Week 11

What happens when we modulate a function in time with a complex exponential?

\begin{align*}
g(n) &= \hat{G}(z) \quad R_1 \lt \abs{z} \lt R_2 \\
h(n) &= z_0^ng(n) \ztp \hat{H}(z) = \mbox{ } ?
\end{align*}

The result must be consistent with our result for the fourier transform version. 

\begin{align*}
\hat{H}(z) &= \sum \limits_{n\in\Z} h(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} z_0^ng(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} g(n) \lr{\frac{z}{z_0}}^{-n} \\
\hat{H}(z) &= \hat{G}\lr{\frac{z}{z_0}} \quad \quad R_1 \lt \abs{\frac{z}{z_0}} \lt R_2
\end{align*}

The $RoC$ simplifies to $\{ z \st \abs{z_0}R_1 \lt \abs{z} \lt \abs{z_0}R_2\}$, such that $RoC(h) = \abs{z_0}Roc(g)$.

\begin{example}

Given the following transform pair, determine what happens to the poles and zeros.

$$
e^{i\omega_0 n} g(n) \ztp  \hat{G}(e^{-i\omega_0 }z)
$$

We know that the $RoC$ is the same as $RoC(g)$, since $\abs{e^{i\omega}} = 1$. Assume that $z=p$ is a pole of $\hat{G}$. Now we have $e^{-i\omega_0} z = p$ is a pole. This implies that when $z = e^{i\omega_0} p$, we have a pole rotated by $\omega_0$ from the position of the original pole.

Suppose that $\omega_0 = \frac{\pi}{4}$, and $g(n) = u(n)$.
\begin{align*}
u(n) &\ztp \hat{U}(z) \\
u(n) &\ztp \frac{1}{1-z^{-1}} = \frac{z}{z-1}\\
e^{i\omega_0}u(n) &\ztp \hat{U}(e^{-i\omega_0}z) \\
e^{i\omega_0}u(n) &\ztp \frac{e^{-i\omega_0}z}{e^{-i\omega_0}z -1} \\
e^{i\omega_0}u(n) &\ztp \frac{z}{z - e^{i\omega_0}} \\
\end{align*}

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Here the $RoC$ stays the same since the magnitudes are equal.

\end{example}

\begin{example}

Determine what happens to the $RoC$ for $z_0 = 2$.

\begin{align*}
h(n) = 2^n u(n) &\ztp \hat{U}\lr{ \frac{z}{2} } \\
h(n) = 2^n u(n) &\ztp \frac{1}{1-2z^{-1}} \\
\end{align*}

The $RoC$ gets scaled:

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\end{example}

\begin{example}

Determine what happens to the $RoC$ for $z_0 = 2e^{-i\frac{\pi}{4}}$.

\begin{align*}
f(n) = \lr{2e^{-i\frac{\pi}{4}}}^n u(n) &\ztp \hat{U}\lr{ \frac{z}{2} } \\
f(n) = \lr{2e^{-i\frac{\pi}{4}}}^n u(n) &\ztp \frac{1}{1-\lr{2e^{-i\frac{\pi}{4}}}z^{-1}} \\
\end{align*}

The $RoC$ gets rotated and scaled, a combination of the above two cases.

\end{example}

Modulation by Complex Exponentials

What happens when we do time-reversal on a function in time?

\begin{align*}
g(n) &= \hat{G}(z) \quad R_1 \lt \abs{z} \lt R_2 \\
h(n) &= g(-n) \ztp \hat{H}(z) = \mbox{ } ?
\end{align*}

\begin{align*}
\hat{H}(z) &= \sum \limits_{n\in\Z} h(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} g(-n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} g(n) z^{n} \mbox{ let } n = -n \\
\hat{H}(z) &= \sum \limits_{n\in\Z} g(n) \lr{\frac{1}{z}}^{-n} \\
\hat{H}(z) &= \hat{G}\lr{z^{-1}} \quad R_1 \lt \abs{z^{-1}} \lt R_2\\
\end{align*}

The $RoC$ is changed, we must take the reciprocal: $RoC(h) = \frac{1}{RoC(g)}$. This gives us $\{ z \st \frac{1}{R_2} \lt \abs{z} \lt \frac{1}{R_1} \}$.


\begin{example}

Find what happens to the $RoC$ for $h(n) = u(-n)$.
\begin{align*}
u(n) &\ztp \frac{1}{1-z^{-1}} \quad \abs{z} \gt 1 \\
h(n) = u(-n) &\ztp \frac{1}{1-z} \quad \abs{z} \lt 1\\
\end{align*}

All poles and zeros are inverted. If $z=p$ is a pole of $\hat{G}(z)$, the $z=\frac{1}{p}$ is a pole of $\hat{H}(z)$. The same is true for every zero.

