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Unformatted text preview: ECE 421 - Sum 2010 Detailed Solutions - Set 4: Digital Equalizers 1 4-1. Let a set of noiseless digital channels have the impulse response h m℄ am u m℄. Consider the three cases: (i) a 0 9, (ii) a 0 5, (iii) a 0 5 . For each of these three digital channels do the following: (a) Plot h m℄ over 0 m 5. (b) Compute HE ´zµ, the z-domain equalizer transfer function. Plot the poles and zeros of HE ´zµ. (c) Compute and plot hE m℄, the impulse response of the equalizer. Plot hE m℄ over 0 m 5. DETAILED SOLUTION : (a) Plot h m℄ over 0 m 5. The plots of each h m℄ are shown below:
1 a = 0.9 ...
.9 ... 1 . .
a = 0.5
.5 1 .
a = −0.5 .59 m=0 m=5 m=0 .... .03 m=0 m=5 .
1 H ´zµ .. ..
−0.5 m=5 −.03 (b) Compute HE ´zµ, the z-domain equalizer transfer function. The inverse ﬁlter equalizer HE ´zµ is given by the reciprocal of the channel transfer function: HE ´zµ With h m℄ am u m℄, we have H ´zµ
´aµ z , in which case equation ´aµ gives the following result: z a HE ´zµ z a z 1 az 1
´bµ Substituting the values of a given in (i)-(iii) into equation ´bµ thus gives a HE ´zµ 09: 1 0 9z 1 a HE ´zµ 05: 1 0 5z 1 a HE ´zµ 0 5 :
1 · 0 5z 1
´cµ ECE 421 - Sum 2010 Detailed Solutions - Set 4: Digital Equalizers 2 4-1. (cont.) From equation ´bµ we can plot the poles and zero for the HE ´zµ corresponding to the three values of a:
a = 0.9 a = 0.5 a = −0.5 x
0.5 −0.5 x (c) Compute and plot hE m℄, the impulse response of the equalizer. Taking the inverse z-transform of equation ´bµ gives the desired impulse response: hE m℄ Z 1 1 az 1 Ò Ó δ m℄ a δ m 1℄ ´d µ Using equation ´d µ the desired impluse repsonses are plotted below for the three values of a:
1 . 1 a = 0.9 . . 1 a = 0.5 .
0.5 . a = −0.5 m=1 m=0 .. m=1 m=0 −.5 .. m=0 m=1 .. −.9 . ...
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This note was uploaded on 09/07/2010 for the course ECE 421 taught by Professor Hallen during the Summer '08 term at N.C. State.
- Summer '08
- Signal Processing