Question

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7. (12 points) A time-domain equalizer is often used in digital communication to reduce the interferences between transmitted signals. The following diagram shows the implementation of the equalizer which is also called digital transversal filter. All the weights or coefficients in the z-domain diagram are considered constant. x[n] z-1 z-1 y[n] a) find the difference equation that represents the digital transversal filter. Is it an IIR or FIR filter? b) If x[n] = 8[n] – 8[n – 1] + 28[n – 2], then y[0] = 1, y[1] = -2, and y[2] = 5. Then find the value of all coefficients: ho, h, and h2.

Answer 1

X(Z).7-1 21 sol xen] o 2 1 V, 2xtz) X(2) { " 2.2 M2) y En] hox(z) 412) - hox(7) thi 77x(z) + h2 z xcz) Y(z) & hoth, zt the Z 2 ) x (2) - H (2) = 1 (2 = hot hit the 22 X(z ) difference eq? 8 h [n] = ho thi 8{n-1] the 8[n-27 (By Inverse z trans). This car is a FIR (finite impube response) also from ean 4 [n] = ho x [n] + hi x[n 1] the x [n-2] b). if x [n] - 8 [n] - 8 [64] + 28[n-2] then y[o] 21 407 = -2, 9 [2] - 5 if x[n] is as given in question. then.

& [n] - S[nl] – S[n-2] + 2 S [n-3] & x[n-2] = 8 [n-2] - S En 3] +25[n- 47. putting & En] x [n ], a[n-2] in y En] we get y [n] = ho (8[n] - S[n] + 28 [n-2] ) thics [n 1] - S[N-2] + 28 (n-37) +h21 stn-2] - SEN-3] + 28 [n-4]) 9 [ng ho $ [n] - 8 [n-17h - 1h) + Su-27( h) & 8 [n-3] (2hi – h2) + 2628 [n-47 Now! Y[0] =) (given) tha Now y[o] . ho sto] - SC-] (no-hi) + SC-2](cho this tha) + SC-3] 24, -42) + 2h28ca-47 8[0]=1 else o. Il = ho y CJ .-2 ( given) 9[1] = 0.–S[o] (ho-hi) to toto - 2 = /ho-b)

11 oly. for 4[4.5 ML] - 6 -6 + Sto1( 2 be -b th.) to to 5 - 2 honhitha 5 - 2 + 1 - (-1) + ha o in s 5 = 2 + 1 tha 225-3 Th222 [ hol hier, h2=2 Ans.
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