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The transfer function of a digital filter, running at a sample rate fs = 10Khz is given by

H(z) = a+(z to the power -1) / 1 +(a*z to the power -1)

I have not got a clue how to work out this question any help most welcome.

2007-03-21 00:19:22 · 1 answers · asked by mdonaghy17 1 in Science & Mathematics Mathematics

1 answers

|H(e^(jωT))| will give the gain. So let's set z=jωT

(a+e^(-jωT))/(1+ae^(-jωT))=

(a+cos(-ωT)+jsin(-ωT))/(1+acos(-ωT)+jasin(-ωT))

The magnitude squared is

((a+cos(ωT))^2+sin^2(ωT))/(1+acos(ωT)^2+(asin(ωT))^2)

expanding
(a^2+2acos(ωT)+cos^2(ωT)+sin^2(ωT))/(1+2acos(ωT)+a^2cos^2(ωT)+a^2sin^2(ωT))

intermediate step
a^2cos^2(ωT)+a^2sin^2(ωT))=a^2

simplifying
(a^2+2acos(ωT)+1)/(a^2+2acos(ωT)+1)

this is magnitude 1

2007-03-21 14:18:59 · answer #1 · answered by Rob M 4 · 0 0

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