡
∂z ⎤
+ 0 ⋅ ⎥ + ⎢2 y + 2 z ⋅ ⎥ = 0
⎢
∂u ⎣ x
∂x ⎦ ∂v ⎣
∂y ⎦
⇒
∂φ ⎡ 1 ⎤ ∂φ
+ [2 y + 2 zq ] = 0
∂u ⎢⎣ x ⎥⎦ ∂v
Where, q =
∂z
∂y
Eliminating
∂φ
∂φ
from equations (2) and (3), we get,
and
∂v
∂u
−y
we get,
(1 + p) 2( x − zp)
=0
(1 + q) 2( y − zq)
⇒
⇒ 2(1 + p) (y – zq) – 2(x – zp) (1 + q) = 0
⇒ (1 + p) (y – zq) – (1 + q) (x – zp) = 0
⇒ y – zq + py – zpq – [x – zp + xq – zpq] = 0
⇒ y – zq + py – zpq – x + zp – xq + zpq = 0
⇒ y – x + p(y + z) – q(x + z) = 0
⇒ p(y + z) – q(x + z) = x – y
This is the required partial differential equation.
Q26. Form the partial differential equation by eliminating
⎛y 2
2
2⎞
φ from φ ⎜ , x + y + z ⎟ = 0.
⎝x
⎠
⎛y
⎞
φ ⎜ , x2 + y2 + z 2 ⎟ = 0
x
⎝
⎠
y
= u and
x
x2 + y2 + z2 = v
⇒ φ(u, v) = 0
... (1)
On differentiating equation (1), partially with respect to
‘x’ by chain rule, we get,
Let,
∂φ ⎡ ∂u ∂u ∂z ⎤ ∂φ ⎡ ∂v ∂v ∂z ⎤
+
+
+
=0
∂u ⎢⎣ ∂x ∂z ∂x ⎥⎦ ∂v ⎢⎣ ∂x ∂z ∂x ⎥⎦
⇒
∂φ ⎡ − y
∂z ⎤ ∂φ ⎡
∂z ⎤
+ 0⋅ ⎥ +
2x + 2 z ⎥ = 0
∂u ⎢⎣ x 2
∂x ⎦ ∂v ⎢⎣
∂x ⎦
⇒
∂φ ⎡ − y ⎤ ∂φ
+ [2 x + 2 zp ] = 0
∂u ⎢⎣ x 2 ⎥⎦ ∂v
Where, p =
∂z
∂x
... (2)
2( x + zp )
x2
=0
1
2( y + zq)
x
2
−2 y
( y + zq ) − ( x + zp ) = 0
2
x
x
−y
1
( x + zp ) = 0
x
x
⇒ – y(y + zq) – x(x + zp) = 0
⇒ – y2 – yzq – x2 – xzp = 0
⇒ x2 + y2 + xzp + yzq = 0
This is the required partial differential equation.
⇒
4.2
Ans: Given that,
... (3)
2
( y + zq ) −
METHOD OF SEPARATION OF VARIABLES
Q27. Explain the stepwise procedure to solve second
order equations using method of separation of
variables.
Ans:
Method of Separation of Variables
Step 1: For the given second order equations, Assume
z = X(x), Y(y)
... (1)
Where,
X(x) denotes a function of x
Y(y) denotes a function of y.
Step 2: Partially differentiate equation (1) with respect to x and
y, substitute in the given equation and write it in the form of
f(X, X', X'' ....) = g(Y, Y', Y''...)
... (2)
Which is separable in X and Y.
Step 3: The LHS of equation (2) is a function of only x and RHS
is a function of only y and both are equal for all values of x and
y. Therefore equating LHS and RHS to same constant (say λ),
which is called separation constant.
∴ f(X, X', X''....) = K
... (3)
g(Y, Y', Y''...) = K
... (4)
Step 4: Solve for X using equation (3)
Step 5: Similarly solve for Y using equation (4).
Step 6: Substitute the corresponding values of X and Y in
equation (1), therefore the required solution is obtained.
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.14
MATHEMATICS-II [JNTU-ANANTAPUR]
Q28. Solve by the method of separation of variables.
ux = 2ut + u where u(x, 0) = 6e–3x
Ans: Given that,
∂u
∂u
= 2 +u
∂x
∂t
Q29. Solve by separation of variables 3ux + 2uy = 0
Model Paper-II, Q8
with u(x, 0) = 4e– x.
Ans: Given that,
3ux + 2uy = 0 with u(x, 0) = 4e– x
Let, u(x, t) = X(x)T(t)
∂u
∂u
=0
+2
∂y
∂x
... (1)
Let, u(x, y) = X(x) Y(y) = XY
... (2)
⇒
... (1)
... (2)
3
∂u
∂u
= XY '
= X 'Y and
∂x
∂y
= XT be a solution of equation (1), then,
∂u
∂u
= X'T and
= XT'
∂x
∂t
On substituting above values in equation (1), we get,
3X 'Y + 2 XY ' = 0
Dividing by XY, we get,
... (3)
On substituting equations (2) and (3) in equation (1), we
get,
X'T = 2XT' + XT ⇒
⎛ X '⎞ ⎛ Y'⎞
3 ⎜ ⎟ + 2⎜ ⎟ = 0
⎝ X ⎠ ⎝Y ⎠
X ′ 2T ′
=
+1
X
T
On equating each side to an arbitrary constant K.
⎛ X'⎞
Y'
3 ⎜ ⎟ = − 2⎛⎜ ⎞⎟ = K
⎝X ⎠
⎝Y ⎠
Equating each side to some arbitrary constant k.
⇒
X ′ 2T ′
=
+1 = k
X
T
⇒
X′
= k ⇒ X' − kX = 0
X
⇒ X = c1e kx and
⇒
T = c2
⎛Y'⎞
− 2⎜ ⎟ = K
⎝Y ⎠
⇒ − 2Y ' = KY
⇒ 3 X ' = KX
⇒
⎛ k −1⎞
2T ′
⎟T = 0
+1 = k ⇒ T ′ = ⎜
T
⎝ 2 ⎠
X' =
KX
3
⇒
On substituting the values of X and T in equation (2), we
get,
u(x, t) = (c1 ekx).c2 e
= c1c2ekx e
⎛ KX ⎞
⎟
⎜
3 ⎠
− KY
2 ⎛ KY ⎞
Y = C2 e
X = C1 e ⎝
−⎜
⎟
⎝ 2 ⎠
⎛ KY ⎞ ⎞
⎛ KX ⎞ ⎞ ⎛
⎛
−⎜
⎟
⎟
⎜
⎜
3 ⎠⎟ ⎜
⎝ 2 ⎠⎟
⎝
C
e
u(x, y) = ⎜ C1 e
2
⎟
⎟⎜
⎟
⎟⎜
⎜
⎠
⎠⎝
⎝
⎛ k −1 ⎞
⎟t
⎜
⎝ 2 ⎠
= C1 C 2
⎛ k −1 ⎞
⎟t
⎜
⎝ 2 ⎠
= B e⎝
∴ u(x, y) = B
⎛ KX ⎞
⎜
⎟
e⎝ 3 ⎠
e
⎛ KY ⎞
−⎜
⎟
e ⎝ 2 ⎠
⎛ KY ⎞
−⎜
⎟
⎝ 2 ⎠
⎛ KY ⎞
−⎜
⎟
e ⎝ 2 ⎠
KX
Be 3
⇒ u(x, 0) =
But given,
u(x, 0) = 4e–X
From equations (4) and (5), we get,
u(x, 0) = Aekx
6e–3x = Aekx
A = 6 and k = –3.
Look for the SIA GROU P LOGO
... (4)
... (5)
KX
K = –3.
u(x, t) = 6e–3x e–2t or
Which is the particular solution of equation (1).
... (3)
4e–X = Be 3
Comparing the terms on both sides, we get, B = 4 and
Since u(x, 0) = 6 e–3x
= 6e– (3x+2t)
[ Q B = C1 C2 ]
Taking, y = 0 in equation (3), we get,
Taking, t = 0 in equation (4),
On comparing the terms both sides, we get,
⎛ KX ⎞
⎟
⎜
e⎝ 3 ⎠
⎛ KX ⎞
⎟
⎜
3 ⎠
⎛ k −1 ⎞
⎟t
⎜
⎝ 2 ⎠
Where, A = c1c2 ... (4)
u(x, t)
Y' =
Substituting these X and Y values in equation (2), we get,
⎛ k −1 ⎞
⎜
⎟t
e⎝ 2 ⎠
= A ekx e
⎛ X'⎞
i.e., 3⎜ ⎟ = K
⎝X ⎠
∴ u(x, y) = 4 e − X e
3Y
2
Which is the required particular solution.
on the TITLE COVER before you buy
4.15
UNIT-4 (Partial Differential Equations)
Integrating equation (9) with respect to ‘t’, we get,
∂ 2u
Q30. Solve
= e − t cosx , given that u = 0 when
∂x∂t
T =
∂u
= 0 when x = 0. Show also that as
∂t
t tends to ∞, u tends to sinx.
1 −t
(e ) (−1) + a2
C
t = 0 and
Ans: To solve
(i)
(ii)
∂ 2u
= e–t cosx
∂x.∂t
Assume that, u(x, t) = e–t cosx
Given conditions are,
u = 0 when t = 0
⇒ u(x, 0) = 0
T=
Substituting equations (10) and (11) in equation (5),
we get,
⎛ − e −t
⎞
+ a2 ⎟
u = (C sin x + a1) ⎜⎜
⎟
⎝ C
⎠
... (3)
∂u
= 0 when x = 0
∂t
From equation (3),
∂
u ( 0, t ) = 0
∂t
X'=
⎛ − e0
⎞
+ a2 ⎟
u(x, 0) = (C sin x + a1) ⎜⎜
⎟
⎝ C
⎠
... (4)
Let the solution of equation (2) be,
u(x, t) = X(x).T(t)
Here,
X is a function of only ‘x’ and
T is a function of only ‘t’.
∴
... (5)
⎛ −1
⎞
+ a2 ⎟
0 = (C sin x + a1) ⎜
⎝C
⎠
⇒
C sin x + a1 ≠ 0,
∂X
∂T
and T ' =
∂x
∂t
∴ a2 =
2
∂ ( XT )
∂X ∂T
=
∂x.∂t
∂x. ∂t
But, XT = u from equation (5),
⇒ X 'T '=
∂2 u
∴ X 'T ' =
∂x. ∂t
... (6)
X'
e −t
= C and
=C
cos x
T'
[Q From equations (8), (9), (10) and (11)]
From equation (4),
⎛ e −t
∂
u ( 0, t ) = (C sin 0 + a1) ⎜
⎜ C
∂t
⎝
⇒
⎛ e −t
0 = (0 + a1) ⎜⎜
⎝ C
⇒
0 = a1
... (9)
Integrating equation (8) with respect to ‘x’, we get,
X = C sin x + a1
⎞
⎛ −t
⎞
⎟ + (C cos x )⎜ − e + a ⎟
2⎟
⎟
⎜ C
⎠
⎝
⎠
⎞
⎛ −t
⎞
⎟ + (C cos 0)⎜ − e + a ⎟
2⎟
⎟
⎜ C
⎠
⎝
⎠
⎞
⎛ −t
⎞
⎟ + C (1)⎜ − e + a ⎟
2⎟
⎟
⎜ C
⎠
⎝
⎠
... (8)
−t
e
C
... (12)
⎛ e −t
= (C sin x + a1 )⎜⎜
⎝ C
e
X'
... (7)
=
cos x
T'
Since x and t are independent variables, equation (7) is
true only if both L.H.S and R.H.S are equal to the same arbitrary
constant C.
T '=
1
C
∂u
= XT' + X'T
∂t
−t
X '= C cos x
−1
+ a2 = 0
C
Since,
On substituting equation (6) in equation (1), we get,
X ' T ' = e–t cosx
On separating the variables ‘x’ and ‘t’, we get,
∴
... (11)
Here, a1 and a2 are constants.
... (1)
... (2)
− e −t
+ a2
C
⇒
... (10)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
a1
⎛ e −t
⎞
e −t
+ C⎜ −
+ a2 ⎟
⎜ C
⎟
C
⎝
⎠
⎛ − e −t
⎞
e −t
+ a2 ⎟
= − C⎜⎜
⎟
C
⎝ C
⎠
SIA GROUP
4.16
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting these values in given equation, we get,
X' Y = 4XY'
e −t
a1
= e–t – Ca2
C
⇒
⇒
a1 e–t = C(e–t – Ca2)
⇒
⎛
⎛ 1 ⎞⎞
a1 e–t = C ⎜⎜ e −t − C ⎜ ⎟ ⎟⎟
⎝ C ⎠⎠
⎝
Since x and y are independent variables.
∴ Each side is equal to the constant, say K
[Q From equation (12)]
⇒
a1
a1 e–t = C(e–t – 1)
X′
=K
X
∴
⎛1
⎞
= C ⎜ t −1⎟
et
⎝e
⎠
And
⎛1− e
= C ⎜⎜ t
e
⎝ e
⇒
1
a1
t
t
4Y ′
=K
Y
log X = Kx + log C1
⎞
⎟
⎟
⎠
∴
t
∴ a1 = C (1 − e )
∴
X ′ 4Y ′
=
X
Y
⇒
... (13)
The required solution is,
u = X.T
⎛ − e −t
⎞
+ a2 ⎟
= (C sinx + a1) ⎜⎜
⎟
⎝ C
⎠
Where X is a function of x and that of y
∂u
∂u
= X' Y and
then
= XY'
∂x
∂y
Look for the SIA GROU P LOGO
Y = 4 C 2 eKy/4
The unique solution of the given equation is,
⇒ u(0, y) = Ce0 + Ky/4
[Q C1 4 C 2 = C]
⇒ 8e–3y = Ce Ky/4
∴
C = 8,
K
= –3 ⇒ K = –12
4
i.e., u = 8 e– 3y
Q32. Solve by the method of separation of variables
∂2z
∂z ∂z
−2
+
= 0.
∂x ∂y
∂x 2
Ans: Here z is a function of X and Y
Let, z = XY
Where, X is a function of x and that of y
∂z
∂2z
∂z
= X' Y,
= XY'
2 =X" Y,
∂y
∂x
∂x
On substituting this value of z in the given equation,
∂u
∂u
where u(0, y) = 8e–3y by the
Q31. Solve
=4
∂x
∂y
method of separation of variables.
Ans: Here u is a function of x and y
Let, u = XY
Y 4 = C2eKy
⇒ u(x, y) = C1 4 C 2 eKx + Ky/4
⎛ − e −t + 1 ⎞
t
⎟
= C[sin x + (1 − e )] ⎜⎜
⎟
C
⎝
⎠
Hence proved.
⇒
u = C1eKx · 4 C 2 eKy/4
[Q From equations (12) and (13)]
When, t → ∞
u(x, ∞) = sinx(1 – 0)
u(x, ∞) = sinx
⇒
Y4
= eKy
C2
∴
⎛ − e −t 1 ⎞
t
+ ⎟
= (C sin x + C (1 − e ))⎜⎜
C
C ⎟⎠
⎝
u(x,t) = [sinx(1 – e–t)] + [(1 – et) (1 – e–t)]
= sinx(1 – e–t)
[Q (1 – et) (1 – e–t) = 0]
X = C1eKx
4 log Y = Ky + logC2
∴
[ Q From equations (10) and (11)]
u( x, t ) = [sin x + (1 − e t )] (1 − e −t )
X
= eKx
C1
Then
we get,
⇒
X" Y – 2X' Y + XY' = 0
(X" – 2X') Y = – XY'
X ′′ − 2 X ′
Y′
=–
X
Y
Since x and y are independent variables.
⇒
on the TITLE COVER before you buy
... (1)
4.17
UNIT-4 (Partial Differential Equations)
∴ Equation (1) can be true if each side is equal to the same
constant (say a),
Since X and T are independent variable we can equate to
it by constant, we get,
∴
X ′′ − 2 X ′ −Y ′
=
=a
X
Y
1 T ′′
X ′′
= 2
=K
X
a T
⇒
X ′′ − 2 X ′
=a
X
∂X
X ′′
= K i.e., X' =
∂x
X
X" – 2X' – aX = 0
⇒
On solving the ordinary linear equation, we get,
⎡
(D2 – 2D – a) X = 0 ⎢Q D =
⎣
Case (i): If K is positive constant, say p2
d⎤
dx ⎥⎦
(D2 – p2)X = 0
A. E D2 – p2 = 0 ⇒ D = ± p
Auxiliary roots are 0, 1 ± 1 + a
∴
The solution is, X = C1 e
And the solution of
⇒
(1+ (1+ a ) x )
(D2 – K)X = 0
+ C2 e
∴
(1− (1+a ) x )
C.I X = C1 e–px + C2epx
Case (ii): If K is negative say ‘–p2’, we get,
−Y ′
=a
Y
(D2 + p2)X = 0
A. E D2 + p2 = 0
– log Y = ay + log C
D = ± ip
⎛Y ⎞
⇒ log ⎜ ⎟ = – ay
⎝C ⎠
∴
C.I X = C3 cos px + C4 sin px
Case (iii): If K is zero, then,
∴
⇒
Y
= e–ay
C
∴
Y = Ce–ay
(D2 – 0)X = 0
A. E D2 = 0
D = 0, 0
The required solution,
z = XY
[
z = C1e (1+
(1+ a ) x )
+ C2 e (1−
(1+ a ) x )
]Ce
∴
C.I X = C5 + C6x and
− ay
∂ 2u
∂ 2u
= a 2 2 by separation of variable
Q33. Solve
2
∂x
∂t
method.
2
( D 1 – a2K)T = 0
i.e., T' =
T ′′
=K
a 2T
dT
dt
Case (i): If K is positive say ‘–p2’, we get,
Ans: The given partial differentiation,
2
( D 1 + a2p2)T = 0
∂ u
∂ u
2
2 =a
∂t
∂x 2
2
2
... (1)
2
A.E D 1 + a2 p2 = 0
The solution of the form u = XT where X is a function
of x and that of T.
∂ u
∂ u
= X"T
2 = XT" and
∂t
∂x 2
2
Then
2
D' = ± ap
∴
Case (ii): If K is negative say ‘–p2’, we get,
Equation (1) becomes,
XT" = a2X"T
⇒
C.I T = C7 e–apt + C8 eapt
X ′′
T ′′
=
2
X
a T
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
2
( D 1 + a2p2)T = 0
2
A. E D 1 + a2 p2 = 0
D1 = ± ap
∴
C.I T = C9 cos (apt) + C10 sin (apt)
SIA GROUP
4.18
MATHEMATICS-II [JNTU-ANANTAPUR]
Case (iii): If K is zero,
2
( D 1 – 0)T = 0
A. E D1 = 0, 0
∴
C.I T = C11 + C12 t
The possible solution of a differential equation is u = XT
u = (C1e–px + C2epx)(C7e–apt + C8eapt)
⇒
u = (C3 cos px + C4 sin px)(C9 cos (apt) + C10 sin (apt))
And u = (C5 + C6 x)(C11 + C12t)
4.3 SOLUTIONS OF ONE-DIMENSIONAL WAVE EQUATION, HEAT EQUATION
Q34. Write complete solutions of wave equation.
Ans: One of the engineering applications of partial differential equations is in wave equations.
The one-dimensional wave equation is,
2
∂2 y
2 ∂ y
c
=
∂t 2
∂x 2
Vibration of a Stretched String (Derivation)
Consider a string (elastic) of length ‘l’, fixed lightly between points O and A and subjected to constant tension T. Neglecting
the effect of gravity. The string at rest is allowed to vibrate with no external force acting on it, assume that each point on the string
OA makes small vibrations at right angles to equilibrium position of string OA, entirely in one plane.
Let end ‘O’ be origin,
OA is along X-axis
OY ⊥lar to X-axis is Y-axis
So that the motion is entirely one-dimensional in XY-plane.
Figure (a) shows the string is at rest.
y
y
T
Q
δy
O
A
x
Figure (a): String is at Rest
T
θ
x
O
P
δ + δθ
y
δx
A
x
Figure (b): String in Vibration at Time ‘t’
Figure (b) shows the string making vibrations perpendicular to X-axis.
At t the string is at position OPA. Consider the motion of two points say P and Q on the string.
P(x, y) and Q(x + δx, y + δy), where the tangents make angle θ and (θ + δθ) with X-axis.
Acceleration of element moving upward =
∂2 y
∂t 2
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
4.19
UNIT-4 (Partial Differential Equations)
Vertical component of force action on this element (PQ)
= T sin(θ + δθ) – Tsinθ
= T [sin(θ + δθ) – Tsinθ]
[Qθ is small]
= T [tan(θ + δθ) – tanθ]
dy ⎤
⎡
⎢Q tan θ = dx ⎥
⎣
⎦
⎡⎧ ∂y ⎫
⎧ ∂y ⎫ ⎤
−⎨ ⎬ ⎥
⎬
⎣⎩ ∂x ⎭ x +δx ⎩ ∂x ⎭ x ⎦
= T ⎢⎨
If m is the mass per unit length of string, then by Newton’s second law of motion.
Force = Mass × Acceleration
∂2 y
=T
m.δx.
∂t 2
⎡⎧ ∂y ⎫
⎧ ∂y ⎫ ⎤
−⎨ ⎬ ⎥
⎢⎨ ⎬
⎣⎩ ∂x ⎭ x +δx ⎩ ∂x ⎭ x ⎦
⎡ ⎧ ∂y ⎫
⎧ ∂y ⎫ ⎤
−⎨ ⎬ ⎥
⎨ ⎬
⎢
∂ y T ⎢ ⎩ ∂x ⎭ x + δx ⎩ ∂x ⎭ x ⎥
=
⎥
δx
∂t 2 m ⎢
⎥
⎢
⎦
⎣
2
Taking limit as Q → P
δx → O
Applying limits we get,
∂2 y T ∂2 y
=
m ∂x 2
∂t 2
f ( x + h ) − f ( x) d
⎡
⎤
=
f ( x) ⎥
⎢Q hlt
→
0
h
dx
⎣
⎦
Let, c2 =
T
m
2
∂2 y
2 ∂ y
=
c
∂t 2
∂x 2
The above partial differential equation gives the transverse vibration of string, which is known as one-dimensional wave
equation.
Q35. If a string of length ‘l ’ is initially at rest in equilibrium position and each of its points is given the
πx
, find the displacement y(x, t).
velocity v0 sin3
l
Ans: Wave equation is,
2
∂2 y
2 ∂ y
a
=
∂x 2
∂t 2
... (1)
Boundary conditions are,
(i) y(0, t) = 0
(ii) y(l, t) = 0
(iii)
∂y
(x, 0) = 0
∂t
(iv)
∂y
⎛ πx ⎞
(x, 0) = v0 sin3 ⎜
⎟
∂t
⎝ l ⎠
y(x, t) = (c1 cos px + c2 sin px) (c3 cos pat + c4 sin pat) which satisfies the boundary conditions .
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
... (2)
4.20
MATHEMATICS-II [JNTU-ANANTAPUR]
By applying condition (i) in equation (2), we get,
y(0, t) = c1(c3 cos pat + c4 sin pat)
Where, c3 cos pat + c4 sin pat ≠ 0
∴c1 = 0
On substituting in equation (2), we get,
y(x, t) = c2 sin px(c3 cos pat + c4 sin pat)
... (3)
By applying condition (ii) in equation (3), we get,
y(l, t) = c2 sin pl(c3 cos pat + c4 sin pat)
0 = c2 sin pl(c3 cos pat + c4 sin pat)
When c2 ≠ 0 and c3 cos pat + c4 sin pat ≠ 0
∴
sin pl = nπ
⇒
pl = nπ
p=
nπ
l
On substituting in equation (3), we get,
y(x, t) = c2 sin
nπat ⎞
nπat
nπx ⎛
+ c4 sin
⎟
⎜ c3 cos
l ⎠
l
l ⎝
... (4)
On differentiating equation (4) with respect to ‘t’, we get,
nπx ⎛
nπa
nπat ⎞
nπa
nπat
∂y
+ c4
sin
cos
⎟
⎜ − c3
( x, t ) = c2 sin
l ⎝
l
l ⎠
l
l
∂t
By applying condition (iii), we get,
∂y
nπx ⎛⎜ c nπa ⎞⎟
(x,0) = c2 sin
4
l ⎠
∂t
l ⎝
0 = c2 sin
nπx
(c4 nπa)
l
nπx
⎡
⎤
⎢Q c2 ≠ 0 and sin l ≠ 0⎥
⎣
⎦
∴c4 = 0
On substituting above value in equation (4), we get,
∴
y(x, t) = c2 sin
nπx
nπ at
c3 cos
l
l
y(x, t) = cn sin
nπx
nπ at
cos
l
l
[Put c2 c3 = cn]
General solution is,
∞
y(x, t) =
∑c
n =1
n
sin
nπx
nπat
cos
l
l
Look for the SIA GROU P LOGO
... (5)
on the TITLE COVER before you buy
4.21
UNIT-4 (Partial Differential Equations)
Given,
1
πx
and sin3 x = (3 sin x − sin 3x )
4
l
y(x, 0) = v0 sin3
sin 3 πx 1
=
(3 sin πx – sin 3πx)
4l
l
... (6)
From equations (5) and (6),
∞
∑c
n
sin
n =1
nπx v0
(3 sin πx – sin 3πx)
=
4l
l
v0
sin πx
sin 2πx
sin 3πx
(3 sin πx − sin 3πx )
+ c2
+ c3
+ .... =
4l
l
l
l
c1
On equating like coefficients, we get,
c1 =
−v0
3v0
, c = 0, c5 = 0
, c = 0, c3=
4l 2
4l 4
On substituting these values in equation (6), we get,
y(x, t) =
3v0
πx
πat v0
3πx
3πat
sin cos
− sin
cos
l
l
l
l
4
4
Q35. A tightly stretched string with fixed end points x = 0 and x = 1 is initially at rest in its equilibrium
position. If it is set on vibrating by giving each of its points a velocity λ x(1–x), find the displacement of
the string at any distance x from end at any time ‘t’.
Model Paper-II, Q9
Ans: Given boundary conditions are,
y(0, t) = y(l, t) = 0
Since,
∞
y(x, t) =
∑ ⎛⎜⎝ a
n
cos
n =1
nπct ⎞
nπx
nπct
+ bn sin
⎟ sin
l ⎠
l
l
… (1)
As the string is at rest initially,
y(x, 0) = 0
∴ From equation (1),
∞
0=
∑a
n
nπx
l
sin
n =1
⇒
an = 0
∞
∴y(x, t) =
∑b
n
sin ⎛⎜ nπct ⎞⎟ sin ⎛⎜ nπx ⎞⎟
⎝ l
n =1
⎠
⎝ l ⎠
… (2)
∞
And
nπc
∂y
bn cos nπct sin nπx
=
∂t n =1 l
l
l
∑
=
πc
l
∞
∑ nb
n
n =1
sin
nπx
l
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.22
MATHEMATICS-II [JNTU-ANANTAPUR]
δy
= λx (l – x) when t = 0
δt
But,
∞
πc
λx (l – x) =
l
∴
2
πc
nbn =
l
l
⇒
∑ nb
n
sin
n =1
nπx
l
1
∫ n(1 − n)
0
1
2λ
πc
.nbn =
l
l
=
⎡
⎛ l3
⎛ − l2
nπx ⎞
nπx ⎞⎤
nπx ⎞
⎛ −l
⎜ 3 3 cos
⎟⎥
⎜
⎟
+
−
(
)
cos
(
2
)
sin
(
2
)
−
l
x
x
l
x
−
−
⎜
⎟
⎢
2 2
⎜
⎜
⎟
l ⎠
l ⎟⎠⎦⎥
l ⎠
⎝ nπ
⎝n π
⎝n π
⎣⎢
0
4λ l 2
4λ l 2
(1
–
cos
n
π
)
=
[1 – (– 1)n]
n 3π 3
n 3π 3
= 0, When n is even
8λ l 2
= 3 3 , When n is odd
n π
8λl 2
, taking n = 2m – 1
π3 (2m − 1) 3
i.e.,
⇒
bn =
8λl 3
cπ ( 2m − 1) 4
4
From equation (2), the required solution is,
8λl 3
y(x, t) =
cπ 4
∞
∑ (2m − 1)
1
m =1
4
(2m − 1)πx
⎛ (2m − 1)πct ⎞
sin ⎜
⎟ sin
⎝
⎠
l
l
Q37. Find the deflection u(x, y, t) of the square membrane with a = b = 1 and c = 1, if the initial velocity is
zero and the initial deflection is f(x, y) = A sin π x sin 2π y.
Model Paper-I, Q9
Ans: Given that,
a = b = 1 and f (x, y) = A sin πx sin 2πy
On substituting in generalised Euler’s formula,
1 1
Amn = 4
∫ ∫ A sin πx sin 2πy sin mπx sin nπy dy dx
0 0
1
⎡1
∫
∫
⎢⎣
= 4A sin πx sin mπx dx ⎢
0
⎛1⎞
= 4A ⎜ ⎟
⎝ 2⎠
0
⎤
sin 2πy sin nπy dy ⎥ = 0
⎥⎦
1
∫
sin 2πy sin nπy dy, for m = 1
0
Look for the SIA GROU P LOGO
for m ≠ 1
⎡ 1
1⎤
⎢Q sin πx sin πx dx = ⎥
2 ⎦⎥
⎣⎢ 0
∫
on the TITLE COVER before you buy
4.23
UNIT-4 (Partial Differential Equations)
1
= 2A
∫
sin 2πy sin nπy dy = 0, for n ≠ 2
0
⎛1⎞
⎝ 2⎠
= 2A × ⎜ ⎟ (for n = 2)
∴ A12 = A
... (1)
2
Also, pmn = πc
⎛ m⎞ ⎛ n⎞
⎜ ⎟ +⎜ ⎟
⎝ a ⎠ ⎝b⎠
2
= π × 1 (1) 2 + ( 2) 2 = 5 π
... (2)
On substituting equations (1) and (2) in u(x, y, t) =
ab
A (A cos pt + Bmnsin pt)
4 mn mn
∴ Required solution is u(x, y, t) = A sin πx sin 2πy cos ( 5 πt)
Q38. The points of trisection of a string are pulled aside through the same distance on opposite sides of the
position of equilibrium and the string is released from rest. Derive an expression for the displacement
of the string at subsequent time and show that the midpoint of the string always remains at rest.
Ans: Let B and C be the points of trisection of the string OA of length l, say initially the string is hold in the form OB' C 'A.
Where,
BB' = CC' = a (say)
The equation of OB' is, y =
3a
x
l
The equation of B' C' is, y – a =
l⎞
− 2a ⎛
⎜x − ⎟
l ⎝
3⎠
3
y
0
⎞
⎛l
B′⎜ , a ⎟
⎝3 ⎠
x
C
B
A'(1, 0)
⎞
⎛ 2l
C′⎜ , − a ⎟
⎠
⎝3
Figure
y–a =
− 6a ⎛
l⎞
⎜x− ⎟
l ⎝
3⎠
⎛ 6⎛
l ⎞⎞
y = a⎜⎜1 − ⎜ x − ⎟ ⎟⎟
3 ⎠⎠
⎝ l⎝
⇒
6x ⎞
⎛ 6x
⎛
⎞
+ 2 ⎟ = a⎜ 3 − ⎟
= a⎜1 −
l
l ⎠
⎝
⎝
⎠
⇒
y=
3a
(l – 2x)
l
The equation of C'A is, y =
3a
(x – l)
l
Here the boundary conditions are,
y(0, t) = y (l, t) = 0
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.24
MATHEMATICS-II [JNTU-ANANTAPUR]
And the initial conditions are
⎧ 3a
l
0≤ x≤
⎪ l x
3
⎪⎪
l
a
3
2l
y(x, 0) = ⎨ (l − 2 x ),
≤x≤
3
3
⎪ l
2l
⎪ 3a
≤ x≤l
⎪⎩ l ( x − l ),
3
and
∂y
= 0 when t = 0
∂t
Therefore, wave equation is,
∞
∑a
y(x, t) =
n =1
n
⎛ nπct ⎞ ⎛ nπx ⎞
⎟
⎟ sin⎜
cos⎜
⎝ l ⎠ ⎝ l ⎠
Where,
2 ⎡ 3a
⎛ nπx ⎞
⎛ n πx ⎞
⎟ dx= ⎢ l x sin⎜ l ⎟dx +
y(x,0) sin⎜
⎠
⎝
l ⎢⎣ 0
⎝ l ⎠
0
l/3
l
2
an =
l
∫
∫
2l / 3
∫
l /3
nπx
3a
dx +
(l − 2 x ) sin
l
l
nπx ⎤
3a
dx ⎥
( x − l ) sin
l
l
⎥
l /3
⎦
l
∫
2l
l
3
⎡
3
⎡
2
2
⎞
⎛
⎞
⎛
⎛
−
π
⎞
π
l
n
x
l
n
x
⎞
⎛
−
π
l
n
x
l
n
x
−
π
6a ⎢ x⎜ cos
⎢
⎟
⎜
⎟
⎜
⎟ − (−2) 2 2 sin
⎟ −1
sin
(l − 2 x)⎜ cos
= 2 ⎢ ⎝ nπ
⎝ nπ
l ⎠ ⎝ n2π 2
l ⎠ +⎢
l ⎠
l ⎠l
⎝n π
l ⎣
0
⎣
3
l
⎛−l
nπx ⎞ ⎛ − l 2
nπx ⎞
⎟
⎟ − 1⎜ 2 2 sin
cos
+ ( x − l )⎜
⎝ nπ
l ⎠ ⎝n π
l ⎠ 2l
=
6a
l2
3
⎡⎛ − l 2
2 nπ 2l 2
2 nπ l 2
2l 2
nπ
l2
nπ ⎞ ⎛ l 2
nπ
nπ ⎞
⎟+⎜
⎟
+ 2 2 sin
cos
cos
sin
sin
cos
−
+
+
⎢⎜⎜
2 2
2 2
⎟
⎜
3 n π
3 ⎠ ⎝ 3nπ
3 ⎟⎠
3
3
3nπ
3 n π
n π
⎣⎢⎝ 3nπ
⎛ l2
l2
2 nπ
2 nπ ⎞ ⎤ 6a 3l 2 ⎛
nπ
nπ
2nπ ⎞ 18a
⎟⎥ =
cos
+ 2 2 sin
−⎜
(1 + (−1) n )
⎟ = 2 2 sin
⎜sin
−
sin
2
2
2
n
3
3
3
π
n π
⎠ ⎦⎥ l n π ⎝
⎝
3
3 ⎠ n π
3
⎛
2nπ
nπ
nπ
⎛ 3nπ − nπ ⎞
nπ ⎞
nπ
⎟ = sin nπ cos
⎜since, sin
⎟
= – (–1)3 sin
= sin ⎜
– cos nπ sin
= 0 – (–1)n sin
3
⎝
3
3 ⎠
⎝
⎠
3
3
3
an = 0, when n is odd
=
36a
nπ
sin
, when n is even
n 2π 2
3
9a
Hence, y(x,t) = 2
π
Also,
⎛ l ⎞ 9a
y⎜ ,t⎟ = 2
⎝2 ⎠ π
∞
∑m
1
m =1
∞
∑m
m =1
2
1
2
i.e.,
9a
2mπ
sin
Taking, n = 2m
m 2π 2
3
sin
2mπct
2mπx
2mπ
cos
sin
l
l
3
sin
2mπct
2mπ
cos
sin mπ
l
3
= 0 since sin m π = 0
The displacement of the midpoint of the string is zero for all the values of t.
Thus the midpoint of the string is always at rest.
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
4.25
UNIT-4 (Partial Differential Equations)
Q39. A tightly stretched unit square membrane starts vibrating from rest and its initial displacement is
k sin 2π x sin π y. Show that the deflection at any instant is k sin 2π x sin π y cos( 5 π ct).
Ans: Given that,
Taking a = b = 1, f(x, y) = k sin 2πx sinδπ y.
1 1
∴
Amn = 4
∫ ∫ k sin 2πx sin πy sin mπx sin nπy dy dx
0 0
⎡1
1
=4k
∫
sin 2πx sin mπx dx ⎢
∫
⎢⎣ 0
0
⎤
sin πy sin nπy dy ⎥ = 0; for n ≠ 1
⎥⎦
1
⎛1⎞
= 4k ⎜ ⎟
⎝ 2⎠
∫
1
sin 2πx sin mπx dx, for n = 1
0
∫
Q sin πx sin πx dx =
0
1
2
1
∫
= 2k sin 2πx sin mπy dx = 0, for m ≠ 2
0
⎛1⎞
⎝2⎠
= 2k × ⎜ ⎟ ; for m = 2
∴ A12 = k
∴
and pmn = πc
m2 + n2
= πc 2 2 + 12 = πc 5
The required solution is obtained by, substituting the values of A12 and P12 in,
∞
u=
∞
⎛ mπx ⎞
⎛ nπy ⎞
⎟ sin ⎜
⎟ [Amn cos pt]
a ⎠
⎝ b ⎠
∑ ∑ sin ⎝⎜
m =1 n =1
∞
∴ f (x, y)
∞
∑∑ A
mn
m =1 n =1
⎛ mπx ⎞ ⎛ nπy ⎞
sin ⎜
⎟ sin ⎜
⎟
⎝ a ⎠ ⎝ b ⎠
u(x, y, t) = Amn f (x, y)cos pt
[Q f (x, y) = Initial deflection]
u(x, y, t) = k sin 2πx sin πy cos πc 5 t
Q40. The ends A and B of a rod 20 cm long have the temperature at 30°C and 80°C until study states prevail.
The temperatures of the ends are changed at 40°C and 60°C respectively. Find the temperature
distribution in the rod at time t.
Ans: Let
∂ 2u
∂u
= α2 2
∂t
∂x
... (1)
The boundary conditions are,
u(0, t) = 30°C
and u(l, t) = 80°C
Where l is the length of the rod. Under steady state conditions
... (2)
... (3)
∂u
=0
∂t
From equation (1),
∂ 2u
= 0 ⇒ u = ax+b
∂x 2
... (4)
and from equation (2), we get,
b = 30
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.26
MATHEMATICS-II [JNTU-ANANTAPUR]
ut(x, 0) = u(x, 0) – us(x)
On substituting b value in equation (4), we get,
⇒
u = ax + 30
80 = la + 30 [Q u(1, t) = 80°]
⇒
a=
50 x
⎛ 20 x
⎞
+ 30 − ⎜
+ 40 ⎟
l
⎝ l
⎠
=
50
l
From equations (8) and (10), we get,
Hence, equation (4) becomes,
⇒
50 x
+ 30
u=
l
⇒
... (5)
Therefore, the transient solution is given by,
The subsequent temperature distribution is determined
from the boundary value problem (1) with boundary conditions,
u(0, t) = 40
and u(l, t) = 60
and the initial temperature is,
50 x
+ 30
u(x, 0) =
l
Let, u(x, t) = us(x, t) + ut(x)
... (6)
ut(x, t) = (A cos px + B sin px) e
... (7)
With boundary conditions
... (8)
and
... (9)
[Q u(x, t) is non zero]
us(x) = px+q
Putting x = 0 and using equation (5), we get q = 40
⇒
... (11)
ut(0, t) = 0
... (12)
ut(l, t) = 0
... (13)
ut(x, 0) =
30 x
− 10
l
... (14)
From equation (11), A = 0
⇒
ut(x, t) = B sin px e
−α 2 p 2t
... (15)
Also, from equation (12), sin px = 0 for all n so that
40 = l
Thus,
−α 2 p 2t
and the initial condition is,
Where us (x) is a solution of (1) satisfying conditions
equations (5) and (6) and is called steady state solution whereas
ut(x, t) is called the transient solution which decreases with the
increase of time.
Let,
30 x
− 10
l
ut(x, 0) =
us(x) = kx+40.
p =
Putting x = l and using equation (6), we get,
nπ
, n = 1, 2, 3,........
l
60 = kl + 40
⇒
k=
20
l
20 x
+ 40
⇒ us(x) =
l
nπx
− α 2 n 2 π2 t
l e
⇒ ut(x, t) = B sin
l2
, n = 1, 2, 3
... (16)
Hence, the general solution is,
... (10)
ut(x, t) = u(x, t) – us(x)
So that,
ut(0, t) = u(0, t) – us(0)
= 40 – 40
= 0 and
ut(l, t) = u(l, t) – us(l)
⎛ 20 x
⎞
+ 40 ⎟
= 60 – ⎜
⎝ l
⎠
=0
Look for the SIA GROU P LOGO
∞
ut(x, t) =
∑b
n
sin
n =1
nπx −α 2 n 2 π2 t
e
l
l2
, n = 1, 2, 3
... (17)
On substituting t = 0 in equation (16) and then comparing
equation (13), we get,
30 x
− 10 =
l
∞
∑b
n
n =1
sin
nπx
l
l
nπx
2 ⎛ 30 x
⎞
− 10 ⎟ sin
dx
⎜
bn =
l 0⎝ l
l
⎠
∫
on the TITLE COVER before you buy
... (18)
4.27
UNIT-4 (Partial Differential Equations)
l
2 ⎡⎛ 30 x
nπx ⎞⎛ l ⎞ 30 ⎛
nπx ⎞ l 2 ⎤
⎞⎛
− 10 ⎟⎜ − cos
⎟⎜ ⎟ − ⎜ − sin
⎟
= ⎢⎜
⎥
l ⎣⎝ l
l ⎠⎝ nπ ⎠ l ⎝
l ⎠ n 2 π2 ⎦ 0
⎠⎝
=−
⎤
2 ⎡⎛ 30l
⎞
− 10 ⎟ cos nπ − (− 10 )cos 0⎥
⎜
⎢
nπ ⎣⎝ l
⎠
⎦
bn = –
20
2
n
20 x(− 1)n + 10 =
2(− 1) + 1 ,
nπ
nπ
[
[
]
]
n = 1, 2,...
On substituting value of bn in equation (16), we get,
∞
ut(x, t) =
⇒ u(x, t) =
nπx −α 2 n 2 π2 t
− 20
e
[( −1) n 2 + 1] sin
nπ
l
n =1
∑
20 x
+ 40 +
l
l2
... (19)
∞
nπx −α 2 n 2 π 2 t
− 20
×e
[( −1) n 2 + 1] sin
nπ
l
n =1
∑
Writing l = 20 then u(x, t) = x + 40 –
20
π
∞
∑
l2
(− 1)n 2 + 1 sin nπx × e − α n π t 20
n =1
2 2 2
n
2
20
Q41. A bar 100 cm long, with insulated sides, has its ends kept at 0°C and 100°C until steady state conditions
prevail. The two ends are then suddenly insulated and kept 80°C. Find the temperature distribution.
Ans: The temperature U(x, t) along the bar satisfies the equation.
∂u
∂ 2u
= c2
∂t
∂x 2
... (1)
By law of heat conduction, the rate of heat flow is proportional to the gradient of the temperature.
Thus, if the ends x = 0 and x = l = 100 cm of the bar are insulated so that no heat can flow through the ends. The boundary
conditions are,
∂u
(0, t ) = 0
∂x
∂u
(l , t ) = 0 for all ‘t’
∂x
Initially under steady state conditions
... (2)
∂ 2u
= 0. Its solution is u = ax + b.
∂x 2
Since u = 0 for x = 0 and u = 100 for x = l.
b = 0 and a = 1.
Thus, the initial condition is
u(x, 0) = x
0<x<1
... (3)
The solution of equation (1) is of the form
u(x, t) = (c1 cos px + c2 sin px) e
− c 2 p 2t
... (4)
On differentiating partially with respect to ‘x’ we get,
∂u
2
= (− c1 p sin px + c2 p cos px ) e −c pt
∂x
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (5)
SIA GROUP
4.28
MATHEMATICS-II [JNTU-ANANTAPUR]
Putting x = 0.
⎛ ∂u ⎞
2 2
⎜ ⎟ = c2 pe − c p t = 0 for all t
⎝ ∂x ⎠ x =0
(QBy equation (2))
c2 = 0
Putting x = l in equation (5)
⎛ ∂u ⎞
− c 2 p 2t
⎜ ⎟
= c1p sin ple
for all t
⎝ ∂x ⎠ x = l
or
(QBy equation (2))
(Q p ≠ 0 either c1 = 0 or sin pl = 0)
c1p sin pl = 0
pl = n π
p=
nπ
, n = 0, 1, 2, ......
l
Hence equation (4) becomes,
nπx −c 2
u(x, t) = c1cos
e
l
n 2 π2
t
l2
The most general solution of equation (1) satisfying the boundary conditions equation (2) is,
∞
u(x, t) =
∑ A cos
n
n =0
nπx −c 2 n 2 π2 t
e
l
∞
= A0 +
l2
nπx −c 2 n 2 π2 t
e
l
∑ A cos
n
n =0
l2
(Where, An = c1)
Putting t = 0.
∞
u(x, 0) = A0+
∑ A cos
n
n =1
nπx
=x
l
(Q by equation (3))
By expanding ‘x’ into a half range cosine series in (0, 1)
Thus,
x=
a0
+
2
∞
∑a
n
cos
n =1
nπx
l
l
Where,
a0 =
2
x dx = l
l 0
and
an =
2l
nπx
2
x cos
dx = 2 2 (cos nπ − 1)
n π
l 0
l
∫
l
∫
⎧ 0
⎪
= ⎨ − 4l
⎪⎩ n 2 π 2
A0 =
when n is even
for n odd
a0 l
= and
2 2
An = an = 0 for n even
=
−4l
for n odd
n 2 π2
Hence equation (6) takes the form,
∞
− 4l
l
nπx −c 2 n 2 π 2 t
+
cos
e
u(x, t) =
2 2
l
2 m =1 n π
∑
=
l 4l
−
2 π2
∞
1
∑ (2n − 1)
n =1
2
cos
l2
(2n − 1)πx e −c
−c 2 ( 2 n−1)2 π2 t l 2
l
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... (6)
4.29
UNIT-4 (Partial Differential Equations)
0oC
Q41. The temperature at one end of a bar is 50 cm long with insulated sides is kept at
and that the other
end is kept at 100oC until steady state condition prevails. The two ends are then suddenly insulated so
that the temperature gradient is zero at each and thereafter. Find the temperature distribution.
Ans: Let the length of the rod be ‘l’ cm
Where, l = 50 cm
The steady state solution is, u = ax + b
and the boundary conditions are,
u(0) = 0 and u(50)= 100
⇒
b = 0 and
∴
u = ax + b
∴
u = 2x.
a =2
One-dimensional heat equation is,
∂ 2u
∂u
= α2 . 2
∂t
∂x
... (1)
Given that the ends are insulated and hence no heat flows through the ends. The boundary conditions are,
⎛ ∂u ⎞
⎜ ⎟
=0
⎝ ∂t ⎠ x = 0
... (A)
⎛ ∂u ⎞
⎜ ⎟ =0
⎝ ∂t ⎠ x =l
... (B)
The initial condition is,
u(x, 0) = 2x , 0 < x < l
The solution of equation (1) is,
u(x, t) = (A cos px + B sin px) e − p
2
α 2t
... (2)
On differentiating equation (2) partially with respect to ‘x’, we get,
∂u
∂x
⇒
= (–Ap sin px + Bp cos px) e − p
2
α2 t
Using equation (A) we get,
Bp e − p
2
α 2t
⇒
=0
(Q cos 0 = 1)
B =0
∴
∂u
2 2
= – Ap sin px e − p α t
∂x
Using equation (B),
–Ap sin pl e −α
⇒
p 2t
=0
(Q B = 0)
sin pl = 0
⇒
pl = nπ
⇒
p =
∴
2
nπ
l
⎛ nπx ⎞ − α2 n2 π 2t /l 2
⎟.e
u(x, t) = A cos ⎜
⎝ l ⎠
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4.30
MATHEMATICS-II [JNTU-ANANTAPUR]
∴ The most general solution is,
∞
∴
u(x, t) =
∑
An cos
n =1
nπx − α2 n2 π 2t /l 2
e
l
∞
∴
u(x, t) = A0+
∑ A cos
n
n =1
nπx − α 2 n 2 π 2t / l 2
e
l
Let t = 0,
∞
⇒
u(x, 0) = A0 +
∑
An cos
n =1
nπx
l
But u(x, 0) = 2x, 0 < x < l. (initial condition)
Expanding u(x, 0) in half-range cosine series,
∞
∑a
a0
u(x, 0) =
+
2
2
a0 =
l
n
n =1
⎛ nπx ⎞
cos ⎜
⎟
⎝ l ⎠
l
l
2 ⎡ 2x2 ⎤
2
2x dx =
⎢
⎥ = . l2
l ⎣ 2 ⎦
l
0
0
∫
a0 = 2l
∴
A0 =
a0
=l
2
2
an =
l
=
∴
l
nπx ⎞ l
nπx ⎞ l 2 ⎤
⎛
nπx
4 ⎡ ⎛
+
sin
1
cos
x
⎜
⎟
⎜
⎟.
⎢
⎥
dx =
2x cos
l ⎠ nπ ⎝
l ⎠ n2 π2 ⎦ 0
l
l ⎣ ⎝
l
∫
0
⎤
4 ⎡ l2
4l
× ⎢ 2 2 [cos nπ − cos 0]⎥ = 2 2 [(–1)n – 1]
l
⎣n π
⎦ n π
An = an =
4l
[(–1)n – 1].
n2 π2
∞
∴ u(x, t) = l +
4l
∑n π
n =1
2
2
[(–1)n – 1]cos
nπx
− α 2 n 2 π 2t / l 2
l e
On substituting, l= 50 in the above equation, we get,
∞
nπx
[(−1) n − 1]
200
cos
e
∴ u(x, t) = 50 – 2 .
2
50
n
π n =1
∑
− α 2 n 2 π 2t
( 50 ) 2
.
Q43. A homogeneous rod of conducting material of length 100 cm has its ends kept at zero temperature and
the temperature initially is
u(x, 0) = x;
0 < x < 50
= 100 – x; 50 < x < 100
Find the temperature u(x, 0) at any t.
Ans: The general solution of the heat equation is,
2
∂u
2 ∂ u
=C
∂t
∂x 2
... (1)
Which describes the temperature distribution u(x, t) subject to the boundary conditions u(0, t) = 0 is given by,
2
∞
⎛ cnπ ⎞
⎛ nπx ⎞
u(x, t) = Σ bn sin⎜
⎟ t
⎟ .C − ⎜
n =1
⎝ 100 ⎠
⎝ 100 ⎠
... (2)
Where 0 < x < 100.
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4.31
UNIT-4 (Partial Differential Equations)
On substituting the initial condition the half-range sine series for u(x, 0) in 0 < x < 100 is obtained as,
∞
∑b
u(x, 0)=
n
n =1
⇒
2
bn =
100
⎛ nπx ⎞
sin ⎜
⎟
⎝ 100 ⎠
100
⎛ nπx ⎞
∫ u ( x, 0) sin⎜⎝ 100 ⎟⎠ dx
0
1 ⎡
⎛ nπx ⎞
⎢ x sin ⎜
⎟ dx +
=
50 ⎢ 0
⎝ 100 ⎠
⎣
50
∫
⎤
100
⎛ nπx ⎞
∫ (100 − x)sin ⎜ ⎟ dx⎥
⎝ 100 ⎠
50
⎥⎦
100
50
100
1 ⎡
⎛ nπx ⎞
⎛ nπx ⎞ ⎤ − x sin ⎛⎜ nπx ⎞⎟ dx
⎢ x sin ⎜
⎟ dx + 100 sin ⎜
⎟ dx ⎥
=
⎝ 100 ⎠
50 ⎢
100 ⎠
100 ⎠ ⎥ 50
⎝
⎝
50
⎦
⎣0
∫
∫
∫
... (3)
Considering,
50
⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
sin ⎜
cos⎜
50
⎟⎥
⎟
⎢
n
x
π
⎞
⎛
⎝ 100 ⎠ ⎥
⎝ 100 ⎠ + 1.
x sin ⎜
⎟ dx = ⎢− x .
nπ
⎢
n2π2 ⎥
⎝ 100 ⎠
0
⎥
⎢
100
100
⎦ x =0
⎣
∫
2
=−
100
Similarly,
−
∫
50
100 ⎡
⎛ nπ ⎞⎤ ⎛ 100 ⎞
⎛ nπ ⎞
⎟ sin ⎜ ⎟
⎢50 cos⎜ 2 ⎟⎥ + ⎜
nπ ⎣
⎝ ⎠⎦ ⎝ nπ ⎠
⎝ 2 ⎠
50
⎛ n πx ⎞
⎛ nπx ⎞
⎟ dx
x sin ⎜
⎟ dx = x sin ⎜
⎝ 100 ⎠
⎝ 100 ⎠
100
∫
50
⎡
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
cos⎜
sin ⎜
⎟⎥
⎟
⎢
100
⎠ + 1.
⎝ 100 ⎠ ⎥
⎝
= ⎢(− x ).
nπ
⎛ nπ ⎞ ⎥
⎢
⎟
⎜
⎢
100
⎝ 100 ⎠ ⎥⎦ x =100
⎣
2
=−
2
100
⎛ nπ ⎞ ⎛ 100 ⎞
⎛ nπ ⎞ 100
cos nπ − 0
.50 cos⎜ ⎟ + ⎜
sin
⎜ ⎟+
⎟
nπ
nπ
⎝ 2 ⎠ ⎝ nπ ⎠
⎝ 2 ⎠
100
⎡
⎛ nπx ⎞ ⎤
− cos⎜
⎟
100
⎢
100 ⎠ ⎥⎥
⎛ nπx ⎞
⎝
⎢
Considering, 100 sin ⎜
⎟ dx = 100
nπ
⎢
⎥
⎝ 100 ⎠
50
⎢
⎥
100
⎣
⎦ x =50
∫
=−
100
100
⎛ nπ ⎞
.100 cos( nπ) + 100
cos⎜ ⎟
nπ
nπ
⎝ 2 ⎠
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.32
MATHEMATICS-II [JNTU-ANANTAPUR]
Adding these equations and substituting in equation (3), we get,
bn =
2
if n is even
⎧ 0
1 ⎡ ⎛ 100 ⎞
⎛ nπ ⎞ ⎤ ⎪
⎟ sin ⎜ ⎟⎥ = ⎨ 400 sin ⎛ n π ⎞ if n is odd
⎢2 .⎜
⎜
⎟
50 ⎣⎢ ⎝ nπ ⎠
⎝ 2 ⎠⎦⎥ ⎪ 2 2
⎝ 2 ⎠
⎩n π
n = 2m – 1
n
1
=m−
2
2
∴
nπ
π
= mπ −
2
2
π⎞
⎛ nπ ⎞
⎛
sin ⎜ ⎟ = sin ⎜ mπ − ⎟ = –cos mπ
2⎠
⎝ 2 ⎠
⎝
= (–1)m–1 0
=
if n is even
400
(− 1)m −1
2 2
n π
if n = 2m – 1
400 ∞ (− 1)
⎧ (2 m − 1)πx ⎫ (c (2 m −1)π 100 )2 t
sin ⎨
u(x, t) = 2
⎬ e
0 < x < 100 .
2
π m =1 (2 m − 1)
⎩ 100 ⎭
m −1
∑
∴
Q44. Find the temperature in a laterally insulated rod of length L whose both ends are insulated and the
initial temperature u(x, 0) = x if 0 < x <
L
L
and L – x if
< x < L.
2
2
Ans: Given that,
Length of the insultated rod = L
L
2
Initial temperature (t = 0) u(x, 0) = x, 0 < x <
= L − x,
L
<x<L
2
... (1)
The equation for conduction of heat is,
2
∂u
2 ∂ u
= C
∂t
∂x 2
... (2)
The boundary conditions are,
u(0, t)= 0 ∀ t
... (3)
u(L, t) = 0 ∀ t
[Q Both ends are insulated ⇒ No heat can flow through end]
... (4)
The solution of equation (2) is,
u(x, t) = (C1cos px + C2 sin px) e −C
2
p 2t
... (5)
Put x = 0 in equation (5), we get,
⇒
⇒
∴
u(0, t) = C1e −C
2
p 2t
0 = C1e −C
2
p 2t
[Q From equation (3)]
C1 = 0
u(x, t)= C 2 sin px.e −C
2
p2t
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... (6)
on the TITLE COVER before you buy
4.33
UNIT-4 (Partial Differential Equations)
Put, x = L in equation (6), we get,
u(L, t) = C 2 sin pL.e −C
−C
0 = C 2 sin pL.e
⇒
2
2
p 2t
p 2t
[Q From equation (4)]
As C2 ≠ 0 ⇒ sin Lp = 0
⇒ Lp = nπ
⇒P=
∴
nπ
L
Equation (6) reduces to,
nπx −
u(x, t) = C 2 sin
e
L
C 2 n 2 π 2t
L2
... (7)
∴ The most general solution of equation (2) is of the form
∞
nπx −
Cn sin
.e
u(x, t) =
L
n =1
∑
C 2 x 2 π 2t
L2
... (8)
Put t = 0 in equation (8), we get,
∞
u(x, 0) =
∑C
n
sin
n =1
nπx
L
... (9)
In order to satisfy the boundary conditions of equation (1),
∞
∑C
n
sin nx = u(x, 0) = x in 0 ≤ x ≤ L/2
n =1
= L – x in L/2 ≤ x ≤ L
... (10)
The expnasion of u(x, 0) in half-range sine series in the interval (0, L) is obtained as,
∞
Since u(x, 0) =
2
Where, An = L
∑ A sin
n
n =1
100
nπx
L
∫ u ( x, 0) sin
0
... (11)
nπx
dx
L
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.34
MATHEMATICS-II [JNTU-ANANTAPUR]
On comparing equations (9) and (11), we get,
Cn = An
∴
2⎡
⎢
Cn = L ⎢
⎣
L/2
∫
x sin
0
2⎡
= ⎢x
L⎢
⎣
L/2
∫
sin
0
L
nπx
nπx ⎤
dx ( L − x ) sin
dx ⎥
L
L
⎥
L/2
⎦
∫
nπx
dx −
L
L/2
L
L
nπx ⎞
nπx
nπx ⎞ ⎤
⎛
⎛
dx ⎟dx + ( L − x ) sin
dx − ⎜ (−1) sin
dx ⎟dx⎥
⎜1. sin
L
L
L
⎠
⎝
⎠ ⎥⎦
⎝
L/2
L/2
∫ ∫
0
∫
∫
∫
L
L/2
⎧⎡ ⎛
⎡
nπx ⎞ ⎫
nπx ⎞ ⎤
nπx ⎞ ⎤
⎛
⎛ nπx ⎞
⎛
⎪⎢ x⎜ − cos
−
cos
L / 2 − cos ⎜
⎟
L ⎜ − cos
⎟
⎟⎥
⎟
⎜
⎥
⎢
2 ⎪⎢ ⎝
L ⎠ ⎪⎪
L
L
L ⎠⎥
⎝
⎠
⎝
⎠
⎝
⎥
⎢
−
dx + ( L − x)
+
dx ⎬
= ⎨
nπ
nπ
nπ
nπ
L ⎪⎢
⎥
⎥
⎢
⎪
0
L/2
⎥
⎥
⎢
L
⎪⎢⎣
L
L
L
⎪⎭
⎦
⎦
⎣
L
/
2
0
⎩
∫
∫
L/2
L
⎧
nπx ⎤
nπx ⎤ ⎫
⎡
⎡
sin
sin
⎪
⎪
nπ ⎞ L ⎢
2⎪ ⎛L L
L ⎥ + 0 + L × L cos nπ − L ⎢
L ⎥ ⎪
− 0⎟ +
cos
= ⎨− ⎜ ×
⎢
⎥
⎢
⎥
⎬
⎠ nπ ⎢ nπ ⎥
L ⎪ ⎝ 2 nπ
2
2 nπ
2 nπ ⎢ nπ ⎥ ⎪
⎣ L ⎦0
⎣ L ⎦ L / 2 ⎪⎭
⎪⎩
=
2
nπ L
L ⎡ ⎡ nπ
nπ L
L ⎡
nπ ⎤ ⎤ ⎫⎪
2 ⎧⎪ L2
⎤ L
cos
cos
sin nπ − sin ⎥ ⎥ ⎬
−
×
+
×
− 0⎥ +
⎢ ⎢sin
⎨−
⎢
2 nπ nπ ⎣⎢ ⎣
2
2 nπ nπ ⎣
2 ⎦ ⎥⎦ ⎪⎭
L ⎪⎩ 2nπ
⎦ 2 nπ
nπ 1
nπ 1
nπ 1
nπ ⎫
1
2 L2 ⎧ 1
+
+ cos
− sin nπ +
sin ⎬
sin
⎨− cos
= ×
L nπ ⎩ 2
nπ
2 ⎭
2 2
2 nπ
2 nπ
=
∴
Cn =
4L
nπ
2L ⎡ 1
nπ ⎤
sin
2. sin ⎥ =
2
⎢
2
( nπ)
2 ⎦
nπ ⎣ nπ
4L
2 2
n π
sin
[Q sin nπ = 0]
nπ
2
When n is even, Cn = 0
4L
When n is odd, Cn =
2 2
n π
( −1) n
On substituting in equation (8), we get,
u(x, t) =
4L
π2
4L
u(x, t) = π 2
∞
∑
(−1) n
∞
( −1) n
n =1,3,5
∑
n =1
(odd)
n2
nπx −
e
sin
L
C 2 n 2 π 2t
L2
( 2n − 1) πx −
e
sin
2
L
( 2 n − 1)
C 2 ( 2 n −1) 2 π 2t
L2
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4.35
UNIT-4 (Partial Differential Equations)
Q45. A rod of length 100 cm length has its ends A
and B kept at 0° and 100° centigrade until steady
state conditions prevail. The temperatures at
the ends are changed to 20°C and 60°C
respectively. Find the temperature in the rod.
Ans: Given that,
Length of rod, L = 100 cm
Initial temperature of the end A, θA1 = 0°C
Initial temperature of the end B, θB1 = 100°C
Final temperature of the end A, θA2 = 20°C
Final temperature of the end B, θB2 = 60°C
After reaching the steady state condition, the
temperature at the ends A and B are now changed to 20°C and
60°C respectively. The boundary conditions are,
u (0, t) = 20°C
... (7)
u (L, t) = 60°C
... (8)
The temperature in the intermediate period is given by,
U(x, t) = Us (x) + Ut (x, t)
... (9)
Where, Us (x) represents the steady state solution as
given by equation (6) and Ut (x, t) represents the transient solution.
Us (x) is evaluated by solving the equation,
Temperature in the rod, at a later time t, u(x, t) = ?
∂ 2u
=0
∂x 2
The one-dimensional heat conduction equation is given by,
∂u
∂ 2u
= C2
∂t
∂x 2
... (1)
Under steady state conditions,
∂u
=0
∂t
∴
Equation (1) becomes,
∂ 2u
=0
∂x 2
⇒
... (2)
The solution of equation (2) is given by,
u = ax + b
Thus, its solution is,
Us (x)= ax + b
... (10)
From equation (7),
x = 0 , u = 20
Substituting the values in equation (10), we get,
20 = a. 0 + b
... (3)
The boundary conditions are,
u (0, t) = 0°C
... (4)
u (L , t) = 100°C
... (5)
And also, from equation (8), we have,
x = L, u = 60
Substituting the values in equation (10), we get,
⇒ 60 = a (L) + 20 [Q b = 20]
⇒ 60 – 20 = aL
⇒ 40 = aL
⇒ a = 40/L
∴ a = 40 / L
From equation (4),
When, x = 0, u = 0
Thus, equation (10) becomes,
On substituting the above values in equation (3), we get,
∴
∴b = 0
From equation (9),
Ut (x, t) = u(x, t) – Us (x)
∴
Ut (0, t) = U (0, t) – Us (0)
= 20 – 20
[From equations (7) and (11)]
=0
Also,
Ut (L, t) = U (L, t) – Us (L)
When, x = L, u = 100
On substituting the values in equation (3), we get,
100 = a (L) + 0
⇒ a = 100 / L
Hence, equation (3) is given as,
∴
⎛ 40 ⎞
Us(x)= ⎜ ⎟ x + 20
⎝ L ⎠
0 = a (0) + b
From equation (5),
∴
b = 20
⎛ 100 ⎞
⎟
u (x, 0) = ⎜
⎝ L ⎠
x
... (6)
Equation (6) represents the initial temperature
distribution in the rod.
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... (11)
⎤
⎡⎛ 40 ⎞
= 60 – ⎢⎜ ⎟ × L + 20⎥
L
⎦
⎣⎝ ⎠
[ From equations (8) and (11)]
= 60 – [40 + 20]
= 60 – 60 = 0
SIA GROUP
4.36
MATHEMATICS-II [JNTU-ANANTAPUR]
Ut(x, 0) = U(x, 0) – Us (x)
⎡⎛ 100 ⎞ ⎤ ⎡⎛ 40 ⎞
⎤
⎟ x ⎥ − ⎢⎜ ⎟ x + 20⎥
= ⎢⎜
⎣⎝ L ⎠ ⎦ ⎣⎝ L ⎠
⎦
=
∴
[From equations (6), (11)]
60
x − 20
L
⎛ 60 ⎞
Ut (x, 0) = ⎜ ⎟ x − 20
⎝ L⎠
... (12)
Thus,
The boundary conditions of the transient solution Ut (x, t) are,
Ut(0, t) = 0
... (13)
Ut(L, t) = 0
... (14)
⎛ 60 ⎞
Ut (x, 0) = ⎜ ⎟ x − 20
⎝ L⎠
... (15)
The solution of
Ut (x, t) is given by,
Ut (x, t) = (C1 cos px + C2 sin px) e −c
2
p 2t
... (16)
Put x = 0 in equation (16), we get,
⇒
Ut (0, t) = (C1 cos 0 + C2 sin 0) e −c
⇒
0 = (C1 + 0) – C2p2t
2
p 2t
[From equation (3) Ut (0, t) = 0]
C1 = 0
⇒
Put x = L in equation (16)
−c
Ut (L, t) = (C1 cos PL + C2 sin PL) e
⇒
0 = (0 + C2 sin PL) e −c
⇒
[C2 sin pL] e −c
⇒
C2 sin pL = 0
⇒
sin pL =0
[Q C2 ≠ 0]
⇒
pL = nπ
[Q sin nπ = 0]
p=
2
p 2t
2
p 2t
2
p 2t
[Q C1 = 0 and Ut (L, t) = 0 from equation (14)]
=0
nπ
L
Thus,
Substituting the values of C1 and p in equation (16), we get,
⎡⎛ nπ ⎞ ⎤ −c
Ut (x, t)= C2 sin ⎢⎜ ⎟ x ⎥ e
⎣⎝ L ⎠ ⎦
⇒
2 ⎡ nπ ⎤
2
⎢ L ⎥ t
⎣ ⎦
⎡ c 2n2π 2 ⎤
⎥t
L2 ⎦⎥
⎡⎛ nπ ⎞⎤ − ⎢⎣⎢
x ⎟⎥ e
Ut (x, t) = C2 sin ⎢⎜
⎣⎝ L ⎠⎦
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... (17)
on the TITLE COVER before you buy
4.37
UNIT-4 (Partial Differential Equations)
The most general solution is given by,
⎡ c 2 n 2 π2 ⎤
⎥t
L2 ⎦⎥
∞
⎛ nπx ⎞ ⎢⎣⎢
Ut(x, t)= ∑ Bn sin⎜
⎟e
⎝ L ⎠
n=1
... (18)
Put t = 0 in equation (18), we get,
∞
nπx ⎞
Ut (x, 0) = ∑ Bn sin ⎛⎜
⎟
n =1
⎝ L ⎠
... (19)
In order to satisfy the boundary condition given by equation (15),
We have,
∞
⎛ nπx ⎞ ⎡ 60 ⎤
∑ Bn SM ⎜
⎟ =
x – 20
⎝ L ⎠ ⎢⎣ L ⎥⎦
n =1
The value of co-efficient Bn is given as,
L
⎤ ⎛ nπ ⎞
2 ⎡⎛ 60 ⎞
⎜ ⎟ x − 20⎥ sin⎜ ⎟ x dx
Bn =
⎢
L ⎣⎝ L ⎠
⎦ ⎝ L⎠
0
∫
L
L
⎤
2 ⎡⎛ 60 ⎞
⎛ nπ ⎞
⎛ nπ ⎞
⎢
⎜ ⎟ x sin⎜ ⎟ x − 20 sin⎜ ⎟ x dx ⎥
=
L ⎢⎝ L ⎠
⎥
⎝ L ⎠
⎝ L⎠
0
0
⎣
⎦
∫
∫
L
L⎫
⎧ ⎡ ⎡
⎤
⎤
⎡
⎤
⎡
π
π
n
n
⎞
⎛
⎞
⎛
⎛ nπ ⎞ ⎤ ⎪
⎪ ⎢ ⎢ − cos⎜ ⎟ x ⎥
−
−
x
x
cos
cos
⎟
⎜
⎟
⎜
⎥
⎢
⎥
⎢
⎥
2 ⎪ 60
⎝ L ⎠ ⎥ − 1× ⎢
⎝ L ⎠ ⎥ dx ⎥ − 20⎢
⎝ L⎠ ⎥ ⎪
= ⎨ ⎢ x⎢
⎬
⎢ nπ / L ⎥ ⎥
⎢ nπ / L ⎥ ⎪
L ⎪ L ⎢ ⎢ nπ / L ⎥
⎥
⎢
⎥ ⎥
⎢
⎥ ⎪
⎪ ⎢⎣ ⎢⎣
⎦
⎣
⎦ ⎦0
⎣
⎦0 ⎭
⎩
∫
L
⎧ ⎡
⎫
⎡ ⎛ nπ ⎞ ⎤ ⎤
⎪ ⎢
⎪
x
sin
L
⎟
⎜
⎥
⎡ 20 L
2 ⎪ 60 ⎢ − xL ⎡ ⎛ nπ ⎞ ⎤ L ⎢⎢ ⎝ L ⎠ ⎥⎥ ⎥
⎛ nπ ⎞ ⎤ ⎪
+⎢
cos⎜ ⎟ x ⎥ ⎬
= ⎨
⎢cos⎜ ⎟ x ⎥ +
L ⎪ L ⎢ nπ ⎣ ⎝ L ⎠ ⎦ nπ ⎢ nπ / L ⎥ ⎥
⎝ L ⎠ ⎦0 ⎪
⎣ nπ
⎢
⎥⎥
⎪ ⎢⎣
⎪
⎣
⎦⎦ 0
⎩
⎭
⎧
L2
2 ⎪ 60 ⎡ − xL ⎡ ⎛ nπ ⎞ ⎤
x
cos
+
⎟
⎜
= ⎨ ⎢
⎢
⎥
L ⎪ L ⎢⎣ nπ ⎣ ⎝ L ⎠ ⎦ n 2 π 2
⎩
L
L⎫
⎡ ⎛ nπ ⎞ ⎤ ⎤
⎡ 20 L
⎛ nπ ⎞ ⎤ ⎪
x
x
sin
cos
+
⎟
⎟
⎜
⎜
⎥
⎢
⎢
⎥
⎥ ⎬
⎝ L ⎠ ⎦0 ⎪
⎣ ⎝ L ⎠ ⎦ ⎥⎦ 0 ⎣ nπ
⎭
L
=
2 ⎡ − 60 x
60 L
20 L
⎛ nπ ⎞
⎛ nπ ⎞
⎛ nπ ⎞ ⎤
cos⎜ ⎟ x ⎥
⎢ π cos ⎜ ⎟ x + 2 2 sin⎜ ⎟ x +
nπ
L⎣ n
n π
⎝ L ⎠
⎝ L⎠
⎝ L ⎠ ⎦0
L
2 ⎡ − 20 x
60 L
⎛ nπx ⎞
⎛ nπ ⎞ ⎤
cos ⎜
⎟[3x − L ]+ 2 2 sin⎜ ⎟ x ⎥
= ⎢
L ⎣ nπ
L
n π
⎠
⎝
⎝ L ⎠ ⎦0
2 ⎧⎪⎡ − 20
60L
60 L
⎡ nπ ⎤ ⎤ ⎡ − 20
⎛ nπ ⎞
⎤ ⎫⎪
= L ⎨⎢ nπ cos⎜ L L ⎟[(3 × L) − L] + 2 2 sin ⎢ L × L ⎥ ⎥ − ⎢ nπ cos 0[(3 × 0) − L] + 2 2 sin 0⎥ ⎬
⎪⎩⎣
n π
n π
⎠
⎦⎦ ⎣
⎣
⎝
⎦ ⎪⎭
=
2 ⎧⎡ − 40 L
⎤ ⎡ 20 L ⎤ ⎫
(−1) n + 0⎥ − ⎢
+ 0⎥ ⎬
⎨⎢
L ⎩ ⎣ nπ
⎦ ⎣ nπ
⎦⎭
[Q sin nπ = 0, sin (0) = 0, cos nπ = (–1)]
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.38
MATHEMATICS-II [JNTU-ANANTAPUR]
∴
=
2 ⎡ − 40 L
20 L ⎤
(−1) n −
L ⎢⎣ nπ
nπ ⎥⎦
=
2 ⎡ − 20 L
⎤
[1 + 2(−1) n ]⎥
L ⎢⎣ nπ
⎦
=
−40
1 + 2( −1) n
nπ
bn =
−40
1 + 2( −1) n
nπ
[
]
[
]
On substituting the value of bn in equation (18), we get,
⎡ c 2n2π 2 ⎤
⎥t
L ⎥⎦
−
∞ − 40
⎛ nπx ⎞ ⎢⎢⎣
[1 + 2(−1) n ] sin⎜
Ut (0, t) = ∑
⎟e
⎝ L ⎠
n=1 nπ
⇒
⎡ c 2 n2π 2 ⎤
⎥t
L ⎦⎥
− 40 ∞ [1 + 2(−1) n ] ⎛ nπx ⎞ − ⎢⎣⎢
sin ⎜
∑
⎟e
Ut (0, t) =
⎝ L ⎠
π n =1
n
... (20)
Hence, substituting equation (11) and equation (17) in equation (9), we get,
⎡ c 2n2π 2 ⎤
⎥t
L ⎥⎦
40 ∞ [1 + 2(−1) n ] ⎛ nπx ⎞ − ⎢⎢⎣
⎛ 40 ⎞
U(x, t) = ⎜ ⎟ x + 20 −
sin⎜
⎟e
∑
π n=1
n
⎝ L⎠
⎝ L ⎠
Put L = 100,
⎡ c 2 n 2 π2 ⎤
⎥t
100 ⎦⎥
−
40 ∞ [1 + 2(−1) n ] ⎛ nπx ⎞ ⎢⎣⎢
⎡ 40 ⎤
∴ U(x, t) = ⎢
e
x
+
−
20
sin
⎟
⎜
∑
⎥
n
π n=1
⎝ 100 ⎠
⎣100 ⎦
⎡ c 2n 2π 2 ⎤
⎥t
100 ⎦⎥
40 ∞ ⎡ 1 + 2( −1) n ⎤ ⎛ nπx ⎞ − ⎢⎣⎢
U ( x, t ) = 0.4 x + 20 −
⎟e
∑ ⎢
⎥ sin ⎜
π n =1 ⎢⎣
n
⎥⎦ ⎝ 100 ⎠
Q46. A bar of length L is laterally insulated with its ends A and B kept at 0° and 100° respectively until steady
state condition is reached. The temperature at A is raised to 30° and that at B is reduced to 80° simulModel Paper-III, Q9
taneously. Find the temperature in the rod at a later time.
Ans: Given that,
Length of the rod = L
Initial temperature at the end A, θ A1 = 0°C
Initial temperature at the end B, θ B1 = 100°C
Final temperature at the end A, θ A2 = 30°C
Final temperature at the end B, θ B2 = 80°C
Temperature in the rod, at a later time t, u(x, t) = ?
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
4.39
UNIT-4 (Partial Differential Equations)
On substituting these values in equation (10), we get,
One-dimensional heat conduction equation is given by,
30 = a(0) + b
2
∂u
2 ∂ u
=c
∂t
∂x 2
Under steady state conditions,
... (1)
∴ b = 30
Using equation (8),
∂u
=0
∂t
If x = L, u = 80
∴ Equation (1) becomes,
∂ 2u
=0
⇒
80 = a[L] + 30
⇒
aL= 50
⇒
∴ a = 50 / L
∴
⎡ 50 ⎤
us(x) = ⎢ ⎥ x + 30
⎣L⎦
... (2)
∂x 2
Solution of equation (2) is given by,
u = ax + b
... (3)
The boundary conditions are,
u(0, t) = 0
... (4)
u(L, t) = 100
... (5)
ut(x, t) = u(x, t) – us(x)
∴ ut(0, t)= u(0, t) – us(0)
i.e., if x = 0, u = 0
= 30 – 30
On substituting these values in equation (3), we get,
Also,
∴b = 0
ut(L, t) = u(L, t) – us(L)
Using equation (5),
⎡⎛ 50 ⎞
⎤
= 80 – ⎢⎜⎝ ⎟⎠ L + 30⎥
L
⎣
⎦
[Q From equations (8) and (11)]
i.e., if x = L, u = 100
100 = a[L] + 0
∴ a = 100 / L
∴
⎡100 ⎤
u(x, 0) = ⎢
⎥x
⎣ L ⎦
[Q From equations (7) and (11)]
=0
0 = a(0) + b
⇒
... (11)
From equation (9),
Using equation (4),
∴
[Q b = 30]
= 80 – 80
=0
... (6)
ut(x, 0) = u(x, 0) – us(x)
⎤
⎡100 ⎤ ⎡⎛ 50 ⎞
x ⎥ − ⎢⎜ ⎟ x + 30⎥
=⎢
⎠
⎝
⎣ L ⎦ ⎣ L
⎦
[Q From equations (6) and (11)]
Equation (6) represents the initial temperature
distribution in the rod.
After reaching steady state condition, the temperature
at the ends A and B are now changed to 30 and 80 respectively.
=
The boundary conditions are,
u(0, t) = 30
... (7)
u(L, t) = 80
... (8)
⎛ 50 ⎞
∴ ut(x, 0) = ⎜ ⎟ x − 30
⎝ L⎠
The temperature in the intermediate period is given by,
u(x, t) = us(x) + ut(x, t)
... (9)
Where us(x) represents the steady state solution as
given by equation (6) and ut(x, t) represents the transient
solution.
50
x − 30
L
... (12)
Thus, the boundary conditions of the transient solution
ut(x, t) are,
∂ u
= 0.
∂x 2
ut(0, t)= 0
... (i)
ut(L, t) = 0
... (ii)
⎛ 50 ⎞
ut(x, 0)= ⎜ ⎟ x − 30
⎝ L⎠
... (iii)
2
us(x) is evaluated by solving the equation
Hence its solution is,
us(x) = ax + b
... (10)
Using equation (7),
If x = 0, x = 30
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
The solution of ut(x, t) is given by,
ut(x, t) = [c1 cos px + c2 sin px] e − c
2
p 2t
... (13)
SIA GROUP
4.40
MATHEMATICS-II [JNTU-ANANTAPUR]
Put x = 0 in equation (13), we get,
ut(0, t) = [c1 cos 0 + c2 sin 0] . e − c
.
0 = [ c1 + 0]e − c
⇒
2
2
p 2t
p 2t
∴ c1 = 0
Put x = L in equation (13), we get,
ut(L, t) = [ c1 cos( pL ) + c 2 sin( pL )]e
0 = [0 + c2 sin( pL)]e −c
⇒
⇒
⇒
⇒
⇒
c2 sin pL = 0
sin pL = 0
pL = sin–1 (0)
pL = nπ
∴p=
2
− c 2 p 2t
p 2t
[Q c1 = 0 and ut(L, t) = 0 from equation (ii)]
[Q c2 ≠ 0]
nπ
L
Thus, substituting the values of c1, c2 and p in equation (13), we get,
⎡ nπ ⎤ −c
ut(x, t) = c2 . sin ⎢ ⎥.x.e
⎣L⎦
⎡ nπ ⎤
ut(x, t) = c 2 . sin ⎢ ⎥.xe
⎣L⎦
⇒
2 ⎡ nπ ⎤
2
⎢ L ⎥ .t
⎣ ⎦
⎡ c 2n 2π 2 ⎤
−⎢
⎥.t
⎢⎣ L ⎥⎦
... (14)
The most general solution is given by,
∞
⎛ nπx ⎞
Bn sin⎜
⎟.e
ut(x, t) =
⎝ L ⎠
n =1
∑
⎡ c 2 n 2 π2 ⎤
−⎢
⎥t
⎣⎢ L ⎦⎥
... (15)
Put t = 0 in equation (15), we get,
∞
⇒
ut(x, 0)=
n
⎛ nπx ⎞
sin⎜
⎟ .e–0
⎝ L ⎠
n
⎛ nπx ⎞
sin⎜
⎟
⎝ L ⎠
∑B
n =1
∞
⇒
ut(x, 0)=
∑B
n =1
... (16)
In order to satisfy the boundary condition given by equation (iii), equating equation (16) and equation (iii), we get,
∞
∑B
n =1
n
⎛ nπx ⎞ ⎡ 50 ⎤
sin⎜
⎟ = ⎢ ⎥ x − 30
⎝ L ⎠ ⎣L⎦
The value of Bn can be obtained as,
L
Bn =
⎤ ⎛ nπx ⎞
2 ⎡⎛ 50 ⎞
⎟dx
⎜ ⎟ x − 30⎥. sin⎜
⎢
L ⎣⎝ L ⎠
L ⎠
⎝
⎦
0
∫
L
L
2 ⎡ 50
⎛ nπx ⎞ ⎤
⎛ nπx ⎞
⎢
x sin ⎜
⎟ dx − 30 sin⎜
⎟dx ⎥
=
L⎢ L
L ⎠
L ⎠ ⎥
⎝
⎝
0
⎣ 0
⎦
∫
∫
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
4.41
UNIT-4 (Partial Differential Equations)
L
⎡ ⎧ ⎡
⎡
⎛ nπx ⎞ ⎫
⎛ nπx ⎞ ⎤ ⎤
⎛ nπx ⎞ ⎤
− cos⎜
− cos ⎜
⎟ ⎪
⎟ ⎥⎥
⎟⎥
⎢ ⎪ ⎢ − cos⎜
⎢
2 ⎢ 50 ⎪ ⎢
⎝ L ⎠ dx ⎪ − 30⎢
⎝ L ⎠ ⎥⎥
⎝ L ⎠⎥ −
.
x
= ⎢ ⎨
⎬
nπ
nπ
nπ
L L ⎪ ⎢
⎥⎥
⎢
⎥
⎪
⎢
⎥⎥
⎥
⎢
⎢
L
L
L
⎪
⎪
⎦ ⎥⎦ 0
⎦
⎣
⎢⎣ ⎩ ⎣
⎭
∫
L
L
2 ⎡ 50 ⎧ − xL
⎛ nπx ⎞ ⎫ 30 L
⎛ nπx ⎞
⎛ nπx ⎞ ⎤
. cos⎜
cos⎜
⎟⎬ +
⎟ + 2 2 sin ⎜
⎟⎥
= ⎢ ⎨
π
L ⎣⎢ L ⎩ nπ
L
L
n
⎠⎭
⎝
⎠ n π
⎝
⎝ L ⎠ ⎦⎥ 0
L
=
2 ⎡ 50 x
⎛ nπx ⎞ 50
⎛ nπx ⎞⎤
⎛ nπx ⎞ 30L
cos⎜
cos⎜
⎟ + 2 2 sin⎜
⎟⎥
⎟+
⎢−
L ⎣ nπ
⎝ L ⎠ n π
⎝ L ⎠⎦ 0
⎝ L ⎠ nπ
L
2 ⎡ 10
50
⎛ nπx ⎞
⎛ nπx ⎞⎤
cos ⎜
⎟.[−5 x + 3L ] + 2 2 sin⎜
⎟⎥
= ⎢−
L ⎣ nπ
L
n π
⎠
⎝
⎝ L ⎠⎦ 0
∴
=
2⎧
− 10
50
50
− 10
⎛ nπL ⎞
⎫
⎛ nπL ⎞⎫ ⎧
cos ⎜
cos 0 + 2 2 sin 0⎬
⎟. + 2 2 sin⎜
⎟⎬ − ⎨[(−5 × 0) + 3L ] ×
⎨(−5L + 3L) ×
π
L
n
L⎩
nπ
L
π
π
n
n
⎝
⎠
⎭
⎝
⎠⎭ ⎩
=
2 ⎧ 20 L
30L ⎫
cos nπ + 0 +
⎨
⎬
nπ ⎭
L ⎩ nπ
=
2 ⎧ 20L
30L ⎫
(−1) n +
⎨
⎬
L ⎩ nπ
nπ ⎭
=
2 ⎧10L
⎫
[2(−1) n + 3]⎬
⎨
L ⎩ nπ
⎭
=
20
[ 2( −1) n + 3]
nπ
Bn =
20
[3 + 2.( −1) n ]
nπ
On substituting the value of Bn in equation (15), we get,
⎡ c 2n 2π2 ⎤
⎥t
L ⎥⎦
∞
−⎢
20
⎡ nπ ⎤
[3 + 2(−1) n ]. sin ⎢ ⎥ x.e ⎢⎣
ut(x, t) =
nπ
⎣L⎦
n =1
∑
⇒
20
ut(x, t) =
nπ
∞
⎡ nπ ⎤
[3 + 2(−1) ]. sin ⎢ ⎥ x.e
⎣L⎦
n =1
∑
n
⎡ c2n 2π 2 ⎤
−⎢
⎥t
⎣⎢ L ⎦⎥
... (17)
Hence, substituting equations (11) and (17) in equation (9), we get,
20
⎛ 50 ⎞
u ( x, t ) = ⎜ ⎟ x + 30 +
nπ
⎝ L⎠
∞
⎡ nπ ⎤
[3 + 2( −1) n ]. sin ⎢ ⎥ x.e
⎣L⎦
n =1
∑
⎡ c 2 n2π 2 ⎤
−⎢
⎥t
⎣⎢ L ⎦⎥
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.42
MATHEMATICS-II [JNTU-ANANTAPUR]
Q47. A homogeneous rod of length L with insulated
sides has its ends A and B are maintained at
temperatures 50°C and 100°C until steady state
conditions prevail. The temperature at end A is
suddenly raised to 90°C and at the same time
the temperature at B is lowered to 60°C. Show
that the temperature at the midpoint of the rod
remains unaltered for all times regardless of the
material of the rod.
Ans: Given that,
Length of the homogeneous rod = L units
On substituting the values in equation (3), we get,
100 = a(L) + b
⇒
a(L) + 50 = 100
⇒
aL = 100 – 50
⇒
aL = 50
⇒
a=
[Q b = 50]
50
L
Hence, the initial condition can be expressed as,
U(x, 0) = ax + b
Initial temperature at the end A, θ A1 = 50°C
⇒
aL = 100 – b = 100 – 50 = 50
Initial temperature at the end B, θB1 = 100°C
∴
a =
⇒
u(x, 0) =
50
L
Final temperature at the end A, θ A2 = 90°C
Final temperature at the end B, θ B2 = 60°C
To prove that,
Temperature at the midpoint of the rod remains unaltered
for all times regardless of the material of the rod.
We know that,
One dimensional heat conduction equation is given by,
∂u
∂ 2u
= α2 2
∂t
∂x
Under this state as t approaches ∞, the temperature again
attains steady state.
Hence,
u(x, t) ≈ u(x)
... (1)
AP = x
The boundary conditions are,
u(0, t) = 50°C
u(L, t) = 100°C
Prior to the temperature change at the ends A and B,
when t = 0, the heat flow was independent of time.
Hence,
Under steady state condition,
As such that,
∂ 2u
∂x 2
=0
The solution remains same as given by equation (3),
∴ u(0, t) = u(0) = 90°C
and u(L, t) = u(L) = 60°C
∴ ax + b = u
⇒
a(0) + b = 90
∴ b = 90°
And,
∂u
=0
∂t
a(L) + b = 60
∴ Equation (1) reduces to,
The solution of equation (2) is given by,
u = ax + b
Using the boundary limits,
If x = 0, u = 50
Substituting the values in equation (3), we get,
50 = a(0) + b
... (4)
When the temperature at the ends A and B are changed
suddenly to 90°C and 60°C respectively.
Consider,
The temperature at any point P of the rod = u(x, t) i.e.,
∂ 2u
=0
∂x 2
50
x + 50
L
⇒
aL + 90 = 60
⇒
aL = 60 – 90
... (2)
∴a =
... (3)
∴ b = 50
If x = L, u = 100
Look for the SIA GROU P LOGO
−30
L
Thus, the solution is,
⎛ − 30 ⎞
⎟ x + 90°
u(x, t) = u(x) = ⎜
⎝ L ⎠
... (5)
Assuming that,
⎡ 30 ⎤
u(x, t) = – ⎢ ⎥ x + 90 + U(x, t)
⎣L⎦
on the TITLE COVER before you buy
... (6)
4.43
UNIT-4 (Partial Differential Equations)
Using u(0, t) = 90°, in equation (6), we get,
The solution of equation (13) is of the form,
U(x, t) = (C1 cospx + C2 sinpx)⋅ e − c
Put x = 0 in equation (15),
⎡ 30 ⎤
90 = – ⎢ ⎥ × 0 + 90 + U(0, t)
⎣L⎦
⇒
90 – 90 = U(0, t)
⇒
U(0, t) = 0
... (7)
⇒
U(0, t) = (C1 cos0 + C2 sin0) ⋅ e − c
⇒
0 = C1 e −c
2
⇒
⎡ 30 ⎤
60 = – ⎢ ⎥ × L + 90 + U(L, t)
⎣L⎦
⇒
U(L, t) = 0
Put t = 0 in equation (6),
U(L, t) = (C1 cos(pL) + C2 sin(pL))⋅ e − c
⇒
0 = 0 + C 2 sin( pL ) e −c
... (9)
⇒
⎡ 30 ⎤
⎡ 50 ⎤
U(x, 0) = ⎢ ⎥ x + ⎢ ⎥ x + 50 − 90
L
⎣L⎦
⎣ ⎦
⇒
⎡ 80 ⎤
U(x, 0) = ⎢ ⎥ x − 40
⎣L⎦
2
p 2t
2
p 2t
[Q U(L, t) = 0
∀ t and C1 = 0]
⇒ c2 sin(pL) = 0
⇒ C2 = 0 or sin(pL) = 0
If C2 = 0, then the solution of U(x, t) given by equation
(15) vanishes,
Hence, C2 ≠ 0
∴ sin(pL) = 0
pL = sin–1 (0)
⇒ pL = nπ
∴p=
Equating equation (4) and (9), we get,
⎡ 30 ⎤
⎡ 50 ⎤
⎢ L ⎥ x + 50 = – ⎢ L ⎥ x + 90 + U(x, 0)
⎣ ⎦
⎣ ⎦
∴ C1 = 0
⇒
... (8)
Further at t = 0, i.e., the instant, the temperature is
changed at both the ends, the initial temperature distribution is
that which is given by equation (4).
⇒
p 2t
Put x = L in equation (15), we get,
∴ U(L, t) = 0 for all t
⎡ 30 ⎤
u(x, 0) = – ⎢ ⎥ x + 90 + U(x, 0)
⎣L⎦
... (15)
[Q U(0, t) = 0 ∀ t from equation (13)]
Using u(L, t) = 60, in equation (6), we get,
60 = – 30 + 90 + U(L, t)
p 2t
p 2t
∴ U(0, t)= 0 for all t
⇒
2
2
nπ
L
Thus, on substituting the values of C1, C2 and p in
equation (15), we get,
⎡ nπx ⎤ −c
U(x, t) = C 2 sin ⎢
⎥ ⋅e
⎣ L ⎦
⎡ nπx ⎤
U(x, t) = C 2 sin ⎢
⎥ ⋅e
⎣ L ⎦
... (10)
Further using equation (1), we get,
2 ⎡ nπ ⎤
2
⎢ L⎥ t
⎣ ⎦
⎡ c 2 n 2 π 2t ⎤
−⎢
⎥
2
⎣⎢ L
⎦⎥
... (16)
The most general solution of equation (12) satisfying
the boundary conditions, equation (12) and equation (13) is
given by,
2
∂U
2 ∂ U
=α
∂t
∂x 2
∞
Thus,
⇒
The conditions can be written as,
⎛ nπx ⎞
Bn ⋅ sin ⎜
⎟e
U(x, t) =
⎝ L ⎠
n =1
∑
⎡ c 2 n 2 π 2t ⎤
−⎢
⎥
2
⎢⎣ L
⎥⎦
... (17)
1 ∂U
∂ U
= 2
α ∂t
∂x 2
2
⇒
... (11)
Put t = 0 in equation (17),
∞
U(0, t) = 0 for all t
... (12)
U(L, t) = 0 for all t
... (13)
⎡ 80 ⎤
U(x, 0) = ⎢ ⎥ x − 40
⎣L⎦
⇒
⎡ nπx ⎤
Bn ⋅ sin ⎢
U(x, 0) =
⎥ ⋅e
⎣ L ⎦
n =1
⇒
U(x, 0) =
∑
∞
... (14)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
∑B
n =1
n
⎡ nπx ⎤
⋅ sin ⎢
⎥
⎣ L ⎦
⎡ c 2 n 2 π 2 ×0 ⎤
−⎢
⎥
L2
⎦⎥
⎣⎢
... (18)
SIA GROUP
4.44
MATHEMATICS-II [JNTU-ANANTAPUR]
Equating equations (18) and (14), we get,
∞
∑B
n
n =1
⎡ nπx ⎤ ⎡ 80 ⎤ x − 40
sin ⎢
⎥ =⎢ ⎥
⎣ L ⎦ ⎣L⎦
The value of Bn can be found as,
L
2 ⎡ 80
⎤ ⎡ nπx ⎤
x − 40⎥ sin ⎢
Bn =
⎥ dx
L ⎢⎣ L
⎦ ⎣ L ⎦
0
∫
L
L
2 ⎡ ⎡ 80 ⎤
⎡ nπx ⎤
⎡ nπx ⎤ ⎤⎥
⎢
x
dx
⋅
sin
40
sin
−
=
⎢ L ⎥
⎢ L ⎥ dx
L ⎢ ⎢⎣ L ⎥⎦
⎣
⎦
⎦ ⎥⎦
⎣
0
⎣0
∫
=
∫
L
L
2 ⎡ 80
⎡ nπx ⎤
⎡ nπx ⎤ ⎤⎥
⎢
x sin ⎢
dx
dx
−
40
sin ⎢
⎥
L⎢ L
L ⎦
L ⎥⎦ ⎥
⎣
⎣
0
⎦
⎣ 0
∫
∫
L
⎡ ⎧ ⎡
⎧
⎡ nπx ⎤ ⎤
⎡ nπx ⎤ ⎫
⎡ nπx ⎤ ⎫⎤
− cos ⎢
− cos ⎢
⎢ ⎪ ⎢ − cos ⎢
⎪
⎥
⎪
⎥
⎥
⎥ ⎪⎥
2 80 ⎪
⎣ L ⎦⎥ −
⎣ L ⎦ ⎪ − 40⎪
⎣ L ⎦ ⎪⎥
= ⎢ ⎨x⎢
⎬
⎨
⎬⎥
nπ
nπ
nπ
⎥
L⎢ L ⎪ ⎢
⎪
⎪
⎪⎥
⎢ ⎪ ⎢
⎥
⎪⎭
⎪⎩
L
L
L
⎪⎭⎥
⎦
⎢⎣ ⎩ ⎣
⎦0
∫
L
⎤
⎡ ⎧
⎡ ⎡ nπx ⎤ ⎤ ⎫
sin ⎢
⎥
⎢ ⎪
⎪
⎥
⎢
⎡L
2 80 ⎪ − L
⎡ nπx ⎤ L ⎢ ⎣ L ⎥⎦ ⎥ ⎪
⎡ nπx ⎤ ⎤ ⎥
+
x cos ⎢
40
cos
+
= ⎢⎢ ⎨
⎢
⎥
⎢ L ⎥⎥ ⎥
nπ ⎥ ⎬⎪
L L ⎪ nπ
⎣ L ⎦ nπ ⎢
⎦⎦
⎣
⎣ nπ
⎥
⎢
⎥
⎢
L
⎪
⎪
⎦⎭
⎣
⎥⎦ 0
⎢⎣ ⎩
L
2 ⎡ − 80
⎛ nπx ⎞ 80 L
⎡ nπx ⎤ ⎤
⎡ nπx ⎤ 40 L
x cos⎜
⎟ + 2 2 sin ⎢
= ⎢
⎥ + nπ cos ⎢ L ⎥ ⎥
L ⎣ nπ
L
L
n
π
⎠
⎝
⎦⎦ 0
⎣
⎣
⎦
L
2 ⎡ 40
80 L
⎡ nπx ⎤
⎡ nπx ⎤ ⎤
(−2 x + L) + 2 2 sin ⎢
= ⎢ cos ⎢
⎥
⎥⎥
L ⎣ nπ
n π
⎣ L ⎦
⎣ L ⎦⎦ 0
⎡⎧ 40
80
80
⎫⎤
⎡ nπL ⎤
⎡ nπL ⎤ ⎫ ⎡⎧ 40
⋅ (−2 L + L) + 2 2 L sin ⎢
= ⎢⎨ cos ⎢
⎬ – ⎢⎨ cos 0 ⋅ (−2(0) + L) + 2 2 L sin 0⎬⎥
⎥
⎥
n π
n π
⎭⎦
⎣ L ⎦
⎣ L ⎦ ⎭ ⎣ ⎩ nπ
⎢⎣⎩ nπ
=
2 ⎡⎧ − 40 L
⎫⎤
⎫ ⎧ 40L
cos nπ + 0⎬ − ⎨
+ 0⎬ ⎥
⎨
π
L ⎢⎣⎩ nπ
n
⎭⎦
⎭ ⎩
=
2 ⎡ − 40 L
⎤
(cos nπ + 1) ⎥
⎢
L ⎣ nπ
⎦
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
4.45
UNIT-4 (Partial Differential Equations)
=
−80
[cos nπ + 1]
nπ
=
−80
[(−1) n + 1]
nπ
If n is odd, Bn = 0
If n is even, let Bn = B2m =
⇒
Bn = B2m =
−160
2 mπ
⇒
Bn = B2m =
−80
mπ
∞
∴ U(x, t) =
− 80
−80
[( −1) 2 m + 1]
2m × π
⎡ 2mπx ⎤
⋅e
L ⎥⎦
∑ mπ sin ⎢⎣
m =1
⎡ ( 2 m ) 2 ⋅ π 2 c 2t ⎤
−⎢
⎥
L2
⎦⎥
⎣⎢
∞
⇒
− 80
1
⎡ 2mπx ⎤
⋅e
sin
U(x, t) =
π m=1 m ⎢⎣ L ⎥⎦
∑
[ From equation (17) ]
−4 m 2 π 2 c 2 t
L2
... (19)
Substituting equation (19) in equation (6), we get,
∞
80
1
⎡ 2mπx ⎤
⎡ − 30 ⎤
sin ⎢
u(x, t) = ⎢
⎥ ⋅e
⎥ x + 90 − π
L
m
⎣ L ⎦
⎦
⎣
m =1
∑
−4 m 2 π 2 c 2t
L2
... (20)
Hence the required temperature distribution is,
∞
80
1
⎡ 2mπx ⎤
⎡ − 30 ⎤
u ( x, t ) = ⎢
sin ⎢
⎥e
⎥ x + 90 − π
m
⎣ L ⎦
⎣ L ⎦
m =1
∑
− 4 m 2 π 2 c 2t
L2
Temperature at Midpoint
Put x =
L
in equation (20), we get,
2
⎛ L ⎞ ⎡ − 30 ⎤ ⋅ ⎡ L ⎤ + 90 − 80 [ 0]
u⎜ , t⎟ = ⎢
⎥ ⎢ ⎥
π
⎝ 2 ⎠ ⎣ L ⎦ ⎣2⎦
⇒
[Q sin mπ = 0]
⎛L ⎞
u ⎜ , t ⎟ = – 15 + 90
⎝2 ⎠
⎛L ⎞
∴u ⎜ , t⎟ = 75
⎝2 ⎠
... (21)
⎛L ⎞
From equation (21), it can be observed that, the temperature distribution at the midpoint of the rod i.e., U ⎜ , t ⎟ is
⎝2 ⎠
independent of t and also independent of the material of the rod as it does not depend on C.
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.46
MATHEMATICS-II [JNTU-ANANTAPUR]
4.4 TWO DIMENSIONAL LAPLACE EQUATION
∴ Equation (6) becomes,
UNDER INITIAL AND BOUNDARY CONDITIONS
2
u(x, y) = C2sin
2
∂ u ∂ u
+
= 0 which satisfies the
∂x 2 ∂y 2
Q48. Solve
u(x, y) = C 2C 3 sin
conditions u(0, y) = u(l, y) = u(x, 0) = 0 and
u(x, a) = sin
nπ
πx
.
l
nπx nπy / l
(e
− e − nπy / l )
l
Replacing C2, C3 by bn, we have,
∂ 2u ∂ 2u
+
=0
Ans: The given equation is
∂x 2 ∂y 2
⇒
nπx
(C3 e nπy/l – C3e –nπy/l)
l
u(x, y) = bn sin
... (1)
The three possible solutions of equation (1) are,
u = (C1epx + C2e–px)(C3 cos py + C4 sin py) ... (2)
u = (C1 cos px + C2 sin px)(C3epy + C4e–py) ... (3)
... (4)
u = (C1x + C2)(C3 y + C4)
= 2bn sin
nπy
nπx
. sinh
l
l
... (7)
Putting, y = a, we get,
u(x, a) = sin
Keeping in view the boundary conditions,
The only possible solution is (3),
u(x, y) = (C1 cos px + C2 sin px)(C3epy + C4e–py)
... (5)
nπx nπy/l
(e
– e –nπy/l)
l
⇒
2bn sinh
nπa
nπx
nπx
=2bn sin
. sinh
l
l
l
nπa
= 1 ⇒ bn =
l
1
2 sinh
nπa
l
Since u(0, y) = 0
Hence, equation (7) reduces to,
0 = C1 (C3epy + C4e–py)
C1 = 0
On substituting C1 = 0 in equation (5), we get,
u(x, y)= C2 sin pl (C3epy + C4e–py)
... (6)
nπx
u(x, y) = sin
l
u(l, y) = 0
∴ 0 = C2 sin pl (C3epy + C4e–py)
⇒
sin pl = 0
⇒
pl = nπ
∴
p=
nπ
, n being integer
l
nπ x
l
u=0
x
Figure
Also u(x, 0) = 0
0 = C2 sin
∂ 2u ∂ 2u
+
=0
∂x 2 ∂y 2
nπx
(C3 + C4)
l
C3 = – C 4
Look for the SIA GROU P LOGO
... (1)
The boundary conditions are,
u(0, y) = 0 for all values of y ≥ 0
C3 + C4 = 0
∴
Q49. A rectangular plate with insulated surface is 10
cm wide and so long compared to its width that
it may be considered infinite in length without
introducing an appreciable error. If the temperature of the short edge y = 0 is given by,
Ans: The temperature u(x, y) at any point p(x, y) satisfies the
equation.
P(Q, Y)
0
Which is the required solution.
u = 20x for 0 ≤ x ≤ 5 and u = 20 (10 – x) for 5 ≤ x
≤ 10 and the two long edges x = 0, x = 10 as
well as the other short edge are kept at 0°C.
Find the temperature u at any point.
y
u = sin
nπy ⎤
⎡
⎢ sinh l ⎥
⎥
⎢
⎢ sinh nπa ⎥
⎢⎣
l ⎥⎦
... (2)
u(10, y) = 0 ∀ y ≥ 0
... (3)
u(x, ∞) = 0 in 0 ≤ x ≤ 10
... (4)
0≤ x ≤5
⎧20 x
⎩20(10 − x ), 5 ≤ x ≤ 10
u(x, 0) = ⎨
on the TITLE COVER before you buy
... (5)
4.47
UNIT-4 (Partial Differential Equations)
Thus the three possible solutions of equation (1) are,
u = (C1epx + C2e–px)(C3 cos py + C4 sin py)
= (C1 cos px + C2 sin px)(C3epy + C4e–py)
= (C1 x + C2)(C3y + C4)
The solution (6) cannot satisfies the condition (2), we get u ≠ 0 for x = 0 ∀ y
The solution (8) cannot satisfies the condition (4). Thus the only possible solution is (7),
∴ u(x, y) = (C1 cos px + C2 sin px)(C3epy + C4e–py)
Since u(0, y) = 0
∴ 0 = C1 (C3epy + C4e–py)
C1 = 0
∴ Equation (9) reduces to,
u(x, y)= C2 sin px (C3epy + C4e–py)
∴ u(10, y) = 0
∴
0 = C2 sin 10p (C3epy + C4e–py)
⇒ sin 10p = 0
i.e., 10p = nπ
p=
... (6)
... (7)
... (8)
... (9)
... (10)
nπ
, Where n being an integer.
10
Also, u(x, ∞) = 0
∴ C3 = 0
Hence from equation (10) a solution satisfying equations (2), (3) and (4) is,
nπx –nπy/10
e
10
Replacing C2, C4 by bn. The most general solution is,
u(x, y) = C2C4 sin
∞
u(x, y) =
∑
bn sin
n =1
nπx
⋅e
10
− nπy
10
... (11)
Putting y = 0, we get,
∞
u(x, 0) =
∑b
n
sin
n =1
nπx
10
5
10
10
1⎡
nπx
nπx ⎤
2
nπx
x
dx
dx ⎥
20
sin
20(10 − x) sin
⎢
+
u ( x, 0) sin
dx =
bn =
5 ⎢0
10
10
10 0
10
⎥⎦
5
⎣
∫
∫
∫
5
⎡
⎡
⎛
⎞⎤
⎛
⎜ − sin nπx ⎟ ⎥
⎢ ⎛⎜ − cos nπx ⎞⎟
⎢
⎜
⎜
⎟⎥
⎢ ⎜
⎢
10
10
⎟ − (1)⎜
+ 4⎢(10 − x )⎜ − cos
= 4 ⎢x
⎟
2
⎥
nπ ⎟
⎜
⎜⎜ ⎛⎜ nπ ⎞⎟ ⎟⎟ ⎥
⎢ ⎜⎝
⎢
⎝
⎠
10
⎢⎣
⎢⎣
⎝ ⎝ 10 ⎠ ⎠ ⎥⎦ 0
10
⎛
⎞⎤
nπx ⎞
⎜ − sin nπx ⎟ ⎥
⎟
10 ⎟ − (−1)⎜
10 ⎟ ⎥
⎜
2 ⎟⎥
nπ ⎟
⎛ nπ ⎞ ⎟ ⎥
⎜
⎜
⎟
⎜ ⎝ ⎠ ⎟
10 ⎠
⎝ 10
⎠ ⎥⎦ 5
2
2
⎡ − 50
⎡ 50
nπ ⎛ 10 ⎞
nπ ⎤
nπ ⎛ 10 ⎞
nπ ⎤ 800
nπ
+
sin
cos
⎥
⎢
+ ⎜ ⎟ sin ⎥ = 2 2 sin
⎜ ⎟
=4
+ 4 ⎢ cos
n
n
π
π
2
2
2 ⎝ nπ ⎠
2 ⎥⎦ n π
2
⎝ ⎠
⎥⎦
⎢⎣
⎢⎣ nπ
⎧ 800
( −1) n −1/2 , when n is odd
⎪
= ⎨ n 2π 2
⎪
0, when n is even
⎩
Hence from equation (11), the required solution is,
800
∴ u(x, y)= 2
π
∞
( −1) n −1 / 2
nπx −nπy /10
sin
e
2
10
n
n =1,3,5
∑
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
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4.48
MATHEMATICS-II [JNTU-ANANTAPUR]
Q50. Infinitely long plane uniform plate is bounded
by two parallel edges and an end at right angles
to them. The breadth is π , this end is maintained
at a temperature µ 0 at all points and other edges
are at zero temperature. Determine the
temperature at any point of the plate in the
steady state.
From condition (iii), u = 0 when y → ∞
On substituting C1,p value and condition (iii) in equation
(1), we get,
0 = C2 sin nx C3
C3 = 0
∴ Equation (3) becomes,
Ans: Steady state temperature u(x, y) at a point p(x, y) is,
u(x, y) = C2 sin nx (C4e–ny)
∂ 2u ∂ 2u
+
= 0.
∂x 2 ∂y 2
∴
u(x, y) = Cn sin nxe–ny
... (4)
[Put C2C4 = Cn]
General solution is,
y
∞
u(x, y) =
0o p(x, y) 0o
∑C
n
sin nx e − ny
... (5)
n
sin nx [Put y = 0]
... (6)
n =1
∞
µo
u(x, 0) =
∑C
n =1
The boundary conditions are,
From condition (iv), we get,
(i)
(ii) u(π, y) = 0
u(0, y) = 0
(iii) u(x, ∞) = 0
u(x, 0) = µo Where 0 < x < π
(iv) u(x, 0) = µo
By half-range sine series, we get,
u = 0 when y → ∞, we have,
∞
u(x, y) = (C1 cos px + C2 sin px) (C3 epy + C4e–py)
... (1)
u(x, 0) =
∑b
n
sin nx
... (7)
n =1
Equating equations (6) and (7), we get,
From above condition (i), u = 0 when x = 0.
Cn = bn
On substituting in equation (1), we get,
π
π
2µ o ⎡ − cos nx ⎤
2
bn =
µ o sin nx dx = π ⎢ n ⎥
⎣
⎦0
π0
∫
0 = (C1 (1) + 0) (C3 epy + C4e–py)
0 = C1 (C3
epy
+ C4
e–py)
⇒
=
∴ C1 = 0
Cn = 0
∴ Equation (1), becomes,
u(x, y) = C2 sin px(C3epy + C4e–py)
... (2)
Cn =
From condition (ii), u = 0 when x = π.
On substituting C1 value and condition (ii) in equation
(1), we get,
0 = C2 sin pπ (C3
epy
+ C4
∴ C2n – 1 =
When n is even
4µ o
When n is odd
nπ
4µ o
( 2n − 1) π
On substituting the above value in equation (5), we get,
e–py)
∞
u(x, y) =
sin pπ = 0
pπ = nπ ⇒
−2µ o
[(–1)n –1]
nπ
∑C
2 n −1 sin( 2 n − 1) xe
− ( 2 n −1) y
n =1
p=n
∞
=
∴ u(x, y) = C2 sin nx (C3eny + C4e–ny)
... (3)
Look for the SIA GROU P LOGO
4µ o
∑ (2n − 1)π sin( 2n − 1) xe
− ( 2 n −1) y
n =1
on the TITLE COVER before you buy
UNIT-4 (Partial Differential Equations)
4.49
Q51. Consider a semi-infinite insulated plate in the first quadrant of the xy plane bounded by y = 0, x = 0,
x = l. Determine the temperature u(x, y) at any point of the plate in the steady state given that, u(x, 0)
= f(x), 0 ≤ x ≤ l and u(0, y) = u(l, y) = 0 for y ≥ 0 and u(x, y) = as y → ∞
Ans: The temperature u(x, y) at any point (x, y) satisfies the equation,
∂ 2u ∂ 2u
+
=0
∂x 2 ∂y 2
... (1)
The boundary conditions are,
u(0, y) = 0
for all values of y
... (2)
u(l, y) = 0
for all values of y
... (3)
u(x, ∞) = 0
for all 0 ≤ x ≤ l
... (4)
u(x, 0) = f (x)
for all 0 ≤ x ≤ l
... (5)
The three possible solutions of equation (1) are,
u(x, y) = (A1epx + B1e–px) (C1 cos py + D1 sin py)
... (6)
u(x, y) = (A2x + B2) (C2y + D2)
... (7)
u(x, y) = (A1 cos px + B1 sin px) (C1 epy + D1 e–py)
... (8)
The solution of equations (6) and (7) doesn’t satisfies condition (4),
Thus, the only possible solution is equation (8).
For the condition (4) assume C1 = 0, otherwise u will not be infinite as y → ∞.
∴
Let
⇒
u(x, y) = (A1 cos px + B1 sin px) D1 e–py
A1 D1= A and B1 D1 = B
u(x, y) = (A cos px + B sin px)e–py
... (9)
Using condition (2), i.e.,
For x = 0, i.e.,
u(x, y) = 0
(A cos 0 + B sin 0) e–py = 0
⇒
Ae–py = 0
⇒
A=0
∴
Equation (5) becomes,
(Q From condition (2))
(Q e–py ≠ 0)
u(x, y) = B sin px e–py
... (10)
Using condition (3) in equation (10),
For x = l, we get,
B sin pl e–py = 0
⇒
sin pl = 0
(Q B ≠ 0 and e–py ≠ 0)
⇒
pl = nπ for n = 1, 2, 3,......
⇒
p=
⇒
Equation (10) becomes,
nπ
for n = 1, 2, 3......
l
u(x, y) = Bn sin
nπx − nπy / l
e
, for n = 1, 2, 3.....
l
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
4.50
MATHEMATICS-II [JNTU-ANANTAPUR]
The most general solution of homogeneous equation (1) satisfying conditions (2) (3) and (4) is,
∞
u(x, y) =
∑B
sin
n
n =1
nπx − nπy / l
e
l
... (11)
To find Bn, put y = 0 in above equation,
∞
u(x, 0) =
n
sin
nπx
= f (x), 0 ≤ x ≤ l (given)
l
n
sin
nπx
= f (x)
l
∑B
n =1
∞
=
∑B
n =1
By half range sine series of f (x) in [0, l].
2
Bn =
l
l
∫ f ( x) sin
0
nπx
dx , for n = 1, 2, 3.....
l
... (12)
∴ Equations (11) and (12) are the required solution.
Q52. Derive the solution of the Laplace equation
∂ 2u ∂ 2 y
+
= 0 . Given that, u(0, y) = u(a, y) = u(x, b) = 0 and
∂x 2 ∂y 2
u(x, 0) = u(x, 0).
Ans: Assuming,
Given that,
∂ 2u ∂ 2u
+
=0
∂x 2 ∂y 2
... (1)
u(x, 0) = x(a – x) = x(a – x)
The boundary conditions are,
u(0, y) = 0
u(a, y)= 0
u(x, b)= 0
and u(x, 0) = x(a – x)
The given equation i.e., equation (1) is of the form,
u(x, y) = (c1 cos px + c2 sin px)(c3 epy + c4 e–py )
... (2)
Now, using the boundary condition u(0, y) = 0 in equation (2), we get,
u(0, y) = [c1 cos p(0) + c2 sin p(0)] (c3 epy + c4 e–py) = 0
⇒
[c1 cos 0 + c2 sin 0] (c3 epy + c4 e–py) = 0
⇒
(c1 + 0) (c3 epy + c4 e–py ) = 0
⇒
c1(c3 epy + c4e–py ) = 0
∴
c1 = 0
... (3)
On substituting value of c1 in equation (2), we get,
u(x, y) = (c2 sin px) (c3 epy + c4 e–py)
Look for the SIA GROU P LOGO
... (4)
on the TITLE COVER before you buy
4.51
UNIT-4 (Partial Differential Equations)
Using the boundary condition u(a, y) = 0 in equation (4), we get,
u(a, y) = (c2 sin pa) (c3 epy + c4 e–py) = 0
⇒
c2 sin pa = 0
⇒
sin pa = 0
⇒
pa = nπ
⇒
p=
nπ
a
On substituting the value of p in equation (4), we get,
⎡
⎛ nπ ⎞⎤
x ⎟⎥(c3 e
u(x, y) = ⎢c2 sin⎜
⎝ a ⎠⎦
⎣
⎛ nπx ⎞
⎟( c2 c3 e
= sin ⎜
⎝ a ⎠
nπy
a
nπy
a
+ c4e
+ c2 c4 e
−
−
nπ
y
a )
nπy
a
)
Let, A = c2 c3 and B = c2 c4
⎛ nπx ⎞
⎟( Ae
u(x, y) = sin ⎜
⎝ a ⎠
∴
nπy
a
+ Be
−
nπy
a
)
... (5)
Using the boundary condition u(x, b) = 0 in equation (5).
∴
∴
nπb
nπb ⎞
−
⎛ nπx ⎞ ⎛⎜
⎟ A ⋅ e a + Be a ⎟ = 0
u(x, b) = sin⎜
⎟
⎝ a ⎠ ⎜⎝
⎠
Ae
nπb
a
+ Be
−
nπb
a
=0
... (6)
Let bn be a constant whose value is given by,
bn = − 2( Ae
⇒
Ae
nπb
a
− Be
−
nπb
a
=–
nπb
a
− Be
−
nπb
a )
1
bn
2
... (7)
Using equations (6) and (7), equation (5) can be written as,
[
⎛ nπx ⎞ 1
n π (b − y ) / a
− e − nπ( b − y ) / a
⎟ ⋅ bn e
u(x, y) = sin ⎜
⎝ a ⎠ 2
u(x, y) = bn sin
]
nπx
nπ(b − y )
sin h
a
a
... (8)
But, the most general solution of equation (1) is,
∞
u(x, y) =
∑b
n
n =1
⎛ nπ(b − y ) ⎞
⎛ nπx ⎞
sin⎜
⎟
⎟ sinh⎜
a
a
⎠
⎝
⎠
⎝
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (9)
SIA GROUP
4.52
MATHEMATICS-II [JNTU-ANANTAPUR]
Using the boundary condition,
u(x, 0) = x(a – x) in equation (9).
∞
∴
u(x, 0) =
∞
∑b
⇒
n
sin h
n =1
∑b
n
sin
n =1
nπx
nπb
= x( a − x)
. sinh
a
a
nπb
nπx
= x ( a − x)
sin
a
a
a
Where, bn sin h
nπb
nπx
2
x( a − x ) sin
.dx
=
a
a
a
∫
0
⎛ − sin nπx / a ⎞
⎛
⎞⎤
π
2⎡
2 ⎛ − cos nπx / a ⎞
⎟ + (−2)⎜ cos n x / a ⎟⎥
⎟ − (a − 2 x)⎜⎜
= ⎢(ax − x )⎜
2
3
⎜
⎟
⎟
nπ / a ⎠
a ⎣⎢
⎝
⎝ (nπ / a) ⎠
⎝ (nπ / a) ⎠⎦⎥
⎡
nπa
⎛
⎢
⎜ − cos
2⎢
2
a
= ⎢( a.a − a )⎜
π
a
⎜
n
⎢
⎜
a
⎝
⎢⎣
⎛
⎞
⎜ − sin nπa
⎟
⎜
a
⎟ − ( a − 2.a )⎜
2
⎟
π
⎜ ⎛n ⎞
⎟
⎟
⎜
⎜
⎠
⎝ ⎝ a⎠
−
=
∴
bn sin h
0
⎞⎤
⎟⎥
⎟⎥
⎟⎥
⎟⎥
⎟⎥
⎠⎦
⎛ − sin 0
2⎡
2 ⎛ − cos 0 ⎞
⎟⎟ − ( a − 2.0)⎜
⎢ ( a.0 − ( 0) )⎜⎜
⎜ ( nπ / a ) 2
a ⎢⎣
n
a
/
π
⎝
⎠
⎝
⎞⎤
⎛
⎞
⎟ + ( −2)⎜ cos 0 ⎟ ⎥
⎟
⎜ (n π / a) 3 ⎟⎥
⎠
⎠⎦
⎝
⎤
2⎡
cos nπ ⎤ 2 ⎡
1
− ⎢0 − 0 − 2.
⎢0 − 0 − 2
3⎥
3⎥
a ⎢⎣
(n π / a ) ⎥⎦ a ⎢⎣
( n π / a ) ⎥⎦
=−
=
⎛
⎞
⎜ cos n πa
⎟
⎜
⎟
a
⎟ + ( −2)⎜
3
π
⎜ ⎛n ⎞
⎟
⎟
⎜
⎜
⎟
⎝ ⎝ a⎠
⎠
a
4a 2
4a 2
cos
π
+
n
n 3 π3
n 3 π3
4a 2
[1 − cos nπ]
n 3 π3
nπb
8a 2
= 3 3 when n is odd.
a
n π
(or)
bn sin h
nπb
= 0 when n is even.
a
Therefore, when ‘n’ is odd, equation (9) can be written as,
u(x, y) =
⇒
u(x, y) =
8a 2
π3
8a 2
π
3
∞
nπx
sin h nπ(b − y ) / a
. sin
3
a
n =1,3,5... n sin h nπb / a
∑
∞
sin h(2 n + 1)π(b − y ) / a
∑ (2n + 1)
n =0
3
sin h (2n + 1) πb/a
. sin
(2n + 1)πx
a
Which is the required solution.
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.1
UNIT-5 (Z-Transform)
UNIT
Z-TRANSFORM
5
PART-A
SHORT QUESTIONS WITH SOLUTIONS
Q1.
Find the Z-transform of the following sequences,
(i)
1
n!
(ii)
1
(n + 1)!
(iii)
1
(n + 2)!
θ
(iv) cos h nθ
n
n
(v) a + b
(vi) πn – 4 sin
nπ
– 7a.
4
Ans:
(i)
1
n!
Given that,
un =
1
n!
∞
Z(un) = Σ un . z–n
n =0
⎛1⎞ ∞ 1
⎟= Σ
. z–n
⎝ n! ⎠ n =0 n!
Z⎜
=1+
1 –1 1 –2 1 –3
z +
z +
z + ..
1!
2!
3!
=1+
1
1
1
+ ...
+
2 +
z
2! z
3! z 3
⎛ 1 ⎞ 1/z
⎟=e
⎝ n! ⎠
∴ Z⎜
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.2
MATHEMATICS-II [JNTU-ANANTAPUR]
−2
⎞
z −3
2⎛ z
⎜
z
+
+ ...⎟⎟
= ⎜
3!
⎝ 2!
⎠
1
(n + 1)!
(ii)
Given that,
un =
⎞
⎛ ⎛ 1 ⎞ ⎛ 1 ⎞ 2 ⎛ 1 ⎞3
⎟
⎜ ⎜ ⎟ ⎜ ⎟
⎜ ⎟
1⎟
z⎠ ⎝z⎠
z⎠
⎝
⎝
2⎜
+
+
+ ... − 1 − ⎟
= z ⎜1 +
z
1!
2!
3!
⎟
⎜
⎟
⎜
⎠
⎝
1
( n + 1)!
∞
Z(un) = Σ un . z–n
n =0
1
⎡ 1 ⎤ ∞
–n
Σ
Z⎢
=
⎥ n =0 ( n + 1)! z
(
n
+
1
)!
⎣
⎦
∞
= Σ
n =0
∞
= Σ
n =0
1⎞
2 ⎛ 1/ z
= z ⎜e −1− ⎟
z⎠
⎝
1
1
z–n × × z
z
( n + 1)!
1
z–n × z–1 × z
( n + 1)!
z − ( n +1)
=z Σ
n = 0 ( n + 1)!
∞
⎛1 1
⎝ z 1!
=z⎜ . +
1 1 1 1
⎞
. + 3 . + ...⎟
2
z 2! z 3!
⎠
⎡ 1 ⎛ 1 ⎞ 2 ⎛ 1 ⎞3
⎤
⎢
⎥
⎜ ⎟
⎜ ⎟
⎝ z ⎠ + ⎝ z ⎠ + ..... − 1⎥
= z ⎢1 + z +
⎢ 1!
⎥
2!
3!
⎢
⎥
⎣⎢
⎦⎥
= z 2 (e1 / z − 1 − z −1 )
(iv)
θ
cosh nθ
Given that,
un = cosh nθ
∞
Z(un) = Σ un . z–n
n =0
⎛ e nθ + e − nθ ⎞
⎟
⎟
2
⎝
⎠
Z(cosh nθ) = Z ⎜⎜
=
1 nθ –nθ
Z(e + e )
2
=
1 nθ 1
Z(e ) + Z(e–nθ)
2
2
=
z
z
1
1
×
×
θ +
2 z − e 2 z − e −θ
= z[e1/ z − 1]
⎛ 1 ⎞
⎟⎟ = z(e1/z –1)
∴ Z ⎜⎜
⎝ (n + 1)! ⎠
(iii)
1
(n+ 2)!
Given that,
un =
z ⎤
z
⎡
an
− an
⎢Q Z (e ) = z − e a and Z (e ) = z − e − a ⎥
⎦
⎣
=
z ⎤
1 ⎡ z
+
θ
⎢
z − e −θ ⎥⎦
2 ⎣z −e
=
z
2
⎡ z − e −θ + z − e θ ⎤
⎢
θ
−θ ⎥
⎣ ( z − e )( z − e ) ⎦
=
z
2
⎡ 2 z − e θ − e −θ ⎤
⎥
⎢ 2
θ
−θ
⎣ z − z (e + e ) + 1 ⎦
=
z
2
⎡ 2 z − (e θ + e − θ ) ⎤
⎥
⎢ 2
θ
−θ
⎣ z − z (e + e ) + 1 ⎦
1
(n + 2)!
∞
Z(un) = Σ un . z–n
n =0
∞
⎡ 1 ⎤
1
1
Z⎢
z −n × 2 × z 2
= Σ
⎥
n
=
0
(n + 2)!
z
⎣ (n + 2)!⎦
z − ( n+ 2)
= z Σ
n = 0 ( n + 2)
2
∞
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.3
UNIT-5 (Z-Transform)
Where,
Z(n) = Z(n2n)
z ⎡ 2 z − 2 cosh θ ⎤
=
⎥
⎢
2 ⎣ z 2 − 2 z cosh θ + 1⎦
∞
Z(n2n) =
(Q 2 cosh θ = e θ + e − θ )
z − cosh θ
⎛
⎞
⎟
2
⎝ z − 2 z cosh θ + 1 ⎠
(vi)
=
=
(ii)
Given that,
un = nπ – 4 sin
⎛
⎝
nπ
– 7a
4
nπ
⎞
− 7a ⎟
4
⎠
⎛
⎝
= π.
=
Q2.
Find the Z-transform of the sequences {x(n)}
where x(n) is,
(i) n.2n
(ii) an2 + bn + c.
Ans:
(i)
n.2n
Given that,
x(n) = n2n
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
2
z2
z ( z − 2) 2
=
2z
( z − 2) 2
d
Z ( n P −1 )
dz
P =2
Z(n2) = − z
Z(n2) =
=
z
z sin( π / 4)
z
– 7a .
–4× 2
2
−1
z
z − 2 z cos(π / 4) + 1
( z − 1)
7az
4z
πz
– 2
–
2
z −1
z 2 − 2z + 2
( z − 1)
−1
Where,
nπ ⎞
⎟ – 7a z(1)
4 ⎠
⎛ 1 ⎞
4 z⎜
⎟
zπ
7 az
2⎠
⎝
−
−
=
2
( z − 1)
⎛ 1 ⎞
( z − 1)
z 2 − 2z⎜
⎟ +1
⎝ 2⎠
2 ⎡ 2⎤
1− ⎥
z ⎢⎣
z⎦
Z(n P ) = − z
nπ ⎞
⎛
⎟ – z(7a)
= z(nπ) – z ⎜ 4 sin
4 ⎠
⎝
= π z(n) – 4z ⎜ sin
3
2
⎛2⎞
⎛2⎞
+ 2⎜ ⎟ + 3⎜ ⎟ + .....
z
⎝z⎠
⎝z⎠
an2 + bn + c
Z(n) = a Z(n2) + b Z(n) + c Z(1)
First consider, Z(n2)
From recurrrence formula, we get,
By applying Z-transform on both sides, we get,
Z(un) = z ⎜ nπ − 4 sin
n
2
⎤
2⎡
⎛2⎞ ⎛2⎞
= ⎢1 + 2⎜ ⎟ + 3⎜ ⎟ + .....⎥
z⎢
⎝z⎠ ⎝z⎠
⎥⎦
⎣
z
z
+
z −a z −b
nπ
π n – 4 sin
– 7a
4
∑
2
=
an + bn
Given that,
un = an + bn
Taking Z-transform on both sides,
Z(un) = Z(an + bn)
n
Z(a + bn) = Z(an) + Z(bn)
=
⎛2⎞
n⎜ ⎟
z
n=0 ⎝ ⎠
n=0
∴ Z(cosh nθ) = z ⎜
(v)
∑
∞
n2n z −n =
d
z
dZ ( z − 1) 2
− z[( z − 1) 2 − 2( z − 1) z ]
[( z − 1) 2 ]2
z2 + z
∞
∴
Z(n) =
... (1)
( z − 1)3
∑ nz
−n
n=0
=
1
2
3
+
+
+ .....
z z 2 z3
=
1⎛
2
3
⎞
⎜1 + + 2 + ..... ⎟
z⎝
z z
⎠
1⎛
1⎞
Z(n) = ⎜1 − ⎟
z⎝
z⎠
−2
=
z
( z − 1) 2
Also,
∞
Z(1) =
∑1 z
−n
n=0
∞
=
∑z
n=0
1
n
SIA GROUP
... (2)
5.4
MATHEMATICS-II [JNTU-ANANTAPUR]
= 1+
=
1
1
+ 2 + .....
z z
1
1−
z
z −1
=
1
z
⇒
⇒
Z[cos3n] =
Q4.
Find the Z-transform of,
(i)
2n + 3
(ii)
cos 3n.
Model Paper-III, Q1(j)
Let, un = 2n + 3
By applying Z-transform on both sides, we get,
Z(un)
1 ⎡ z ( z − cos 3) ⎤ 3 ⎡ z ( z − cos 1) ⎤
+
4 ⎢⎣ z 2 − 2 z cos 3 + 1 ⎥⎦ 4 ⎢⎣ z 2 − 2 z cos 1 + 1 ⎥⎦
State and prove the linearity property of
Z-transform.
Model Paper-I, Q1(i)
Ans:
Proof: From the definition of Z-transform,
= Z(2n + 3)
∞
Z(x un + yvn – zwn) = Σ (xun + yvn – zwn) z–n
= Z(2n) + Z(3)
n =0
= 2.Z(n) + 3 Z(1)
∞
∞
= 2.
=
(ii)
z
2z
+
( z − 1) 2
n =0
= xZ(un) + yZ(vn) – zZ(wn)
∞
3z
z −1
−n
Since, Z (un ) = Σ un z
n=0
∴ Z(xun + yvn – zwn) = xZ(un) + yZ(vn) – Z(wn)
cos 3n = 4cos3n – 3cosn
Q5.
⇒ 4cos3n = cos3n + 3cosn
Damping Rule (or) Change of Scale Property
If Z(un) = u (z ) then Z(a–n un) = u (az)
1.
1
3
cos 3n + cos n
4
4
∞
Z(un)=
On applying Z-transform on both sides, we get,
3
⎡1
⎤
Z[cos3n] = Z ⎢ cos 3n + cos n ⎥
4
⎣4
⎦
⇒
As, Z[cos nθ] =
n=0
Z(a–n un) =
∑
n=0
n
z − n = u (z )
∞
a − n u n z −n =
∑ u (az)
−n
n
n=0
= u (az )
1
3
= Z [cos 3n]+ Z [cos n]
4
4
⇒
∑u
∞
⎡1
⎤
⎡3
⎤
= Z ⎢ cos 3n ⎥ + Z ⎢ cos n ⎥
⎣4
⎦
⎣4
⎦
1
3
Z[cos3n] = Z [cos 3n] + Z [cos n]
4
4
State and prove damping rule.
Ans:
cos 3n + 3 cos n
⇒ cos3n =
4
⇒ cos3n =
∞
= x Σ un z–n + y Σ vn z–n – z Σ wn z–n
n =0
n =0
z
+3
z −1
( z − 1) 2
... (3)
z − 2 z cos 1 + 1
2
Linearity Property: If x, y, z are any constants and un, vn, wn
are any distinct functions, then
Z(xun + yvn – zwn) = xZ(un) + yZ(vn) – Z(wn)
Ans:
(i)
z ( z − cos 1)
Z[cos n] =
On substituting equations (2) and (3) in equation (1),
we get,
⎡ z2 + z ⎤
⎡ z ⎤
⎡ z ⎤
+ c⎢
+ b⎢
Z(n) = a ⎢
⎥
3⎥
2⎥
⎢⎣ ( z − 1) ⎥⎦
⎢⎣ ( z − 1) ⎥⎦
⎣ z − 1⎦
Q3.
... (2)
2
For, Z[cos n], θ = 1
... (3)
On substituting the values from equations (1), (2) and
(3) in the given equation, we get,
z ( z − cos 3)
z − 2 z cos 3 + 1
Z[cos 3n] =
Hence, the Z-transform of a– n un is equal to the Z-transform
of un when z is replaced by az.
... (1)
2.
⎛z⎞
If Z(un) = u (z ) , then Z(an un) = u ⎜ ⎟
⎝a⎠
Z ( Z − cos θ)
∞
2
Z − 2 Z cos θ + 1
For, Z[cos 3n], θ = 3
Look for the SIA GROU P LOGO
Z(un)=
∑u
n
z −n
n= 0
on the TITLE COVER before you buy
5.5
UNIT-5 (Z-Transform)
∞
Z(anun) =
∑
n= 0
a n un z −n =
∞
⎛z⎞
un ⎜ ⎟
=
⎝a⎠
n =0
∑
Q6.
Show that, Z(nan) =
−n
∞
⎛a⎞
un ⎜ ⎟
⎝z⎠
n= 0
∑
n
az
.
(z − a) 2
∴
Model Paper-II, Q1(i)
− z.
d
X ( z)
dz
z
nx(n) ←⎯→ – z
d
X(z)
dz
In this property, ROC does not change.
Division by n
Statement
z
az
⎧ f (n) ⎫
−1
If Z{f(n)} = f(z) then z ⎨
⎬ = − z f ( z ) dz
n
⎭
⎩
0
∫
(z - a) 2
Proof
Consider,
z
= U(z)
( z − 1) 2
⎧ f (n) ⎫
z⎨
⎬=
⎩ n ⎭
Applying scaling property,
i.e., If Z(un) = U(z) then Z(an. un) = U(z/a)
∞
∑
n = −∞
∞
z
Where, U(z) =
( z − 1) 2
∑
=
n = −∞
f ( n) − n
z
n
⎤
⎡ z
f (n) ⎢− z n −1dz ⎥
⎥
⎢ 0
⎦
⎣
∫
z/a ⎤
2⎥
⎣ ( z a − 1) ⎦
−1
=− z
z
× a2
a × ( z − a) 2
−1
= − z f ( z )dz
⎡
∞
z
∫ ∑ f ( n) z
∴ Z(n.an) = ⎢
=
−n
= z[(n x (n))]
⎛z⎞
= u⎜ ⎟
⎝a⎠
Let, un = n.an
Z(n) =
∑ n . x ( n) z
n = −∞
Ans: To prove,
Z(nan) =
∞
−1
= −z .
0
−n
dz
n = −∞
z
∫
0
∴ Z(n.an) =
Q7.
az
( z − a) 2
z
Write the properties of multiplication by n and
Model Paper-III, Q1(i)
division by n of Z-transforms.
Ans:
Multiplication by n
∞
X(z) =
∑ x (n ) z
⎧ f (n) ⎫
−1
∴ z⎨
⎬ = − z f ( z )dz
⎩ n ⎭
0
∫
Q8.
−n
n = −∞
Discuss about the initial and final value
theorems of Z-transforms.
OR
State and prove the initial and final value
theorems of Z-transforms.
Ans:
On differentiating on both sides with respect to ‘z’, we get,
(i)
Initial Value Theorem of Z-transform
If Z(f (n)) = f (z), then
∞
⎤
d
d ⎡
( X ( z )) =
x( n ) z − n ⎥
⎢
dz
dz ⎢⎣ n = −∞
⎥⎦
∑
∞
Z[f(n)] = f(z) =
Lt f (z) = f (0)
z →∞
∑ f ( n) z
−n
n =0
⇒
d
( X ( z )) =
dz
∞
= f (0) + f (1)z–1 + f (2)z–2 + ...
∑
d
x( n ) ( z − n )
dz
n = −∞
= f(0) + f (1).
∞
=
∑ x( n ) − n . ( z
− n −1
)
n = −∞
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
As z → ∞,
1
1
+ f (2) 2 +...
z
z
Lt f (z) = f (0) ⇒ f (0) = Lt f (z)
z →0
z →∞
SIA GROUP
5.6
MATHEMATICS-II [JNTU-ANANTAPUR]
f (1) =
f (2) =
f (3) =
(ii)
Lt z[f (z) – f (0)]
z →∞
Lt z2[f (z) – f (0) – f (1) z–1]
z →∞
Lt z3[f (z) – f (0) – f (1) z–1 – f (1)2 z–2] ....
z →∞
Final Value Theorem of Z-Transform
If Zf (n) = f (z), then nLt
f (n) =
→∞
Lt (z – 1)f (z)
z →1
∞
Z[f(n + 1) – f (n)] =
∑ [ f (n + 1) − f (n)] z
–n
n=0
∞
⇒ Z[f (n + 1)] – z[f (n)] =
∑ [ f ( n + 1) − f ( n) ]z
–n
n =0
∞
⇒ Z[f (z) – f (0)] – f(z) =
∑ [ f (n + 1) − f (n)]z
n= 0
∞
⇒ f (z) (z – 1) – z.f (0) =
−n
∑ [ f (n + 1) − f (n)]z
−n
n= 0
Taking the limits as z → 1, we get,
Lt [f (z) (z – 1)] – f (0) =
z →1
∞
∑ [ f (n + 1) − f (n)]
n=0
[
= Lt ( f (1) − f (0)) + ( f (2) − f (1)) + ... + ( f (n + 1) − f (n))]
n →∞
[ f ( n + 1) − f (0)]
= nLt
→∞
∴
f (n) − f (0)
Lt ( z − 1) f ( z ) − f (0) = nLt
→∞
z →1
So, Lt f (n) =
n→∞
Q9.
Lt ( z − 1) f ( z )
z →1
⎧⎪ z 2 + 2 ⎫⎪
Find Z–1 ⎨
⎬.
⎪⎩ (z − 1) 2 ⎪⎭
Model Paper-II, Q1(J)
Ans: Given that,
⎧⎪ z 2 + 2 ⎫⎪
Z −1 ⎨
⎬
⎪⎩ ( z − 1) 2 ⎪⎭
Considering,
B
A
z2 + 2
+
=
2
(
z
−
1
)
(
z
−
1) 2
( z − 1)
z2 + 2
( z − 1)
2
=
... (1)
A( z − 1) + B
( z − 1) 2
⇒ z2 + 2 = A(z – 1) + B
Look for the SIA GROU P LOGO
... (2)
on the TITLE COVER before you buy
5.7
UNIT-5 (Z-Transform)
On substituting Z = 1 in equation (2), we get,
∴B = 3
On comparing constant terms, we get,
2 =–A+B
2 =–A+3
2–3 =–A
–1 = – A
∴ A=1
On substituting the values of A and B in equation (1), we get,
z2 + 2
( z − 1)
2
=
1
3
+
( z − 1)
( z − 1) 2
⎡ z2 + 2 ⎤
⎛ 3
⎛ 1 ⎞
–1
⎟ + Z–1 ⎜⎜
Z–1 ⎢
2
2⎥=Z ⎜
⎝ z −1 ⎠
⎢⎣ ( z − 1) ⎥⎦
⎝ ( z − 1)
⎞
⎟
⎟
⎠
⎛ 1
⎛ 1 ⎞
⎟ + 3Z–1 ⎜⎜
= Z–1 ⎜
2
⎝ z −1 ⎠
⎝ ( z − 1)
⎞
⎟
⎟
⎠
⎤
⎡
n −1
−1 ⎛ 1 ⎞
⎟ = a u ( n − 1) and
⎥
⎢Q Z ⎜
⎝ z−a⎠
⎥
⎢
⎥
⎢
⎞
⎛
1
⎟ = ( n − 1) a n − 2 u ( n − 2) ⎥
⎢ Z −1 ⎜
⎜ ( z − a) 2 ⎟
⎥⎦
⎢⎣
⎠
⎝
= 1n–1 u(n – 1) + 3((n–1)1n–2 u(n – 2))
= u(n – 1) + 3(n – 1) u(n – 2)
Q10. State and prove convolution theorem.
Ans:
Model Paper-I, Q1(j)
Statement
If Z–1[f (z)] = f (α) and Z–1[g(z)] = g(α)
Then,
α
Z–1[f(z).g(z)] = f (α)* g(α) = Σ f(β) g(α – β)
β=0
∞
∞
α =0
α =0
f(z) = Σ f(α) z–α, g(z) = Σ g(α) z–α
Proof :
∴
f(z) g(z)= [f (0) + f (1) z–1 + f (2) z–2
+ ..... + f (α) z–α +.....] × [g(0) + g(1) z–1 + g(2) z–2 +.....+g(α) z–α +.....]
= [f(0) g(0) + f(0) g(1) z–1 + ..... + f(0) g(α) z–α + ..... + f(α) g(0) z–α + f(α) g(1) z–α + f(α) g(α) z–α]
∞
= Σ [f (0) g(α) + f (1) g(α–1) + f (2) g(α–2) +.... + f (α) g(0) ] z–α
α =0
Applying transform on both sides, we get
f(z) g(z) = Z [f (0) g(α) + f (1) g(α–1) +....+ f (α) g(0) ]
= Z–1 [f(z).g(z)] = f(0) g(α) + f(1) g(α – 1) +..... + f(α) g(0)
= f(0) g(α–0) + f(1) g(α–1) +......+f(α) g(α– α)
α
= Σ f(β) g(α–β).
β=0
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.8
MATHEMATICS-II [JNTU-ANANTAPUR]
PART-B
ESSAY QUESTIONS WITH SOLUTIONS
5.1 INTRODUCTION
⎡
⎤
1
Q11. Find Z ⎢
⎥
⎣ (n + 2)(n − 1) ⎦
Ans: Given equation is,
⎞
1
⎟⎟
⎝ (n + 2)(n − 1) ⎠
⎛
To find Z ⎜⎜
Consider,
1
( n + 2)(n − 1)
... (1)
On solving equation (1) using partial fractions, we get,
1
A
B
=
+
( n + 2)(n − 1)
( n + 2) ( n − 1)
1
A(n − 1) + B (n + 2)
=
( n + 2)(n − 1)
( n + 2)(n − 1)
1 = A(n – 1) + B(n + 2)
On substituting n = –2 in equation (2), we get,
1 = –3A ⇒ A = –
... (2)
1
3
On substituting n = 1 in equation (2), we get,
1 = –3B ⇒ B =
∴ Equation (2) changes as
1
3
1⎡ 1
1 ⎤
−
⎢
3 ⎣ (n − 1) n + 2 ⎥⎦
On applying Z-transform, we get,
∞
Z(un) =
−n
n
n =0
=
1 ⎡ 1
1 ⎤
Z⎢
−
3 ⎣ n − 1 n + 2 ⎥⎦
=
1⎧ ⎛ 1 ⎞
⎛ 1 ⎞⎫
⎟ − Z⎜
⎟⎬
⎨Z ⎜
3 ⎩ ⎝ n −1⎠
⎝ n + 2 ⎠⎭
⎛ 1 ⎞
⎟=
⎝ n −1 ⎠
Z⎜
∑u z
... (3)
∞
z −n
n −1
n =0
∑
⎛
⎞
z +2 z +3
= z–1 ⎜ z .1 −
+
+ .....⎟
⎜
⎟
2
3
⎝
⎠
⎛ z ⎞
= z–1 log ⎜ −1 ⎟
⎝z ⎠
Look for the SIA GROU P LOGO
... (4)
on the TITLE COVER before you buy
5.9
UNIT-5 (Z-Transform)
⎛ 1 ⎞
⎟=
Z⎜
⎝n+2⎠
=
∞
z −n
n+2
n =0
∑
1 −αi −1
[e z ]
2i
Consider,
1
1
[ e − α i z −1 ] =
{1 + (e–αi z–1) + (e–αi z–1)2 + ....}
2i
2i
1 z −1 z −2 z −3
+
+
+
+ .....
2
3
4
5
=
⎛ 1 z −1 z −2
⎞
+
+ .....⎟
= ⎜⎜ +
⎟
3
4
⎝2
⎠
⎛ z ⎞
= z2 log ⎜ −1 ⎟ – z
⎝z ⎠
... (5)
On substituting the values of equations (4) and (5) in
equation (3), we get,
⎞ 1 ⎡ −1 ⎛ z ⎞ 2 ⎛ z ⎞ ⎤
1
⎟⎟ = ⎢ z log⎜
⎟ − z log⎜ −1 ⎟ + z ⎥
⎝ z −1 ⎠
⎝z ⎠ ⎦
⎝ (n − 1)(n + 2) ⎠ 3 ⎣
⎛
kz
k =0
∞
Z{sin αk}=
=
1
2i
⎧⎪ 1 − e −αi z −1 − 1 + e αi z −1 ⎫⎪
⎨
⎬
⎪⎩ (1 − e αi z −1 )(1 − e −αi z −1 ) ⎪⎭
1
=
2i
⎧⎪
⎫⎪
(e αi − e − αi ) z −1
⎨
αi
− αi
−1
−2 ⎬
⎪⎩1 − (e + e ) z + z ⎪⎭
[Taking LCM in denominator]
−k
∑ sin(αk ) z
1
1
1 ⎧
⎫
−
⎨
αi −1
− αi −1 ⎬
2i ⎩1 − e z
1− e z ⎭
⎛ e αi − e −αi ⎞ −1
⎟z
⎜
⎟
⎜
2i
⎠
⎝
=
⎛ e αi + e −αi ⎞ −1 1
⎟z +
1 − 2⎜⎜
⎟
2
z2
⎠
⎝
Q12. Find the Z-transform of sin α k, k ≥ 0.
Ans: Given that,
U(k) = sin αk, k ≥ 0
∑U
=
On multiplying and dividing (−e αi + e − αi ) by ‘2’, we get,
Z ⎜⎜
Z{U(k)} =
... (3)
On substituting equations (2) and (3) in equation (1), we get,
⎛ 1 z −1 z −2
⎞
= z2 ⎜ +
+
+ ..... − z ⎟
⎜2
⎟
3
4
⎝
⎠
∞
1
⎤
1 ⎡
⎢
− αi −1 ⎥
2i ⎣ 1 − e z ⎦
⎛ sin α ⎞
⎟
⎜
(sin α ) Z
⎝ Z ⎠
= 2
= 2
Z − 2 cos α + 1
z − 2 cos αz + 1
z2
−k
k =0
∞
⇒ Z{sin αk} =
⎧⎪ e iαk − e −iαk ⎪⎫ − k
⎬z
2i
⎪⎭
k =0
∑ ⎨⎪⎩
∞
=
iαk
∑
∑
− iαk
e
e
z −k −
z −k
i
i
2
2
k =0
k =0
∞
=
∞
=
∞
e iαk − k
e −iαk − k
z −
z
2i
2i
k =0
k =0
∑
1
=
2i
∑
∞
∑
(e iα z −1 ) k −
k =0
∞
∑
1
(e −iα z −1 ) k ... (1)
2i k = 0
Consider,
1
2i
1
2i
∞
∑
∞
∑ (e
z sin α
2
z − 2 cos α + 1
Q13. Find the Z-transforms of:
(i) (n – 1)2
θ.
(ii) e–an cosnθ
Ans:
(i)
(n – 1)2
Z[(n – 1)2] = Z[n2 – 2n + 1]
= Z[n2] – 2Z[n] + Z[1]
Consider,
Z[1] = Z[f(n)]
By the definition of Z-transform,
i α −1 k
∞
z )
Z[f(n)] =
k =0
∑ f (n).z
−n
n =0
1
(e z ) =
{1 + eαi z–1 + (eαi z–1)2 + .... }
2
i
k =0
iα −1 k
=
1 ⎧
1
⎫
⎬
⎨
2i ⎩1 − e αi z −1 ⎭
∞
∴
... (2)
1 ⎫
⎧
2
⎬
⎨Q1 + a + a + ... + ∞ =
1− a ⎭
⎩
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
Z[1] =
∑1.z
−n
n =0
= 1 + z–1 + z–2 + z–3 + ......
= 1+
1 1
1
+ 2 + 3 + .....
z z
z
SIA GROUP
5.10
MATHEMATICS-II [JNTU-ANANTAPUR]
⎡ z − 1 − 2z ⎤
= − z ( z − 1) ⎢
4 ⎥
⎣⎢ ( z − 1) ⎦⎥
1
Z[1] =
1
1−
z
⎡ − z −1 ⎤
z2 + Z
z
−
⎢
=
3
3⎥ =
⎣⎢ ( z − 1) ⎦⎥ ( z − 1)
a
. Here a = 1
Since, Sum of geometric series is
1− r
and r =
1
z
∴ Z[(n – 1)2 = Z[n2] – 2Z[n] + Z[1]
∴
z
Z[1] =
z −1
=
⎡ z ⎤ ⎡ 2 ⎤
+
− 2⎢
⎥
2⎥ ⎢
( z − 1)
⎣⎢ ( z − 1) ⎦⎥ ⎣ z − 1⎦
=
⎤
z ⎡ z +1
z
−
+ 1⎥
⎢
2
( z − 1) ⎣⎢ ( z − 1)
z − 1 ⎦⎥
=
z ⎡ z + 1 − 2( z − 1) + ( z − 1) 2 ⎤
⎥
⎢
z − 1 ⎣⎢
( z − 1) 2
⎦⎥
The recurrence formula of Z-transform is given by,
Z[np] = − z
d
[ z ( n p −1 )]
dz
... (1)
On substituting p = 1 in equation (1), we get,
Z[n] = − z
= −z
= −z
d
[ Z ( n 0 )]
dz
=
d
[Z (1)]
dz
d ⎡ z ⎤
dz ⎢⎣ z − 1 ⎥⎦
=
z ⎤
⎡
⎢Q z (1) = z − 1 ⎥
⎣
⎦
⎡ ( z − 1).1 − z (1 − 0) ⎤
z
= − z⎢
⎥ =
2
2
( z − 1)
⎢⎣
⎥⎦ ( z − 1)
∴
Z[n] =
3
z
( z − 1) 3
z
( z − 1) 3
[z + 1 − 2z + 2 + z
[z
2
]
− 3z + 4 =
2
]
− 2z + 1
z 3 − 3z 2 + 4 z
( z − 1) 3
z 3 − 3z 2 + 4 z
∴ Z[(n – 1) ] =
( z − 1)3
2
(ii)
θ]
[e–an cos nθ
Z[e–an cos nθ] = Z [(e − a ) − n . cos nθ]
Since,
z
( z − 1)
z2 + z
2
Z[cos(θ)] =
On substituting p = 2 in equation (1), we get,
z ( z − cos θ)
z 2 − 2 z cos θ + 1
From the change of scale property or the damping rule,
Z[n2] = − z
d
[ z ( n 2−1 )]
dz
If Z[x(n)] = X(z)
Then,
Z[a– n.x(n)] = X(az)
d
[ z ( n)]
= −z
dZ
Here,
x(n) = cos nθ
d
= −Z
dZ
⎡ Z ⎤
⎢
2⎥
⎣ (Z − 1) ⎦
⎡
z ⎤
⎢Q Z (n) =
⎥
( z − 1) 2 ⎥⎦
⎢⎣
⎡ ( z − 1) 2 .1 − z.2( z − 1).(1) ⎤
= − z⎢
⎥
( z − 1) 4
⎣⎢
⎦⎥
Look for the SIA GROU P LOGO
And
a = e– a
∴ Z[(ea)–n.cos(nθ)] =
=
(e a .z )[(e a .z ) − cos θ]
(e .z ) 2 − 2(e a .z ). cos θ + 1
a
ze a .[ z.e a − cos θ]
z 2 .e 2a − 2 ze a cos θ + 1
on the TITLE COVER before you buy
5.11
UNIT-5 (Z-Transform)
⎧ 1 ⎫
z
⎬ = z log
.
Q14. Show that Z ⎨
n
+
1
⎭
⎩
z −1
Ans: To prove,
z
⎧ 1 ⎫
Z⎨
⎬ = z log
z −1
⎩ n +1⎭
5.2 INVERSE Z-TRANSFORM
⎛ z 2 − 3z ⎞
⎟.
Q15. Find Z–1 ⎜⎜
⎟
⎝ (z + 2)(z − 5) ⎠
Consider L.H.S,
⎧ 1 ⎫
Z⎨
⎬ = z{f(n)}
⎩ n +1⎭
Z{f(n)} =
∑ f (n).z
−n
⎧ 1 ⎫
Z⎨
⎬ =
⎩ n +1⎭
∞
Az( z − 5) − Bz ( z + 2)
z 2 − 3z
=
( z + 2)( z − 5)
( z + 2)( z − 5)
⇒
z2 – 3z = Az2 – 5Az – Bz2 – 2Bz
⇒
z2 – 3z = z2(A – B) – z(5A + 2B)
⇒
A – B =1
... (1)
⇒
5A + 2B = 3
... (2)
From equation (1), we get,
n =0
∴
⇒
On comparing L.H.S and R.H.S, we get,
By the definition of Z-transform,
∞
Az
Bz
z 2 − 3z
−
=
z + 2 z −5
( z + 2)( z − 5)
Ans: Let,
A= 1 + B
⎡ 1 ⎤
∑ ⎢⎣ n + 1⎥⎦.z
On substituting A = 1 + B in equation (2), we get,
−n
5(1 + B) + 2B = 3
⇒ 5 + 5B + 2B = 3
n= 0
⇒
1 −1 1 − 2 1 −3
= 1 + .z + .z + .z + ...
2
4
3
5 + 7B = 3
⇒
7B = 3 – 5
⇒
7B = – 2
∴B =
1
1
1
+ 2 + 3 + ...
= 1+
2 z 3z
4z
⎡1
On substituting B =
2
3
4
⎤
1⎛1 ⎞
1⎛1⎞ 1 ⎛1⎞
f ⎟ + ⎜ ⎟ + ⎜ ⎟ + ...⎥
2⎝ z ⎠
3⎝ z ⎠
4⎝ z⎠
⎥⎦
= z⎢ + ⎜
⎣⎢ z
⎡
⎛ 1 ⎞⎤
= z ⎢− log⎜1 − ⎟⎥
⎝ z ⎠⎦
⎣
⇒
⎡
⎤
x 2 x3 x 4
+
+
+ ... = − log(1 − x )⎥
⎢Q x +
2
3
4
⎢⎣
⎥⎦
A+
2
=1
7
A =1–
∴A =
5
2
=
7
7
5
7
⎡ z 2 − 3z ⎤
⎡ 5 ⎛ z ⎞ 2 ⎛ z ⎞⎤
⎟+ ⎜
⎟⎥
⎥ = z–1 ⎢ ⎜
∴ z–1 ⎢
⎣ 7 ⎝ z + 2 ⎠ 7 ⎝ z − 5 ⎠⎦
⎣⎢ ( z + 2)( z − 5) ⎦⎥
⎡ ⎛ z ⎞⎤
⎟⎥
= z ⎢log⎜
⎣ ⎝ z −1 ⎠⎦
∴
−2
in equation (1), we get,
7
A – B =1
⇒
⎡
⎛ z − 1 ⎞⎤
⎟⎥
= z ⎢− log⎜
⎝ z ⎠⎦
⎣
−2
7
⎛ z ⎞
⎧ 1 ⎫
⎟
Z⎨
⎬ = z log ⎜
⎩ n +1⎭
⎝ z −1 ⎠
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
=
⎛ z ⎞
5 –1 ⎛ z ⎞
2
⎟ + z–1 ⎜
⎟
z ⎜
7
⎝ z+2⎠ 7
⎝ z −5⎠
=
5
2
(–2)n + (5)n
7
7
=
1
[5(–2)n + 2(5)n]
7
SIA GROUP
5.12
MATHEMATICS-II [JNTU-ANANTAPUR]
Q16. Find the inverse Z-transform of
Ans:
3z 2 + z
.
(5z − 1)(5z − 2)
⎛ −8⎞
⎛ 11 ⎞
⎟z
⎜
⎜ ⎟z
⎝ 5 ⎠
⎝5⎠
x(z) =
+
2⎞
1
⎞
⎛
⎛
5⎜ z − ⎟ 5⎜ z − ⎟
5⎠
5⎠
⎝
⎝
Model Paper-III, Q10
Given that,
Taking inverse Z-transform on both sides, we get,
3z 2 + z
x(z) =
(5 z − 1)(5 z − 2)
n
x(n) =
z (3z + 1)
x(z) = (5 z − 1)(5 z − 2)
⎡ 1 ⎤
Q17. Find Z–1 ⎢
when | z | > 5. Determine the
3⎥
⎣ (z − 5) ⎦
Model Paper-I, Q10
region of convergence.
Ans: Given that,
(3 z + 1)
x( z)
= (5 z − 1)(5 z − 2)
z
By partial fractions, we get,
(3 z + 1)
A
B
x( z)
= (5 z − 1)(5 z − 2) = (5z − 1) + (5 z − 2)
z
A(5z – 2) + B(5z – 1) = (3z + 1)
⎡ 1 ⎤
Z–1 ⎢
3⎥
⎣⎢ ( z − 5) ⎦⎥
... (1)
... (2)
Let, U(z) =
2
in equation (2), we get,
On substituting z =
5
⎛ ⎛2⎞
⎞ ⎛ ⎛ ⎛ 2 ⎞⎞ ⎞ ⎛ ⎛ 2 ⎞ ⎞
A⎜⎜ 5⎜ ⎟ − 2 ⎟⎟ + ⎜⎜ B⎜⎜ 5⎜ ⎟ ⎟⎟ − 1⎟⎟ = ⎜⎜ 3⎜ ⎟ + 1⎟⎟
⎝ ⎝5⎠
⎠ ⎝ ⎝ ⎝ 5 ⎠⎠ ⎠ ⎝ ⎝ 5 ⎠ ⎠
=
11
∴B =
5
⇒
⇒
... (3)
(1 − 5 z −1 ) 3
= z–3(1 – 5z–1)–3
( n + 1)(n + 2)
⎤
(5 z −1 ) n + ......⎥
2
⎦
⎧ ( n + 1)(n + 2) n ⎫ − n
5 ⎬z
= z–3 ⎨
2
⎩
⎭
On substituting ‘n’ as n – 3, we get,
⎡ [( n − 3) + 1][( n − 3) + 2] ( n −3) ⎤ − ( n −3) − 3
5
U(z) = ⎢
⎥z
2
⎣
⎦
... (4)
On substituting equations (3) and (4) into equation (1),
we get,
⎡ ( n − 3 + 1)(n − 3 + 2) n −3 ⎤ −n
5 ⎥z
U(z) = ⎢
2
⎣
⎦
11
−8
x( z)
5
5
=
+
( 5 z − 2)
(5z − 1)
z
=
z −3
3
⎧ ( n + 1)( n + 2) n ⎫ −n −3
5 ⎬z
=⎨
2
⎭
⎩
−8
5
⎛ −8⎞
⎟
⎜
⎝ 5 ⎠
1⎞
⎛
5⎜ z − ⎟
5⎠
⎝
⎛ 5⎞
z 3 ⎜1 − ⎟
⎝ z⎠
⎫
⎧ ( n + 1)(n + 2)
(5 z −1 ) n ⎬
= z–3 ⎨
2
⎭
⎩
⎛ 11 ⎞
5A + 5 ⎜ ⎟ = 3
⎝5⎠
5A + 11 = 3
5A = 3 – 11
∴A =
1
+ ...... +
On comparing the coefficients of z, we get,
5A + 5B = 3
⇒
=
[
⎛ 6+5⎞
⎟
B=⎜
⎝ 5 ⎠
⇒
1
( z − 5) 3
= z–3 1 + 3(5 z −1 ) + 6(5z −1 ) + 9(5 z −1 ) 3
⎛6 ⎞
A(0) + B(1) = ⎜ +1⎟
⎝5 ⎠
⇒
n
−8 ⎛ 1 ⎞
11 ⎛ 2 ⎞
⎜ ⎟ u (n) + ⎜ ⎟ u (n)
25 ⎝ 5 ⎠
25 ⎝ 5 ⎠
+
⎛ 11 ⎞
⎜ ⎟
⎝5⎠
2⎞
⎛
5⎜ z − ⎟
5⎠
⎝
Look for the SIA GROU P LOGO
⎡ ( n − 2)(n − 1) n −3 ⎤ −n
5 ⎥z
=⎢
2
⎣
⎦
⎧ (n − 1)(n − 2) n −3
⎪
5 , when n ≥ 3
∴ Z [U(z)] = [U(n)] = ⎨
2
⎪⎩ 0,
when n < 3
–1
on the TITLE COVER before you buy
5.13
UNIT-5 (Z-Transform)
Q18. Find the inverse Z-transform of
z2 − z
.
(z + 1)2
Ans: Given function is,
⎡ z2 − z ⎤
⎡ z ⎤
–1
–1
⇒ Z–1 ⎢
2 ⎥ = Z ⎢ z − ( −1) ⎥ + 2Z
⎣
⎦
⎣⎢ (z + 1) ⎦⎥
= (–1)n + 2n (–1)n
z ( z − 1)
= (–1)n (1 + 2n)
( z + 1) 2
⇒
z2 − z
( z + 1)
2
=
⎧⎪
⎫⎪
z3
Q19. Find Z–1 ⎨
⎬.
⎪⎩ (z + 1)(z − 1)2 ⎪⎭
z ( z − 1)
( z + 1)
z −1
2
B
A
... (1)
+
( z + 1)
( z + 1)
( z + 1) 2
⇒ z – 1 = A (z + 1) + B
⇒ z–1=Az+A+B
Equating the coefficients of z on both sides, we get,
Consider,
2
=
Ans: Given function is,
z3
( z + 1)( z − 1) 2
∴A =1
Let, F(z) =
Equating the constants on both sides, we get,
A + B = –1
On substituting A = 1 in the above equation, we get,
1 + B = –1
B = –1 –1 = – 2
∴
⇒
z −1
( z + 1)
2
=
−2
1
+
z + 1 (z + 1)2
=
1
2
–
z + 1 (z + 1)2
F(z) =
(z + 1)2
⇒
z2 − z
( z + 1)
2
z2
( z + 1)( z − 1)
⎞
⎟
⎟
⎠
⇒
⇒
⎡ z2 − z ⎤
–1
⎢
2⎥ =Z
⎢⎣ (z + 1) ⎥⎦
⇒
⎡ z
2z ⎤
−
⎢
2⎥
⎣⎢ z + 1 ( z + 1) ⎦⎥
⎡ z ⎤
= Z–1 ⎢
⎥ – z–1
⎣ z +1 ⎦
= Z–1
2
A
B
C
+
+
z + 1 z − 1 ( z − 1) 2
=
z2 = A[(z –1)2] + B[(z + 1) (z –1)] + C [z +1]
z2 = A [(z – 1)2] + B [(z2 –1)] + C [z + 1]
... (2)
On substituting z = 1 in equation (2), we get,
12 = 0 + 0 + C [1 +1]
2z
z
–
z + 1 ( z + 1) 2
By applying inverse z-transforms on both sides, we get,
Z–1
... (1)
Taking partial fractions of the R.H.S term of equation (1),
we get,
⎛ 1
2
−
= z ⎜⎜
+
1
z
+
(
z
1) 2
⎝
=
z3
( z + 1)( z − 1) 2
F (z)
z2
=
z
( z + 1)( z − 1) 2
On multiplying by ‘z’ on both sides, we get,
z ( z − 1)
( z + 1)( z − 1) 2
⎧⎪
⎫⎪
z3
Z–1{F(z)} = Z–1 ⎨
2⎬
⎪⎩ ( z + 1) ( z − 1) ⎪⎭
On substituting A = 1 and B = –2 is equation (1), we get,
z −1
z3
Z–1[F(z)] = ?
∴B = −2
( z + 1) 2
⎡ (−1) z ⎤
⎢
2⎥
⎣⎢ ( z − (−1)) ⎦⎥
∴C =
⎡ z ⎤
⎢
2⎥
⎣⎢ (z + 1) ⎦⎥
⎡ az ⎤
⎡ z ⎤
n
= an and Z–1 ⎢
As, Z–1 ⎢
2 ⎥ = n.a
⎥
⎢⎣ ( z − a) ⎦⎥
⎣z −a⎦
1
2
On substituting z = – 1 in equation (2), we get,
(–1)2 = A [(–1–1)2] + 0 + 0
⎡ 2z ⎤
⎢
2⎥
⎣ ( z + 1) ⎦
⎡ z ⎤
⎢ z +1⎥ – 2Z–1
⎣
⎦
1 = 2C
⇒
1 = 4A
∴A =
1
4
Comparing the coefficients of z2 on both sides of equation (2), we get,
1 =A+B
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⇒
B = 1 –A
SIA GROUP
5.14
MATHEMATICS-II [JNTU-ANANTAPUR]
⇒
B=1–
On substituting z = – 3 in equation (3), we get,
1 = A(– 3 + 8) + B( – 3 + 3)
⇒
1 = A(5) + 0
1
4
3
4
∴B =
1
5
On substituting z = – 8 in equation (3), we get,
1 = A(– 8 + 8) + B(– 8 + 3)
⇒
1 = A(0) + B(– 5)
⇒ A =
∴
F ( z)
1/ 4 3/ 4
1/ 2
+
+
=
z + 1 z − 1 ( z − 1) 2
z
⇒
F (z) 1 ⎡ 1 ⎤ 3 ⎡ 1 ⎤ 1 ⎡ 1 ⎤
+
+ ⎢
= ⎢
⎥
z
4 ⎣ z + 1 ⎥⎦ 4 ⎢⎣ z − 1⎥⎦ 2 ⎢⎣ ( z − 1) 2 ⎥⎦
⇒
On substituting A =
⇒
1⎡ z ⎤ 3⎡ z ⎤ 1⎡ z ⎤
+
+ ⎢
F(z) = ⎢
⎥ ... (3)
4 ⎣ z + 1 ⎥⎦ 4 ⎢⎣ z − 1⎥⎦ 2 ⎢⎣ ( z − 1) 2 ⎥⎦
−1
5
1 −1 ⎡ z ⎤ 3 −1 ⎡ z ⎤ 1 −1 ⎡ z ⎤
Z–1 [F(z)] = Z ⎢
⎥ + Z ⎢ z − 1⎥ + 2 Z ⎢
2⎥
4
⎣ z + 1⎦ 4
⎦
⎣
⎣⎢ ( z − 1) ⎦⎥
⎛ −1 ⎞
⎛1⎞
⎜ ⎟
⎜ ⎟
F (z)
⎝5⎠ + ⎝ 5 ⎠
=
( z + 3) ( z + 8)
z
⎤
⎡
⎡ z ⎤
⎡ z ⎤
−1 ⎡ z ⎤
n
= −a n , Z −1 ⎢
= a n , Z −1 ⎢
=
⎥
⎢Q Z ⎢
⎥
⎥
⎥
2
⎣z + a⎦
⎣z − a⎦
⎥⎦
⎢⎣
⎣ ( z − 1) ⎦
(−1) n 3 n
+ +
4
4 2
1 z
1 z
.
− .
5 z +3 5 z +8
Applying inverse Z-transform on both sides of above
equation, we get,
⇒
F(z) =
z
1 z ⎤
−1 ⎡ 1
− .
Z–1|F(z)| = z ⎢ .
⎥
⎣5 z + 3 5 z + 8 ⎦
⎧⎪
⎫⎪ (−1) n 3 n
z3
+ +
–1 ⎨
⎬=
∴Z
4
4 2
⎪⎩ ( z + 1)( z − 1) 2 ⎪⎭
⇒
z
⎫
⎧
⎬.
Q20. Evaluate Z–1 ⎨ 2
⎩ z + 11z + 24 ⎭
1 −1 ⎡ z ⎤ 1 −1 ⎡ z ⎤
z
⎡
⎤
z −1 ⎢ 2
⎥ = 5 .z ⎢⎣ z + 3 ⎥⎦ − 5 .z ⎢⎣ z + 8 ⎥⎦
⎣ z + 11z + 24 ⎦
=
Ans: Given function is,
z
⎫
⎧
Z −1 ⎨ 2
⎬
⎩ z + 11z + 24 ⎭
z
=
z + 11z + 24
Let,
⇒
∴
z
1
[( −3) n − ( −8) n ]
5
z
⎡
⎤ 1
n
n
z −1 ⎢ 2
⎥ = [( −3) − ( −8) ]
⎣ z + 11z + 24 ⎦ 5
2
z + 11z + 24
1
F ( z)
=
z
+
z + 8)
(
3
)(
z
Q21. Find the inverse Z-transform of
... (1)
1
A
B
+
=
( z + 3)( z + 8) ( z + 3) ( z + 8)
1= A(z + 8) + B(z + 3)
f(z) =
... (2)
... (3)
Look for the SIA GROU P LOGO
z
2
z + 8z + 15
Ans: Given function is,
Applying partial fractions on R.H.S, we get,
⇒
1
1
.( −3) n − .(−8) n
5
5
⎧ z ⎫
z
⎬ = an]
⇒ z–1 ⎨
z−a
⎩z − a⎭
[Q z[an] =
2
F(z) =
... (4)
On substituting equation (4) in equation (1), we get,
1
3 n 1
n
= ( −1) + (1) + n
4
4
2
Consider,
1
−1
,B=
in equation (2), we get,
5
5
⎛ −1 ⎞
⎛1⎞
⎜ ⎟
⎜ ⎟
1
⎝5⎠ + ⎝ 5 ⎠
=
( z + 3)( z + 8) ( z + 3) ( z + 8)
Applying inverse Z-transform on equation (3), we get,
=
B=
⇒
z
z + 8 z + 15
2
1
f (z)
= 2
z
z + 8 z + 15
on the TITLE COVER before you buy
.
5.15
UNIT-5 (Z-Transform)
⇒
1
f (z)
= 2
z
z + 5 z + 3 z + 15
Q22. Find the inverse Z-transform of
⇒
1
f (z)
=
( z + 5)( z + 3)
z
Ans: Let, F(z) =
z2 + z
(z − 1)2
.
z2 + z
( z − 1) 2
Taking partial fractions of the R.H.S term, we get,
1
A
B
+
=
( z + 5)( z + 3) z + 5 z + 3
⇒
1= A[z + 3] + B[z + 5]
F(z) =
⇒
z +1
F (z )
=
z
( z − 1) 2
On substituting z = – 3, we get,
1 = A[0] + B[– 3 + 5]
⇒
( z − 1) 2
Taking partial fractions of R.H.S term, we get,
A
B
z +1
+
2 =
z − 1 ( z − 1) 2
( z − 1)
1 = 2B
∴B =
z ( z + 1)
⇒
1
2
⇒
On substituting z = – 5, we get,
⇒
1 = A[– 5 + 3] + B[0]
⇒
1 = – 2A
z + 1 = A[z – 1] + B
... (1)
On substituting z = 1, we get,
⇒
1 + 1 = A[0] + B
∴B = 2
Comparing the constants of equation (1), we get,
1
∴A = −
2
–A+B = 1
A =B–1
−1
1
f (z)
2 + 2
∴
=
z +5 z +3
z
⇒
⇒
A =2–1
∴A=1
−1 ⎛ 1 ⎞ 1 ⎛ 1 ⎞
f (z)
=
⎜
⎟+ ⎜
⎟
2 ⎝ z + 5⎠ 2 ⎝ z + 3⎠
z
∴
F ( z)
1
2
+
=
z
z − 1 ( z − 1) 2
−1 ⎛ z ⎞ 1 ⎛ z ⎞
⎜
⎟+ ⎜
⎟
2 ⎝ z + 5⎠ 2 ⎝ z + 3⎠
⇒
F(z) =
f(z) =
1 ⎛ z ⎞⎤
−1 ⎡ − 1 ⎛ z ⎞
∴ Z–1[f(z)] = z ⎢ ⎜
⎟+ ⎜
⎟⎥
⎝
⎠
+
2
5
2 ⎝ z + 3⎠ ⎦
z
⎣
⇒
⇒
− 1 −1 ⎛ z ⎞ 1 −1 ⎛ z ⎞
z
⎤
⎡
z ⎜
⎟+ z ⎜
⎟
=
Z–1 ⎢ 2
⎥
⎝ z + 5⎠ 2
⎝ z + 3⎠
2
⎣ z + 8 z + 15 ⎦
=
1
−1
( −5) n + ( −3) n
2
2
⎡ −1 ⎛ z ⎞
n⎤
⎟ = (−a) ⎥
⎢Q z ⎜⎝
z + a⎠
⎣
⎦
=
1
[(−3) n − (−5) n ]
2
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎡ z ⎤
z
+ 2⎢
2⎥
z −1
⎣ ( z − 1) ⎦
Hence,
⎡ z
⎡ z ⎤⎤
Z–1[F(z)]= Z–1 ⎢ z − 1 + 2 ⎢
2 ⎥⎥
⎣ ( z − 1) ⎦ ⎦⎥
⎣⎢
⎡ z ⎤
⎡
z
⎤
−1
−1
=Z ⎢
⎥ + 2Z ⎢ ( z − 1) 2 ⎥
⎣ z − 1⎦
⎦
⎣
⎤
⎡
−1 ⎡ z ⎤
n
⎥
⎢Q Z ⎢ z a ⎥ = (a )
⎣ − ⎦
⎥
⎢
⎥
⎢
⎡ z
n⎤
⎥
⎢ Z −1 ⎢
(
)
n
a
=
⎥
2
⎢⎣
⎦ ⎥⎦
⎣ ( z − a)
= (1)n + 2n (1)n = [1 + 2n](1)n
SIA GROUP
5.16
MATHEMATICS-II [JNTU-ANANTAPUR]
5.3 PROPERTIES – DAMPING RULE – SHIFTING RULE
Q23. Find the Z-transform of the following,
(i) (n + 1)2
(ii) sin(3n + 5).
Ans:
(i)
(n + 1)2
Given function is,
(n + 1)2
By the definition of Z-transform,
Z(n + 1)2 = z(n2 + 2n + 1)
∞
= Σ (n2 + 2n + 1) z–n
n =0
∞
= Σ
n =0
∞
2 Σ
n2 z–n +
n =0
⎡
z ⎤
z
z2 + z
2
=
=
z
z
n
;
(
)
;
(
1
)
⎢Q z (n ) =
⎥
z − 1 ⎥⎦
( z − 1) 3
( z − 1) 2
⎣⎢
∞
nz–n
+ Σ 1
n =0
z–n
By linearity property, we get,
Z(n + 1)2 = Z(n2) + 2Z(n) + Z(1)
z2 + z
z
z 2 + z + 2 z ( z − 1) + z ( z − 1) 3
z
=
+ 2.
+
=
( z − 1) 3
( z − 1) 3
( z − 1) 2 z − 1
∴
(ii)
Z(n +
1) 2
z 2 ( z + 1)
=
( z − 1) 3
sin(3n + 5)
Given that,
un = sin(3n + 5)
By the definition of Z-transforms,
Z[sin (3n + 5)]= Z(sin 3n cos 5 + cos 3n sin 5)
∞
∞
n =0
n =0
[Q sin (A + B) = sin A cos B + cos A sin B]
= cos 5 Σ sin 3n z–n + sin 5 Σ cos 3n z–n
By linearity property,
Z[sin (3n + 5)] = cos 5 Z(sin3n) + sin 5 Z(cos3n)
Using formula of Z(sin nθ), Z(cos nθ), we get,
Z[sin (3n + 5)]= cos 5
z sin 3
z ( z − cos 3)
+ sin 5. 2
z − 2 z cos 3 + 1
z − 2 z cos 3 + 1
2
Q24. Find the linearity property of Z-transforms,
(i)
⎛ nπ π ⎞
+ ⎟
cos ⎜
4⎠
⎝ 2
(ii)
⎛ nπ
⎞
+ θ⎟ .
cosh ⎜
⎝ 2
⎠
Ans:
(i)
Model Paper-I, Q11
⎛ nπ π ⎞
cos ⎜
+ ⎟
⎝ 2 4⎠
Given that,
⎛ nπ π ⎞
+ ⎟
⎝ 2 4⎠
cos ⎜
By the definition of Z-transforms,
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.17
UNIT-5 (Z-Transform)
⎛
nπ
nπ
π
π⎞
⎛ nπ π ⎞ ⎞ ⎛
+ ⎟ ⎟⎟ = Z ⎜ cos cos − sin sin ⎟
2
4
2
4⎠
⎝ 2 4 ⎠⎠ ⎝
[Q cos( A + B) = cos A cos B − sin A sin B ]
Z ⎜⎜ cos⎜
⎝
∞
⎛
⎝
= Σ ⎜ cos
n =0
nπ
nπ
π
π⎞
cos − sin sin ⎟ z–n
2
4
4
4⎠
= cos
π ∞
nπ –n
π ∞
nπ –n
Σ cos
Σ sin
z – sin
z
4 n =0
2
4 n =0
2
= cos
nπ ⎞
π ⎛
π
⎟ – sin
z ⎜ cos
2
4 ⎝
4
⎠
⎛
z ⎜ sin
⎝
nπ ⎞
⎟
2 ⎠
Using Z(sin nα) and Z(cos nα), we get,
⎛ ⎛
⎞
π⎞
π
⎜ z ⎜ z − cos ⎟
⎟
z sin
⎛ ⎛ nπ π ⎞ ⎞
2⎠
⎝
1
⎜
⎟
2
−
+ ⎟ ⎟⎟ =
Z ⎜⎜ cos⎜
⎟
⎜
π
π
⎝ ⎝ 2 4 ⎠⎠
2 ⎜ z 2 − 2 z cos + 1 z 2 − 2 z cos + 1 ⎟
2
2
⎝
⎠
2
1 ⎜⎛ z − z ⎞⎟
=
2
⎟=
⎜ 2
2 ⎝ z +1 z +1⎠
(ii)
⎡Q cos 90° = 0⎤
⎢ sin 90° = 1 ⎥
⎣
⎦
z ( z − 1)
2 ( z 2 + 1)
⎞
⎛ nπ
cosh ⎜
+ θ⎟
⎠
⎝ 2
Given that,
⎛ nπ
⎞
+ θ⎟
⎝ 2
⎠
u n = cosh ⎜
By the definition of Z-transforms,
⎛
nπ
⎛ nπ
⎞
⎞
−
+θ
⎜ e 2 + θ + e ⎜⎝ 2 ⎟⎠ ⎟
⎛
⎛ nπ
⎞⎞
+ θ ⎟ ⎟⎟ = Z ⎜
⎟
Z ⎜⎜ cosh ⎜
2
⎝ 2
⎠⎠
⎟
⎜
⎝
⎟
⎜
⎝
⎠
⎡
e x + e− x ⎤
x
cosh
=
Q
⎥
⎢
2 ⎦
⎣
⎛ nπ ⎞ ⎞
⎛ nπ
− nπ
nπ
−
+θ
∞
∞
⎤
⎜ e 2 + θ + e ⎜⎝ 2 ⎟⎠ ⎟
1 ⎡e θ Σ
2 z − n + e −θ Σ e 2 z − n ⎥
–n
e
⎢
⎟
⎜
= Σ
z =
n= 0
2
n =0 ⎜
⎟
2 ⎣⎢ n =0
⎦⎥
⎟
⎜
⎝
⎠
∞
=
∴
⎛ nπ ⎞
⎜ 2 ⎟
Z ⎜e ⎟ =
⎝
⎠
1
2
⎡ θ ⎛ nπ ⎞ − θ ⎛ − nπ ⎞⎤
⎢e Z ⎜ e 2 ⎟ + e Z ⎜ e 2 ⎟⎥
⎟⎥
⎜
⎟
⎜
⎠⎦
⎝
⎠
⎝
⎣⎢
z
π
z −e2
z ⎤
⎡
n
⎢Q Z (a ) = z − a ⎥
⎣
⎦
⎛ − nπ ⎞
z
Z⎜e 2 ⎟ =
⎟
⎜
z − e −π 2
⎝
⎠
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.18
MATHEMATICS-II [JNTU-ANANTAPUR]
⎛
⎛
⎛ nπ
⎞⎞ 1 ⎜
+ θ ⎟ ⎟⎟ = ⎜ eθ .
Z ⎜⎜ cosh ⎜
⎝ 2
⎠⎠ 2 ⎜
⎝
⎝
=
z
π
z −e2
+e
⎞
⎟ z ⎡ eθ
e− θ ⎤
+
⎟ = ⎢
⎥
π/2
z − e − π / 2 ⎦⎥
⎟ 2 ⎢⎣ z − e
⎠
z
−θ
z −e
−
π
2
z ⎡ eθ ( z − e − π / 2 ) + e − θ ( z − e π / 2 ) ⎤
z ⎡ ze θ − e θ e − π / 2 + ze − θ − e − θ e π / 2 ⎤
⎢
⎢
⎥
⎥
=
2 ⎢⎣ z 2 − z (e π / 2 + e − π / 2 ) + 1 ⎥⎦ 2 ⎢⎣
z 2 − z (e π / 2 + e − π / 2 ) + 1
⎥⎦
⎛π ⎞⎞⎞
⎛
⎛
⎜
⎜ (π 2 −θ ) −⎜⎝ 2 − θ ⎟⎠ ⎟ ⎟
θ
−θ
+
−
+
z
e
e
e
e
(
)
⎜
⎜⎜
⎟⎟ ⎟ z 2 cosh θ − z cosh ⎛⎜ π − θ ⎞⎟
z ⎜
⎝
⎠⎟
⎠
⎝2
⎟=
= ⎜
π
π
⎛
⎞
π
⎞
⎛
2⎜
−
⎟
z 2 − 2 z cosh ⎜ ⎟ + 1
z 2 − z⎜ e 2 + e 2 ⎟ + 1
⎜
⎟
⎜
⎟
⎝2⎠
⎜
⎟
⎝
⎠
⎝
⎠
Q25. Find the Z-transform of,
θ
(i) cos(n + 1)θ
sinh
(ii)
nπ
.
2
Ans:
θ
(i)
cos(n + 1)θ
Given that,
un = cos(n + 1)θ
Applying Z-transform on both sides, we get,
Z[un] = Z[cos(n + 1)θ]
Q cos (A + B) = cos A cos B – sin A sin B
= Z[cos nθ cos θ – sin nθ sin θ]
∞
=
∑[cos nθ cos θ − sin nθ sin θ]Z
−n
∞
Q Z [an ] =
∞
∑
cos nθZ − n − sin θ
n =0
−n
n
n =0
n =0
= cos θ
∑a Z
∞
∑sin nθZ
−n
n =0
= cos θ Z[cos nθ] – sin θ Z[sin nθ]
Since,
Z[cos nα] =
Z [ Z − cos α]
Z − 2 Z cos α + 1
2
Z sin α
and Z[sin nα] =
2
Z − 2 Z cos α + 1
Then,
Z[un] = cos θ.
Z ( Z − cos θ)
Z − 2 Z cos θ + 1
2
– sin θ.
Z sin θ
Z − 2 Z cos θ + 1
2
=
Z 2 cos θ − Z cos 2 θ
Z sin 2 θ
− 2
2
Z − 2Z cos θ + 1 Z − 2Z cos θ + 1
=
Z 2 cos θ − Z cos 2 θ − Z sin 2 θ
Z 2 − 2Z cos θ + 1
=
Z 2 cos θ − Z (cos 2 θ + sin 2 θ)
Z 2 − 2Z cos θ + 1
Look for the SIA GROU P LOGO
Q cos 2 θ + sin 2 θ = 1
on the TITLE COVER before you buy
5.19
UNIT-5 (Z-Transform)
=
∴ Z (cos( n + 1)θ) =
(ii)
Z ( Z cos θ − 1)
Z 2 cos θ − Z
= 2
2
Z − 2 Z cos θ + 1
Z − 2Z cos θ + 1
π
−π
⎡
Z ⎢
e2 −e 2
=
π
π
π π
2 j ⎢⎢ 2
−
−
2
2
2 2
⎣ Z − Ze − Ze + e
Z ( Z cos θ − 1)
Z 2 − 2 Z cos θ + 1
⎡Q e0 = 1⎤
⎣
⎦
⎛ nπ ⎞
sinh ⎜ ⎟
⎝ 2⎠
⎤
⎡
⎛π⎞
⎥
⎢
2 j sin h⎜ ⎟
⎥
Z ⎢
2
⎝ ⎠
⎥
⎢
=
π
π
2j ⎢
⎛
− ⎞
⎥
2
2 ⎟
⎜ 2
⎢ Z − Z ⎜ e + e ⎟ + 1⎥
⎝
⎠ ⎦⎥
⎣⎢
Given that,
⎛ nπ ⎞
un = sinh⎜ ⎟
⎝ 2 ⎠
Applying Z-transform on both sides, we get,
⎛
⎛ nπ ⎞ ⎞
Z(un) = Z ⎜⎜ sinh ⎜ ⎟ ⎟⎟
⎝ 2 ⎠⎠
⎝
− nπ
⎡ nπ
2
⎢e − e 2
= Z⎢
2j
⎢
⎣
⎤
⎥
⎥
⎥
⎦
⎡Q 2 j sinh θ = eθ − e −θ ⎤
⎣
⎦
⎤
⎥
⎥
⎥
⎦
⎤
⎡
⎛π⎞
2 j sin h⎜ ⎟
⎥
⎢
Z
⎝2⎠
⎥
⎢
=
2j ⎢ 2
⎛π⎞ ⎥
⎢ Z − Z 2 cos h⎜ 2 ⎟ + 1⎥
⎝ ⎠ ⎦
⎣
nπ
− nπ ⎤
1 ⎡ 2
2 ⎥
⎢
−
Z
e
e
=
2j ⎢
⎥⎦
⎣
⎡Q 2 cosh θ = eθ + e −θ ⎤
⎣
⎦
nπ
− nπ
1 ⎛⎜ 2 ⎞⎟ 1 ⎛⎜ 2 ⎞⎟
Z
e
Z
e
−
=
⎟
⎟ 2j ⎜
2 j ⎜⎝
⎠
⎝
⎠
⎤
⎡
⎛π⎞
2 j sin h⎜ ⎟
⎥
⎢
Z
⎝2⎠
⎥
⎢
=
2j ⎢ 2
⎛π⎞ ⎥
⎢ Z − 2Z cos h⎜ 2 ⎟ + 1⎥
⎝ ⎠ ⎦
⎣
⎛
⎞
⎛
⎞
Z
1 ⎜ Z ⎟ 1 ⎜
⎟
−
=
⎜
nπ ⎟
− nπ ⎟
j
2 j ⎜⎜
2
⎟
⎜
⎟
⎝Z −e 2 ⎠
⎝ Z −e 2 ⎠
⎛ π⎞
Z sin h ⎜ ⎟
⎝2⎠
=
⎛ π⎞
Z 2 − 2Z cos h⎜ ⎟ + 1
⎝2⎠
Z
⎡
⎤
an
⎢Q Z (e ) =
a ⎥
Z −e ⎥
⎢
Z ⎥
⎢
− an
⎢⎣ Z (e ) = Z − e− a ⎥⎦
⎛
1 ⎜ Z
Z
−
=
⎜
π
−π
2j ⎜
⎝ Z − e2 Z − e 2
⎞
⎟
⎟
⎟
⎠
⎛
⎞
Z ⎜ 1
1 ⎟
−
=
π
−π ⎟
2 j ⎜⎜
⎟
⎝ Z −e2 Z −e 2 ⎠
⎡
⎤
π
π ⎥
⎢
−
Z ⎢ Z −e 2 −Z +e2 ⎥
⎢
⎥
=
π ⎞⎛
−π ⎞
2 j ⎢⎛
⎥
⎢ ⎜⎜ Z − e 2 ⎟⎟ ⎜⎜ Z − e 2 ⎟⎟ ⎥
⎠⎝
⎠ ⎦⎥
⎣⎢ ⎝
⎛ π⎞
Z sin h ⎜ ⎟
⎡
⎛ nπ ⎞ ⎤
⎝2⎠
Z ⎢sin h⎜ ⎟⎥ =
2
⎝ ⎠⎦
⎛ π⎞
⎣
Z 2 − 2Z cos h⎜ ⎟ + 1
⎝2⎠
=
Z (1)
Z 2 − 2Z (0) + 1
⎡Q cos h 90o = 0⎤
⎢
⎥
o
⎣⎢ sin h 90 = 1 ⎦⎥
⎛
Z
⎛ nπ ⎞ ⎞
∴ Z ⎜⎜ sin h ⎜ ⎟ ⎟⎟ = 2
2
⎝ ⎠⎠ Z + 1
⎝
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.20
MATHEMATICS-II [JNTU-ANANTAPUR]
Q26. Show that, Z(n2 an) =
az 2 + a 2 z
(z − a)
3
Z[cos nθ – i sin nθ]=
.
