ecen325: electronics spring 2021laplace transform circuit example • note that the transient...
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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
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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
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[McGillem]
Laplace Transform of Operations
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[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
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sVsV
tvtv
sHi
o
i
o LL
RC Transfer Function
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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
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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
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123100010tan
10010tan
1010tan
110tan
101000101001010101
1041.110100010100
101010110100010100
1010101
101000101001010101
1111
32222
2222
jjjj
jjjj
jjjj
Numerical Example
[Silva]
Inverse Tangent Function
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• 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
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dB log20dB log10 102
102 jHjHjH
dB
• Phase is typically plotted in degrees
jHjHjH
ReImtan 1
RC Bode Plot Example
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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
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(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
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-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
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+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
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+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
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
37
Qk
Qj
kjH 120
202
0
201
0
Next Time• OpAmp Circuits• Reading
• Razavi 8.1,2,4,5
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