Digital Signal Processing with Field Programmable Gate Arrays

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Exercise. This is the solution to exercise 5.1 in the book.

Solution. We must have that:

F(z) + F(z) = C

or

1 + zd + 1 + (z)d = C

We have that:

(z)d = (1)dzd

hence

F(z) + F(z) = 2 + zd + (1)dzd = C

If d is odd then

F(z) + F(z) = 2 + zd zd = 2 = C

Hence, the requirement is that d is odd. Then both:

F(z) + F(z) = C

and

F(z) F(z) = Czd

are satisfied and

f[0] = 1

and

f[k] = 0

for all k even and k0.