Transfer Functions RLC Circuits - Part of Part 3.
Resource: Solutions & Problems of Control Systems, 2nd ed - AK Jairath.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Source of study material: Electric Circuits 6th Ed., Nahvi & Edminister. Engineering Circuit Analysis, Hyatt & Kimmerly 4th Ed. McGrawHill.
Karl S. Bogha.
Solved Problems In Transfer Functions of RLC circuits.
Resource: Solutions & Problems of Control Systems, 2nd ed - AK Jairath.
Level: Intermediate.
Apologies for any errors and omissions.
August 2020.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
I selected AK Jairath textbook because it goes back to 1992, when this engineer first
published this book. 2nd edition in 1994, and reprinted in 1996.
Solutions & Problems in Control System. May not be in circulation now. I ts a small
book. Concise similar to Schaums (Supplementary), its not a main textbook. Chapter
1 is Transfer Functions. All the problems in chapter 1 are are made up of R L C
components. So this was in line with my/our starting plan to stay within the electric
circuits corridor. First keep things simple. So if you asked why, thats the reason I
selected this chapter. We did some theory-examples in transfer function at end of
Part B, so its best to do them first since these are fresh in minds.
Got an oppurtunity to work with RLC components in the transfer function and
secondly control systems context, why waste it. So I did these few example problems.
AK Jairath: The transfer function of a system is the ratio of Lapalce transforms of
the output and input quantities, initial conditions being zero. When a physical system
is analysed, a mathematical model is prepared by writing differential equations with
the help of various laws. An equation describing a physical system has integrals and
differentials. The response can be obtained by solving such equations.
The steps involved in obtaining the transfer function are:
1. Write differential equations of the system.
d
2. Replace terms involvingƀby s and ϣ
Ϥ dt by 1/ s. < - - - Applies to L & C.
dt
L and C from RLC was worked in
electric circuits.
See notes bottom next page.
3. Eliminate all but the desired variable.
See figure next page.
v (t ) = e
st
OR
і di љ
L їƀњ L : sL
ј dt ћ
1
ƀϣ
Ϥ i (t ) dt
C
i (t )
= e
st
di
ƀ: I ( s))
dt
1
1
ƀ: ƀƀϣ
Ϥ i (t ) dt
C
sC
st
st
st
d іјe љћ
1 ϣ st
1
L ƀƀ= sL Έ e
Έe
ƀϤ e d t = ƀƀ
C
sC
dt
^
l Here* .
^
l Here* .
Inductor current derivative of i(t) - time domain.
Its equivalent frequency domain: I(s).
: I ( s)) Capacitor current integral over a
limit t - time domain. Its equivalent
frequency domain: I(s).
* Figure and notes below for reference.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Where the sL and 1/ sC came from?
v (t ) = Vme
s =
v (t ) =
=
=
Inductor:
v (t )
K ļ st
Re іјV m e e љћ
st
Re іјV m e љћ
=
st
cos(( ō U+ ļ))
i (t )
= I me
Ň+ K ō
ŇU
V m e cos(( ō U+ ļ))
st
Re іјV m e љћ =
Vme
ŇU
cos(( ō U+ Ř))
s =
i (t ) =
Taking the real part of v(t)/ i(t)
See notes in Part 3 A and B.
=
=
st
d іјI m e љћ
st
L ƀƀƀ
= sL I m e
dt
sL I m e
ŇU
st
st
=
sL I m e
Vm
=
sL I m
V
=
sL Έ I
V ( s))
=
sL Έ I ( s))
< ----
Capacitor:
Ň+ K ō
ŇU
I m e cos(( ō U+ Ř))
K ļ st
Re іјI m e e љћ
st
Re іјI m e љћ
st
1 ϣ
= ƀϤ I m e d t
C
st
st
1
Re іјV m e љћ =
I me
ƀƀ
sC
st
st
1
Vme
=
I me
ƀƀ
sC
1
Vm
=
Im
ƀƀ
sC
1
I
V
=
ƀƀ
sC
1
V ( s))
=
I ( s))
ƀƀ
sC
v (t )
st
1
= ƀƀ
I me
sC
< ----
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Derivative and Integral substitues for s and 1/s for the component L and C respectively.
Inductor:
d
ƀ---> s
dt
di
vL ( t ) = L Έ ƀ
dt
---> V L ( s)) = Ls Έ I ( s)) < ---
1
ϣ . d t ---> ƀ
Ϥ
s
1 ϣ
Capacitor: v C ( t ) = ƀϤ i d t
C
1
---> V C ( s)) = ƀƀ
Έ I ( s)) < --sC
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1-1:
Derive the transfer function
of the circuit shown in
figure to the left.
Solution:
First thing is its a series circuit. We do a voltage conservation. Meaning
the sum of voltages add to zero. You call that Kickoff's OR Kickout's Law.
The output is across the capacitor terminals.
The input is supply voltage for the resistor and capacitor.
v_i ( t )
Set
=
v_o ( t )
v_i ( t )
=
R Έ i ( t ) + v_C ( t )
=
v_C ( t ) =
i(t) is the circuit's current.
1
ƀϣ
Ϥ i dt
C
R Έ i ( t ) + v_o(( t )
Now we convert the expression above to the s-domain.
Which in control systems textbook they say 'Taking the Laplace transform'.
Laplace Transforms starts with transfer functions in the s-plane or in terms of
complex frequency. So, thats why we used a Controls textbook. Same.
V i ( s))
=
RI ( s)) + V 0 ( s))
Vo(s) is that voltage across the capacitor C terminals, which we can
set this in the s-domain of the capacitor.
V o ( s))
=
1
Έ I ( s))
ƀƀ
sC
< - - - C: 1/ sC, and i(t): I(s).
Its more than forming a loop equation, we want to all the required variables
in the expression so we can form that Vo(s)/Vi(s).
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
How do we know what all terms and their forms we need before we can
get to forming a transfer function?
Keep working in more and more example problems, partially looks like a
guess, but after a few examples we get the general idea.
The Electrical Engineering expressions for defining components are formed in
such a way that they have a future in advanced math where they can be
manipulated in various ways to take full benefit of the math resulting in some
output that serves a circuit's purpose - Karl Bogha.
V o ( s))
=
1
Έ I ( s))
ƀƀ
sC
I ( s))
=
V o ( s)) Έ sC
V i ( s))
=
RI ( s)) + V 0 ( s))
V i ( s))
=
R іјV o ( s)) Έ sCљћ + V o ( s))
=
sRC іјV o ( s)) љћ + V o ( s))
=
V o ( s)) Έ ( sRC + 1))
=
V o ( s)) Έ ( 1 + sRC))
V i ( s))
V o ( s))
ƀƀ
V i ( s)
=
< --- Lets plug in or if you prefer substitute the
expression we got into this expression we
formed earlier.
< --- How would we had known that?
Surely had to work examples.
Keep clear of people and peers who say dont do
the example go to the end of chapter problems,
they lie so they have the edge - Engineer.
1
ƀƀƀ Answer.
1 + sRC
I n the work place you never ever get problems
to solve like hard end of chapter problems in
hard core engineering textbooks, fake, it rarely
help, most time you got all the time in the world Karl Bogha.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1-2:
We seek the transfer
function I (s)/ Vi(s) ?
Solution:
First thing is its a series circuit. We do a voltage conservation, meaning the
sum of voltages add to zero. You call that Kickoff's Law!
v_i ( t )
=
R Έ i ( t ) + v_C ( t )
i(t) is the circuit's current.
v_C ( t ) =
v_i ( t )
=
V i ( s))
=
=
=
I ( s))
ƀƀ
V i ( s)
=
I ( s))
ƀƀ
V i ( s)
=
I ( s))
ƀƀ
V i ( s)
=
1
ƀϣ
Ϥ i dt
C
1
= ƀƀ
I ( s))
sC
R Έ i ( t ) + v_C ( t )
1
RI ( s)) + ƀƀ
I ( s))
sC
1
RI ( s)) + ƀƀ
I ( s))
sC
і
1 љ
I ( s)) їR + ƀƀ
sC њћ
ј
1
ƀƀƀ
і
1 љ
R
+
ƀƀ
їј
sC њћ
Simplify this term, multiply by sC/R.
і sC љ
їјƀƀƀ
1 + sCR њћ
Answer.
і sC љ
і sC љ
їјƀƀ
њ
їјƀƀ
і sC љ
љ sC
і
1
R ћ
R њћ
ƀƀƀƀƀ = ƀƀƀ = їƀƀƀ
ƀƀ = їƀƀƀ
њ
R
sC і
1 љ
1
1
ј sCR + 1 њћ
R
+
sC
+
sC
+
ƀƀ
ƀƀ
ƀ
ƀ
ї
њ
R їј
sC њћ
R
Rћ
ј
Good if we can work the final form of expression like
this instead of the one a few steps before. I t takes
some extra effort to get it in a neat form that is more
electric circuit friendly and meaningful.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1-3:
We seek the transfer
function Vo(s)/ Vi(s) ?
Solution:
Conservation of voltage means something else?
I am not sure, when conserved it would remain the same.
So the sum equal zero in a loop. That for me is conserved.
Maybe they used it for something else. Usuall I am not the first.
We kickoff with the voltage conservation.
Vi (t )
V i ( s))
=
=
1
Ri ( t ) + ƀϣ
Ϥ i (t ) dt
C
1
ƀϣ
Ϥ i (t ) dt
C
1
= ƀƀ
Έ I ( s))
sC
I ( s))
RI ( s)) + ƀƀ
sC
Our circuit identifies voltage across resistor terminals as Vo(t)
which now becomes? Vo(s) for the frequency domain.
V o ( s))
=
RI ( s))
I ( s))
=
V o ( s))
ƀƀ
R
V i ( s))
=
і V o ( s)) љ і V o ( s)) љ 1
R їƀƀ
њ + їƀƀњ Έ ƀƀ Isolate Vo(s)
ј R ћ ј R ћ sC
V i ( s))
=
і
1 љ
V o ( s)) Έ ї1 + ƀƀ
sCR њћ
ј
Substitute in here: V i ( s))
=
I ( s))
RI ( s)) + ƀƀ
sC
Next for the required transfer function:
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V o ( s))
ƀƀ
V i ( s)
1
= ƀƀƀƀ
і
1 љ
їј1 + ƀƀ
sCR њћ
< --- This can be simplified.
Its awkward, that is why we simplify
these awkward terms.
Multiply by sCR:
V o ( s))
ƀƀ
V i ( s)
1
sCR
sCR
= ƀƀƀƀ
Έ ƀƀ = ƀƀƀƀ Answer.
і
( sCR + 1)
1 љ sCR
їј1 + ƀƀ
њ
sCR ћ
Chp 1 Problem 1-4:
We seek the transfer function,
Vo(s)/Vi(s), of the electrical
network shown to the left in
phase lead form ?
Solution:
Z1 is the parallell of C and R1:
1
ƀƀ=
Z1
1
1
1
1
+ ƀƀ = ƀƀ
+ ƀƀ s = Ň + K ō
ƀƀ
R1
R1
1
1
ƀƀ
ƀƀ Ň = 0
K ō$
sC
s = K ō
1
ƀƀ=
Z1
1
+ sC
ƀƀ
R1
1
ƀƀ=
Z1
R1 sCR1
+ ƀƀ =
ƀƀ
R1
1
R1 + sCR1
ƀƀƀƀ =
1 Έ R1
R1 + sCR1
ƀƀƀƀ
R1
1
ƀƀ=
Z1
R1 sCR1
+ ƀƀ =
ƀƀ
R1
R1
sCR1
1 + ƀƀ
R1
1 + sCR1
ƀƀƀ
R1
=
We are concerned with
frequency, so we can set
sigma = 0.
1
sC
+ ƀƀmultiply by R1
ƀƀ
R1
1
=
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Z1
R1
ƀƀƀ
1 + sCR1
=
After inverting.
We kickoff with the voltage conservation.
vi ( t )
=
Z1I ( s)) + R2I ( s))
vo ( t )
=
R2I ( s))
Taking the Laplace Transform of the above 2 equation:
V i ( s))
=
Z1I ( s)) + R2I ( s))
Vo (t )
=
R2I ( s))
I ( s))
Plug in equation above
=
V o ( s))
ƀƀ Plug in equation above
R2
V i ( s))
=
і V o ( s)) љ
Z1 їƀƀ
њ + V o ( s))
ј R2 ћ
V i ( s))
=
љ
і Z1
V o ( s)) Έ їƀƀ
+ 1њ
ј R2
ћ
V i ( s))
=
=
Plug in Z1
љ
іі
љ
R1
ї їјƀƀƀ
њ
њ
1 + sCR1 ћ
V o ( s)) Έ їƀƀƀƀ+ 1њ
R2
ј
ћ
іі
љ
љ
R1
ї їјƀƀƀ
њ
1 + sCR1 ћ R2 њ
V o ( s)) Έ їƀƀƀƀ+ ƀƀ
њ
R2
R2 ћ
ј
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, I ndia, Pakistan, UK, and other Common Wealth Count ry engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
=
V i ( s))
љ
іі
љ
R1
R2
+
ƀƀƀ
ї їј 1 + sCR1 њћ
њ Next rearrange and 1 + sCR1
V o ( s)) Έ їƀƀƀƀƀƀ
ƀƀƀ
њ multiply by --->
R2
ј
ћ
1 + sCR1
=
V o ( s)) іі
љ R2 Έ ( 1 + sCR1)) љ
R1
ƀƀΈ їїƀƀƀњ + ƀƀƀƀƀњ
( 1 + sCR1) ћ
R2 јј 1 + sCR1 ћ
=
V o ( s)) і R1 + R2 + sCR1R2 љ
ƀƀΈ їƀƀƀƀƀƀњ
R2 ј
1 + sCR1
ћ
=
і R1 + R2 љ і 1 + sCR1R2 љ
V o ( s)) їƀƀƀ
ƀƀƀƀ
ј R2 њћ їј 1 + sCR1 њћ
=
V i ( s))
ƀƀ
V o ( s)
=
V o ( s))
ƀƀ
V i ( s)
=
V o ( s))
ƀƀ
V i ( s)
=
...not finished yet in this expression.
1
ƀƀƀ
і
sCR1R2 љ
1 + ƀƀƀњ Place R1 + R2
in there so it cancels
і R1 + R2 љ ї
R1 + R2
V o ( s)) їƀƀƀ
the
middle
term
(R1+
R2)/ R2
ƀƀƀƀƀ
ї
њ
ј R2 њћ ј 1 + sCR1 ћ when multiplied.
і
sCR1R2 љ
1 + ƀƀƀњ
ї
і R1 + R2 љ
R1 + R2
їƀƀƀƀƀ
њ Next invert both sides.
їјƀƀƀ
њ
R2 ћ ј 1 + sCR1 ћ
і R2 љ і 1 + sCR1 љ
As provided in textbook.
ƀƀƀƀƀ
їјƀƀƀ
R1 + R2 њћ ї
sCR1R2 њ
ї 1 + ƀƀƀњ
R1 + R2 ћ
ј
љ
і R2 љ і
1 + sCR1
ƀƀƀƀƀƀƀ
їјƀƀƀ
њ
ї
њ
і R2 љ
R1 + R2 ћ
sCR1
ї 1 + їƀƀƀ
њ
ј
ј R1 + R2 њћ
ћ
Transfer function.
We can simplify a little. Make RC the time constant in a series circuit = tau,
and make the constant R2/ (R1+ R1) = a. OR just any constant T.
T
=
CR1
a
=
R2
ƀƀƀ
R1 + R2
V o ( s))
ƀƀ
V i ( s)
=
і 1 + sT љ
a їƀƀƀ
ј 1 + asT њћ
V o ( s))
ƀƀ
V i ( s)
=
a Έ ( 1 + sT))
ƀƀƀƀ Answer.
( 1 + B T ň) 5
Comment: Previous example problems used T
for RC in the final transfer functions.
I left it out because my aim was the approach
on how to get the transfer functions. T is not
necessarily a time constant for this circuit.
You can verify. We could use P of Q but since
its RLC, T or tau makes more sense.
Took time with the algebra otherwise
a good easy example for most.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, I ndia, Pakistan, UK, and other Common Wealth Count ry engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chap 1 Problem 1.7 :
I jump to problem 1.7 because its the same circuit. This provides a continuity and
not having to return later after several problems.
Derive the transfer function of the circuit shown (same circuit of problem 1.4).
I f v_i(t) = 8 sin(10t) V, R1 = 50 k Ohms, R2 = 5 k Ohms and C = 1 uF.
Calculate the output voltage in magnitude and phase angle relative to input voltage?
Solution:
Gain
G ( s)) =
k ̗ 10
R1 ̗ 50 k
3
V o ( s))
ƀƀ
V i ( s)
M ̗ 10
R2 ̗ 5 k
6
=
і R2 љ і
љ
1 + sCR1
ƀƀƀƀƀƀƀ
їјƀƀƀ
њ
ї
њ
і R2 љ
R1 + R2 ћ
sCR1 њ
ї 1 + їƀƀƀ
ј
ј R1 + R2 њћ
ћ
˕6
u ̗ 10
C̗1 u
Substitute into transfer function:
G ( s)) =
V o ( s))
ƀƀ
V i ( s)
G ( s))
=
( 1 + sCR1))
R2 Έ ƀƀƀƀƀƀƀ
( R1 + R2) + ( sCR1R2)
=
і 1 + 0.05 s љ
5000 їƀƀƀƀƀ
Divide numerator and
ј 55000 + 250 s њћ denominator by 55,000.
=
і 1 + 0.05 s љ
0.091 їƀƀƀƀњ
ј 1 + 0.0045 s ћ
=
( 1 + 0.05 s))
0.01 ƀƀƀƀƀ Constant 0.091 rounded off to 0.01
( 1 + 0.0045 s)
Zero:
( 1 + 0.05 s))
( 1 + 0.0045 s))
Pole:
We are interested in s = 0 + jw, where sigma = 0.
s =
Hence we can analyse the frequency response.
Substitute s for jw in transfer function.
s =
Now we have 1+ 0.05s and 1+ 0.0045s, this gives us
the magnitude and angle for both. Since we have a
real and imaginary part.
) =
G (K ō
Ň+ K ō
Ň =
0
0+ K ō =
0
( 1 + 0.05 K ō
)
0.01 ƀƀƀƀƀ
( 1 + 0.0045 K ō
)
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, I ndia, Pakistan, UK, and other Common Wealth Count ry engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Before we can calculate the angles we need the value of w ?
v (t )
8 si n ( 10 Έ t )
=
ō =
Zero:
Pole:
---> Asi n ( ō U)
10
( 1 + 0.05 j 10))
( 1 + 0.0045 j 10))
=
=
1 + 0.5j
1 + 0.045j
і 0.5 љ
Z_Ang_G_s ̗ at an їƀƀ
= 26.5651 deg
ј 1 њћ
і 0.045 љ
P_Ang_G_s ̗ at an їƀƀњ = 2.5766 deg
ј 1 ћ
Ang_G ( s)) =
26.565 ˕ 2.577 = 23.988
degrees. Answer.
Now for the magnitude of the transfer function,
here is where the constant 0.1 is applied.
Magnitude of zero:
2
2
ƠƠƠƠƠƠƠ
1 + 0.5 = 1.118
Magnitude of pole:
2
2
ƠƠƠƠƠƠƠƠƠ
1 + 0.045 = 1.001
Magnitude of G(s):
і 1.118 љ
( 0.1)) Έ ƀƀ = 0.1117
їј 1.001 њћ
The input signal is vi(t) = 8 sin (wt)
From which we can obtain the amplitude is 8 V maximum.
