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Dr. Mohammed Safiuddin
Chapter 1
REVIEW OF MATHEMATICS FOR ENGINEERS
Dedicated to engineering faculty everywhere
Ó 2020
1
Dr. Mohammed Safiuddin
PREFACE
As a full time employee at the Buffalo Divisions of Westinghouse Electric Corporation, I accepted
the assignment to teach an introductory industrial electronics course in the Millard Fillmore College, the
Evening Division of University of Buffalo, in the Fall semester of 1962. Initially, I was to teach the
Feedback Control Systems course but, just before the classes were to begin, Chairman of Electrical
Engineering Department informed me that the course had been assigned to another part-time instructor
from a local aerospace research company, and that instead I was to teach the industrial electronics course.
However, as the semester was ending, the Chairman asked me if I would be able to teach the Feedback
Control Systems course to be re-offered in the Spring semester because all the students had dropped the
course after the first Quiz. I accepted the offer. As I started the course, I asked the class as to how many
of them have had a course in Calculus and/or Advanced Mathematics. All twenty students raised their
hands. Then I asked them as to how many had it ten years ago. Most of them raised their hands. Then I
asked again as to how many had it five years ago. A few hands were raised. Then, I showed the students
the prescribed text book and told them that we would be spending first three weeks to Review Mathematics
before we get to the text book material. As the thyristor powered motor drive systems were being
commercialized in the mid 1960s, Westinghouse Electric’s Industrial Drive Systems Division in Buffalo
developed a two-week intensive short course offered annually to its own application engineers and those of
its licensees around the world from 1967-1977. Again the course started with review of mathematics
fundamentals. Since then my Industrial Control and Motor Drive Systems courses always started with first
three weeks set aside for the review of mathematics. Hence this Chapter with the title REVIEW OF
MATHEMATICS FOR ENGINEERS.
When you ask engineering students to define ZERO and ONE or UNITY, they are puzzled and
can’t really answer it. They were taught in the elementary school that multiplication of numbers is like a
recurrent addition, but when you ask them to explain why the product of two negative numbers is positive,
they are again at a loss. So, I start this chapter with first defining a set of universal elementary mathematical
operations consisting of eight steps, which include definitions of ZERO and ONE/UNITY for all number
systems used in formulating engineering problems. A dimensional space is introduced for expressing all
types of engineering measurements from REAL variables, as single dimensional on a line from -¥ to +¥
passing through ZERO in the middle, to multi-dimensional vector spaces. As the Table of Content lists,
the topics covered range from scalars and vectors to computer arithmetic, matrix algebra, linear
simultaneous equations, differentiation & integration, time-domain differential equations, Laplace
transforms, and probability & statistics. The topic of Z-transforms is not covered since the technique is no
longer in use by practicing engineers. In my own twenty five year career at Westinghouse Electric, with
responsibility to design some of the most complex industrial control systems, I never used it for analysis or
synthesis. Both analog and digital computer modeling were found to be more than adequate.
I also had an opportunity to teach mathematics in the 1970s for the review course for the New York
State Professional Engineers License Exam. A selected set of twenty five problems from these exams are
included in the Appendix, with their respective solutions by UB graduate student, Sabbir S. Moochalla. His
time and efforts are acknowledged with appreciation.
Mohammed Safiuddin
September 2020
(The year of the COVID-19)
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CHAPTER I
REVIEW OF MATHEMATICS FOR ENGINEERS
Elementary Mathematical Operations
1.1 Scalars & Vectors
1.2 Computer Arithmetic
1.3 Matrix Algebra
1.4 Linear Simultaneous Equations
1.5 Differentiation & Integration
1.6 Linear Differential Equations (Time Domain)
1.7 Laplace Transforms
1.8 Probability & Statistics
Appendix A – Sample Problems from New York State PE License Exams
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CHAPTER I: REVIEW OF MATHEMATICS FOR ENGINEERS
Elementary Mathematical Operations
Mathematics is a language that we use to express relationships amongst various data, so as to
understand the system behavior, in order to solve everyday problems, we encounter in our lives. These data
comprise quantitative measurements of variables for which we need a ‘Number System’. For example,
real, complex, binary, hexadecimal etc. are typical number systems used in solving engineering problems.
Each of these number systems consist of a set of elements, such as 0,1, 2, 3……………to infinity in REAL
number system.
Common to all number systems, we first define a set of " Elementary Mathematical Operations ".
While these operations may appear to be different for different systems, they follow a general pattern
consisting of the steps described below:
1.
Define rule for ADDITION of two elements A and B such that
A + B = C (a third element)
2.
Search for existence of a ZERO (O) element such that when it is added to any element A of the
system using the rule defined in step (1) it does not alter its value. That is,
A + O = A?
3.
If a ZERO element exists, then search for an element A' such that when it is added to element A the
resultant yields the ZERO element. That is,
A + A' = O?
If such an element exists then define the element A' as negative (complement) of the element A.
(A' = - A)
4.
Define SUBTRACTION as an operation of ADDITION of one element to the complement of
another element. That is,
A - B = A + B'
5.
Define rule for MULTIPLICATION of two elements A and B such that
(A) x (B) = C (a third element)
6.
Search for the existence of a UNITY `U', as an element such that when multiplied to another element
A, using the rule of multiplication in step (5), it does not alter the value of the element A. That is,
(U) x (A) = (A)?
7.
If the unity element exists, then search for an element A-1 such that when it is multiplied by the
element A the resultant yields the unity element `U'. That is,
(A) x (A-1) = U?
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If such an element exists, then define the element A-1 as inverse (reciprocal) of the element A. That
is,
A-1 = Inverse of A
8.
Define DIVISION of an element A by another element B as MULTIPLICATION of the element A
with that of the inverse of B. That is,
𝐴 ÷ 𝐵 = (A) x (B-1)
These elementary mathematical operations are further governed by three properties with respect to
the order in which the elements are selected for these elementary operations.
1. COMMUTATIVE, if
A+B=B+A
(A) x (B) = (B) x (A)
(A) x (B-1) = (B-1) x (A) etc.
2. ASSOCIATIVE, if
A + (B + C) = (A + B) + C
(A) x [(B)x(C)] = [(A)x(B)] x (C)
3. DISTRIBUTIVE, if
(A) x (B+C) = (A) x (B) + (A) x (C)
When variables in control system problems have more than one quantitative attribute, they are often
expressed by multi-dimensional quantities called VECTORS. For example,
Complex numbers, sinusoidal waveforms, force and velocity of bodies in motion.
Scalars with positive and negative values can be treated as vectors of single dimension, expressed by
magnitude and an angle of zero radians for the positive numbers and an angle of p radians for the negative
numbers with respect to the reference axis, along the REAL LINE. On the other hand, so called COMPLEX
numbers are expressed as vectors in a 2-dimensional plane defined by their components along the
HORIZONTAL (REAL) and the VERTICAL (IMAGINERY) axes.
When the values of a number of variables are needed to describe the condition (STATE) of a given
system, these variables can be treated as components of a single variable, and expressed as a
multi-dimensional vector variable. For Example, air gap flux, armature current and armature voltage for a
DC motor can be treated as three components of a vector variable which describes the `state' of the motor
at the specified instant of time.
All the elementary operations are applicable to the single and two-dimensional vectors. However,
the same is not true for the three and higher dimensional vectors. Step (6) for search of a UNITY element
for the set of three-dimensional vectors results in the conclusion that no such vector exists. No inverse of
three-dimensional vectors exist and hence no division of two vectors is possible in the three-dimensional
space. The vector notations and the elementary operations for the three-dimensional vectors are defined to
facilitate expression of relationships and physical phenomena observed experimentally. For Example,
Power in AC circuits can be expressed as a DOT product of voltage and current because it is a scalar
quantity. Similarly, electro-magnetic force can be expressed as a CROSS product of current and flux
density because it is vector quantity with direction of motion perpendicular to the plane of the current and
the flux vectors.
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1.1
Scalars & Vectors
We define scalars as 'single-dimensional' vectors along the real-line and treat them as a sub-set
of the so called "Complex Numbers". That is, we define all positive scalars as vectors at zero radian
angle with respect to the real-axis and with the length representing their value.
1. Addition: Add scalars A & B by placing the start of scalar B to the end of scalar A on the
REAL line
A+B=C
The total length represents the resultant C.
2. Zero:
Scalar of length zero.
3. Complement (negative): Scalars along the real axis pointing in the opposite direction and of
length equal to their respective positive scalars. That is, vectors along the real-line with an
angle of π radians as opposed to the vectors with the radian angle of zero.
4. Subtraction: Subtract scalar B from scalar A by adding the complement of B to A.
A - B = A + (B') = C if A > B and C' if A< B
5. Multiplication:
Multiply scalars A and B by multiplying their magnitudes and by adding
their directional angles of ‘0' (for positive scalars) and 'π' (for negative scalars).
(A) x (B) = |A| x |B| Ð (0 + 0) = AB Ð0 = AB
(A) x (-B) = (A) x (B') = |A| x |B| Ð (0 + π) = AB Ð π = - AB
(-A) x (-B) = (A') x (B') = |A| x |B| Ð (π + π) = AB Ð 2π = AB
6. Unity: Scalar of one-unit length pointing in the direction of positive scalars. [U = 1 Ð0]
(A) x (U) = |A| .1. Ð(0 + 0) = A
7. Inverse:
Inverse of scalar A is a scalar with the magnitude such that its product with the
magnitude of A equals the unity scalar and its angle is same as the angle of A. NOTE: The
Zero scalar has no inverse.
(A) x (A-1) = |A| . [1/|A|] Ð (0 + 0) = 1 Ð 0 = ‘U'
(-A) x (-A-1) = |A| . [1/|A|] Ð (π + π) = 1 Ð 2π = 'U'
8. Division:
Divide scalar A by scalar B, by multiplying the scalar A by inverse of scalar B.
𝐴 ÷ 𝐵 = (A) x (B-1) = |A| . [1/|B|] Ð (0 + 0) = A/B
When a scalar is multiplied by itself, like (A) x (A), it is raised to powers by use of exponents.
Certain basic rules apply when adding and multiplying such scalars. For example: Consider positive
scalars (A) and (B) and positive or negative exponents (x) and (y).
(B)-x = 1/(B)x ; (B)x. (B)y = (B)(x+y) ; (A.B)x = (A)x.(B)x ; (B)x.y = [(B)x]y ; (B)x+(B)x= 2(B)x
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Logarithms
Logarithm answers the simple question of how many times can a given number be multiplied
to get another number. For scalars, b, y and x, if by = x for positive b and x, then logb x = y, where b
is defined as ‘Base’ of the logarithm. Two base values are commonly used.
b = 10 for common or Briggsian logarithms
e = 2.718 for Natural, Naperian or Hyperbolic logarithms [logs]
Rules in use of logarithms are:
log ab = log a + log b
log a/b = log a – log b
log (a)n = n log a
,
log *√𝑎 = - log 𝑎
Antilogarithms or inverse logarithms are also defined so that transformed operations can be
performed in the logarithmic domain and final solutions can be obtained in the original number
domain.
Since most numbers are irrational powers of 10, a common logarithm, in general, consists of
an integer called the CHARACTERISTIC and an endless decimal called the MANTISSA.
Example:
log 256 = 2.40824 ® Mantissa (from the Log table)
¯
Characteristic (1 less than the number of digits of the number left of the decimal point)
log 256,000,000 = log (256 x 106) = log 256 + log 10 6 = log 256 +6 log 10 = 2.40824+6 = 8.40824
log 0.00000256 = log (2.56 x 10-6 ) = log 2.56 +log 10 -6 = log 2.56 -6log 10 = 0.40824-6 = -5.59176
Since the hand calculators can instantly give us results for multiplication problems, the use
of logarithms for simple multiplications is no longer needed. However, numerical data over a large
range can be compressed for presentation on a “semi-log” or “log-log” graph papers. Also, the
technique is very valuable in frequency domain analysis of control systems.
Complex Numbers
Complex numbers are not really complex, and √−1
does exist. It doesn’t exist on the REAL line, but on the line
perpendicular to it, referred to in most books as the j axis. These
are two-dimensional numbers that can be represented in four
different forms.
* Rectangular Coordinate Representation Z = a + j b
* Polar Coordinate Representation Z = |Z| Ðq
* Trigonometric Representation Z = |Z| (Cos q + j Sin q)
* Exponential Representation Z = |Z| e jqwhere, j =
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1. Addition: Add real components for the real part of the sum and j components for the j part of
the sum
Z 1 = a1 + j b1 ; Z 2 = a 2 + j b2
Z 1 + Z 2 = ( a1 + a 2 ) + j ( b1 + b2 )
2. Zero: Both components α and b equal to zero; |Z| = 0
Z 1 + 0 = ( a1 + 0 ) + j ( b1 + 0 ) = Z 1
3. Complement (Negative): Both components a and b in opposite direction. Add 180 degrees to
the polar form and others.
Z 1 = a1 + j b1 ; Z 1¢ = - a1 - j b1
Z 1 + Z 1¢ = ( a1 - a1 ) + j ( b1 - b1 ) = 0
4.
Subtraction: Subtract Z 2 from Z 1 by adding complement of Z 2 [ Z 2¢ ] to Z 1 .
Z 1 - Z 2 = Z 1 + Z 2¢ = ( a1 - a 2 ) + j ( b1 - b2 )
5. Multiplication: Multiply Z 1 and Z 2 by multiplying their magnitudes |Z1 | and |Z2 |and adding
their angles q1 and q2.
(Z1) x (Z2) = |Z1| . |Z2| ; Ð (q1 + q2)
6. Unity: Complex number of unit length along the real axis. Also Z with a = 1 and b = 0.
(Z1) x (U) = |Z1| . 1 ; Ð(q1+0) = Z1
7. Inverse: Inverse of a complex number Z is the complex number
with magnitude [1/|Z|] and angle negative of the angle of Z .
(Z1) x (Z1)-1 = |Z1| . [1/|Z1|] = 1; Ð (q1 + (- q1) = 0; That is 'U'
8. Division: To divide complex number Z 1 by Z 2 , multiply Z 1
by the inverse of Z 2 .
(Z1) ÷ (Z2) = (Z1) x (Z2)-1 = |Z1| . 1/|Z2|; Ð (q1 - q2)
Example 1.
