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The digital filter with system function H(z)=0.5zz0.5 Finds the output noise power from the digital filter when input is quantized to have 8- bits.
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Solution:

σ2e=22B3B=8σ2e=22×83σ2e=5.08×106σ2eoi=σ2eNi=1[[zPi)H[z]H[z1]z1]|z=Pi

H[z]H[z1]z1=0.5zz0.50.5z1z10.5z1=(0.5)(0.5)z1(z0.5)(z10.5)=(0.5)(0.5)z(z0.5)(10.5z)=0.5(z0.5)(2z)=0.5(z0.5)(z2)

Here one pole P1=0.5 is inside unit circle.

σ2e0i=σ2e(z0.5)[0.5(z0.5)(z2)]|z=00=σ2e(0.5)(0.52)=5.08×106[0.51.5]σ2e0i=1.6933×106

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