Question

3: (Practice Problem)Consider the representation of the process of sampling followed by reconstruction shown below oce=nt) C) Assume that the input signal is Ia(t) = 2 cos(100nt – /4) + cos(300nt + 7/3) -0<t< The frequency response of the reconstruction filter is H.(12) = {T 121</T 10 1921 > A/T (a) Determine the continuous-time Fourier transform X (12) and plot it as a function of N. (b) Assume the fs = 1/T = 500 samples/sec and plot the Fourier transform X (22) as a function of 2 for -27/T<N<27/T. What is the output r(t) in this case. (c) Now, assume that fs = 1/T = 250 samples/sec. Repeat part (b) for this condition.

Answer 1

sit-nt) talt anty | Halast un Replt) = xalt) x Set-no) n = 2 Tourier transoron Xalas n im 2 alt) x y(t) = se-nt) = a (x (2) * y(J] ss & scn-nous) 15 - ?* n=-2 n=-2 xalt) = halt) x E s(t-nT) - 18xlr-nous). n=-d TA Halt) = 2 cos (100at - ^/4) + Cos(30016 +1/3) A Cos (20+) = Ar le Msca-so) te fscata.)

a Xar) jala cae 2x rejala 75 -3004 -1000 3007 100 * jal4 1 Xpla) jala IST 27 7007 Post -3000 1000 100k 300k 500k Tood ws 13000 13007 Xp (6) is replicated at multiples of Ws. ^ 2X500 – 10001 We = 20 ↑ Hocal 지 지 - 5000 = -5000 a f Xx (A) Holes) Xp(2) 2007 gal jal 1000 1007 -3000 3000

(C) When fs = 17 - 250 7 Ws = 500k a Xp (2) 4007 -5007 - 100k - 3000 - 2002 Similiar Components enist for negative side 10001 2000 3007 Wr a goon 25000 This is a part of replicated Component 9) Bounwanted Ox spurious components > Overlapping of replicated components takes place - Reconstruction not possible Nyquist criterion. as in positive side of frequencies so by 2 des g> 6000 mannanimum frequency o f XA(). = 3002 &