Lecture 10 Sinusoidal steady-state and frequency response

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) = ( 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