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70 config/local.js
@@ -0,0 +1,70 @@
+/*************************************************************
+ *
+ * MathJax/config/local/local.js
+ *
+ * Include changes and configuration local to your installation
+ * in this file. For example, common macros can be defined here
+ * (see below). To use this file, add "local/local.js" to the
+ * config array in MathJax.js or your MathJax.Hub.Config() call.
+ *
+ * ---------------------------------------------------------------------
+ *
+ * Copyright (c) 2009 Design Science, Inc.
+ *
+ * Licensed under the Apache License, Version 2.0 (the "License");
+ * you may not use this file except in compliance with the License.
+ * You may obtain a copy of the License at
+ *
+ * http://www.apache.org/licenses/LICENSE-2.0
+ *
+ * Unless required by applicable law or agreed to in writing, software
+ * distributed under the License is distributed on an "AS IS" BASIS,
+ * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
+ * See the License for the specific language governing permissions and
+ * limitations under the License.
+ */
+
+
+MathJax.Hub.Register.StartupHook("TeX Jax Ready",function () {
+ var TEX = MathJax.InputJax.TeX;
+
+ // place macros here. E.g.:
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View
13 cs191.html
@@ -1,4 +1,4 @@
-<script src='file:///home/steve/hackerspace/mathjax/unpacked/MathJax.js?config=default'></script>
+<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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@@ -130,9 +130,11 @@
not going to be a mechanism unlike anything else.</p>
<p>Part of understanding QM is coming to terms psychologically with this
superposition of states, the existence in more than one state
-simultaneously.</p>
-<h1>CS 191: Qubits, Quantum Mechanics and Computers</h1>
-<h2>Qubits, Superposition, &amp; Measurement -- January 19, 2012</h2>
+simultaneously.
+CS 191: Qubits, Quantum Mechanics and Computers
+===============================================
+Qubits, Superposition, &amp; Measurement -- January 19, 2012</p>
+<hr />
<h2>Quantization:</h2>
<ul>
<li>Atomic orbitals: Electrons within an atom exist in quantized energy
@@ -1346,8 +1348,7 @@
<p><mathjax>$F_{jk} = e^{ijk2\pi/n}$</mathjax></p>
<p>Primitive square root of unity: <mathjax>$e^{i2\pi n}$</mathjax>. When we do <mathjax>$F_n$</mathjax>, it's the
same, but with phases.</p>
-<p>Showing that <mathjax>$\braket{F_i}{F_j} = \delta_{ij}$</mathjax>...</p>
-<h2>Properties of the Fourier Transform</h2>
+<h2>Showing that <mathjax>$\braket{F_i}{F_j} = \delta_{ij}$</mathjax>...Properties of the Fourier Transform</h2>
<p>This is a proper circuit, and in fact, you can implement it very very
efficiently.</p>
<p>The QFT is very efficient to implement by a quantum circuit.</p>
View
94 ee105.html
@@ -1,4 +1,4 @@
-<script src='file:///home/steve/hackerspace/mathjax/unpacked/MathJax.js?config=default'></script>
+<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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@@ -179,8 +179,7 @@
εA/W = ε/√(εV₀/q (1/N{A} + 1/N{D})⁻¹) = √(εq/(2V₀) N{A}N{D}/(N{A} +
N{D})). Freshman physics.</p>
<p>All of this stays the same, except we replace V₀ with V₀ - V{D}.</p>
-<p>AC{j0}/√(1 - stuff)</p>
-<p>reverse bias:</p>
+<p>AC{j0}/√(1 - stuff)reverse bias:</p>
<pre><code>V{D} &lt; 0, increasing E field, increasing W, decreasing C,
decreasing diffusion, increasing barrier height.
</code></pre>
@@ -468,8 +467,7 @@
| |active | |
+---+-------+----------+</p>
<p>N^+PN^+, wouldn't care which way you ran. Work just the same. Curves would
-be the same. Called reverse active.</p>
-<p>We talked about this last time: let's say I increase V{be} by \delta
+be the same. Called reverse active.We talked about this last time: let's say I increase V{be} by \delta
V{be}by 26mV. The 26mV here means that I increase I{c} by some \delta I{c}.
That will cause a decrease in voltage. \partial V{ce}/\partialV{bd}.</p>
<p>Operating point: where you first bias a device All of the extra \delta s
@@ -853,11 +851,9 @@
\\ \frac{1}{R_x} v_o - g_m v_{be - \frac{1}{r\pi}v_{be} = 0
\\ \frac{1}{R_x}v_o = \left(g_m + \frac{1}{r_\pi}\right)i_i r_\pi
\\ r_{in} = r_\pi + (\beta + 1)R_x
-$$</mathjax>
-EE 105: Devices &amp; Circuits
-==========================
-Monday, March 5, 2012</p>
-<hr />
+$$</mathjax></p>
+<h1>EE 105: Devices &amp; Circuits</h1>
+<h2>Monday, March 5, 2012</h2>
<p>Last time: real amplifiers; idealized model.</p>
<p>Result is you get two voltage dividers.</p>
<p>Similarly, you'll get a voltage divider on the output.</p>
@@ -903,11 +899,9 @@
<p>On the homework this week, you'll see that the same circuit for bipolar, if
the source degeneration is 0, if you vary process (<mathjax>$I_s$</mathjax>, <mathjax>$V_A$</mathjax>, <mathjax>$\beta$</mathjax>,
<mathjax>$\mu_n$</mathjax>, <mathjax>$C_{ox}$</mathjax>, <mathjax>$V_{th}$</mathjax>, <mathjax>$\lambda$</mathjax>), voltage (<mathjax>$V_{CC}$</mathjax>, <mathjax>$V_{DD}$</mathjax>),
-temperature (also, heating) -- "PVT".
-EE 105: Devices &amp; Circuits
-==========================
-Wednesday, March 7, 2012</p>
-<hr />
+temperature (also, heating) -- "PVT".</p>
+<h1>EE 105: Devices &amp; Circuits</h1>
+<h2>Wednesday, March 7, 2012</h2>
<p>potatoes</p>
<p><mathjax>$G_m$</mathjax>, <mathjax>$G_o$</mathjax>, <mathjax>$R_o$</mathjax>, <mathjax>$A_v$</mathjax>.</p>
<p>Graph of <mathjax>$I_C$</mathjax> vs <mathjax>$V_{CE}$</mathjax>.</p>
@@ -969,11 +963,9 @@
<mathjax>$-g_mR_c$</mathjax>. Once we add in a resistance, we have <mathjax>$R_o = R_c \parallel r_o(1
+ g_mR_E)$</mathjax>; <mathjax>$G_m = \frac{g_m}{1+g_mR_E}$</mathjax>; <mathjax>$A_v = -G_mR_o = -\frac{R_c}
{R_E}$</mathjax>.</p>
-<p>Gain from base to collector is going to be -10.
