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Commits on Mar 21, 2012
 steveWang more updates to notes and building 03d9fe7
Commits on Mar 22, 2012
 steveWang more notes 19e714e steveWang cs191.html 99bbee3 steveWang update notes 9a860ac
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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. - -
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.
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.
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 $$+ + 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". + + 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 - + + 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. + + 82 sp2012/ee105/26.html  @@ -0,0 +1,82 @@ + + + + + + +26.md + + + + + + + + + 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} + + 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) 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. 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}. 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)
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. -
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) -
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. -
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. -
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 $\omega_0-\omega_s = -\frac {\omega_0}{2}$. -
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 $\frac{G}{T} nH_d \parens{ \omega T}$ -
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 $x_c(kT_s)$. -
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 $\infty. Similarly, for left-sided signals, you'd then exclude 0. 1  sp2012/ee120/17.md  @@ -154,3 +154,4 @@ Note that radius of convergence isn't changing. Will leave it to you to figure out what the transform of$r^n \cos(\omega_0 n) u(n)$is. + 2  sp2012/ee120/18.md  @@ -0,0 +1,2 @@ +EE 120: Signals and Systems +=========================== 1  sp2012/ee120/2.md  @@ -97,4 +97,3 @@$y(n) = \alpha y(n-1) + x(n)$. causal,$y(-1) = 0$. (geometric sum; BIBO stability depends on magnitude of \alpha being less than 1.) - 3  sp2012/ee120/4.md  @@ -59,7 +59,7 @@$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 @@$G(\omega) = \int g(t)e^{-i\omega t}dt \\ = 1/(i\omega - (-\alpha ))\sqrt{\omega ^2 + \alpha ^2}$- 1  sp2012/ee120/5.md  @@ -116,4 +116,3 @@ synthesis equation, analysis equation:$x(n) = \sum X_k e^{ik\omega_0n}X(n) = \sum x_k e^{-ik\omega_0n}$- 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. - 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! - 1  sp2012/ee120/8.md  @@ -63,4 +63,3 @@ Ideal discrete-time Low-pass Filter Impulse response$h(n) = \frac{1}{2\pi}\int H(\omega)e^{i\omega n}d\omega$.$\frac{B}{\pi n} \sin(An)$- 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. - 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 +===========================
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
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
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)
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