Question

X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)?
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Answer 1

Answer:

X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)? Therefore, we get -z\frac{dX(z)}{dz} = Z{nx(n)}.

Answer 2

Answer:

The $z-$ transform of the signal $nx(n)$ is $-z\frac{dX(z)}{dz}$.

Step-by-step explanation:

By using the $z-$ transform definition that is the $z-$transform converts a discrete-time signal, which is a sequence of real or complex numbers, into a complex frequency-domain representation, we have

$X(z)=\sum_{n=-\infty}^{\infty} x(n)z^{-n}$

Now, differentiating both sides, we have

$\begin{aligned}\frac{dX(z)}{dz}&=\sum_{n=-\infty}^{\infty}(-n)x(n)z^{-n-1}\\ &=-z^{-1}\sum_{n=-\infty}^{\infty}nx(n)z^{-n}\\ &=-z^{-1}Z\{nx(n)\}\end{aligned}$

Hence, we get $-z\frac{dX(z)}{dz}=Z\{nx(n)\}$.

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