,
If 𝑓 (𝑧) = 𝑧 4 + 6𝑧 + 7 , find f(3j)
Solution:
,
(;),
;
,
𝑓 (3𝑗 ) = (3𝑗 )4 + 6(3𝑗 ) + :; = 9𝑗 4 + 18𝑗 + (:;); = −9 + 18𝑗 + :; > = −9 + 18𝑗 − : 𝑗
Simplifying the above, we get 𝑓 (3𝑗) = −9 +
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𝑗
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Dr. Mohammed Safiuddin
Example 2.
Capacitive and inductive elements with impedances (in ohms) 7-12j and 3+26j,
respectively, are connected in parallel. Find the circuit impedance in rectangular
and polar forms.
Solution:
Let 𝑧, = 7 − 12 𝑗 and 𝑧4 = 3 + 26 𝑗
7C7>
The total impedance z for z1 and z2 in parallel, 𝑍 =
𝑍=
:::D,JG;
,ID,J;
:::D,JG; (,IF,J;)
= K ,ID,J; L (,IF,J;) =
7CD7>
(EF,4;)(:D4G;)
= (EF,4;)D(:D4G;) =
:::IFJGG4;D,JGIMF4IJJ ; >
,IID,NG
4,D,H4;F:G;D:,4
,ID,J;
= 18.16 − 10.82𝑗 Ω
Example 3. The total impedance in a circuit is given by (Z = Z1 + Z2)
Where, Z1 = 4Ð 0 0 and Z 2 = 8Ð45 0 , find the total impedance in polar form.
Solution:
Z = Z1 + Z2 = 4Ð00 + 8Ð450 = 4(Cos 0 + j Sin 0) + 8(Cos 45 + j Sin 45)
𝑍 = (1 + 𝑗 0) + 8 R
,
√4
+𝑗
,
S = R4 +
√4
𝑍 = V(9.6574 + 0.7074 ) Ð tan-1R
H
S+𝑗
√4
I.EIE
N.?GE
,
√4
= 9.657 + 𝑗 0.707
S = 9.683 Ð4.19 W
Multi-dimensional Space Vectors
Force and velocity measurements in mechanical systems require a vector representation in a threedimensional space. They are represented with three components along the three space axes X, Y, and
Z with symbols i, j, and k. For example,
A = a1 i + a2 j + a3 k; the linear length of the vector |A| = V𝑎,4 + 𝑎44 + 𝑎:4
1. Addition: Add respective components along the three axes of the two elements to be added.
A = a1 i + a2 j + a3 k; and B = b1 i + b2 j + b3 k; then
A + B = (a1+b1) i + (a2+b2) j + (a3+b3) k = C = c1 i + c2 j + c3 k
2. Zero: All three components are zero length. Z = 0
A + Z = (a1+0) i + (a2+0) j + (a3+ 0) k = A
3. Complement (Negative): All three components along axes X, Y and Z in opposite direction.
A + A¢ = (a1-a1) i + (a2-a2) j + (a3-a3) k = Z = 0
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4. Subtraction: Subtract B from A by adding complement of B to A.
A - B = A + B¢ = (a1-b1) i + (a2-b2) j + (a3-b3) k = C = c1 i + c2 j + c3 k
5. Multiplication: There two types of multiplication rules for three-dimensional vectors.
CROSS Product of A and B: The resultant is also a vector (C), which is perpendicular to the
plane of vectors A and B.
𝑖
A x B = W𝑎,
𝑏,
𝑗
𝑎4
𝑏4
𝑘
𝑎
𝑎: W = [ 4
𝑏4
𝑏:
𝑎:
𝑎,
[
[
𝑖
−
𝑏:
𝑏,
𝑎:
𝑎,
[
[
𝑗
+
𝑏:
𝑏,
𝑎4
𝑏4 [ 𝑘
Some properties of this multiplication rule are:
1) A x B = - B x A
2) A x (b B + g C) = b (A x B) + g (A x C); where b and g are scalars.
3) A x A = - (A x A) = 0
DOT Product of A and B: The resultant is a scalar
A . B = |A| . |B| Cos q ; Where |A| and |B| are the lengths of the vectors and q is the
angle between them. Note that while the Cross Product of a vector A by itself is zero,
the Dot product is |A|2. These multiplication rules are actually mathematical representations
of phenomena in physical systems. For example, interaction of current I and flux f in an
electromagnetic field produces a force F proportional to their product but perpendicular to the
plane of the current and flux vectors. On the other hand, the work done (W) by an object
moving at a velocity V under an applied force F is a scalar quantity proportional to the product
of the velocity and component of the force along the direction of the velocity vector.
6. Unity: No unity vector exists in multi-dimensional vector space since no vector, using either
Cross or Dot product, can produce the vector it multiplies with.
CAUTION: Vector [ i + j + k ] with unit lengths along the X, Y, and Z axes is NOT a Unity
vector.
7. Inverse: Since no Unity vector exists, there can be no pair of vectors whose product would
result in a unity vector.
8. Division: Since no inverse exists, division of two vectors is not defined.
Example 1:
V1: V2 ; V3 ; and V4 are three dimensional vectors as follows:
V1 = 2 i + 2 j + 2 k; V2 = 4 i + 3 j + k; V3 = 4 i + 3 j ; V4 = 2 j + 3 k
Compute: (V1 + V2); (V1 – V4); (V1 x V3); (V3 x V1); (V2 * V4); (V4 * V2);
(V1 x V2) x V4 ; V1 x (V2 x V4)
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Solution:
(V1 + V2) = (2+4) i + (2+3) j + (2+1) k = 6 i + 5 j + 3 k
(V1 – V4) = (2-0) i +(2-2) j + (2-3) k = 2 i – k
𝑖 𝑗
(V1 x V3) = \2 2
4 3
𝑘
𝑖
]
\
=
6
i
+
8
j
–
2
k
;
(V
x
V
)
=
2
3
1
4
0
2
𝑗
3
2
𝑘
0] = 6 i – 8 j + 2 k
2
Note: (V1 x V3) = - (V3 x V1)
(V2 * V4) = (4x0) + (3x2) + (1x3) = 9 = (V4 * V2)
(V1 x V2) x V4 : First compute (V1 x V2) then multiply it with V4
𝑖
(V1 x V2) = \2
4
𝑗 𝑘
𝑖
𝑗
2 2] = - 4i+6j-2k; (V1 x V2) x V4 = \−4 6
3 1
0 2
𝑘
−2]= 22i+12j-8k
3
V1 x (V2 x V4): First compute (V2 x V4) then pre-multiply with V1
𝑖 𝑗
(V2 x V4) = \4 3
0 2
𝑘
𝑖
𝑗
𝑘
1] = 7i-12j+8k; V1 x (V2 x V4) = \2
2
2] = 40i-2j-38k
3
7 −12 8
Note: (V1 x V2) x V4 ¹ V1 x (V2 x V4)
Example 2: For a three-dimensional space with i , j , and k as unit vectors along the X, Y and Z
axes respectively,
a.
Determine the value of α such that A = -2i + aj- 2k and B = 3i + 4j +19k are
perpendicular
b.
Prove that |AxB|2 + |A.B|2 = |A|2 |B|2
Solution:
a. The dot product of A and B; A*B = |A| |B| Cos j.
For A and B to be perpendicular, Cos (90) = 0
The dot product of A = -2i + aj- 2k and B = 3i + 4j +19k = (-2x3) + 4a - 19x2 = 0; a = 11
b. |A*B| = |A||B| Cos j and |AxB| = |A||B| Sin j
|AxB|2 + |A.B|2 = |A|2|B|2 Cos2 j + |A|2|B|2 Sin2 j = |A|2|B|2 (Cos2 j + Sin2 j ) = |A|2|B|2
Example 3: The Hay Bridge is an AC circuit used in electrical measurements. It is governed by the
following equation. Express Rx and Lx in terms of the other circuit constants.
^𝑅, − 𝑗
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b (𝑅c + 𝑗𝜔 𝐿c ) = 𝑅4 𝑅:
𝜔 𝐶,
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Dr. Mohammed Safiuddin
Solution:
,
R𝑅, − 𝑗 e f S (𝑅c + 𝑗𝜔 𝐿c ) = 𝑅4 𝑅:
(1)
C
g
i
𝑅, 𝑅c + fh + 𝑗 R𝜔𝐿c 𝑅, − efh S = 𝑅4 𝑅:
C
(2)
C
Equating the real and imaginary parts:
g
𝑅, 𝑅c + fh = 𝑅4 𝑅:
(3)
C
i
i
𝜔𝐿c 𝑅, − efh = 0; 𝐿c = e> fh i
And,
C
C C
(4)
Substituting the value of Lx in from (4) into (3)
𝑅, 𝑅c +
𝑅c =
1.2
𝑅𝑥
2
𝜔 𝐶21 𝑅1
1
= 𝑅2 𝑅3 ; 𝑅𝑥 ^ 𝑅1 + 2 2 b = 𝑅2 𝑅3
𝜔 𝐶1𝑅1
(efC )>iC i> il
; and
(efC iC )>D,
f i i
𝐿c = (efC i> )>lD,
C C
Computer Arithmetic
Binary number system is a 'Base 2' system consisting of only two base integers of the scalar system;
the Zero and the Unity. Numbers larger than 2 are represented by digits to the left in a manner similar to
the decimal number system. Though numbers up to the infinity can be represented in the binary system, the
number of digits (called BITS in the computer language) available in a given computing system limit the
maximum decimal number that can be handled.
Elementary operations in a four-bit system:
There are 16 elements in this binary system ranging from 0000 to 1111. The “Hexadecimal” system
is “Base 16” system, represented by a single digit. After the numbers 0- 9, A, B, C, D, E and F are used to
represent numbers 10, 11, 12, 13, 14, and 15, as shown in the circle diagram below.
1. Addition:
Rule 1: 1 + 1 = 0 with one (1) carry to the left bit.
Rule 2: 1 + 0 = 0 + 1 = 1
Rule 3: 0 + 0 = 0
Example:
1110
+ 1010
1 1000
Any carry beyond the bit size of the system is ignored by the computer.
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2. Zero: All four bits are zero. That is Zero = 0000 since it will not alter the value of any of the
sixteen numbers, if added to it by the foregoing rules.
3. Complement (Negative): Since there are only sixteen numbers in this four-bit system, they can
be divided up into two groups of eight numbers each requiring only three bits for their
representation. Refer to the circle diagram. Note: Just like the general scalar system, the zero
is treated as a positive number. The complement of number 0101 (decimal 5) is 1011 (decimal
11) because the sum of these two numbers is a five-bit binary number 10000 (decimal 16)
which wrap arounds to 0000.
4.
Subtraction: Take the two's complement of the subtrahend and add it to the minuend.
Decimal:
Binary:
7-3= 4
0111 - 0011 = 0111 + 1101 = [1]0100
5. Multiplication: Repetitive addition can be used to perform the multiplication. However, the
number of bits available in a system restrict the operands to those for which the product remains
within the available number of bits unless two sets of bits are used for storing the resultant.
The multiplicand is repetitively added (multiplier-1) times to itself.
Or, multiplier is
repetitively added (multiplicand - 1) times to itself, whichever yields lower number of
operations.
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Decimal:
3x2 = 2x3 =6
Binary:
(0011) x (0010) = (0010)x(0011) = 0011 + 0011 = 0110
= 0010 + 0010 + 0010 = 0110
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Dr. Mohammed Safiuddin
6. Unity: The unity (0001) exists since any number added zero (unity - 1) times to itself would
not alter the original multiplicand.
Decimal:
5x1=1x5 = 5
Binary:
(0101) x (0001) = (0001) x (0101) = 0101
7. Inverse: Since binary numbers are integers, their inverses would have to be less than one so
that the product of number and its inverse would yield a resultant equal to the unity. The
inverses, therefore, do not exist.
8. Division: Since inverses do not exist, division of two binary numbers is carried out by
repetitive subtraction. The divisor is subtracted from the dividend until there is no remainder.
It can also be carried out in the same way as division of decimal numbers. The four general
rules are:
0 ÷ 0 and 1÷ 0 have no meaning; 0 ÷ 1 = 0 and 1÷ 1 = 1; Decimal:
4
= 2
2
Binary: (0100) ÷ (0010) = 0010
1.3
Matrix Algebra
Matrix is an array of elements (quantities or operators) arranged in rows and columns.
Examples:
é2 - 3 1ù
ê
ú
ê 1 1 - 2ú
ê 3 5 1ú
ë
û
;
é s
ê
ê
ê 2
ê2 s
êë
s ù
1+ s ú
ú
ú
2ú
s úû
;
[( 3 + j 6 ) ( 5 + j 2)]
The number of rows and columns determine the dimensions of a given matrix. That is, a matrix A
é a11 a12 a13
ê
ê a 21 a 22 a 23
ê
A = [ aij ] mn = ê .
.
.
ê
.
.
ê .
ê 1
am 2 am 3
ëa m
. .
. .
. .
. .
. .
a1n ù
ú
a 2n ú
ú
.ú
ú
.ú
ú
a mn û
of dimension (m x n) is said to have m number of rows and n number of columns.
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Here, aij represents the element in the ith.row and the jth.column.
Elementary mathematical operations
1.
ADDITION:
A = [ aij ]mn
;
B = [ bij ]pq then
A + B = C is defined only if m = p
Ù
n=q
C = [ cij ]mn ; where cij = aij + bij
Addition is commutative and associative
A+ B = B+ A
2.
Ù
A+(B+C) = ( A+ B)+C
ZERO or NULL MATRIX:
O = [ a ij ] mn where each element a ij = 0
A + O = A because each element of the sum is aij + 0 = aij
3.
NEGATIVE (COMPLEMENT)
A¢ = [ - A ]mn that is, each element of A¢ is negative of the corresponding element of A
4.
SUBTRACTION
A - B = A + B¢ = C ; that is, element cij = aij + (bij)¢ = aij - bij
5.
MULTIPLICATION
If A = [ a ij ]mn and B = [ bij ]pq then,
(A) (B) = Cmq is defined only for matrices with ( n = p ) and
(B) (A) = Dpn is defined only for matrices with ( q = m )
where,
n
cij = å aik bkj
k =1
6.
UNITY (IDENTITY)
[U] or [I] of order (mxn) exists only if m = n .
[U] = [I] = {uij} where, uij = 1 for i =j and uij = 0 for i ¹ j
That is, (A) (U) = (U) (A) = A
7.
INVERSE (RECIPROCAL)
A -1 for A exists only if m = n and, det [A] ¹ 0. Then [A] [A] -1 = [A] -1[A] = [U]
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Calculation of the inverse of matrix A requires following steps.
i.
Calculate the determinant of [A] to make sure det [A] ¹ 0
ii. Replace each element of A by its Cofactor to obtain the Adjugate of A
iii. Transpose the resulting matrix to obtain Adjoint of A [ adj ( A) ]
𝐴F, =
1
(𝑎𝑑𝑗 𝐴)
det 𝐴
iv. Divide each element of adj ( A) by the det A
For given i, j, the Cofactor of the element aij of A is obtained by calculating the determinant of
[A] after striking out the ith. row and jth. column, multiplied by ( - 1 )i + j .
Another method of finding inverse of a matrix is the Gauss-Jordan reduction technique. An
‘Identity’ matrix [ I ] is appended to the matrix to be inverted and then linear operations are
performed on the rows of the appended matrix until the two interchange their positions within the
appended matrix. See example below.
Example: Find inverse of
é 3 - 2 - 1ù
ê
ú
A = ê- 1 3 2 ú
ê 1 1 1ú
ë
û
é 3 - 2 -1 |
ê
[ A_I] = ê- 1
3 2 |
ê
1 1 |
êë 1
0ù
ú
1 0ú
ú
0 1úû
1 0
0
0
Subtract 2 times the last row from the first and add third row to the second.
é 1 - 4 - 3 | 1 0 - 2ù
ê
ú
ê0 4 3 | 0 1 1ú
ê
ú
1úû
ëê 1 1 1 | 0 0
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Add second row to the first.
é 1 0 0 | 1 1 - 1ù
ê
ú
ê0 4 3 | 0 1 1ú
ê
ú
ëê 1 1 1 | 0 0 1úû
Subtract first row from the last
é 1 0 0 | 1 1 - 1ù
ê
ú
ê0 4 3 | 0 1 1ú
ê
ú
êë0 1 1 | - 1 - 1 2 úû
Subtract 3 times the last row from the second Subtract 3 times the last row from the second.
é 1 0 0 | 1 1 - 1ù
ê
ú
ê0 1 0 | 3 4 - 5 ú
ê
ú
ëê0 1 1 | - 1 - 1 2 úû
Subtract the second row from the last.
The inverse of A is:
é 1 1 - 1ù
ê
ú
ê 3 4 - 5ú
ê- 4 - 5 7 ú
ë
û
As a check, we compute [A] [A] -1 to confirm that it equals the identity matrix I .
é 1 0 0 | 1 1 - 1ù
ê
ú
ê0 1 0 | 3 4 - 5 ú
ê
ú
ëê0 0 1 | - 4 - 5 7 úû
8.
=
[ I | A-1 ]
DIVISION
Division of matrix A by matrix B is obtained by either pre-multiplying or post-multiplying the
matrix A by the inverse of matrix B if it exists. Following conditions for B be must be met.
i.
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B must be a square matrix with non-zero determinant.
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Dr. Mohammed Safiuddin
ii. Dimensions of B must equal the number of columns in A for A. B-1 product.
and dimensions of B must equal the number of rows in A for B-1.A product.
Terms and definitions associated with matrices
é 3 2 4ù
ê
ú
A = ê 5 11 1ú
ê2 9 - 3ú
ë
û
1. Singular matrix: A matrix whose determinant does not exist or is zero.
det A = 3(-33-9) - 2(-15-2) + 4(45-22) = 0
2. Non-singular matrix: A matrix whose determinant is non-zero.
3. Rank of a matrix: The dimension of the largest non-singular submatrix of A mn
Rank
of
Rank
| 4ù
é3 2
ê
ú
ê 5 11 | 1ú
A = ê
ú = 2
ê- - - - - - - - ú
ê
ú
| 3û
ë2 9
of
4ù
é3 2
ê
ú
B = ê2 9 - 3ú
ê 5 0 - 10ú
ë
û
=
3
because the det [B] ¹ 0
4. Diagonal matrix: A matrix with all its off-diagonal elements equal to zero.
é1
0
0ù
ê
ú
A = ê0 (-2 + 3 j )
0ú
ê
ú
0 ( - 2 - 3 j )ûú
ëê0
[ A ]nn = ( a ij )nn is diagonal if aij = 0 for all i ¹ j
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5. Diagonalizable matrix: A matrix that can be represented by a non-singular matrix P and a
diagonalizable matrix L such that A P = P L ; that is A = P L P-1
Example:
4ù
é- 3 - 2
ê
ú
A = ê 7 7 13ú ;
ê 5 6 - 10ú
ë
û
é 3 - 2 - 1ù
ê
ú
P = ê- 1 3 2 ú ;
ê 1 1 1ú
ë
û
é- 1 0 0 ù
ê
ú
L = ê 0 - 2 0ú
ê 0 0 - 3ú
ë
û
6. Similar matrices: Matrices A and B are said to be similar if there is a non-singular matrix P
such that
A = P -1 B P .
é 3 - 1 1ù
ê
ú
P * = ê- 2 3 1ú
ê - 1 2 1ú
ë
û
is transpose of
é 3 - 2 - 1ù
ê
ú
P = ê- 1 3 2 ú
ê 1 1 1ú
ë
û
7. Transposed matrix: A matrix obtained by interchanging rows and columns of another matrix
8. Symmetric matrix:
A square matrix which is equal to its transposed matrix. That is,
*
A = A . For example,
1
sù
é(s + 1)
ê
ú
A = ê
1 (s + 2)
2ú
ê
ú
s
2 (s + 3)úû
êë
9. Characteristic polynomial of a matrix: The polynomial in λ obtained from [det (A-lI)] for a
given square matrix A . For example,
é - 3 1 0ù
ê
ú
A = ê- 2 0 1ú ; [ A - lI ] =
ê 0 0 0ú
ë
û
1 0ù
é- ( 3 + l )
ê
ú
-2 -l
1ú
ê
ê
ú
0
0 - l úû
ëê
Characteristic polynomial is det[𝐴 − 𝜆 𝐼] = −(3 + 𝜆) 𝜆4 − 2𝜆 = −[ 𝜆: + 3 𝜆4 + 2 𝜆]
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10. Eigenvalues of a matrix: The roots of the characteristic equation [det (A-lI) = 0] are called the
eigenvalues of matrix A . For example,
l 3 + 3 l 2 + 2l = 0 ; l ( l + 1 ) ( l + 2 ) = 0
l 1 = 0 ; l 2 = - 1 ; l 3 = - 2 are eigenvalues of the matrix A .
11. Eigenvectors of a matrix:
The eigenvectors x i of an n-dimensional square matrix A are
vectors such that [ l i I - A ] x i = 0 ;
i = 1,2,.... where, l i are the eigenvalues of A .
For example, for l 1 = 0 , and the A matrix of (9) and (10) above,
é 3 - 1 0 ù éx 11ù
ê
ú ê ú
2
0
1
ê
ú êx 12 ú =
ê0 0 0 ú ê ú
ë
û êëx 13 úû
é0 ù
ê ú
ê0 ú ;
ê0 ú
ë û
éx 11ù
é 1ù
ê ú
ê ú
x 1 = êx 12 ú = a 1 ê 3ú
ê ú
ê 2ú
êëx 13úû
ë û
[ l1 I - A ] x 1 = - A x 1 = 0
x 1 is the eigenvector for l 1 = 0 . NOTE: Since the last equation results in 0 = 0, the
first two are solved for x 12 and x 13 by setting x 11 = a 1.
12. Adjoint of a square matrix [A]:
A matrix obtained by taking the transpose of matrix of
cofactors of the original matrix A .
Example:
é 1 - 2 - 1ù
ê
ú
A = ê 3 - 1 0ú
ê0 1 2 ú
ë
û
Cofactors of
é- 2 - 6 3ù
ê
ú
A = ê 3 2 - 1ú
ê - 1 - 3 5ú
ë
û
é- 2 3 - 1ù
ê
ú
Adj(A) = ê- 6 2 - 3ú
ê 3 - 1 5ú
ë
û
13. Unitary matrix: If the product of a matrix and its adjoint is equal to a Unity (Identity) matrix,
then the matrix is called a unitary matrix. That is, if the determinant of a matrix is equal to 1
then it is unitary.
14. Bordering of a matrix: It is the process of building a matrix of dimension n+1 from an nxn
matrix A by adding a vector U, transpose of a vector V and a scalar α.
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Dr. Mohammed Safiuddin
A ! n +1
| Un ù
é An
ê
ú
= ê - - - - - -ú
ê T
ú
| a úû
êëV n
For the matrices A and B find A2, BA-1, and (A - B).
Example:
−3 2
1
\
𝐴 = 2 −3 −2]
−1 −2 0
−1 −1 −2
\
𝐵= 2
3 −3]
−3 −4 2
Solution:
−3 2
1
−3 2
𝐴 = \ 2 −3 −2] \ 2 −3
−1 −2 0
−1 −2
4
1
12 −14 −7
−2] = \−10 17
8]
0
−1
4
3
To calculate [A]-1: det A = 9;
−4
2 −7
−4 −2 −1
1 −8]; Adjoint of 𝐴 = \ 2
1 −4]
−1 −4 5
−7 −8 5
−0.444 −0.222 −0.111
,
𝐴F, = uvw(x) 𝐴𝑑𝑗𝑜𝑖𝑛𝑡 (𝐴) = \ 0.222
0.111 −0.444]
−0.777 −0.888 0.555
Adjugate of 𝐴 = \−2
−1 −1 −2 −0.444
𝐵∗ 𝐴F, = \ 2
3 −3] \ 0.222
−3 −4 2 −0.777
−0.222
0.111
−0.888
−0.111
1.778
]
\
=
−0.444
2.111
0.555
−1.111
1.889
2.556
−1.556
−0.556
−3.222]
−3.222
−2 3
3
\
𝐴 − 𝐵 = 0 −6 1 ]
2
2 −2
Determinants
While matrices are an array of numbers arranged in rows and columns, a determinant is a
number. It is a convenient way to express the number [ ad-bc] in the form [
𝑎
𝑏
𝑐
[. Number of rows and
𝑑
columns are always equal in the array. It is denoted by the symbol D or letter D. This 2x2 array is called a
‘Second Order’ determinant. A ‘Third Order’ determinant would be an array of 3x3. [a,b,c,d,] are called
elements, [ad-bc] is the expansion or the value of the determinant, [a and b] make the first column while
[a and c] make the first row. [a] is the leading element, [a, d] lie on the ‘Leading’ or ‘Principal’ diagonal.
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D¢ is transpose of the determinant with rows and columns switched. Some of the properties of determinants
are:
1) D and D¢ have identical expansions.
2) A common factor of the elements of a row (column) is a factor of D, for example;
~
𝑎𝑏𝑥
𝑏𝑦
𝑎𝑧
~ = 𝑎𝑏𝑥𝑡 − 𝑎𝑏𝑧𝑦 = 𝑎𝑏(𝑥𝑡 − 𝑧𝑦), so you can remove common factors before expanding
𝑡
3) Interchange of two rows (or columns) changes D in to -D
𝑎 𝑐
𝑐 𝑎
[
[ = (𝑎𝑑 − 𝑏𝑐) but [
[ = (𝑏𝑐 − 𝑎𝑑)
𝑏 𝑑
𝑑 𝑏
𝑎 𝑏
4) If two rows or columns are identical then D = 0. [
[ = (𝑎𝑏 − 𝑎𝑏) = 0.
𝑎 𝑏
5) If two rows or columns are proportional then D = 0
6) The product of any row (column) and any arbitrary constant may be added to any other row
(column) without changing the value of the determinant.
𝑎
𝑏
D =[
𝑐
(𝑎 + 𝑘𝑐)
[ = (𝑎𝑑 − 𝑏𝑐)and, D = ~
𝑑
(𝑏 + 𝑘𝑑)
𝑐
~ = [(𝑎 + 𝑘𝑐)𝑑 − (𝑏 + 𝑘𝑑)𝑐] = 𝑎𝑑 − 𝑏𝑐
𝑑
7) The value of a triangular determinant is equal to the product of all elements on its principal
diagonal. Triangular determinant has all elements on one side of the diagonal equal to zero.
𝑎,, 𝑎,4 𝑎,:
∆ = W 0 𝑎44 𝑎4: W = 𝑎,, . 𝑎44 . 𝑎::
0
0 𝑎::
Evaluation of Determinants
1. ajk is an element in the jth row and kth column.
2. ajk Minor written Mjk is a determinant with jth row and kth column eliminated.
3. ajk Cofactor = [(-1) j+k . Mjk] is also written as Djk
4. A determinant of any order can be expanded about any row or column by summing the products of
the elements of rows(columns) and their respective Cofactor.
5. An alternate method is to reduce the given determinant in to a triangular determinant and
then multiply the elements on the principal diagonal for the answer.
The Binomial Theorem
Sometimes expressions of two unknown variables involve their sum raised to powers of two or
higher in the form (a + b)n. The Binomial Theorem, known to Islamic and Chinese mathematicians in
tenth and eleventh centuries, offers the formula for its expansion as follows.
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(a + b)2 = a2 + 2ab + b2; (a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)n = an + n a(n-1)b +[n(n-1)/2]a(n-2)b2 +[n(n-1)(n-2)/2.3]a(n-3)b3......+bn
Note: 1. The sum of the exponents of each term is equal to the exponent on the left-hand side
2. The coefficient of the first term is 1, and the exponent of a decreases by 1 to zero in the last term
3. The exponent of b increases by 1 from zero in the first term to the exponent on the left-hand side
4. The coefficient of the next term is obtained by multiplying the exponent of a by its coefficient
and then dividing it by the total number of existing terms.
𝑛
𝑟
It is represented by the formula R S =
-!
(-Fƒ)!ƒ!
where n is the exponent and r is the term number.
Note that 0! is equal to 1. The coefficients of the right-hand terms can also be computed from Pascal’s
Triangle, introduced by the 17th century French mathematician Blaise Pascal. The row number is equal to
the exponent n on the left-hand side of the
equation.
Row starts and ends with 1 as
coefficients of the first and the last term. The
coefficients in the middle are computed by
adding the two numbers of the row above it.
1.4
Linear Simultaneous Equations
In control system problems we are concerned with the determination of unknown variables
from known relationships among the unknown themselves or among some unknown variables and known
quantities. These relationships are expressed by equations, such as [4 x + 3 y = 10]. Methods of
determining the unknowns depend upon the nature of relationships. The algebraic relationships, expressed
by equations, can be ‘Linear’ or ‘Non-linear’. We define a non-linear relationship as the one in which the
coefficient(s) of relationship is(are) not constant for all values of the variable(s). For example, [ y = 4 x2
= (4 x) x, and y = 4 x + 2 xy]. When multiple unknown variables are expressed by a set of linear equations,
as shown below, the variables can only be solved for by a simultaneous solution for all the equations.
a11 x1 + a12 x2 + a13 x3 + ……. + a1n xn = y1
a21 x1 + a22 x2 + a23 x3 + ……. + a2n xn = y2
…………………………………………………..
am1 x1 + am2 x2 + am3 x3 + ……. + amn xn = ym
This set of equations can be expressed as a single matrix equation of the form A X = Y, where A
is the coefficient matrix, X is the column vector of unknown variables x1.....xm, and Y is the column vector
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Dr. Mohammed Safiuddin
of y knowns. This set of equations can be characterized as ‘Consistent’ or ‘Inconsistent’, and ‘Linearly
Dependent’ or ‘Linearly Independent’. An inconsistent set of equations is one which cannot be satisfied
simultaneously by any value of unknowns. For example, [4x + 3y =10, and 4x + 3y =12] is a set of two
equations in two unknowns. However, it is an inconsistent set because no values of x and y can satisfy both
equations simultaneously. A linearly dependent set of equations is one in which one or more equations can
be obtained from a linear combination of other equations.
For example:
2x+3y=8
3 x + 4 y = 10
4 x + 6 y = 16
5 x + 7 y = 18
This is a set of four equations in two unknowns which consists of two linearly independent
equations and two linearly dependent ones. The third equation is obtained by multiplying the first
one by 2 and the fourth equation is a sum of first two equations. We note from these two examples
that first step in solving a set of simultaneous equations is to check of consistency and linear
independence of the equations.
In general, A X = Y is a matrix representation of m equations with n unknowns.
A mn is coefficient matrix of m rows and n columns
X n1 is the unknown column vector of n rows
Y m1 is the known column vector of m rows
There are three possibilities.
(i) m > n; more equations than unknowns
(ii) m = n; equal number of equations and unknowns
(iii) m < n; fewer equations than unknowns
There are two conditions.
(i) Is the set linearly independent?
(ii) Is the set consistent?
Case (i) m > n :
If the set of m equations is consistent, then there exists a set of n linearly independent equations.
The remaining (m-n) equations are linearly dependent upon the n equations. However, we must check the
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Dr. Mohammed Safiuddin
set for consistency.
If rank of A = rank of [ A| Y ], the set is Consistent
If rank of A ¹ rank of [ A| Y ], the set is Inconsistent
Case (ii) m = n
An explicit solution exists for all n unknowns if, and only if, the set is consistent and linearly
independent. We check the rank of matrix [A]. That is, if its determinant is not zero, then the rank is full
at m=n. We then have a linearly independent and consistent set and it can be solved by computing X =
[A]-1Y. On the other hand, if the rank of matrix [A] is only of the order m1 then only m1 of the m equations
are linearly independent, then m1 < m and hence less than n. We then check for consistency of the set, and
if is consistent, then it falls under the third case of m < n.
Example 1. Determine the consistency and linear independence of the following equations.
x + y + z = 4; 2x – y + z = 0; 4x + y + 3z = 9
Solution:
1 1
Check for the rank of the coefficient matrix 𝐴 = \2 −1
4 1
1
1]
3
det A = 1(-3-1)-1(6-4)+1(2+4) = 0; Rank of A< 3
Check for the rank of [A|Y]
1 1 1 4
1 1 4
\
The matrix [𝐴|𝑌] = \2 −1 1 0]; [𝐴|𝑌]†‡ƒˆ
=
−1
1 0]
:c:
4 1 3 9
1 3 9
†‡ƒˆ
det [𝐴|𝑌]:c: = 1(9)-1(-9)+4(-3-1) = 34 ¹ 0
The rank of A is not equal to the rank of [A|Y], the set is Inconsistent.
Example 2.
Determine the consistency and linear independence of the following equations.
x + y +z =13; 2x -y + z =6; 4x +y +3z = 32
1
Check for the rank of coefficient matrix 𝐴 = \2
4
1 1
−1 1];
1 3
det A = 1(-3-1)-1(6-4)+1(2+4) = 0; Rank of A< 3
1 1 1 13
1 1 13
\
The [𝐴|𝑌] = \2 −1 1 6 ] [𝐴|𝑌]†‡ƒˆ
=
−1
1 6]
:c:
4 1 3 32
1 3 32
†‡ƒˆ
det [𝐴|𝑌]:c: = 1(32-18) +1(6+32) + 13(-3-1)= 14+38-52 = 0
The rank of [A|Y] is also less than 3 and it is 2.
Similarly, [A|Y] matrix of first, second and fourth column also has a rank of 2.
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This set is CONSISTENT but LINEARLY DEPENDENT.
The third equation is linear
combination of the first two equations. Two times the first equation added to the second yields the
third equation. This set can be solved for two of the variables in terms of the third variable.
Case (iii) m < n:
The set can be solved for m of the unknowns in terms of the remaining (n-m) unknowns if, and
only if, the set is consistent.
There are four techniques for solving simple simultaneous equations of scalar variables.
1. Solution by Elimination of Variables
2. Solution by Determinants [Cramer’s Rule]
3. Solution by Matrix Method
4. Solution by Signal Flow Graph and Mason’s Gain Formula
Let’s use an example to review first three techniques.
Method I – Elimination of variables
3 x + 2 y + 4 z = 3.................i
2 x + 9 y - 3 z = 0.................ii
5x
From equation (iii),
𝑥=
+ 10 z = -5.................iii
F?F,I7
?
= −1 − 2𝑧
Substitute this value of x in (i) and (ii), the resulting equations are:
-3 -6z + 2 y + 4 z = 3.................ia
-2 – 4z + 9 y - 3 z = 0.................iia
That is,
2 y – 2z = 6.............. ib
9 y – 7z = 2.............. iib
Multiplying equation (ib) by 9 and (iib) by 2, we get:
18 y – 18z = 54.............. ic
18 y – 14z = 4.............. iic
Subtracting (iic) from (ic), we get:
-4z = 50; z = -12.5;
And, from (ib);
2 y = 6 + 2z = 6 + 2(-12.5) = -19;
x = -1- 2(-12.5) = 24
y = -9.5
Therefore, the solution is: x = 24; y = -9.5; and z = -12.5
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Dr. Mohammed Safiuddin
Method II – Solution by Determinants
3 x +2 y + 4 z = 3.................i
2 x + 9 y - 3 z = 0.................ii
5x
+ 10 z = -5.................iii
3 2 4 3
9
∆ = W2 9 −3W 2 = 3 [
0
5 0 10 5
−3
−3 2
2 9
[ + 2[
[ +4[
[
10
10 5
5 0
= 3(90) +2(-15-20) +4(-45) = 20
Obtain D1 by replacing first column by column of coefficients on the right-hand side.
3 2 4
9
∆, = W0 9 −3W = 3 [
0
5 0 10
−3
2
[ − 5[
10
9
4
[
−3
= 3(90) -5(-6-36) = 270 + 210 = 480;
∆
JHI
𝑦 = ∆> =
∆
F,NI
∆
F4?I
𝑥 = ∆C = 4I = 24
3 3 4
−3 2
4 3
∆4 = W2 0 −3W = 3 [
[ − 5[
[
10 5
−3 2
5 5 10
= 3(-15-20) -5(8+9) = -190;
4I
= −9.5
3 2 3
2 9
3 2
∆: = W2 9 0 W = 3 [
[ −5[
[
5 0
2 9
5 0 −5
= 3(-45) -5(27-4) = -250;
𝑧 = ∆l =
4I
= −12.5
Method III – Solution by matrix
3 x +2 y + 4 z = 3.................i
2 x + 9 y - 3 z = 0.................ii
5x
+ 10 z = -5.................iii
Matrix representation of the equations – AX = Y, and X = A-1 Y
3
𝐴 = \2
5
F,
𝐴
9/2
= \−7/4
−9/4
−1
1/2
1/2
2
9
0
4
𝑥
3
𝑦
−3] and, X = ‰ Š; Y = \ 0 ]
𝑧
−5
10
−21/10
9/2
−1 −21/10 3
𝑥
24
17/20 ] ; X = ‰𝑦Š = \−7/4 1/2 17/20 ] \ 0 ] = \ −9.5 ]
𝑧
−9/4 1/2 23/20 −5
23/20
−12.5
A number of websites, such as [https://matrixcalc.org/en/], and hand calculators, offer computations for
basic operations of matrices, including inversion.
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1.5
Differentiation & Integration
In engineering problems, we deal with relationships among the various measurements.
In
mathematical terms, we refer to these relationships as functions and the measurements as the variables.
If the functional relationships involve the variables only, we get the ordinary algebraic equations. However,
if the functional relationships also involve rates of change of variables, then the resulting equations become
differential equations. For example, an electrical circuit comprising DC voltage sources and a network of
resistors requires an algebraic equation for analysis. On the other hand, a differential equation is needed to
analyze the circuit if capacitors and/or inductors are also present.
Functions and functional relationships are often expressed by equations with the unknown
variable(s) designated as the dependent variable(s), placed on the left-hand side of the equate symbol [ =
], and the known values as the independent variable(s) on the right-hand side.
For example, in the equation [𝑦 = 3 𝑥, + 2 𝑥4 + 4 𝑥: ], the value of y depends upon the
independent variables x1, x2, and x3. But in the equation [3 𝑥, + 2 𝑥4 + 4 𝑥: = 3 𝑦, − 2 𝑦4 ], the x
variables are the dependent variables and the y variables are considered to be the independent ones. The
algebraic functional relationships are grouped into Linear and Non-linear functions. The equation of a
straight line [𝑦 = 𝑚𝑥 + 𝑏] is the simplest example of a linear function. The equation of a parabola [𝑦 4 =
2 𝛼 𝑥] is a good example of a non-linear function. The distinguishing property of a linear function is that
the principle of superposition holds true. That is,
f ( m x + n y ) = m. f ( x ) + n. f ( y )
Where, m and n are constants, and x and y are independent variables. If the function fails to meet this test,
then it is non-linear.
Besides the simple example of the parabolic function, following non-linear functions are often
encountered in engineering problems.
- A unit step function
- An exponential function
- All trigonometric functions (Sines, Cosines etc.)
- All polynomial functions of the second and higher orders
When the exact value of a dependent variable cannot be computed directly for certain value (s) of
the independent variable by direct substitution, the concept of "the Limit" is used for determining the
approximate value of the function as the independent variable (s) reaches that value. For example,
lim
2
s + s+3
= 0
s ® ¥ s ( s2 + 5 )
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Using the concept of this limit of a function, we define the rate of change in the value of a function
of a single or of multi-variables, due to incremental changes in the independent variable(s), as its derivative.
For example, let us consider the formula for the area "A" of a square of sides "x" given by the simple
equation " A = x 2 ". If the lengths of the sides are changed by an incremental amount "Δx", then the area
"A" is changed by an amount "ΔA" as follows.
A + DA = ( x + D x )2 = x 2 + 2 x D x + ( D x )2
DA = 2 x D x + ( D x )2
Now if Δx is made small enough, or "approaches zero", then the last term on the right-hand side ( D x )2
becomes negligible as compared to the ( 2 x D x ) term. Symbolically this is written using the symbol dx
in place of Δx to indicate the results of the limiting process,
lim
Dx ® 0
DA =
lim
Dx ® 0
[ 2x D x + ( D x )2 ]
we get:
dA = 2 x dx ; That is:
dA
= 2x
dx
In general, for any function f (x) of the variable x , the derivative is defined as follows
lim
df
f ( x+D x ) - f ( x )
=
D x ®0
dx
Dx
The derivative of a function at a specified point x1 is, in essence, the slope of the tangent at that point on
the curve displaying the functional relationship between f(x) and x. Higher ordered derivatives are also
defined and are important in control theory, especially with variables which are functions of time t. For
example, the second order derivative of f(t) with respect to the time t, is defined by
2
lim y ( t + 2 D t ) - 2 y ( t + D t ) + y ( t )
d y(t)
=
Dt ®0
d t2
D t2
If we look at the graph of an arbitrary function and determine the slope [ derivative] of the
function at some point " t0 ", then we may use this information to predict or anticipate the value of
function at some nearby point "t1" as follows.
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Ž•
𝑦(𝑡, ) = 𝑦(𝑡I ) + ∆𝑡 ( Žˆ )ˆ•ˆ‘ ; where Δt = ( t1-t0 )
Example: Let us suppose
y ( t ) = 3 t 3 + 5( t - 1 ) is an unknown function of time t and that we only know that its first
derivative
Ž•
Žˆ
𝑎𝑡 𝑡 = 2.0 is equal to 41 and that y (2.0) = 29.
We can predict the value of "y" at t = 2.1 to be [𝑦 (2.1) = 29.0 + (0.1)𝑥(41) = 33.1] while the
actual value of "y" from the function is 33.283. The error in our estimate (prediction) is about 0.5%.
This prediction is not exact, because higher ordered derivatives have not been considered though
they are non-zero. In theory, approximation can be made better by including more and more higher ordered
derivatives. The higher order approximation is:
y ( t1 ) = y ( t0 ) + D t (
2
dy
1
d y
) _ t = t0 +
( Dt )2 (
) _ t = t0 + ........
dt
2!
d t2
An application of this mathematical technique is found in control systems employing op-amps
and/or computer software for PID and Lead-Lag controllers which provide anticipatory feedbacks for the
controlled variable(s).
Some of the rules for differentiation are considered quite useful in solving engineering problems.
d (au) = a. du [a is a constant]; d (u + v – w) = du + dv – dw; d (uv) = u dv + v du
’
𝑑 R“ S =
“ Ž’F’ Ž“
“>
; d (un) = n un-1 du; d (eu) = eu du; d (eau) = a eau du; d (au) = au(loge a)du
Positive values of the derivative of a variable indicate that the value of dependent variable increases
as independent variable increases. On the other hand, a negative value indicates the opposite. So, when
derivative goes to zero for a given value of the independent variable, the dependent variable reaches either
a Maxima or a Minima depending upon its value within its vicinity. If the derivative reaches zero from a
positive value then it approaches Maxima, otherwise it approaches Minima. This property of the derivatives
is very useful in many engineering problems.
