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==Question 1==
 
==Question 1==
  
−
Pick a note frequency <span class="texhtml">''f''<sub>0</sub> = 392''H''''z''</span>  
+
Pick a note frequency <span class="texhtml">''f''<sub>0</sub> = 392''H'''z''</span>  
  
 
{|
 
{|
 
|-
 
|-
−
| <span class="texhtml">''x''(''t'') = ''c''''o''''s''(2π''f''<sub>0</sub>''t'') = ''c''''o''''s''(2π * 392''t'')</span>
+
| <span class="texhtml">''x''(''t'') = 'c''o''s'(2π''f''<sub>0</sub>''t'') = 'c''o''s'(2π * 392''t'')</span>
 
|-
 
|-
 
| <math>a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000}</math>
 
| <math>a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000}</math>
Line 98: Line 98:
 
==Question 2==
 
==Question 2==
  
−
(1) <math>x[n]=a^{n+1}u[n-1],\ a>0</math>
+
<math>(1)\ x[n]=a^{n+1}u[n-1],\ a>0</math>
 +
 
 +
Compute Z transform
  
 
<math>\begin{align}
 
<math>\begin{align}
Line 105: Line 107:
 
&= a\sum_{n=1}^{\infty} a^{n}z^{-n} \\
 
&= a\sum_{n=1}^{\infty} a^{n}z^{-n} \\
 
&= \frac{a^2z^{-1}}{1-az^{-1}}
 
&= \frac{a^2z^{-1}}{1-az^{-1}}
 +
\end{align}</math>
 +
 +
with ROC: <math>|z|>a</math>
 +
 +
Compute Inverse Z transform
 +
 +
The power series expansion of the given function is
 +
 +
<math>\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}</math>
 +
 +
Substitute n=m-1
 +
 +
<math>\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}</math>
 +
 +
<math>\begin{align}
 +
x[n] &= a^{n+1} u[n-1]
 +
\end{align}</math>
 +
 +
----
 +
 +
<math>(2)\ x[n]=-a^{n}u[-n-1],\ a>0</math>
 +
 +
Compute Z transform
 +
 +
<math>\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}</math>
 +
 +
Substitute m=-n
 +
 +
<math>\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}</math>
 +
 +
with ROC: <math>|z|<a</math>
 +
 +
Compute Inverse Z transform
 +
 +
<math>\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}</math>
 +
 +
The power series expansion of the given function is
 +
 +
<math>\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}</math>
 +
 +
Substitute n+1=-m
 +
 +
<math>\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}</math>
 +
 +
<math>\ x[n]=-a^{n}u[-n-1],\ a>0</math>
 +
 +
----
 +
<math>(3) x[n]=u[n+1]-u[n-1]</math>
 +
 +
Compute Z transform
 +
 +
<math>\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}</math>
 +
 +
with ROC: <math>z\in R,\ z\neq 0</math>
 +
 +
Compute Inverse Z transform
 +
 +
<math>\text{Since }z^k=\delta[n-k]z^n</math>
 +
 +
<math>\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}</math>
 +
 +
Substitute n=-m
 +
 +
<math>\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}</math>
 +
 +
<math>\begin{align}
 +
x[n] &= \delta[-n+1]+\delta[-n]+\delta[-n-1] \\
 +
&= u[n+1]-u[n-1]
 +
\end{align}</math>
 +
 +
----
 +
<math>(4)\ (cos\omega_0 n)u[n]</math>
 +
 +
Compute Z transform
 +
 +
<math>\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}</math>
 +
 +
with ROC: <math>|z|>|e^{j\omega_0}|,\ \text{and }|z|>|e^{-j\omega_0}|</math>
 +
 +
i.e. <math>|z|>1</math>
 +
 +
Simplify the answer
 +
 +
<math>\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{2-2(cos\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}</math>
 +
 +
 +
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 of partial fraction expansion)
 +
 +
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}
 +
X(z) &= \frac{1-\frac{e^{j\omega_0}+e^{-j\omega_0}}{2}z^{-1}}{1-(e^{j\omega_0}+e^{-j\omega_0})z^{-1}+z^{-2}},\ |z|>a \\
 +
&= \frac{1}{2}\frac{2-2(cos\omega_0)z^{-1}}{(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})} \\
 +
&= \frac{1}{2}[\frac{1}{1-e^{j\omega_0}z^{-1}} + \frac{1}{1-e^{-j\omega_0}z^{-1}}] \\
 +
&= \frac{1}{2}[\sum_{n=0}^{\infty}e^{j\omega_0n}z^{-n}  + \sum_{n=0}^{\infty}e^{-j\omega_0n}z^{-n}] \\
 +
&= \sum_{n=0}^{\infty} \frac{e^{j\omega_0n}+e^{-j\omega_0n}}{2}z^{-n} \\
 +
&= \sum_{n=-\infty}^{\infty} cos(\omega_0 n)u[n]z^{-n},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n}
 +
\end{align}</math>
 +
 +
<math>\begin{align}
 +
x[n] &= (cos\omega_0 n)u[n]
 +
\end{align}</math>
 +
 +
----
 +
(5) Compute Z transform of
 +
 +
<math>x[n]=nu[n]</math>
 +
 +
<math>\begin{align}
 +
X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\
 +
&= \sum_{n=-\infty}^{\infty} nu[n]z^{-n} \\
 +
&= \sum_{n=0}^{\infty} nz^{-n}
 +
\end{align}</math>
 +
 +
Let <math>k=\frac{1}{z}</math>
 +
 +
<math>\begin{align}
 +
X(k) &= \sum_{n=0}^{\infty} nk^{n} \\
 +
&= k\sum_{n=0}^{\infty} nk^{n-1} \\
 +
\end{align}</math>
 +
 +
We know that
 +
 +
<math>\sum_{n=0}^{\infty} k^{n} = \frac{1}{1-k},\ |k|<1</math>
 +
 +
Compute derivative with respect to k on both side we have
 +
 +
<math>\sum_{n=0}^{\infty} nk^{n-1} = \frac{1}{(1-k)^2},\ |k|<1</math>
 +
 +
Therefore
 +
 +
<math>X(k)=\frac{k}{(1-k)^2},\ |k|<1</math>
 +
 +
<math>X(z)=\frac{z^{-1}}{(1-z^{-1})^2},\ \text{with ROC }|z|>1</math>
 +
 +
----
 +
(6) Compute inverse Z transform of
 +
 +
<math>X(z) = \log \left( 1+z \right), \quad |z|<1 </math>.
 +
 +
expand the function into a power series using either the Taylor series formula or a [[PowerSeriesFormulas|table of power series formulas]].
 +
 +
The power series expansion of the given function is:
 +
 +
<math>\begin{align}
 +
X(z) &= \sum_{n=1}^{\infty} (-1)^{n+1} \frac{z^n}{n}, \ -1 < z \le 1 \\
 +
&= \sum_{n=-\infty}^{\infty} (-1)^{n+1} u[n-1] \frac{z^n}{n}
 +
\end{align}</math>
 +
 +
Substitute n = -k
 +
 +
<math>\begin{align}
 +
X(z) &= \sum_{k=-\infty}^{\infty} (-1)^{-k+1} u[-k-1] \frac{z^{-k}}{-k} \\
 +
&= \sum_{k=-\infty}^{\infty} \frac{(-1)^{-k+1}}{-k} u[-k-1]z^{-k} \\
 +
&= \sum_{k=-\infty}^{\infty} \frac{(-1)^{-k}(-1)}{-k} u[-k-1] z^{-k} \\
 +
&= \sum_{k=-\infty}^{\infty}\frac{(-1)^{-k}}{k} u[-k-1]z^{-k}, \text{ and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n}
 +
\end{align}</math>
 +
 +
<math>\begin{align}
 +
x[n] &= \frac{(-1)^{-n}}{n} u[-n-1] \\
 +
&= \frac{(-1)^{n}}{n} u[-n-1]
 