-EE 105: Devices &amp; Circuits
-==========================
-Wednesday, March 14, 2012</p>
-<hr />
+<p>Gain from base to collector is going to be -10.</p>
+<h1>EE 105: Devices &amp; Circuits</h1>
+<h2>Wednesday, March 14, 2012</h2>
<p>PNP, PMOS</p>
<p>For the PNP, we have</p>
<pre><code>E P+ (highest doping)
@@ -1211,4 +1203,64 @@
<p>For Bode plot, straight line approximation; fine. However, accuracy demands
that we consider the actual numerical values (not going to be drawn): just
examine the unit circle and reference triangles, and use small signal
-approximation.</p>
+approximation.
+EE 105: Devices &amp; Circuits
+==========================
+Wednesday, March 21, 2012</p>
+<hr />
+<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>. And in
+particular, in a MOS, it's very common that you'd see the output between
+two MOSFETs, so you'd get something like <mathjax>$\frac{g_m r_o}{1 + g_m r_o}$</mathjax>. If
+you're lucky, this looks something like <mathjax>$0.9 - 0.999$</mathjax>.</p>
+<p>All right. So here's a couple of good ones to think about. (various
+practice circuits)</p>
View
121 ee120.html
@@ -1,4 +1,4 @@
-<script src='file:///home/steve/hackerspace/mathjax/unpacked/MathJax.js?config=default'></script>
+<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">
MathJax.Hub.Register.StartupHook("TeX Jax Ready",function () {
var TEX = MathJax.InputJax.TeX;
@@ -110,11 +110,13 @@
* Causality
+ Basically, no dependence on future values.
- If two inputs are identical to some point, then their outputs
- must also be identical to that same point.</p>
+ must also be identical to that same point.
+EE 120: Signals and Systems
+===========================
+January 19, 2012.</p>
+<hr />
</li>
</ul>
-<h1>EE 120: Signals and Systems</h1>
-<h2>January 19, 2012.</h2>
<h2>More on causality</h2>
<p>(proof by contradiction of previous example)</p>
<p>(input zero up to a certain point, but the output is nonzero before said
@@ -176,9 +178,11 @@
<p><mathjax>$y(n) = \alpha y(n-1) + x(n)$</mathjax>. causal, <mathjax>$y(-1) = 0$</mathjax>.</p>
<p>(geometric sum; BIBO stability depends on magnitude of \alpha being less
than
-1.)</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>January 24, 2012.</h2>
+1.)
+EE 120: Signals and Systems
+===========================
+January 24, 2012.</p>
+<hr />
<h1>LTI Systems and Frequency Response</h1>
<p>Freq. response: <mathjax>$H(\omega)$</mathjax>
(continuous-time?)</p>
@@ -268,7 +272,7 @@
<mathjax>$y(-N) = ... = y(n) = 0$</mathjax></p>
<p><mathjax>$(\abs{\alpha} &lt; 1)$</mathjax></p>
<p><mathjax>$$h_{N}(n) = \alpha h_{N}(n-N) + \delta(n) = !(n % N) &amp;&amp; \alpha^{n/N}.
-H_{N}(\omega)\\ = \sum\alpha^{k}e^{-i\omega Nk} (k \from 0 \to \infty)
+H_{N}(\omega)\\ = \sum_{k=0}^\infty\alpha^{k}e^{-i\omega Nk}
\sum(\alpha e^{-i\omega N})^k = 1/(1-\alpha e^{-i\omega N})
\\ \abs{H_{N}(\omega)} = \abs{e^{i\omega N}/(e^{i\omega N} - \alpha)}
$$</mathjax></p>
@@ -286,9 +290,11 @@
<p><mathjax>$G(\omega) = \int g(t)e^{-i\omega t}dt
\\ = 1/(\alpha + i\omega ).
\\ = 1/(i\omega - (-\alpha ))$</mathjax></p>
-<p><mathjax>$\sqrt{\omega ^2 + \alpha ^2}$</mathjax></p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>January 31, 2012.</h2>
+<p><mathjax>$\sqrt{\omega ^2 + \alpha ^2}$</mathjax>
+EE 120: Signals and Systems
+===========================
+January 31, 2012.</p>
+<hr />
<p>Homework should be out today. Yes, there are homeworks in this class.</p>
<p>analog system procedure for figuring out frequency response of
systems.</p>
@@ -365,9 +371,11 @@
\frac{1}{p} \braket{x}{\psi_l}$</mathjax>.</p>
<p>synthesis equation, analysis equation:</p>
<p><mathjax>$x(n) = \sum X_k e^{ik\omega_0n}$</mathjax>
-<mathjax>$X(n) = \sum x_k e^{-ik\omega_0n}$</mathjax></p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 2, 2012.</h2>
+<mathjax>$X(n) = \sum x_k e^{-ik\omega_0n}$</mathjax>
+EE 120: Signals and Systems
+===========================
+February 2, 2012.</p>
+<hr />
<h1>DTFS</h1>
<p><mathjax>$x(n) = \sum X_k e^{ik\omega_0n}$</mathjax>. The complex exponentials form a basis
for <mathjax>$\mathbb{C}^2$</mathjax>. The way we define inner products for signals is exactly
@@ -377,9 +385,11 @@
<p><mathjax>$x(n) = cos(n)$</mathjax> not periodic in discrete time <mathjax>$\implies$</mathjax> no DTFS.</p>
<p>Evidently there will be a quiz on Tue, since pset is due on Wed.</p>
<p>Consider what it means to send discrete-time periodic signals through
-LTI systems.</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 7, 2012.</h2>
+LTI systems.
+EE 120: Signals and Systems
+===========================
+February 7, 2012.</p>
+<hr />
<p>CTFS:</p>
<p>periodic if <mathjax>$x(t+p) = x(t) \exists p \in R$</mathjax>. Smallest positive <mathjax>$p$</mathjax> is
called the fundamental period. Fundamental frequency <mathjax>$\omega_0 \equiv
@@ -440,9 +450,11 @@
<mathjax>$\frac{\sin(k\omega_0\Delta/2)}{pk\pi\Delta/2}$</mathjax></p>
<p>What happens if I want to approximate a signal that has finite energy? What
should the coefficients <mathjax>$\alpha_k$</mathjax> be?</p>
-<p>orthogonal projection! Least squares!</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 9, 2012.</h2>
+<p>orthogonal projection! Least squares!
+EE 120: Signals and Systems
+===========================
+February 9, 2012.</p>
+<hr />
<h1>Discrete-time Fourier transform</h1>
<p>Discrete aperiodic signals. The DTFT can also handle discrete-time periodic
signals, provided that we make use of Dirac deltas in the frequency
@@ -485,9 +497,11 @@
variable <mathjax>$\omega$</mathjax> which happen to be periodic with period <mathjax>$2\pi$</mathjax>.</p>
<h1>Ideal discrete-time Low-pass Filter</h1>
<p>Impulse response <mathjax>$h(n) = \frac{1}{2\pi}\int H(\omega)e^{i\omega n}d\omega$</mathjax>.</p>
-<p><mathjax>$\frac{B}{\pi n} \sin(An)$</mathjax></p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 16, 2012.</h2>
+<p><mathjax>$\frac{B}{\pi n} \sin(An)$</mathjax>
+EE 120: Signals and Systems
+===========================
+February 16, 2012.</p>
+<hr />
<h2>Discrete-Time Fourier Transform: continued</h2>
<p>Recall the ideal low-pass filter. In time-domain, it had some
discontinuities. In freq domain, it was NOT absolutely summable. However,
@@ -586,9 +600,11 @@
<p>easy way of carrying out circular convolution: take one of the two
functions, keep one replica (e.g. the one from <mathjax>$-\pi$</mathjax> to <mathjax>$\pi$</mathjax>), and then
do a regular convolution with the other function. advice: choose the
-flipped + shifted signal for elimination of replicas.</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 21, 2012.</h2>
+flipped + shifted signal for elimination of replicas.
+EE 120: Signals and Systems
+===========================
+February 21, 2012.</p>
+<hr />
<h1>Circular convolution wrap-up</h1>
<p><mathjax>$h(n) = \frac{B}{\pi n} \sin(An) \fourier H(\omega) = u(n+A)-u(n-A)$</mathjax>.
(ideal low-pass filter and its impulse response)</p>
@@ -692,9 +708,11 @@
\frac{1}{\Delta^2 + t^2}$</mathjax>. Turns out <mathjax>$\alpha=1$</mathjax>.</p>
<p>This is a Cauchy probability density function.</p>
<p>(names for dirac delta: generalized function, distribution. Look up theory
- of distributions)</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 23, 2012.</h2>
+ of distributions)
+EE 120: Signals and Systems
+===========================
+February 23, 2012.</p>
+<hr />
<h2>Continuous-time Fourier transform</h2>
<p>Recall:
<mathjax>$$X(\omega) = \int_{-\infty}^\infty x(t)e^{-i\omega t}dt
@@ -761,9 +779,11 @@
signal is the same as the transmitted signal. Whole area of communication
theory that deals with deterioration. Other assumption is that the
oscillator at the receiver that can generate the exact same frequency as
-the transmitter oscillator and at the same phase.</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>February 28, 2012.</h2>
+the transmitter oscillator and at the same phase.
+EE 120: Signals and Systems
+===========================
+February 28, 2012.</p>
+<hr />
<h1>AM Continued</h1>
<p>(review of what we just did)</p>
<h2>Recall:</h2>
@@ -802,9 +822,11 @@
<p>Each player is allocated a piece of real estate along the frequency axis.</p>
<h2>Quadrature multiplexing</h2>
<p>The way we can do this is by exploiting the orthogonality of cosine and
-sine. What's being transmitted is the sum of the two.</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>March 1, 2012.</h2>
+sine. What's being transmitted is the sum of the two.
+EE 120: Signals and Systems
+===========================
+March 1, 2012.</p>
+<hr />
<h1>Sampling of CT Signals</h1>
<p>Now we have to differentiate between frequency in radians per second
(<mathjax>$\omega$</mathjax>), and frequency in radians per sample (<mathjax>$\Omega$</mathjax>). Our sampling
@@ -928,9 +950,11 @@
<mathjax>$\omega_0$</mathjax>. The sampling signal <mathjax>$Q(\omega)$</mathjax> will have an impulse train
separated by <mathjax>$\omega_s$</mathjax>. When we convolve this and apply a low-pass filter,
we have just one remaining frequency at <mathjax>$\omega_0-\omega_s = -\frac
-{\omega_0}{2}$</mathjax>.</p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>March 6, 2012.</h2>
+{\omega_0}{2}$</mathjax>.
+EE 120: Signals and Systems
+===========================
+March 6, 2012.</p>
+<hr />
<h1>Sampling Cont'd</h1>
<p>We are still in the first of three blocks (where we take a continuous-time
signal and create a discrete-time signal).</p>
@@ -954,9 +978,11 @@
your reconstruction period is the same as your sampling period.</p>
<p>Then you know that your output is equal <mathjax>$\frac{G}{T} H_d\parens{ \omega T}
X_c(\omega)$</mathjax>. The equivalent LTI filter is simply <mathjax>$\frac{G}{T} nH_d \parens{
-\omega T}$</mathjax></p>
-<h1>EE 120: Signals and Systems</h1>
-<h2>March 8, 2012.</h2>
+\omega T}$</mathjax>
+EE 120: Signals and Systems
+===========================
+March 8, 2012.</p>
+<hr />
<p>Spoke about sampling, impulse train, conditions under which signal can be
recovered -- if band-limited, and frequency high enough, we can recover
with a low-pass filter.</p>
@@ -1025,9 +1051,11 @@
functions.</p>
<p>It turns out these coefficients are the values of the signal at the sampled
points. That isn't obvious yet. After some algebraic manipulations, we can
-finally see that this whole thing is just <mathjax>$x_c(kT_s)$</mathjax>.</p>
-<h1>EE 120: Signals and Systems</h1>
-<h1><mathjax>$\mathcal{Z}$</mathjax> Transform</h1>
+finally see that this whole thing is just <mathjax>$x_c(kT_s)$</mathjax>.
+EE 120: Signals and Systems
+===========================
+<mathjax>$\mathcal{Z}$</mathjax> Transform
+=======================</p>
<p>Something we've been brushing under the carpet for a while is the set of
signals in which the Fourier transform is not defined.</p>
<p>So the Z transform is defined for discrete-time LTI systems. (discrete
@@ -1187,7 +1215,7 @@
<p>We have a few minutes, so let me talk about the distinctions between causal
signals and right-sided signals (and also anticausal / left-sided).</p>
<p>So let's say we take a right-sided but not causal signal. Now the RoC is
-outside of radius <mathjax>$\alpha$</mathjax>, but now you have to exclude <mathjax>$\infinity$</mathjax>.</p>
+outside of radius <mathjax>$\alpha$</mathjax>, but now you have to exclude $\infty.</p>
<p>Similarly, for left-sided signals, you'd then exclude 0.
EE 120: Signals and Systems
===========================
@@ -1296,4 +1324,5 @@
this (albeit in parts).</p>
<p>Note that radius of convergence isn't changing.</p>
<p>Will leave it to you to figure out what the transform of <mathjax>$r^n \cos(\omega_0
-n) u(n)$</mathjax> is.</p>
+n) u(n)$</mathjax> is.</p>
+<h1>EE 120: Signals and Systems</h1>
View
11 phys112.html
@@ -1,4 +1,4 @@
-<script src='file:///home/steve/hackerspace/mathjax/unpacked/MathJax.js?config=default'></script>
+<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">
MathJax.Hub.Register.StartupHook("TeX Jax Ready",function () {
var TEX = MathJax.InputJax.TeX;
@@ -507,8 +507,7 @@
<p>What is chemical potential? It is a measure of the concentration! (verify
by working out -τ ∂σ/∂N). A lot of work for a very simple result. But this
is much more powerful. We will use this to look at batteries, equilibrium
-in various systems.</p>
-<p>MIDTERM: Friday the 24th. 8:10-9:00. Why Friday and not Monday? Turns out
+in various systems.MIDTERM: Friday the 24th. 8:10-9:00. Why Friday and not Monday? Turns out
there is an important workshop on dark matter in UCLA. There is a potential
problem: Monday the 20th is President's day, so no class, no office
hours. Proposition: Tuesday the 21st, we have extended office hours from
@@ -614,9 +613,9 @@
<p>C{V} = (∂Q/∂τ){V} = T(∂σ/∂τ){V}
C{p} = (∂Q/∂τ){p} = T(∂σ/∂τ){p}</p>
<p>Thus C{v} = C{p} + τ(∂σ/∂P·∂P/∂τ)</p>
-<p>lpha ≡ 1/V ∂V/∂τ. "volume coefficient of expansion"</p>
+<p>\alpha ≡ 1/V ∂V/∂τ. "volume coefficient of expansion"</p>
<p>κ ≡ -1/V ∂V/∂p "isothermal compressibility"</p>
-<p>Therefore C{p} - C{v} = Vτ(lpha²/κ)</p>
+<p>Therefore C{p} - C{v} = Vτ(\alpha²/κ)</p>
<p>What's needed to calculate entropy</p>
<p>Use the relations shown in the previous part. All we need are C{v} and
equation of state. But C{v} is also computable using equation of state at a
@@ -652,7 +651,7 @@
<p>Clausius statement for 2nd law of thermodynamics: effectively, Carnot
engine is maximally efficient, given τ₁, τ₂. (q₂/q₁ ≤ τ₂/τ₁)</p>
<p>Joule-Thomson process: H = U + PV ⇒ H₁ = H₂ if dQ = 0.
-μ ≡ V/C{p} (τlpha - 1)
+μ ≡ V/C{p} (τ\alpha - 1)
Physics 112: Statistical Mechanics
==================================
Chemical Potential. Feb 22, 2012</p>
View
2  phys137a.html
@@ -1,4 +1,4 @@
-<script src='file:///home/steve/hackerspace/mathjax/unpacked/MathJax.js?config=default'></script>
+<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">
MathJax.Hub.Register.StartupHook("TeX Jax Ready",function () {
var TEX = MathJax.InputJax.TeX;
View
2  sp2012/cs191/1.md
@@ -136,5 +136,3 @@ not going to be a mechanism unlike anything else.
Part of understanding QM is coming to terms psychologically with this
superposition of states, the existence in more than one state
simultaneously.
-
-
View
5 sp2012/cs191/16.md
@@ -38,10 +38,7 @@ $F_{jk} = e^{ijk2\pi/n}$
Primitive square root of unity: $e^{i2\pi n}$. When we do $F_n$, it's the
same, but with phases.
-Showing that $\braket{F_i}{F_j} = \delta_{ij}$...
-
-
-Properties of the Fourier Transform
+Showing that $\braket{F_i}{F_j} = \delta_{ij}$...Properties of the Fourier Transform
-----------------------------------
This is a proper circuit, and in fact, you can implement it very very
efficiently.
View
7 sp2012/cs191/cs191.md
@@ -136,8 +136,6 @@ not going to be a mechanism unlike anything else.
Part of understanding QM is coming to terms psychologically with this
superposition of states, the existence in more than one state
simultaneously.
-
-
CS 191: Qubits, Quantum Mechanics and Computers
===============================================
Qubits, Superposition, & Measurement -- January 19, 2012
@@ -1653,10 +1651,7 @@ $F_{jk} = e^{ijk2\pi/n}$
Primitive square root of unity: $e^{i2\pi n}$. When we do $F_n$, it's the
same, but with phases.
-Showing that $\braket{F_i}{F_j} = \delta_{ij}$...
-
-
-Properties of the Fourier Transform
+Showing that $\braket{F_i}{F_j} = \delta_{ij}$...Properties of the Fourier Transform
-----------------------------------
This is a proper circuit, and in fact, you can implement it very very
efficiently.
View
2  sp2012/ee105/19.md
@@ -82,3 +82,5 @@ v_i - v_o = i_i r_\pi = v_{in}
\\ \frac{1}{R_x}v_o = \left(g_m + \frac{1}{r_\pi}\right)i_i r_\pi
\\ r_{in} = r_\pi + (\beta + 1)R_x
$$
+
+
View
2  sp2012/ee105/20.md
@@ -69,3 +69,5 @@ On the homework this week, you'll see that the same circuit for bipolar, if
the source degeneration is 0, if you vary process ($I_s$, $V_A$, $\beta$,
$\mu_n$, $C_{ox}$, $V_{th}$, $\lambda$), voltage ($V_{CC}$, $V_{DD}$),
temperature (also, heating) -- "PVT".
+
+
View
3  sp2012/ee105/21.md
@@ -35,4 +35,5 @@ Remember $A_v = -G_m R_o$ whether it's bipolar or MOS, which is equal to
$-\frac{g_m}{1 + g_M R_s}\parens{R_o\parallel r_o(1 + g_m R_s)}$. So if
$R_s \gg \frac{1}{g_m}$, we get $A_v \approx -\frac{R_D}{R_S} = -\frac{R_C}
{R_E}$. So you'll often make five resistors next to each other, and that
-
+
+
View
2  sp2012/ee105/22.md
@@ -44,3 +44,5 @@ $-g_mR_c$. Once we add in a resistance, we have $R_o = R_c \parallel r_o(1
{R_E}$.
Gain from base to collector is going to be -10.
+
+
View
82 sp2012/ee105/26.html
@@ -0,0 +1,82 @@
+<?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">
+MathJax.Hub.Register.StartupHook("TeX Jax Ready",function () {
+var TEX = MathJax.InputJax.TeX;
+var PREFILTER = TEX.prefilterMath;
+TEX.Augment({
+ prefilterMath: function (math,displaymode,script) {
+ math = "\\displaystyle{"+math+"}";
+ return PREFILTER.call(TEX,math,displaymode,script);
+ }
+ });
+});
+</script>
+
+<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>
View
78 sp2012/ee105/26.md
@@ -0,0 +1,78 @@
+EE 105: Devices & Circuits
+==========================
+Wednesday, March 21, 2012
+-------------------------
+
+Midterm Friday; OBONNS.
+
+Today: Graphing and calculating $G_m$, $R_o$, $A_v$, keep talking about
+frequency response.
+
+Graphings
+=========
+So, the thing to do: you know you're going to have (if I've got
+some $A_0\sin(\omega t) \mapsto \abs{H(j\omega)}A_0\sin(\omega t + \angle
+H(\omega))$. If I give you a plot of angle or frequency, one way or
+another, you've got to find the angle of $H(j\omega)$. 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.
+
+If it's greater than 0, it's called **"phase lead"**. You want to find and
+label and mark zero-crossings and min/max in time (which is really in
+phase). Points move *left*.
+
+If it's less than 0, it's called **"phase lag"**, and points move right.
+
+Calculating $G_m$, $R_o$
+========================
+
+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
+$r_o$; when you're looking into an emitter or a source, and the
+collector/drain are at some AC ground, you're going to see $\frac{1}{g_m}$,
+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 ($G_m \equiv \frac{g_m}{1 + g_m R_s}$ for MOS,
+$R_s \equiv R_E \parallel r_\pi$); if I'm looking into the source/emitter,
+then this looks like $\frac{1}{g_m}\parens{1 + \frac{R_D}{r_o}}$. Note that
+for a BJT, you have $r_\pi$ in parallel with this whole thing.
+
+For example: steps for finding gain: $A_v = -G_m R_o$. In general, this is
+how you'll find gain. $G_m = \frac{I_o}{V_i}\bigg|_{v_i=0}$. Straight
+taylor approximation: this is why we set $V_o = 0$ when calculating $G_m$
+and $V_i = 0$ when calculating $R_o$.
+
+When doing $G_m$ calculation, we just don't care about anything above the
+junction.
+
+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
+$\frac{1}{g_m}$ to ground, and now I can draw my small signal model." And
+so the current coming out is roughly $g_m v_i$. The only time you look up
+is if there's a current source up there that depends on one of the lower
+node voltages.
+
+The reason why we call it $g_m$ is because it's **mutual**
+transconductance.
+
+So now we're not so sure. What I see is $\frac{1}{g_m}\parens{1 + \frac{1 /
+g_m}{r_o}}$ when looking into source.
+
+So now we've figured out what our $g_m v_i$ is. Current dividers.
+
+So I wanted to do a couple more examples. For example, source follower
+(recall, measure output at source). We know $A_v = -G_m R_o$. So what is
+$R_o$? It's $\frac{1}{g_m} \parallel R_s$. $G_m$ is a little tricky. Draw
+the small signal model, then do KCL at the output node. Note that since
+we're grounding $v_s$, $v_{gs} = v_i$, and no current is flowing through
+the source resistance (or the output resistance, even!). Thus $G_m \equiv
+-g_m$. What does this mean? We're shoving current out, instead of shoving
+current in.
+
+So finally $A_v = -G_m R_o = \frac{g_m R_s}{1 + g_m R_s}$. And in
+particular, in a MOS, it's very common that you'd see the output between
+two MOSFETs, so you'd get something like $\frac{g_m r_o}{1 + g_m r_o}$. If
+you're lucky, this looks something like $0.9 - 0.999$.
+
+All right. So here's a couple of good ones to think about. (various
+practice circuits)
View
5 sp2012/ee105/5.md
@@ -24,10 +24,7 @@ N{D})). Freshman physics.
All of this stays the same, except we replace V₀ with V₀ - V{D}.
-AC{j0}/√(1 - stuff)
-
-
-reverse bias:
+AC{j0}/√(1 - stuff)reverse bias:
V{D} < 0, increasing E field, increasing W, decreasing C,
decreasing diffusion, increasing barrier height.
View
5 sp2012/ee105/9.md
@@ -33,10 +33,7 @@ BC In |Rev|Cutoff |Forward |
+---+-------+----------+
N^+PN^+, wouldn't care which way you ran. Work just the same. Curves would
-be the same. Called reverse active.
-
-
-We talked about this last time: let's say I increase V{be} by \delta
+be the same. Called reverse active.We talked about this last time: let's say I increase V{be} by \delta
V{be}by 26mV. The 26mV here means that I increase I{c} by some \delta I{c}.
That will cause a decrease in voltage. \partial V{ce}/\partialV{bd}.
View
97 sp2012/ee105/ee105.md
@@ -232,10 +232,7 @@ N{D})). Freshman physics.
All of this stays the same, except we replace V₀ with V₀ - V{D}.
-AC{j0}/√(1 - stuff)
-
-
-reverse bias:
+AC{j0}/√(1 - stuff)reverse bias:
V{D} < 0, increasing E field, increasing W, decreasing C,
decreasing diffusion, increasing barrier height.
@@ -595,10 +592,7 @@ BC In |Rev|Cutoff |Forward |
+---+-------+----------+
N^+PN^+, wouldn't care which way you ran. Work just the same. Curves would
-be the same. Called reverse active.
-
-
-We talked about this last time: let's say I increase V{be} by \delta
+be the same. Called reverse active.We talked about this last time: let's say I increase V{be} by \delta
V{be}by 26mV. The 26mV here means that I increase I{c} by some \delta I{c}.
That will cause a decrease in voltage. \partial V{ce}/\partialV{bd}.
