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Compute Inverse Z transform
 
Compute Inverse Z transform
  
The power series expansion of the given function is
+
We can use partial fraction expansion to rewrite the z transform in a form similar to (1), (2). (See [Partial_Fraction_Expansion|here] for a general review)
 +
 
 +
Then we can use power series expansion (in this case: geometric series) and by comparison, we can obtain its z inverse transform.
  
 
<math>\begin{align}
 
<math>\begin{align}

Revision as of 04:10, 14 September 2011

Homework 2, ECE438, Fall 2011, Prof. Boutin


Question 1

Pick a note frequency f0 = 392Hz

x(t) = 'cos'(2πf0t) = 'cos'(2π * 392t)
$ a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000} $
$ 2f_0<\frac{1}{T_1}, \ No\ aliasing\ occurs. $

$ \begin{align} x_1(n) &=x(nT_1)=cos(2\pi *392nT_1)=cos(2\pi *\frac{392}{1000}n) \\ &=\frac{1}{2}\left( e^{-j2\pi *\frac{392}{1000}n} + e^{j2\pi *\frac{392}{1000}n} \right) \\ \end{align} $

$ 0<2\pi *\frac{392}{1000}<\pi $
$ -\pi<-2\pi *\frac{392}{1000}<0 $

$ \begin{align} \mathcal{X}_1(\omega) &=2\pi *\frac{1}{2} \left[\delta (\omega -2\pi *\frac{392}{1000}) + \delta (\omega + 2\pi *\frac{392}{1000})\right] \\ &=\pi \left[\delta (\omega -2\pi *\frac{392}{1000}) + \delta (\omega + 2\pi *\frac{392}{1000})\right] \\ \end{align} $

Xw1 singleperiod.jpg

$ for\ all\ \omega $
$ \mathcal{X}_1(\omega)=\pi* rep_{2\pi} \left[\delta (\omega -2\pi *\frac{392}{1000}) + \delta (\omega + 2\pi *\frac{392}{1000})\right] $

Xw1 multiperiod.jpg

In this situation, no aliasing occurs. In the interval of [ − π,π], which represents one period, the frequcy spectrum remains the same as Fig a-1.
$ b.\ Assign\ sampling\ period\ T_2=\frac{1}{500} $
$ 2f_0>\frac{1}{T_2}, \ Aliasing\ occurs. $

$ \begin{align} x_2(n) &=x(nT_2)=cos(2\pi *392nT_2)=cos(2\pi *\frac{392}{500}n) \\ &=\frac{1}{2}\left( e^{-j2\pi *\frac{392}{500}n} + e^{j2\pi *\frac{392}{500}n} \right) \\ \end{align} $

$ \pi<2\pi *\frac{392}{500}<2\pi $
$ -2\pi<-2\pi *\frac{392}{500}<\pi $
$ \mathcal{X}_2(\omega)=\pi \left[\delta (\omega -2\pi *\frac{392}{500}) + \delta (\omega + 2\pi *\frac{392}{500})\right] $
$ X_2(f)=\frac{1}{2}\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right] $

Xw2 singleperiod.jpg

$ for\ all\ \omega $
$ \mathcal{X}_2(\omega)=\pi* rep_{2\pi} \left[\delta (\omega -2\pi *\frac{392}{500}) + \delta (\omega + 2\pi *\frac{392}{500})\right] $
$ X_2(f)=\frac{1}{2}rep_2\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right] $

Xw2 multiperiod.jpg

In this situation, aliasing DO occurs. In the interval of [ − π,π], which represents one period, the frequcy spectrum is different from Fig b-1.

Xf2 multiperiod.jpg


Question 2

$ (1)\ x[n]=a^{n+1}u[n-1],\ a>0 $

Compute Z transform

$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} a^{n+1} u[n-1]z^{-n} \\ &= a\sum_{n=1}^{\infty} a^{n}z^{-n} \\ &= \frac{a^2z^{-1}}{1-az^{-1}} \end{align} $

with ROC: $ |z|>a $

Compute Inverse Z transform

The power series expansion of the given function is

$ \begin{align} X(z) &= a^2 z^{-1}\sum_{n=0}^{\infty} a^n z^{-n},\ |z|>a \\ &= a\sum_{n=0}^{\infty} a^{n+1}z^{-n-1} \end{align} $

Substitute n=m-1

$ \begin{align} X(z) &= a\sum_{m=1}^{\infty} a^{m}z^{-m} \\ &= \sum_{m=-\infty}^{\infty} a^{m+1}u[m-1]z^{-m},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $

$ \begin{align} x[n] &= a^{n+1} u[n-1] \end{align} $


$ (2)\ x[n]=-a^{n}u[-n-1],\ a>0 $

Compute Z transform

$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= -\sum_{n=-\infty}^{\infty} a^{n} u[-n-1]z^{-n} \\ &= -\sum_{n=-\infty}^{-1} a^{n}z^{-n} \\ \end{align} $

Substitute m=-n

$ \begin{align} X(z) &= -\sum_{n=1}^{\infty} a^{-n}z^{n} \\ &= -\frac{a^{-1}z}{1-a^{-1}z} \\ &= \frac{1}{1-az^{-1}} \end{align} $

with ROC: $ |z|<a $

Compute Inverse Z transform

$ \begin{align} X(z) &= \frac{1}{1-az^{-1}} \\ &= \frac{a^{-1}z}{a^{-1}z-1} \\ &= -a^{-1}z\frac{1}{1-a^{-1}z} \end{align} $

The power series expansion of the given function is

$ \begin{align} X(z) &= -a^{-1}z\sum_{n=0}^{\infty} a^{-n}z^{n} \\ &= -\sum_{n=0}^{\infty} a^{-n-1}z^{n+1} \\ \end{align} $

Substitute n+1=-m

$ \begin{align} X(z) &= -\sum_{m=-1}^{-\infty} a^{m}z^{-m} \\ &= -\sum_{m=-\infty}^{\infty} a^{m}u[-m-1]z^{-m},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $

$ \ x[n]=-a^{n}u[-n-1],\ a>0 $


$ (3) x[n]=u[n+1]-u[n-1] $

Compute Z transform

$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} (u[n+1]-u[n-1])z^{-n} \\ &= \sum_{n=-1}^{1} z^{-n} \\ &= 1+z^{-1}+z^1 \end{align} $

with ROC: $ z\in R,\ z\neq 0 $

Compute Inverse Z transform

$ \text{Since }z^k=\delta[n-k]z^n $

$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty}\sum_{k=-1}^{1} \delta[n-k] z^{n},\ z\in R,\ z\neq 0 \\ &= \sum_{n=-\infty}^{\infty} (\delta[n+1]+\delta[n]+\delta[n-1])z^{n} \\ \end{align} $

Substitute n=-m

$ \begin{align} X(z) &= \sum_{m=-\infty}^{\infty} (\delta[-m+1]+\delta[-m]+\delta[-m-1])z^{-m},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $

$ \begin{align} x[n] &= \delta[-n+1]+\delta[-n]+\delta[-n-1] \\ &= u[n+1]-u[n-1] \end{align} $


$ (4)\ cos(\omega_0 n)u[n] $

Compute Z transform

$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} cos(\omega_0 n)u[n]z^{-n} \\ &= \sum_{n=0}^{\infty} \frac{e^{j\omega_0n}+e^{-j\omega_0n}}{2}z^{-n} \\ &= \frac{1}{2}[\sum_{n=0}^{\infty}e^{j\omega_0n}z^{-n} + \sum_{n=0}^{\infty}e^{-j\omega_0n}z^{-n}] \\ &= \frac{1}{2}[\frac{1}{1-e^{j\omega_0}z^{-1}} + \frac{1}{1-e^{-j\omega_0}z^{-1}} ] \end{align} $

with ROC: $ |z|>|e^{j\omega_0}|,\ \text{and }|z|>|e^{-j\omega_0}| $

i.e. $ |z|>1 $

Simplify the answer

$ \begin{align} X(z)&= \frac{1}{2}\frac{1-e^{j\omega_0}z^{-1} + 1-e^{-j\omega_0}z^{-1}}{(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})} \\ &= \frac{1}{2}\frac{2-(e^{j\omega_0}+e^{-j\omega_0})z^{-1}}{1-(e^{j\omega_0}+e^{-j\omega_0})z^{-1}+z^{-2}} \\ &= \frac{1}{2}\frac{2-2cos(\omega_0)z^{-1}}{1-(2cos\omega_0)z^{-1}+z^{-2}} \\ &= \frac{1-(cos\omega_0)z^{-1}}{1-(2cos\omega_0)z^{-1}+z^{-2}} \end{align} $


Compute Inverse Z transform

We can use partial fraction expansion to rewrite the z transform in a form similar to (1), (2). (See [Partial_Fraction_Expansion|here] for a general review)

Then we can use power series expansion (in this case: geometric series) and by comparison, we can obtain its z inverse transform.

$ \begin{align} X(z) &= a^2 z^{-1}\sum_{n=0}^{\infty} a^n z^{-n},\ |z|>a \\ &= a\sum_{n=0}^{\infty} a^{n+1}z^{-n-1} \end{align} $

Substitute n=m-1

$ \begin{align} X(z) &= a\sum_{m=1}^{\infty} a^{m}z^{-m} \\ &= \sum_{m=-\infty}^{\infty} a^{m+1}u[m-1]z^{-m},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $

$ \begin{align} x[n] &= a^{n+1} u[n-1] \end{align} $


Back to Homework2

Back to ECE438, Fall 2011, Prof. Boutin

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood