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22. The input-output relationship for a linear, time-invariant system is described by differential equation y") +5y'()+6y(1)=2x'()+x(1) This system is excited from rest by a unit-strength impulse, i.e., X(t) = 8(t). Find the corresponding response y(t) using Fourier transform methods. 23. A signal x(1) = 2 + cos (215001)+cos (210001)+cos (2.15001). a) Sketch the Fourier transform X b) Signal x() is input to a filter with impulse response (1) given below. In each case, sketch the associated frequency response (i.c., transfer function H D ) and determine the minimum rateſ, at which y(t) can be sampled and then recovered exactly from its sample values. i. h(t) = sinc 20001 ii. h(t) = sinc 6001 iii. h(t) = sinc' 10001

Answer 1

differential equation. time invaraint Given. Linear y'ut) + 5y'Ct) + 6964) - 2 x'(t) +ex(+). Taking Fourier Transform both sides. (jw) Ycjw) +5(jw) Ycjus) +6 Yljw) - 2 jw x(jw)+ ax(jw) . (ET Ww?xjws) Yljw) [uw' +506) +6] - (2jw+1)xcju). Given, XCE )= $(1) = x(jw) = 1 rjw) = (2jW+1) 2jw t o Gw)? 7 5 jw +6 (ju +2) (jwt3). Yljwjut 3 - 3 Yuw)=5 3+jw at jw atw @ Taking Inverse Fourier Transform ( ea tute) & 1 y(t) = 5.2 tuct) -3.8 2!uct) Given , xt) = 2 + Cos(217500+) +cos (21T1000)+cos(21715001) Ac cos 2Tfct FIT > Ac [46-50) +$(878) Ac & Fil Ac-s(f). Taking Fourier Transform X*)=2.$(1) ++ ($(5=500) +$($9500) + a[96-100) +$(51000) 44/45-1952)+(810) AXA) -1500 -1000 -500 500 1000 1500

6 Given. h(t)= sinc 2000t. sinat & FT that -a to a systems sinec 200 = sentit Sinc 2000t - Sin 100011 H(f) 20000 TTT 72000 Sin 2000 t 1000 hlt) -1000 LT x0) het) - yct) YU Y() = X(f).(1) HGH) Μιοστό Y Yomo yuoc 1 Yupon Vuooo yooo ! 42000 - Yo Yo Y2 1160 SUD 1000f 1 Y2 Y2 Y2 | - 1000 -1000-30 f 1000 500 1000 1500 STU-LOW_STO O Taking ya) se). I. Sin3TT 500t +1. Sin 2T1000 0 2000 0 1000 2000 sampling frequency &s , 2 Imad max. frequency pre fs > 2 x 1000 fs > 2000 Hz HA sinc boot. 7600 Min. sampling rate fs= 2000 Hz (11) hit 300 -300 X(t) bit) Yet) when & above ht) is a low pass tilter unth cen tiller with cutoff frequency 600t2. wt) is passed through net). prequencies below Boom' passes attenuates to above BooH . Y(t) = toote sad) oltz f572fmaz → fs lesen Min sampling for the fs=0Hz.

© ht) = sinc`1000t >> SIDEJO H() = Fit {suncioot?. F. 7{sinclooot? H2 V1000 - 1000 1000 a(t) ht) YCt). 1000 H2 passes. H(f) frequencies below through when &(t) passes y(t) = 16oo + Loco Sin (211 500+) 1000 fs 2 fmax. fs > 2X500 -> fs > 1000 Hz Mon. Sampling rate = 1000 Hr.
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