@@ -1108,6 +1102,8 @@ v_i - v_o = i_i r_\pi = v_{in}
\\ \frac{1}{R_x}v_o = \left(g_m + \frac{1}{r_\pi}\right)i_i r_\pi
\\ r_{in} = r_\pi + (\beta + 1)R_x
$$
+
+
EE 105: Devices & Circuits
==========================
Monday, March 5, 2012
@@ -1179,6 +1175,8 @@ On the homework this week, you'll see that the same circuit for bipolar, if
the source degeneration is 0, if you vary process ($I_s$, $V_A$, $\beta$,
$\mu_n$, $C_{ox}$, $V_{th}$, $\lambda$), voltage ($V_{CC}$, $V_{DD}$),
temperature (also, heating) -- "PVT".
+
+
EE 105: Devices & Circuits
==========================
Wednesday, March 7, 2012
@@ -1216,7 +1214,8 @@ Remember $A_v = -G_m R_o$ whether it's bipolar or MOS, which is equal to
$-\frac{g_m}{1 + g_M R_s}\parens{R_o\parallel r_o(1 + g_m R_s)}$. So if
$R_s \gg \frac{1}{g_m}$, we get $A_v \approx -\frac{R_D}{R_S} = -\frac{R_C}
{R_E}$. So you'll often make five resistors next to each other, and that
-
+
+
EE 105: Devices & Circuits
==========================
Friday, March 9, 2012
@@ -1263,6 +1262,8 @@ $-g_mR_c$. Once we add in a resistance, we have $R_o = R_c \parallel r_o(1
{R_E}$.
Gain from base to collector is going to be -10.
+
+
EE 105: Devices & Circuits
==========================
Wednesday, March 14, 2012
@@ -1581,3 +1582,81 @@ For Bode plot, straight line approximation; fine. However, accuracy demands
that we consider the actual numerical values (not going to be drawn): just
examine the unit circle and reference triangles, and use small signal
approximation.
+EE 105: Devices & Circuits
+==========================
+Wednesday, March 21, 2012
+-------------------------
+
+Midterm Friday; OBONNS.
+
+Today: Graphing and calculating $G_m$, $R_o$, $A_v$, keep talking about
+frequency response.
+
+Graphings
+=========
+So, the thing to do: you know you're going to have (if I've got
+some $A_0\sin(\omega t) \mapsto \abs{H(j\omega)}A_0\sin(\omega t + \angle
+H(\omega))$. If I give you a plot of angle or frequency, one way or
+another, you've got to find the angle of $H(j\omega)$. 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.
+
+If it's greater than 0, it's called **"phase lead"**. You want to find and
+label and mark zero-crossings and min/max in time (which is really in
+phase). Points move *left*.
+
+If it's less than 0, it's called **"phase lag"**, and points move right.
+
+Calculating $G_m$, $R_o$
+========================
+
+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
+$r_o$; when you're looking into an emitter or a source, and the
+collector/drain are at some AC ground, you're going to see $\frac{1}{g_m}$,
+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 ($G_m \equiv \frac{g_m}{1 + g_m R_s}$ for MOS,
+$R_s \equiv R_E \parallel r_\pi$); if I'm looking into the source/emitter,
+then this looks like $\frac{1}{g_m}\parens{1 + \frac{R_D}{r_o}}$. Note that
+for a BJT, you have $r_\pi$ in parallel with this whole thing.
+
+For example: steps for finding gain: $A_v = -G_m R_o$. In general, this is
+how you'll find gain. $G_m = \frac{I_o}{V_i}\bigg|_{v_i=0}$. Straight
+taylor approximation: this is why we set $V_o = 0$ when calculating $G_m$
+and $V_i = 0$ when calculating $R_o$.
+
+When doing $G_m$ calculation, we just don't care about anything above the
+junction.
+
+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
+$\frac{1}{g_m}$ to ground, and now I can draw my small signal model." And
+so the current coming out is roughly $g_m v_i$. The only time you look up
+is if there's a current source up there that depends on one of the lower
+node voltages.
+
+The reason why we call it $g_m$ is because it's **mutual**
+transconductance.
+
+So now we're not so sure. What I see is $\frac{1}{g_m}\parens{1 + \frac{1 /
+g_m}{r_o}}$ when looking into source.
+
+So now we've figured out what our $g_m v_i$ is. Current dividers.
+
+So I wanted to do a couple more examples. For example, source follower
+(recall, measure output at source). We know $A_v = -G_m R_o$. So what is
+$R_o$? It's $\frac{1}{g_m} \parallel R_s$. $G_m$ is a little tricky. Draw
+the small signal model, then do KCL at the output node. Note that since
+we're grounding $v_s$, $v_{gs} = v_i$, and no current is flowing through
+the source resistance (or the output resistance, even!). Thus $G_m \equiv
+-g_m$. What does this mean? We're shoving current out, instead of shoving
+current in.
+
+So finally $A_v = -G_m R_o = \frac{g_m R_s}{1 + g_m R_s}$. And in
+particular, in a MOS, it's very common that you'd see the output between
+two MOSFETs, so you'd get something like $\frac{g_m r_o}{1 + g_m r_o}$. If
+you're lucky, this looks something like $0.9 - 0.999$.
+
+All right. So here's a couple of good ones to think about. (various
+practice circuits)
View
1  sp2012/ee120/1.md
@@ -102,4 +102,3 @@ Basic Properties of Systems
+ Basically, no dependence on future values.
- If two inputs are identical to some point, then their outputs
must also be identical to that same point.
-
View
1  sp2012/ee120/10.md
@@ -137,4 +137,3 @@ This is a Cauchy probability density function.
(names for dirac delta: generalized function, distribution. Look up theory
of distributions)
-
View
1  sp2012/ee120/11.md
@@ -96,4 +96,3 @@ signal is the same as the transmitted signal. Whole area of communication
theory that deals with deterioration. Other assumption is that the
oscillator at the receiver that can generate the exact same frequency as
the transmitter oscillator and at the same phase.
-
View
1  sp2012/ee120/12.md
@@ -63,4 +63,3 @@ Quadrature multiplexing
The way we can do this is by exploiting the orthogonality of cosine and
sine. What's being transmitted is the sum of the two.
-
View
1  sp2012/ee120/13.md
@@ -163,4 +163,3 @@ $\omega_0$. The sampling signal $Q(\omega)$ will have an impulse train
separated by $\omega_s$. When we convolve this and apply a low-pass filter,
we have just one remaining frequency at $\omega_0-\omega_s = -\frac
{\omega_0}{2}$.
-
View
1  sp2012/ee120/14.md
@@ -37,4 +37,3 @@ your reconstruction period is the same as your sampling period.
Then you know that your output is equal $\frac{G}{T} H_d\parens{ \omega T}
X_c(\omega)$. The equivalent LTI filter is simply $\frac{G}{T} nH_d \parens{
\omega T}$
-
View
1  sp2012/ee120/15.md
@@ -85,4 +85,3 @@ functions.