We next multiply the magnitude of G(s) to 8V for the maximum output voltage.
Am pli t ude ̗ 8.0
M ag_G ( s)) ̗ 0.1117
V o ̗ Am pl i t ude Έ M ag_G ( s)) = 0.894
V. Answer.
Good example. Can be found in most
circuits and all controls textbook.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, I ndia, Pakistan, UK, and other Common Wealth Count ry engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chap 1 Problem 1.8 :
Problem 1.8 is next here because it works on the same transfer function of problem 1.4.
This is indicated in the problem statement, exact same circuit.
If C = 1uF in the circuit of problem 1.4.
What values of R1 and R2 will give T = 0.6 sec, and a = 0.1
Solution:
љ
і R2 љ і
1 + sCR1
G ( s)) = їƀƀƀ
ƀƀƀƀƀƀƀ
њ
ї
њ T = CR1
і R2 љ
ј R1 + R2 ћ
sCR1 њ
ї 1 + їƀƀƀ
ј
ј R1 + R2 њћ
ћ
a Έ ( 1 + sT))
G ( s)) = ƀƀƀƀ
( 1 + asT )
C̗1 u F
CR1 = 0.6, solve for R1:
CR1 = 0.6
T ̗ 0.6
R2
a = ƀƀƀ
R1 + R2
a ̗ 0.1
( 1 uF)) R1 = 0.6
R1
R2
a = ƀƀƀ --->
R1 + R2
=
5
0.6
= 6 Έ 10 Ohm. = 0.6 Έ M
ƀƀ
1Έu
R2
0.1= ƀƀƀƀ --->
600000 + R2
60000 + 0.1 R2
= R2
0.9 R2
=
60000
R2
=
60000
= 66666.7
ƀƀƀ
0.9
=
Ohm. Answer.
0.1 ( 600000 + R2))
= R2
0.066 M Ohms. Answer.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1-5:
We seek the transfer
function, Vo(s)/Vi(s),
of the electrical
network shown to
the left in phase lead
form ?
Solution:
Current at node:
i (t )
=
i 1 (t ) + i 2 ( t )
Voltage conservation in loop at left side:
vi ( t )
=
di
L ƀ+ Ri 1 ( t )
dt
Next, in a cleaver way, we pull in the v_o(t) relationship thru the capacitor
voltage, where C is voltage across resistor R, and we know v_o(t) is the
voltage across the capacitor.
vo ( t )
=
Ri 1 ( t )
1
= ƀϣ
Ϥ i 2 (t ) dt
C
vi ( t ) =
di
L ƀ+ Ri 1 ( t )
dt
V i ( s)) =
sL I ( s)) + RI 1 ( s))
RI 1 ( s)) =
1
1
I 2 ( s))
ƀϣ
Ϥ i 2 ( t ) d t = ƀƀ
C
sC
RI 1 ( s)) =
V o ( s))
=
I 2 ( s))
ƀƀ
sC
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Voltage across R:
V o ( s))
I 1 ( s)) = ƀƀ
R
I 2 ( s))
V o ( s)) = ƀƀ = RI 1 ( s)) thus
sC
V i ( s)) =
We update our I(s)expression here
sL I ( s)) + RI 1 ( s))
i (t )
Substitute voltage across C for R:
=
i 1 (t ) + i 2 ( t )
I ( s)) =
I 1 ( s)) + I 2 ( s))
V i ( s)) =
sL ( I 1 ( s)) + I 2 ( s)) ) + RI 1 ( s))
V i ( s)) =
sL ( I 1 ( s)) + I 2 ( s)) ) + V o ( s))
V o ( s)) =
I 2 ( s))
ƀƀ
sC
sCV o ( s)) =
I 2 ( s))
љ
і V o ( s))
sL їƀƀ
+ sCV o ( s)) њ + V o ( s))
ј R
ћ
V i ( s))
=
V i ( s))
=
љ
і V o ( s))
V o ( s)) + sL їƀƀ
+ sCV o ( s)) њ
ј R
ћ
V i ( s))
=
љ
і sL
V o ( s)) + V o ( s)) Έ їƀ + sCsL њ
јR
ћ
V i ( s))
=
љ
і
2
sL
V o ( s)) Έ ї1 + ƀ + s LC њ
R
ј
ћ
V o ( s))
ƀƀ
V i ( s)
=
V o ( s))
ƀƀ
V i ( s)
=
1
ƀƀƀƀƀƀ
і
љ
2
sL
+ s LC њ
їј1 + ƀ
R
ћ
1
ƀƀƀƀƀƀ
і 2
љ
sL
+ 1њ
їјs LC + ƀ
R
ћ
Answer. Lots of substitutions.
A compact answer below.
The Engineer makes the expression simpler in appearance, quadratic expression,
thru the use of variable T1 and T2. T1 = L/R maybe a time contant but not here.
T2 = CR which is NOT a time constant, you verify should it be of concern.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
T1
L
ƀ
R
=
T1T 2
=
V o ( s))
ƀƀ
V i ( s)
=
T2
іL љ
( CR)) =
їјƀ
R њћ
CR
=
LC
1
ƀƀƀƀƀƀ
2
іјT 1T 2s + T 1s + 1љћ
Answer.
Chp 1 Problem 1-6:
We seek the transfer function, Vo(s)/ Vi(s), of the electrical network shown above ?
Solution:
Set up the impedance Z for each component:
Z1
=
R1
Z2
=
1
ƀƀ
sC1
Z3
=
R2
Z4
=
1
ƀƀ
sC2
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Voltage mesh/loop equations in Laplace:
Left loop:
V i ( s))
=
Z1I 1 ( s)) + Z2 ( I 1 ˕ I 2))
V i ( s))
=
I 1 ( s)) ( Z1 + Z2)) ˕ Z2I 2
...Eq 1
Right loop:
0
=
Z2 ( I 2 ˕ I 1)) + Z3I 2 ( s)) + Z4I 2 ( s))
0
=
˕Z2I 1 + I 2 ( s)) ( Z2 + Z3 + Z4))
...Eq 2
Next we form an expression for Vo:
V o ( s))
Z4I 2 ( s))
=
...Eq 3
If I am correct, from these few examples we seen, we want to place
one expression for current, into the the other equation, then work
towards the transfer function, provided we have Vo(s) and Vi(s) in
that expression to work with.
Here, I1(s) looks the better simpler choice to place in Eq 2.
Because we do not have a voltage source on the RHS.
Then we set Vo(s) for Z4I2(s).
Then work with the equation which can fit-in Vo, Vi, and I1 and I2 in it.
If we dont have it yet continue re-hashing.
What you think, that's the plan? Of course!
=
˕Z2I 1 + I 2 ( s)) ( Z2 + Z3 + Z4))
I 1 ( s)) Z2
=
I 2 ( s)) ( Z2 + Z3 + Z4))
I 1 ( s))
=
I 2 ( s)) ( Z2 + Z3 + Z4))
ƀƀƀƀƀƀƀ
Z2
V i ( s))
=
I 1 ( s)) ( Z1 + Z2)) ˕ Z2I 2
V i ( s))
=
I 2 ( s)) ( Z2 + Z3 + Z4)) Έ ( Z1 + Z2))
ƀƀƀƀƀƀƀƀƀƀ˕ Z2I 2 ( s)) Fix for Z2 at very right
Z2
of numerator.
V i ( s))
=
I 2 ( s)) Έ ( ( Z2 + Z3 + Z4)) Έ ( Z1 + Z2)) ) Z2 Z2Z2I 2 ( s))
˕ ƀƀƀƀ
ƀƀƀƀƀƀƀƀƀƀƀƀ
Z2
Z2
0
...Eq 2
...Eq 1, substitute I1(s)
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V i ( s))
=
V i ( s))
ƀƀ
I 2 ( s)
=
2
I 2 ( s)) Έ іј( ( Z2 + Z3 + Z4)) Έ ( Z1 + Z2)) ) Z2 ˕ Z2 љћ
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
Z2
( ( Z2 + Z3 + Z4)) Έ ( Z1 + Z2)) ) Z2 ˕ Z2
ƀƀƀƀƀƀƀƀƀƀƀƀ
Z2
2
For me this is new, not a twist but certainly new I dont remember doing a
substitution on the LHS! Ok Not typical. Hope I am gaining skills here.
V o ( s))
=
I 2 ( s))
=
Z4I 2 ( s))
...Eq 3
V o ( s))
ƀƀ
Z4
Substitute this in the expression Vi(s)/I2(s)
V i ( s))
=
ƀƀƀ
і V o ( s) љ
їƀƀњ
ј Z4 ћ
V i ( s))
ƀƀ
V o ( s)
=
іј( Z2 + Z3 + Z4)) Έ ( Z1 + Z2)) ˕ Z2 2 љћ
ƀƀƀƀƀƀƀƀƀƀƀ
Z2
іј( Z1 + Z2)) Έ ( Z2 + Z3 + Z4)) ˕ Z2 2 љћ
ƀƀƀƀƀƀƀƀƀƀƀ
Z2 Έ Z4
Invert the expression so we get Vo(s) in the numerator.
V o ( s))
ƀƀ
V i ( s)
=
Z2 Έ Z4
ƀƀƀƀƀƀƀƀƀƀƀ
іј( Z1 + Z2) Έ ( Z2 + Z3 + Z4) ˕ Z2 2 љћ
Lets expand the denominator expression:
( Z1 + Z2)) Έ ( Z2 + Z3 + Z4))
=
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z2 + Z2Z3 + Z2Z4
Now for the full denominator expression:
V o ( s))
ƀƀ
V i ( s)
=
=
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z2 + Z2Z3 + Z2Z4 ˕ Z2Z2
=
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z3 + Z2Z4
Z2 Έ Z4
ƀƀƀƀƀƀƀƀƀƀƀƀ The transfer function.