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1.6
Linear Differential Equations (Time Domain)
Differential equations are used for describing the relationships among the variables of a physical
system. If these relationships are linear and the coefficients of these relationships are constant, the system
can be described by linear differential equations with constant coefficients. Feedback control systems
comprise passive energy storage devices such as inductors, capacitors, inertias etc., which often give rise
to linear differential equations with constant coefficients in the time domain. Study of linear differential
equations with constant coefficients, where time (t) is the independent variable, is essential to the study and
understanding of industrial control and drive systems.
EXAMPLES
I
Given, L (di/dt) + R i(t) = 0;
Choose solution:
Find. i(t)
i(t) = Io ext
L. d(Io ext)/ dt + R (Io ext) = 0
Then,
L. Io. d( ext)/dt + R (Io e xt) = 0
L Io x ext + R Io e xt = 0
That is,
Io e xt ( L x + R ) = 0
Either
Io = 0; Or, (L x + R) = 0 since ext ¹ 0
For Io ¹ 0, x = - (R/L) and i (t) = Io e –(R/L) t. satisfies the given differential equation.
i (t) = Io e –(R/L) t
That is,
Is a solution to the first order differential equation [L (di/dt) + R i(t) = 0]
II
Given,
L (di2/dt2) + R (di/dt) + 1/C i(t) = 0,
Again,
assume i(t) = Io ext and substitute it in the given equation
Find i(t)
L. d 2 (Io ext)/dt 2 + R d (Io ext)/dt + 1/C (Io ext) = 0
L. I0 x2 ext + R I0 x ext + (Io ext)/C = 0
Therefore,
(Io ext )(L x2 + R x + 1/C) = 0
With I0 ¹ 0, we set (L x2 + R x + 1/C) = 0
There are two roots of this quadratic equation
x1 = [ -R + Ö{ R2 – (4L)/C}]/(2L) and x2 = [ -R - Ö{ R2 – (4L)/C}]/(2L)
If R2 ³ (4L)/C both x1 and x2 are Real;
If R2 £ (4L)/C both x1 and x2 are Complex: (x1 = s1+ jw1) and (x2 = s1 - jw1)
The equation in x [(L x2 + R x + 1/C) = 0] is called the Characteristic Equation
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If there are two solutions to a differential equation, then their sum is also a solution to it. That is,
the solution to the given second order differential equation is: i(t) = I0 ex1t + I0 ex2t
Io e x t = Io e (s + jw ) t = Io es t e jw t
That is, i(t) = Io es t [e jw t + e - jw t ] is the solution to the given second order differential
equation.
III
Given L (di/dt) + R i(t) = E;
Find i(t)
i(t) = A (t) ext
Assume solution:
di/dt = ext [d A(t)/dt]+ A(t) x ext
Substituting in the given equation:
L.ext.d A(t) /dt + [L.x.ext A(t) + R. A(t).ext] = E
We have two unknowns A(t) and x but only one equation. To simplify the solution, we choose x
by setting the expression in the bracket to zero
That is,
(Lx + R) A (t) ext = 0
Then, either
A (t) ext = 0, or Lx + R = 0; So, x = -(R/L)
Therefore;
Le-R/Lt . [d A(t)/dt ] = E
Integrating;
A(t) = [(E/R) e (R/L)t + A1]. And,
i (t) = A (t) ext = E/R +A1e-(R/L)t
We solve for A1 by using the initial conditions, i = 0 at t = 0
0 = E/R + A1 e –(R/L) 0 = E/R + A1
Therefore,
A1 = - E/R
Substituting for A1,
”
—
”
š
𝑖(𝑡) = i •1 − 𝑒 FR ˜ Sˆ ™ = i K1 − 𝑒 F› L
For t > 0, the right-hand side function is a step of magnitude E/R at t = 0. The solution is response
of the function to step input applied at the instant t = 0. The ratio [L/R] is defined as the time constant t of
the circuit.
There are two parts to the solution; "Transient" and "Steady-State". The nature of the transient
response is completely described by the solution of the algebraic equation (the characteristic equation). On
the other hand, the steady-state portion of the response has same form as the input (forcing) function.
IV
Given L (di/dt) + R i(t) = E Sin w t
Again, choose solution:
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Find i(t)
i(t) = A (t) ext
34
Dr. Mohammed Safiuddin
di/dt = ext [d A(t)/dt]+ A(t) x ext
Substituting in the given equation, L.ext.d A(t) /dt + [L.x.ext A(t) + R.A(t).ext] = E Sin wt
L.ext.d A(t) /dt + (L.x + R) A(t).ext = E Sin wt
That is,
Set
(Lx + R) A (t) ext = 0
Then, either A (t) ext = 0, or (Lx + R) = 0; x = -(R/L)
Therefore;
L e-(R/L) t . [d A(t)/dt ] = E Sin wt
d A(t)/dt = (E/L) e (R/L) t Sin wt
That is,
A (t) = (E/L) ò e (R/L) t. Sin wt. dt
Integrating by parts
é ( R )t
ù
ê e L ( R sin wt - cos wt ) ú
E ê
ú
wL
A(t ) =
ú + A1
2
wL ê
R
ì
ü
ê
ú
í
ý +1
êë
úû
î wL þ
And,
i (t ) =
E
wL
é
ù
ê R
ú
sin wt - cos wt ) ú
ê (
1
ê wL
ú + A1 e - ( R / L )t
2
2
ê
ú
ì R ü
ì R ü
ú
í ý +1 ê
í ý +1
êë
úû
î wL þ
î wL þ
1
Sin f =
Cos f =
2
ì R ü
í ý +1
î wL þ
( R / wL )
2
ì R ü
í ý +1
î wL þ
Sin (A –B) = Sin A Cos B - Sin B Cos A
Steady-state term is given by
iss =
E
wL
1
2
Sin (w t - Tan -1j )
é R ù
ê wL ú + 1
ë û
Response has same frequency as the forcing function, and the response differs in amplitude and
phase with respect to the input function.
FOR LINEAR SYSTEMS, THE OUTPUT RESPONSE DIFFERS ONLY IN AMPLITUDE AND
PHASE NOT IN FREQUENCY WITH RESPECT TO THE INPUT
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1.7
Laplace Transforms
Periodic functions of time can be represented by a set of sinusoidal functions using the "Fourier
Series" expansion.
f(t) = a0 /2 + a1 cos wt + a2 cos 2wt + a3 cos 3wt + … + b1 sin wt + b2 sin 2 wt + b3 sin 3wt +
Where,
an = w/p ò f(t) cos nwt dt
; n = 0,1,2,3,… and bn = w/p ò f(t) sin nwt dt
; n = 0,1,2,3,…
Any non-periodic arbitrary function can also be represented by a periodic wave over the region of interest.
If the period of the wave ( region of interest ) is expanded
(i) Fourier series will represent the non-periodic function over a wider and wider interval
(ii) Fourier series will contain more frequencies
As the interval gets larger, the fundamental frequency gets lower and the harmonics get closer spaced.
For 1 second period, f1 = 1 Hz and f2 = 2 Hz , and for 5 second period, f1 = 0.2 Hz and f2 = 0.4 Hz
If the interval is expanded to include the entire time region of -¥ to +¥ for a single period,
(i)
Fourier series becomes Fourier Integral
(ii) Discrete frequency spectrum becomes continuous, and
(iii) The function f(t) is expressed by
f(t) = ò [ A (w) cos wt + B(w) sin wt] d w
(iv) The coefficients become
A(w) = 1/ p ò [ f(t) cos wt ] dt
B(w) = 1/ p ò [ f(t) Sin wt ] dt
Fourier Integral Theorem leads to Fourier transforms
f(t) = 1/(2 p) ò [ F(w) ej wt ] d w
Where,
F(w) = ò [f (t) e - j wt ] dt
F(ω) is called the Fourier transform of f(t) and f(t) is called the Inverse Fourier transform
of F(ω).
By this transformation, a function of time can be expressed as a function of frequency.
Many of the common functions used in the study of control systems do not have Fourier transforms.
For example, the step function. If the integrand does not approach zero fast enough, the result of integration
required for the computation of the transform is not finite. A convergence factor [e - σ t] (σ real) is used for
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Dr. Mohammed Safiuddin
multiplying the integrand which makes it go to zero faster and thus keeps the resultant of integration finite.
Use of the convergence factor [ e - σ t] makes it necessary that the limits of integration be restricted to the
time interval with positive values of ' t ' only.
Combining the convergence factor [ e - σ t] with the factor [e - jω t ] results in the kernel of the
Laplace transforms [e - s t ] , where s = ( s + j w t ) is a complex variable. This is, of course, applicable
to only those functions whose value is zero for all t < 0.
Laplace transform provides a means for expressing functions f(t) of the time domain into functions
F(s) of the complex s domain.
L [f(t) ] = F(s) =∫ 𝑒 •ˆ 𝑓(𝑡) 𝑑𝑡
The inverse transform is obtained from; L -1 [F(s)] = 𝑓 (𝑡) =
,
4ž;
∫ 𝐹(𝑠) 𝑒 •ˆ 𝑑𝑠 ; c> s0
Properties of Laplace transforms.
1. Transforms and their Inverse are Linear
L [a1f1(t) + a2f2(t)] = a1F1 (s) + a2F2(s), if L [f1(t)] = F1 (s) and L [f2(t)] = F2 (s)
L -1 [b1 F1(s) + b2 F2(t)] = b1 f1(t) + b2 f2(t),
if
L -1 F1(s) = f1(t) and L -1 F2(s) = f2(t)
2. Transformation of the derivative df / dt
L [df / dt ] = sF(s) – f(0+).
Where, f(0+) = f(t) evaluated at t = 0+
3. Transformation of the integral.
L ò f (t)dt = ¡ ¢(•) + 𝑙𝑖𝑚
∫[𝑓(𝜏)]𝑑𝜏/𝑠
ˆ→I
4. Initial Value Theorem
f (0+) = lim f(t) = 𝑙𝑖𝑚 𝑠 𝐹(𝑠)
t→0
•→¦
t>0
5. Final Value Theorem
f (¥) = lim f(t) = 𝑙𝑖𝑚 𝑠 𝐹(𝑠)
t→∞
•→I
6. Time Scaling
f (t/a) = a F (a s) where L[f(t)] = F(s)
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Dr. Mohammed Safiuddin
7. Frequency Scaling
L -1[F(s/a)] = a f (a t) where, L -1[F(s)] = f(t)
8. Frequency Scaling
L -1[F(s/a)] = a f (a t) where, L -1[F(s)] = f(t)
9. Complex translation
L [e-at f(t)] = F (s + a) where, L [f(t)] = F(s)
10. Higher ordered derivatives of f(t)
L [d n f(t) / dt n] = s n F(s) – s n-1 f (0+) – s n-2 f 1(0+) ... f n-1(0+)
where, F(s) = L [f(t)] and f n-1(0+) is the value of (n-1) th derivative of the function at t = 0+
11. Higher ordered integrals of f(t)
L ò ò ò f(t) dt n = F(s) / s n + f-1 (0) / s n + f-2(0)/s n-1 + … + f- n (0) /s
f -1(0) = lim ∫ 𝑓(𝜏)𝑑𝜏
ˆ→I
12. Complex convolution integral
L [f1(t) . f2(t)] = 1/(2p j) ò F1(w) F2 (s- w) dw
where, F1(s) = L [ f1(t) ] and F2(s) = L [ f2(t)]
13. Convolution integral
L -1[F1(s) . F2(s)] = ò f1(t) f2(t - t)dt = ò f2(t) f1(t - t)dt
Example:
Find Laplace transform of f(t) = t e-a t using the basic definitions
L (t e- a t) = ∫I¦ 𝑡. 𝑒 Fªˆ . 𝑒 F•ˆ 𝑑𝑡 = ∫I¦ 𝑡. 𝑒 F(•Dª) ˆ 𝑑𝑡
Integrating by parts
¦
u = t; dv = e-(s+a)t dt;
« ¬(-®¯)š
and
¦ « ¬(-®¯)š
∫I 𝑡. 𝑒 F(•Dª) ˆ 𝑑𝑡 = [ 𝑡. F(•Dª) ]¦
I − ∫I
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du = dt; v = e-(s + a)t / - (s+ a)
F(•Dª)
𝑑𝑡
38
Dr. Mohammed Safiuddin
« ¬(-®¯)š ¦
1
𝑒−(𝑠+𝛼)𝑡 ∞
= [ 𝑡.
] + [ . ( ) ]0
F(•Dª) I
𝑠+𝛼 − 𝑠+𝛼
= [−
°ˆ.« ¬(-®¯)š ± ¦
,
]I − [(•Dª)> 𝑒 F(•Dª)ˆ ]¦
I
•Dª
In the first term, e-(s+a) t approaches 0 faster than t approaches ¥ as t ® ¥
L (t e - at ) = (•Dª)>
,
Therefore,
Frequency shifting property can be used as follows
L [𝑓(𝑡) 𝑒 •‘ 𝑡 ]= F(s-so)
f(t) = t; F(s) = 1/s2 ; so = - a That is, L (t e - at ) =
1.8
,
(•Dª)>
Probability & Statistics
Engineering and social sciences problems involving random processes, or variables, are defined as
stochastic (non-deterministic). Instead of deterministic values, the measurements are expressed by the
“Mean” of the distribution (𝑋³) and its variance s 2. Smaller the variance, closer the random values of the
signal/variable to the ‘Mean’, and closer it is to a deterministic system behavior.
Probability of an event taking place under specified conditions is expressed in per-unit ranging
from 0 (zero) for occurrence of an impossible outcome to 1 (one) for occurrence of an outcome with
certainty (full assurance).
There are 6 general rules when we are dealing with probabilities of single and multiple events.
Rule 1.
Probability of an impossible outcome of an event ......P{0} = 0
Rule 2.
If Ai [ i = 1, 2, .... n] are n possible outcomes of a process, then
P{A1 or A2 or .......An } = P{A1} + P{A2} + ........+P{An}
Rule 3.
If Ai, Bi, .... Zi are independent processes with multiple outcomes having probabilities of P{Ai},
P{Bi}, ...... P{Zi}, then P{Ai and Bi and ..... Zi} = P{Ai} p{Bi}..........P{Zi}
Rule 4
Probability of an outcome not occurring is complement of the probability of its occurrence.
P{not A1} = 1- P{A1}
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Dr. Mohammed Safiuddin
Rule 5
P{Ai or Bi} = P{Ai} + P{Bi} - P{Ai} P{Bi}
Rule 6
Probability that A will occur given that B has already occurred, where the two event are dependent.
