Sam PalermoAnalog & Mixed-Signal Center

Texas A&M University

ECEN325: ElectronicsSpring 2021

Lecture 2: Linear Circuit Analysis Review

Announcements• Reading

• Chapter 1 (Razavi)• Fundamentals of Circuit Analysis (Dr. Silva)

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Agenda• Laplace Transform• Passive Circuit s-Domain Models• Transfer Functions• Sinusoidal Steady-State Response• Poles & Zeros• Bode Plots• Second-Order Systems

3

References• Continuous & Discrete Signal & System Analysis,

3rd Ed., C. McGillem and G. Cooper, Saunders College Publishing, 1991.

• Feedback Control of Dynamic Systems, 3rd Ed., G. Franklin, J. Powell, and A. Emami-Naeini, Addison-Wesley, 1994.

• Design of Analog Filters, R. Schaumann and M. Van Valkenburg, Oxford University Press, 2001.

4

Motivation Example

5

00Given ov

4510sin2

12110sin

2110cos

21

21

l.not trivia is thisNote, equation. thissolvecan weclass, Eq. Diff.out from anythingremember weif Now,

10110sin

101

0101

10sin

at KCL a Write

5105510

5

5

55

tettetv

nFkt

nFktv

dttdv

dttdvnF

kttv

v

tto

oo

oo

o

transient response(can go to zero quickly)

sinusoidal steady-state response

• Now, let’s look at Laplace Transforms to make this easier

Laplace Transform• Laplace transforms are useful for solving

differential equations• One-Sided Laplace Transform

6

0

dtetxsXtx stL

where s is a complex variable

• s has units of inverse seconds (s-1)

(rad/s)frequency angular theis and 1 Note,

j

js

Laplace Transform of Signals

7

[McGillem]

Laplace Transform of Operations

8

[McGillem]

Resistor s-Domain Equivalent Circuit

9

tvR

ti

tRitv1

sRIsV sVR

sI 1

Time-domain Representation:

Complex Frequency Representation:

Capacitor s-Domain Equivalent Circuit

10

01

0vdi

Ctv

dttdvCti

t

011 vs

sIsC

sV

Time-domain Representation:

Complex Frequency Representation:

0CvsCsVsI

Inductor s-Domain Equivalent Circuit

11

tidv

Lti

dttdiLtv

001

0LisLsIsV 011 is

sVsL

sI

Time-domain Representation:

Complex Frequency Representation:

s-Domain Impedance w/o I.C.

12

RsZ

RsIsV

sC

sZ

sCsIsV

1

1

sLsZ

sLsIsV

Transfer Function

• The transfer function H(s) of a network is the ratio of the Laplace transform of the output and input signals when the initial conditions are zero

• This is also the Laplace transform of the network’s impulse response

13

sVsV

tvtv

sHi

o

i

o LL

RC Transfer Function

14

sVsRC

sV

sCR

sCsVZZ

ZsV inininCR

Co

1

11

1

sRCsVsV

sHin

o

11

AC Transfer Function, H(s)

Laplace Transform Circuit Example

15

Convert to Laplace Domain

00Given ov

4510sin2

12110sin

2110cos

21

21v

Transform Laplace inversewith 10

1021

1021

1021

expansionfraction partialwith 10

1010

10

1010

101

11

1

5105510o

252

5

2525

252

5

5

5

5

5

5

55

tettet

ss

s

ssV

sssVsHsV

sssRCsVsV

sH

tt

o

io

in

o

Laplace Transform Circuit Example

• Note that the transient response decays very quickly!16

4510sin2

1v

21v

v4510sin2

121v

response state-steady and transientsit' intooutput thedecomposecan We

5ss

10tr

ss510

o

5

5

tt

et

ttvtet

t

trt

Sinusoidal Steady-State Response

17

jHtAjHtv

tAtv

tv

ss

i

i

cos

be loutput wil state-steady The

cos

sinusoidal is input If

• Here we have evaluated the transfer function H(s) with s=jω• The magnitude of H(jω) scales the input amplitude• The phase of H(jω) shifts the input phase