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z3 + Z2Z4
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Next we substitute the values of impedances Z1...Z4:
V o ( s))
ƀƀ
V i ( s)
Z1
V o ( s))
ƀƀ
V i ( s)
=
Z2 Έ Z4
ƀƀƀƀƀƀƀƀƀƀƀƀ
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z3 + Z2Z4
= R1
=
Z2
1
= ƀƀ
sC1
Z3
= R2
Z4
1
= ƀƀ
sC2
1
1
Έ ƀƀ
ƀƀ
sC1 sC2
ƀƀƀƀƀƀƀƀƀƀƀƀ
R1
R1
R2
1
+ R1R2 + ƀƀ
+ ƀƀ
+ ƀƀƀ
ƀƀ
2
sC1
sC2 sC1 s C1C2
As usual these types expressions are made simpler, especially in electric circuits.
It helps in building the physical circuit. Which I almost forgot the true purpose here.
We are building circuits and components are to be put together on a bread board for
testing. Hello?..true purpose? Why not?
=
=
=
=
V o ( s))
ƀƀ
V i ( s)
=
і
љ 1
1
1
Έ ƀƀ
Έ ƀƀ
їƀƀƀƀƀƀƀƀƀƀƀƀ
њ
sC1 sC2
R1
R1
R2
1
+ R1R2 + ƀƀ
+ ƀƀ
+ ƀƀƀњ
ї ƀƀ
2
sC2 sC1 s C1C2 ћ
ј sC1
1
Multiplied by sC1 sC2
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
2
R1sC2 + R1R2 Έ s C1C2 + R1sC1 + R2sC2 + 1 top and bottom.
1
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
2
sR1C2 + s R1R2 Έ C1C2 + sR1C1 + sR2C2 + 1
1
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
2
sR1C2 + sR1C1 + sR2C2 + 1 + s R1R2 Έ C1C2
1
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
2
1 + s ( R1C2 + R1C1 + R2C2) + s R1R2 Έ C1C2
Answer.
The denominator is a neat 2nd order expression.
The circuit is also a practical circuit for application in electric
circuits, electronics and other electrical/electronic applications.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1.11:
Problem 1.11 comes here because this problem has a similar transfer function to
problem 1.6. As indicated in the problem statement of 1.11. The changes being
only in the arrangement of components, that being the swap between R and C.
Determine the transfer function relation Vo(s) to Vi(s) for the circuit.
Calculate output voltage t> > 0 for a unit step voltage input at t= 0.
Solution:
In 1.6 we used the impedance Z to construct the transfer function. Later we
plugged in the values for Z's. So thats why this transfer function is relevant.
V o ( s))
ƀƀ
V i ( s)
=
Z2 Έ Z4
ƀƀƀƀƀƀƀƀƀƀƀƀ < --- From problem 1.6
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z3 + Z2Z4
< --- This is the circuit
for problem 1.8.
u ̗ 10
˕6
C1 ̗ 1 u
M ̗ 10
F
R1 ̗ 1 M
The Z impedance circuit becomes:
6
C2 ̗ 0.5 u F
R2 ̗ 1 M
We make 10^ 6 the common
multiplier for resistors and
10^ -6 for capacitor. Now we
only need work with the
simple numbers.
1
Z1 = ƀ
s
Z2 = 1
1
Z3 = ƀƀ Z4 = 1
0.5 s
V o ( s))
ƀƀ
V i ( s)
=
=
Z2 Έ Z4
ƀƀƀƀƀƀƀƀƀƀƀƀ
Z1Z2 + Z1Z3 + Z1Z4 + Z2Z3 + Z2Z4
1Έ1
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
і1љ і 1 љ і1љ
і1љ
і 1 љ
( 1) + ( 1) ƀƀ + ( 1) ( 1)
1
Έ
+
+
ƀ
ƀ
ƀƀ
ƀ
їј s њћ
їј s њћ їј 0.5 s њћ їј s њћ
їј 0.5 s њћ
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
1
ƀƀƀƀƀƀƀƀƀƀ
1 і 1 љ і1љ і 1 љ
+ ƀƀƀ
+ ƀ+ ƀƀ + 1
ƀ
s їј 0.5 s2 њћ їј s њћ їј 0.5 s њћ
=
2
V o ( s))
ƀƀ
V i ( s)
=
s
=
ƀƀƀƀƀƀ
2
s+ 2 + s+ 2 s+ s
=
sΈ s
ƀƀƀƀ
2
s + 4 s+ 2
2
Multiply by s^ 2
s
=
ƀƀƀƀ
2
2 + 4 s+ s
2
s
ƀƀƀƀ
2
s + 4 s+ 2
Unit step voltage comes on at t= 0 and is of unit value, ie 1. Vi(s) must equal 1.
V o ( s))
V i ( s)) Έ s Έ s
=
ƀƀƀƀ
2
s + 4 s+ 2
=
sz1 ̗ 1
2
ax + bx + c
V o ( s))
s
ƀƀƀƀ
2
s + 4 s+ 2
=
2
: s + 4 s+ 2
˕b ˕ ƠƠƠƠƠƠƠ
b ˕ 4 ac
ƀƀƀƀƀƀ =
2a
˕4 ˕ ƠƠƠƠƠƠƠƠ
4 ˕4 1 2
= ˕3.4142
ƀƀƀƀƀƀ
21
2
˕b + ƠƠƠƠƠƠƠ
b ˕ 4 ac
ƀƀƀƀƀƀ =
2a
2
˕4 + ƠƠƠƠƠƠƠƠ
4 ˕4 1 2
= ˕0.5858
ƀƀƀƀƀƀ
21
2
s1
=
s2
=
1 Έ sΈ s
ƀƀƀƀ
2
s + 4 s+ 2
2
We solved the denominator for the poles. Which math wise were the
roots but electrical wise these are the poles.
2
V o ( s))
=
s
ƀƀƀƀƀƀƀ
( s + 3.414) Έ ( s + 0.586)
The poles going back in the transfer
function with the opposite sign.
For the pole to be maximum s1 and s2? ˕3.414 and ˕0.586
What about the numerator what any value to solve?
I ts NOT the numerator its the COEFFI CI ENTS of Vo(s) and those same for time domain.
At t< 0 Vo(< 0) = 0, and t> 0 Vo(> 0) = 0, but for t> > 0 Vo(> > 0) = 1u(t).
At -0 its near same as 0+ equal 0. So we use continuity here?
No, basically math. To solve for coefficients using the?
Method of proper fractions OR Equating coefficients of like powers.
Next calculate the coefficients.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
2
V o ( s))
=
s
ƀƀƀƀƀƀƀ
( s + 3.414) Έ ( s + 0.586)
sΈ s
ƀƀƀƀƀƀƀ
( s + 3.414) Έ ( s + 0.586)
=
Split LHS to solve for coefficients.
A
B
+ ƀƀƀƀ
ƀƀƀƀ
( s + 3.414) ( s + 0.586)
A ( s + 0.586)) + B ( s + 3.414)) =
Arrange like terms:
As + B s
0.586 A + 3.414 B
2nd order eq.
As + 0.586 A + B s + B3.414
s below is numerator term in transfer function - s* s split to s* s.
One 's' for 1 equation (As+ Bs) = 1 < ---coefficient of s = 1. Like terms.
=
=
s --->
A+ B
0 ---> 0.586 A + 3.414 B
=
=
1
0
Eq 1
Eq 2
0.586 A + 0.586 B
0.586 A + 3.414 B
=
=
0.586
0
Eq 1 x 0.586...Eq 3
Eq 2
=
0.586
Eq 3 - 2
=
0.586
( 0.586 ˕ 3.414)) B
( 0.586 ˕ 3.414)) = ˕2.828
˕2.828 B
B =
A+ B
A ˕ 0.207
A
=
=
=
0.586
= ˕0.2072
ƀƀƀ
˕2.828
1
1
1 + 0.207 = 1.207
V o ( s))
=
A
B
+ ƀƀƀƀ
ƀƀƀƀ
( s + 3.414) ( s + 0.586)
V o ( s))
=
1.21
0.21
˕ ƀƀƀƀ
ƀƀƀƀ
( s + 3.414) ( s + 0.586)
The general form of v_o(t): Ae
˕s1 Έ t
The circuit s-domain:
vo ( t )
= 1.21 e
+ Be
˕s1 Έ t
˕3.414 Έ t
˕ 0.21 e
˕0.586 Έ t
Answer.
Interesting solution math
wise. What math can do for
determining coefficients by
'equating coefficients of like
powers'.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1.9:
Find the transfer function of the network shown in figure above.
Plot its poles and zeros for R1 = R2 = 1, and C1 = C2 = 1.
Solution:
Current equation at node:
i (t )
=
i 1 (t ) + i 2 ( t )
Note: Current thru R1 and C1 is i(t).
Voltage mesh equations:
vi ( t )
=
1 ϣ
R1i ( t ) + ƀƀ
Έ Ϥ i d t + R2 Έ ( ( i 1 ( t ) ˕ i 2 ( t ) ) )
C1
R2 Έ ( ( i 1 ( t ) ˕ i 2 ( t ) ) ) we neglect i2(t) leaving ---> R2i 2 ( t )
Shown later.
Voltage across R2 is Vo(t).
Form the voltage mesh equation using Vo(t).
Deviation here:
vo ( t )
=
1 ϣ
R2 Έ ( ( i 2 ( t ) ˕ i 1 ( t ) ) ) + ƀƀ
Έ Ϥ i 2 (t ) dt
C2
We may not need this mesh
equation.
We did this just so we see the time domain equations, we could
have started with to s-domain as we did in other example(s).
Now for converting to s-domain,
in other words taking the Laplace Transform:
I ( s))
=
I 1 ( s)) + I 2 ( s))
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V i ( s))
=
1
R1I ( s)) + ƀƀ
I ( s)) + R2 Έ ( I 1 ( s)) ˕ I 2 ( s)) )
sC1
V o ( s))
=
1
R2 Έ ( I 2 ( s)) ˕ I 1 ( s)) ) + ƀƀ
Έ I 2 ( s))
sC2
We may not need this equation.
The voltage across C2 is the same across R2.
This is the voltage v_o(t) or Vo(s).
We can use this voltage expression and plug into the Vi(s) equation.
Obviously we want to plug in for R2I 1(s).
vo ( t )
=
1 ϣ
ƀƀ
Ϥ i 2 (t ) d t
C2
=
R2 Έ i 1 ( t )
Here we do not do a mesh method on the current thru R2.
We simply identify it to i1(t), since its the voltage across the
resistor terminals equated to v_o(t).
Their Laplace transform:
=
1
I 2 ( s))
ƀƀ
sC2
=
R2 Έ I 2 ( s))
V i ( s))
=
1
R1I ( s)) + ƀƀ
I ( s)) + R2 Έ ( I 1 ( s)) ˕ I 2 ( s)) )
sC1
The main equation now.
V i ( s))
=
1
R1I ( s)) + ƀƀ
I ( s)) + R2 Έ I 1 ( s)) ˕ R2I 2 ( s))
sC1
Plug in Vo at R2I 1(s)
V i ( s))
=
1
R1I ( s)) + ƀƀ
I ( s)) + V o ( s)) ˕ R2I 2 ( s))
sC1
Vo ( s))
Mesh or voltage loop problem, stated earlier below.
Deviation here: R2 Έ ( ( i 1 ( t ) ˕ i 2 ( t ) ) ) we neglect i2(t) leaving ---> R2i 2 ( t )
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Few attempts to find a substitute for R2I2(s) was not obtained.
The equation, voltage conservation, by the author-engineer did not include the i2(t)
expression for R2. The engineer is taking i1(t) as a known current or on its own.
So there is no need for R2(i1(t) - i2(t)), rather just R2i1(t).
The engineer's solution stated the assumption current distribution as shown below.
I did it taking two loops, mesh equations, until I knew why. Otherwise the
assumption would not been clear to me. Thus I leave it as it is, with correction
continued below.
The improved or updated voltage equation becomes:
vi ( t )
=
vi ( t )
=
V i ( s))
=
1 ϣ
R1i ( t ) + ƀƀ
Έ Ϥ i d t + R2 Έ ( ( i 1 ( t ) ˕ i 2 ( t ) ) )
C1
1 ϣ
R1i ( t ) + ƀƀ
Έ Ϥ i d t + R2 Έ i 1 ( t )
C1
1
R1I ( s)) + ƀƀ
I ( s)) + R2I 1 ( s))
sC1
Figure to left is
the voltage
loop given i1(t)
and i2(t) are
known values.
Vo (t )
=
R2 Έ I 1 ( s))
V i ( s))
=
1
R1I ( s)) + ƀƀ
I ( s)) + V o ( s))
sC1
I ( s))
V i ( s))
=
Plug in equation above.
=
I 1 ( s)) + I 2 ( s)) Plug in equation below.
1
( I 1 ( s)) + I 2 ( s)) ) + V o ( s))
R1 ( I 1 ( s)) + I 2 ( s)) ) + ƀƀ
sC1
Rearranging:
V i ( s))
=
і
1 љ
( I 1 ( s)) + I 2 ( s)) ) R1 + ƀƀ+ V o ( s))
їј
sC1 њћ
I cannot find a Vo(s)/Vi(s) from the above expression.
Cleaver engineer does a substitution for I1(s) and I2(s).
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V o ( s))
=
R1I 1 ( s))
I 1 ( s))
=
V o ( s))
ƀƀ
R1
V o ( s))
=
1
I 2 ( s))
ƀƀ
sC2
I 2 ( s))
=
sC2 V o ( s))
Substitute the expressions for I1(s) and I 2(s) into the Vi(s) equation.
і
1 љ
( I 1 ( s)) + I 2 ( s)) ) R1 + ƀƀ+ V o ( s))
їј
sC1 њћ
V i ( s))
=
V i ( s))
=
і V o ( s))
љі
1 љ
+ V o ( s))
їƀƀ+ sC2 V o ( s)) њ їR1 + ƀƀ
sC1 њћ
ј R1
ћј
V i ( s))
=
љі
і 1
1 љ
V o ( s)) Έ їƀƀ
+ sC2њ їR1 + ƀƀ
+ V o ( s))
sC1 њћ
ј R1
ћј
Another new trick, maybe not, but not common divide by Vo(s)
V i ( s))
ƀƀ
V o ( s)
=
і 1
љі
1 љ
+1
+ sC2њ їR1 + ƀƀ
їјƀƀ
R1
sC1 њћ
ћј
R1
1
sC2
+ ƀƀƀ
+ sC2R1 + ƀƀ
Multiply the parenthesis: ƀƀ
R1 sC1R1
sC1
Note: C1 = C2
1
+ sC2R2 + 1
= 1 + ƀƀƀ
sC1R1
V i ( s))
ƀƀ
V o ( s)
=
1
1 + ƀƀƀ
+ sC2R2 + 1 + 1
sC1R1
V i ( s))
ƀƀ
V o ( s)
=
1
+ sC2R2 + 3
ƀƀƀ
sC1R1
Invert for Vo(s)/Vi(s):
V o ( s))
ƀƀ
V i ( s)
=
1
ƀƀƀƀƀƀƀ
1
+ sC2R2 + 3
ƀƀƀ
sC1R1
Next simplify this expression for the purpose of attaining an expression in s form.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
The s form of expression we seek where we can identify zeros and poles.
V o ( s))
ƀƀ
V i ( s)
=
=
=
1
ƀƀƀƀƀƀƀ
1
+ sC2R2 + 3
ƀƀƀ
sC1R1
Multiply top and bottm by sC1R1
sC1R1
ƀƀƀƀƀƀƀƀƀƀƀ
1 + ( sC1R1) ( sC2R2) + 3 ( sC1R1)
sC1R1
ƀƀƀƀƀƀƀƀƀƀ
2
1 + іјs Έ C1C2R1R2љћ + 3 sC1R1
Let C = C1 = C2 = 1
R = R1 = R2 = 1
V o ( s))
ƀƀ
V i ( s)
=
V o ( s))
ƀƀ
V i ( s)
=
s
=
ƀƀƀƀ
2
1+ s + 3 s
Zero:
0
Answer.
sCR
ƀƀƀƀƀƀƀƀ
2
2
2
1 + іјs Έ C Έ R љћ + 3 sCR
s
Answer for transfer function.
ƀƀƀƀ
2
s + 3 s+ 1
Pole(s): Solve quadratic equation
2
As + B s + C
Since R1= R2= C1= C2= 1
We substitute for 1.
2
s + 3 s+ 1
2
ƠƠƠƠƠƠƠƠƠ
s1 s2 = ˕B
_____________________
+/B ˕ 4 AC
2A
˕1
˕3 + ƠƠƠƠƠƠƠƠƠƠ
3 ˕ ( 4 Έ 1 Έ 1))
= ˕4 Έ 10
ƀƀƀƀƀƀƀ
2Έ1
2
s1 =
˕3 ˕ ƠƠƠƠƠƠƠƠƠƠ
3 ˕ ( 4 Έ 1 Έ 1))
= ˕3
ƀƀƀƀƀƀƀ
2Έ1
2
s2 =
Poles:
˕0.382 and ˕2.618 Answer.
The manual plot is easy, real x-axis and imaginary y-axis.
Here all the zero and ploes are on the x-axis at 0, -0.382, and -2.618.
Lets try plotting the functions, numerator and denominator.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Or i gi n ( 1 , 1))
Set start of matrix at 1,1.
ќ ˕0.382 0 џ
P̗ѝ
ў ˕2.618 0 Ѡѡ
Z ̗[ 0 0]
Using matriz Z for zero and P for poles.
Έ ƾ y
0
Z
P
P
Έ ƾ y
-4
-4
-3
-3
-2
-2
- 1 Έ 0ƾ y Έ ƾ1y
Z
P
P
2
2
3
3
4
1,2
1,2
2,2
4
1,1
1,1
2,1
Answer. Plot above.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1.10:
Write the differential equations for the electrical circuit above.
Solution:
I kickoff with the sum of voltage around a loop equal zero.
I do an equation for each loop.
Loop i1(t):
vi ( t )
=
і di 1 ( t ) љ
1 ϣ
1 ϣ
L1 їƀƀƀ
+ R1i 1 ( t ) + ƀƀ
Έ Ϥ i 1 ( t ) d t ˕ ƀƀ
Έ Ϥ i 2 (t ) dt
њ
C1
C1
ј dt ћ
Loop i2(t):
0
=
і di 2 ( t ) љ
1 ϣ
1 ϣ
L2 їƀƀƀ
+ R2i 2 ( t ) + ƀƀ
Έ Ϥ i 2 ( t ) d t ˕ ƀƀ
Έ Ϥ i 1 (t ) dt
њ
C1
C1
ј dt ћ
Answer.
Not part of the question but how if I did a s-domain on the time domain, what the
typical controls engineering course will say is taking the Laplace transform? You
verify.
V i ( s))
=
1
1
sL 1 Έ I 1 ( s)) + R1 Έ I 1 ( s)) + ƀƀ
Έ I 1 ( s)) ˕ ƀƀ
Έ I 2 ( s))
sC1
sC1
0
=
1
1
sL 2 Έ I 2 ( s)) + R2 Έ I 2 ( s)) + ƀƀ
Έ I 2 ( s)) ˕ ƀƀ
Έ I 2 ( s))
sC1
sC1
Answer.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Chp 1 Problem 1.12:
Determine the transfer function relating Vo(s) to Vi(s) for network above.
Calculate the output voltage, t> > 0, for a unit step voltage input at t= 0, when
C1 = 1 uF, R = 1 M Ohm, C2 = 0.5 uF and R2 = 1 M Ohm.
Solution:
Circuit re-sketched for applying sum of voltage in a loop method.
Kickoff's Voltage Law, KVL, usually what the electrical engineer calls.
Amplifier gain e2(t)/e1(t) = 1. Therefore e1(t) = e2(t).
The circuit has a voltage input v_i(t), and to the output side of the
amplier is a voltage gained e2(t) this is similar to supplying voltage to the
circuit to the right of the ampliffier.
We proceed with KVLoop on the left and right, and we equate the
resistor R1 voltage for e1(t).
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
vi ( t )
=
1 ϣ
ƀƀ
Ϥ i 1 ( t ) d t + R1i 1 ( t )
C1
e2 ( t )
=
1 ϣ
ƀƀ
Ϥ i 2 ( t ) d t + R2i 2 ( t )
C2
e1 ( t )
=
R1i 1 ( t )
Amplifier left side voltage.
vo ( t )
=
R2i 2 ( t )
Amplifier right side voltage.
This being the voltage output v_o(t)
Now we take the Lapalce transforms of the expressions above.
Call it what you want, La Place or No Place, its converting to s-domain.
V i ( s))
=
I 1 ( s))
ƀƀ+ R1 Έ I 1 ( s))
sC1
Eq 1
E2 ( s))
=
I 2 ( s))
ƀƀ+ R2 Έ I 2 ( s))
sC2
Eq 2
E1 ( s))
=
R1 Έ I 1 ( s))
Eq 3
V o ( s))
=
R2 Έ I 2 ( s))
Eq 4
METHOD 1:
This by building interconnected relationship, as I done in the past problems here.
The long way and the answer is same as the textbook answer.
If I had not done this then it may remain a mystery!
You may verify.
Method 2 is easy, which was my first re-action to the problem. Just place Vo/Vi,
after forming their expression without the usual inter-related quations.
I will do method 2 after method 1 completion.
Rearrange Eq 1:
V i ( s))
=
љ
і 1
I 1 ( s)) Έ їƀƀ
+ R1њ
ј sC1
ћ
Eq 5
E2 ( s))
=
љ
і 1
I 2 ( s)) Έ їƀƀ
+ R2њ
ј sC2
ћ
Eq 6
I 2 ( s))
=
V o ( s))
ƀƀ
R2
Eq 7...substitute in Eq 6
Rearrange Eq 2:
Rearrange Eq 4:
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
E2(s) = E1(s):
V o ( s)) і 1
љ
+ R2њ
ƀƀΈ їƀƀ
R2 ј sC2
ћ
E2 ( s))
=
E1 ( s))
= R1 Έ I 1 ( s))
Eq 8
= E2 ( s))
Next substitute E1(s) for E2(s) in Eq 8.
E1 ( s))
= E2 ( s))
= R1 Έ I 1 ( s))
V o ( s)) і 1
љ
+ R2њ
= ƀƀΈ їƀƀ
R2 ј sC2
ћ
Eq 9
Substitute Eq 9 for R1I1(s) in Eq 1.
V i ( s))
=
љ
I 1 ( s)) V o ( s)) і 1
+ R2њ
ƀƀ+ ƀƀΈ їƀƀ
sC1
R2 ј sC2
ћ
Eq 10
How do I substitute for I1(s), try Eq 5, then substitute into eq 10:
V i ( s))
=
I 1 ( s))
=
V i ( s))
=
љ
і 1
I 1 ( s)) Έ їƀƀ
+ R1њ
ј sC1
ћ
V i ( s))
ƀƀƀƀ
і 1
љ
+ R1њ
їјƀƀ
sC1
ћ
Eq 5
Eq 11.....substitute in Eq 10.
V i ( s))
ƀƀƀƀ
і 1
љ
+
R1
ƀƀ
їј sC1
њћ V o ( s)) і 1
љ
+ R2њ
ƀƀƀƀ+ ƀƀΈ їƀƀ
sC1
R2 ј sC2
ћ
Eq 12...looks messy may do it.
V i ( s))
ƀƀƀƀ
і 1
љ
+
R1
ƀƀ
їј sC1
њћ
V o ( s)) і 1
љ
Έ їƀƀ
V i ( s)) ˕ ƀƀƀƀ = ƀƀ
+ R2њ
sC1
R2 ј sC2
ћ
V i ( s))
1
V i ( s)) ˕ ƀƀƀƀ
Έ ƀƀ
і 1
љ sC1
+ R1њ
їјƀƀ
sC1
ћ
V o ( s)) і 1
љ
= ƀƀΈ їƀƀ
+ R2њ
R2 ј sC2
ћ
V o ( s)) і 1
љ
љ
і
1
= ƀƀΈ їƀƀ
+ R2њ
V i ( s)) Έ ї1 ˕ ƀƀƀƀƀƀ
њ
і 1
љ
R2 ј sC2
ћ
+
Έ sC1 њ
R1
ƀƀ
ї
ї
њ
ј
ј sC1
ћ
ћ
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
љ
і 1
= V o ( s)) Έ їƀƀƀ
+ 1њ
ј sC2R2
ћ
і
љ
1
V i ( s)) Έ ї1 ˕ ƀƀƀƀњ
( 1 + sC1R1) ћ
ј
V o ( s))
ƀƀ
V i ( s)
V o ( s))
ƀƀ
V i ( s)
V o ( s))
ƀƀ
V i ( s)
Let
і
љ
1
їј1 ˕ ƀƀƀƀ
( 1 + sC1R1)) њћ
ƀƀƀƀƀƀ
љ
і 1
+
1
ƀƀƀ
їј sC2R2
њћ
=
і ( 1 + sC1R1)
љ
1
˕ ƀƀƀƀњ
їјƀƀƀƀ
( 1 + sC1R1)) ( 1 + sC1R1)) ћ
=
ƀƀƀƀƀƀƀƀƀƀ
і 1 + sC2R2 љ
їјƀƀƀƀ
sC2R2 њћ
=
і ( sC1R1)) љ і sC2R2 љ
Έ ƀƀƀƀ
їјƀƀƀƀ
( 1 + sC1R1) њћ їј 1 + sC2R2 њћ
=
іјs2 Έ C1C2R1R2љћ
љ
і
їƀƀƀƀƀƀƀƀƀƀƀƀ
њ
2
ј іј1 + sC2R2 + sC1R1 + s Έ C1C2R1R2љћ ћ
=
іјs2 Έ C1C2R1R2љћ
ƀƀƀƀƀƀƀƀƀƀƀƀƀ
іј1 + s ( C1R1 + C2R2) + s2 Έ ( C1C2R1R2) љћ
B = C1R1
іјA Έ s2 љћ
ƀƀƀƀƀƀƀ
іј1 + ( B + C ) Έ s + A Έ s2 љћ
=
C1 ̗ 1 Έ 10
˕6
C2 ̗ 0.5 Έ 10
˕6
=
і1љ 2
Έs
їјƀ
2 њћ
ƀƀƀƀƀƀ
і3љ
і1љ 2
1 + їƀ
Έ s + їƀ
Έs
њ
ј2ћ
ј 2 њћ
C = C2R2
One Transfer Function - METHOD 1.
R1 ̗ 1 Έ 10
1
A ̗ C1 Έ C2 Έ R1 Έ R2 = 0.5 Or fraction: ƀ
2
V o ( s))
ƀƀ
V i ( s)
і ( sC1R1) љ
їјƀƀƀƀ
( 1 + sC1R1)) њћ
ƀƀƀƀƀ
і 1 + sC2R2 љ
їјƀƀƀƀ
sC2R2 њћ
=
A = C1C2R1R2
V o ( s))
ƀƀ
V i ( s)
Transfer function. Need simplifying.
6
R2 ̗ 1 Έ 10
6
B ̗ C1 Έ R1 + C2 Έ R2 = 2
Multiply by 2.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V o ( s))
ƀƀ
V i ( s)
V o ( s))
ƀƀ
V i ( s)
=
і1љ 2
2 їƀ
Έs
ј 2 њћ
ƀƀƀƀƀƀƀƀ
і
і3љ
і1љ 2љ
2 Έ ї1 + їƀ
Έ s + їƀ
Έs њ
њ
ј
ј2ћ
ј 2 њћ
ћ
2
=
2
s
=
ƀƀƀƀ
2
2 + 3 Έ s+ s
s
Answer. SAME AS TEXTBOOK!
ƀƀƀƀ
2
s + 3 Έ s+ 2
Calculate the output voltage, t> > 0, for a unit step voltage input at t= 0:
Since its unit step voltage input the initial conditions for t< 0 = 0.
So i(-0) = i(0+ ...just near 0) = 0 and
v(-0) = v(0+ ...just near 0) = 0 v(+ + ) = 1
Comment: How do I get the numerator (zero) = 1 for t> > 0 so the Vi(s) = 1
or greater; u(t= 0 or t> 0) = 1 or u(t) = 1x Constant. But NOT equal 0.
V o ( s))
ƀƀ
V i ( s)
2
2
s
= V i ( s)) Έ ƀƀƀƀƀ
( s + 2) ( s + 1)
s
= ƀƀƀƀƀ V o ( s))
( s + 2) ( s + 1)
sΈ s
A
B
= 1 Έ ƀƀƀƀƀ = ƀƀƀ
+ ƀƀƀ
( s + 2) ( s + 1)
( s + 2) ( s + 1)
V o ( s))
To solve for coefficients using the? Method of proper fractions OR Equating coefficients
of like powers.
s
=
A ( s + 1)) + B ( s + 2))
=
As + A + B s + 2 B
s
=
s(( A + B ) + ( A + 2 B))
Arrange for like terms:
s
:
s(( A + B )
--->
A+ B
=
1
Eq 1
0
:
( A + 2 B))
--->
A+ 2 B
=
0
Eq 2
B
=
˕1
Eq 2-1
Substitute B in Eq 1.
A+ B
A ˕1
A
=
=
=
1
1
2
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V o ( s))
=
A
B
+ ƀƀƀ =
ƀƀƀ
( s + 2) ( s + 1)
2
1
˕ ƀƀƀ
ƀƀƀ
( s + 2) ( s + 1)
Now with the coefficients, zeros, and poles I can form the voltage output in time domain.
This will be an exponential equation because the voltage source is a step function, unity,
or constant.
s1t
V o ( s))
=
Ae
V o ( s))
=
˕2 e
+ Be
˕2 t
s2t
˕1 e
˕1 t
Now to convert from s-domain to time domain:
vo ( t )
=
˕2 e
˕2 t
˕1 e
vo ( t )
=
˕2 e
˕2 t
˕e
˕1 t
˕t
Answer. Same as textbook.
Please verify the solution steps and reasoning
on the voltage output equation
where Vi(s) = 1.
METHOD 2:
Now for Method 2, the supposed to be simpler and shorter solution.
V i ( s))
=
I 1 ( s))
ƀƀ+ R1 Έ I 1 ( s))
sC1
Eq 1
E2 ( s))
=
I 2 ( s))
ƀƀ+ R2 Έ I 2 ( s))
sC2
Eq 2
E1 ( s))
=
R1 Έ I 1 ( s))
Eq 3
V o ( s))
=
R2 Έ I 2 ( s))
Eq 4
Set up the transfer function, Vo(s)/ Vi(s) based on their respective equations directly:
V o ( s))
ƀƀ
V i ( s)
=
I 1 ( s))
=
R2 Έ I 2 ( s))
I 2 ( s)) Έ R2
ƀƀƀƀƀƀ = ƀƀƀƀƀƀƀ
і
I 1 ( s)
1 љ
I 1 ( s) Έ їR1 ( s) + ƀƀ
ƀƀ+ R1 Έ I 1 ( s)
sC1
sC1 њћ
ј
E1 ( s))
ƀƀƀ I 2 ( s))
R1
=
E2 ( s))
ƀƀƀƀ
і
1 љ
їјR2 + ƀƀ
sC2 њћ
Eq 5...maybe I 2(s)
and I 1(s) substituion
may help.
From Eq 2 above.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
I 2 ( s))
ƀƀ
I 1 ( s)
=
E2 ( s))
ƀƀƀƀ
і
1 љ
їјR2 + ƀƀ
sC2 њћ
ƀƀƀƀ
E1 ( s)
ƀƀƀ
R1
E2 ( s))
R1
ƀƀƀƀΈ ƀƀƀ
і
1 љ E1 ( s)
їјR2 + ƀƀ
sC2 њћ
=
Gain = 1, E2(s)/ E1(s) = 1, therefore E1(s) = E2(s).
E1 ( s))
=
E2 ( s))
I 2 ( s))
ƀƀ
I 1 ( s)
Now the current ratio equation becomes:
R1
= ƀƀƀƀ
і
1 љ
їјR2 + ƀƀ
sC2 њћ
Returning to Eq 5 substitute for I 2(s)/ I 1(s):
V o ( s))
ƀƀ
V i ( s)
=
=
=
Let:
A
B
D
I 2 ( s)) Έ R2
ƀƀƀƀƀƀ
і
1 љ
I 1 ( s) Έ їR1 + ƀƀ
sC1 њћ
ј
Eq 5
R1
R2
ƀƀƀƀΈ ƀƀƀƀ
і
1 љ і
1 љ
R1 + ƀƀ
їјR2 + ƀƀ
њ
ї
sC2 ћ ј
sC1 њћ
R1R2
ƀƀƀƀƀƀƀƀƀƀ
R2
R1
1
R1R2 + ƀƀ
+ ƀƀ
+ ƀƀƀ
2
sC1 sC2 s C1C1
= R1 Έ R2 = 1 Έ 10
12
12
R2
= ƀƀ
= 1 Έ 10
C1
12
1
= 2 Έ 10
= ƀƀƀ
C1 Έ C2
C
12
R1
= ƀƀ
= 2 Έ 10
C2
1 Έ 10
ƀƀƀƀƀƀƀƀƀƀƀ
12
12
12
12
1 Έ 10
2 Έ 10
2 Έ 10
1 Έ 10 + ƀƀƀ
+ ƀƀƀ
+ ƀƀƀ
2
s
s
s
12
V o ( s))
ƀƀ
V i ( s)
=
V o ( s))
ƀƀ
V i ( s)
=
1
ƀƀƀƀƀ
1 2
2
1 + ƀ+ ƀ
+ƀ
s s s2
=
1
ƀƀƀƀ
3
2
1+ ƀ
+ƀ
s s2
Multiply by s^ 2
top and bottom.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
V o ( s))
ƀƀ
V i ( s)
V o ( s))
ƀƀ
V i ( s)
=
іјs2 љћ Έ 1
ƀƀƀƀƀƀ
3
2љ
іјs2 љћ Έ і1 + ƀ
+ ƀњ
ї
s s2 ћ
ј
2
=
s
ƀƀƀƀ
2
s + 3 s+ 2
Answer.
Same method used by engineer the faster method.
The short method may give the impression there is no relationship with the
components like that established in the longer method. However, the transfer
function's definition is just that, output divided by input. Do consider the circuit's
components and connections, and carefully construct the equations.
The remaining part on the output voltage same as completed following the long
method of the transfer functions.
Chp 1 Problem 1.13:
Determine the transfer function of the electrical network above:
Solution:
C1 = C2, the question did not show C1 and C2, rather C.
To ease tracking the solution they were made into C1 and C2.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
The basic steps we first started with provided here again, these steps were
much the same to what we did in the previous problems.
The steps involved in obtaining the transfer function are:
1. Write differential equations of the system.
d
2. Replace terms involvingƀby s and ϣ
Ϥ dt by 1/ s, for inductor and capacitor respectively.
dt
3. Eliminate all but the desired variable.
Step 1:
Check current flow direction. Coming
out of C1 -ve terminal -ve voltage.
v_i(t) :
1 ϣ
v i ( t ) = ˕ƀƀ
Έ Ϥ i (t ) dt
C1
1 ϣ
Έ Ϥ (i 1 ( t ) ˕ i ( t ) ) dt
ƀƀ
C1
Check current flow direction. Coming out of C1 ve terminal -ve voltage (left loop).
=
1 ϣ
Έ Ϥ i 1 ( t ) d t ˕ vi ( t )
ƀƀ
C1
This can be written as:
˕v i ( t )
1 ϣ
= ƀƀ
Έ Ϥ i 1 (t ) dt
C1
OR
1 ϣ
v i ( t ) + ƀƀ
Έ Ϥ i 1 ( t ) d t = 0 Sum of voltages,
C1
next vi(t) to the
LHS, resulting in
1 ϣ
the same.
v i ( t ) = ˕ƀƀ
Έ Ϥ i 1 (t ) dt
C1
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Step 1:
v_i(t) :
1 ϣ
v i ( t ) = ˕ƀƀ
Έ Ϥ i (t ) dt
C1
Check current flow direction. Coming
out of C1 -ve terminal -ve voltage.
1 ϣ
v i ( t ) = ƀƀ
Έ Ϥ i 1 (t ) dt
C1
Check current flow direction. Coming
out of C1 -ve terminal -ve voltage.
This can be written as:
˕v i ( t )
1 ϣ
= ƀƀ
Έ Ϥ i 1 (t ) dt
C1
OR
1 ϣ
v i ( t ) + ƀƀ
Έ Ϥ i 1 ( t ) d t = 0 Sum of voltages,
C1
next vi(t) to the
LHS, resulting in
1 ϣ
the same.
v i ( t ) = ˕ƀƀ
Έ Ϥ i 1 (t ) dt
C1
Centre loop:
і di 1 ( t ) љ
1 ϣ
1 ϣ
0 = ƀƀ
Έ Ϥ i 1 ( t ) d t + R1i 1 ( t ) + L 1 їƀƀƀ
+ ƀƀ
Έ Ϥ (i 1 ( t ) ˕ i 2 ( t ) ) dt
њ
C1
ј dt ћ C2
˕v i ( t )
1 ϣ
= ƀƀ
Έ Ϥ i 1 (t ) dt
C1
Substitute in equation above.
і di 1 ( t ) љ
1 ϣ
0 = ˕v i ( t ) + R1i 1 ( t ) + L 1 їƀƀƀ
+ ƀƀ
Έ Ϥ ( i 1 (t ) ˕ i 2 ( t ) ) dt
њ
ј dt ћ C2
vi ( t )
=
і di 1 ( t ) љ
1 ϣ
R1i 1 ( t ) + L1 їƀƀƀ
+ ƀƀ
Έ Ϥ (i 1 ( t ) ˕ i 2 ( t ) ) d t
њ
ј dt ћ C2
Eq 1
v_o(t):
і di 2 љ
v o ( t ) = R2 Έ i 2 ( t ) + L2 їƀƀ
ј dt њћ
Eq 2
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Right loop:
0
і di 2 ( t ) љ
1 ϣ
= ƀƀ
Έ Ϥ ( i 2 ( t ) ˕ i 1 ( t ) ) d t + R2i 2 ( t ) + L2 їƀƀƀ
ј dt њћ
C2
Eq 3
Substitute v_o(t) into Right Loop.
Step 2:
Assuming all initial conditions for L and C are zero.
