Practice Question on "Digital Signal Processing"

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] \ $

(Write enough intermediate steps to fully justify your answer.)


Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too! No need to write your name: we can find out who wrote what by checking the history of the page.


Answer 1

Green26 ece438 hmwrk3 power series.png

$ 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!

Answer 2

Muhammad Syafeeq Safaruddin

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.

Answer 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.


Answer 4

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.


Answer 5

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.

Back to ECE438 Fall 2013 Prof. Boutin


Answer 6

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? .

Answer 7

$ 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.


Answer 8

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? .

Answer 9

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!

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn