That should be relatively straightforward. Something like this should work for the matrix R: [U,S,V]= svd(R) keepInd = find(diag(S)>0.01*S(1)); Rnew = U(:,keepInd)*S(keepInd,keepInd)*V(:,keepInd)'; The matrix Rnew is now very close/similar to R, but it doesn't use any singular values that are less than 0.01 percent of the largest singular value S(1). Hope that helps, Adam

The question asked here is only related to the number of bits in the ADC. If an ADC has B bits, then it breaks up the signal into 2^B ADC levels. Those ADC levels are spread over the full dynamic range of 2*xmax. This means the space inbetween each levels is deltaV = 2*xmax/L. The largest error will then be when you're inbetween levels, or 0.5*2*xmax/L = xmax/L. So, we just need to find the value of B such that xmax/2^B is less than 10^-3*xmax. The "Pulse Coded Modulation" video linked on the page below covers these concepts in better detail. Hope that helps, Adam https://www.adampanagos.org/courses/members/dt/samplingpart02

The code I provide in the speechQuantizationExample code on the website does something very similar to that; the only difference is it does just a few different values, not all values in a FOR loop. It also doesn't use LSBs,it actually just re-quantizes to different levels. Did you look at that specific example? If they want you to quantize by blanking out LSBs, you'll just need to represent each sample as a binary number, e.g. 1101011011101001 and then slowly turn this into something like: 1101011011101000 .... 1101011011 ... 11010 ... 11 etc

The "short answer" is there's probably a variety of different ways to end up at the correct answer in either of these cases. Check out this page, in particular, Theorem 5.31: https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/5%3A_Set_Theory/5.5%3A_Indexed_Families_of_Sets There's a nice result that computes the intersection of a set B, with the union of a family of sets. And also, another result that computes the union of a set B, with the intersection of a family of sets. Those are some key "building blocks" that might be helpful. Let me know how that goes.

Here's the table I typically use in the class.

Ok, I think I know what I had wrong and it should be fixed now. Try again and let me know.....

Yes, since N = 6, then its fundamental frequency is omega0 = 2*pi/6 = pi/3 as you noted. You should end up with the summation of 2 sine functions whose frequencies are multiples of the fundamental frequency omega0. Check out Problem 2 in the attached exam. It's very similar to your problem but with just an extra term. Since how to group the terms together? Let me know if that example helps.

That is a periodic discrete-time signal so you need to compute the Discrete-Time Fourier Series (DTFS) of the signal. The videos in the link below introduce the DTFS and work several examples of how to compute the DTFS: https://www.adampanagos.org/dt-fourieranalysispart01 When you setup your DTFS summation I would sum from -2 to 2 so you can group terms and take advantage of the odd-symmetry of the signal. Let me know how it goes, Adam

Yes, I have a video here: https://www.adampanagos.org/courses/ala/orthAndLeastSquaresPart01 It's the 10th video of the playlist. Let me know if that works. Adam

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

I'm not sure if I understand your question. Can you provide a little more information or context on what you're asking about? Thanks, Adam

I'm not sure if I follow your question exactly. The slides typically introduce "theory" and "definitions", and then detailed examples are worked out "by hand" in the videos. So you're correct, as such, the slides don't typically contain the details of the examples.All the details are in the videos.

Ok, I justed added a "Lecture Material" section to the spreadsheet and put the Lecture Summaries document as well as Lecture Notes zip file there. Hope that's what you were looking for. Let me know, Adam

Are you referring to the "Lecture Summaries" document? Yes, I do have a DT version of that, I'll get it posted in a few minutes......

Excellent, idea. Just added it. Thanks, Adam

Sorry for the delay in reply. I usually get an email notification when a post is made to a forum and I didn't get one this time for some reason (or I overlooked it). Currently, I don't have much material on filter design. I have the Butterworth filter content you noted. I'm planning on making some additional videos to conclude Chapter 7 of the DT signals and systems class. These videos will describe a "time-domain technique" and a "frequency-domain technique" for designing filters. That's the next set of videos I'll be working on so hopefully they'll be done in the next month or so. As far as the other methods you noted, I would love to get to those at some point but not sure when I will at the moment. Thanks, Adam

Awesome, good luck! Just curious, where are you studying at? I wrapped up my PhD in 2006 and have working since; grad school was a good time.

The poles of a function H(z) are defined as the values of z that make H(z) = plus or minus infinity. If H(z) = 1-0.9z^-1, the only point that will be an issue is z = 0. At this point, we have H(0) = 1-0.9/0 = 1 - infinity = -infinity. So, z = 0 is a pole of H(z). Essentially, we're just looking for values of z that make H(z) of "blow up". That's not a mathematically rigorous definition, but that's how I think of it. Hope that helps, Adam

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

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