It turns out these coefficients are the values of the signal at the sampled
points. That isn't obvious yet. After some algebraic manipulations, we can
finally see that this whole thing is just $x_c(kT_s)$.
-
View
2  sp2012/ee120/16.md
@@ -220,6 +220,6 @@ We have a few minutes, so let me talk about the distinctions between causal
signals and right-sided signals (and also anticausal / left-sided).
So let's say we take a right-sided but not causal signal. Now the RoC is
-outside of radius $\alpha$, but now you have to exclude $\infinity$.
+outside of radius $\alpha$, but now you have to exclude $\infty.
Similarly, for left-sided signals, you'd then exclude 0.
View
1  sp2012/ee120/17.md
@@ -154,3 +154,4 @@ Note that radius of convergence isn't changing.
Will leave it to you to figure out what the transform of $r^n \cos(\omega_0
n) u(n)$ is.
+
View
2  sp2012/ee120/18.md
@@ -0,0 +1,2 @@
+EE 120: Signals and Systems
+===========================
View
1  sp2012/ee120/2.md
@@ -97,4 +97,3 @@ $y(n) = \alpha y(n-1) + x(n)$. causal, $y(-1) = 0$.
(geometric sum; BIBO stability depends on magnitude of \alpha being less
than
1.)
-
View
3  sp2012/ee120/4.md
@@ -59,7 +59,7 @@ $y(-N) = ... = y(n) = 0$
$(\abs{\alpha} < 1)$
$$h_{N}(n) = \alpha h_{N}(n-N) + \delta(n) = !(n % N) && \alpha^{n/N}.
-H_{N}(\omega)\\ = \sum\alpha^{k}e^{-i\omega Nk} (k \from 0 \to \infty)
+H_{N}(\omega)\\ = \sum_{k=0}^\infty\alpha^{k}e^{-i\omega Nk}
\sum(\alpha e^{-i\omega N})^k = 1/(1-\alpha e^{-i\omega N})
\\ \abs{H_{N}(\omega)} = \abs{e^{i\omega N}/(e^{i\omega N} - \alpha)}
$$
@@ -88,4 +88,3 @@ $G(\omega) = \int g(t)e^{-i\omega t}dt
\\ = 1/(i\omega - (-\alpha ))$
$\sqrt{\omega ^2 + \alpha ^2}$
-
View
1  sp2012/ee120/5.md
@@ -116,4 +116,3 @@ synthesis equation, analysis equation:
$x(n) = \sum X_k e^{ik\omega_0n}$
$X(n) = \sum x_k e^{-ik\omega_0n}$
-
View
1  sp2012/ee120/6.md
@@ -16,4 +16,3 @@ Evidently there will be a quiz on Tue, since pset is due on Wed.
Consider what it means to send discrete-time periodic signals through
LTI systems.
-
View
1  sp2012/ee120/7.md
@@ -87,4 +87,3 @@ What happens if I want to approximate a signal that has finite energy? What
should the coefficients $\alpha_k$ be?
orthogonal projection! Least squares!
-
View
1  sp2012/ee120/8.md
@@ -63,4 +63,3 @@ Ideal discrete-time Low-pass Filter
Impulse response $h(n) = \frac{1}{2\pi}\int H(\omega)e^{i\omega n}d\omega$.
$\frac{B}{\pi n} \sin(An)$
-
View
1  sp2012/ee120/9.md
@@ -140,4 +140,3 @@ easy way of carrying out circular convolution: take one of the two
functions, keep one replica (e.g. the one from $-\pi$ to $\pi$), and then
do a regular convolution with the other function. advice: choose the
flipped + shifted signal for elimination of replicas.
-
View
21 sp2012/ee120/ee120.md
@@ -102,7 +102,6 @@ Basic Properties of Systems
+ Basically, no dependence on future values.
- If two inputs are identical to some point, then their outputs
must also be identical to that same point.
-
EE 120: Signals and Systems
===========================
January 19, 2012.
@@ -202,7 +201,6 @@ $y(n) = \alpha y(n-1) + x(n)$. causal, $y(-1) = 0$.
(geometric sum; BIBO stability depends on magnitude of \alpha being less
than
1.)
-
EE 120: Signals and Systems
===========================
January 24, 2012.
@@ -335,7 +333,7 @@ $y(-N) = ... = y(n) = 0$
$(\abs{\alpha} < 1)$
$$h_{N}(n) = \alpha h_{N}(n-N) + \delta(n) = !(n % N) && \alpha^{n/N}.
-H_{N}(\omega)\\ = \sum\alpha^{k}e^{-i\omega Nk} (k \from 0 \to \infty)
+H_{N}(\omega)\\ = \sum_{k=0}^\infty\alpha^{k}e^{-i\omega Nk}
\sum(\alpha e^{-i\omega N})^k = 1/(1-\alpha e^{-i\omega N})
\\ \abs{H_{N}(\omega)} = \abs{e^{i\omega N}/(e^{i\omega N} - \alpha)}
$$
@@ -364,7 +362,6 @@ $G(\omega) = \int g(t)e^{-i\omega t}dt
\\ = 1/(i\omega - (-\alpha ))$
$\sqrt{\omega ^2 + \alpha ^2}$
-
EE 120: Signals and Systems
===========================
January 31, 2012.
@@ -483,7 +480,6 @@ synthesis equation, analysis equation:
$x(n) = \sum X_k e^{ik\omega_0n}$
$X(n) = \sum x_k e^{-ik\omega_0n}$
-
EE 120: Signals and Systems
===========================
February 2, 2012.
@@ -502,7 +498,6 @@ Evidently there will be a quiz on Tue, since pset is due on Wed.
Consider what it means to send discrete-time periodic signals through
LTI systems.
-
EE 120: Signals and Systems
===========================
February 7, 2012.
@@ -592,7 +587,6 @@ What happens if I want to approximate a signal that has finite energy? What
should the coefficients $\alpha_k$ be?
orthogonal projection! Least squares!
-
EE 120: Signals and Systems
===========================
February 9, 2012.
@@ -658,7 +652,6 @@ Ideal discrete-time Low-pass Filter
Impulse response $h(n) = \frac{1}{2\pi}\int H(\omega)e^{i\omega n}d\omega$.
$\frac{B}{\pi n} \sin(An)$
-
EE 120: Signals and Systems
===========================
February 16, 2012.
@@ -801,7 +794,6 @@ easy way of carrying out circular convolution: take one of the two
functions, keep one replica (e.g. the one from $-\pi$ to $\pi$), and then
do a regular convolution with the other function. advice: choose the
flipped + shifted signal for elimination of replicas.
