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Do u have a good reference table for FT and DTFT?
In DT Signals and Systems
Hi Adam, Do you have video on orthogonal complement? Can't seem to view it. Thanks in advance
In Linear Algebra
Hi Adam, Do you have video on orthogonal complement? Can't seem to view it. Thanks in advance
In Linear Algebra
POLE ZERO PLOT QUESTION
In DT Signals and Systems
Adam Panagos
Mar 15, 2020
This video: https://youtu.be/cXbmdrxo9L4 would be good to watch. Basically, there’s a nice relationship between the discrete-time frequency Omega and the continuous-time frequency omega. Omega = omega*T where T is the sampling period of the system. In the pole-zero plot, if we stay on the unit circle, as we move from Omega = 0 to Omega = pi, we’re moving in frequency from 0 rad/s up to omegaS/2 where omegaS is the sampling rate of the system. Remember, when we sample a signal, we can only “see” frequencies from -fs/2 to fs/2. So, if we want the filter to have zero response at some frequency, we just need to find the frequency, and then figure out the corresponding angle/Omega. We then want to place a zero on the unit circle at this location. In the plot you provided, the first tone we want to reject is at frequency bin 1.408e5. Since we used N = 1960000 samples at a rate of fs =22050, we can figure out what frequency this corresponds to. The frequency increment is delF = fs/N = 22050/1960000 = 0.0113. So, the 1.408e5 bin is at frequency 1.408e5 * 0.0113 = 1591.04 Hz. Now, we just need to figure what Omega this is. 1591.04 Hz is 253.22 rad/s. So, omega = 253.22. Now, we have Omega = 253.22*T = 253.22*(1/fs) = 253.22/22050 = 0.0114839 rad/sample. So, we know that we want a zero in our filter response at this angle. So, part of H(z) will include a term in the numerator of (z-0.0114839). You can do similar things for the other desired zeros. When you’re done, the numerator will look something like (z-0.01114839)(z-z2)(z-z3) where z2 and z3 are the other zero locations you’ll find. Hope that gets you started in the right direction, Adam
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Do you have Lecture Notes for the DT course, similar to the notes you have for the CT course?
In DT Signals and Systems
Do you have Lecture Notes for the DT course, similar to the notes you have for the CT course?
In DT Signals and Systems
Missing Zip file for Basics
In Matlab
M/M/1 queues
In Random Processes
Adam Panagos
Oct 04, 2018
It's been quite a while since I did this M/M/1 queue stuff but here are some thoughts. First, if there are two independently operating queues, I would say that there are N1 messages in the first queue and N2 messages in the second, and that the total number of messages is N = N1 + N2. Obviously, N1 and N2 are independent random variables (since the problem tells us they are operating independently) and thus N is also a random variable. As you noted, if we were talking about just one queue, figuring out the probability of having m messages is easy, it's just p(m) = (1-p)p^m as your already wrote down. Now that we have two queues, I think this essentially turns into a "counting problem". For example, what is the probability of having 3 total messages in the system? Well, there are variety of ways that can happen 0 messages in Q1, 3 messages in Q2 1 message in Q1, 2 messages in Q2 2 messages in Q1, 1 message in Q2 3 messages in Q1, 0 messages in Q2 For each of these cases you can compute the probability by multiplying the probability. Recall, since these are independent events we're essentially "ANDing" things in probability which means we need to multiply the probabilities. So, for the cases listed above I think we'd have the following probabilities: p(0)*p(3) p(1)*p(2) p(2)*p(1) p(3)*p(0) After we add everything up we can write: Probability of 3 messages in the system = [p(0)*p(3) + p(1)*p(2) + p(2)*p(1) + p(3)*p(0)] Obviously, you now need to generalize this approach to be able to write the probability of N messages in the system. It looks like your final answer will be something like Probability of N messages = sum(k = 0 to N) p(k)p(N-k) Again, it's been quite some time since I've worked a problem like this but I think my thoughts above should get you going in the right direction. Hope that helps, Adam
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I need help!!!
In DT Signals and Systems
Adam Panagos
Oct 03, 2018
If H(z) = 1/(1-0.9z^-1), then we have h[n] = (0.9)^nu[n]. We would call h[n] a right-sided signal since it starts at time n = 0, and then continues for all positive time (e.g. it goes to the right). You can show that a right-sided circle must have an ROC that is the outside of some circle.  To figure out the radius of the circle you need to determine the poles of H(z).  If z = 0.9, then H(z) = 1/0, and thus z = 0.9 is a pole of H(z).  The ROC cannot contain poles so the ROC for this single signal is |z| > 0.9. Now, consider what happens when you time-reverse the signal.  The signal h[-n] is a left-sided signal since it exists for all negative time and then turns off at time n = 0. To figure out  the z-transform of h[-n], you can use the property you noted, i.e. just replace z with 1/z.  So, the z-transform of time reversed signal is Hrev(z) = 1/(1-0.9z). We must also take care to determine the ROC of this signal.  You can show that a left-sided signal must have an ROC that is inside of some circle.  To figure out the radius of the circle you need to determine the poles of Hrev(z).  If z = 1/0.9, then Hrev(z) = 1/0, and thus z = 1/0.9 is a pole of Hrev(z).  The ROC cannot contains poles, and the ROC must be INSIDE of a circle, so the ROC for this individual signal is |z| < 1/0.9. Now, finally we can discuss what the ho[n] = h[n]*h[-n].  Since ho[n] is the convolution of a right-sided signal with a left-sided signal, the result is a signal ho[n] that exists for all time.  I call this a double-sided signal. In the z-domain, we have Ho(z) = H(z)*Hrev(z).  You can show that a double-sided signal must have an ROC that is a ring.  In this case, the ring is defined by the two pieces we already examined.  The ring is defined by |z| > 0.9 and |z| < 1/0.9 Since 1/0.9 is about 1.11, this region in the complex plane consists of all points |z| > 0.9 and |z| < 1.11. I have a few examples like this on the channel.  I'd recommend checking out the following videos: Z-Transform Region of Convergence Details - Z-Transform Part 1 (10/18) https://youtu.be/ZGQLBIDTAPc Z-Transform Region of Convergence (ROC) of a Two-Sided Signal - Z-Transform Part 1 (11/18) https://youtu.be/m-hTWznOZu4 Hope that helps! Adam
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