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
y
H
¡
jωt
sinusoidal input u(t) = cos(ωt) = e + e
Z t
h(τ ) cos(ω(t − τ )) dτ
output is y(t) =
−jωt
¢
/2
0
let’s write this as
Z
Z ∞
h(τ ) cos(ω(t − τ )) dτ −
y(t) =
0
∞
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 the
transient
Sinusoidal steady-state and frequency response
10–2
if system is stable, sinusoidal steady-state response can be expressed as
ysss(t) =
Z
∞
0
h(τ ) cos(ω(t − τ )) dτ
= (1/2)
Z
= (1/2)e
∞
³
h(τ ) ejω(t−τ ) + e−jω(t−τ )
0
jωt
Z
´
dτ
∞
h(τ )e
−jωτ
dτ + (1/2)e
−jωt
0
Z
∞
h(τ )ejωτ dτ
0
= (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 is
asymptotically sinusoidal, with the same frequency as the input, and with
magnitude & 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.,
Z ∞
h(t)e−jωtdt
H(jω) =
0
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)
1
0.5
0
−0.5
PSfrag replacements
−1
0
5
10
15
20
t
Sinusoidal steady-state and frequency response
10–5
more generally, if system is stable and the input is asymptotically
sinusoidal, i.e.,
¡ jωt¢
u(t) → < Ue
as t → ∞, then
as t → ∞, where
¡
y(t) → yss(t) = < Ysse
jωt
¢
Yss = H(jω)U
Yss
H(jω) =
U
for a stable system, H(jω) gives ratio of phasors of asymptotic sinusoidal
output & 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 with
frequnecy ω is large
• H(0) = 2 means asymptotic response to a constant signal is twice the
input value
• H(jω) small for large ω means the asymptotic output for high
frequency 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 (called
Nyquist 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
1Ω
1
Y (s) =
U (s)
1+s
rag replacements
u(t)
y(t)
1F
H(jω) =
0
−5
−15
−20
H(jω)
−10
−30
6
20 log10 |H(jω)|
0
replacements
1
1 + jω
−60
−25
−30
PSfrag replacements
−35
−40 −2
10
−1
10
0
10
ω
1
10
Sinusoidal steady-state and frequency response
2
10
−90 −2
10
−1
10
0
10
ω
1
10
2
10
10–10
example: suspension system of page 8-6 with m = 1, b = 0.5, k = 1,
0.5s + 1
,
H(s) = 2
s + 0.5s + 1
(−0.75ω 2 + 1) − j0.5ω 3
H(jω) =
ω 4 − 1.75ω 2 + 1
replacements
0
H(jω)(degrees)
0
−10
−20
−30
PSfrag replacements
−40
−50 −2
10
6
20 log10 |H(jω)|
10
−1
10
0
10
ω
1
10
Sinusoidal steady-state and frequency response
2
10
−20
−40
−60
−80
−100
−120
−2
10
−1
10
0
10
ω
1
10
2
10
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 magnitudes
ratio 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◦ (or
0, 360◦); it is called unwrapped if multiple of 360◦ is chosen to make
phase plot continuous
Sinusoidal steady-state and frequency response
10–12
Bode plots of products
consider product of transfer functions H = F G:
PSfrag replacements
u
G
F
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
0
ag replacements
F (jω)
−25
−30
−35
PSfrag replacements
−40
−45 −2
10
−1
10
0
10
ω
1
10
−90
−2
10
2
10
−1
10
0
10
ω
1
10
2
10
0
35
G(jω)
30
25
20
15
−45
6
20 log10 |G(jω)|
40
ag replacements
−45
6
20 log10 |F (jω)|
−20
10
PSfrag replacements
5
0
−2
10
−1
10
0
10
ω
1
10
Sinusoidal steady-state and frequency response
2
10
−90
−2
10
−1
10
0
10
ω
1
10
2
10
10–14
Bode plot of
s+1
1 + 1/s
=
H(s) = F (s)G(s) =
s + 10
s(s + 10)
−30
−40
0
H(jω)
replacements
10
−50
6
20 log10 |H(jω)|
20
−70
PSfrag replacements
−80
−10
−20
−30
−40
−2
10
−1
10
0
10
ω
1
10
Sinusoidal steady-state and frequency response
2
10
−60
−90 −2
10
−1
10
0
10
ω
1
10
2
10
10–15
Bode plots from factored form
rational transfer function H in factored form:
(s − z1) · · · (s − zm)
H(s) = k
(s − p1) · · · (s − pn)
20 log10 |H(jω)| = 20 log10 |k| +
−
6
H(jω) =
6
n
X
i=1
k+
m
X
i=1
20 log10 |jω − zi|
20 log10 |jω − pi|
m
X
i=1
6
(jω − zi) −
n
X
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)|
s = jω
Qm
dist(jω, zi)
|H(jω)| = |k| Qni=1
i=1 dist(jω, pi )
since for u, v ∈ C, dist(u, v) = |u − v|
placements
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)
s = jω
6
H(jω) = 6 k+
m
X
i=1
6
(jω−zi)−
n
X
6
(jω−pi)
i=1
placements
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,
0
H(jω)
0
−50
6
20 log10 |H(jω)|
50
−180
−360
−100
replacements
PSfrag replacements
−150 −1
10
0
10
ω
Sinusoidal steady-state and frequency response
1
10
−540
−1
10
0
10
ω
1
10
10–19
All-pass filter
(s − 1)(s − 3)
(s + 1)(s + 3)
0
0.5
−90
H(jω)
1
0
−180
6
20 log10 |H(jω)|
H(s) =
−0.5
−270
replacements
PSfrag replacements
−1 −3
10
−2
10
−1
10
0
10
ω
1
10
2
10
3
10
−360 −3
10
−2
10
−1
10
0
10
ω
1
10
2
10
3
10
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
PSfrag replacements
π/(2n)
π/n
H(s) =
1
(s − p1)(s − p2) · · · (s − pn)
n stable poles, equally spaced
on the unit circle
p1
p2
π/n
π/n
pn−1
pn
π/n
π/(2n)
Sinusoidal steady-state and frequency response
10–21
magnitude plot for n = 2, n = 5, n = 10
PSfrag replacements
20 log |H(jω)|
0
−20
−40
n=2
−60
n=5
−80
−100 −2
10
Sinusoidal steady-state and frequency response
n = 10
−1
10
0
10
ω
1
10
2
10
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):
Sinusoidal steady-state and frequency response
6
= 90◦, slope is +20dB/decade
10–24
Real poles and zeros
H(s) = 1/(s − p) (p < 0 is stable pole; p > 0 is unstable pole)
p
magnitude: |H(jω)| = 1/ ω 2 + 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):
Bode Diagrams
0
−10
−20
Phase (deg); Magnitude (dB)
−30
−40
−50
−60
0
−20
−40
−60
−80
0.001
0.01
0.1
1
10
100
1000
Frequency (rad/sec)
Sinusoidal steady-state and frequency response
10–26
Bode plot for H(s) = 1/(s − 1) (unstable pole):
Bode Diagrams
0
−10
−20
Phase (deg); Magnitude (dB)
−30
−40
−50
−60
−100
−120
−140
−160
−180
0.001
0.01
0.1
1
10
100
1000
Frequency (rad/sec)
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
Bode Diagrams
60
50
40
Phase (deg); Magnitude (dB)
30
20
10
0
80
60
40
20
0
0.001
0.01
0.1
1
10
100
1000
Frequency (rad/sec)
Sinusoidal steady-state and frequency response
10–29
example:
104
H(s) =
(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:
Bode Diagrams
50
Phase (deg); Magnitude (dB)
0
−50
0
−50
−100
−150
−200
−250
0
10
1
10
2
10
3
10
4
10
5
10
Frequency (rad/sec)
Sinusoidal steady-state and frequency response
10–31
example:
(s − 1)(s − 3) s2 − 4s + 3
=
H(s) =
(s + 1)(s + 3) s2 + 4s + 3
|H(jω)| = 1 (obvious from graphical interpretation)
(called all-pass filter or phase filter since gain magnitude is one for all
frequencies)
Sinusoidal steady-state and frequency response
10–32
Bode plot:
Bode Diagrams
1
0.5
Phase (deg); Magnitude (dB)
0
−0.5
−1
0
−100
−200
−300
0.001
0.01
0.1
1
10
100
1000
Frequency (rad/sec)
Sinusoidal steady-state and frequency response
10–33
High frequency slope
b0 + · · · b m s m
H(s) =
a0 + · · · + a n sn
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