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steveWang committed Aug 17, 2012
2 parents a63ddaf + b6ef4a3 commit 1223466778b0834ad92b9c2c3845f9d61df3b847
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@@ -2560,7 +2560,7 @@ <h2>April 26, 2012</h2>
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@@ -856,7 +856,7 @@ <h1>Friday, April 27</h1>
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-<?xml version="1.0" encoding="UTF-8" ?>
-<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
- "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
-
-<html xmlns="http://www.w3.org/1999/xhtml">
-
-<head>
-<title>26.md</title>
-
-</head>
-
-<body>
-
-<script src='https://d3eoax9i5htok0.cloudfront.net/mathjax/latest/MathJax.js?config=default,http://steveWang.github.com/Notes/config/local'></script>
-<script type="text/x-mathjax-config">
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- math = "\\displaystyle{"+math+"}";
- return PREFILTER.call(TEX,math,displaymode,script);
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-
-<h1>EE 105: Devices &amp; Circuits</h1>
-<h2>Wednesday, March 21, 2012</h2>
-<p>Midterm Friday; OBONNS.</p>
-<p>Today: Graphing and calculating <mathjax>$G_m$</mathjax>, <mathjax>$R_o$</mathjax>, <mathjax>$A_v$</mathjax>, keep talking about
-frequency response.</p>
-<h1>Graphings</h1>
-<p>So, the thing to do: you know you're going to have (if I've got
-some <mathjax>$A_0\sin(\omega t) \mapsto \abs{H(j\omega)}A_0\sin(\omega t + \angle
-H(\omega))$</mathjax>. If I give you a plot of angle or frequency, one way or
-another, you've got to find the angle of <mathjax>$H(j\omega)$</mathjax>. Also, find max/min
-voltages (determined by gain at various stages). Should make sure that those
-points are marked (circled, even) so Pister can see that we know what's up.</p>
-<p>If it's greater than 0, it's called <strong>"phase lead"</strong>. You want to find and
-label and mark zero-crossings and min/max in time (which is really in
-phase). Points move <em>left</em>.</p>
-<p>If it's less than 0, it's called <strong>"phase lag"</strong>, and points move right.</p>
-<h1>Calculating <mathjax>$G_m$</mathjax>, <mathjax>$R_o$</mathjax></h1>
-<p>There have been some questions about this on Piazza. There's the basics,
-like when you're looking into a collector or a drain, you're going to see
-<mathjax>$r_o$</mathjax>; when you're looking into an emitter or a source, and the
-collector/drain are at some AC ground, you're going to see <mathjax>$\frac{1}{g_m}$</mathjax>,
-and then there's the blocks that you know and ought to recognize/use in
-circuits, which are: what happens if I have some emitter degeneration or
-some source degeneration (<mathjax>$G_m \equiv \frac{g_m}{1 + g_m R_s}$</mathjax> for MOS,
-<mathjax>$R_s \equiv R_E \parallel r_\pi$</mathjax>); if I'm looking into the source/emitter,
-then this looks like <mathjax>$\frac{1}{g_m}\parens{1 + \frac{R_D}{r_o}}$</mathjax>. Note that
-for a BJT, you have <mathjax>$r_\pi$</mathjax> in parallel with this whole thing.</p>
-<p>For example: steps for finding gain: <mathjax>$A_v = -G_m R_o$</mathjax>. In general, this is
-how you'll find gain. <mathjax>$G_m = \frac{I_o}{V_i}\bigg|_{v_i=0}$</mathjax>. Straight
-taylor approximation: this is why we set <mathjax>$V_o = 0$</mathjax> when calculating <mathjax>$G_m$</mathjax>
-and <mathjax>$V_i = 0$</mathjax> when calculating <mathjax>$R_o$</mathjax>.</p>
-<p>When doing <mathjax>$G_m$</mathjax> calculation, we just don't care about anything above the
-junction.</p>
-<p>Mentally, what people like Razavi do is say "aha! I've got a bias voltage
-on that gate; this thing now is one of my basic elements! Aha! I've got a
-<mathjax>$\frac{1}{g_m}$</mathjax> to ground, and now I can draw my small signal model." And
-so the current coming out is roughly <mathjax>$g_m v_i$</mathjax>. The only time you look up
-is if there's a current source up there that depends on one of the lower
-node voltages.</p>
-<p>The reason why we call it <mathjax>$g_m$</mathjax> is because it's <strong>mutual</strong>
-transconductance.</p>
-<p>So now we're not so sure. What I see is <mathjax>$\frac{1}{g_m}\parens{1 + \frac{1 /
-g_m}{r_o}}$</mathjax> when looking into source.</p>
-<p>So now we've figured out what our <mathjax>$g_m v_i$</mathjax> is. Current dividers.</p>
-<p>So I wanted to do a couple more examples. For example, source follower
-(recall, measure output at source). We know <mathjax>$A_v = -G_m R_o$</mathjax>. So what is
-<mathjax>$R_o$</mathjax>? It's <mathjax>$\frac{1}{g_m} \parallel R_s$</mathjax>. <mathjax>$G_m$</mathjax> is a little tricky. Draw
-the small signal model, then do KCL at the output node. Note that since
-we're grounding <mathjax>$v_s$</mathjax>, <mathjax>$v_{gs} = v_i$</mathjax>, and no current is flowing through
-the source resistance (or the output resistance, even!). Thus <mathjax>$G_m \equiv
--g_m$</mathjax>. What does this mean? We're shoving current out, instead of shoving
-current in.</p>
-<p>So finally <mathjax>$A_v = -G_m R_o = \frac{g_m R_s}{1 + g_m R_s}$</mathjax></p>
-</body>
-</html>
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-<?xml version="1.0" encoding="UTF-8" ?>
-<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
- "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
-
-<html xmlns="http://www.w3.org/1999/xhtml">
-
-<head>
-<title>18.md</title>
-
-</head>
-
-<body>
-
-<script src='https://d3eoax9i5htok0.cloudfront.net/mathjax/latest/MathJax.js?config=default,https://raw.github.com/steveWang/Notes/master/config/local.js'></script>
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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>
-</body>
-</html>
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