-
EE 120: Signals and Systems
===========================
February 21, 2012.
@@ -941,7 +933,6 @@ This is a Cauchy probability density function.
(names for dirac delta: generalized function, distribution. Look up theory
of distributions)
-
EE 120: Signals and Systems
===========================
February 23, 2012.
@@ -1040,7 +1031,6 @@ signal is the same as the transmitted signal. Whole area of communication
theory that deals with deterioration. Other assumption is that the
oscillator at the receiver that can generate the exact same frequency as
the transmitter oscillator and at the same phase.
-
EE 120: Signals and Systems
===========================
February 28, 2012.
@@ -1106,7 +1096,6 @@ Quadrature multiplexing
The way we can do this is by exploiting the orthogonality of cosine and
sine. What's being transmitted is the sum of the two.
-
EE 120: Signals and Systems
===========================
March 1, 2012.
@@ -1272,7 +1261,6 @@ $\omega_0$. The sampling signal $Q(\omega)$ will have an impulse train
separated by $\omega_s$. When we convolve this and apply a low-pass filter,
we have just one remaining frequency at $\omega_0-\omega_s = -\frac
{\omega_0}{2}$.
-
EE 120: Signals and Systems
===========================
March 6, 2012.
@@ -1312,7 +1300,6 @@ your reconstruction period is the same as your sampling period.
Then you know that your output is equal $\frac{G}{T} H_d\parens{ \omega T}
X_c(\omega)$. The equivalent LTI filter is simply $\frac{G}{T} nH_d \parens{
\omega T}$
-
EE 120: Signals and Systems
===========================
March 8, 2012.
@@ -1400,7 +1387,6 @@ functions.
It turns out these coefficients are the values of the signal at the sampled
points. That isn't obvious yet. After some algebraic manipulations, we can
finally see that this whole thing is just $x_c(kT_s)$.
-
EE 120: Signals and Systems
===========================
$\mathcal{Z}$ Transform
@@ -1623,7 +1609,7 @@ We have a few minutes, so let me talk about the distinctions between causal
signals and right-sided signals (and also anticausal / left-sided).
So let's say we take a right-sided but not causal signal. Now the RoC is
-outside of radius $\alpha$, but now you have to exclude $\infinity$.
+outside of radius $\alpha$, but now you have to exclude $\infty.
Similarly, for left-sided signals, you'd then exclude 0.
EE 120: Signals and Systems
@@ -1782,3 +1768,6 @@ Note that radius of convergence isn't changing.
Will leave it to you to figure out what the transform of $r^n \cos(\omega_0
n) u(n)$ is.
+
+EE 120: Signals and Systems
+===========================
View
5 sp2012/phys112/10.md
@@ -51,10 +51,7 @@ the H theorem, the chemical potential is equal in all accessible states.
What is chemical potential? It is a measure of the concentration! (verify
by working out -τ ∂σ/∂N). A lot of work for a very simple result. But this
is much more powerful. We will use this to look at batteries, equilibrium
-in various systems.
-
-
-MIDTERM: Friday the 24th. 8:10-9:00. Why Friday and not Monday? Turns out
+in various systems.MIDTERM: Friday the 24th. 8:10-9:00. Why Friday and not Monday? Turns out
there is an important workshop on dark matter in UCLA. There is a potential
problem: Monday the 20th is President's day, so no class, no office
hours. Proposition: Tuesday the 21st, we have extended office hours from
View
4 sp2012/phys112/12.md
@@ -49,11 +49,11 @@ C{p} = (∂Q/∂τ){p} = T(∂σ/∂τ){p}
Thus C{v} = C{p} + τ(∂σ/∂P·∂P/∂τ)
-lpha ≡ 1/V ∂V/∂τ. "volume coefficient of expansion"
+\alpha ≡ 1/V ∂V/∂τ. "volume coefficient of expansion"
κ ≡ -1/V ∂V/∂p "isothermal compressibility"
-Therefore C{p} - C{v} = Vτ(lpha²/κ)
+Therefore C{p} - C{v} = Vτ(\alpha²/κ)
What's needed to calculate entropy
View
2  sp2012/phys112/13.md
@@ -33,4 +33,4 @@ Clausius statement for 2nd law of thermodynamics: effectively, Carnot
engine is maximally efficient, given τ₁, τ₂. (q₂/q₁ ≤ τ₂/τ₁)
Joule-Thomson process: H = U + PV ⇒ H₁ = H₂ if dQ = 0.
-μ ≡ V/C{p} (τlpha - 1)
+μ ≡ V/C{p} (τ\alpha - 1)
View
11 sp2012/phys112/phys112.md
@@ -671,10 +671,7 @@ the H theorem, the chemical potential is equal in all accessible states.
What is chemical potential? It is a measure of the concentration! (verify
by working out -τ ∂σ/∂N). A lot of work for a very simple result. But this
is much more powerful. We will use this to look at batteries, equilibrium
-in various systems.
-
-
-MIDTERM: Friday the 24th. 8:10-9:00. Why Friday and not Monday? Turns out
+in various systems.MIDTERM: Friday the 24th. 8:10-9:00. Why Friday and not Monday? Turns out
there is an important workshop on dark matter in UCLA. There is a potential
problem: Monday the 20th is President's day, so no class, no office
hours. Proposition: Tuesday the 21st, we have extended office hours from
@@ -821,11 +818,11 @@ C{p} = (∂Q/∂τ){p} = T(∂σ/∂τ){p}
Thus C{v} = C{p} + τ(∂σ/∂P·∂P/∂τ)
-lpha ≡ 1/V ∂V/∂τ. "volume coefficient of expansion"
+\alpha ≡ 1/V ∂V/∂τ. "volume coefficient of expansion"
κ ≡ -1/V ∂V/∂p "isothermal compressibility"
-Therefore C{p} - C{v} = Vτ(lpha²/κ)
+Therefore C{p} - C{v} = Vτ(\alpha²/κ)
What's needed to calculate entropy
@@ -878,7 +875,7 @@ Clausius statement for 2nd law of thermodynamics: effectively, Carnot
engine is maximally efficient, given τ₁, τ₂. (q₂/q₁ ≤ τ₂/τ₁)
Joule-Thomson process: H = U + PV ⇒ H₁ = H₂ if dQ = 0.
-μ ≡ V/C{p} (τlpha - 1)
+μ ≡ V/C{p} (τ\alpha - 1)
Physics 112: Statistical Mechanics
==================================
Chemical Potential. Feb 22, 2012

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