𝑃{𝐴 𝑔𝑖𝑣𝑒𝑛 𝐵} =
¹{x ‡-Ž º}
¹{º}
Probability density functions are mathematical functions describing probabilities of numerical events.
Numerical events are outcomes which have numerical values expressed by real numbers. There are four
such types of density functions.
1. Binomial
n = the number of trials
x = the number of successes desired
P = the probability of a success in a single trial
q = (1-P) the probability of failure
The binomial coefficient is given by
𝑛!
𝑃 c 𝑞(-Fc)
(𝑛 − 𝑥)!
The probability of obtaining x successes in n trials is given by
𝑛
𝑃 (𝑥) = K L 𝑃 c 𝑞(-Fc)
𝑥
The mean of the binomial distribution is nP and the variance of the distribution is nPq
2. Poisson
If an event occurs, on the average, λ times per period, the probability that it will occur x times per
period is given by f [ x ] = e
-l
lx
x!
3. Exponential
If the mean of the exponential distribution is 1/u, then
𝑓[𝑥] = 𝑢(𝑒 F’c )
The variance is ( 1/ u )2 and the probability F[x] of x or less occurring is (1 − 𝑒 F’c ).
4. Normal (Gaussian)
Although f[x] may be expressed mathematically for the normal distribution, tables are used to
evaluate F[x] since f[x] cannot be easily integrated. Since the x-axis of the normal distribution
Ó 2020
40
Dr. Mohammed Safiuddin
will seldom correspond to actual sample variables, the sample values need to be normalized for use
of the standard tables. Given the mean (u), and the standard deviation (σ), the normalized variable
is
𝑧=
(𝑠𝑎𝑚𝑝𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝑢)
𝜎
Then, the probability of a sample exceeding the given sample value is equal to the area in the tail
past point z.
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41
Dr. Mohammed Safiuddin
Appendix A – Sample Problems from New York State PE License Exams
[Solutions by Sabbir S. Moochalla, Graduate Student Elec. Eng. Dept., University at Buffalo, 1976]
1.
A rectangle lies in the first quadrant, has two of its sides on the x and y axes, one vertex at the origin,
and the opposite vertex on the curve x2 + 2y2 = 9. Find either coordinate of the latter vertex so the
area of the rectangle is a maximum.
Solution:
Equation of the curve: x2+2y2-9 =0.
Let the coordinates of the unknown vertex be (x1 , y1) on the given curve.
Then,
x12 + 2y12 -9 =0 ....................(1)
Area of the rectangle
A = x1.y1 ..................................(2)
From (1),
x1 = (9-2y12)1/2;
A = x1.y1 = (9-2y12)1/2. y1
Area A reaches its maximum when the rate of change of A with respect to y reaches zero.
Therefore,
1
𝑑𝐴
1/2
−
= (9 − 2𝑦21 )
− 2𝑦21 ¿9 − 2𝑦21 À 2 = 0
𝑑𝑦
:
\
9 – 2 y12 - 2 y12 = 0; 𝑦, = 4
Substituting in (1); 𝑥, = V(9 − 2𝑦,4 = Á
N
4
:
=
√4
:
:
\ (𝑥, , 𝑦, ) = (√4 , 4 )
2.
Brine containing 1 lb. of salt per gallon, runs in to a 200-gallon tank, initially full of brine containing
3 lbs. of salt per gallon, at the rate of 4 gals/minute. If the mixture runs out at the rate of 5 gals/minute,
when will the concentration of the salt in the tank reach 1.01 lbs./gal?
Solution:
Let x = Amount of salt in tank at time t
C = Concentration of salt at time t
Then, amount of salt at instant t, x = C (200 – 5t + 4t);
c
That is, x = C (200 – t); and 𝐶 = (4IIFˆ)
The rate of change of salt in the tank
Substituting for x,
Ó 2020
Žc
Žˆ
Ž {f(4IIFˆ)}
Žˆ
= 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑔𝑎𝑖𝑛 − 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑙𝑜𝑠𝑠 = 4 −
=4−
?f(4IIFˆ)
(4IIFˆ)
?c
(4IIFˆ)
= 4 − 5𝐶
42
Dr. Mohammed Safiuddin
Taking the derivative on the left-hand side; (200 − 𝑡).
ŽÃ
That is;
J(,Ff)
=
ŽÃ
Žˆ
− 𝐶 = 4 − 5𝐶
Žˆ
(4IIFˆ)
Integrating both sides with limits of C from 3 lbs./gal to 1.01 lbs/gal
,.I,
ˆ
:
I
𝑑𝑐
𝑑𝑡
Ä
= Ä
4(1 − 𝐶)
(200 − 𝑡)
1
[𝑙𝑛(1 − 𝐶)],.I,
= [𝑙𝑛(200 − 𝑡)]ˆI
:
4
¼ [ ln 0.01 – ln 2] = ln (200-t) – ln 200
,
I.I,
4IIFˆ
Æ
ln 4 = ln 4II ; √0.005 =
J
3.
4IIFˆ
4II
; t = 147 minutes
A closed circular cylindrical can is to have a volume of V. If the cost of the material per square inch
in the top and bottom is twice that used in the curved surface find the ratio of height H, to diameter
D, for minimum cost.
Solution:
Volume of a cylinder V = p r2 H. That is, 𝐻 =
È
žƒ >
Surface area of the side: S1 = 2p r H; Surface area of top and bottom S2 = 2 p r2
Let the cost of top and bottom be x, then the total cost C = (x/2). 2p r H + x. 2 p r2
c
È
Substituting for H in the above equation for cost, 𝐶 = 4 . 2𝜋𝑟 žƒ> + 𝑥. 2𝜋𝑟 4 =
È.c
ƒ
+ 𝑥. 2𝜋𝑟 4
To minimize the cost C, we take its derivative with respect to radius r and set it equal to zero.
ŽÃ
That is:
Žƒ
È
= 𝑥. R− > + 4𝜋 𝑟S = 0
ƒ
È
Since x ¹ 0, R− ƒ> + 4𝜋 𝑟S = 0. Therefore, 4𝜋 𝑟 =
È
ƒ>
Substituting the value of r in the expression for H, we get 𝐻 =
That is,
Ó 2020
Ì
Í
=
Ì
4ƒ
=
È
Ê
Ê
4ž( )>/l ( )C/l
ÆË
ÆË
=
È
Ê
4ž( )
ÆË
È
𝑜𝑟 𝑟 = (Jž ),/:
=2
È
ž(
Ê >/l
)
ÆË
Ì
Í
=2
43
Dr. Mohammed Safiuddin
4.
A parabolic arch has a span of 120’ and height of 25’. Derive the equation of the parabola and
compute heights of the arch at points 10’, 20’ and 40’ from the center.
Parabolic Arch with X and Y Axes
Solution:
General equation for a parabola is: x2 = -2py
At x = 60, y = -25, Therefore, 𝑝 =
c>
The equation for the Parabolic Arch is:
5.
:GII
F4•
= ?I = 72
x2 = -144y
,II
i.
At x = 10, 𝑦 = F,JJ = −0.695 𝑓𝑡; H1 = 25-0.695 = 24.305 ft.
ii.
At x = 20, 𝑦 = F,JJ = −2.78 𝑓𝑡;
iii.
At x = 40, 𝑦 = F,JJ = −11.11 𝑓𝑡; H3 = 25-11.11= 13.89 ft.
JII
H2 = 25-2.78 = 22.22 ft.
,GII
An airplane headed N 300 E, climbing on a 200 incline at a speed of 240 mph passes one mile directly
over an automobile headed south on a flat
road traveling at the speed of 60 mph. At
what rate of speed are they separating after
10 minutes?
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44
Dr. Mohammed Safiuddin
Solution:
This is a time and space vector problem in three dimensions. Need to determine three time dependent
components of the velocity vector along the x [East-west], y [North-south], and z [Elevation] axes.
The component î of the distance vector along the x-axis is: di = [240 t (Cos 20) (Sin30)]
The component 𝚥̂ of the distance vector along the y-axis is: dj = [60 t + 240 t (Cos 20) (Cos 30)]
The component 𝑘Ð of the distance vector along the z-axis is: dk = [1+ 240 t (Sin 20)]
That is: D = d1 î + d2 𝚥̂ + d3 𝑘Ð
D2 = [240 t (Cos 20) (Sin30)]2 + [60 t + 240 t (Cos 20) (Cos 30)]2 + [1+ 240 t (Sin 20)]2
Differentiating with respect to t:
2𝐷
ŽÍ
Žˆ
= 2𝑡 [240 (Cos 20) (Sin30)]4 + 2𝑡 [60 + 240 (Cos 20) (Cos 30)]4 +
2 𝑡 [240 Sin 20]4 + 2(240 Sin 20)
2𝐷
ŽÍ
Žˆ
= 2 𝑡 {[112.76]4 + [60 + 195]4 + [82]4 } + 164
,I
ŽÍ
At 𝑡 = GI ℎ𝑜𝑢𝑟𝑠, 𝐷 Žˆ =
,I
GI
{[112.76]4 + [60 + 195]4 + [82]4 } +
,I
,I
,I
,GJ
4
= 14,159
,I
And, D2 = [240 GI (Cos 20) (Sin30)]2 + [60 GI+240 GI(Cos 20) (Cos 30)]2 + [1+ 240 GI (Sin 20)]2
D = √353 + 1810 + 215.5 = 48.77 miles
ŽÍ
,J,?N
= JH.EE ≅ 290 mph.
Žˆ
6.
Given coordinates of points A (3,5) and B (-1,3), Find
a)
Slope of line AB
b)
Distance from the x intercept at 8 to point B
c)
Equation of the line through A that is perpendicular
to line AB
d)
Equation of line through C (2,0) parallel to AB
e)
A in Ax + 3y = 4 so that it is parallel to line AB
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45
Dr. Mohammed Safiuddin
Solution:
(• F• )
?F:
,
a) Slope of a straight line between two points is 𝑚 = (cC Fc> = :D, = 4
C
>)
b) Distance between two points 𝑑 = V(𝑦,F 𝑦4 )4 + (𝑥, − 𝑥4 )4
𝑑 = V(3 − 0)4 + (8 + 1)4 = √90 = 9.486
c) Slope of line AB = m = ½;
Slope of line perpendicular to AB is
Therefore, equation of the line perpendicular to line AB: −2 =
F,
×
= −2
•F?
cF:
-2x + 6 = y – 5; Or, y + 2x = 11
That is,
,
•FI
d) Equation of the line through C (2,0) and parallel to line AB: 𝑚 = 4 = cF4
2y – x + 2 = 0
e)
7.
Given:
Fx
,
:
A x + 3y = 4; Slope of the line m = -A/3; Therefore, : = 4 ; 𝐴 = − 4
A farmer decides to build a rectangular feed lot with semi-circular pen at one end. If he has five
hundred feet of fencing, find the radius of the pen which would give a maximum area for the feed lot.
Y
r
X
Solution:
Length of the perimeter of the feedlot and pen = 2 (Y+ X) + 𝜋 𝑟 = 500; Y = 2 r
Therefore,
2 (2 r+ X) + 𝜋 𝑟 = (4 + 𝜋 ) r + 2 X =500; 𝑋 =
Area of the rectangle feedlot A= X.Y =
Ó 2020
?IIF(JDž)ƒ
4
?IIF(JDž)ƒ
4
. 2 𝑟 = 500 𝑟 − 8𝑟 4 − 2𝜋 𝑟 4
46
Dr. Mohammed Safiuddin
To maximize the area, we compute rate of change of area A with respect to radius r
𝑑𝐴
= 500 − 8𝑟 − 2𝜋 𝑟 = 0
𝑑𝑟
500
𝑟=
= 35 𝑓𝑡.
8 + 2𝜋
8.
The cross section of a 10 ft long tank is an equilateral triangle, as shown. Water runs in to the tank at
the rate of 2 cu ft./minute. At what rate is the depth of water increasing when the depth is 6 inches?
Solution:
Surface area at height h: A = 10´ x Surface Width w
Ù/,4
Ù/,4
Surface width 𝑤 = fÚ• :I ; Then 𝐴 = 10 𝑥 fÚ• :I
𝑑ℎ
𝐹𝑙𝑜𝑤 𝐼𝑛 𝐹
=
𝑑𝑡
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝐴
𝑑ℎ
2 𝐶𝐹𝑀
2 𝐶𝑜𝑠 30
|Ù•G Ü-. =
=
= 0.3464 𝐹𝑃𝑀
ℎ/12
𝑑𝑡
5
10 𝑥
𝐶𝑜𝑠 30
Rate of increase in depth at 6 inches = 4.1568 inches/minute
9.
A porous material dries in the open air at a rate proportional to its moisture content. If a slab hung in
the wind loses half its free moisture in one hour, how long will it take to lose 95%, weather conditions
remaining the same? Let “a” be the original amount of moisture in the slab, and “x” be the amount
of moisture removed in time “t”.
Solution:
Amount of moisture left at time t, Mt = a – x
Then,
Žc
Žˆ
Žc
= 𝑘 (𝑎 − 𝑥); That is, (‡Fc) = 𝑘 𝑑𝑡
Integrating both sides,
I.?‡ FŽc
− ∫I
(‡Fc)
,
= ∫I 𝑑𝑡
− ln(𝑎 − 𝑥 )]I.?‡
= 𝑘 [𝑡],I
I
‡
ln I.?‡ = 𝑘 = ln 2 = 0.693/ℎ𝑟
I.NN‡ Ž‡
ˆ
= 0.693 ∫I 𝑑𝑡
‡Fc
Then to remove 99% moisture ∫I
− ln(𝑎 − 𝑥 )]I.NN‡
= 0.693 𝑡
I
‡
,
ln I.I,‡ = 0.693 𝑡, ∴ 𝑡 = I.GN: ln 100 = 6.65 ℎ𝑜𝑢𝑟𝑠
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47
Dr. Mohammed Safiuddin
10.
A radioactive substance decomposes according to the law
Ž×
Žˆ
= −16 𝑚, where m is the mass of
,E
the substance at time ‘t’. If m =17 at t = 0, find the time when 𝑚 = « > .
Solution:
𝑑𝑚
= −16 𝑚
𝑑𝑡
Ž×
,
− ∫ ,G× = ∫ 𝑑𝑡 ; − ,G ln 𝑚 = 𝑡 − 𝑐
Integrating both sides,
,
Since m=17 at t = 0; 𝑐 = ,G ln 17; Therefore,
𝑚=
𝑡=
11.
,
,G
,E
ln × = 𝑡
17
; 𝑡 =?
𝑒4
1
17
1
1
ln à á =
ln 𝑒 4 =
17
16
16
8
𝑒4
Plans for a supermarket require a floor area of 15,000 sq. ft. The building will be rectangular with
three solid brick walls and a glass front. However, the heat loss across this glass front is 7 times that
across the brick walls, which is 1 Btu/ft, and the glass costs 2 times as much as brick walls per lineal
foot, which costs $ 15/ft. We shall neglect the heat losses through floor and ceiling. The cost of the
market is to be amortized over 10 years and the heat loss is figured in Btu/year at $ 2.00/Btu/year.
What must the dimensions of the supermarket be so as to minimize the total cost (walls & heating)
over this 10-year period?
Solution:
Let x and y be the floor dimensions
Cost of the Btu loss of wall over 10 years
Cw = (2 y + x). 1. 10. 2
Cost of wall plus glass Cw&g = (2y+x) .15+ 30x
Cost of the Btu loss of glass over 10 years Cg = (7x).20 = 140 x
Total Cost T = Cw + Cg + Cw&g = (2 y + x). 20 +140 x + (2y+x) .15+ 30x = 70 y + 205 x
Floor area A = xy = 15,000 sq. ft., Therefore, 𝑥 =
Substituting value of x in total cost, 𝑇 = 70 𝑦 +
,?III
•
4I?c,?III
•
For minimization of the total cost T, we take its derivative with respect to y and set it equal to zero
and solve for y.
Ó 2020
48
Dr. Mohammed Safiuddin
𝑑𝑇
70 𝑦 4 − (205). 15000
=
=0
𝑑𝑦
𝑦4
,?III
70 y2 = (205).15000; y = 209.6 ft. and 𝑥 = 4IN.G = 71.56 𝑓𝑡.
Glass front side = 71.56 ft. and the total walls = (209.6)2 + 71.56 = 490.76 ft.
12.
A force F, is proportional to x2 and is equal to 20 pounds when x = 2 ft. Find the work done as it
moves an object from x=2 ft. to x=5 ft.
F = k x2; at x =2, F=20; k = 5
Solution:
?
?
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒, 𝑊 = Ä 𝐹. 𝑑𝑥 = Ä 5 𝑥 4 . 𝑑𝑥 =
4
13.
4
5 : ?
625 40
𝑥 ]4 =
−
= 𝟏𝟗𝟓 𝒇𝒕. 𝒍𝒃𝒔
3
3
3
A dam has a vertical, trapezoidal upstream face with dimensions as shown in the figure. Assuming
water to have a density of 62 lbs./cu.ft., find the force of the water on the face of the dam.
Solution:
Force on the dam dF = p. dA; where p is the pressure per square ft. p = r h
Area of the strip dh wide; dA = L. dh; where L is the length of the strip. L = 100 - 2h
dA = (100 – 2h) dh, and dF = r h (100 – 2h) dh
Therefore,
4I
4I
∴ 𝐹 = Ä 𝑟ℎ (100 − 2ℎ )𝑑ℎ = Ä (100 𝑟ℎ − 2𝑟 ℎ 4 )𝑑ℎ
I
∴
𝐹 = [ 50 𝑟 ℎ4 −
I
2 : 4I
2
𝑟 ℎ ]I = •50(400) − (8000)™ . 𝑟
3
3
Substituting for r and simplifying, we get F = 909,354 lbs.
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49
Dr. Mohammed Safiuddin
14.
The resistance in an electrical circuit is determined from the well-known formula 𝑅 =
”
ì
, where I is
the current measured in amperes, E the electromotive force in volts, and resistance R in ohms. If
there is an error of 0.1 of an ampere in reading I and 0.05 volt error in reading E, determine:
(a) What is the maximum approximate error in R if the readings are I = 15 amps, and E = 110 volts?
(b) What is the percentage of error in this case?
Solution:
,
R = E/I, then error 𝑑𝑅 =
”
𝑑𝐸 − ì> 𝑑𝐼
ì
𝐸
110
=
= 7.33 Ω
𝐼
15
1
110
(0.05) +
(0.1) = 𝟎. 𝟎𝟓𝟐𝟐 Ω
𝑑𝑅 =
15
225
𝑅=
Percentage error =
15.
I.I?44
E.::
𝑥 100 = 𝟎. 𝟕𝟏𝟐 %
A particle sliding on a certain inclined plane is subject to an acceleration downward of 4 ft./sec.2. It
started upward from the bottom of the plane with a velocity of 6 ft./sec.
a. Find the distance moved after “t” seconds.
b. How far will it go before sliding backwards?
c. If the inclined plane is 20 ft. long, find the necessary initial speed in order that the particle may
just reach the top.
Solution:
The net velocity of the particle is initial velocity minus the negative velocity of downward accel.
V(t) = u – at = 6 – 4t.
ˆ
a. The distance moved after “t” seconds, 𝐷 = ∫I (𝑢 − 𝑎𝑡)𝑑𝑡
b. The velocity reaches zero when velocity due to downward acceleration reaches 6 ft./sec.
V = 0 = 6 – 4t; t= 1.5 sec.
To calculate the distance reached, we integrate the velocity function for the period of 1.5 sec.
ˆ
𝐷 = Ä (𝑢 − 𝑎𝑡)𝑑𝑡 = [𝑢𝑡 −
I
c.
𝑎 4 ,.?
4
𝑡 ]I = •6𝑥1.5 −
1.54 ™ = 9 − 4.5 = 4.5 𝑓𝑡.
2
2
We have two equations in two unknowns, u and t.
’
t(u-2t) = 20 and u-4t = 0; so 𝑡 = J
Substituting for “t” in the first equation, we get:
’
J
Ó 2020
’
R𝑢 − 2 J S = 20; 𝑢4 = 160; 𝑢 = √160 = 12.65 ft/sec.
50
Dr. Mohammed Safiuddin
16.
In the figure shown, two towns are
located as indicated. A water filter plant
is to be constructed at a point C on the
river to service the two towns. Where
must this plant be located so as to
minimize the total amount of pipeline
needed?
Solution:
For the triangle ABF, AF2 =AB2 -BF2 = 25 – 9
Therefore, the distance DE = AF = 4 miles
Let the distance CD = x, then CE = (4-x) miles
Total length of the pipeline P = AC + BC
From the right-angled triangle ACD, 𝐴𝐶 = √1 + 𝑥 4
And, from right-angled triangle BCE, 𝐵𝐶 = V16 + (4 − 𝑥)4
𝑃 = V1 + 𝑥 4 + V16 + (4 − 𝑥)4
We differentiate P with respect to x and set it equal to zero to find x for minima.
𝑑𝑃
1
= 𝑥(1 + 𝑥 4 )F,/4 + (𝑥 4 − 8𝑥 + 32)F,/4 (2𝑥 − 8) = 0
𝑑𝑥
2
𝑥(1 + 𝑥 4 )F,/4 = (4 − 𝑥) (𝑥 4 − 8𝑥 + 32)F,/4
∴ 15 𝑥 4 + 8𝑥 − 16 = 0; (5𝑥 − 4)(3𝑥 + 4) = 0
Since distance can’t be negative, the distance DC = 4/5 = 0.8 miles.
17.
A 200-pound vertical load is supported by the concrete
column as shown. The radius of the column at the top is 4
inches but varies such that the load per unit area is same at
all cross-sections. If the height of the column is 8 feet and
concrete weighs 150 lbs./cubic foot, determine the radius at
the base of the column. (Note: do not neglect the weight of
the column.)
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51
Dr. Mohammed Safiuddin
Solution:
dy » ds
Kc = 150 lbs./cu.ft.
•
Weight of the column at any height 𝑊Ã = 𝐾Ã ∫I 𝜋 𝑟 4 𝑑𝑦,
At any height, load per unit area is same, therefore;
200
200 + 𝑊Ã
4 =
𝜋𝑟 4
𝜋𝑟I
𝑦
200
1
2
4 = 𝑟 4 [200 + 150 Ä 𝜋 𝑟 𝑑𝑦1 ]
𝑟I
0
JII ƒ
Differentiating,
ƒ‘>
Žƒ
ƒ
𝑑𝑟 = 150 𝜋 𝑟 4 𝑑𝑦
,?I
:
= JII 𝜋 𝑟I4 𝑑𝑦; 𝑙𝑛 𝑟 = H 𝜋 𝑟I4 𝑦 + 𝑐
𝑟I = 1ó3 𝑓𝑡. , ln 𝑟 = 4J 𝑦 + 𝑐 ; 𝑜𝑟, 𝑟 = 𝑘 𝑒 ¿ ó4JÀ•
At y = 0, r = 1/3, therefore, k = 1/3; 𝑟 =
At y = 8, 𝑟 =
18.
ž
ž
For
,
:
,
ž
𝑒 ¿ ó4JÀ•
:
ž
𝑒 ó: ; 𝒓 = 𝟎. 𝟗𝟓 𝒇𝒕. = 𝟏𝟏. 𝟒 𝒊𝒏𝒄𝒉𝒆𝒔
A farmer wishes to fence-in a field located in a corner of his property. A stream flows along one side
of the field and his property line meets the stream at an angle of 450. If no fence is required along
the stream side of the field, what are the
dimensions of a and b of the largest area
that can be enclosed with 1,000 feet of
fence?
Solution:
Total length of the fence L = (a + b + c)
= 1000
Using the right-angled triangle identity,
‡
𝑆𝑖𝑛𝑒 45I = Ã ; 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑐 = √2 𝑎 , and ¿√2 + 1À𝑎 + 𝑏 = 1000;
𝑏 = 1000 − ¿√2 + 1À𝑎
Area to be fenced A is sum of the areas of the rectangle and the triangle;
𝐴 = 𝑎𝑏 +
Ó 2020
1 4
1
𝑎 = 𝑎 û1000 − ¿√2 + 1À𝑎ü + 𝑎4
2
2
52
Dr. Mohammed Safiuddin
Differentiating area A with respect to the length a and setting it equal to zero for maxima;
𝑑𝐴
1000
= 1000 − ¿2√2 + 1À𝑎 = 0; 𝑎 =
= 261.2 𝑓𝑡.
𝑑𝑎
¿2√2 + 1À
𝑏 = 1000 − ¿√2 + 1À
1000
¿2√2 + 1À
= 369.4 𝑓𝑡.
Total length L = (261.2 + 369.4 + 369.4) = 1000
Note: Length b is equal to length c
19.
The parabolic reflector of an automobile headlight is 12” in diameter and 4” deep. What is its surface
area?
Solution:
The equation of a parabolic curve is y2 = k x
When x = 4, y = 6; Therefore, k = 9 and y2 = 9 x
The surface area of the ring of radius y and width ds is
dA = 2 p y. ds, where ds2 = dy2 + dx2
𝑑𝑠 = V𝑑𝑦 4 + 𝑑𝑥 4 = ý1 +
𝑑𝑦4
𝑑𝑥
𝑑𝑥
Ž•
N
N
Differentiating y2 = 9 x, we get 2 y dy = 9 dx and Žc = 4•; Therefore, 𝑑𝑠 = Á1 + (4•)4
N
Surface area of the infinitesimal strip 𝑑𝐴 = 2𝜋𝑦. Á1 + (4•)4
For the full surface area of the headlight, we integrate dA over the range of x from 0 to 4”.
J
J
I
I
9
81
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝐴 = 2𝜋 Ä 𝑦ý1 + ( )4 𝑑𝑥 = 2𝜋 Ä ý𝑦 4 +
𝑑𝑥
2𝑦
4
2
Substituting for y = 9 x in the integrand, and integrating, we get:
J
H,
J
N
J
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝐴 = 2𝜋 ∫I Á9 𝑥 + J 𝑑𝑥 = 6𝜋 ∫I Á𝑥 + J 𝑑𝑥 = 3𝜋 ∫I √4𝑥 + 9 𝑑𝑥
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝐴 =
Ó 2020
:
𝜋
[(4𝑥 + 9)4 ]JI = 49 𝜋 = 154 𝑠𝑞. 𝑖𝑛𝑐ℎ𝑒𝑠
2
53
Dr. Mohammed Safiuddin
20.
A spherical shape mothball loses mass at the rate proportional to its surface area. Its diameter is
observed to decrease by a factor of 2 in 200 days. How long will it take for the mass to decrease to
one-half of its initial value?
Solution:
Surface area of a sphere of diameter D, S = p D 2
,
Mass of the sphere of diameter D, 𝑀 = 𝜌 G 𝜋 𝐷: , where r is the density per cubic volume.
Rate of change of mass 𝑑𝑀 =
:ž
G
ž
𝜌 𝐷4 𝑑𝐷 = 4 𝜌 𝐷4 𝑑𝐷 ,
The rate of change of mass as a function time is proportional to its surface area; dM = -k p D2 dt
Equating the two expressions for rate of change of mass, we get
𝜋
−𝑘 𝜋𝐷4 𝑑𝑡 = 𝜌 𝐷4 𝑑𝐷
2
ŽÍ
ÿ
= −2 ! ; Therefore, 𝐷 =
Žˆ
F4ÿ
!
𝑡+𝐶
When t = 0, D = D0, \ C = D0, and 𝐷 =
At t = 200, 𝐷 =
Í‘
4
1
𝑀,
2 I
!
𝑡 + 𝐷I
!Í
∴ 𝑘 = HII‘
Substituting for k in the expression for D, we get: 𝐷 = 𝐷I −
𝑀" =
F4ÿ
Í‘
JII
Í
ˆ
𝑡, 𝑇ℎ𝑎𝑡 𝑖𝑠 Í = 1 − JII
‘
𝑀"
1
𝜋𝜌 𝐷:
𝐷
𝑡
= =
= lV1/2 = 0.794 = 1 −
: ;
𝑀I
2
𝐷I
400
𝜋𝜌 𝐷I
Therefore, t = 82.4 days
21.
Half-inch thick sheet steel is used to construct a cylindrical tank 10 ft. in diameter and 20 ft. high,
closed at top and bottom. How many square feet of sheet steel would be saved if it were so
proportioned that total sheet steel used is minimized?
Solution:
Total area of sheet steel used, A = 2p R2 + 2p RH, where R is the radius of the tank and H is height.
Volume of the tank, V = p R2H = p (5)2 . (20) = 1,570 cu. ft.
