Question

Problem 4 (30 pts) This problem explores the sampling theorem and its consequences. Consider the system shown in the figure below, where the two input signals are given to be xi(t) = sinc(10t) and x2(t) = sinc(6t). ylt) z(t) LTI →2;(+) ht) Xalt) P(+) - E81-nt) a) State the sampling theorem. Be sure to include all conditions for its validity. (5 pts) b) What is the minimum frequency at which y(t) must be sampled such that it could be completely reconstructed from its samples? (9 pts) c) Suppose that the impulse response of the LTI system whose input is y(t) and output is z(t) is given by h(t) = sinc(9t - 4), and that the signal z(t) is sampled via impulse train sampling as shown to obtain z(t). Determine the maximum value of the sampling period T such that z(t) can be completely recovered from its samples z.(t). (5 pts) d) Is Z(jo), the Fourier transform of the sampled signal, periodic? Explain your answer, and if periodic, determine the fundamental period. (5 pts) e) Suppose that the sampling period T is chosen such that z(t) can be completely recovered from its samples z(t). Design a system that can be used to recover z(t) from z(t). Draw a complete block diagram for your design. If you employ any LTI systems in your design, you may specify either the impulse response or the frequency response, and be sure to specify the values of any parameters needed for this design. (6 pts)

Answer 1

e) when Tc Tman 14 to > Ismina Zs(t) is identical to ZCF) with amplitude saaled by yt: Zs(f), HI(4) (f) HICA: VTT f -4.5 45 CS Scanned with CamScanner

2 x 9 manimum value the q Sampling peliod a such that z ct) cah be completely recoveled from it Samples 2s(t) is (Ts) man zitem = 1 (Tuman ㅗ sec. 9 d) Digital signal proces essing conversion Time Domain Frequency Domain Periodic > Discrele a periodic Continons Discrete periodic Continons aperiodic As we know after Samping z ct) is time domain so will be periodic in frequency domain and its findam ental period will be fs. 2 fm CS Scanned with Cañs her discrele ofs = 91.2

b) rate or a=6 ylt)a Ni(t).02 (t) To frequency domain Y61) - XI (+)* X264) *44-1) = we know for sinc (at), sampling sampling frequency of te PS Sin clat) Ma: fs H(E)=sinc (10t) a:10 fs 10 Hz 212 (t) = - Sin c (6t) ; fs2 - 6 HZ YH): dilt). 12 (7) we know the propelly of Nyquist sampling rate (or frequency &s) for multiplication is given by fs . 15 mi(t) m2 (t) fsg fsz = mi (t).m2lt) fsi + 4s2 fs = 1076 16 H2 - Minimum Sampling frequently for c) Zlt)-ylt) *hlt) (1)-46) HCF) het) sinccat-4) - Sac[2(t-4] H(A) = a red (the Ilmited to 9/2t2 and yef) is and 2(F) is multiplication limited to 8H2 q Ylf) and H (f), so ZCF) will be limited to 9/2H2 ts) milt) fs e J2 5144 9 is band H() band band OS Scanned with CamScanner

Given System Xi(t). Y(+). 2(t) LTT >2slt) X₂ (t) hlt). to NE) - Ed(t-nt) no Hi(t) = sin cclot) 12 (+) = sinc (6t) the field of a time for a rala that to Sampling theorem states that "In Processing a digital signal, which serves as bridge between continous time signal & discrete signal. It establishes a sufficient condition Sample permite a discrele sequence samples q capture all the information from a continons time signal of finite bandwidth let fs be sampling frequency fmy frequency of signal (man) then Valid conditions fs >2fm " Signal should be q finite fequency 2) Signal is band to tm thaen Sampling interval should be less than 1/2 fm. will be - limited Sampling interval Ts ste 2 fm sampling frequency fs>, 2fm CS Scanned with CamScanner