=
z + e iθ
z
×
z − e −iθ
z − e −iθ
=
z ( z − e iθ )
z 2 − z (e iθ + e −iθ ) + 1
=
z ( z − cos θ − i sin θ)
z 2 − 2 z cos θ + 1
Ans: To prove,
Z(n2 an) =
az 2 + a 2 z
( z − a)3
Let,
un = n2 an
z2 + z
2
Z(n ) =
= U(z)
( z − 1) 3
z ( z − cos θ)
z sin θ
–i 2
z − 2 z cos θ + 1
z − 2 z cos θ + 1
Equating real and imaginary parts, we get,
Z[cos nθ – i sin nθ] =
Applying scaling property,
i.e., if Z(un) = U(z) then Z(an. un) = U(z/a)
Here, U(z) =
z2 + z
( z − 1) 3
Z(n2.an) =
=
z ( z − cos θ)
z − 2 z cos θ + 1
2
Q28. State and prove shifting rule.
Ans:
(i)
Shifting un to the Right
Statement
If Z(un) = u(z), then
Z(un – r) = z–r u(z) (k > 0)
Proof
By definition,
(z a ) + (z a )
⎛z ⎞
⎜ − 1⎟
⎝a ⎠
2
∴Z[cos nθ] =
2
∴
z
z − e −iθ
3
z 2 + az
× a3
a 2 ( z − a)3
∞
Z(un – r) =
a ( z + az )
Z(n2.an) =
( z − a )3
2
∴
θ) =
Q27. Show that, Z(cos nθ
z(z − cos θ)
z 2 − 2zcos θ + 1
.
Ans: To prove,
Z(cos nθ) =
Z(1) =
z ( z − cosθo
z − 2az cosθ + 1
2
z
= U(z)
z −1
Consider,
Z(e–inθ) = Z[(e–inθ) .1]
Z[(e–iθ)n .1] is in the form of Z[an. un]
Where, a = e–iθ and un = 1.
Applying scaling property,
i.e., if Z(un) = U(z)
Then Z(an . un) = U(z/a)
∴ Here U(z) = U(1) =
z
z −1
− iθ
∴ Z[(e–iθ)n.1] =
z e
z e −i θ − 1
Look for the SIA GROU P LOGO
∑u
n −k
z −n
n =0
= u0 z–k + u1 z–(k + 1) +.....
= u0 z–k + u1 z–k – 1 + u2 z–k –2 + u3 z–k – 3 +.....
= u0 z–k + u1 z–k z–1 + u2 z–k z–2 + u3z–k z–3 + ...
= z–k [u0 + u1 z– 1 + u2 z–2 + u3 z– 3 +...]
⎤
⎡∞
−k
−n
= z ⎢ u n z ⎥ = z–k u(z)
⎦⎥
⎣⎢ n =0
∑
(ii)
Shifting un to the Left
Statement
If, Z(un) = u(z), then
Z(un + k) = zk[u(z) – u0 – u1 z–1 – u2 z–2 –...... – uk – 1 z–(k – 1)]
Proof
By definition,
∞
Z(un + k) =
∑u
n+k
z −n
n =0
∞
k
= z
∑u
n+k
z −( n + k )
n =0
k −1
⎤
⎡ ∞
k
−n
z
u
z
u n z −n ⎥
−
⎢
n
=
n= 0
⎦⎥
⎣⎢ n =0
∑
∑
∴ Z(un + k) = zk[u(z) – u0 – u1 z–1 – u2 z–2 –....... –uk – 1 z–(k – 1)]
on the TITLE COVER before you buy
5.21
UNIT-5 (Z-Transform)
n
n+3
Q29. Find Z(2.3 + 5.n) deduce Z[2.3
ing shifting theorem.
Ans: To find,
Z(2.3n + 5.n)
n
∴ Z(2.3 + 5.n) = Z(2.3n) + Z(5.n)
= 2Z(3n) + 5Z(n)
= 2.
+5(n + 3)] us-
u ⎤
⎡
z2 ⎢U ( z ) − u 0 − 1 ⎥
u2 = zLt
→∞
z⎦
⎣
⎤
⎡ 2 z 2 + 4 z + 12
2
− 0 − 0⎥
Lt
= z →∞ z ⎢
4
⎦
⎣ ( z − 1)
1
Z
+ 5.
Z −3
( Z − 1) 2
... (1)
⎡ 2⎛
4 12 ⎞ ⎤
⎢ z ⎜2 + + 2 ⎟⎥
z z ⎠⎥
z2 ⎢ ⎝
= z Lt
4
→∞
⎢
⎛ 1⎞ ⎥
⎢ z 4 ⎜1 − ⎟ ⎥
⎝ z ⎠ ⎦⎥
⎣⎢
1
⎞
⎛
⎟
⎜Q Z (a n ) =
−
Z
a
⎟
⎜
⎜
Z ⎟
(
)
=
Z
n
⎟
⎜
( Z − 1) 2 ⎠
⎝
Applying shifting theorem,
2 + 4 z + 12 z
= z Lt
→∞
i.e., Z(un) = U (Z )
⇒ Z(un-k)
= z–k
(1 − 1 z )4
U (Z )
2
=2
From equation (1), we get,
Z
2
= Z(2.3n–(–3) + 5(n – (–3)))
+ 5.
Z + ( −3)
( Z + (−1)) 2
= Z(2.3n+3 + 5(n + 3)).
5.4
Q31. If U(z) =
2z 2 + 5z + 14
find u2 and u3 .
(z − 1) 4
Ans: Given that,
INITIAL AND FINAL VALUE THEOREMS
Q30. If Z(un) =
2z 2 + 4z + 12
(z − 1) 4
find u2.
Model Paper-II, Q10
U(z) =
2 z 2 + 5 z + 14
( z − 1) 4
Ans: Given that,
u0 = zLt
U(z)
→∞
2 z 2 + 4 z + 12
Z(un) =
( z − 1) 4
= z Lt
→∞
From initial value theorem,
u0 = zLt
U(z) = zLt
Z(un)
→∞
→∞
z 2 ( 2 + 5 z + 14 z 2 )
= zLt
→∞
4
4
2 z 2 + 4 z + 12
= zLt
→∞
( z − 1) 4
z (1 − 1 z )
(
4 12 ⎞
⎛
z2 ⎜ 2 + + 2 ⎟
z z ⎠
⎝
= zLt
4
→∞
⎛ 1⎞
z 4 ⎜1 − ⎟
⎝ z⎠
= z Lt
→∞
2 + 4 z + 12 z
(1 − 1 z )
4
2
1 . 2 + 5 z + 14 z
= zLt
→∞
z
(1 − 1 z )
)
4
2
=0
u1 = z Lt
[z[u(z) – u0]]
→∞
2
.
1
=0
z
u 1 = zLt
z[U(z) – u0]
→∞
⎡ 2 z 2 + 4 z + 12
⎤
Lt
z
− 0⎥
= z →∞ ⎢
4
⎦
⎣ ( z − 1)
= zLt
→∞
2 z 2 + 5 z + 14
( z − 1) 4
2 + 4 z + 12 z 2 1
(1 − 1 z )4 . z = 0
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎡ 2 z 2 + 5 z + 14
⎤
−
0
⎥
4
⎣ ( z − 1)
⎦
z⎢
= z Lt
→∞
(
z × z 2 + 5 z + 14 z
= zLt
4
→∞
4
2
2
z (1 − 1 z )
(
)
)
1 2 + 5 z + 14 z 2
Lt
= z →∞
=0
z
(1 − 1 z )4
SIA GROUP
5.22
MATHEMATICS-II [JNTU-ANANTAPUR]
[z2(u(z) – u0 – u1 z–1)]
u2 = zLt
→∞
⎡ 2 ⎛ 2 z 2 + 5 z + 14
⎞⎤
− 0 − 0 ⎟⎟⎥
⎢ z ⎜⎜
= zLt
4
→∞
⎢⎣ ⎝ ( z − 1)
⎠⎥⎦
z × z ( 2 + 5 z + 14 z )
=2
= z Lt
4
4
→∞
2
2
2
z (1 − 1 z )
z3 [U(z) – u0 – u1 z–1 – u2 z–2]
u 3 = z Lt
→∞
⎡ 2 z 2 + 5 z + 14
⎡ 2 z 4 + 5 z 3 + 14 z 2 − 2( z − 1) 4 ⎤
2⎤
−
−
3
3
0
0
Lt
Lt
⎢
⎥
⎥
= z →∞ z
=
z ⎢
4
z 2 ⎦ z →∞ ⎣
z 2 ( z − 1) 4
⎣ ( z − 1)
⎦
⎡ 2 z 4 + 5z 3 + 14 z 2 − 2( z 4 − 4 z 3 + 6 z 2 − 4 z + 1) ⎤
3
Lt
⎥
= z →∞ z ⎢
z 2 ( z − 1) 4
⎣
⎦
⎡ 2 z 4 + 5z 3 + 14 z 2 − 2 z 4 + 8 z 3 − 12 z 2 + 8z − 2 ⎤
⎥
z 2 ( z − 1) 4
⎣
⎦
= z Lt
z3 ⎢
→∞
⎡13 z 3 + 2 z 2 + 8z − 2 ⎤
⎡13 + 2 / z + 8 / z 2 − 2 / z 3 ⎤
3
3
3
Lt
Lt
⎢
⎥
⎥ = 13
= z →∞ z
=
z ×z ⎢
2
4
4
z 2 ( z − 1) 4
⎣
⎣ z × z (1 − 1 / z )
⎦ z →∞
⎦
Q32. If Z {Un} =
z
z
+ 2
, find the Z-transform of Un+2.
z −1 z +1
Ans: Given that,
z
z
+ 2
z −1 z +1
Z{ Un+2} = ?
Z{Un} =
Let,
U(z) =
⇒
U(z) =
⇒
U(z) =
U(z) =
z
z
+
z −1 z 2 +1
z ( z 2 + 1) + z ( z − 1)
( z − 1)( z 2 + 1)
z3 + z + z2 − z
( z − 1)( z 2 + 1)
z3 + z2
( z − 1)( z 2 + 1)
By shifting theorem, we get,
Z{Un+2} = z2 {U(z) –U(0) –U (1) z–1}
By initial value theorem,
If, Z {Un} = U(z), then,
... (1)
lim U(z) = U(0)
z →∞
∴ U(0) = zlim
U(z)
→∞
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.23
UNIT-5 (Z-Transform)
And U (1) = zlim
z [U(z) – U(0)]
→∞
U (0) = zlim
→∞
= zlim
→∞
= zlim
→∞
=
∴
z2 + z3
( z − 1)( z 2 + 1)
z 3 (1 / z + 1)
1 ⎞
⎛ 1 ⎞⎛
z 3 ⎜1 − ⎟⎜1 + 2 ⎟
⎝ z ⎠⎝ z ⎠
1+1/ z
1 ⎞
⎛
(1 − 1 / z )⎜1 + 2 ⎟
⎝ z ⎠
1+ 0
=1
(1 − 0)(1 + 0)
U(0) = 1
U(1)= Lim
z [U (z) – U (0)]
z→∞
⎤
⎡ z 2 + z3
− 1⎥
Lim
= z →∞ z ⎢
2
⎢⎣ ( z − 1)( z + 1) ⎥⎦
[Q U (0) = 1]
⎡ z 2 + z 3 − ( z − 1)( z 2 + 1) ⎤
= Lim
z
⎥
z→∞ ⎢
( z − 1)( z 2 + 1)
⎦⎥
⎣⎢
⎡ z 2 + z 3 − ( z 3 + z − z 2 − 1) ⎤
= Lim
z
⎥
z →∞ ⎢
( z − 1)( z 2 + 1)
⎦⎥
⎣⎢
⎡ z 2 + z 3 − z 3 − z + z 2 + 1⎤
= Lim
z
⎥
z →∞ ⎢
( z − 1)( z 2 + 1)
⎥⎦
⎢⎣
2
⎤
⎡
= Lim
z 2z − z + 1 ⎥
z →∞ ⎢
2
⎣⎢ ( z − 1)( z + 1) ⎦⎥
⎡ 2z 3 − z 2 + z ⎤
= Lim
⎥
z →∞ ⎢
2
⎣⎢ ( z − 1)( z + 1) ⎦⎥
1 1 ⎞
⎛
z3⎜ 2 − + 2 ⎟
z z ⎠
⎝
= Lim
z →∞
1⎞⎛
1 ⎞
3⎛
z ⎜1 − ⎟ ⎜1 + 2 ⎟
z
⎝
⎠⎝ z ⎠
1 1
+
z z2
= Lim
z→∞ ⎛
1⎞⎛
1 ⎞
⎜1 − ⎟ ⎜1 + 2 ⎟
⎝ z⎠⎝ z ⎠
2−
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.24
MATHEMATICS-II [JNTU-ANANTAPUR]
=
2−0+0
(1 − 0)(1 + 0)
=2
∴
U(1) = 2
From equation (1), we get,
⎧⎪ z 2 + z 3
⎫
−1 ⎪
−
−
1
2
z
Z {Un+ 2} = z2 ⎨
⎬
⎪⎭
⎪⎩ ( z − 1)( z 2 + 1)
⎧⎪ z 2 + z 3
2 ⎫⎪
− − 1⎬
= z2 ⎨
2
⎪⎩ ( z − 1)( z + 1) z ⎪⎭
⎧⎪ ( z 2 + z 3 ) × z − 2[( z − 1)( z 2 + 1)] − 1[ z ( z − 1)( z 2 + 1)] ⎫⎪
= z2 ⎨
⎬
⎪⎩
⎪⎭
z ( z − 1)( z 2 + 1)
⎧⎪ ( z 3 + z 4 ) − 2[ z 3 + z − z 2 − 1] − {z ( z 3 + z − z 2 − 1)} ⎫⎪
=z⎨
⎬
⎪⎩
⎪⎭
( z − 1)( z 2 + 1)
⎧⎪ z 3 + z 4 − 2 z 3 − 2 z + 2 z 2 + 2 − z 4 − z 2 + z 3 + z ⎫⎪
⎧⎪ z 2 − z + 2 ⎫⎪
=z⎨
=
z
⎬
⎨
⎬
⎪⎩
⎪⎭
⎪⎩ ( z − 1)( z 2 + 1) ⎪⎭
( z − 1)( z 2 + 1)
∴
Q33. If
⎧⎪ z 2 − z + 2 ⎫⎪
Un+2 = z ⎨
⎬
⎪⎩ ( z − 1)( z 2 + 1) ⎪⎭
3z 2 − 4z + 7
is the Z-transform of f(n), then find f(0), f(1) and f(2).
(z − 1) 3
Ans: Given that,
Z[f(n)] =
3z 2 − 4 z + 7
( z − 1)3
= F(z)
To find,
f(0) = ?
f(1) = ?
f(2) = ?
F(z) =
⇒
⎡ 4 7⎤
z 2 ⎢3 − + 2 ⎥
⎣ z z ⎦
⎡ 1⎤
z 3 ⎢1 − ⎥
⎣ z⎦
3
⎡
4 7
⎢3− +
1⎢
z z2
F(z) = ⎢
z ⎛ 1 ⎞3
⎢ ⎜1 − ⎟
⎣⎢ ⎝ z ⎠
⎤
⎥
⎥
⎥
⎥
⎦⎥
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.25
UNIT-5 (Z-Transform)
From initial value theorem,
F (z )
f(0) = Lim
z →∞
⎡
4 7
⎢3 − + 2
1⎢
z z
= Lim ⎢
3
z →∞ z
1
⎢ ⎛⎜1 − ⎞⎟
⎢⎣ ⎝ z ⎠
⎤
⎥
⎥ =0
⎥
⎥
⎥⎦
∴ f ( 0) = 0
z[ F ( z ) − f (0)]
f(1)= Lim
z→∞
⎡ ⎛
⎢ ⎜ 3− 4 + 7
1⎜
z z2
z ⎢⎢ ⎜
= Lim
3
z→∞
z
⎢ ⎜ ⎛⎜1 − 1 ⎞⎟
⎢⎣ ⎜⎝ ⎝ z ⎠
⎞ ⎤
4 7
⎟ ⎥
3− + 2
⎟ ⎥
3− 0+ 0
z z
⎟ − 0⎥ = Lim
=
=3
3
(1 − 0) 3
⎟ ⎥ z →∞ ⎛ 1 ⎞
⎜1 − ⎟
⎟ ⎥
z⎠
⎠ ⎦
⎝
∴ f (1) = 3
2
−1
f(2) = Lim z [ F ( z ) − f (0) − f (1) z ]
z →∞
⎡ ⎛
⎢ ⎜ 3− 4 + 7
z z2
2 ⎢1 ⎜
= Lim z ⎢ ⎜
3
z→∞
z
⎢ ⎜ ⎛⎜1 − 1 ⎞⎟
⎢⎣ ⎜⎝ ⎝ z ⎠
⎡
4 7
⎢3 − + 2
z ⎢
z z
= Lim
3
z →∞ z ⎢
1
⎢ ⎛⎜1 − ⎞⎟
⎣⎢ ⎝ z ⎠
2
⎡
4 7
⎢3 − + 2
z z
z⎢
3
= Lim
z →∞ ⎢
1
⎢ ⎛⎜1 − ⎞⎟
z⎠
⎢⎣ ⎝
3z − 4 +
= Lim
z →∞
⎤
⎞
⎟
⎥
⎟
−1 ⎥
⎟ − 0 − 3z ⎥
⎟
⎥
⎟
⎥⎦
⎠
⎤
⎥
2
⎥ − 3z
⎥
z
⎥
⎦⎥
⎤
⎥
⎥ − 3z
⎥
⎥
⎥⎦
7
⎡ 1⎤
− 3z ⎢1 − ⎥
z
⎣ z⎦
⎛ 1⎞
⎜1 − ⎟
⎝ z⎠
3
3
2
3
⎡
7
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎤
3z − 4 + − 3z ⎢1 − 3⎜ ⎟ + 3⎜ ⎟ − ⎜ ⎟ ⎥
z
⎝ z ⎠ ⎝ z ⎠ ⎝ z ⎠ ⎥⎦
⎢⎣
Lim
= z→∞
3
⎛ 1⎞
⎜1 − ⎟
z⎠
⎝
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.26
MATHEMATICS-II [JNTU-ANANTAPUR]
7
9
⎛ 1 ⎞
3 z − 4 + − 3 z + 9 − + 3⎜ 2 ⎟
z
z
⎝z ⎠
= Lim
3
z →∞
⎛ 1⎞
⎜1 − ⎟
z⎠
⎝
2 3
5− + 2
z z
5−0+0
Lim
3
→
∞
z
=
1 ⎞ = (1 − 0) 3 = 5
⎛
⎜1 − ⎟
z⎠
⎝
Q35. Using convolution theorem, find the inverse Ztransform of
CONVOLUTION THEOREM
⎧⎪
z2
⎪⎫
⎬ using convolution
Q34. Evaluate Z–1 ⎨
⎪⎩ (z − 1) (z − 3) ⎪⎭
Model Paper-II, Q11
theorem.
⎫⎪
⎧⎪
z2
⎬
Ans: To find Z–1 ⎨
⎪⎩ ( z − 1) ( z − 3) ⎪⎭
Z(aα) =
U(z) =
Considering,
Z–1
= 1α = 1 and
1⎞
⎟
z ⎟⎠
=
1
⎛
1 ⎞⎛
1 ⎞
⎜⎜1 − ⎟⎟⎜⎜1 − ⎟⎟
⎝ 2 z ⎠⎝ 4 z ⎠
1
1
= (2 z − 1)(4 z − 1) = (2 z − 1)(4 z − 1)
2 z.4 z
8z 2
8z 2
(2 z − 1)(4 z − 1)
8z
U ( z)
=
z
−
(
2
1
)(4 z − 1)
z
∑ 1 .3
β
α–β
α
∑3
α–β
β=0
= 3α + 3α–1 + 3α–2 + .... +30
A
B
+
(2 z − 1) (4 z − 1)
4A + 2B = 8
2A + 2B = 0
– –
–
The above expression is in G.P series.
3α +1 (3α +1 − 1)
2.3α +1
... (1)
8z = A(4z – 1) + B(2z – 1)
8z = 4Az – A + 2Bz – B
... (2)
Comparing the coefficients of ‘z’ equation (2), we get,
8 = 4A + 2B
... (3)
Comparing the constant terms in equation (2), we get,
0=–A–B
⇒ A + B =0
Multiplying, the above equation by 2, we get,
⇒ 2A + 2B = 0
... (4)
On solving equations (3) and (4), we get,
⎧⎪ z
z ⎫⎪
⎨
⎬
⎪⎩ z − 1 z − 3 ⎪⎭
β=0
=
1
⎛ 1 1 ⎞⎛ 1
⎜⎜1 −
⎟⎟⎜⎜1 −
⎝ 2 z ⎠⎝ 4
=
α
=
1
⎛ 1 −1 ⎞⎛ 1 −1 ⎞
⎜⎜1 − z ⎟⎟⎜⎜1 − z ⎟⎟
⎝ 2
⎠⎝ 4
⎠
8z
U ( z)
=
(2 z − 1)(4 z − 1)
z
⎧ z ⎫
Z–1 ⎨
⎬ = 3α
⎩ z − 3⎭
=
.
By applying partial fractions, we get,
⎧ z ⎫
⎬
⎨
⎩ z −1⎭
⎧⎪
⎫⎪
z2
∴ Z–1 ⎨
⎬ = Z–1
⎪⎩ ( z − 1) ( z − 3) ⎪⎭
=
U(z) =
⎧ z ⎫
z
(or) Z–1 ⎨
⎬ = aα
z−a
⎩z − a⎭
⎛ 1 −1 ⎞ ⎛ 1 −1 ⎞
⎜1− z ⎟ ⎜1− z ⎟
⎠⎝ 4
⎝ 2
⎠
Ans: Let,
∴ f ( 2) = 5
5.5
1
⎡
a(1 − r ) ⎤
⎢Q S n =
⎥
1 − r ⎥⎦
⎢⎣
n +1
1
= (3α+1 –1)
2
Look for the SIA GROU P LOGO
2A = 8 ⇒ A =
8
=4
2
∴A=4
On substituting the value of A in equation (4), we get,
2(4) + 2B = 0
8 + 2B = 0
on the TITLE COVER before you buy
5.27
UNIT-5 (Z-Transform)
2B = –8
B=
−8
2
∴ B = −4
Equation (1) can now be written as,
8z
U ( z)
=
(2 z − 1)(4 z − 1)
z
4
(− 4)
U ( z)
+
=
2z − 1 4z − 1
z
U(z) =
4z
4z
−
2z − 1 4z − 1
4z
4z
−
⎛
⎞
⎛
1⎞
1
2⎜⎜ z − ⎟⎟ 4⎜⎜ z − ⎟⎟
4⎠
2⎠
⎝
⎝
=
z
2z
−
1
1
z−
z−
2
4
=
⎛
⎞ ⎛
⎞
⎜
⎟ ⎜
⎟
z
z
⎟−⎜
⎟
U(z)= 2⎜
1⎟ ⎜
1⎟
⎜
⎜ z− ⎟ ⎜z− ⎟
2⎠ ⎝
4⎠
⎝
Applying Z–1 on both sides,
⎛
⎞
⎛
⎜
⎟
⎜
z
−1
⎟ − z −1 ⎜ z
Z–1[U(z)]= 2 z ⎜
1⎟
1
⎜
⎜
⎜z− ⎟
⎜z−
2⎠
4
⎝
⎝
n
⎛1⎞ ⎛1⎞
= 2⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟
⎝2⎠ ⎝4⎠
⎞
⎟
⎟
⎟
⎟
⎠
⎤
⎡
−1 ⎛ z ⎞
⎟⎟ = a n ⎥
⎢Q z ⎜⎜
⎝ z−a⎠
⎦⎥
⎣⎢
n
⎡
⎤
n
n
⎢
⎥
⎛1⎞ ⎛1⎞
1
⎥
−1 ⎢
z
= 2⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟
⎢⎛ 1
1 −1 ⎞ ⎥
⎝2⎠ ⎝4⎠
−1 ⎞⎛
⎢ ⎜⎜1 − z ⎟⎟⎜⎜1 − z ⎟⎟ ⎥
⎢⎣ ⎝ 2
⎠⎝ 4
⎠ ⎥⎦
⇒
⎡
⎤
z2
–1
Q36. Using convolution theorem evaluate Z–1 ⎢
⎥ and deduce Z
(z
−
a)(z
−
b)
⎣⎢
⎦⎥
⎡
⎤
z2
⎢
⎥.
⎣⎢ (z − 4)(z − 5) ⎦⎥
⎤
z2
z ⎤
⎡ z
.
⎥ = Z–1 ⎢
⎥
z
a
z
b
(
−
)(
−
)
⎣z −a z −b⎦
⎦
⎣
⎡
Ans:
Z–1 ⎢
⎡ z ⎤
⎥
⎣z − a⎦
Z–1 ⎢
⎡ z ⎤
⎥ = bn
⎣z −b⎦
= an and Z–1 ⎢
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.28
∴
Z–1
MATHEMATICS-II [JNTU-ANANTAPUR]
⎡
⎤
z2
⎢
⎥ = anbn
⎣ ( z − a)( z − b) ⎦
n
= Σ am.bn–m
[From convolution theorem]
m =0
⎛a⎞
Σ ⎜ ⎟
m=0 ⎝ b ⎠
n
=
bn
=
bn ⎢1 + ⎜
⎡
⎢⎣
m
2
3
⎛a⎞
⎛a⎞ ⎛a⎞ ⎛a⎞
⎟ + ⎜ ⎟ + ⎜ ⎟ + ... + ⎜ ⎟
⎝b⎠
⎝b⎠ ⎝b⎠ ⎝b⎠
n
⎤
⎥
⎥⎦
[Which is in a geometric series]
n +1
⎛a⎞
⎜ ⎟ −1
⎝b⎠
n
=b ×
⎛a⎞
⎜ ⎟ −1
⎝b⎠
=
a n +1 − b n +1
a −b
Consider, a = 4 and b = 5,
⎤ 4 n +1 − 5 n +1
z2
⎥=
−1
⎣ ( z − 4)( z − 5) ⎦
⎡
Z–1 ⎢
= 5n+1 – 4n+1
Q37. Using convolution theorem evaluate,
3
3
⎛ z ⎞
z ⎞
⎟ .
⎜
⎟ and deduce Z–1 ⎜
⎝ z − 1⎠
⎝z−a⎠
⎛
Z–1
Ans: Given function is,
⎛ z ⎞
⎜
⎟
⎝ z−a⎠
3
3
⎛ z ⎞
⎟ =?
⎝ z−a⎠
Z–1 ⎜
3
⎛ z ⎞
⎟ =?
⎝ z −1⎠
Z–1 ⎜
Since,
⎛ z ⎞ n
⎟= a
⎝z−a⎠
Z–1 ⎜
2
z ⎞
⎛ z
⎛ z ⎞
.
⎟
⎟ = Z–1 ⎜
⎝ z−a z−a⎠
⎝ z−a⎠
Z–1 ⎜
= an.an
n
= Σ am.an–m
m=0
[From convolution theorem]
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.29
UNIT-5 (Z-Transform)
n
= an Σ am . a–m
m=0
⎡
⎤
Z2
Z ⎤
⎡ Z
.
⎥ = Z −1 ⎢
⎥
−
−
(
3
)(
4
)
Z
Z
⎣⎢
⎣Z −3 Z − 4⎦
⎦⎥
−1
∴ Z ⎢
= an (1 + 1 + 1 + ... +1)
⎡ Z ⎤
−1 ⎡ Z ⎤
= Z −1 ⎢
⎥ * Z ⎢Z − 4⎥
⎣ Z − 3⎦
⎣
⎦
= an(n + 1)
3
⎛ z2
z ⎞
⎛ z ⎞
⎟
.
⎟ = Z–1 ⎜⎜
2
⎟
z
−
a
−
(
)
z
a
z
−
a
⎝
⎠
⎝
⎠
= 3n * 4n
Z–1 ⎜
n
=
= [an (n + 1)] an
∑3
m
.4 n − m
m =0
n
= Σ am (m + 1) an–m
n
m=0
=
∑3
m
.4 n .4 − m
m =0
n
= an Σ (m + 1)
n
n
=4
m=0
n
3
∴
⎛ z ⎞
⎟ = an . 1 (n + 1) (n + 2)
⎝ z−a⎠
2
= 4n
Z–1 ⎜
Consider, a = 1,
m
⎛3⎞
⎜ ⎟
4
m =0 ⎝ ⎠
∑
m
n
⎡ ⎛ 3 ⎞1 ⎛ 3 ⎞ 2
⎛3⎞ ⎤
n
= 4 ⎢1 + ⎜ ⎟ + ⎜ ⎟ + ...... + ⎜ ⎟ ⎥
⎝ 4 ⎠ ⎥⎦
⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠
3
⎛ z ⎞
1
⎟ = 1n . (n + 1) (n + 2)
⎝ z −1⎠
2
Z–1 ⎜
⎡ ⎛ ⎛ 3 ⎞ n +1 ⎞ ⎤
⎢1× ⎜1 − ⎜ ⎟ ⎟ ⎥
⎢ ⎜⎝ ⎝ 4 ⎠ ⎟⎠ ⎥
n
4
⎥
= ⎢
3
⎢
⎥
1−
⎢
⎥
4
⎣⎢
⎦⎥
1
= (n + 1) (n + 2)
2
Q38. State convolution theorem and using it find the
inverse Z-transform of
∑4
m=0
= an(1 + 2 + 3 + .... + (n +1))
3m
z2
.
(z − 3)(z − 4)
⎡
a(1 − r n +1 ) ⎤
⎥
⎢Q Sum of n terms in G . P is, S n =
1− r
⎥⎦
⎣⎢
Ans:
Convolution Theorem
⎡ 3n +1 ⎤
n +1
= 4 ⎢1 − n +1 ⎥
⎢⎣ 4 ⎥⎦
Statement
If Z–1[f(z)] = f(n) and Z–1[g(z)] = g(n)
Then, Z–1[f(z). g(z)] = f(n) * g(n)
n +1
=4 −
3n +1
4 n +1
.4 n +1 = 4 n +1 − 3 n +1
n
=
∑ f (m).g (n − m)
SOLUTION OF DIFFERENCE EQUATION
BY Z-TRANSFORMS
Q39. Solve the difference equation, using Z-transform
y(k + 2) – 5y(k + 1) + 6y(k) = 5n, given y(0) = 0, y(1) = 0.
To find,
⎡
⎤
Z2
Z −1 ⎢
⎥ using convolution theorem.
⎣⎢ ( Z − 3)( Z − 4) ⎦⎥
Since,
∴
5.6
n =0
Ans: Given Z-transform is,
yn+2 – 5yn+1 + 6yn = 5n
⇒
⎡ Z ⎤
n
Z −1 ⎢
⎥ =a
⎣Z −a⎦
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
z
⎡
y ⎤
z2 ⎢ y ( z ) − y 0 − 1 ⎥ –5z [y(z) – y0] + 6y(z) =
z
−
5
z⎦
⎣
Given, y0 = 0, y1 = 0
SIA GROUP
5.30
⇒
⇒
MATHEMATICS-II [JNTU-ANANTAPUR]
Q40. Solve the difference equation using Z-transform
x(n+2) – 3xn+1 – 10xn = 0, given x(0) = 1, x(1) = 0.
Ans: Given that,
z
y(z) [z –5z + 6] =
z −5
2
z
y(z) =
xn+2 – 3xn+1 – 10xn = 0
( z − 5)( z 2 − 5 z + 6)
z
=
( z − 5)( z − 2)( z − 3)
B
A
C
+
+
z − 2 ( z − 3) ( z − 5)
=
A( z − 3) ( z − 5) + B( z − 2) ( z − 5) + C ( z − 2) ( z − 3)
= (z2 – 8z + 15) A + B (z2 – 7z + 10) + C(z2 – 5z + 6)
= z2 (A + B + C) + z (–8A – 7B – 5C) + (15A + 10B + 6C)
On comparing,
A + B + C = 0.
