2
Fourier Series and Fourier Transform
2.1
INTRODUCTION
Fourier series is used to get frequency spectrum of a time-domain signal, when signal is a periodic function of
time. We have seen that the sum of two sinusoids is periodic provided their frequencies are integer multiple
of a fundamental frequency, w0 .
2.2
TRIGONOMETRIC FOURIER SERIES
Consider a signal x(t), a sum of sine and cosine function whose frequencies are integral multiple of w0
x(t) = a0 + a1 cos (w0t) + a2 cos (2w0t) + · · ·
b1 sin (w0t) + b2 sin (2w0t) + · · ·
∞
x(t) = a0 + ∑ (an cos (nw0t) + bn sin (nw0t))
(1)
n=1
a0 , a1 , . . . , b1 , b2 , . . . are constants and w0 is the fundamental frequency.
Evaluation of Fourier Coefficients
To evaluate a0 we shall integrate both sides of eqn. (1) over one period (t0 ,t0 + T ) of x(t) at an arbitrary
time t0
tZ
0 +T
x(t)dt =
t0
Since
R t0 +T
t0
tZ
0 +T
∞
a0 dt + ∑ an
n=1
t0
tZ
0 +T
∞
cos (nw0t)dt + ∑ bn
n=1
t0
tZ
0 +T
sin (nw0t)dt
t0
cos (nw0 dt) = 0
tZ
0 +T
sin (nw0 dt) = 0
t0
1
a0 =
T
tZ
0 +T
x(t)dt
t0
To evaluate an and bn , we use the following result:
tZ
0 +T
cos (nw0t) cos (mw0t)dt =
t0
94
0
m 6= n
T /2 m = n 6= 0
(2)
96
• Basic System Analysis
Multiply eqn. (1) by sin (mw0t) and integrate over one period
tZ
0 +T
x(t) sin (mw0t)dt = a0
tZ
0 +T
∞
sin (mw0t)dt + ∑ an
n=1
t0
t0
cos (nw0t) sin (mw0t)dt +
t0
∞
∑ bn
n=1
2
bn =
T
tZ
0 +T
tZ
0 +T
sin (mw0t) sin (nw0t)dt
t0
tZ
0 +T
x(t) sin (nw0t)dt
(4)
t0
Example 1:
1.0 −
−3
0 1
−2 −1
2
3
− −1.0
Fig. 2.1.
T → −1 to 1
a0 =
1
2
Z1
−1
T =2
w0 = π
x(t) = t, −1 < t < 1
1
t dt = (1 − 1) = 0
4
an = 0
bn =
Z1
t sin πntdt =
−1
−t cos πnt cos πnt
−
nπ
nπ
1
−1
−1
1
[t cos πnt + cos πnt]1−1 = − [2 cos π + cos π − cos π]
nπ
nπ
−2
2 −(−1)n
bn =
cos nπ =
nπ
π
n
=
b1
2
π
b2
b3
b4
b5
b6
−2
2
−2
2
−2 · · ·
2π 3π 4π 5π
6π
∞
n
2 −(−1)
sin nπt
x(t) = ∑
n
n=1 π
2
1
1
1
=
sin πt − sin 2πt + sin 3πt − sin 4πt + · · ·
π
2
3
4
Fourier Series and Fourier Transform
Example 2:
1.0
−2π
0
2π
4π
6π
t
Fig. 2.2.
x(t) =
t
2π
1
a0 =
T
T = 2π
w0 =
2π
=1
T
Z2π
0
2
an = 2
4π
1 1 2 2π 1
=
t
x(t)dt = 2
4π 2 0
2
Z2π
0
1 t sin t sin nt 2π
+
t cos ntdt = 2
2π
n
n
0
1 2π sin 2nπ sin 2nπ
=0
+
= 2
2π
n
n
2
bn = 2
4π
Z2π
t sin ntdt =
0
=
bn =
−1 h t cos nt cos nt i2π
+
2π2
n
n
0
−1 2π cos 2nπ cos 2nπ 1
+
−
2π2
n
n
n
−1
nπ
∞ −1
1
1 1 ∞ 1
x(t) = + ∑
sin nt = + ∑ cos (nt + π/2)
2 n=1 nπ
2 π n=1 n
sin 2t sin 3t
1 1
sin t +
+
+···
= −
2 π
2
3
Example 3:
A x(t)
−T/2
−T/4
T/4
Fig. 2.3. Rectangular waveform
T/2
t
•
97
98
• Basic System Analysis
Figure shows a periodic rectangular waveform which is symmetrical to the vertical axis. Obtain its F.S.
representation.
∞
x(t) = a0 + ∑ (an cos nw0t + bn sin nw0t)
n=1
∞
x(t) = a0 + ∑ an cos (nw0t) bn = 0
n=1
−T
−T
<t <
2
4
−T
T
+ A for
<t <
4
4
T
T
0 for < t <
4
2
x(t) = 0
for
1
a0 =
T
T /4
Z
Adt =
T /4
Z
A cos (nw0t)dt =
A
2
−T /4
an =
2
T
−T /4
2A
T
T
sin nw0 + sin nw0
T nw0
4
4
nπ 2A
nπ 4A
=
sin
sin
2πn
2
πn
2
4A 2A
a1 =
=
2π
π
an =
w0 =
2π
T
a2 = 0
3π 2A
−2A
2A
sin
=
(−1) =
3π
2
3π
3π
A 2A
1
1
x(t) = +
cos w0t − cos 3w0t + cos 5w0t + · · ·
2
π
3
5
a3 =
Example 4: Find the trigonometric Fourier series for the periodic signal x(t).
x(t)
1.0
−9
−7
−5
−3
−1
0 1
T
Fig. 2.4.
3
5
7
9
11
t
Fourier Series and Fourier Transform
S OLUTION:
bn = 0
a0 =
1
T
x(t) =
Z3
(
1
−1
x(t)dt =
−1
−1 < t < 1
1<t <3

1
T
Z1
dt +
Z3
t
−1

(−1)dt 
T =4
1
= [2 − 2] = 0
T


Z1
Z3
2
an =  cos (nw0t)dt + cos (nw0t)dt 
T
−1
∴
w0 =
2π 2π π
=
=
T
4
2
1
h
2
πn i
3nπ
nπ
=
2 sin
− sin
− sin
2πn
2
2
2
nπ
3nπ
nπ
3nπ
nπ 1
3 sin
sin
= − sin
− sin
= sin π +
=
nπ
2
2
2
2
2
nπ 4
an =
sin
nπ
2

n = even

0

 4
n = 1, 5, 9, 13
an =
nπ



 −4 n = 3, 7, 11, 15
nπ
π 4
4
4
3π
5π
7π
4
cos
t +
cos
t −
cos
t +···
x(t) = cos t −
π
2
3π
2
5π
2
7π
2
π
1
1
3π
5π
4
cos t − cos
t + cos
t ···
x(t) =
π
2
3
2
5
2
Example 5: Find the F.S.C. for the continuous-time periodic signal
x(t) = 1.5
0≤t <1
= −1.5
with fundamental freq. w0 = π
1.5
0
1≤t <2
x(t)
1
2
3
−1.5
Fig. 2.5.
4
5
•
99
100
•
Basic System Analysis
S OLUTION:
T=
2π
= 2, w0 = π
w0
a0 = an = 0
bn =
Z1
0
1.5 sin nπtdt −
Z2
1.5 sin nπtdt
1
o
1.5 n
[− cos nπ + 1] + [cos 2nπ − cos nπ]
nπ
3
bn =
[1 − cos nπ]
nπ
2
2
3
2 sin πt + sin 3πt + sin 5πt + · · ·
x(t) =
π
3
5
6
1
1
sin πt + sin 3πt + sin 5πt + · · ·
π
3
5


Z1
Z2
1
1.5dt − 1.5 dt  = 0
C0 =
2
=
0
OR
1
By using complex exponential Fourier series


Z1
Z2
1
Cn =  1.5e− jnπt dt − 1.5 e− jnπt dt 
2
0
Cn =
=
=
3  − jnπt
e
−4 jnπ
2
1
0
−e− jnπt
1



−3 − jnπ
e
− 1 − e− j2nπ + e− jnπ
4 jnπ
3
3 1 − e− jnπ =
[1 − cos nπ]
2 jnπ
2 jnπ
∞
x(t) =
1

