lecture 10 sinusoidal steady-state and frequency responseweb.stanford.edu/~boyd/ee102/freq.pdfs....
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S. Boyd EE102
Lecture 10
Sinusoidal steady-state and frequency response
• sinusoidal steady-state
• frequency response
• Bode plots
10–1
Response to sinusoidal input
convolution system with impulse response h, transfer function H
PSfrag replacements
u yH
sinusoidal input u(t) = cos(ωt) =(
ejωt + e−jωt)
/2
output is y(t) =
∫ t
0
h(τ) cos(ω(t− τ)) dτ
let’s write this as
y(t) =
∫ ∞
0
h(τ) cos(ω(t− τ)) dτ −∫ ∞
t
h(τ) cos(ω(t− τ)) dτ
• first term is called sinusoidal steady-state response
• second term decays with t if system is stable; if it decays it is called thetransient
Sinusoidal steady-state and frequency response 10–2
if system is stable, sinusoidal steady-state response can be expressed as
ysss(t) =
∫ ∞
0
h(τ) cos(ω(t− τ)) dτ
= (1/2)
∫ ∞
0
h(τ)(
ejω(t−τ) + e−jω(t−τ))
dτ
= (1/2)ejωt∫ ∞
0
h(τ)e−jωτ dτ + (1/2)e−jωt
∫ ∞
0
h(τ)ejωτ dτ
= (1/2)ejωtH(jω) + (1/2)e−jωtH(−jω)
= (<H(jω)) cos(ωt)− (=H(jω)) sin(ωt)
= a cos(ωt+ φ)
where a = |H(jω)|, φ = 6 H(jω)
Sinusoidal steady-state and frequency response 10–3
conclusion
if the convolution system is stable, the response to a sinusoidal input isasymptotically sinusoidal, with the same frequency as the input, and withmagnitude & phase determined by H(jω)
• |H(jω)| gives amplification factor, i.e., RMS(yss)/RMS(u)
• 6 H(jω) gives phase shift between u and yss
special case: u(t) = 1 (i.e., ω = 0); output converges to H(0) (DC gain)
frequency response
transfer function evaluated at s = jω, i.e.,
H(jω) =
∫ ∞
0
h(t)e−jωtdt
is called frequency response of the system
since H(−jω) = H(jω), we usually only consider ω ≥ 0
Sinusoidal steady-state and frequency response 10–4
Example
• transfer function H(s) = 1/(s+ 1)
• input u(t) = cos t
• SSS output has magnitude |H(j)| = 1/√
2, phase 6 H(j) = −45
u(t) (dashed) & y(t) (solid)
0 5 10 15 20−1
−0.5
0
0.5
1
PSfrag replacements
t
Sinusoidal steady-state and frequency response 10–5
more generally, if system is stable and the input is asymptoticallysinusoidal, i.e.,
u(t) → <(
Uejωt)
as t→∞, theny(t)→ yss(t) = <
(
Yssejωt
)
as t→∞, whereYss = H(jω)U
H(jω) =Yss
U
for a stable system, H(jω) gives ratio of phasors of asymptotic sinusoidaloutput & input
Sinusoidal steady-state and frequency response 10–6
thus, for example (assuming a stable system),
• H(jω) large means asymptotic response of system to sinusoid withfrequnecy ω is large
• H(0) = 2 means asymptotic response to a constant signal is twice theinput value
• H(jω) small for large ω means the asymptotic output for highfrequency sinusoids is small
Sinusoidal steady-state and frequency response 10–7
Measuring frequency response
for ω = ω1, . . . , ωN ,
• apply sinusoid at frequency ω, with phasor U
• wait for output to converge to SSS
• measure Yss
(i.e., magnitude and phase shift of yss)
N can be a few tens (for hand measurements) to several thousand
Sinusoidal steady-state and frequency response 10–8
Frequency response plots
frequency response can be plotted in several ways, e.g.,
• <H(jω) & =H(jω) versus ω
• H(jω) = <H(jω) + j=H(jω) as a curve in the complex plane (calledNyquist plot)
• |H(jω)| & 6 H(jω) versus ω (called Bode plot)
the most common format is a Bode plot
Sinusoidal steady-state and frequency response 10–9
example: RC circuit
PSfrag replacements
1Ω
1Fu(t) y(t)Y (s) =
1
1 + sU(s)
H(jω) =1
1 + jω
10−2
10−1
100
101
102
−40
−35
−30
−25
−20
−15
−10
−5
0
PSfrag replacements
ω
20log
10|H(jω)|
10−2
10−1
100
101
102
−90
−60
−30
0
PSfrag replacements
ω
6H(jω)
Sinusoidal steady-state and frequency response 10–10
example: suspension system of page 8-6 with m = 1, b = 0.5, k = 1,
H(s) =0.5s+ 1
s2 + 0.5s+ 1, H(jω) =
(−0.75ω2 + 1)− j0.5ω3
ω4 − 1.75ω2 + 1
10−2
10−1
100
101
102
−50
−40
−30
−20
−10
0
10
PSfrag replacements
ω
20log
10|H(jω)|
10−2
10−1
100
101
102
−120
−100
−80
−60
−40
−20
0
PSfrag replacements
ω
6H(jω)(
deg
rees
)
Sinusoidal steady-state and frequency response 10–11
Bode plots
frequency axis
• logarithmic scale for ω
• horizontal distance represents a fixed frequency ratio or factor:ratio 2 : 1 is called an octave; ratio 10 : 1 is called a decade
magnitude |H(jω)|• expressed in dB, i.e., 20 log10 |H(jω)|• vertical distance represents dB, i.e., a fixed ratio of magnitudesratio 2 : 1 is +6dB, ratio 10 : 1 is +20dB
• slopes are given in units such as dB/octave or dB/decade
phase 6 H(jω)
• multiples of 360 don’t matter
• phase plot is called wrapped when phases are between ±180 (or0, 360); it is called unwrapped if multiple of 360 is chosen to makephase plot continuous
Sinusoidal steady-state and frequency response 10–12
Bode plots of products
consider product of transfer functions H = FG:PSfrag replacements
G Fu y
frequency response magnitude and phase are
20 log10 |H(jω)| = 20 log10 |F (jω)|+ 20 log10 |G(jω)|
6 H(jω) = 6 F (jω) + 6 G(jω)
here we use the fact that for a, b ∈ C, 6 (ab) = 6 a+ 6 b
so, Bode plot of a product is the sum of the Bode plots of each term(extends to many terms)
Sinusoidal steady-state and frequency response 10–13
example. F (s) = 1/(s+ 10), G(s) = 1 + 1/s
10−2
10−1
100
101
102
−45
−40
−35
−30
−25
−20
PSfrag replacements
ω
20log
10|F(jω)|
10−2
10−1
100
101
102
−90
−45
0
PSfrag replacements
ω
6F(jω)
10−2
10−1
100
101
102
0
5
10
15
20
25
30
35
40
PSfrag replacements
ω
20log
10|G(jω)|
10−2
10−1
100
101
102
−90
−45
0
PSfrag replacements
ω
6G(jω)
Sinusoidal steady-state and frequency response 10–14
Bode plot of
H(s) = F (s)G(s) =1 + 1/s
s+ 10=
s+ 1
s(s+ 10)
10−2
10−1
100
101
102
−40
−30
−20
−10
0
10
20
PSfrag replacements
ω
20log
10|H(jω)|
10−2
10−1
100
101
102
−90
−80
−70
−60
−50
−40
−30
PSfrag replacements
ω
6H(jω)
Sinusoidal steady-state and frequency response 10–15
Bode plots from factored form
rational transfer function H in factored form:
H(s) = k(s− z1) · · · (s− zm)
(s− p1) · · · (s− pn)
20 log10 |H(jω)| = 20 log10 |k|+m∑
i=1
20 log10 |jω − zi|
−n∑
i=1
20 log10 |jω − pi|
6 H(jω) = 6 k +
m∑
i=1
6 (jω − zi)−n∑
i=1
6 (jω − pi)
(of course 6 k = 0 if k > 0, and 6 k = 180 if k < 0)
Sinusoidal steady-state and frequency response 10–16
Graphical interpretation: |H(jω)|
PSfrag replacements
s = jω
|H(jω)| = |k|∏m
i=1 dist(jω, zi)∏n
i=1 dist(jω, pi)
since for u, v ∈ C, dist(u, v) = |u− v|
therefore, e.g.:
• |H(jω)| gets big when jω is near a pole
• |H(jω)| gets small when jω is near a zero
Sinusoidal steady-state and frequency response 10–17
Graphical interpretation: 6 H(jω)
PSfrag replacements
s = jω
6 H(jω) = 6 k+
m∑
i=1
6 (jω−zi)−n∑
i=1
6 (jω−pi)
therefore, 6 H(jω) changes rapidly when a pole or zero is near jω
Sinusoidal steady-state and frequency response 10–18
Example with lightly damped poles
poles at s = −0.01± j0.2, s = −0.01± j0.7,s = −0.01± j1.3,
10−1
100
101
−150
−100
−50
0
50
PSfrag replacements
ω
20log
10|H(jω)|
10−1
100
101
−540
−360
−180
0
PSfrag replacements
ω
6H(jω)
Sinusoidal steady-state and frequency response 10–19
All-pass filter
H(s) =(s− 1)(s− 3)
(s+ 1)(s+ 3)
10−3
10−2
10−1
100
101
102
103
−1
−0.5
0
0.5
1
PSfrag replacements
ω
20log
10|H(jω)|
10−3
10−2
10−1
100
101
102
103
−360
−270
−180
−90
0
PSfrag replacements
ω
6H(jω)
called all-pass filter since gain magnitude is one for all frequencies
Sinusoidal steady-state and frequency response 10–20
Analog lowpass filters
analog lowpass filters: approximate ideal lowpass frequency response
|H(jω)| = 1 for 0 ≤ ω ≤ 1, |H(jω)| = 0 for ω > 1
