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<h1>EE 120: Signals and Systems</h1>
<h1>Upsampling property</h1>
<p><mathjax>$$
x(n) \mapsto \uparrow N \mapsto y(n) = \begin{cases}
x(n/N) &amp; \text{if } n \equiv 0 (\mod N)
\\ 0 &amp; \rm{e/w}
\end{cases}
$$</mathjax></p>
<p>i.e. we have the same values, but now interspersed with more zeroes. Take
the axis and dilate by three. So see if you can come up with an expression
for the Z-transform of the upsampled signal. We should just have
<mathjax>$\hat{X}(z^N)$</mathjax>.</p>
<p>This should not surprise you. When you upsampled in the time domain, what
happened in frequency? We contracted in the frequency domain. You get that
even from here. If I remind you of an example we did eons ago, <mathjax>$y(n) =
\alpha y(n-1) + x(n)$</mathjax> had a frequency response of <mathjax>$G(\omega) = \frac{1}{1 -
\alpha e^{-i\omega}}$</mathjax>. If I change this to the parameters of <mathjax>$y(n) = \alpha
y(n-N) + x(n)$</mathjax>, <mathjax>$H(\omega) = \frac{1}{1 - \alpha e^{i\omega N}$</mathjax>. But if you
compare <mathjax>$g(n) = \alpha^n u(n)$</mathjax> with <mathjax>$h(n) = \alpha^{nN} u(n)$</mathjax>, we've
already seen this.</p>
<p>So when you upsample, you have the <mathjax>$z$</mathjax> raised to the <mathjax>$n^{th}$</mathjax> power.</p>
<p>What's the RoC? Should bring up two more questions: what happens to the
poles and zeroes? We take the <mathjax>$n^{th}$</mathjax> root of everything (i.e. the inverse
function), so everything moves closer to 1. (rationale: <mathjax>$z_p^N = p \implies
z_p = ?$</mathjax>. We get <mathjax>$N$</mathjax> times as many poles, in fact, since we have <mathjax>$N$</mathjax> roots
of <mathjax>$p$</mathjax>. ibid for zeros.</p>
<p>Going back to the question for the region of convergence for y: if the RoC
for x is <mathjax>$R_1 &lt; \abs{z} &lt; R_2$</mathjax>, the RoC for y is <mathjax>$R_1 &lt; \abs{z}^N &lt; R_2$</mathjax>,
so <mathjax>$R_y = R_x^{1/N}$</mathjax>.</p>
<p>So let's do the example given earlier: <mathjax>$y(n) = \alpha y(n-N) + x(n)$</mathjax>.
<mathjax>$\hat{H}_1(z) = \frac{1}{1 - \alpha z^{-1}}$</mathjax>. <mathjax>$\hat{H}_4 = \frac{1}{1 -
\alpha z^{-4}}$</mathjax>. Draw pole-zero diagrams, region of convergence?</p>
<p>(note that we've got degeneracy -- multiplicity. Must denote with a number
in parentheses if you've got multiplicity greater than 2; if multiplicity
is 2, you can use a double-circle or double-x).</p>
<h1>Differentiation</h1>
<p>Another property that's actually very important is differentiation in Z. So
suppose you've transformed <mathjax>$x \ztrans \hat{X}(z)$</mathjax>. What is <mathjax>$\deriv{\hat{X}}
{z}$</mathjax>? <mathjax>$-z\deriv{\hat{X}}{z} \ztrans n(x(n)$</mathjax>.</p>
<h2>Example:</h2>
<p><mathjax>$g(n) = n\alpha^n u(n) \ztrans \hat{G}(z) = ?$</mathjax></p>
<p>If you want to make this look like the original form, just multiply top and
bottom by <mathjax>$z^{-2}$</mathjax>. Very important point: extension to higher derivatives.</p>
<p>So what happens as we increase this? What does this mean?</p>
<p>We can decompose any rational z transform into a linear composition of
lower-order terms. Fundamental theorem of
algebra. Proposition: suppose we've got a transfer function. We've got a
numerator over a denominator. We can factor the numerator and
denominator. You also learned that whenever you do this, you can break
apart the ratio in terms of a sum.</p>
<p>Note that this starts breaking when you have degeneracy (i.e. systems with
duplicate poles). So from this qualitative argument, it should not surprise
you if I tell you that the only way you can get a rational Z-transform is
if the system is the sum of one-sided exponentials multiplied by
some polynomial.</p>
<p>We'd also have to include the left-sided versions of these.</p>
<p>We can make a general statement: a Z-transform expression <mathjax>$\hat{X}(z)$</mathjax> is
rational iff x(n) is a linear combination of terms <mathjax>$n^k \alpha^n u(n)$</mathjax>,
<mathjax>$n^k\beta^n u(-n)$</mathjax>. Shifted versions will certainly also work.</p>
<p>Using partial fractions is one of the methods of doing an inverse
transform. We're not going to learn a formal inverse Z-transform; we're
just going to use various heuristics (not unlike solving differential
equations).</p>
<p>In general, inverse z-transform requires a contour integral (complex
analysis) and thus is not a required in this class.</p>
<p>Now, if you believe this, we've got several things: <mathjax>$n^k\alpha^n u(n)$</mathjax>,
LCCDEs, and rational Z-transforms. They form a family.</p>
<h1>LCCDEs and Rational Z Transforms</h1>
<p>Suppose I've got an input, an impulse response, and an output. You know
this is a convolution of x and h, so <mathjax>$\hat{Y} = \hat{X}\hat{Z}$</mathjax>, which
means the transfer function of an LTI system is the ratio of the transform
of the output to the transform of the input (for LTI systems).</p>
<p>Frequency response of the filter gives you the Fourier transform of the
output.</p>
<p>We can write our difference equation as <mathjax>$\sum_{k=0}^N a_k y(n-k) = \sum_m^M
b_m x(n-m)$</mathjax>. We've seen this.</p>
<p>One way to get the transfer function is to take the z-transforms of both
sides. If they're equal in the time domain, their z-transforms must also be
equal in the frequency domain. Time-shift property. Just considering the
ratio <mathjax>$\hat{H} \equiv \frac{\hat{Y}}{\hat{X}}$</mathjax>, we have our transfer
function.</p>
<p>Familiarize yourself with going from the LCCDE to the transfer function by
inspection.</p>
<p>Now, for the end of the lecture: irrational Z-transform.</p>
<h2>Example</h2>
<p>This is a standard example in practically any signal-processing book you'll
find. <mathjax>$\hat{X} = \log(1 + \alpha z^{-1}$</mathjax>. Determine <mathjax>$x(n)$</mathjax>.</p>
<p>Using the differentiation property, <mathjax>$-z\deriv{\hat{X}}{z} \ztrans nx(n)$</mathjax>.</p>
<p><mathjax>$\frac{\alpha z^{-1}}{1 + \alpha z^{-1}} \ztrans $</mathjax></p>
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