Latest revision as of 12:34, 9 September 2013

HW2_Solution_ECE438F13

Question 1

Pick a note frequency $f 0 = 392 H z$

x (t) = 'c o s'(2π f 0 t) = 'c o s'(2π \cdot 392 t)

a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000}

2f_0<\frac{1}{T_1}, \ No\ aliasing\ occurs.

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

0<2\pi \cdot\frac{392}{1000}<\pi

-\pi<-2\pi \cdot\frac{392}{1000}<0

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

for\ all\ \omega

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

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.

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

\pi<2\pi \cdot\frac{392}{500}<2\pi

-2\pi<-2\pi \cdot\frac{392}{500}<\pi

\mathcal{X}_2(\omega)=\pi \left[\delta (\omega -2\pi \cdot\frac{392}{500}) + \delta (\omega + 2\pi \cdot\frac{392}{500})\right]

X_2(f)=\frac{1}{2}\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right]

for\ all\ \omega

\mathcal{X}_2(\omega)=\pi\cdot rep_{2\pi} \left[\delta (\omega -2\pi \cdot\frac{392}{500}) + \delta (\omega + 2\pi \cdot\frac{392}{500})\right]

X_2(f)=\frac{1}{2}rep_2\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right]

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.

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 2013 Fall ECE 438 Boutin

Difference between revisions of "HW2 Solution ECE438F13" - Rhea

Latest revision as of 12:34, 9 September 2013

HW2_Solution_ECE438F13

Question 1

Question 2

Alumni Liaison

@@ Line 4: / Line 4: @@
+----
-Put your content here . . .
+==Question 1==
+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>
+|-
+| <math>a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000}</math>
+|-
+| <math>2f_0<\frac{1}{T_1}, \ No\ aliasing\ occurs.</math>
+|}
+<div align="left" style="padding-left: 0em;">
+<math>
+\begin{align}
+x_1(n) &=x(nT_1)=cos(2\pi \cdot 392nT_1)=cos(2\pi \cdot\frac{392}{1000}n) \\
+&=\frac{1}{2}\left( e^{-j2\pi \cdot \frac{392}{1000}n} + e^{j2\pi \cdot\frac{392}{1000}n} \right) \\
+\end{align}</math>
+</div>
+{|
+|-
+| <math>0<2\pi \cdot\frac{392}{1000}<\pi</math>
+|-
+| <math>-\pi<-2\pi \cdot\frac{392}{1000}<0</math>
+|}
+<div align="left" style="padding-left: 0em;">
+<math>
+\begin{align}
+\mathcal{X}_1(\omega) &=2\pi \cdot\frac{1}{2} \left[\delta (\omega -2\pi \cdot\frac{392}{1000}) + \delta (\omega + 2\pi \cdot\frac{392}{1000})\right] \\
+&=\pi \left[\delta (\omega -2\pi \cdot\frac{392}{1000}) + \delta (\omega + 2\pi \cdot\frac{392}{1000})\right] \\
+\end{align}</math>
+</div>
+[[Image:Xw1 singleperiod.jpg]]
+{|
+|-
+| <math>for\ all\ \omega</math>
+|-
+| <math>\mathcal{X}_1(\omega)=\pi\cdot rep_{2\pi} \left[\delta (\omega -2\pi \cdot\frac{392}{1000}) + \delta (\omega + 2\pi \cdot\frac{392}{1000})\right]</math>
+|}
+[[Image:Xw1 multiperiod.jpg]]
+{|
+|-
+| In this situation, no aliasing occurs. In the interval of <span class="texhtml">[ − π,π]</span>, which represents one period, the frequcy spectrum remains the same as Fig a-1.
+|}
+{|
+|-
+| <math>b.\ Assign\ sampling\ period\ T_2=\frac{1}{500}</math>
+|-
+| <math>2f_0>\frac{1}{T_2}, \ Aliasing\ occurs.</math>
+|}
+<div align="left" style="padding-left: 0em;">
+<math>
+\begin{align}
+x_2(n) &=x(nT_2)=cos(2\pi \cdot 392nT_2)=cos(2\pi \cdot\frac{392}{500}n) \\
+&=\frac{1}{2}\left( e^{-j2\pi \cdot\frac{392}{500}n} + e^{j2\pi \cdot\frac{392}{500}n} \right) \\
+\end{align}</math>
+</div>
+{|
+|-
+| <math>\pi<2\pi \cdot\frac{392}{500}<2\pi</math>
+|-
+| <math>-2\pi<-2\pi \cdot\frac{392}{500}<\pi</math>
+|-
+| <math>\mathcal{X}_2(\omega)=\pi \left[\delta (\omega -2\pi \cdot\frac{392}{500}) + \delta (\omega + 2\pi \cdot\frac{392}{500})\right] </math>
+|-
+| <math>X_2(f)=\frac{1}{2}\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right]</math>
+|}
+[[Image:Xw2 singleperiod.jpg]]
+{|
+|-
+| <math>for\ all\ \omega</math>
+|-
+| <math>\mathcal{X}_2(\omega)=\pi\cdot rep_{2\pi} \left[\delta (\omega -2\pi \cdot\frac{392}{500}) + \delta (\omega + 2\pi \cdot\frac{392}{500})\right]</math>
+|-
+| <math>X_2(f)=\frac{1}{2}rep_2\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right]</math>
+|}
+[[Image:Xw2 multiperiod.jpg]]
+{|
+|-
+| In this situation, aliasing DO occurs. In the interval of <span class="texhtml">[ − π,π]</span>, which represents one period, the frequcy spectrum is different from Fig b-1.
+|}
+[[Image:Xf2 multiperiod.jpg]]
+----
+==Question 2==
+<math>(1)\ x[n]=a^{n+1}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+1} u[n-1]z^{-n} \\
+&= a\sum_{n=1}^{\infty} a^{n}z^{-n} \\
+&= \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>
+----
+[[HW2ECE438F13|Back to Homework2]]
 [[ 2013 Fall ECE 438 Boutin|Back to 2013 Fall ECE 438 Boutin]]