by a rational transfer function (which can be synthesized using R, L, C)
example nth-order Butterworth filter
H(s) =1
(s− p1)(s− p2) · · · (s− pn)
n stable poles, equally spacedon the unit circle
PSfrag replacements
p1
p2
pn−1
pn
π/(2n)
π/n
π/n
π/n
π/n
π/(2n)
Sinusoidal steady-state and frequency response 10–21
magnitude plot for n = 2, n = 5, n = 10
10−2
10−1
100
101
102
−100
−80
−60
−40
−20
0
PSfrag replacements
ω
20log|H(jω)|
n = 2
n = 5
n = 10
Sinusoidal steady-state and frequency response 10–22
Sketching approximate Bode plots
sum property allows us to find Bode plots of terms in TF, then add
simple terms:
• constant
• factor of sk (pole or zero at s = 0)
• real pole, real zero
• complex pole or zero pair
from these we can construct Bode plot of any rational transfer function
Sinusoidal steady-state and frequency response 10–23
Poles and zeros at s = 0
the term sk has simple Bode plot:
• phase is constant, 6 = 90k
• magnitude has constant slope 20kdB/decade
• magnitude plot intersects 0dB axis at ω = 1
examples:
• integrator (k = −1): 6 = −90, slope is −20dB/decade
• differentiator (k = +1): 6 = 90, slope is +20dB/decade
Sinusoidal steady-state and frequency response 10–24
Real poles and zeros
H(s) = 1/(s− p) (p < 0 is stable pole; p > 0 is unstable pole)
magnitude: |H(jω)| = 1/√
ω2 + p2
for p < 0, 6 H(jω) = − arctan(ω/|p|)
for p > 0, 6 H(jω) = ±180 + arctan(ω/p)
Sinusoidal steady-state and frequency response 10–25
Bode plot for H(s) = 1/(s+ 1) (stable pole):
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−60
−50
−40
−30
−20
−10
0
0.001 0.01 0.1 1 10 100 1000
−80
−60
−40
−20
0
Sinusoidal steady-state and frequency response 10–26
Bode plot for H(s) = 1/(s− 1) (unstable pole):
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−60
−50
−40
−30
−20
−10
0
0.001 0.01 0.1 1 10 100 1000−180
−160
−140
−120
−100
magnitude same as stable pole; phase starts at −180, increases
Sinusoidal steady-state and frequency response 10–27
straight-line approximation (for p < 0):
• for ω < |p|, |H(jω)| ≈ 1/|p|• for ω > |p|, |H(jω)| decreases (‘falls off’) 20dB per decade
• for ω < 0.1|p|, 6 H(jω) ≈ 0
• for ω > 10|p|, 6 H(jω) ≈ −90
• in between, phase is approximately linear (on log-log plot)
Sinusoidal steady-state and frequency response 10–28
Bode plots for real zeros same as poles but upside down
example: H(s) = s+ 1
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
0
10
20
30
40
50
60
0.001 0.01 0.1 1 10 100 10000
20
40
60
80
Sinusoidal steady-state and frequency response 10–29
example:
H(s) =104
(1 + s/10)(1 + s/300)(1 + s/3000)
(typical op-amp transfer function)
DC gain 80dB; poles at −10, −300, −3000
Sinusoidal steady-state and frequency response 10–30
Bode plot:
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−50
0
50
100
101
102
103
104
105
−250
−200
−150
−100
−50
0
Sinusoidal steady-state and frequency response 10–31
example:
H(s) =(s− 1)(s− 3)
(s+ 1)(s+ 3)=s2 − 4s+ 3
s2 + 4s+ 3
|H(jω)| = 1 (obvious from graphical interpretation)
(called all-pass filter or phase filter since gain magnitude is one for allfrequencies)
Sinusoidal steady-state and frequency response 10–32
Bode plot:
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−1
−0.5
0
0.5
1
0.001 0.01 0.1 1 10 100 1000
−300
−200
−100
0
Sinusoidal steady-state and frequency response 10–33
High frequency slope
H(s) =b0 + · · · bmsma0 + · · ·+ ansn
bm, an 6= 0
for ω large, H(jω) ≈ (bm/an)(jω)m−n, i.e.,
20 log10 |H(jω)| ≈ 20 log10 |bm/an| − (n−m)20 log10 ω
• high frequency magnitude slope is approximately −20(n−m)dB/decade
• high frequency phase is approximately 6 (bm/an)− 90(n−m)
Sinusoidal steady-state and frequency response 10–34
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