Topic: Computing a z-transform

## Question

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

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

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

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

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

Let k = n+3:

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

Using the geometric series property:

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

TA's comment: Simple and clear derivation!

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

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

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

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

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

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

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

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

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

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

By geometric series formula,

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

X(z) = diverges, else

So,

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

TA's comment: You could just use geometric series to calculate the summation in step 4, and dont'need to change variable. ROC means region of CONVERGENCE. So ROC should be|z|>3.

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

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

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

$X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{3}{z})^{-3} + (\frac{3}{z})^{-2} + (\frac{3}{z})^{-1}$

$X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})$

if |3/z|<1,i.e z<-3 or z>3,

$(z) = (\frac{1}{1-(\frac{3}{z})}) + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3})$

if -3<z<3,

X(z) diverges

TA's comment: You can get a closed form expression by applying the geometric series property to the whole equation rather than split them into to parts.

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

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

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

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

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

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

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

or diverges else.


TA's comment: The ROC of the Z transform is the intersection of each part's ROC, which is |z|>3. You should state that explicitly.

Yixiang Liu

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

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

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

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

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

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

using geometric series formula

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

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

TA's comment: Once you drop the unit step function u[k] in step 4, the summation should be from 0 to infinity.

Xi Wang

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

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


if z > 3

  $X[z] = (\frac{1}{1-(\frac{3}{z})})$


if z < 3

  D'i'v'e'r'g'e's


TA's comment: Once you drop the unit step function u[k], the summation should be from 0 to infinity. The last step is wrong. Where is the (3/z)^3? .

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

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

With that we obtain,

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

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

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

Thus, ROC is |z| > 3 because it is restricted by geometric series.

TA's comment: Once you drop the unit step function u[k], the summation should be from 0 to infinity.

Cary Wood

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) = 3nznf'o'r'a'l'l'n > − 3

and

X(z) = 0,e'l's'e

Thus, we re-write X(z) as...

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

By the geometric series formula,

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

X(z) = diverges, elsewhere

ROC, |z| > 3

TA's comment: In step 3: X(z) is not a function of n, so it's not proper say it that way. You probably apply the geometric series in a wrong way. If the summation doesn't start from 0, you should multiply the sum by the first term (3/z)^3? .

Shiyu Wang

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

when  |3/z| < 1, |z| > 3

$X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) = (\frac{z^3}{27}) (\frac{z}{z-3})$ , for |z|>3; else, diverges.

TA's comment: Simple and clear derivation!

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