Question

Consider a mysterious continuous-time LTI system that has a one-sided decaying exponential im- pulse response given by h(t) = exp(-at)u(t); the mystery lies in exactly what a is. Suppose all we have to test our system with is an old, broken signal generator that's stuck generating a cosine wave with phase 0, an amplitude of 200, and a frequency of 40 radians. (Yes, that's 40 radians, not 407T radians; we're making it exactly 40 radians so the math in part (a) works out nicely.) When given this input, the output is a cosine wave with amplitude 4 and a frequency of 40 radians (a) Find the value of a using the information given in the problem. (b) Find thee phase of the output cosine. (We sometimes try to set up our homework and quiz problems so that you can do them without a calculator, but for this one, you'll need to go ahead and pull out your calculator.)

PLEASE SHOW ALL YOUR WORK

Answer 1

Ans:- Gifven that, ht) e at utt. for decau ing function a is positive * 又しも)= A,CoSs (wt te) i.e aro - 300 COS (40t) (input) ct) A2 CO S(nt +02) 4 cos (4ot +02) toutput) yt) = Htw) A Cos (tut t 0,+ [H(w)) 200 COs (uot +1H(o)) 0 By comparing y a00 H(w)4 4 COS (40t t02) oith eqn & O2= LACW) Continuous time Fourier Transfom of ht) HlLo) /tan (u/a) 14 jo+a H(ta)= and LHCw)tan (a) eat ut) CTFT atiuo foraro 4 =(5o-ub) (40ta = 50 a Vuo ta 200 a30 for decaying choose a t30 exptnen tial function Cayo) 30 11 fM phase of the Output cosine 0, = /H(w) 6aH(lo)tan(a) ()53.13 -tan 30 - 53.13 = -53.130 J0ttAe tt)A cos (uot-53.13).