Question

(a) Given the following signals: z(t) = { ={ex? exp(-kt) t> 0 0 t<0 sin(Ot) g(t) = **(t) art (i) Explain what the symbol * means in this context and write down the expression for the function y(t). (ii) Compute the energy of the signal x(t) in the time domain. (iii) Using the formulae 1 F[2(t)]() = k + 2ris F(II(t)](s) = sinc(s) It > 1/2 II(t) It < 1/2 sin(TTS) sinc(s) ITS compute the energy of the signal y(t) in the frequency domain. (iv) Find the relation between k and 0 such that the energy of the signal y(t) is half the energy of the signal z(t). (b) The Fourier transform of the following signal (t) = cos (5)n() is 47 cos(2) X(S) 772 – 167232 where II is the rectangle function defined in A2 (a)(iii). Determine the Fourier transform of the function 4 cos (27) y(t) 72 – 167242

Answer 1

(a)(i) The symbol means convolution, i.e.,

.

© aC) *b(!)= facu) b(e-und -

Hence

y (2) = Sin (et) TE * x(2-) Sin cou) *x(2-4) du = ere 2T Thu E -k (t-u) Sin cou) e du = 2T TC U since (t-u) 30 => ust

(ii) The energy of the signal can be computed as

.

СО akt di- E[x] - - Skecelar sek -∞ ب kl = Ге: -2 k ak O

(iii) Let us write the Fourier transform of a function as

.

F [ace](S) = A(9)

Now, we know that

.

A(S) B(5 = act) b(2-) = *

Hence

9 = d()* 2-) = P(S) X$) where act) = Sin Col) TE - Now, given, X (5)= k+anis and F [t1 (2)] (5) Sin (TS) TIS Simcns> ] (6) TI (2) F [ Sin (7.b) ] (0-7(-5) now, d(L) Sint OL T o T. eE T TO Sin (12+) [= = 0 T2 TE :: DC) = 0 F C Sin Crete) T(-s') [s'= 0 os TI TC2 Now, TI(t) = 0,12)? 1/2 =1,Iti< 112 .(-5) = 0, 151>112 =1, IsiZ 1/2 Now, y = D(s) X (s) TI(S) kiaris . The energy of y(t) is co 2 E[y] (+(-5')]2 ds (k-anis) (k+anis) TA * ) 88 [?(-5')]? as k²+40252 72 dole [ac-soglas' k247A A az siz 42 ds' 3 2 TL -1/2 k² 459 2 1/2 as [oc=27 20 73 k²t«²s2 12 ds - 20 7²f² ras k z=ocs dlak ķ dz = 20 s 73 k² a 1+22 O alak 20 Stan 'z. ) O tka tan? (a ak 7² k RT 2 bant (2) RAK since &=21/6,

(iii)

E[y] - Ž E[x] 2 cs, tant ( o ) = 1 / 2 24 Tak TO A k mo, tan-l 60%) = ate 0 2 24 462 TL 9 m T =tan G ok 482 on, k= t lot (24)