Topic: Computing a z-transform

Question

Compute the z-transform (including the ROC) of the following DT signal:

$x[n]=3^n u[-n+3] \$

x[n] = 3nu[-n + 3]

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

$X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n}$

Let k = -n+3, n = -k+3

$X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{-k+3}$

$X(z) = (\frac{3}{z})^{3} \sum_{k=0}^{+\infty} (\frac{z}{3})^{k}$

$X(z) = (\frac{27}{z^3}) \sum_{k=0}^{+\infty} (\frac{z}{3})^{k}$

By geometric series formula,

$X(z) = (\frac{27}{z^3}) (\frac{1}{1-(\frac{z}{3})})$ ,for |z| < 3

X(z) = diverges, else

So,

$X(z) = (\frac{3}{3-z})$ with ROC, |z| < 3

Grader's comment: Obtain the Fourier transform of the signal

$X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n}$

Let k=-n+3, n=3-k, then

$X(z) = \sum_{k=-\infty}^{+\infty} (3)^{n-k}u[k](z)^{-3+k}$

$X(z) = (\frac{3}{z})^{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^{k}$

$X(z) = \left\{ \begin{array}{l l} (\frac{3}{z})^3 \frac{1}{1-\frac{z}{3}} &, if \quad |z| < 3\\ \text{diverges} &, \quad \text{otherwise} \end{array} \right.$

$\mathcal{F}(x[n]r^{-n}) = X(3e^{jw}) = \mathcal{X}(w) = \frac{\frac{3}{3e^{jw}}}{1-e^{jw}}$

Grader's comment: The Fourier Transform calculated is wrong

Kyungjun Kim

$X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n}$

Let l=-n+3, n=3-l, then

$X(z) = \sum_{l=-\infty}^{+\infty} (3)^{n-l}u[k]z^{-3+l}$

$X(z) = (\frac{3}{z})^{3}\sum_{l=0}^{+\infty} (\frac{z}{3})^{l}$

$X(z) = (\frac{3}{z})^3 \frac{1}{1-\frac{z}{3}}$ if |z| < 3

Grader's comment: Obtain the Fourier transform of the signal

$X[Z] = \sum_{n=-\infty}^{+\infty} 3^{n}u[n+3] Z^{-n}$

$X[Z] = \sum_{n=-3}^{+\infty} 3^{n}Z^{-n}$

$X[Z] = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}$

$X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n}$

$for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, no effect, because this converges everywhere on plane.$

$for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n}) = \frac{1}{1-\frac{3}{z}}, if |\frac{3}{z}|<1, |z|>3$

or diverges else.


for the DTFT for this signal,

$for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \frac{1}{1-\frac{3}{z}}, |z|>3, so it is impossible to have e^{j\omega}, because ROC is bigger at 3$

$for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, the DTFT is follow:$

$\sum_{n=-3}^{n=-1} (\frac{3}{e^{j\omega}})^{n}$

for all, this signal can't have DTFT.

Grader's comment: You copied the question wrongly

x[n] = 3nu[-n + 3]

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

$X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n}$

$X(z) = \sum_{-\infty}^{3} 3^n z^{-n}$

Let k = -n,

$X(z) = \sum_{k=-3}^{+\infty} 3^{-k} z^k$

$X(z) = \sum_{k=-3}^{+\infty} (\frac{z}{3})^{k}$

For |z| < 3, we have, by geometric series, that:

$X(z) = (\frac{27 z^{-3}}{1+(\frac{z}{3})})$

By simplification,

$X(z) = (\frac{-81 z^{-4}}{1-3 z^{-1}})$, ROC |z| < 3.

Since the ROC contains the unit circle, there exists a Fourier transform representation for the signal. Therefore, to find the signal`s Fourier representation, we just need to replace z be $e^{j w}$

So,

$X(\omega) = -(\frac{81 e^{-j 4 w}}{1-3 e^{-j w}})$

$x[n] = 3^n u[-n+3]$

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

$= \sum_{n=-\infty}^{\infty} 3^n u[-n+3] z^{-n}$

let k = -n+3 => n = -k+3

$X(z) = \sum_{k=0}^{\infty} 3^{-k+3} z^{k-3}$

$= \frac{3^3}{z^3} \sum_{k=0}^{\infty} {\frac{z}{3}}^k$

$= \frac{3^3}{z^3} \frac{1}{1-\frac{z}{3}}, \left |\frac{z}{3} \right | < 1$

$X(z) = \frac{3^3}{z^3} \frac{1}{1-\frac{z}{3}}, \left | z \right | < 3$

diverges , else

$F{{x[n]r^{-n}}} = X(3e^{jw}) = X(\omega) = \frac{e^{-j \omega 3}}{1-e^{jw}}$

Grader's comment: Replace Z by e^jw

Xiang Zhang

In order to get Z-transform, we can first apply the basic transformation equation.

$X_(z) = \sum_{n = -\infty}^{ \infty} x[n] z^{- n}$

Substitute in x[n], we can get that,

$X_(z) = \sum_{n = -\infty}^{ \infty} 3^n u[-n+3] z^{- n}$

let's use variable substitution by k = -n+3 hence n = 3-k. It can make our life beautiful and easier!

$X_(z) = \sum_{n = +\infty}^{- \infty} 3^{3-k} u[k] z^{k - 3}$

Now we can use u[k] to change the lower boundary to

$X_(z) = \sum_{n = 0 }^{ \infty} 3^{3-k} z^{k-3}$

We can take out some terms from Z.

$X_(z) = \frac{27}{ z^3} \sum_{n = 0 }^{ \infty} {(\frac{z}{3})}^{k}$

Now we can apply geometric series formula into the transform by letting q = z/3

$\frac{1}{1-q} = \sum_{ n = 0 }^{+ \infty} q^k for |q| < 1$

$X_(z) = \frac{27}{ z^3} (\frac{1}{1-\frac{z}{3}}) for |z| < 3$

Hence, the final x(z) is

$X(z) = \left\{ \begin{array}{l l} \frac{27}{ z^3} (\frac{1}{1-\frac{z}{3}}) & \quad when \quad |z| < 3\\ diverges & \quad \text{else} \end{array} \right.$

Grader's comment: Obtain the Fourier transform of the signal

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