\end{align}</math>
 
\end{align}</math>
  

Latest revision as of 04:33, 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{2-2(cos\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 here for a general review of partial fraction expansion)

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) &= \frac{1-\frac{e^{j\omega_0}+e^{-j\omega_0}}{2}z^{-1}}{1-(e^{j\omega_0}+e^{-j\omega_0})z^{-1}+z^{-2}},\ |z|>a \\ &= \frac{1}{2}\frac{2-2(cos\omega_0)z^{-1}}{(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})} \\ &= \frac{1}{2}[\frac{1}{1-e^{j\omega_0}z^{-1}} + \frac{1}{1-e^{-j\omega_0}z^{-1}}] \\ &= \frac{1}{2}[\sum_{n=0}^{\infty}e^{j\omega_0n}z^{-n} + \sum_{n=0}^{\infty}e^{-j\omega_0n}z^{-n}] \\ &= \sum_{n=0}^{\infty} \frac{e^{j\omega_0n}+e^{-j\omega_0n}}{2}z^{-n} \\ &= \sum_{n=-\infty}^{\infty} cos(\omega_0 n)u[n]z^{-n},\ \text{and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $

$ \begin{align} x[n] &= (cos\omega_0 n)u[n] \end{align} $

(5) Compute Z transform of

$ x[n]=nu[n] $

$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n]z^{-n} \\ &= \sum_{n=-\infty}^{\infty} nu[n]z^{-n} \\ &= \sum_{n=0}^{\infty} nz^{-n} \end{align} $

Let $ k=\frac{1}{z} $

$ \begin{align} X(k) &= \sum_{n=0}^{\infty} nk^{n} \\ &= k\sum_{n=0}^{\infty} nk^{n-1} \\ \end{align} $

We know that

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

Compute derivative with respect to k on both side we have

$ \sum_{n=0}^{\infty} nk^{n-1} = \frac{1}{(1-k)^2},\ |k|<1 $

Therefore

$ X(k)=\frac{k}{(1-k)^2},\ |k|<1 $

$ X(z)=\frac{z^{-1}}{(1-z^{-1})^2},\ \text{with ROC }|z|>1 $

(6) Compute inverse Z transform of

$ X(z) = \log \left( 1+z \right), \quad |z|<1 $.

expand the function into a power series using either the Taylor series formula or a table of power series formulas.

The power series expansion of the given function is:

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

Substitute n = -k

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

$ \begin{align} x[n] &= \frac{(-1)^{-n}}{n} u[-n-1] \\ &= \frac{(-1)^{n}}{n} u[-n-1] \end{align} $

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Back to ECE438, Fall 2011, Prof. Boutin

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