S. Boyd EE102 Lecture 10 Sinusoidal steady-state and frequency response sinusoidal steady-state ² frequency response ² Bode plots ² 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(!t) = ej!t + e¡j!t =2 t ¡ ¢ output is y(t) = h(¿) cos(!(t ¿)) d¿ Z0 ¡ let's write this as 1 1 y(t) = h(¿) cos(!(t ¿)) d¿ h(¿) cos(!(t ¿)) d¿ Z0 ¡ ¡ Zt ¡ ¯rst term is called sinusoidal steady-state response ² second term decays with t if system is stable; if it decays it is called the ² transient Sinusoidal steady-state and frequency response 10{2 if system is stable, sinusoidal steady-state response can be expressed as 1 ysss(t) = h(¿) cos(!(t ¿)) d¿ Z0 ¡ 1 = (1=2) h(¿) ej!(t¡¿) + e¡j!(t¡¿) d¿ Z0 ³ ´ 1 1 = (1=2)ej!t h(¿)e¡j!¿ d¿ + (1=2)e¡j!t h(¿)ej!¿ d¿ Z0 Z0 = (1=2)ej!tH(j!) + (1=2)e¡j!tH( j!) ¡ = ( H(j!)) cos(!t) ( H(j!)) sin(!t) < ¡ = = a cos(!t + Á) where a = H(j!) , Á = 6 H(j!) j j 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(j!) H(j!) gives ampli¯cation factor, i.e., RMS(y )=RMS(u) ² j j ss 6 H(j!) gives phase shift between u and y ² ss special case: u(t) = 1 (i.e., ! = 0); output converges to H(0) (DC gain) frequency response transfer function evaluated at s = j!, i.e., 1 H(j!) = h(t)e¡j!tdt Z0 is called frequency response of the system since H( j!) = H(j!), we usually only consider ! 0 ¡ ¸ Sinusoidal steady-state and frequency response 10{4 Example transfer function H(s) = 1=(s + 1) ² input u(t) = cos t ² SSS output has magnitude H(j) = 1=p2, phase 6 H(j) = 45± ² j j ¡ 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) Uej!t ! < as t , then ¡ ¢ ! 1 y(t) y (t) = Y ej!t ! ss < ss as t , where ¡ ¢ ! 1 Yss = H(j!)U Y H(j!) = ss U for a stable system, H(j!) 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(j!) large means asymptotic response of system to sinusoid with ² frequnecy ! is large H(0) = 2 means asymptotic response to a constant signal is twice the ² input value H(j!) small for large ! means the asymptotic output for high ² frequency sinusoids is small Sinusoidal steady-state and frequency response 10{7 Measuring frequency response for ! = !1; : : : ; !N , apply sinusoid at frequency !, with phasor U ² wait for output to converge to SSS ² measure Y ² 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(j!) & H(j!) versus ! ² < = H(j!) = H(j!) + j H(j!) as a curve in the complex plane (called ² Nyquist plot< ) = H(j!) & 6 H(j!) versus ! (called Bode plot) ² j j the most common format is a Bode plot Sinusoidal steady-state and frequency response 10{9 example: RC circuit 1­ 1 PSfrag replacements Y (s) = U(s) 1 + s u(t) 1F y(t) 1 H(j!) = 1 + j! 0 0 −5 j ) ! −10 j −30 ) ( −15 ! H j 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 ! ! 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 ( 0:75!2 + 1) j0:5!3 H(s) = ; H(j!) = ¡ ¡ s2 + 0:5s + 1 !4 1:75!2 + 1 ¡ 10 0 j 0 −20 ) ! j −10 −40 ( H j −20 (degrees) −60 ) 10 ! −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 ! ! Sinusoidal steady-state and frequency response 10{11 Bode plots frequency axis logarithmic scale for ! ² horizontal distance represents a ¯xed frequency ratio or factor: ² ratio 2 : 1 is called an octave; ratio 10 : 1 is called a decade magnitude H(j!) j j expressed in dB, i.e., 20 log H(j!) ² 10 j j vertical distance represents dB, i.e., a ¯xed ratio of magnitudes ² ratio 2 : 1 is +6dB, ratio 10 : 1 is +20dB slopes are given in units such as dB/octave or dB/decade ² phase 6 H(j!) multiples of 360± don't matter ² phase plot is called wrapped when phases are between 180± (or ² 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(j!) = 20 log F (j!) + 20 log G(j!) 10 j j 10 j j 10 j j 6 H(j!) = 6 F (j!) + 6 G(j!) here we use the fact that for a; b C, 6 (ab) = 6 a + 6 b 2 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 j ) ! −25 j ) ( ! F −30 j 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 ! ! 40 0 j 35 ) ! 30 j ( 25 ) ! G j 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 ! ! 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 j 10 −40 ) ! j −50 0 ) ( ! H j 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 ! ! 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(j!) = 20 log k + 20 log j! z 10 j j 10 j j 10 j ¡ ij Xi=1 n 20 log j! p ¡ 10 j ¡ ij Xi=1 m n 6 H(j!) = 6 k + 6 (j! z ) 6 (j! 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(j!) j j s = j! m i=1 dist(j!; zi) H(j!) = k n j j j jQi=1 dist(j!; pi) Q since for u; v C, dist(u; v) = u v 2 j ¡ j PSfrag replacements therefore, e.g.: H(j!) gets big when j! is near a pole ² j j H(j!) gets small when j! is near a zero ² j j Sinusoidal steady-state and frequency response 10{17 Graphical interpretation: 6 H(j!) s = j! m n 6 H(j!) = 6 k+ 6 (j! z ) 6 (j! p ) ¡ i ¡ ¡ i Xi=1 Xi=1 PSfrag replacements therefore, 6 H(j!) changes rapidly when a pole or zero is near j! 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 j ) 0 ! j ) −180 ( ! H j 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 ! ! Sinusoidal steady-state and frequency response 10{19 All-pass ¯lter (s 1)(s 3) H(s) = ¡ ¡ (s + 1)(s + 3) 1 0 j ) 0.5 ! −90 j ) ( ! H j 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 ! ! called all-pass ¯lter since gain magnitude is one for all frequencies Sinusoidal steady-state and frequency response 10{20 Analog lowpass ¯lters analog lowpass ¯lters: approximate ideal lowpass frequency response H(j!) = 1 for 0 ! 1; H(j!) = 0 for ! > 1 j j · · j j by a rational transfer function (which can be synthesized using R, L, C) example nth-order Butterworth ¯lter PSfrag replacements ¼=(2n) ¼=n p1 p2 1 ¼=n H(s) = (s p )(s p ) (s p ) ¡ 1 ¡ 2 ¢ ¢ ¢ ¡ n n stable poles, equally spaced ¼=n pn¡1 on the unit circle pn ¼=n ¼=(2n) Sinusoidal steady-state and frequency response 10{21 magnitude plot for n = 2, n = 5, n = 10 0 −20 j ) ! j ( −40 H j n = 2 log −60 n = 5 PSfrag replacements 20 −80 n = 10 −100 −2 −1 0 1 2 10 10 10 10 10 ! Sinusoidal steady-state and frequency response 10{22 Sketching approximate Bode plots sum property allows us to ¯nd Bode plots of terms in TF, then add simple terms: constant ² factor of sk (pole or zero at s = 0) ² real pole, real zero ² complex pole or zero pair ² 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± ² magnitude has constant slope 20kdB=decade ² magnitude plot intersects 0dB axis at ! = 1 ² examples: integrator (k = 1): 6 = 90±, slope is 20dB=decade ² ¡ ¡ ¡ di®erentiator (k = +1): 6 = 90±, slope is +20dB=decade ² 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(j!) = 1= !2 + p2 j j p for p < 0, 6 H(j!) = arctan(!= p ) ¡ j j for p > 0, 6 H(j!) = 180± + arctan(!=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