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 sinusoidal input u(t) = cos(ωt) = ejωt + e−jωt /2

t ¡ ¢ output is y(t) = h(τ) cos(ω(t τ)) dτ Z0 − let’s write this as ∞ ∞ y(t) = h(τ) cos(ω(t τ)) dτ h(τ) cos(ω(t τ)) dτ Z0 − − Zt − 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) = h(τ) cos(ω(t τ)) dτ Z0 −

∞ = (1/2) h(τ) ejω(t−τ) + e−jω(t−τ) dτ Z0 ³ ´

∞ ∞ = (1/2)ejωt h(τ)e−jωτ dτ + (1/2)e−jωt h(τ)ejωτ dτ Z0 Z0

= (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(y )/RMS(u) • | | ss 6 H(jω) gives phase shift between u and y • ss 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) = h(t)e−jωtdt Z0 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., u(t) Uejωt → < as t , then ¡ ¢ → ∞ y(t) y (t) = Y ejωt → ss < ss as t , where ¡ ¢ → ∞ Yss = H(jω)U

Y H(jω) = ss 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 Y • ss (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 PSfrag replacements Y (s) = U(s) 1 + s u(t) 1F y(t) 1 H(jω) = 1 + jω

0 0

−5 | )

ω −10

j −30 ) ( −15 ω H j |

−20 ( 10 −25 H 6 −60 log −30 20 PSfrag replacements −35 PSfrag replacements

−40 −2 −1 0 1 2 −90 −2 −1 0 1 2 10 10 10 10 10 10 10 10 10 10 ω ω

Sinusoidal steady-state and frequency response 10–10 example: suspension system of page 8-6 with m = 1, b = 0.5, k = 1,

0.5s + 1 ( 0.75ω2 + 1) j0.5ω3 H(s) = , H(jω) = − − s2 + 0.5s + 1 ω4 1.75ω2 + 1 −

10 0

| 0 −20 ) ω

j −10 −40 ( H | −20 (degrees) −60 ) 10 ω

−30 j

( −80 log H 20 PSfrag replacements −40 PSfrag replacements6 −100

−50 −120 −2 −1 0 1 2 −2 −1 0 1 2 10 10 10 10 10 10 10 10 10 10 ω ω

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 log H(jω) • 10 | | 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 log H(jω) = 20 log F (jω) + 20 log G(jω) 10 | | 10 | | 10 | |

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 −20 | )

ω −25 j ) ( ω

F −30 j | −45 ( 10 −35 F 6 log PSfrag replacements −40 PSfrag replacements 20

−90 −45 −2 −1 0 1 2 −2 −1 0 1 2 10 10 10 10 10 10 10 10 10 10 ω ω

40 0

| 35 )

ω 30 j ( 25 ) ω G | 20 j −45 ( 10

15 G 6

log 10

PSfrag replacements 20 5 PSfrag replacements

0 −90 −2 −1 0 1 2 −2 −1 0 1 2 10 10 10 10 10 10 10 10 10 10 ω ω

Sinusoidal steady-state and frequency response 10–14 Bode plot of

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

−30 20

| 10 −40 ) ω

j −50 0 ) ( ω H j |

−10 ( −60 10 H

−20 6 −70 log

PSfrag replacements 20 −30 PSfrag replacements −80 −40 −2 −1 0 1 2 −90 −2 −1 0 1 2 10 10 10 10 10 10 10 10 10 10 ω ω

Sinusoidal steady-state and frequency response 10–15 Bode plots from factored form rational transfer function H in factored form: (s z ) (s z ) H(s) = k − 1 · · · − m (s p ) (s p ) − 1 · · · − n

m 20 log H(jω) = 20 log k + 20 log jω z 10 | | 10 | | 10 | − i| Xi=1 n 20 log jω p − 10 | − i| Xi=1

m n 6 H(jω) = 6 k + 6 (jω z ) 6 (jω p ) − i − − i Xi=1 Xi=1

(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 m i=1 dist(jω, zi) H(jω) = k n | | | |Qi=1 dist(jω, pi) Q since for u, v C, dist(u, v) = u v ∈ | − |

PSfrag replacements

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

m n 6 H(jω) = 6 k+ 6 (jω z ) 6 (jω p ) − i − − i Xi=1 Xi=1

PSfrag replacements

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, § − § − §

50 0 | ) 0 ω j

) −180 ( ω H j |

−50 ( 10 H

6 −360 log −100

PSfrag replacements 20 PSfrag replacements

−540 −150 −1 0 1 −1 0 1 10 10 10 10 10 10 ω ω

Sinusoidal steady-state and frequency response 10–19 All-pass filter

(s 1)(s 3) H(s) = − − (s + 1)(s + 3)

1 0 | ) 0.5

ω −90 j ) ( ω H j |

0 ( −180 10 H 6 log −0.5 −270

PSfrag replacements 20 PSfrag replacements

−1 −360 −3 −2 −1 0 1 2 3 −3 −2 −1 0 1 2 3 10 10 10 10 10 10 10 10 10 10 10 10 10 10 ω ω

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 p1

p2 1 π/n H(s) = (s p )(s p ) (s p ) − 1 − 2 · · · − n n stable poles, equally spaced π/n pn−1 on the unit circle

pn π/n π/(2n)

Sinusoidal steady-state and frequency response 10–21 magnitude plot for n = 2, n = 5, n = 10

0

−20 | ) ω j ( −40 H

| n = 2

log −60 n = 5 PSfrag replacements 20

−80 n = 10

−100 −2 −1 0 1 2 10 10 10 10 10 ω

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 | | p 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

−30

−40

−50

−60

0

−20 Phase (deg); Magnitude (dB)

−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

−30

−40

−50

−60

−100 Phase (deg); Magnitude (dB) −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

30

20

10

0

80 Phase (deg); Magnitude (dB) 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

0

−50

0

−50 Phase (deg); Magnitude (dB) −100

−150

−200

−250

0 1 2 3 4 5 10 10 10 10 10 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

0

−0.5

−1

0

Phase (deg); Magnitude (dB) −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

b + b sm H(s) = 0 · · · m a + + a sn 0 · · · n b , a = 0 m n 6 for ω large, H(jω) (b /a )(jω)m−n, i.e., ≈ m n 20 log H(jω) 20 log b /a (n m)20 log ω 10 | | ≈ 10 | m n| − − 10

high frequency magnitude slope is approximately 20(n m)dB/decade • − − high frequency phase is approximately 6 (b /a ) 90(n m)◦ • m n − −

Sinusoidal steady-state and frequency response 10–34