∑
Cn e− jnπt
n=−∞
∞
3 1 − e− jnπ e jnπt
n=−∞ 2 jnπ
∑
∞
=
3 jnπt
e
− e jnπt cos πn
2
jnπ
n=−∞
∑
102
•
Basic System Analysis
for n = 1
A
=
2π
Zπ
A
sin t sin tdt =
2π
0
=
Zπ
0
(1 − cos 2t)dt
A
A
[π] =
2π
2
When n is even
2A
2
2
A
=
−
=
2π n + 1 1 − n
π(1 − n2 )
Example 7:
x(t)
b
2
−3 −2
−1
0
1
2
3
t
−2
a
T
Fig. 2.7.
S OLUTION:
2π
=π
T = 2 w0 =
T
2t − 1 < t < 1
x(t) =
0
Point (a) (−1, −2)
Point (b) (1, 2)
y − (−2) =
2 − (−2)
(x − (−1))
1 − (−1)
4
y + 2 = (x + 1)
2
y + 2 = 2x + 2
y = 2x
x(t) = 2t
Since function is an odd function
1
T
Z1
1
×0 = 0
2
−1

Z1
2
2 −t cos nπt
bn =
t sin (nπt)dt = 
T
T
nπ
an = 0, a0 =
−1
2tdt =
1
+
−1
1
cos nπt
n2 π2
1
−1


104
•
2.3
CONVERGENCE OF FOURIER SERIES – DIRICHLET CONDITIONS
Basic System Analysis
Existence of Fourier Series: The conditions under which a periodic signal can be represented by an F.S.
are known as Dirichlet conditions. F.P. → Fundamental Period
(1) The function x(t) has only a finite number of maxima and minima, if any within the F.P.
(2) The function x(t) has only a finite number of discontinuities, if any within the F.P.
(3) The function x(t) is absolutely integrable over one period, that is
ZT
x(t) dt < ∞
0
2.4
PROPERTIES OF CONTINUOUS FOURIER SERIES
(1) Linearity: If x1 (t) and x2 (t) are two periodic signals with period T with F.S.C. Cn and Dn then F.C. of
linear combination of x1 (t) and x2 (t) are given by
FS [Ax1 (t) + Bx2 (t)] = ACn + BDn
Proof: If z(t) = Ax1 (t) + Bx2 (t)
1
an =
T
tZ
0 +T
[Ax1 (t) + Bx2 (t)] e− jnw0 t =
t0
A
T
Z
x1 (t)e− jnw0 t dt +
T
B
T
Z
x2 (t)e− jnw0 t dt
T
an = ACn + BDn
(2) Time shifting: If the F.S.C. of x(t) are Cn then the F.C. of the shifted signal x(t − t0 ) are
FS [x(t − t0 )] = e− jnw0 t0 Cn
Let t − t0 = τ
dt = dτ
Bn =
=
(3) Time reversal:
1
T
Z
x(t − t0 )e− jnw0 t dt
1
T
Z
x(τ)e− jnw0 (t0 +τ) dτ =
T
T
1
T
Bn = e− jnw 0 t 0 ·Cn
Z
T
x(τ)e− jnw0 τ dτ · e− jnw0 t0
FS[x(−t)] = C−n
Z
Z
1
1
x(−t)e− jnw0 t dt =
x(−t)e− j(−n)w0 T dt
Bn =
T
T
T
T
−t = τ
dt = −dτ
=
1
T
Z
−T
x(τ)e− j(−n)w0 τ dτ = C−n
106
•
Basic System Analysis
Example 8: Compute the exponential series of the following signal.
x(t)
2.0
1.0
−3 −2 −1 0
−5 −4
1
3
2
6
5
4
t
T
Fig. 2.8.
S OLUTION:
T =4
C0 =
1
T
w0 =
ZT
π
2
x(t)dt =
0
Cn =

Z1
1
4
0

Z1
1
4
2dt +
0
Z2
1

dt  =
3
4

πt
Z2
− jn
πt
2 dt + e− jn 2 dt 
2e
1




π
h
i
π
1  −4  − jn
2
2 − 1 −
e− jnπ − e− jn 2
e
=

4  jnπ
jnπ
=

−1 
2e
2 jnπ
− jnπ
2



π
π
π
−
jn
−
jn
−
jn
−1 
2=
2 +e
2 − 2
− 2 + e− jnπ − e
e
2 jnπ
1
1
1 − jn π
n
2
=−
1 − (−1) − e
jnπ
2
2
∞
3
1
1
1
jn π2
n jn π2
x(t) = + ∑
e − (−1) e −
4 n=−∞ jnπ
2
2
Example 9:
x(t)
1.0
b
↓
↓
a
−2 −1
0
1
2
3
Fig. 2.9.
4
5
6
7
t
Fourier Series and Fourier Transform
• 107
S OLUTION:
2π
T = 5 w0 =
5

 t + 2 − 2 < t < −1
x(t) = 1.0
−1 < t < 1

2−t
1<t <2
(−2, 0)(−1, 1)
−1
(y − 1) =
(x + 1)
−1
y = t +2
(a)
(b)
(1, 1)(2, 0)
y−0 =
1
(x − 2)
−1
y = −x + 2 = −t + 2


Z−1
Z1
Z2
1
C0 =
(t + 2)dt + dt + (2 − t)dt 
5
−2
C0 =
−1
1
3
5


 −1

Z
Z1
Z2

2nπ
2nπ
2nπ
1


Cn =  (t + 2)e− j 5 dt + e− j 5 dt + (2 − t)e− j 5 dt 


5
−2

−1
1
{z
}
{z
} | {z } |
|
B
A
A=
Z−1
− j 2nπ
5 t
e
−2
dt +
Z−1
2e− j
2nπ t
5
C
dt
−2


Z−1
Z−1
Z−1

2 j 2nπ
1 − jφ
1
− jφ
e
te
+
+ e
A=−
 φ2
jφ 
− jφ
5
−2
−2
−2
2nπ
4nπ
25 2nπ
10
5
−e j 5 + 2e j 5 + 2 2 e j 5 − e j 5 −
=
j 2nπ
4n π
2nπ j
2nπ
2nπ
4nπ
4nπ
5
25
A=
−e j 5 + 4e j 5 + 2 2 e j 5 − e j 5
j2nπ
4n π
4nπ
2nπ
2nπ
5 j 2nπ
− e− j 5
5 − e− j 5
e
=
j 2nπ
j 2nπ
5
4nπ
n2π
2nπ
10 − j 4nπ
5 − j 2nπ
25
25
−10 − j 4nπ
e 5 − e− j 5 +
e 5 −
e 5 − 2 2 e− j 5 + 2 2 e j 5
C=
j 2nπ
j 2nπ
j 2nπ
4n π
4n π
B=
ej
2nπ
5
108
•
Basic System Analysis
j4nπ
j2nπ
25 j2nπ
25 − j4nπ
1
5 −e 5
5 − e− 5
e
e
−
5 n2 4π2
4n2 π2
5
4πn
2πn
Cn = 2 2 cos
− cos
2n π
5
5
Cn =
Example 10: For the continuous-time periodic signal
5π
2π
t + 4 sin
t
x(t) = 2 + cos
3
3
Determine the fundamental frequency w0 and the Fourier series coefficients Cn such that
∞
x(t) =
∑
Cn e jnw0 t
n=−∞
S OLUTION:
Given
2π
5π
x(t) = 2 + cos
t + 4 sin
t
3
3
The time period of the signal cos 2π
3 t is
T1 =
The time period of the signal sin 5 π2 t is
2π
2π
= π = 3 sec
w1
23
T2 = 2
π
6
2π
= π = sec
w2
53
5
5
3
T1
= 6 = ratio of two integers, rational number, hence periodic.
T2
2
5
2T1 = 5T2
The fundamental period of the signal x(t) is
T = 2T1 = 5T2 = 6 sec
and the fundamental frequency is
2π 2π π
=
=
T
6
3
2π
5π
x(t) = 2 + cos
t + 4 sin
t
3
3
w0 =
= 2 + cos (2w0t) + 4 sin (5w0t)
4 e j5w0 t − e− j5w0 t
e j2w0 t + e− j2w0 t
+
= 2+
2
2j
= 2 + 0.5 e j2w0 t + e− j2w0 t − 2 j e j5w0 t − e− j5w0 t
x(t) = 2 je+ j(−5)w0t + 0.5e+ j(−2)w0t + 2 + 0.5e+ j2w0 t − 2 je+ j5w0 t
110
•
Basic System Analysis
2
an =
T
Zπ
−π
4
x(t) cos ntdt =
T
Z0 −π
2t
+ 1 cos nt dt
π


Z0

4  2t
sin nt
2
=
sint +
−
sin nt dt

2π  nπ
n
π
=

1  2t
ππ
−π
sin nt + sin nt +
2
cos nt
n2 π

Z0 
−π

(
)
o
2
4
2
4 n
2
1
−
cos
nπ
= 2 2 (1 − (−1)n )
+
cos
nt
=
=
2
2
2
2
π n π n π
n π
n π
(
0
n even 2, 4, 6, 8, · · ·
an =
8
n odd 1, 3, 5, 7, · · ·
n 2 π2
2.5
FOURIER TRANSFORM
2.5.1 Definition
Let x(t) be a signal which is a function of time t. The Fourier transform of x(t) is given as
X ( jw) =
Fourier transform
X (i f ) =
Z∞
−∞
Z∞
x(t)e− jwt dt
(1)
or
x(t)e− j2π f t dt
(2)
−∞
Since w = 2π f
Similarly, x(t) can be recovered from its Fourier transform X( jw) by using Inverse Fourier transform
x(t) =
1
2π
x(t) =
Z∞
Z∞
X( jw)e jwt dw
(3)
−∞
X(i f )e j2π f t dt
(4)
−∞
Fourier transform X( jw) is the complex function of frequency w. Therefore, it can be expressed in the
complex exponential form as follows:
X( jw) = |X( jw)|e j
Here |X( jw)| is the amplitude spectrum of x(t) and
|X( jw)
|X( jw)
is phase spectrum.
For a real-valued signal
(1) Amplitude spectrum is symmetric about vertical axis c (even function.)
(2) Phase spectrum is anti-symmetrical about vertical axis c (odd function.)
Fourier Series and Fourier Transform
2.5.2
• 111
Existence of Fourier transform (Dirichlet’s condition)
The following conditions should be satisfied by the signal to obtain its F.T.
(1)
(2)
(3)
(4)
The function x(t) should be single valued in any finite time interval T .
The function x(t) should have at the most finite number of discontinuities in any finite time interval T .
The function x(t) should have finite number of maxima and minima in any finite time interval T .
The function x(t) should be absolutely integrable, i.e.
Z∞
−∞
|x(t)|dt < ∞
• These conditions are sufficient, but not necessary for the signal to be Fourier transformable.
• A physically realizable signal is always Fourier transformable. Thus, physical realizability is the
sufficient condition for the existence of F.T.
• All energy signals are Fourier transformable.
j
d
X( jw) = FT (tx(t))
dw
d
FT (tx(t)) = j X( jw)
dw
Example 12: Obtain the F.T. of the signal e−at u(t) and plot its magnitude and phase spectrum.
S OLUTION:
x(t) = e−at u(t)
X( f ) =
Z∞
− j2π f t
x(t)e
dt =
−∞
Z∞
e−(a+ j2π f )t dt
0
1
X( f ) =
a + j 2π f
To obtain the magnitude and phase spectrum:
a − j 2π f
2π f
a
|X( f )| = 2
A− j 2
B
=
a + (2π f )2
a2 + 4π2 f 2
a + 4π2 f 2
p
1
1
=√
|X( f )| = A2 + B2 = p
a2 + w2
a2 + 4π2 f 2
w
−2π f
= − tan−1
|X( f )| = tan−1
a
a
1
|X( f )
for a = 1, |X( f )| = √
= − tan−1 w
,
2
1+w
w
0
1
2
3
4
5
10
15
25
8
|X(w)|
1
.707
0.447
0.316
0.242
0.196
0.09
0.066
0.03
0
| X(w)
0
45◦
−63.4
−71.5
−75.9
−78.6
−84.2
−86.2
−87.7
−90◦
Fourier Series and Fourier Transform
(ii) x(t) = e−a|t| =
• 113
e−at t > 0
eat t > 0
e−a|t|
t
Fig. 2.13. Graphical representation of e−a|t|
1
1
2a
+
= 2
a + jw a − jw a + w2
2
for a = 1 X(w) =
1 + w2
2
|X(w)
=0
|X(w)| =
2
1+w
x(w) =
w (in radians)
|X(w)|
−∞
−10
−5
−3
−2
−1
0
1
2
3
4
5
10
∞
0
0.019
0.0769
0.2
0.4
1
2
1
0.4
0.2
.1176
0.0769
0.019
0
|X(w)|
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
−10 . . . . −5
−4
−3
−2
−1
0
1
2
3
Fig. 2.14. Magnitude plot
(iii)
x(t) = e−a|t|
sgn(t)
x(t) = e−a |t| sgn(t)
1.0
t
−1.0
Fig. 2.15. Graphical representation of e−a|t| sgn(t)
4
5 . . . . 10 w
Fourier Series and Fourier Transform
• 115
x(t) = 1
(ii)
X(w) =
Z∞
e− jwt dt = ∞
−∞
This means Dirichlet condition is not satisfied. But its F.T. can be calculated with the help of duality
property.
FT
δ(t) ←→ 1
FT
Duality property states that: x(t) ←→ X(w) then
Here X(t) = 1, then
FT
X(t) ←→ 2πx(−w)
x(−w) will be
x(t) = δ(t);
X(w) = 1
FT
then X(t) = 1; 1 ←→ 2πδ(−w)
We know that δ(w) will be an even function of w, since it is impulse function.
Hence, δ(−w) = δ(w). Then above equation becomes
FT
1 ←→ 2πδ(−w)
Thus, if x(t) = 1, then X(w) = 2πδ(w)
1
(iii) x(t) = sgn(t)
sgn(t) =
−1
t >0
t <0
sgn(t)
1
t
0
−1
Fig. 2.17. Graphical representation of sgn(t)
x(t) = 2u(t) − 1
Differentiating both the sides
d
d
x(t) = 2 u(t) = 2δ(t)
dt
dt
Taking the F.T. of both sides
F
d
x(t) = 2F[δ(t)]
dt
jwX(w) = 2
2
X(w) =
jw
X(w) =
Z∞
0
e− jwt dt −
Z0
−∞
e− jwt dt
116
•
Basic System Analysis
x(t) = u(t)
(iv)
sgn(t) = 2u(t) − 1
2u(t) = 1 + sgn(t)
Taking F.T. of both sides
2F[2u(t)] = F(1) + F[sgn(t)] = 2πδ(w) +
FT
2u(t) ←→ 2πδ(w) +
FT
u(t) ←→ πδ(w) +
2
jw
2
jw
1
jw
Properties of unit impulse:
(1)
Z∞
x(t)δ(t) = x(0)
−∞
(2) x(t)δ(t − t0 ) = x(t0 )δ(t − t0 )
(3)
Z∞
−∞
x(t)δ(t − t0 )dt = x(t0 )
(4) δ(at) =
(5)
Z∞
−∞
1
|a| δ(t)
x(τ)δ(t − x)dt = x(t)
(6) δ(t) =
d
dt u(t)
Example 15: Obtain the F.T. of a rectangular pulse shown in Fig. 2.18.
x(t)
−T/2
1
0
T/2
t
Fig. 2.18. Rectangular pulse
S OLUTION:
T
X(w) =
Z2
e− jwt dt =
−T
2
X(w) = T
sin π wT
2π
π wT
2π
i 2
T
−1 h − jw T
wT
2 − e jw 2
= sin
e
jw
w
2
wT
= sin c
2π
=T
sin π wT
2π
π wT
2π
• 117
Fourier Series and Fourier Transform
Sampling function or interpolating function or filtering function denoted by Sa (x) or sin c(x) as shown in
figure.
sin πx
sin c(x) =
πx
(1) sin c(x) = 0 when x = ±nπ
(2) sin c(x) = 1 when x = 0 (using L’Hospital’s rule)
(3) sin c(x) is the product of an oscillating signal sin x of period 2π and a decreasing signal 1x . Therefore,
sin c(x) is making sinusoidal of oscillations of period 2π with amplified decreasing continuously as 1x .
sin c(x)
1.0
−4π
−3π −2π
−π
0
π
4π
2π 3π
5π
x
Fig. 2.19. Sine function
sin πx
;
πx
sin π
sin c(1) =
= 0;
π
sin c(2) = 0;
sin cx =
sin c(0) =
0
= 1 L’Hospital rule
0
sin c(−1) = 0
sin c(−2) = 0
sin c(1/4) = 0.9
sin c(−1/4) = 0.9
sin c(2/4) = .6366
sin c(−0.5) = .6366
sin c(3/4) = 0.3
sin c(−7.5) = .3
sin c(1.5) = −.2122
sin c(−1.5) = −.2122
sin c(2.5) = .1273
sin c(2.5) = .1273
.9
.8
.7
.6
.5
.4
.3
.2
.1
−3.5 −3
−2.5
−2
−1.5
−1 −.75
−.5 −.25
0 .25
.5
.75
Fig. 2.20. Sine function
1
1.5
2
2.5
3
3.5
t
Fourier Series and Fourier Transform
Example 17: Obtain F.T. and spectrums of following signals:
(i) x(t) = cos w0t
(ii) x(t) = sin w0t
S OLUTION:
(i)
1
1
x(t) = cos w0t = e jw0 t + e− jw0 t
2
2
FT
1 ←→ 2πδ(w);
1 FT
←→ πδ(w)
2
FT
Frequency shifting property states that e jβt x(t) ←→ X(w − β)
1 jw0 t FT
e
←→ πδ(w − w0 )
2
1 − jw0 t FT
e
←→ πδ(w + w0 )
2
1 jw0 t 1 − jw0 t
F [x(t)] = FT
e
+ e
2
2
X(w) = π [δ(w − w0 ) + δ(w + w0 )]
| X(w) |
π
−w0
w0
w
Fig. 2.22. Magnitude plot of cos w0t
(ii)
x(t) = sin w0t
X(w) =
π
[δ(w − w0 ) − δ(w + w0 )]
j
| X(w) |
π
−ω0
w0
w
π
Fig. 2.23. Magnitude plot of sin w0t
• 119
120
•
Basic System Analysis
Example 18: Obtain the F.T. of
x(t) = te−at u(t)
d
X(w)
from property of Fourier transform FT[tx(t)] = j dw
1
a + jw
d
d
(1) − 1 dw
(a + jw)
(a + jw) dw
d
1
1
−at
FT(te ) = j
=j
=
dw a + jw
(a + jw)2
(a + jw)2
FT e−at =
Inverse Fourier Transform: (IFT)
Example 19: Find the IFT of
(i) X(w) =
2 jw+1
( jw+2)2
by partial fraction expansions
(ii) X(w) =
1
(a+ jw)2
by convolution property
(iii) X(w) = e−|w|
(iv) X(w) = e−2w u(w)
S OLUTION:
(i)
A
B
+
; 2 jw + 1 = A( jw + 2) + B A = 2
jw + 2 ( jw + 2)2
3
2
−
X(w) =
jw + 2 ( jw + 2)2
X(w) =
2A + B = 1 B = −3
x(t) = 2e−2t u(t) − 3te−2t u(t)
(ii)
1
1
=
= X1 (w)X2 (w)
2
(a + jw)
(a + jw)(a + jw)
1
1
, X2 (w) =
X1 (w) =
a + jw
a + jw
x1 (t) = e−at u(t), x2 (t) = e−at u(t)
X(w) =
Using convolution property
x(t) = x1 (t)∗ x2 (t)
FT
x(t) ←→ X(w)
FT
x1 (t)∗ x2 (t) ←→ X1 (w)X2 (w)
x(t) =
=
Z∞
−∞
Zt
0
−at
e
−a(t−τ)
u(t)e
u(t − τ)dτ
e−at dτ = te−at u(t)
(
u(τ) = 1 τ ≤ 0
u(t − τ) = 1 t ≤ τ
122
•
Basic System Analysis
Example 20: Find the F.T. of the function
x(t − t0 ) = e−(t−t0 ) u(t − t0 )
S OLUTION:
If F[x(t)] = X(w)
then FT[x(t − t0 )] = e− jwt0 X(w)
1
F e−t u(t) =
1 + jw
h
i e− jwt0
F e−(t−t0 ) u(t − t0 ) =
1 + jw
Example 21: Find the F.T. of the function
x(t) = [u(t + 1) − u(t − 1)] cos 2πt
S OLUTION:
j2πt
e
+ e− j2πt
FT(cos 2πt) = FT
2
FT[1] = 2πδ(w)
FT[e jw0 t ] = 2πδ(w − w0 )
F[cos 2πt] = πδ(1w − 2π) + πδ(w + 2π)
F[u(t + 1) − u(t − 1)] =
Z1
−1
e− jwt dt = −
(1)
2 sin w
1 − jw
e
− e jw =
jw
w
F[x(t)] = F[{u(t + 1) − u(t − 1)} cos 2πt]
x(t) is multiplication of (1) and (2), so by using multiplication property
FT
x(t)y(t) ←→
1
1
X1 (w)∗Y1 (w) =
2π
2π
Z∞
−∞
X(τ)Y (w − τ)dτ


Z∞
1  2 sin τ
X(w) =
πδ(w − 2π − τ) + δ(w + 2π − τ) dτ
2π
τ
X(w) =
Since
Z∞
−∞
Z∞
−∞
−∞
sin τ
δ(w − 2π − τ)dτ +
τ
Z∞
−∞
sin τ
δ(w + 2π − τ)dτ
τ
x(t)δ(t − t0 )dt = x(t0 )
X(w) = sin(w − 2π)/(w − 2π) + sin(w + 2π)/(w + 2π)
(2)
Fourier Series and Fourier Transform
Example 22: Determine the Fourier transform of a triangular function as shown in figure.
x(t)
A
−T
T
t
Fig. 2.24. Triangular pulse
S OLUTION:
x(t)
(0, A)
a→
→
(−T,0)
Equation of line (a) is
x(t) = A
Equation of line (b) is
b
(T,0)
t
T
t
+1
t
x(t) = A 1 −
T
Mathematically, we can write x(t) as
t
t
[u(t) − u(t − T )]
+ 1 [u(t + T ) − u(t)] + A 1 −
x(t) = A
T
T
x(t) =
x(t) =
x(t) =
=
=
X( jw) =
A
A
(t + T )[u(t + T ) − u(t)] + (T − t)[u(t) − u(t − T )]
T
T
o An
o
An
(t + T )u(t + T ) − (t + T )u(t) +
[(T − t)u(t) − (T − t)u(t − T )]
T
T
o An
o
An
r(t + T ) − tu(t) − Tu(t) +
Tu(t) − tu(t) + r(t − T )
T
T
n
o
o
An
A
r(t + T ) − r(t) − Tu(t) +
Tu(t) − r(t) + r(t − T )
T
T
hn
oi
A
r(t + T ) − 2r(t) + r(t − T )
T
A e jwT
2
e− jwT
−
+
T ( jw)2 ( jw)2 ( jw)2
• 123
Fourier Series and Fourier Transform
Π(t) = rect(t) =
rect(t − 5) =
rect(t − 5) =
X( jw) =
1 − 21 < t <
0 otherwise
1 − 12 ≤ t − 5 <
0 otherwise
(
1
9
2
≤t ≤
0 otherwise
Z∞
− jwt
x(t)e
dt =
11/2
Z
e− jwt dt =
e−
Z∞
−∞
9/2
=
1
2
11
2
−∞
=
1
2
j11w
2
− e−
− jw
9 jw
2
e− jwt
− jw
rect(t − 5)e− jwt dt
11/2
9/2
w
e−9 j 2 −e
=
jw
−11 j w
2
2e−5 jw e jw/2 − e− jw/2
e−5 jw e jw/2 − e−5 jw e− jw/2
=
=
jw
w2 j
sin w2
w
2e−5 jw
sin = e−5 jw
=
w
w
2
2
w
X( jw) = e−5 jw Sa
2
2.6
PROPERTIES OF CONTINUOUS-TIME FOURIER TRANSFORM
(1) Linearity
If FT (x1 (t)) = X1 ( jw)
and FT (x2 (t)) = X2 ( jw)
Then linearity property states that
FT(Ax1 (t) + Bx2 (t)) = AX1 ( jw) + BX2 ( jw)
where A and B are constants.
Proof:
Let r(t) = Ax1 (t) + Bx2 (t)
FT(r(t)) = R( jw) =
=
Z∞
−∞
Z∞
r(t)e− jwt dt
−∞
(Ax1 (t) + Bx2 (t)) e− jwt dt
• 125
Fourier Series and Fourier Transform
=
Z∞
x(τ)e− j(−w)τ dτ
−∞
F(x(t)) = X(− jw)
(4) Time shifting
If FT (x(t)) = X( jw)
then FT (x(t − t0 )) = e− jwt0 X( jw)
Proof:
Let r(t) = x(t − t0 )
R( jw) =
Z∞
r(t)e− jwt dt =
−∞
−∞
R( jw) = FT(x(t − t0 )) =
Let t − t0 = τ
dt = dτ
FT (x(t − t0 )) =
=
Z∞
Z∞
−∞
x(t − t0 )e− jwt dt
x(t − t0 )e− jwt dt
Z∞
x(τ)e− jw(t0 +τ) dτ
Z∞
x(τ)e− jwt e− jwt0 dτ
−∞
−∞
− jwt0
=e
Z∞
x(τ)e− jwτ dτ
−∞
FT (x(t − t0 )) = e− jwt0 X( jw). Similarly, FT (x(t + t0 )) = e jwt0 X( jw)
So FT (x(t ± t0 )) = e± jwt0 X( jw)
(5) Frequency shifting
If FT (x(t)) = X( jw)
FT (e jw 0 t x(t)) = X( j(w − w0 ))
Let r(t) = e jw 0 t x(t)
FT (r(t)) = FT e
jw0 t
Z∞
x(t) = R( jw) = e jw0 t x(t)e− jwt dt
−∞
FT e
jw0 t
∞
Z
x(t) = x(t)e− j(w−w0 )t dt
−∞
• 127
•
128
Basic System Analysis
Let w − w0 = w0
=
Z∞
0
x(t)e− jw t dt
−∞
FT e jw0 t x(t) = X( jw0 ) = X( j(w − w0 ))
Similarly, FT(e− jw0 t x(t)) = X( j(w + w0 ))
We can write as FT e± jw0 t x(t) = X( j(w ∓ w0 ))
(6) Duality or symmetry property
If FT (x(t)) = X( jw)
then FT (x(t)) = 2πx(− jw)
Proof:
We know that x(t) =
1 R∞
jwt dw
2π −∞ X( jw)e
Replacing t by −t, we get
1
x(−t) =
2π
2π x(−t) =
2π x(−t) =
2π
2π
Z∞
Z∞
X( jw)e− jwt dw
Z∞
X( jw)e− jwt dw
−∞
−∞
X( jw)e− jwt dw
−∞
Interchanging t by jw
2π x(− jw) =
Z∞
X(t)e− jwt dt
−∞
2π x(− jw) = FT(X(t))
(7) Convolution in time domain
If FT (x1 (t)) = X1 ( jw) and FT (x2 (t)) = X2 ( jw)
then FT (x1 (t)∗ x2 (t)) = X1 ( jw)X2 ( jw)
i.e., convolution in time domain becomes multiplication in frequency domain.
Fourier Series and Fourier Transform
Proof:
∗
r(t) = x1 (t) x2 (t) =
FT(r(t)) = R( jw) =
=
=
=
Z∞


Z∞
−∞ −∞
Z∞ Z∞
−∞ −∞
Z∞
Z∞
=
=
∗
Z∞
−∞
Z∞
−∞
Z∞
x1 (τ)x2 (t − τ)dτ
r(t)e− jwt dt

x1 (τ)x2 (t − τ)dτ e− jwt dt
x1 (τ)x2 (t − τ)dτ e− jwt dt
x1 (τ)dτ
FT [x1 (t)∗ x2 (t)] =
−∞
−∞
−∞
Let t − τ = ∝ so dt = d ∝
Z∞
Z∞
−∞
x2 (t − τ) e− jwt dt
Z∞
x1 (t)dτ
−∞
Z∞
x1 (τ)dτ
x2 (∝) e− jw(∝+τ) d ∝
x2 (∝) e− jw∝ e− jwτ d ∝
−∞
− jwτ
x1 (τ) e
−∞
dτ
FT [x1 (t) x2 (t)] = X1 ( jw) X2 ( jw)
Z∞
x2 (∝) e− jw∝ d ∝
−∞
(8a) Integration in time domain
If FT (x(t)) = X( jw)
Rt
1
× ( jw)
x(τ)dτ = jw
then FT −∞
R
t
Proof: Let r(t) = −∞
x(τ)dτ
Differentiating w.r.t. t
dr(t)
d
= x(t) ⇒ FT(x(t)) = FT
r(t)
dt
dt
From differentiation in time domain
X( jw) = jwX( jw)
R( jw) =
1
X( jw)
jw

FT(r(t)) = FT 
Zt
−∞

x(τ)dτ =
1
X( jw)
jw
• 129
130
•
Basic System Analysis
(8b) Differentiation in time domain
If FT (x(t)) = X( jw)
then dtd x(t) = jw × ( jw)
1
Proof: We know that x(t) =
2π
Z∞
X( jw)e jwt dw. Differentiating both sides w.r.t. t
−∞
1
d
x(t) =
dt
2π
1
=
2π
Z∞
d jwt
dw
e
X( jw)
dt
Z∞
jwX( jw)e jwt dw
−∞
−∞
1
=j
2π
Z∞
(wX( jw))e jwt dw
−∞
d
x(t) = j FT−1 (wX( jw))
dt
n
yields FT dtd x(t) = jwX( jw). On generalizing we get FT dtd n x(t) = ( jw)n X( jw)
(9) Differentiation in frequency domain
If FT (x(t)) = X( jw)
d
X( jw)
then FT (tx(t)) = j dw
∞
Proof: We know that X( jw) = −∞
x(t)e− jwt dt
On differentiating both sides w.r.t. w
Z∞
Z∞
d
d − jwt
X( jw) = x(t)
e
dt = − j t x(t)e− jwt dt
dw
dw
R
−∞
−∞
Multiplying both sides by j
j
d
X( jw) =
dw
Z∞
(tx(t))e− jwt dt
−∞
since j2 = −1 or − j2 = 1
d
X( jw) = FT [t x(t)]
j
dw
d
X( jw)
FT [t x(t)] = j
dw
(10) Convolution in frequency domain (multiplication in time domain (multiplication theorem))
If FT(x1 (t)) = X1 ( jw) and FT [x2 (t)] = X2 ( jw)
1
FT(x1 (t)x2 (t)) =
(X1 ( jw)∗ X2 ( jw))
2π
132
•
Basic System Analysis
Proof:
E=
Z∞
x(t) dt =
1
2π
Z∞
2
−∞
We know that x(t) =
1
2π
So x∗ (t) =
−∞
Z∞
Z∞
x(t)x∗ (t)dt
(1)
−∞
X( jw)e+ jwt dw
X( jw)e− jwt dw
(2)
−∞
on putting (1)
=
Z∞
−∞
=
1
2π
=
1
2π
=
Z∞

1
x(t) 
2π
Z∞
−∞
Z∞
Z∞
−∞
X ∗ ( jw)