We proceed with taking the? Laplace Transform. Convert to s-domain.
vi ( t )
=
і di 1 ( t ) љ
1 ϣ
R1i 1 ( t ) + L1 їƀƀƀ
+ ƀƀ
Έ Ϥ (i 1 ( t ) ˕ i 2 ( t ) ) d t
њ
ј dt ћ C2
V i ( s))
=
1
R1I 1 ( s)) + sL 1 I 1 ( s)) + ƀƀ
Έ ( I 1 ( s)) ˕ I 2 ( s)) )
sC2
V i ( s))
=
1
R1I 1 ( s)) + sL 1 I 1 ( s)) + ƀƀ
Έ ( I 1 ( s)) ˕ I 2 ( s)) ) Eq 4 < ---Same as textbook.
sC2
Eq 1
і di 2 љ
v o ( t ) = R2 Έ i 2 ( t ) + L2 їƀƀ
Eq 2
ј dt њћ
V o ( t ) = R2 Έ I 2 ( s)) + sL2I 2 ( s))
Eq 5 < ----Same as textbook.
0
і di 2 ( t ) љ
1 ϣ
= ƀƀ
Έ Ϥ ( i 2 ( t ) ˕ i 1 ( t ) ) d t + R2i 2 ( t ) + L2 їƀƀƀ
Eq 3
C2
ј dt њћ
0
1
( I 2 ( s)) ˕ I 1 ( s)) ) + R2I 2 ( s)) + sL2 Έ I 2 ( s))
= ƀƀ
sC2
Eq 6
Step 3:
Vo (t )
ƀƀ
V i ( s)
=
Vo (t )
ƀƀ
V i ( s)
=
R2 Έ I 2 ( s)) + sL2I 2 ( s))
ƀƀƀƀƀƀƀƀƀƀƀƀƀƀSame as textbook.
1
Past this point you have to
R1I 1 ( s) + sL1 I 1 ( s) + ƀƀ
Έ ( I 1 ( s) ˕ I 2 ( s) )
sC2
solve it for the best
possible form.
I 2 ( s)) Έ ( R2 + sL2))
ƀƀƀƀƀƀƀƀƀƀƀƀ
і
1 љ
1
I 1 ( s) Έ їR1 + sL1 + ƀƀ
˕ ƀƀ
I 2 ( s)
њ
sC2 ћ sC2
ј
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Find a substitute for I2 in terms of I1 for the denominator.
0
1
( I 2 ( s)) ˕ I 1 ( s)) ) + R2I 2 ( s)) + sL2 Έ I 2 ( s))
= ƀƀ
sC2
Eq 6
Solve for I2 above:
1
I 1 ( s))
I 2 ( s)) + R2I 2 ( s)) + sL2 Έ I 2 ( s)) = ƀƀ
ƀƀ
sC2
sC2
I 2 ( s)) + R2I 2 ( s)) sC2 + sL2 Έ I 2 ( s)) sC2
2
I 2 ( s)) Έ іј1 + sC2R2 + s C2L2љћ
I 2 ( s))
=
Vo (t )
ƀƀ
V i ( s)
=
=
Multiply by sC2.
= I 1 ( s))
I 1 ( s))
Eq 7
I 1 ( s))
ƀƀƀƀƀƀƀƀ
іј1 + sC2R2 + s2 C2L 2љћ
I 2 ( s)) Έ ( R2 + sL2))
ƀƀƀƀƀƀƀƀƀƀƀƀ
і
1 љ
1
I 1 ( s) Έ їR1 + sL1 + ƀƀ
˕ ƀƀ
I 2 ( s)
њ
sC2 ћ sC2
ј
Substitute I2(s) in denominator above.
Vo (t )
ƀƀ
V i ( s)
I 2 ( s)) Έ ( R2 + sL2))
= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
і
1 љ
1
I 1 ( s)
I 1 ( s) Έ їR1 + sL1 + ƀƀ
˕
Έ
ƀƀ
ƀƀƀƀƀƀƀƀ
sC2 њћ sC2 іј1 + sC2R2 + s2 C2L2љћ
ј
Vo (t )
ƀƀ
V i ( s)
I 2 ( s)) Έ ( R2 + sL2))
= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
іі
љљ
1 љ і
1
I 1 ( s) Έ їїR1 + sL1 + ƀƀ
˕
ƀƀƀƀƀƀƀƀƀƀ
sC2 њћ їј sC2 + s2 C2 2 R2 + s3 C2 2 L 2 њћњћ
јј
Vo (t )
ƀƀ
V i ( s)
і I 2 ( s)) љ
( R2 + sL 2))
= їƀƀњ Έ ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
љ
ј I 1 ( s) ћ і
1 љ і
1
R1
+
sL
1
+
˕
ƀƀ
ƀƀƀƀƀƀƀƀƀƀ
їј
sC2 њћ їј sC2 + s2 C2 2 R2 + s3 C2 2 L2 њћ
Find an equation for I 2(s)/ I1(s) .... Eq 7 below.
2
I 2 ( s)) Έ іј1 + sC2R2 + s C2L2љћ
=
I 1 ( s))
Eq 7
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
I 2 ( s))
ƀƀ
I 1 ( s)
=
1
ƀƀƀƀƀƀƀƀ
іј1 + sC2R2 + s2 C2L 2љћ
I 2 ( s))
Substitute ƀƀ
I 1 ( s)
in transfer function
Vo (t )
ƀƀ
V i ( s)
љі
љ
і
( R2 + sL2))
1
= їƀƀƀƀƀƀƀ
њ їƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
њ
2
љ
1 љ і
1
ј 1 + sC2R2 + s C2L 2 ћ ї іR1 + sL1 + ƀƀ
˕ їƀƀƀƀƀƀƀƀƀƀ
њ
їј
њ
њ
2
2
3
2
sC2 ћ ј sC2 + s C2 R2 + s C2 L2 ћ ћ
ј
Vo (t )
ƀƀ
V i ( s)
( R2 + sL2))
= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
2
1 љ і іј1 + sC2R2 + s C2L2љћ љ
іј1 + sC2R2 + s2 C2L 2љћ Έ іR1 + sL1 + ƀƀ
˕
їƀƀƀƀƀƀƀƀƀњ
їј
sC2 њћ ј sC2 іј1 + s C2R2 + s2 C2 L2љћ ћ
Set C1 = C2 = C, as given.
Vo (t )
ƀƀ
V i ( s)
( R2 + sL2))
= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
1 љ і 1 љ
іј1 + sCR2 + s2 CL2љћ Έ іR1 + sL1 + ƀƀ
˕ ƀƀ
їј
sC њћ їј sC њћ
Expand the left side terms at the bottom, and set equal to A.
Then the bottom right side term's denominator set to B.
1 љ
іј1 + sCR2 + s2 CL2љћ Έ іR1 + sL 1 + ƀƀ
=
їј
sC њћ
2
2
3
1
R1 + sL 1 + ƀƀ
+ sCR1R2 + s CR2L1 + R2 + s CR1L2 + s CL1L 2 + sL 2
sC
=
A
љ
і
2
3
1
s ( CL1L 2)) + s Έ ( CR2L1 + CR1L 2)) + s їL1 + ƀƀ+ CR1R2 + L2њ + ( R1 + R2))
2
ј
ћ
s C
= A
і
3
2
1 љ
(
s ( L1L2)) + s Έ C ( R2L 1 + R1L2)) + s їL1 + L2 + CR1R2 + ƀƀ
њ + R1 + R2))
2
ј
s Cћ
= A
1
ƀƀ= B
sC
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.
Chapter 6 Part C. Solving Problems (Examples and Exercises). Source of study material: 1). Electric Circuits 6th Ed Schaums - Nahvi &
Edminister. 2). Solutions & Problems of Control System - AK Jairath. 3). Engineering Circuits Analysis - Hyat & Kemmerly.
My Homework. This is a pre-requisite study for Laplace Transforms in circuit analysis.
Karl S. Bogha.
Vo (t )
ƀƀ
V i ( s)
=
( R2 + sL2))
ƀƀƀƀ
і1љ
A ˕ їƀњ
јB ћ
і
3
2
1 љ
(
s ( L1L2)) + s Έ C ( R2L 1 + R1L2)) + s їL1 + L2 + CR1R2 + ƀƀ
њ + R1 + R2))
2
s Cћ
ј
=
A
In my a equation above there is (1/ s^ 2C) this is not in the textbook anwer.
Textbook answer below does not have B term (1/sC) maybe this was negligible to the
overall function because it becomes huge in the denominator, and when it divides
the numerator its small or negligible. Usually C is in microFarad units. This may also be
the case for (1/ s^ 2C) in the A term. Except for this my result is the same.
Vo (t )
R2 + sL2
ƀƀ= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
і
і 1 љ
3
2
V i ( s)
1 љ
(
)
s ( L1L2) + s Έ C ( R2L1 + R1L2) + s їL1 + L 2 + CR1R2 + ƀƀ
њ + R1 + R2 ˕ їјƀƀ
2
sC њћ
s Cћ
ј
Neglecting (1/ sC) and (1/s^ 2 C):
Vo (t )
ƀƀ
V i ( s)
R2 + sL2
= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
3
2
s ( L1L2) + s Έ C ( R2L1 + R1L2) + s ( L1 + L2 + CR1R2) + ( R1 + R2)
My Answer.
You can verify this answer correct it, or present
your own. Here this is as far as I am going.
Textbook Answer:
V o ( s))
ƀƀ
Vi ( s)
R2 + sL2
= ƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀƀ
3
2
s CL1L2 + s C ( R1L2 + L1R2) + s ( L1 + L2 + CR1R2) + ( R1 + R2)
Transfer function above does look tidy!
You solve it for yourself if you see a need.
You can sort it with your local lecturer/engineer.
Apologies for any errors and omissions.
This brings to end the 13 example problems.
Next Schaum's Chapter 8 Solved Problems.
Engineering college year 2 course of 4 year program OR year 1 of 3 year program. Re-fresher OR Self Study. Graduate Study Review.
May be used in New Zealand, US, Malaysia, India, Pakistan, UK, and other Common Wealth Country engineering colleges.
Any errors and omissions apologies in advance.