RC Circuit Sinusoidal Steady-State Response

18

RCjjH

sRCsVsV

sHjs

in

o

11

11

RCjRCj

jHjHjH

1

11

1* 21

1RC

jH

RCRCjH

jH

DenDen

NumNum

jHjHjH

111

111

tan1

tan10tan

ofr Denominato Den andNumerator Num where

ReImtan

ReImtan

ReImtan

RCjH 1tan

Output Magnitude

Output Phase

RC Circuit Sinusoidal Steady-State Response Example

19

4510sin2

1

451tan10

21

2110

1110

10with

101

1

5

15

5

5

5

5

ttv

jH

jH

jjH

jjs

ssH

ss

Complex Numbers Properties

20

123100010tan

10010tan

1010tan

110tan

101000101001010101

1041.110100010100

101010110100010100

1010101

101000101001010101

1111

32222

2222

jjjj

jjjj

jjjj

Numerical Example

[Silva]

Inverse Tangent Function

21

• For small values approximately 0• For large values saturates at /2 or 90°• Between 0.1 and 10 can be approximated as

changing with a slope of 45° per decade

45/dec.(pi/4)/dec.

Poles & Zeros

22

n

m

pspspszszszsAsH

......

21

21

• Poles are the roots of the denominator (p1, p2, … pn) where H(s)∞• Zeros are the roots of the numerator (z1, z2, … zm) where H(s)0

sradsp

s

ssH

/10

010

1010 :1 Example

51

5

5

5

510 :2 Example

sssH

sradsp

s

sradsz

/10

010

/0

51

5

1

15005015100 :3 Example 2

ssssH

sradjsp

ss

sradsz

s

/6.29252

6000250050

0150050

/15

015

2,12,1

2

1

Bode Plots• Technique to plot the Magnitude (squared) and

Phase response of a transfer function• Magnitude is plotted in Decibels (dB), which is a power

ratio unit

23

dB log20dB log10 102

102 jHjHjH

dB

• Phase is typically plotted in degrees

jHjHjH

ReImtan 1

RC Bode Plot Example

24

rad/s 10 where,1

1101

1101

11011

11

51

1

5

55

p

pjj

jH

jssRCsVsVsH

js

in

o

25101025

1010 101log201log20101

1log20log20

jH

51 10tanPhase jH

Magnitude Squared (dB):

Phase:

RC Bode Plot Example

25

(rad/s) |H(j)| |H(j)|2 20log10|H(j)| (dB) Phase (H(j)) ()103 0.9999 0.9999 ~0 ~0104 0.995 0.990 -0.043 -5.71

5x104 0.894 0.800 -0.969 -26.6105 0.707 0.500 -3.01 -45.0

5x105 0.196 0.039 -14.2 -78.7106 0.100 0.010 -20.0 -84.3107 10-2 10-4 -40.0 -89.4108 10-3 10-6 -60.0 -89.9

251010252

2

1010 101log201log20101

1log20log20

jH

51 10tanPhase jH

Magnitude:

Phase:

~20log10 (1)= 0dB

~-20log10 (10-5)= -20dB/dec

-45/dec

51011

jjH

RC Bode Plot Example

26

-20dB/dec

-45/dec

Max Error = 3.01dB

Max Error = 5.71

Transient Response

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= 103 rad/s = -p1/100 = 105 rad/s = -p1 = 106 rad/s = 10*p1

0 Shift Phase

1tvo

-45Shift Phase

21tvo

3.84Shift Phase

1.0tvo

Bode Plot Algorithm - Magnitude1. Where is a good starting point?

a. Calculate DC value of |H(j)|b. If not a reasonable value, I like to calculate |H(j)| at equal

to the lowest non-zero value of p1/10 or z1/102. Where to end?

a. Calculate |H(j)| as ∞3. Where are the poles and zeros?

a. Beginning at each pole frequency, the magnitude will decrease with a slope of -20dB/dec

b. Beginning at each zero frequency, the magnitude will increase with a slope of +20dB/dec

4. Note, the above algorithm is only valid for real poles and zeros. We will discuss complex poles later.

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Bode Plot Algorithm - Magnitude

29

+20dB/dec.

+20dB/dec.-20dB/dec.

-20dB/dec.

-20dB/dec.

100 ,10 ,1

0 Magnitude HF

2010 Magnitude DC

1001

101

11010010110

211

4

ppz

dB

dB

sss

ssssH

2210

2110

21010

2221

2

1010

101log20101log201log2010log20

101101

110log20log20

jH

Bode Plot Algorithm - Phase1. Calculate low frequency value of Phase(H(j))

a. An negative sign introduces -180 phase shiftb. A DC pole introduces -90 phase shiftc. A DC zero introduces +90 phase shift

2. Where are the poles and zeros?a. For negative poles: 1 dec. before the pole freq., the phase will

decrease with a slope of -45/dec. until 1 dec. after the pole freq., for a total phase shift of -90

b. For negative zeros: 1 dec. before the zero freq., the phase will increase with a slope of +45/dec. until 1 dec. after the zero freq., for a total phase shift of +90

c. Note, if you have positive poles or zeros, the phase change polarity is inverted

3. Note, the above algorithm is only valid for real poles and zeros. We will discuss complex poles later. 30

Bode Plot Algorithm - Phase

31

+45/dec.-45/dec.

-45/dec.

100 ,10 ,1

180 Phase LF

1001

101

11010010110

211

4

ppz

sss

ssssH

100tan

10tan

1tan180 111 jH

+45/dec.-90/dec.

-45/dec.

Second-Order Systems: Real or Complex Poles?

32

20

200

21

20

02

201

22 , poles 2

QQpp

Qss

ksH

5.0 if poles conjugatecomplex 2

5.0 if poles real 2

Q

Q

Second-Order Systems – Real Poles (1)

• If poles are spaced by more than 2 decades, there are 2 distinct regions of -45/dec phase slope

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-20dB/dec.

-40dB/dec.

-45/dec.

-45/dec.

1000110

1000100110 4

2

4

sssssH

032.0 Note,

1000 ,1 :poles 2 21

Q

pp

Second-Order Systems – Real Poles (2)

• If poles are spaced by less than 2 decades, there is a region of -90/dec phase slope• Watch out for system stability!

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-20dB/dec.

-40dB/dec.

101100

1011100

2

sssssH

-45/dec.

-45/dec.

-90/dec.

287.0 Note,

10 ,1 :poles 2 21

Q

pp

Second-Order Systems – Complex Poles

35

occurs! peakingfrequency i.e. value,frequency low theexceeds magnitude then the1 if Note,

?near ly particular middle, in the happensWhat

sfrequenciehigh at slope -40dB/dec.

magnitude?frequency high theisWhat 0

magnitude?frequency low theisWhat

120

202

0

201

0

0

2

201

1

20

02

201

Q

Qk

Qj

kjH

kjH

kjH

Qss

ksH

Frequency Peaking w/ Complex Poles

36

QQk

Q

QkT

QQ

Q

kdd

djHd

pk

pk

pk

largefor

411

is peak value the,At

largefor 2

11

0

frequency?peak theis Where

1

2

1

020

20222

0

40

21

2

For k1=1 and 0=1

• Note, phase always crosses -90 at 0

[Schaumann]

Second-Order Systems’ Bode Plots Summary

• 2 real poles Plot with standard Bode plot techniques

• 2 complex poles Approximate as 2 real poles at 0• Past 0 the magnitude decreases at -40dB/dec• From 0.10 to 100 the phase slope is -90°/dec

• A more exact plot of second order systems can be obtained by calculating Q and using the reference plots on the previous slide

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Qk

Qj

kjH 120

202

0

201

0

Next Time• OpAmp Circuits• Reading

• Razavi 8.1,2,4,5

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