Therefore, 𝐻 =
,?EI
ž i>
,?EI
; Total area A = 2p R2 + 2p R. ž i> = 2p R2 +
:,JI
i
Differentiating A with respect to R and equating to zero, we get:
Žx
= 4𝜋𝑅 −
Ži
:,JI
i>
= 0; 𝑅: =
𝑇ℎ𝑎𝑡 𝑖𝑠, 𝑅 = 6.3 𝑓𝑡. ; 𝑎𝑛𝑑 𝐻 =
Ó 2020
:,JI
Jž
= 250; ;
1570
1570
=
= 12.6𝑓𝑡.
4
𝜋𝑅
𝜋(6.3)4
54
Dr. Mohammed Safiuddin
Sheet metal area A = 2p R2 + 2p RH = 2p (6.3)2 + 2p (6.3)(12.6) = 748.2 sq. ft.
Original sheet metal area A = 2p R2 + 2p RH = 2p (5)2 + 2p (5)(20) = 785.4 sq. ft.
Therefore, sheet metal saved = 785.4 – 748.2 = 37.2 sq. ft.
22.
A hydro-electric power plant on a river bank supplies power to a load center on the other side and
three miles downstream. If the river is 2 miles wide and the power line costs $ 4,000/mile on land
and $ 5,000/mile under water, what alignment would be least cost and what would it be?
Solution:
Let the length of power line on land = (3-x)
Length of power line under water = √4 + 𝑥 4
Total cost of the power line:
𝐶 = 4000 (3 − 𝑥) + 5000 V4 + 𝑥 4
Differentiating C with respect to x and equating it to zero.
𝑑𝐶
5000 𝑥
= −4000 +
= 0; 5000 𝑥 = 4000 (V4 + 𝑥 4 ; 25 𝑥 4 = 16 (4 + 𝑥 4 )
𝑑𝑥
√4 + 𝑥 4
𝑥=
8
𝑚𝑖𝑙𝑒𝑠
3
1/3 mile on land and 3 and 1/3 under water, at a total cost of 𝐶 =
23.
The differential equation
Ž> c
Žˆ >
J,III
:
+
?I,III
:
= $ 18,000
Žc
+ 5 Žˆ + 6𝑥 = 0 governs the motion of a system.
If dx/dt = 0, and x =12 at t = 0, find x at t = 1.
Solution:
Ž> c
Žc
Let’s assume solution 𝑥(𝑡) = 𝐶𝑒 ‡ˆ , then Žˆ = 𝐶 𝑎 𝑒 ‡ˆ 𝑎𝑛𝑑 Žˆ > = 𝐶𝑎4 𝑒 ‡ˆ
Substituting in the given equation, we get
𝐶 𝑒 ‡ˆ [ 𝑎4 + 5𝑎 + 6] = 0,
∴ 𝑎4 + 5𝑎 + 6 = 0,
(𝑎 + 3)(𝑎 + 2) = 0
𝑥 (𝑡) = 𝐶, 𝑒 F:ˆ + 𝐶4 𝑒 F4ˆ ;
At t = 0, x = 12, \ C1 + C2 = 12, and at t = 0, dx/dt = 0, \ 3 C1 + 2 C2 = 0
Solving for C1 and C2 from the above two equations, we get C1 = -24, and C2. = 36
𝑥 (𝑡) = −24 𝑒 F:ˆ + 36 𝑒 F4ˆ
:G
4J
At t=1, 𝑥 (1) = « > − « l = 4.87 − 1.19 = 3.68
Ó 2020
55
Dr. Mohammed Safiuddin
24.
A flower bed is to be in the shape of circular sector of radius “r” and a central angle “q” radians (i.e.
piece of pie shape). The perimeter P and area A are given by 𝑃 = 2𝑟 + 𝑟q and 𝐴 =
,
4
𝑟 4 𝜃 . Find
values of r and q such that the perimeter is minimized for the sector area of 50 sq.ft.
Solution:
,
𝜃=
,II
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑃 = 2 𝑟 + 𝑟 𝜃 = 2 𝑟 +
,II
𝐴𝑟𝑒𝑎 𝐴 = 4 𝑟 4 𝜃 = 50 ∴
ƒ>
ƒ
We take the derivative of P with respect to r and set it equal to zero.
Ž¹
,II
,
,
= 2 − ƒ > = 0; 𝑟 = √50 = 7.07; 𝐴𝑟𝑒𝑎 𝐴 = 4 𝑟 4 𝜃 = 50 = 4 (50)𝜃
Žƒ
\ q = 2 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 =
25.
,HI
ž
2 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 = 114.6 I
A trapezoidal shaped channel with side slopes of 1:1 has a required cross-sectional area of 100 sq.ft.
What must be the depth of flow, d, and the bottom width, b, to minimize the wetted perimeter? The wetted
perimeter is the base plus the two sides.
Solution:
Area A = 100 = bd + d2
∴ 𝑏=
,IIFŽ>
Ž
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑃 = 𝑏 + 2𝑐 =
,IIFŽ>
Ž
+ 2 √2 𝑑 =
,II
Ž
+ ¿2√2 − 1À𝑑
Differentiating and equating to zero,
Ž¹
ŽŽ
,II
,II
= − Ž> + ¿2√2 − 1À = 0, 𝑑 = Á(4√4F, ) = √54.7 = 7.4 𝑓𝑡.
𝑏=
Ó 2020
,II
Ž
,II
− 𝑑 = E.J − 7.4 = 6.11 𝑓𝑡.
56
Dr. Mohammed Safiuddin
Appendix B- Excercise Problems
Indicate the choice of your answer for the following multiple-choice statements. (10 points)
1.
If Z is a complex number 1 + j and Z* is its conjugate, then [Z2/Z*] is given by
(a) 1 + j
2.
(b) 8
(b) [Z*]2
3+ j
If Z is a complex number
d)
3 (1 - j )
2 + j 2 and Z* is its conjugate, then [Z*/Z] is equal to:
b) +j
c) 1+j
b) V1 x V2 = V2 x V1 c) V1 • V2 = -V2 • V1
b) 2i
b) 1 + j/4 - k
d)
2
d) V1 x V2 = -V2 x V1
c) -1
d) none of these
c) 1/j
d) undefined
For the system of equations AX=Y where A is (m x n) , if m < n, the set can be solved for m
unknowns in terms of (m-n) unknowns if and only if:
b) rank of A = rank of [A|Y]
c) rank of A ¹ rank of [A|Y]
d) det [A]= m
A is a (4x4) matrix with a non-zero determinant. Matrix A is
(a) Singular (b) Non-Singular
11.
c) - j 3
3
If V1 = i + j + k , V 2 = i - j - k and V3 = 4 j , then. V1 ÷ (V1 x V2) =?
a) det [A]= n
10.
p
If V1 = i + j + k and V 2 = i - j - k , then V1 • V 2 = ?
a) –4
9.
(d) none of these
If V1 and V2 are three-dimensional vectors then:
a) –2j- 2k
8.
(c) Conjugate of [Z*]2
b)
a) V1 • V2 = V2 • V1
7.
(d) j4
j
a) –j
6.
(c) 4 + j4
If Z a complex number Z = 3e 2 , then the complex conjugate of Z is:
a)
5.
(d) 2 + j2
If Z is a complex number and Z* is its conjugate, then Z2 is equal to
(a) Z.Z*
4.
(c) -1 + j
If Z is a complex number 2 + j2 and Z* is its conjugate, then Z.Z* is
(a) 2 + j2
3.
(b) 1 - j
(c) Zero matrix (d) Symmetric
If A is a 3x4 matrix and B is a 3x5 matrix, then
(a) AB = BA (b) A + B = B + A (c) B T A = AB T (d) A T B is 4x5 matrix
Ó 2020
57
Dr. Mohammed Safiuddin
12.
For the system of equations A X = Y, A is a (4 x 4) matrix with a rank of 3 and [ A|Y ] matrix
has a rank of 4. This system of equation is
(a) consistent
13.
(b) Inconsistent
(c) Non-linear (d) None of these
For the system of equations AX = Y, A is a 3 x 3 matrix with a rank of 2 and [A|Y] also has a
rank of 2. The system of equations is
(a) Non-Linear
(b) Linearly Independent. (c) Inconsistent
(d) Consistent and Linearly Dependent
14.
Matrix A is m x q and B is n x q, then if m ¹ q ¹ n
b) [AB]T=BTAT
a) A+B = B+A
15.
c) A-1B = B-1A
d) ABT is an (m x n) matrix
In Laplace transform terminology, the expression 1/ S 2 represents
(a) Step Function (b) Ramp Function (c) Parabolic Function (d) None of these
16.
The Final Value Theorem of Laplace transform states that if L {f(t)}=F(s), then the final value of
f(t) as t ® ¥ is given by
(a) lim sF ( s )
(b) lim F ( s )
s®0
17.
b
2
s +b
2
_a_
s +a
b)
s +a
2
(s + a ) + b
(b)
s _
s +a
The inverse Laplace transform of
(a) 3 e -3t sin(t )
2
20.
s ®¥
2
b
c)
2
(s + a ) + b
d)
2
s +a
2
s +b2
The Laplace transform of 1-e-at is
(a)
19.
(d) lim F ( s )
s ®¥
The Laplace transform of e-a t cos b t is
a)
18.
(c) lim sF ( s )
s®0
Ó 2020
(d) None of the above
( s 2 + 5)
is given by
( s 2 - 1)( s + 3)
(b) 3 e -t sin(3t ) + 7 e 3t
4
The inverse Laplace transform of
a) - 14 e -5t + 14 t + 1 t 2
25
25 5
14
14
1
c)
e 5t +
+ t
25
25 5
(c) ___ a __
s(s + a)
(c) - 3 e -t + 3 e t + 7 e -3t
2
4
4
4
3S + 1 is given by:
S 2 ( S + 5)
b) - 14 e -5t + 14 + 1 t
25
25
5
d) - 14 e -5t + 14 + 1 e t
25
25 5
58
Dr. Mohammed Safiuddin
21.
a) 5e-4t- 3e-7t
22.
2s + 23
is
( s + 4)( s + 7)
b) 5e4t- 3e-7t c) 5e-4t- 3e7t
The inverse transform of
d) 5e4t- 3e7t
The following system transfer function represents which differential equation
C(s)
(2S + 1)
= 3
R(s) ( S + 3 S 2 + 4S + 1)
a.) !c!! + 3 c!! + 4 c! + c = 2 r! + r b.) !c!! - 3 c!! - 4 c! - c = 2 r! + r
c.) !c!! + 3 c!! + 4 c! + c = - 2 r! - r d.) Cannot be determined from information given.
Ó 2020
59
Dr. Mohammed Safiuddin
Reference Books & Lecture Notes:
1. John M. H. Olmsted; “Real Variables-An Introduction to the Theory of Functions”; AppletonCentury-Crofts, Inc. Publication; Library of Congress Card: 60-5076; 1959
2. Murray R. Spiegel; “Theory and Problems of Advanced Calculus”; Schaum’s Outline Series;
McGraw-Hill Book Company; ISBN 07-060229-8; December 1962.
3. Herbert S. Wilf; “Mathematics for the Physical Sciences”; John Wiley & Sons, Inc.; Library of
Congress Card: 62-15193; 1962
4. Robert C. Weast; Editor-in-Chief; “Standard Mathematical Tables”; Thirteenth Edition; CRC
Publications; Library of Congress Card: 30-4052; 1964
5. Darl C. Washburn Jr.; “Review of Mathematics”; Lecture Notes-Seminar on Feedback Control &
Thyristor Drive Systems; Westinghouse Electric, Industrial Systems Division-Buffalo, NY;
unpublished; 1967
6. Robert W. Hornbeck; “Numerical Methods”; Prentice-Hall/Quantum Publication; ISBN: 0-13626614-2; 1975
7. Michael O’Nan; “Linear Algebra”; Harcourt Brace Jovanovich, Inc. Publication; ISBN:0-15518560-8; 1976
8. Sabbir S. Moochalla; “Solutions to Sample PE Exam Problems”; PE License Review Course;
Graduate Student, EE Department, University at Buffalo; Unpublished; 1976
9. William E. Boyce & Richard C. DiPrima; “Elementary Differential Equations & Boundary Value
Problems”; John Wiley & Sons Publication; Third Edition; ISBN: 0-471-09334-3; 1977
10. Michael D. Greenberg; “Advanced Engineering Mathematics”; Prentice-Hall Publication; Second
Edition; ISBN: 0-13-321431-1; 1998
11. Mohammed Safiuddin; “Industrial Control Systems”; Lecture Notes- Course EE 419/519; Elec.
Eng’g. Dept.; School of Eng’g. & Applied Sciences; Univ. at Buffalo; Fall Semester 2009
Ó 2020
60
Dr. Mohammed Safiuddin
Dr. Mohammed Safiuddin is Research Professor Emeritus in the
Electrical Engineering Department at
University at Buffalo, and President of
STS International, Plainfield, IL, USA.
He started as Junior Engineer, Andhra
Pradesh State Electricity Board (India) in
1958.
Joined
Systems
Control
Department, Westinghouse Electric,
Buffalo, New York in June 1960 as
Associate Engineer. Progressed through
the ranks of Engineer, Senior Engineer,
Fellow Engineer positions to Manager,
Product/Strategic
Planning
(1982),
Power Electronics and Drive Systems
Division; Technical Advisor, Marketing Department (1984). He was parttime instructor in the early sixties, Adjunct Associate Professor (‘77- ‘91),
Research Professor (’91- ’10), at the University at Buffalo, USA. His areas
of technical interests cover industrial control systems, renewable energy,
Smart Grid power systems, and Technology Management. Has been
awarded 10 patents in this field and has dozens of published papers and
conference presentations. He is Life Fellow of the IEEE, was awarded
Roscoe Allen Gold Medal (1957) by Osmania University, was recipient of
"IUSD Award of Merit" (1992) by the IEEE-Industry Applications Society
for contributions to industrial control technologies, and service to the IAS,
was recognized for meritorious achievement in continuing education by
the IEEE-EAB award for the year 2000. He is a member of the Pi Mu
Epsilon (Mathematics), Beta Gamma Sigma (Business) and an "Eminent
Engineer" member of Tau Beta Pi. He was recognized with International
Business and Academic Excellence 2019 “Distinguished Educator
Award” in 2019 by GISR Foundation and American College in Dubai.
University at Buffalo honored him with the “2020 Engineer of the Year”
award.
Ó 2020
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