lecture 10 - web.stanford.edu · lecture 10 sinusoidal steady-state and frequency response •...

34
EE 102 spring 2001-2002 Handout #19 Lecture 10 Sinusoidal steady-state and frequency response sinusoidal steady-state frequency response Bode plots 10–1

Upload: others

Post on 02-Jun-2020

9 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

EE 102 spring 2001-2002 Handout #19

Lecture 10Sinusoidal steady-state and frequency response

• sinusoidal steady-state

• frequency response

• Bode plots

10–1

Page 2: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Response to sinusoidal input

convolution system with impulse response h, transfer function H

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

Page 3: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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 = H(jω)

Sinusoidal steady-state and frequency response 10–3

Page 4: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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)

• 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

Page 5: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Example

• transfer function H(s) = 1/(s + 1)

• input u(t) = cos t

• SSS output has magnitude |H(j)| = 1/√

2, phase H(j) = −45

u(t) (dashed) & y(t) (solid)

0 5 10 15 20−1

−0.5

0

0.5

1

t

Sinusoidal steady-state and frequency response 10–5

Page 6: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

more generally, if system is stable and the input is asymptoticallysinusoidal, i.e.,

u(t) → (Uejωt

)as t → ∞, then

y(t) → yss(t) = (Ysse

jωt)

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

Page 7: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 8: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 9: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Frequency response plots

frequency response can be plotted in several ways, e.g.,

• H(jω) & H(jω) versus ω

• H(jω) = H(jω) + jH(jω) as a curve in the complex plane (calledNyquist plot)

• |H(jω)| & H(jω) versus ω (called Bode plot)

the most common format is a Bode plot

Sinusoidal steady-state and frequency response 10–9

Page 10: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

example: RC circuit

1Fu(t) y(t)Y (s) =

11 + s

U(s)

H(jω) =1

1 + jω

10−2

10−1

100

101

102

−40

−35

−30

−25

−20

−15

−10

−5

0

ω

20lo

g10|H

(jω)|

10−2

10−1

100

101

102

−90

−60

−30

0

ω

H

(jω)

Sinusoidal steady-state and frequency response 10–10

Page 11: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

ω

20lo

g10|H

(jω)|

10−2

10−1

100

101

102

−120

−100

−80

−60

−40

−20

0

ω

H

(jω)(

deg

rees

)

Sinusoidal steady-state and frequency response 10–11

Page 12: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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 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

Page 13: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Bode plots of products

consider product of transfer functions H = FG:

G Fu y

frequency response magnitude and phase are

20 log10 |H(jω)| = 20 log10 |F (jω)| + 20 log10 |G(jω)|

H(jω) = F (jω) + G(jω)

here we use the fact that for a, b ∈ C, (ab) = a + 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

Page 14: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

example. F (s) = 1/(s + 10), G(s) = 1 + 1/s

10−2

10−1

100

101

102

−45

−40

−35

−30

−25

−20

ω

20lo

g10|F

(jω)|

10−2

10−1

100

101

102

−90

−45

0

ω

F

(jω)

10−2

10−1

100

101

102

0

5

10

15

20

25

30

35

40

ω

20lo

g10|G

(jω)|

10−2

10−1

100

101

102

−90

−45

0

ω

G

(jω)

Sinusoidal steady-state and frequency response 10–14

Page 15: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Bode plot of

H(s) = F (s)G(s) =1 + 1/ss + 10

=s + 1

s(s + 10)

10−2

10−1

100

101

102

−40

−30

−20

−10

0

10

20

ω

20

log

10|H

(jω)|

10−2

10−1

100

101

102

−90

−80

−70

−60

−50

−40

−30

ω

H

(jω)

Sinusoidal steady-state and frequency response 10–15

Page 16: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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|

H(jω) = k +m∑

i=1

(jω − zi) −n∑

i=1

(jω − pi)

(of course k = 0 if k > 0, and k = 180 if k < 0)

Sinusoidal steady-state and frequency response 10–16

Page 17: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Graphical interpretation: |H(jω)|

s = jω

|H(jω)| = |k|∏m

i=1 dist(jω, zi)∏ni=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

Page 18: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Graphical interpretation: H(jω)

s = jω

H(jω) = k+m∑

i=1

(jω−zi)−n∑

i=1

(jω−pi)

therefore, H(jω) changes rapidly when a pole or zero is near jω

Sinusoidal steady-state and frequency response 10–18

Page 19: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

ω

20

log

10|H

(jω)|

10−1

100

101

−540

−360

−180

0

ω

H

(jω)

Sinusoidal steady-state and frequency response 10–19

Page 20: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

ω

20

log

10|H

(jω)|

10−3

10−2

10−1

100

101

102

103

−360

−270

−180

−90

0

ω

H

(jω)

called all-pass filter since gain magnitude is one for all frequencies

Sinusoidal steady-state and frequency response 10–20

Page 21: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

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

Page 22: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

magnitude plot for n = 2, n = 5, n = 10

10−2

10−1

100

101

102

−100

−80

−60

−40

−20

0

ω

20lo

g|H

(jω)|

n = 2

n = 5

n = 10

Sinusoidal steady-state and frequency response 10–22

Page 23: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 24: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

Poles and zeros at s = 0

the term sk has simple Bode plot:

• phase is constant, = 90k

• magnitude has constant slope 20kdB/decade

• magnitude plot intersects 0dB axis at ω = 1

examples:

• integrator (k = −1): = −90, slope is −20dB/decade

• differentiator (k = +1): = 90, slope is +20dB/decade

Sinusoidal steady-state and frequency response 10–24

Page 25: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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, H(jω) = − arctan(ω/|p|)for p > 0, H(jω) = ±180 + arctan(ω/p)

Sinusoidal steady-state and frequency response 10–25

Page 26: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 27: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 28: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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|, H(jω) ≈ 0

• for ω > 10|p|, H(jω) ≈ −90

• in between, phase is approximately linear (on log-log plot)

Sinusoidal steady-state and frequency response 10–28

Page 29: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 30: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 31: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 32: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

example:

H(s) =(s − 1)(s − 3)(s + 1)(s + 3)

=s2 − 4s + 3s2 + 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

Page 33: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

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

Page 34: Lecture 10 - web.stanford.edu · Lecture 10 Sinusoidal steady-state and frequency response • sinusoidal steady-state • frequency response • Bode plots 10–1. Response to sinusoidal

High frequency slope

H(s) =b0 + · · · bmsm

a0 + · · · + ansn

bm, an = 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 (bm/an) − 90(n − m)

Sinusoidal steady-state and frequency response 10–34