–8A – 7B – 5C = 1.
⇒ –3A – 2B – [5A + 5B + 5C] = 1
⇒ –3A – 2B – 5(A + B + C) = 1
⇒ –3A – 2B = 1.
... (1)
Similarly,
15A + 10B + 6C = 0.
9A + 4B + 6 (A + B + C) = 0.
⇒ 9A + 4B = 0.
... (2)
Multiplying equation (1) by ‘2’ and adding it to equation
(2), we get,
−6 A − 4 B = 2
9 A + 4B = 0
3A =
2
∴A=
3
2
On substituting the value of A in equation (2) we get.
⎛2⎞
9⎜⎜ ⎟⎟ + 4B = 0
⎝3⎠
∴B =
⇒
⇒
⎡
x ⎤
z2 ⎢ x ( z ) − x 0 − 1 ⎥ – 3z[x(z) – x0]–10x(z) = 0
z⎦
⎣
⇒
x(z) [z2 – 3z – 10] – x0[z2 – 3z] – zx1 = 0
−3
2
5
6
On substituting the above values in the equation (1),
we get,
x(z) [z2 – 3z – 10] – [z2 – 3z] – 0 = 0
⇒
x(z) [z2 – 3z – 10] = z2 – 3z
⇒
x(z) =
⇒
x(z) =
2 n 3 n 5 n
(2) − (3) + (5)
3
2
6
Look for the SIA GROU P LOGO
z 2 − 3z
z 2 − 3z − 10
z ( z − 3)
z 2 − 3 z − 10
=
z ( z − 3)
( z − 5)( z + 2)
z−3
x( z)
=
( z − 5) ( z + 2)
z
By partial fraction expansion, we get,
z −3
x( z )
A
B
+
=
=
z
−
(
5
) ( z + 2)
z
( z − 5)
( z + 2)
(z – 3) = A(z + 2) + B(z – 5)
On substituting z = –2, we get,
⇒ –2 –3 = A(–2 + 2) + B(–2 – 5)
⇒
–5 = 0 – 7B
⇒
–5 = –7B
∴B =
5
7
On substituting z = 5 in equation (3), we get,
⇒
⇒
5 – 3 = A(5 + 2) + B(5 – 5)
2 = 7A + 0
2 = 7A
∴A=
∴
2
7
Equation (2) becomes,
2
5
x( z)
7
7
+
=
z
( z − 5)
( z + 2)
2 ⎡ z ⎤ 3 ⎡ z ⎤ 5⎡ z ⎤
–
+
y(z) =
3 ⎢⎣ z − 2 ⎥⎦ 2 ⎢⎣ z − 3 ⎥⎦ 6 ⎢⎣ z − 3 ⎥⎦
y(z ) =
... (1)
Given, x0 = 1 and x1 = 0
2 3
– +C=0
3 2
∴C =
∴
z(xn+2) – 3z(xn+1) – 10z(xn) = 0
18
+ 4B = 0
3
Since, A + B + C = 0
⇒
Applying Z-transform to the above equation,
⇒
x(z) =
2 z
5 z
+
7 z −5
7 z+2
on the TITLE COVER before you buy
... (2)
... (3)
5.31
UNIT-5 (Z-Transform)
Applying
Z–1
on both sides, we get,
On substituting z = 2 in equation (3), we get,
1 = A(2 + 1) + B(2 – 2)
⇒
1 = A(3) + B(0)
On substituting z = –1 in equation (3), we get,
1 = A(–1 + 1) + B(–1 –2)
⇒
1 = A(0) + B(–3)
1 = B (–3)
⇒
2 n 5
5 + (–2)n
7
7
xn =
∴ x( z ) =
2.5 n + 5(−2) n
7
Q43. Solve the difference equation, using Z-transforms
un +2 – un = 2n where u0 = 0, u1 = 1.
Ans:
⇒
B = –1/3
On substituting A, B values in equation (1), we get,
Model Paper-III, Q11
Given that,
⇒
u n + 2 – u n = 2n
1
−1
u( z)
3
=
+ 3
z
( z − 2) ( z + 1)
Applying Z-transform on both sides,
−1
1
u( z )
=
+
z
3( z − 2) 3( z + 1)
Z[un+2 – un] = z[2n]
⇒ Z(un+2) – z(un) = z[2n]
⇒ z2(u(z) – u0 – u1z–1) – u(z) =
⇒
z
z−2
⇒ u(z) =
Applying Z–1 on both sides, we get,
un = Z–1[uz]
z
u(z) {z –1} – u0 (z ) – u1(z) =
z−2
2
2
⇒ u(z) {z2–1} – 0 – z =
z
z−2
=
[Q Given u0 = 0, u1 = 1]
⇒ u(z) {z2 – 1} =
z
+z
z−2
⇒ u(z) {z2 – 1} =
z + z 2 − 2z
z−2
⇒ u(z) {z2 – 1} =
z2 − z
z−2
∴
u(z) =
=
1
⇒
un =
1 n 1
2 – (–1)n
3
3
⇒
un =
1 n
{2 – (–1)n}
3
⎛1⎞
⎜ ⎟
⎝4⎠
n
⎛ ⎛ 1 ⎞n ⎞
⎜
⎟
1
Z(un+1) + Z(un) = z ⎜⎜ ⎜⎜ 4 ⎟⎟ ⎟⎟
4
⎝⎝ ⎠ ⎠
... (1)
Let, Z(un) = u (z)
Then, Z(un+1) = Z( u (z) – u0) and
... (2)
Z(an) =
∴ Z[ u (z) – u0] +
... (3)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
n
⎛ 1⎞
⎜ ⎟ ,
⎝4⎠
Applying Z-transform on both sides, we get,
A( z + 1) + B ( z − 2)
( z − 2)( z + 1)
1 = A(z +1) + B (z – 2)
⇒
1
un+1 + un =
4
Solving equation (1) using partial fractions,
⇒ ( z − 2)( z + 1) =
⎧⎪ z ⎫⎪ 1
⎧ z ⎫
⎨
⎬ – Z–1 ⎨
⎬
⎪⎩ (z − 2) ⎪⎭ 3
⎩ z +1⎭
n ≥ 0, u0 = 0, u1 = 1 using Z-transforms.
Ans: Given that,
z ( z − 1)
( z − 2)( z + 1)( z − 1)
1
A
B
+
=
( z − 2)( z + 1)
z−2
z +1
1 –1
Z
3
1
Q42. Solve the difference equation un+1 + un =
4
z2 − z
( z − 2)( z 2 − 1)
1
u( z)
=
z
( z − 2)( z + 1)
⇒
z
z
–
3( z − 2) 3( z + 1)
z
z−a
1
z
u (z) =
1
4
z−
4
SIA GROUP
5.32
MATHEMATICS-II [JNTU-ANANTAPUR]
Z[yn + 1] = Z[y(z) – y0]
1
1
⇒ z u (z) + u (z) =
1
4
1 − z −1
4
[Q z(u0) = 0]
Z[yn] = y(z)]
Using u0 = 0 and u1 = 1
1
⎛
1⎞
u (z) ⎜⎜ z + ⎟⎟ =
1
4 ⎠ 1 − z −1
⎝
4
u (z) =
1
1
×
1 −1
1
1− z
z+
4
4
1
1
×
=
1 −1
⎛
1 ⎞
1− z
z⎜⎜1 + ⎟⎟
4
4
z⎠
⎝
−1
=
1
z
×
1 −1
1 −1
1+ z
1− z
4
4
1
] – 3z[u(z) – 0] + 2u(z) = 0
z
⇒
z2[u(z) – 0 –
⇒
z2[u(z) –
⇒
(z2 – 3z + 2) u(z) – z = 0
⇒
z(z2 – 3z – 2) u(z) = z2
⇒
(z2 – 3z – 2) u(z) = z
⇒
(z2 – 2z – z – 2) u(z) = z
⇒
[(z – 2) (z – 1)] u(z) = z
⇒
u(z) =
⇒
1
u( z)
=
( z − 1)( z − 2)
z
By partial fractions, we get,
1
] – 3z u(z) + 2u(z) = 0
z
z
( z − 1)( z − 2)
... (1)
Taking partial fractions on equation (1), we get,
2
2
+
u (z) = –
1 −1
1 −1
1− z
1+ z
4
4
1
A
B
+
=
( z − 1)( z − 2)
z −1 z − 2
⇒
Applying Z–1 on both sides, we get,
−2 ⎤
Z–1( u (z)) = Z–1 ⎡⎢
+ Z–1
1 −1 ⎥
⎢1 + z ⎥
⎣ 4
⎦
n
1 = A[1 – 2] + B[0]
– A =1
∴A = −1
On substituting z = 2 in equation (2), we get,
⇒
1 = 0 + B[2 – 1]
n
∴B =1
Hence,
−1
1
u( z)
+
=
z
−
z
−
1
2
z
⎡⎛ 1 ⎞ n ⎛ − 1 ⎞ n ⎤
= 2 ⎢⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ ⎥ , where (n ≥ 0)
⎢⎝ 4 ⎠ ⎝ 4 ⎠ ⎥
⎦
⎣
Q43. Solve the difference equation, un+2 – 3un+1 + 2 un=
0 given u0 = 0 and u1 = 1.
Ans: Given that,
un + 2 – 3un + 1 + 2un = 0 with u0 = 0 and u1 = 1
Applying Z-transform to the above equation, we get,
Z[un + 2] – 3Z[un + 1] + 2Z[un] = 0
Z2[u(z) – u0 –
u1
] – 3z[u(z) – u0] + 2u(z) = 0
z
[Q Z[yn + 2] = z2[y(z) – y0 –
... (2)
On substituting z = 1 in equation (2), we get,
⎡ 2 ⎤
⎢ 1 −1 ⎥
⎢1 − z ⎥
⎣ 4
⎦
−1 ⎤
−1
⎡ ⎛ 1
⎡
1 −1 ⎞ ⎤
−1 ⎞
−1 ⎛
un = Z–1 ⎢− 2⎜⎜1 + z ⎟⎟ ⎥ + z ⎢2⎜⎜1 − z ⎟⎟ ⎥
⎢ ⎝ 4
⎢ ⎝ 4
⎠ ⎥⎦
⎠ ⎥⎦
⎣
⎣
⎛ −1 ⎞
⎛1⎞
= (–2) ⎜⎜ ⎟⎟ + 2⎜⎜ ⎟⎟
⎝ 4 ⎠
⎝4⎠
1 = A[z – 2] + B[z – 1]
⎡ z ⎤ ⎡ z ⎤
⇒ U(z) = − ⎢
⎥+⎢
⎥
⎣ z − 1⎦ ⎣ z − 2 ⎦
Applying inverse Z-transform on equation (2), we get,
−1 ⎡ z ⎤
−1 ⎡ z ⎤
Z–1[U(z)] = − Z ⎢
⎥ + Z ⎢ z − 2⎥
−
1
z
⎦
⎣
⎦
⎣
Un = – (1)n + (n)n
y1
]
z
Look for the SIA GROU P LOGO
... (3)
⎡
−1 ⎛ z ⎞
n⎤
⎢Q Z ⎜⎝ z − a ⎟⎠ = a ⎥
⎣
⎦
∴Un = 2n − (1)n
on the TITLE COVER before you buy
5.33
UNIT-5 (Z-Transform)
n
Q44. Solve un+2 + 4un+1 + 3un = 3 with u0 = 0; u1 = 1 using Z-transforms.
Ans: Given that,
un+2 + 4un+1 + 3un = 3n
Applying Z-transform on both sides,
Z[un+2 + 4un+1 + 3un] = z[3n]
⇒
Z[un+2] + 4Z[un+1] + 3Z[un] = Z[3n]
⇒
Z2[u(z) – u0 – u1z–1] + 4z[u(Z) – u0] + 3u(z) =
⇒
u(z){z2 + 4z + 3} – u0[z2 + 4z] – u1(z) =
⇒
u(z){z2 + 4z + 3} – 0 – z =
⇒
u(z){z2 + 4z + 3} – z =
⇒
u(z){z2 + 4z + 3} =
z
+z
z−3
⇒
u(z){z2 + 4z + 3} =
z + z 2 − 3z
z−3
⇒
u(z){z2 + 4z + 3} =
z2 − 2z
z −3
∴
u(z) =
=
z
z −3
z
z−3
z
z −3
Q Given u0 = 0, u1 = 1
z
z −3
z 2 − 2z
( z − 3)( z 2 + 4 z + 3)
z ( z − 2)
( z − 3)( z + 1)( z + 3)
( z − 2)
u(z)
=
( z − 3)( z + 1)( z + 3)
z
... (1)
Solving equation (1) using partial fractions,
( z − 2)
u( z)
=
z
( z − 3)( z + 1)( z + 3)
=
⇒
A
B
C
+
+
( z − 3) ( z + 1) ( z + 3)
... (2)
A( z + 1)( z + 3) + B( z − 3)( z + 3) + C ( z − 3)( z + 1)
( z − 2)
=
( z − 3)( z + 1)( z + 3)
( z − 3)( z + 1)( z + 3)
⇒
(z – 2) = A(z+ 1)(z+ 3) + B(z – 3)(z + 3) + C(z – 3)(z + 1)
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
... (3)
SIA GROUP
5.34
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting z = 3 in equation (3), we get,
(3 – 2) = A(3 + 1)(3 + 3) + B(3 – 3)(3 + 3) + C(3 – 3)(3 + 1)
1= A(4)(6) + B(0)(6) + C(0)(4)
1= A(24)
∴A=
1
24
On substituting z = –1 in equation (3), we get,
(–1 –2) = A(–1 + 1)(–1 + 3) + B(–4)(2) + C(–4)(0)
⇒
–3 = A(0) + B(–8) + C(0)
⇒
–3 = B(–8)
∴B=
3
8
On substituting C = –3 in equation (3), we get,
–3 – 2 = A(–3 + 1)(–3 + 3) + B(–3 + 1)(–3 + 3) + C(–3 – 3)(–3 + 1)
⇒
– 5 = 0 + 0 + C(–6)(–2)
∴C =
−5
12
On substituting A, B and C values in equation (2), we get,
3
−5
1
u(z)
8
24 +
+ 12
=
z
( z − 3) ( z + 1) ( z + 3)
⇒
u(z) =
1 ⎛ z ⎞ 3⎛ z ⎞ 5 ⎛ z ⎞
⎟
⎟− ⎜
⎟+ ⎜
⎜
24 ⎝ z − 3 ⎠ 8 ⎝ z + 1 ⎠ 12 ⎝ z + 3 ⎠
Applying Z–1 on both sides, we get,
un = Z–1[uz]
=
1 −1 ⎧ z ⎫ 3 −1 ⎧ z ⎫ 5 −1 ⎧ z ⎫
z ⎨
⎬
⎬− z ⎨
⎬+ z ⎨
24
⎩ z + 3⎭
⎩ z + 1 ⎭ 12
⎩ z − 3⎭ 8
=
1 n 3
5
3 + (−1) n − (−3) n
24
8
12
∴ un =
1 n 3
5
3 + (−1) n − (−3) n
24
8
12
Q45. Solve un+2 – 2un+1 + un = 3n + 5 using Z-transforms.
Ans: Given that,
un+2 – 2un+1 + un = 3n + 5
Applying Z-transform on both sides, we get,
Z[un+2] – 2Z[un+1] + Z[un] = Z[3n + 5]
⎛ z ⎞ ⎛ z ⎞
u ⎤
⎡
⎟ + 5⎜
z 2 ⎢u ( z ) − u 0 − 1 ⎥ − 2 z[u ( z ) − u0 ]+ u ( z ) = 3⎜⎜
⎟
2 ⎟
z⎦
⎣
⎝ ( z − 1) ⎠ ⎝ z − 1 ⎠
Look for the SIA GROU P LOGO
⎛
y1 ⎤ ⎞
2 ⎡
⎜Q z [ y n + 2 ] = z ⎢ y ( z ) − y 0 − z ⎥ ⎟
⎣
⎦⎟
⎜
⎜ z [ y + 1] = z [ y − ( z − y 0 )]
⎟
⎜
⎟
=
z
y
y
z
[
]
(
)
n
⎜
⎟
⎝⎜
⎠⎟
on the TITLE COVER before you buy
5.35
UNIT-5 (Z-Transform)
u1
3z + 5 z 2 − 5z
– 2zu(z) + 2zu0 + u(z) =
z
( z − 1) 2
⇒
z2u(z) – z2u0 – z2
⇒
(z2 – 2z + 1)u(z) + (2z – z2)u0 – z2
u1
5z 2 − 2z
=
z
( z − 1) 2
⇒
(z – 1)2 u(z) + (2z – z2)u0 – u1z =
5z 2 − 2z
( z − 1) 2
⇒
(z – 1)2 u(z) =
⇒
u(z) =
5z 2 − 2 z
( z − 1)
5z 2 − 2 z
( z − 1)
+
4
2
[Q (a – b)2 = a2 – 2ab + b2]
− (2 z − z 2 )u0 + u1 z
( z 2 − 2z )
( z − 1)
2
u0 +
u1 z
... (1)
( z − 1) 2
Let,
U1 =
U2 =
U3 =
5z 2 − 2 z
( z − 1) 4
z2 − 2z
( z − 1) 2
z
( z − 1) 2
u0
u1
Thus, the equation (1) can be written as,
u(z) = U1 + U2 + U3
Taking partial fractions to express. U1 and U2 in terms of z(1), z(n), z(n2) and z(n3).
Let, U1 =
... (2)
A( z 3 + 4 z 2 + z ) B( z 2 + z )
Cz
Dz
5z 2 − 2 z
+
+
+
=
4
3
2
4
− 1)
z
(
( z − 1)
( z − 1)
( z − 1)
( z − 1)
⎛
⎜Q z ( a n ) = 1
z−a
⎜
⎜
z
z
, z (n ) =
z (1) =
⎜
1
z
−
(
1) 2
z
−
⎜
⎜
2
3
2
⎜ z ( n 2 ) = z + z , z (n 3 ) = z + 4 z +
( z − 1) 3
( z − 1) 4
⎝⎜
⇒ 5z2 – 2z = A(z3 + 4z2 + z) + B(z2 + z) (z – 1) + Cz(z – 1)2 + Dz(z – 1)3
⇒ 5z2 – 2z = A(z3 + 4z2 + z) + B(z2 + z) + C(z3 – 2z2 + z) + D(z4 – 3z3 + 3z2 – z)
On substituting z = 1 in equation (3), we get,
5(1) – 2 = A(13 + 4(1)2 + 1)
3 = A(6)
∴A =
... (3)
1
2
On comparing the coefficients of z4, we get,
∴D = 0
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
⎞
⎟
⎟
⎟
⎟
⎟
⎟
z⎟
⎠⎟
SIA GROUP
5.36
MATHEMATICS-II [JNTU-ANANTAPUR]
Comparing the coefficients of z3, we get,
A + B + C – 3D = 0
1
+ B+C −0 =0
2
⇒
B+C=
−1
2
... (4)
Similarly, comparing the coefficients of z2, we get,
4A – 2C + 3D = 5
⎛1⎞
4⎜ ⎟ − 2C + 0 = 5
⎝2⎠
2 – 2C = 5
–2C = 5 – 2
∴C =
−3
2
On substituting the value of C in equation (4), we get,
B=
−1 3
+
2 2
∴B =1
5z 2 − 2 z
1 ⎡ z 3 + 4z 2 + z ⎤ z 2 + z 3 ⎡ z ⎤
− ⎢
⎥
⎥+
⎢
4 =
2 ⎢⎣ ( z − 1) 4 ⎦⎥ ( z − 1) 3 2 ⎣⎢ ( z − 1) 2 ⎦⎥
( z − 1)
... (5)
Similarly,
U2 =
z 2 − 2z
( z − 1)
2
=
z
z
−
( z − 1) ( z − 1) 2
... (6)
Applying inverse Z-transform on equation (1), we get,
⎧ 2
u1 ⎫⎪
( z 2 − 2z)
−1 ⎪ 5 z − 2 z
+
+
u
Un = Z–1(u(z)) = z ⎨
⎬
0
⎪⎩ ( z − 1) 4
( z − 1) 2
( z − 1) 2 ⎪⎭
⎧ 5 z 2 − 2 z ⎫ −1 ⎧ z 2 − 2 z ⎫
⎧ z ⎫
u
u + z −1 ⎨
+z ⎨
2⎬ 1
4 ⎬
2⎬ 0
⎩ ( z − 1) ⎭
⎩ ( z − 1) ⎭
⎩ ( z − 1) ⎭
−1
= z ⎨
... (7)
On substituting equation (5) and (6) in equation (7), we get,
3 ⎞
⎛1
U n = ⎜ n 3 + n 2 − n ⎟ + u0 (1 − n) + u1 (n)
2 ⎠
⎝2
⎡ −1 ⎡ z ⎤
⎤
= an
⎢Q z ⎢
⎥
⎥
⎣z −a⎦
⎢
⎥
⎢
⎥
⎡
⎤
az
⎢ z −1 ⎢
= n.a n ⎥
2⎥
⎢⎣
⎥⎦
⎣ ( z − a) ⎦
Q46. Solve un+2 – 6un + 1 + 9un = 0.
Ans: Given that,
un+2 – 6un+1 + 9un = 0
Applying Z-transform to equation (1), we get,
... (1)
Z[un+2] – 6Z[un+1] + 9Z[un] = 0
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.37
UNIT-5 (Z-Transform)
u ⎤
⎡
z 2 ⎢u ( z ) − u 0 − 1 ⎥ − 6 z[u ( z ) − u 0 ] + 9u ( z ) = 0
Z⎦
⎣
∴
⎡
y1 ⎤ ⎤
2 ⎡
⎢Q z[ y n + 2 ] = z ⎢ y ( z ) − y0 − ⎥ ⎥
z ⎦⎥
⎣
⎢
⎢ z[ y n +1 = z ] y ( z − y 0 )
⎥
⎢
⎥
⎢ z[ y n ] = y ( z )
⎥
⎢
⎥
⎣
⎦
(z2 – 6z + 9)u(z) – (z2 – 6z)u0 – zu1 = 0
⇒
(z2 – 6z + 9)u(z) = (z2 – 6z)u0 + zu1
⇒
u(z) =
⇒
⇒
⇒
u(z) =
( z − 3)
2
⇒
⎡ 1
3 ⎤
1⎡ 3 ⎤
u( z)
u + ⎢
−
=⎢
⎥u1
2⎥ 0
3 ⎢⎣ ( z − 3) 2 ⎥⎦
z
⎢⎣ z − 3 ( z − 3) ⎥⎦
⎡ z
3z ⎤
1 ⎡ 3z ⎤
u + ⎢
−
⎥u1
⇒ u(z) = ⎢
2⎥ 0
3 ⎣⎢ ( z − 3) 2 ⎦⎥
⎣⎢ z − 3 ( z − 3) ⎦⎥
( z 2 − 6 z )u0 + zu1
Applying inverse Z-transform on equation (3), we get,
z − 6z + 9
⎡
⎛ z ⎞ −1 ⎛⎜ 3 z
⎟− z ⎜
2
⎝ z −3⎠
⎝ ( z − 3)
−1
Z–1[u(z)] = ⎢ z ⎜
z ( z − 6)u 0 + zu1
⎢⎣
z2 − 6z + 9
Taking partial fractions of
( z − 3)
... (3)
2
⇒
=
z−6
( z − 3) 2
⎞⎤
⎡
⎤
⎟⎥u 0 + 1 z −1 ⎢ 3 z ⎥u1
2
⎟
3
⎢⎣ ( z − 3) ⎥⎦
⎠⎥⎦
1
n
n
n
un = [(3) − n.3 ]u 0 + [ n.3 ].u1
3
⎡
⎤
−1 ⎡ Z ⎤
= an ⎥
⎢Q Z ⎢
⎥
⎣Z −a⎦
⎢
⎥
⎢
⎥
⎡
⎤
aZ
n
⎢ Z −1 ⎢
⎥
=
n
a
.
2⎥
⎢⎣
⎥⎦
⎣ ( Z − a) ⎦
u( z)
( z − 6)
1
u0 +
u1
=
z
( z − 3) 2
( z − 3) 2
2
−3
1
+
z − 3 ( z − 3) 2
⎡ 1
3 ⎤
1
u( z )
−
u +
u1
=⎢
2⎥ 0
z
( z − 3) 2
⎢⎣ z − 3 ( z − 3) ⎥⎦
( z − 6)u 0 + u1
u( z )
=
z
( z − 3) 2
z−6
=
Hence,
2
2
2 ⎛ u1 ⎞
⇒ z u ( z ) − z u 0 − z ⎜ ⎟ − 6 zu ( z ) + 6 zu 0 + 9u ( z ) = 0
⎝ z ⎠
⇒
z−6
, i.e.,
u n = (1 − n).3n u0 + n.3n −1.u1
A
B
+
z − 3 ( z − 3) 2
z – 6 = A[z – 3] + B[1]
... (2)
On substituting z = 3 in equation (2), we get,
Q47. By using Z-transforms, solve the difference
equation yn + 2 – 3yn + 1 + 2yn = 0 subject to the
conditions y0 = – 1 and y1 = 2.
Ans: Given that,
yn + 2 – 3yn + 1 + 2yn = 0, y0 = – 1 and y1 = 2
3 – 6 = A[0] + B
∴B = −3
Applying Z-transform, we get,
Z[yn + 2] – 3Z[yn + 1] + 2Z[yn] = 0
On comparing the constant terms of equation (2), we get,
– 6 = – 3A + B
⇒ – 3A = – 6 – B
⇒
y ⎤
⎡
z 2 ⎢ y ( z ) − y 0 − 1 ⎥ – 3 z[y(z) – y ] + 2 y(z) = 0
0
z⎦
⎣
⇒
y(z)[z2 – 3z + 2] – z2 y0 – (z)2
⇒ – 3A = – 6 – (– 3)
⇒ – 3A = – 3
∴A =1
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
y1
+ 3zy0 = 0
z
Given that, y0 = – 1 and y1 = 2
SIA GROUP
5.38
MATHEMATICS-II [JNTU-ANANTAPUR]
⇒
⎛ 2⎞
y(z)[z2 – 3z + 2] – z2 (– 1) – z2 ⎜ ⎟ + 3z(–1) = 0
⎝z⎠
⇒
y(z)[z2 – 2z – z + 2] + z2 – 2z – 3z = 0
⇒
y(z)[(z – 2) (z – 1)] + z2 – 5z = 0
⇒
y(z)[(z – 2) (z – 1)] = 5z – z2
⇒
⇒
y(z) =
5z − z 2
( z − 1)( z − 2)
5− z
y( z)
=
( z − 1)( z − 2)
z
By partial fraction expansion, we have,
⇒
5− z
A
B
y ( x)
+
=
=
( z − 1)( z − 2)
z −1 z − 2
z
... (1)
⇒
5 – z = A[z – 2] + B[z – 1]
... (2)
On substituting z = 2, in equation (2), we get,
⇒
5 – 2 = A[0] + B[2 – 1]
∴B = 3
On substituting z = 1 in equation (2), we get,
⇒
⇒
5 – 1 = A[1 – 2] + B[0]
–A=4
∴ A = −4
∴
3
−4
y( z)
+
=
z −1 z − 2
z
⎡ z ⎤ ⎡ z ⎤
y(z) = – 4 ⎢
⎥ + 3⎢
⎥
⎣ z −1⎦ ⎣ z − 2 ⎦
Applying inverse Z-transform on both sides, we get,
−1 ⎡ z ⎤
−1 ⎡ z ⎤
Z–1[y(z)] = – 4 z ⎢
⎥ + 3 z ⎢ z − 2⎥
−
z
1
⎣
⎦
⎣
⎦
= – 4 (1)n + 3(2)n
⎡
−1 ⎡ z ⎤
n⎤
⎢Q Z ⎢
⎥=a ⎥
−
z
a
⎣
⎦
⎣
⎦
= – 4 + 3.2n
= 3(2n) – 4
Q48. Solve the difference equation yn+2 + 3yn+1 + 2yn = 0, y0 = 1, y1 = 2 by Z-transform.
Ans: Given difference equation is,
yn + 2 + 3yn + 1 + 2yn = 0
... (1)
y0 = 1, y1 = 2
Look for the SIA GROU P LOGO
on the TITLE COVER before you buy
5.39
UNIT-5 (Z-Transform)
On applying z-transform to equation (1), we get,
Z[yn + 2] + 3Z[yn + 1] + 2z[yn] = 0
⇒
⇒
y ⎤
⎡
z 2 ⎢ y ( z ) − y0 − 1 ⎥ + 3z[ y ( z ) − y0 ] + 2 y ( z ) = 0
z⎦
⎣
⇒
z 2 y ( z ) − z 2 y0 −
y1 ⎤
⎡
2
⎢Q z[ y n + 2 ] = z [ y ( z ) − y 0 − z ⎥
⎥
⎢
⎥
⎢ z[ yn +1 ] = z[ y ( z ) − y0 ]
⎥
⎢ z[ y ] = y ( z )
n
⎥
⎢
⎦
⎣
z 2 y1
+ 3zy( z ) − 3zy0 + 2 y ( z ) = 0
z
y ( z )[ z 2 + 3z + 2] − z 2 y0 − z 2
y1
− 3zy 0 = 0
z
... (2)
On substituting the values of ‘y0’ and y1 in equation (2), we get,
( 2)
− 3z (1) = 0
z
⇒
y(z)[z2 + 3z + 2] – z2 (1) – z2
⇒
y(z) [z2 + 3z + 2] – z2 – 2z – 3z = 0
⇒
y(z) [(z + 2) (z + 1)] – z2 – 5z = 0
⇒
y(z) [(z + 2) (z + 1)] = z2 + 5z
⇒
z 2 + 5z
y(z) =
( z + 1)( z + 2)
⇒
z+5
y( z)
=
( z + 1)( z + 2)
z
Applying partial fractions, we get,
⇒
z+5
y( z)
A
B
+
=
=
(
1
)(
2
)
z
+
z
+
z
z +1 z + 2
... (3)
z + 5 = A[z + 2] + B[z + 1]
... (4)
On substituting z = – 2 in equation (4), we get,
⇒
– 2 + 5 = A[0] + B[– 2 + 1]
3=–B
∴B = −3
On substituting z = – 1 in equation (4), we get,
⇒ – 1 + 5 = A[– 1 + 2] + B[0]
4 =A
∴A = 4
SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
SIA GROUP
5.40
MATHEMATICS-II [JNTU-ANANTAPUR]
On substituting the values of A and B in equation (3), we get,
y( z)
4
(−3)
=
+
z
z +1 z + 2
⎡ z ⎤ ⎡ z ⎤
y (z ) = 4⎢
⎥ − 3⎢
⎥
⎣ z + 1⎦ ⎣ z + 2 ⎦
... (5)
On applying inverse Z-transform to the above equation, we get,
−1 ⎡ z ⎤
−1 ⎡ z ⎤
Z–1 [y(z)] = 4 z ⎢
⎥ − 3z ⎢ z + 2 ⎥
+
1
z
⎦
⎣
⎣
⎦
= 4(– 1n) – 3 (– 2n)
∴
⎡
−1 ⎡ z ⎤
n⎤
⎢Q z ⎢
⎥ = (− a ) ⎥
z
a
+
⎣
⎦
⎦
⎣
Z–1[y(z)] = 4(–1)n – 3 (–2)n
Look for the SIA GROU P LOGO
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