Continuous-Time Frequency Response (II) Lecture 28: EECS 20 N April 2, Laurent El Ghaoui

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1 EECS 20 N April 2, 2001 Lecture 28: Continuous-Time Frequency Response (II) Laurent El Ghaoui 1

2 annoucements homework due on Wednesday 4/4 at 11 AM midterm: Friday, 4/6 includes all chapters until chapter 8, included no labs during midterm week still, go to lab sessions for problem discussions a review lecture will take place on Wed 4/4 it is recommended to attend both discussion sessions (they might be slightly different) 2

3 what you should know some keywords related to part II of this course LTI systems, state-space, responses impulses (discrete- and continuous-time) impulse response convolution frequency response 3

4 outline this lecture is a review of lecture 27 (we did not finish it) frequency response of continuous-time systems why bother about continuous-time? 4

5 frequency response for continuous-time systems from lecture 27: let (Sxω) be the zero-state output response of a continuous-time LTI system S to a complex exponential input: xω(t) = e iωt define H(ω) = (Sxω)(0) then: t Reals, (Sxω)(t) = H(ω)xω(t) the function ω H(ω) is called the frequency response of S H(ω) is a complex number 5

6 relationship with impulse response: discrete-time case in discrete-time, we have seen that frequency and impulse responses are related by the discrete-time Fourier transform (DTFT) H(ω) = + m= h(m)e iωm proof: zero-state output to input x(n) = e iωn is given by the convolution sum y(n) = (h x)(n) = + m= h(m)e iω(n m) = e iωn ( + m= h(m)e iωm ) analog in continuous-time? 6

7 continuous-time case we have a similar relationship in continuous-time the zero-state output to input xω(t) = e iωt is given by the convolution integral y(t) = (x h)(t) = = + + xω(t τ)h(τ) dτ e iω(t τ) h(τ) dτ + = e iωt e iωτ h(τ) dτ = e iωt H(ω) }{{} independent of t 7

8 frequency and impulse response in continuous-time we obtain y(t) = H(ω)xω(t), where H(ω) = + e iωt h(t) dt is the continuous-time Fourier transform of the impulse response h we ll see later that this relationship is one-to-one, that is, given the frequency response H we can uniquely determine the impulse response h 8

9 example 1delay the T -second delay has impulse response h(t) = δ(t T ) (since the impulse response is by definition the response to an impulse) we have H(ω) = = = h(t)e iωt dt = (x ω δ)(t ) = e iωt δ(t T )e iωt dt δ(t t)e iωt dt 9

10 another way to look at it the delta Diract function has the following property: for every signal x, δ(t)x(t) dt = x(0) likewise, the delayed impulse satisfies: δ(t T )x(t) dt = x(t ) this immediately tells us that the frequency response for the T -delay is H(ω) = = e iωt δ(t T )e iωt dt 10

11 proof try with δ replaced with its approximation 1/2ɛ if t [ ɛ ɛ] δɛ(t) = 0 otherwise and obtain δɛ(t)x(t) dt = 1 2ɛ ɛ ɛ x(t) dt this is the average of x over the interval [ ɛ ɛ], and tends to x(0) as ɛ 0 11

12 average limit coincides with function value x(t) x(0) average of x(t) over [ ɛ ɛ] PSfrag replacements ɛ +ɛ t ude of frequency response 12

13 example 2: moving average consider the length-three moving average system y(t) = 1 3 t t 3 x(τ) dτ impulse response? by changing τ in t τ in the integral, we obtain 3 y(t) = 1 x(t τ) dτ 3 0 = (h x)(t), where h(t) = 1/3 if t [0 3] 0 otherwise frequency response? 13

14 frequency response of moving average by Fourier transform formula, H(ω) = = h(t)e iωt dt e iωt dt = 1 iω (1 e 3iωt ) = sin(3ωt) 3ω i 1 cos(3ωt) 3ω 14

15 15 as expected, H(ω) vanishes at ω/2π = k/3hz, k Ints frequency ω (in Hertz) PSfrag replacements 0.3 H(ω) magnitude of frequency response magnitude plot

16 interpretation the fact that H(ω) vanishes at f = k/3, k Ints means that if we average over 3 seconds a sine waveform x(t) = sin(2πkt/3) where k Ints, then we get zero (same for cosine) 16

17 17 time (seconds) PSfrag replacements x(t) moving average moving average of waveform is zero

18 why bother with continuous-time discrete-time systems seem to be more useful: Fourier series of discrete-time periodic signals involve finite sums only convolution is a sum, not an integral the Fourier transform is a sum, not an integral LTI systems involve difference equations, not differential equations with derivatives hence, they seem to be more amenable to computers impulse signals are more easily defined and manipulated continuous-time systems seem to correspond to a limit case so, why bother? 18

19 case for continuous-time systems continuous-time signals do arise in natural phenomena: body temperature signal strength at the receiving end of a wireless channel births and deaths in a population continuous-time systems also arise Newton s law of physics (f = mγ) Maxwell electromagnetic equations circuit equations (see lecture 20) 19

20 could we always discretize? we ve seen that we can always discretize a continuous-time system for example, we can replace the convolution integral (h x)(t) = + x(τ)h(t τ) dτ by a convolution sum (h x)(n) = k=+k k= K T x(kt )h(t kt ) dτ where K is large and T is small 20

21 frag replacements time (seconds) x(t) sampled waveform how often should we sample? time (seconds) x

22 some issues the resulting discrete-time system depends on how we sample! (in the previous example, the sampled signal has a quite different behavior) clearly, the smaller the sampling period T, the better the approximation we don t know a priori how often we should sample it is cumbersome to take T > 0, apply discrete-time domain analysis, and then pass to the limit T 0 22

23 motivations for continuous-time systems summary: continuous-time signals and systems arise in many phenomena although one can always sample (discretize) these signals, it is often better to keep the continuous-time framework and develop analysis tools for it even if we ultimately want to sample, learning about continuous-time systems helps understanding the issues of discretization and sampling 23