# Digital Signal Processing with Field Programmable Gate Arrays

Chapter05 - Home
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Exercise. This is the solution to exercise 5.1 in the book.

Solution. We must have that:

$F\left(z\right)+F\left(-z\right)=\mathsc{C}$

or

$1+{z}^{-d}+1+{\left(-z\right)}^{-d}=\mathsc{C}$

We have that:

${\left(-z\right)}^{-d}={\left(-1\right)}^{d}{z}^{-d}$

hence

$F\left(z\right)+F\left(-z\right)=2+{z}^{-d}+{\left(-1\right)}^{d}{z}^{-d}=\mathsc{C}$

If $d$ is odd then

$F\left(z\right)+F\left(-z\right)=2+{z}^{-d}-{z}^{-d}=2=\mathsc{C}$

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

$F\left(z\right)+F\left(-z\right)=\mathsc{C}$

and

$F\left(z\right)-F\left(-z\right)=\mathsc{C}{z}^{-d}$

are satisfied and

$f\left[0\right]=1$

and

$f\left[k\right]=0$

for all $k$ even and $k\ne 0$.