Sir, about the linear phase spectrum, is it really just a random straight line that you chose, or is it based on something rigorous? and i suppose the steeper the line, the smoother the resulting magnitude of the IFFT is, right? and is it ever possible to achieve a perfect "blue line" IFFT like one you mentioned in 1:25? completely disregarding how messy the phase spectrum would be, is that ever possible?
The phase spectrum calculation is shown at the very end of the video where some matlab code is provided. I haven't investigated the impact of increasing the steepness of the line so can't comment at this point in time. In theory, it is possible to achieve any frequency response but the trade off is an infinite number of filter coefficients, which isn't practical!
i did some learning, and i found out that the negative of the slope of the phase spectrum (unwrapped) corresponds to the group delay introduced into the signal. in other words, the steeper down the (unwrapped) phase shift will be, the further back in time the peak will be. Wikipedia has an article on phase delay and group delay. i confirmed this using libreoffice calc; i just applied it in practice. and it works!
@@daa2622 Ah yes! That makes perfect sense. I did a video on linear phase filters that might be of interest ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-xPTe7ZWLVhQ.htmlsi=WiqiA7-z8QENey7K
You're correct (kind of!). The signal z^-n does grow exponentially for values of z that lie inside the unit circle. However, the z-transform multiplies the signal z^-n by the signal we are interested in analysing for stability e.g. x[n] (and then sums the resulting multiplied terms). If x[n] is growing exponentially and z^-n is also growing (which is the case for all values of z within the unit circle) then the sum of the products of x[n] and z[n] will never converge within the unit circle. On the other hand when x[n] is decaying and z^-n is growing then there will a set of values of z that lie within the unit circle for which the product of x[n] and he signal z^-n will converge. I suspect it's quite difficult to interpret these previous sentences - if so you could take a look at page 83 of www.researchgate.net/publication/370660126_The_z-transform_A_practical_overview or take the following video ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-dEJp46SFgV4.html. Hopefully, one of these could provide some insight.
@@ddorran I read the paper that you have linked, however I didnt understand one thing, you say that the region of convergence lies outside the circle with radius equal to that of the pole of the system. In your example you used -0.5. My question is why is it that you called -0.5 the pole? Wasn't that the impulse response? Also all of this implies that the region of stability/convergence is > Pole value circle, not that region of stability is <1?
@@AbhinavRao-te9co "My question is why is it that you called -0.5 the pole?" The answer is because H(z) goes toward infinity when z= -0.5 (poles exist at values of z for which H(z) goes toward infinity). "Also all of this implies that the region of stability/convergence is > Pole value circle, not that region of stability is <1?". Almost correct. The region of convergence (not stability) is > magnitude of pole value furthest from the origin. Systems are stable if the unit circle lies in the region of convergence. Having said that, regions outside the unit circle are associated with instability because any pole outside the unit circle will cause the system to be unstable (since the unit circle will not lie in the region of convergence).
@@ddorran "Having said that, regions outside the unit circle are associated with instability because any pole outside the unit circle will cause the system to be unstable (since the unit circle will not lie in the region of convergence)." Is there any proof for this or should i just learn this as a given?
@@AbhinavRao-te9co I don't have a formal proof to hand but there must be one out there! There are a couple of aspects to look at: A formal proof that the region of convergence lies outside the pole furthest from the origin (or proof that the region of convergence cannot contain poles, is another way of saying this); A formal proof that an unstable system contains poles that lie outside the unit circle. If both of these can be proven then it follows that the unit circle will not lie in the region of convergence, for unstable causal systems.
I binged this whole series overnight studying for my DSP midterm. Dude none of this connected until i watched YOUR videos. I had no idea what the Z transform was even supposed to represent in terms of a system or what the magnitude meant or anything. This all cleared it up so much!
I'm about to graduate with my degree in Electrical Engineering at Oregon State University. I've had to watch hundreds(yes, literally hundreds) of videos to understand that concepts i've had to learn since i'm a slow learner, and i can say with complete confidence that this is the absolute best and most intuitive description of the Z transform i've seen.
Our teacher taught us this and I was so confused what the purpose was. I was also confused why you were solving for h at first. But after you said I could get the output signal just by the impulse response of the system. Everything clicked right there.
Hey, there is a smale mistake in the vid, around 4:20 you say that the values in the series are decreasing and the series there for converges. This however is not always the case, the best example of this is the harmonic series that diverges to infinity. In most cases however you are right and they wil converge but it is not the case that by definition when the values in the series are decreasing that the series converges. (excuse my bad english pls)
Your videos truly have helped me a ton as a newbie. I'm a little slow on the uptake... so this alaising catches signals from the waveform at different points rather than changing the sample rate directly (i.e frequency)? I also suppose each of these points has many different signals/sounds within them. This must be where the distortion comes from.
No - that's not really it. Aliasing occurs if you do not sample at a suitably high rate. Your sampling rate has to be at least twice the highest frequency present in a signal, otherwise aliasing will occur. For example, a speech signal has loads of different frequencies present but (for most speech) the highest frequency will be about 8000 Hz, so to avoid aliasing you would need to sample at 16 kHz. If you don't sample at a high enough rate, then the signal you measure won't 'seem' correct. For example, imagine you had an audio signal that contained a 1000 Hz beep sound. Let's say you sampled at 1200 Hz (which is less than twice the highest frequency), If you played that sampled signal back it would sound like a 200 Hz beep rather than a 1000 Hz beep (this apparent change in frequency after sampling is aliasing). If you sampled the 1000 Hz beep sound at a rate above 2000 Hz then the sampled beep would sound like a 1000 Hz beep.
5:22-5:26 Did example of the signal z^-n (for z=0.9e^jx=-0.9+0j) plot correctly? z = (-0.9)^-n for n={0, 1, 2, 3, 4, ...} z = {(-0.9)^0, (-0.9)^-1, (-0.9)^-2, (-0.9)^-3, (-0.9)^-4, ...} z = (1; -1.11; 1.23; -1.37; 1,52; ...}
Why do you multiply like this? drums(44100*10) to get 10 secs worth of data? What is the significance of the 44100? Is this like a nTs where Ts is sampling rate and n=10 means 10th sample? But that still does not make sense because 44100 is the frequency of the audio range, right?
drums (1 : 44100*10) selects 10 seconds of data because the sampling rate is 44100 samples per second (44.1 kHz). 10 seconds of this audio recording equates to 441000 samples i.e. 44100*10.