\end{example}

Time Reversal

Determine the Z-transform and $RoC$ for $h(n) = \cos(\omega_0 n) u(n)$.

\begin{align*}
h(n) = \cos(\omega_0 n) u(n) &\ztp \hat{H}(z) \\
h(n) = \frac{1}{2} e^{i\omega_0 n}u(n)  + \frac{1}{2} e^{-i\omega_0 n} u(n) &\ztp \frac{1}{2} \hat{U}\lr{\frac{z}{e^{-i\omega_0}}} + \frac{1}{2} \hat{U}\lr{\frac{z}{e^{i\omega_0}}} \\
\end{align*}

We can use the fact that $\hat{U}(z) = \frac{z}{z - 1}$.
\begin{align*}
\hat{H}(z) &= \frac{1}{2} \lr{\frac{z}{z - e^{i\omega_0}}} + \frac{1}{2} \lr{\frac{z}{z - e^{-i\omega_0}}} \\
\hat{H}(z) &= \frac{z}{2}\lr{\frac{2z - e^{-i\omega_0} - e^{i\omega_0}}{\lr{z-e^{i\omega_0}}\lr{z-e^{-i\omega_0}}}} \\
\hat{H}(z) &= \frac{z\lr{z - \cos\omega_0}}{\lr{z-e^{i\omega_0}}\lr{z-e^{-i\omega_0}}} \\
\end{align*}

Lets use $\omega_0 = \frac{\pi}{4}$.

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We have zeros at 0, and $z = \cos\omega_0 = \frac{1}{\sqrt{2}}$. We have poles at $e^{\pm i \omega_0}$. There no fourier transform since the unit circle is not included.



Trigonometric Example

Lets see what happens when we differentiate in the Z domain for $g(n) \ztp \hat{G}(z)$:

\begin{align*}
\hat{G}(z) &= \sum \limits_{n\in\Z} g(n) z^{-n} \\
\frac{d}{dz}\hat{G}(z) &= -\sum \limits_{n\in\Z} ng(n) z^{-n-1} \\
\frac{d}{dz}\hat{G}(z) &= -z^{-1}\sum \limits_{n\in\Z} ng(n) z^{-n} \\
\end{align*}

This gives us the differentiation property of the Z-transform:

\begin{nicebox}
\begin{align*}
g(n) &\ztp \hat{G}(z) \\
ng(n) &\ztp -z \frac{d}{dz} \hat{G}(z) \\
\end{align*}
\end{nicebox}

We can see that differentiation of a rational function will produce a rational Z-transform. If you recall, the set of signals that produce rational Z-transforms are obtained from, and only from, linear combinations of $\alpha^n u(n)$, $\alpha^n u(-n)$, $n^k \alpha^n u(n)$, and $n^k \alpha^n u(-n)$, where $k \in \N$, and shifted versions thereof.


Differentiation in the Z Domain

What if we have a signal that upsamples another signal by dialating it in the time domain?

$$ h(n) = \begin{cases} g\lr{\frac{n}{N}} & n\bmod N=0 \\ 0 & \mbox{otherwise} \\ \end{cases} $$

The input signal becomes dialated in the time domain by a factor of $N$---the system inserts $N-1$ zeros between samples.

\begin{align*}
\hat{H}(z) &= \sum \limits_{n\in\Z} h(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in N\Z} g\lr{\frac{n}{N}} z^{-n} \\
\end{align*}

I want to mention that we performed a \emph{dilation} of the integers. A dilation of a set $X$ is a subset $cX = \{ cx \st x \in X \}$, and $c \in \Z$. Here we have successfully dilated the integers such that we are summing over the set $\{Nn \st n \in \Z \}$.

\begin{align*}
\hat{H}(z) &= \sum \limits_{mN\in N\Z} g(m) z^{-mN} \quad \mbox{ let } m = \frac{n}{N} \\
\hat{H}(z) &= \sum \limits_{mN\in N\Z} g(m) z^{-mN} \\
\hat{H}(z) &= \sum \limits_{m\in \Z} g(m) z^{-mN} \\
\end{align*}

The reason that we can now let $m$ sum over all integers is because every $mN$ element in the set $N\Z$ is factoring out $m$ implicitly, so the $N$ drops out of our limits in the sum.
 
\begin{align*}
\hat{H}(z) &= \sum \limits_{m\in \Z} g(m) \lr{z^{N}}^{-m}\\
\hat{H}(z) &= \hat{G}(z^N) \\
\end{align*}

Lets look at the $RoC$ for a dilated signal. 

\begin{align*}
RoC(g) &: R_1 \lt \abs{z} \lt R_2 \\
RoC(h) &: R_1 \lt \abs{z^N} \lt R_2 \\
RoC(h) &: \sqrt[n]{R_1} \lt \abs{z} \lt \sqrt[n]{R_2} \\
\end{align*}

We find that $RoC(h) = \sqrt[n]{RoC(g)}$. But what happens to the poles and the zeros? Assume $z=p$ is a pole of $\hat{G}(z)$, then $\hat{H}(z)$ has $N$ poles at the $N^{th}$ roots of $p$. For demonstrative purposes, assume that $N = 3$, and $\frac{1}{8}$ is a pole of $\hat{G}(z)$. 

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\begin{align*}
z = \frac{1}{8} = \frac{1}{8}e^{i2\pi k}
\end{align*}

We can take any number and multiply it by 1. Then we can take the root.

\begin{align*}
z &= \sqrt[3]{\frac{1}{8}e^{i2\pi k}} \\
z &= \frac{1}{2}e^{i\frac{2\pi}{3} k} \\
\end{align*}

This gives us the set $\{ \frac{1}{2}, \frac{1}{2}e^{i2\pi/3}, \frac{1}{2}e^{i4\pi/3} \}$. Notice that the $RoC$ has been pushed out to $\frac{1}{2}$ and now we have 3 roots:

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(1) If $g(n)$ is stable, what can you say about $H(\omega)$? 


If $g(n)$ is stable, then $G(\omega) = \left. \hat{G}(z) \right|_{z = e^{i\omega}}$, and it is absolutely summable. We aren't adding any new values of $G$ to $H$. We are simply interspersing the same values of $G$, which implies that $H$ is also absolutely summable. We are inserting zeros between adjacent samples in $h$, so we don't change its stability.

$$ \sum \limits_{n \in Z} \abs{g(n)} \lt \infty \Rightarrow \sum \limits_{n \in Z} \abs{h(n)} \lt \infty $$

$H(\omega)$ is also well-defined.

\begin{align*}
\hat{H}(z) &= \hat{G}(z^N) \\
H(\omega)  &= \hat{G}(e^{i\omega N}) = G(\omega N) \\
H(\omega)  &=  G(\omega N) \\
\end{align*}

Dilation

Given the following information and pole-zero diagram, find an explicit formula for $\hat{H}(z)$ and the value of $h(0)$.

\begin{enumerate}
    \item $\hat{H}(z)$ is rational
    \item $\hat{H}(z)$ is causal
    \item given $x(n) = 1$, $y(n) = -4/3$
\end{enumerate}

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Looking at the pole-zero diagram, we can write an explicit formula for $\hat{H}(z)$:

\begin{align*}
\hat{H}(z) &= H_0 \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
\end{align*}

We also can determine the $RoC$ of $H$ because it is causal. This tells us $RoC(h) = \{ z \st \abs{z} \gt \frac{3}{4} \}$. Lets find $H_0$. Recall from the fourier transform days that we found $x(n) = 1 = e^{i 0 n}$, and $y(n) = H(0)e^{i 0 n}$. Here we can't use this exact trick. Recall that when the input of a system is a complex exponential, we get $y(n) = z_0^n\hat{H}(z)$. The trick here is to let $z_0 = 1$ (note that $\hat{H}(1) = H(0)$):

\begin{align*}
y(n) &= z_0^n \hat{H}(z) \\
-\frac{4}{3} &= 1^n \hat{H}(1) \\
-\frac{4}{3} &= \hat{H}(1) \\
\end{align*}

Now we can use the $\hat{H}(1) = -\frac{4}{3}$ to find $H_0$:

\begin{align*}
\hat{H}(z) &= H_0 \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
\hat{H}(1) &= H_0 \frac{ \lr{ 1- \frac{3}{2} } } { \lr{ 1 + \frac{1}{2}} \lr{ 1 - \frac{3}{4} } } \\
\hat{H}(1) &= H_0 \frac{ \lr{ - \frac{1}{2} } } { \lr{ \frac{3}{2}} \lr{ \frac{1}{4} } } \\
-\frac{4}{3} &= -H_0\frac{4}{3} \\
H_0 &= 1 \\
\end{align*}

This tells the exact formula for $\hat{H}$:
$$
\hat{H}(z) = \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
$$

How can we find $h(0)$? If $h$ is causal, then $h(n) = 0$ for $n \lt 0$. 

\begin{align*}
\hat{H}(z) &= \sum \limits_{n \in \Z} h(n) z^{-n} \\
\hat{H}(z) &= h(0) + \frac{h(1)}{z} + \frac{h(2)}{z^2} + \cdots  \\
\end{align*}

This brings us to the \emph{Initial Value Theorem}:

$$ 
h(0) = \lim_{z \to \infty} \hat{H}(z)  \\
$$

The initial value theorem applies when the system is causal. There is a corresponding theorem called the final value theorem for anti-caual systems. 

\begin{align*}
h(0) &= \lim_{z \to \infty} \hat{H}(z)  \\
h(0) &= \lim_{z \to \infty}  \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
h(0) &= 0 \\
\end{align*}


Initial Value Theorem

There are quite a few methods two find inverse Z-transforms, a few of which we will discuss here.

I) Identify familiar forms. 


We have used particular forms involving Z-transforms over and over, so we can use pattern recognition:

\begin{example}

\begin{align*}
\hat{X}(z) &= \frac{1}{1 - \alpha Z^{-1}} \quad  \abs{\alpha} \lt \abs{z} \\
x(n) &= \alpha^n u(n)
\end{align*}


\end{example}

\begin{example}

\begin{align*}
\hat{X}(z) &= z^{-5} \\
x(n) &= \delta(n-5)
\end{align*}


II) Power Series Expansions


Recall the formula
$$
\sum \limits_{n=0}^{\infty} \beta^n = \frac{1}{1 - \beta} \quad  \abs{\beta} \lt 1
$$

\end{example}

\begin{example}

\begin{align*}
\hat{X}(z) &= \frac{1}{1 - \alpha z^{-1}} \quad  \abs{\alpha} \lt \abs{z} \\
\hat{X}(z) &= \sum \limits_{n=0}^{\infty} \lr{\alpha z^{-1}}^n \\
\hat{X}(z) &= \sum \limits_{n=0}^{\infty} \alpha^n z^{-n}\\
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} \alpha^n u(n) z^{-n}\\
x(n) &= \alpha^n u(n) \\
\end{align*}

Recall the expansion for $\log(1 + \beta)$:

$$
\log(1 + \beta) = \sum \limits_{n=1}^{\infty} \frac{ (-1)^{n+1} \beta^n } { n} \quad \abs{\beta} \lt 1 
$$

\end{example}

\begin{example}

\begin{align*}
\hat{X}(z) &= \log( 1 + \alpha z^{-1}) \quad \abs{\alpha} \lt \abs{z} \\
\hat{X}(z) &= \sum \limits_{n=1}^{\infty} \frac{ (-1)^{n+1} \lr{\alpha z^{-1}}^n } { n} \\
\hat{X}(z) &= \sum \limits_{n=1}^{\infty} \lr{\frac{ (-1)^{n+1} \alpha^n } { n}} z^{-n} \\
\end{align*}

This tells us that
$$x(n) = \begin{cases} 0 & n\leq 0 \\ \frac{(-1)^{n+1}\alpha^n}{n} & n \geq 1 \\ \end{cases}$$

Hence, $x(n) = \frac{(-1)^{n+1}\alpha^n}{n} u(n-1)$. Note that $\log( 1 + \alpha z^{-1})$ is not rational in $Z$. $x(n) = \frac{(-1)^{n+1}\alpha^n}{n} u(n-1)$ is not a linear combination of shifted one-sided exponentials multiplied by polynomials.


III) Differentiation


From the above example, let $y(n) = nx(n) = (-1)^{n+1}\alpha^n u(n-1)$.

\begin{align*}
y(n) &= (-1)^{n+1} \alpha^n u(n-1) \\
y(n) &= -(-1)^n \alpha^n u(n-1) \\
y(n) &= -(-\alpha)^n u(n-1) \\
y(n) &= \alpha (-\alpha)^{n-1} u(n-1) \\
\end{align*}

We have $y(n)$ is a shifted and scaled version of $\hat{G}(z) = \frac{1}{1 + \alpha z^{-1}}$.

\begin{align*}
\hat{Y}(z) = \frac{\alpha z^{-1}}{1 + \alpha z^{-1}} \\
\end{align*}

Or, we could use the differentiation property:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) \\
n x(n) &\ztp -z \frac{d}{dz} \hat{X}(z) \\
\end{align*}
\end{nicebox}

\begin{align*}
\hat{X} &= \log(1 + \alpha z^{-1}) \\
\frac{d}{dz}\hat{X} &= \frac{1}{1 + \alpha z^{-1}} \lr{-\alpha z^{-2}}  \\
-z \frac{d}{dz}\hat{X} &= \frac{\alpha z^{-1}}{1 + \alpha z^{-1}} \\
\end{align*}

We arrive at the same answer.


IV) Complex Integration (outside the scope of this course)


$$
x(n) = \frac{1}{2\pi i} \oint_c \hat{X}(z)z^{n-1}dz \ztp \hat{X}(z) = \sum \limits_{n\in\Z} x(n) z^{-n}
$$

Where $c$ is the closed contour within the $RoC$.

\end{example}


11.1 Properties of the Z-transform