X ∗ ( jw)e− jwt dw dt
Z∞
x(t)e− jwt dt dw
−∞
X( jw)X ∗ ( jw)dw
−∞
x(t)2 dt =
−∞
1
2π
Z∞
2
X( jw) dw
−∞
Relation between Laplace Transform and Fourier Transform
Fourier transform X( jw) of a signal x(t) is given as
X( jw) =
Z∞
x(t)e− jwt dt
(1)
−∞
F.T. can be calculated only if x(t) is absolutely integrable
=
Z∞
x(t) dt < ∞
(2)
Z∞
(3)
−∞
Laplace transform X(s) of a signal x(t) is given as
X(s) =
x(t)e−st dt
−∞
We know that s = σ + jw
X(s) =
X(s) =
Z∞
−∞
Z∞
−∞
x(t)e−(σ+ jw)t dt
x(t)e−σt e− jwt dt
(4)
Fourier Series and Fourier Transform
• 133
Comparing (1) and (4), we find that L.T. of x(t) is basically the F.T. of [x(t)e−σt ].
R∞
If s = jw, i.e. σ = 0, then eqn. (4) becomes X(s) = −∞
x(t)e− jwt dt = X( jw)
Thus, X(s) = X( jw) when σ = 0 or s = jw
This means L.T. is same as F.T. when s = jw. The above equation shows that F.T. is special case of L.T.
Thus, L.T. provides broader characterization compared to F.T., s = jw indicates imaginary axis in complex
s-plane.
2.7
APPLICATIONS OF FOURIER TRANSFORM OF NETWORK ANALYSIS
Example 24: Determine the voltage Vout (t) to a current source excitation i(t) = e−t u(t) for the circuit shown
in figure.
i(t) ↑
+
1
F Vout(t)
2
1Ω
−
Fig. 2.26.
S OLUTION:
↓ i1(t)
1Ω
i(t) ↑
↓ i2(t)
1
2F
+
Vout(t)
−
i(t) = i1 (t) + i2 (t)
i(t) =
(
Vout (t) 1 dVout (t)
+
1
2 dt
V
R
i = c dv
dt
since i =
and
e−t u(t) = Vout (t) +
or v =
1
c
1 dVout (t)
2 dt
R
idt
On taking the z-transform on both sides
1
(2 + jw)
jw
=
= Vout ( jw) 1 +
Vout ( jw)
1 + jw
2
2
A
B
2
=
+
Vout ( jw) =
(1 + jw)(2 + jw) 1 + jw 2 + jw
2
2
Vout ( jw) =
−
1 + jw 2 + jw


A(2 + jw) + B(1 + jw) = 2


 2A + B = 2
 A + B = 0 s0 A = −B


 2A − A = 2; A = 2, B = −2
(1)
Fourier Series and Fourier Transform
V0 ( jw) =
V0 ( jw) =
2
2
(6 jw + 1)( jw + 1)
=
6( jw)2 + 7( jw) + 1
1/3
=
( jw + 1/6)( jw + 1)
V0 ( jw) =
5
Taking inverse Fourier transform, we get
1
6
1
6
B
A
+
+ jw 1 + jw
2
2
−
5(1 + jw)
+ jw
V0 (t) =
• 135
(5)
2 −t/6
e
− e−t u(t)
5
(6)
Example 26: Determine the response of current in the network shown in Fig. 2.28(a) when a voltage having
the waveform shown in Fig. 2.28(b) is applied to it by using the Fourier transform.
v(t)
1Ω
v(t)
∼
1F
π
0
(a)
wt
(b)
Fig. 2.28.
S OLUTION:
Waveform V (t) is defined as
V (t) = sint(u(t) − u(t − π))
(1)
1Ω
u(t)
∼
a
i(t)
1F
Let i(t) be the current in the loop. Applying KVL in loop
1
V (t) = 1 · i(t) +
1
Zt
0
i(t)dt = i(t) +
Zt
i(t)dt
0
On taking Fourier transform of
1
e− jπw
+
( jw)2 + 1 ( jw)2 + 1
1
FT sintu(t) =
( jw)2 + 1
e− jπw
FT sintu(t − π) =
( jw)2 + 1
V ( jw) =
Since
(2)
•
136
Basic System Analysis
Solve using F.T. formula
V ( jw) =
1 + e− jπw
( jw)2 + 1
(3)
1
I( jw)
jw
1
jw + 1
V ( jw) = 1 +
I( jw) =
I( jw)
jw
jw
V ( jw) = I( jw) +
I( jw) =
jw
V ( jw)
jw + 1
(1 + e− jπw )
jw
·
From (3)
jw + 1 (( jw)2 + 1)
jw
e− jπw
1
I( jw) =
·
+
jw + 1
( jw)2 + 1 ( jw)2 + 1
I( jw) =
1
jw
1
jw
·
+
·
· e− jπw
( jw + 1) (( jw)2 + 1) ( jw + 1) (( jw)2 + 1)
|
{z
}
I2 ( jw)
I1 ( jw)
=
I1 ( jw) =
=
B jw + c
A
+
jw + 1 (( jw)2 + 1)
1
( jw + 1)
−1/2
+ 2
( jw + 1) (( jw)2 + 1)
1
1
1
1
i1 (t) = − e−t u(t) + costut + sintδt + sintu(t)
2
2
2
2
Since IFT
n
so IFT
1
( jw)2 +1
o
jw
( jw)2 +1
= sintu(t)
=
d
dt
sintu(t)
Using differential in time domain property
jw
= costu(t) + sintδ(t)
IFT
( jw)2 + 1
I2 ( jw) =
jw
1
·
· e− jπw
( jw + 1) (( jw)2 + 1)
I2 ( jw) = I3 ( jw) · e− jπw
Since
so
I3 = I1 ( jw)
1
1
1
1
i3 (t) = − e−t u(t) + costu(t) + sintδ(t) + sintu(t)
2
2
2
2
(4)
Fourier Series and Fourier Transform
• 137
From time shifting property FT (x(t ± t0 )) = e± jwt0 × ( jw)
i2 (t) = i3 (t − π)
so
1
1
1
1
= − e−(t−π) u(t − π) + cos(t − π)u(t − π) + sin(t − π)δ(t − π) + sin(t − π)u(t − π)
2
2
2
2
i
1
1h
1 i(t) = − −e−t + cost + sint u(t) + sintδ(t) + −e−(t−π) + cos(t − π) + sin(t − π) u(t − π)+
2
2
2
1
sin(t − π)δ(t − π)
2
Example 27: For the RC circuit shown in figure.
R
i(t)
x(t)
C
1
y(t)
Fig. 2.29.
(a) Determine frequency response of the circuit.
(b) Find impulse response.
(c) Plot the magnitude and phase response for RC = 1.
S OLUTION:
Applying KVL in loop (1)
1
x(t) − Ri(t) −
C
x(t) = Ri(t) +
Zt
i(t)dt = 0
Zt
i(t)dt
−∞
1
C
(1)
−∞

Since


VR = iR

R

Vc = C1 i(t)dt
and y(t) =
1
C
Zt
−∞
i(t)dt
(2)
Fourier Series and Fourier Transform
• 139
A = B −C
1
H( jw) = √
1 + w2
(8)
H( jw) = 1 − (1 + jw)
= tan−1
0
− tan−1 w = − tan−1 w
1
For different values of w, we find H( jw) and H( jw)
S. No
w
|H( jw)|
H( jw)
1−
2−
3−
4−
5−
6−
7−
8−
9−
10−
11−
12−
13−
14−
15−
−∞
−50
−20
−10
−5
−2
−1
0
1
2
5
10
20
50
∞
0
0.0199
0.0499
0.099
0.196
0.447
0.707
1
0.707
0.447
0.196
0.099
0.0499
0.0199
0
90◦
88.9◦
87.1◦
84.3◦
78.7◦
63.4◦
45◦
0
−45◦
−63.4◦
−78.7◦
−84.3◦
−87.1◦
−88.9◦
−90◦
| H(jw) |
1
−50 −40 −30 −20 −10 0
10 20 30 40 50 w
Fig. 2.30. Magnitude plot frequency response of the circuit
(9)
140
•
Basic System Analysis
∠H(jw)
90°
45°
0 10 20 30 40 50 w
−50 −40 −30 −20 −10
−45°
−90°
Fig. 2.31. Phase plot
Example 28: For the circuit shown in figure, determine the output voltage V0 (t) to a voltage source excitation
Vi(t) = e−t u(t) using Fourier transform
2Ω
Vin(t) +
−
1H
1
+
V0(t)
−
Fig. 2.32.
S OLUTION:
Since Vin(t) = e−t u(t)
(1)
1
1 + jw
(2)
Vin( jw) =
Applying KVL in loop (1)
Vin(t) = 2i(t) + 1 ·
Vin(t) = 2i(t) +
V0 (t) = 1 ·
V0 (t) =
di(t)
dt
di(t)
dt
(3)
di(t)
dt
di(t)
dt
(4)
142
•
Basic System Analysis
Q3: (i) State and prove the following properties of Fourier series:
(a) Time shifting property
(b) Frequency shifting property
(ii) What are Dirichlet’s conditions?
Q4: Find the fundamental period T , the fundamental frequency w0 and the Fourier series coefficients an of
the following periodic signal;
x(t)
1
t
−1 −0.5
t
0 0.5
−1
Fig. 2.3 P.
Q5: Obtain the Fourier series component of the periodic square wave signals.
x(t)
1
−T/2
−T/4
0
T/4
T/2
−1
Fig. 2.4 P.
Q6: Determine the Fourier transform of the Gate function
x(t)
A
−T/2
T/2 t
Fig. 2.5 P.
Q7: Determine the Fourier series representation of the signal
t − t 2 for − π ≤ t ≤ π
x(t) =
0 elsewhere
t
Fourier Series and Fourier Transform
• 143
Q8: For the continuous-time periodic signal
x(t) = 2 + cos[2πt/3] + 4 sin[5πt/3]
determine the fundamental frequency w0 and the Fourier series coefficients Cn such that
∞
x(t) =
∑
Cn e jnw0 t
n=−∞
Q9: Find the Fourier transform of the following signals:
(a) x(t) = δ(t)
(b) x(t) = 1
(c) x(t) = sgn (t)
(d) x(t) = u(t)
(e) x(t) = exp(−at)u(t)
(f) x(t) = cos [w0t] sin [w0t]
Q10: Show that the Fourier transform of rect (t − 5) is Sa(w/2) exp( j5w). Sketch the resulting amplitude
and phase spectrum.
Q11: Find the inverse Fourier transform of spectrum shown in figure.
∠X(w)
π/2
| X(w) |
w0
1
−w0
w
−π/2
−w0
w0 w
(a)
(b)
Fig. 2.6 P.
Q12: Find the Fourier transform of the following waveform.
x(t)
1
−b
−a
0
a
b
t
Fig. 2.7 P.
Q13: State and prove duality property of CTFT.
Q14: Determine the Fourier transform of the signal
x(t) = {tu(t)∗ [u(t) − u(t − 1)]}, where u(t) is unit step function and ∗ denotes the convolution operation.
Q15: Show that the frequency response of a CTLTIS is Y (w) = H(w)X(w)
where X(w) = Fourier transform of the signal x(t)
H(w) = Fourier transform of LTIS response h(t)
144
•
Basic System Analysis
Q16: Find the Fourier transform of the signal x(t) shown in figure below.
x(t)
A
0
T
t
2T
Fig. 2.8 P.
Q17: Determine the frequency response H( jw) and impulse response h(t) for a stable CTLTIS characterized
by the linear constant coefficient differential equation given as
d 2 y(t)/dt 2 + 4dy(t)/dt + 3y(t) = dx(t)/dt + 2x(t)
Q18: Find the Fourier transform of the signal x(t) shown in figure below.
x(t)
K
−T
0
T
t
Fig. 2.9 P.
Q19: If g(t) is a complex signal given by g(t) = gr (t) + jgi (t) where gr (t) and gi (t) are the real and
imaginary parts of g(t) respectively. If G( f ) is the Fourier transform of g(t), express the Fourier transform
of gr (t) and gi (t) in terms of G( f ).
Q20: Find the coefficients of the complex exponential Fourier series for a half wave rectified sine wave
defined by
A sin (w0t), 0 ≤ t ≤ T0 /2
x(t) =
0, T0 /2 ≤ t ≤ T0
with x(t) = x(t + T0 )
Q21: (a) Show that the Fourier transform of the convolution of two signals in the time domain can be given
by the product of the Fourier transform of the individual signals in the frequency domain.
(b) Determine the Fourier transform of the signal
1
1
1
x(t) =
δ(t + 1) + δ(t − 1) + δ t +
δ+ t −
2
2
2
146
•
Basic System Analysis
1 − (−1)n
n2 π2
1
bn =
nπ
an =
Q3:
x(t)
1
−1 −0.5
0 0.5
t
1
−1
T =1
w0 = 2π rad/ sec
y − y1 =
y2 − y1
(x − x1 )
x2 − x1
x(t) = −2t + 1
an =
tZ
0 +T
2
T
x(t) cos n w0t dt
t0
an = 0
Q4:
1.0
−T/2
x(t)
−T/4
T/4
T/2
−1.0
8π
T
3T
T
2π
=
− −
; w0 = 3T =
2
4
4
3T
4
(
1 − T4 ≤ t ≤ T4
x(t) =
−1 T4 ≤ t ≤ T2

 T
T


Z4
Z2


4 T
1
1
dt + (−1)dt =
=
a0 = 3T


3T
4
3

4  T
T
−4
4
t
Fourier Series and Fourier Transform

 T
T


Z4
Z2

8nπ
8nπ 
8
cos
dt − cos
tdt
an =

3T 
3T
3T


− T4
T
4
1
2nπ
4nπ
an =
3 sin
− sin
nπ
3
3
bn = 0, since even function
1 1
2π
4π 3
4π 1
8π
x(t) = +
3 sin
− sin
+ sin
− sin
+···
3 π
3
3
2
3
2
3
Q5:
x(t)
A
−T/2
x(t) =
(
T/2 t
A − T2 ≤ t ≤
0
T
2
elsewhere
T
X( jw) = A
Z2
e− jwt dt =
− T2
wT
2A
wT
AT
sin
= wT sin
w
2
2
2
X(i f ) = AT sin c f T
Q6:
T0 = 2π;
w0 = 1;
a0 =
1
2π
1
an =
π
bn =
1
π
Zπ
−π
Zπ
−π2
t − t 2 dt =
3
−π
−4(−1)n
t − t 2 cos nt dt =
n2
−π
−2(−1)n
t − t 2 sin nt dt =
n
Zπ
• 147
Fourier Series and Fourier Transform
Taking inverse Fourier transform
1
x1 (t) =
2π
Zw0
0
x2 (t) =
1
2π
− j e jwt dw =
Z0
j e jwt dw =
−w0
x(t) = x1 (t) + x2 (t) =
=
1 − e jw0 t
2πt
1 − e− jw0 t
2πt
1
(1 − e jw0 t + 1 − e− jw0 t )
2πt
2 sin2 w20 t
1
(2 − 2 cos w0t) =
2πt
πt
Q11:
x(t)
1.0
−b
−a
x(t) =



X( jw) =


0
t+b
b−a
1
t−b
a−b
a
b
t
for − b < t < −a
for − a < t < a
for
a<t <b
2
(cos wa − cos wb)
w2 (b − a)
Q12:
x(t) = tu(t)∗ [u(t) − u(t − 1)]
x1 (t) = tu(t)
Differentiating in frequency domain property
x2 (t) = u(t) − u(t − 1)
d
X( jw)
dw
1
X1 ( jw) =
( jw)2
FT(tx(t)) = j
X2 ( jw) =
Z1
0
1.e− jwt dt =
1
(1 − e− jw )
jw
X( jw) = X1 ( jw)X2 ( jw) =
1
(1 − e− jw )
( jw)3
• 149
•
150
Basic System Analysis
Q13: Prove convolution in time domain property.
Q14:
x(t)
(T,A)
A
0
(0,0)
x(t) =
0<t <T
A T < t < 2T
X( jw) =
A
T
ZT
T
2T t
A
Tt
te− jwt dt + A
A te− jwt
X( jw) = 
T − jw
e− jwt dt
T
0

Z2T
ZT
0
−
ZT − jwt
e
0

− jwt
e
2T

dt + A
jw
− jw T
− j2wT
A Te jwt
1 − jwT
e
− e− jwT
=
+
e
−1 +A
T − jw w2
− jw
− jwt
A
1
e
+
e− jwT − 1 − e− jwT e− jwT − 1
=A
− jw w2 T
jw
Ae− jw T
A
A
A
A
+ 2 e− jwT − 2 − e− j 2wT + e− jwT
jw
w T
w T
jw
jw
A
1 − jwT 1
−2 jwT
=
e
− + jTe
wT w
w
=
Q15:
dy(t)
dx(t)
d 2 y(t)
+4
+ 3y(t) =
+ 2x(t)
dt 2
dt
dt
(1)
Taking Fourier transform on both sides
( jw)2Y ( jw) + 4( jw)Y ( jw) + 3Y ( jw) = ( jw)X( jw) + 2X( jw)
( jw)2 + 4( jw) + 3 Y ( jw) = (( jw) + 2) X( jw)
Frequency response H( jw) =
2 + jw
Y ( jw)
=
X( jw) ( jw)2 + 4 jw + 3
H( jw) =
2 + jw
A
B
=
+
(3 + jw)(1 + jw) 3 + jw 1 + jw
(2)
(3)
152
•
Basic System Analysis
x(t) = A sin w0t for 0 ≤ t ≤
=0
for
T0
2
T0
≤ t ≤ T0
2
T0
1
C0 =
T0
Z2
A
A sin w0tdt =
T0
0
T0
2
− cos w0t
w0
0
A
A
T0
A
cos w0 · − 1 = − [cos π − 1] =
=−
2π
2
2π
2
T0 · T0
T0
1
Cn =
T0
Z2
A sin w0te− jnw0 t dt
0
T0
=
A
2 jT0
Z2
0
(e jw0 t − e− jnw0 t )e− jnw0 t dt
T0
A
=
2 jT0
Z2 0
e jw0 t(1−n) − e− jw0 t(n+1) dt
e− jw0 t(n+1) T20
e jw0 t(1−n)
−
jw0 (1 − n) − jw0 (n + 1) 0
A
=
2 jT0
A
=
2 jT0 w0
T0
!
T0
e jw0 (1−n) 2
e− jw0 (n+1) 2
1
1
+
−
−
1−n
(n + 1)
1−n n+1
!
"
#
A e jπ(1−n) e− jπ(n+1)
1
1
=−
+
−
−
4π 1 − n
n+1
1−n n+1
e jπ e− jnπ e− jnπ · e− jπ
1
1
+
−
−
1−n
n+1
1−n n+1
− jnπ
A −e
e− jnπ
1
1
=−
−
−
−
4π
1−n
n+1 1−n n+1
A 2e− jnπ
2
=
+
4π 1 − n2 1 − n2
A
=−
4π
=
A
(e− jnπ + 1)
2π(1 − n2 )
Since e jπ = −1
Fourier Series and Fourier Transform
Q19:
1
1
1
δ(t + 1) + δ(t − 1) + δ t +
+δ t −
2
2
2
Taking Fourier transform on both sides
x(t) =
X( jw) =
Z∞
x(t)e− jwt dt
Z∞
1
1
1
δ(t + 1) + δ(t − 1) + δ t +
+δ t −
e− jwt dt
2
2
2
(1)
−∞
X( jw) =
−∞
X( jw) =

1
2
Z∞
δ(t + 1)e− jwt dt +
−∞
+
Since FT(δ(t)) = 1
Z∞ δ t+
−∞
1 − jwt 
δ t−
e
dt
2
So FT(δ(t ± t0 )) = e± jwt0 dt
X( jw) =
−∞
δ(t − 1)e− jwt dt +

Z∞ −∞
Z∞
{using time shifting property}
w
w
1 jw
e + e− jw + e j 2 + e− j 2
2
w
w
e jw + e− jw e j 2 + e− j 2
X( jw) =
+
2
2
w
X( jw) = cos w + cos
2
OBJECTIVE TYPE QUESTIONS
Q1: If the Fourier transform of a function x(t) is X( jw), then X( jw) is defined as
R∞
R ∞ dx(t) − jwt
e
dt
(a) −∞
x(t)e jwt dt
(b) −∞
R∞
R ∞ dt − jwt
(c) −∞ x(t)dt
(d) −∞ x(t)e
dt
Q2: If X( jw) be the Fourier transform of x(t), then
1 R∞
1 R∞
− jwt dw
(a) x(t) = 2π
X( jw)e jwt dw
(b) x(t) = 2π
−∞
−∞ X( jw)e
R
∞
1 R∞
1
(c) x(t) = 2π 0 X( jw)e jwt dw
(d) x(t) = 2π −∞ X( jw)e− jwt dw
Q3: Fourier transform of x(t) = 1 is
(a) 2π δ(w)
(b) π δ(w)
(c) 3π δ(w)
Q4: Fourier transform of x(t − t0 ) is
(a) e− jwt 0 X( jw)
(b) e jwt 0 X( jw)
(c)
(d) 4π δ(w)
1
t0 X( jw)
(d) t0 e− jwt 0 X( jw)
1 − jwt
e
dt
2
• 153
Fourier Series and Fourier Transform
• 155
Q19: The trigonometric Fourier series of a periodic time function have
(a) sine terms
(b) cosine term
(c) both (a) and (b)
(d) DC term
Q20: Fourier series is defined as
∞
x(t) = a◦ + ∑ (an cos nw0t + bn sin w0t)
n=1
(a) True
(b) False
Answers: (1) d
(2) a
(6) c
(7) a
(11) c (12) b
(16) a (17) e
(3) a
(8) a
(13) b
(18) c
(4) a
(9) a
(14) a
(19) c
(5) b
(10) d
(15) a
(20) a
UNSOLVED PROBLEMS
Q1: Show that the Fourier transform of x(t) = δ(t + 2) + δ(t) + δ(t − 2) is (1 + 2 cos 2w).
Q2: Show that the inverse Fourier transform of X( jw) = 2πδ(w) + πδ(w − 4π) + πδ(w + 4π) is x(t) =
1 + cos 4πt.
2
Q3: Calculate the Fourier transform of te−|t| , using the F.T. pair, FT e−|t| = 1+w
2 . Also find the Fourier
4t
transform of (1+t 2 )2 using duality property.
Q4: X( jw) = δ(w) + δ(w − π) + δ(w − 5); find IFT x(t) and show that x(t) is non-periodic.
Q5: Find the Fourier transform of the triangular pulse as shown in figure.
x(t)
1
−T/2
0
T/2
t
Fig. 2.10 P.
Ans. X( jw) =
Q6: Find the Fourier transform of x(t) = rect(t/2).
T
2
sin c2 ( wt
4 )
Ans. X( jw) = 2 sin cw
Q7: Find the Fourier transform of the signal x(t) = cos w0t by using the frequency shifting property.
Ans: X( jw) = π[δ(w − w0 ) + δ(w + w0 )]
Q8: Show that FT [sin w0tu(t)] =
w0
w20 −w2
+ π2j [δ(w + w0 ) − δ(w − w0 )].
Q9: Find inverse Fourier transform of X( jw) =
jw
(1+ jw)2
Ans. x(t) =
d
dt
[te−t u(t)]
156
•
Basic System Analysis
Q10: Sketch and then find the Fourier transform of following signals
x1(t)
(a) x1 (t) = π t + 23 +π t − 23
−1
Ans. (a)
−2 −1
X1 ( jw) = 2 sin c w2 cos 3 w2
2
1
t
Fig. 2.11 P.
2
(b) x2 (t) = π
t
4
+π
t
2
x2(t)
−1
Ans. (b)
−2 −1
X2 ( jw) = 4 sin c2w+2 sin cw
1
2
t
Fig. 2.12 P.
Q11: Find the frequency response x( jw) of the RC circuit shown in figure. Plot the magnitude and phase
response for RC = 1
x( jw) =
↑
y( jw)
1
=
x( jw) 1 + jwRC
R
x(t)
↑
−
−C
↑
y(l)
↑
Fig. 2.13 P.
Ans. |x( jw)| = √
1
1 + w2
x( jw) = − tan−1 w
Q12: Find the Fourier series of the waveform shown in figure.
x(t) =
2A
for n = 1, 3, 5, 7
jnπ
158
•
Basic System Analysis
2
1 1
+ cos 5πt −
cos 10
π 2
3π
2
−
cos 15πt
8π
x(tπ)t =
Q15: The output of a system is given by
x(t) =
A sin w0t for
0
for
Determine trigonometric form of Fourier series of x(t) "
0≤t ≤π
π ≤ t ≤ 2π
∞
2A
π
A A
cos nt
Ans. x(t) = + cos(nt − ) + ∑
π 2
2
π(1
− n2 )
n=2
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