(5 intermediate revisions by the same user not shown)
Line 5: Line 5:
 
==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} $

Back to Homework2

Back to ECE438, Fall 2011, Prof. Boutin

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=Hw2_ECE438F11sln&oldid=47183"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang