Thanks man, i needed this before signals and systems unit
@DhaferThyab29 күн бұрын
Tremendous video .. nice work 🎉
@MDNQ-ud1tyАй бұрын
Just to further simplify for those that don't understand it that well. The maximum of two "values" when multiplied is when they are the same. a*b is a maximum when a = b when then gives a^2 = b^2. Hence when you pointwise multiply two functions you are just multiplying their values and their maximum is always when they are the same. E.g., suppose you have a = 3 and b = 5 then you have 3*5 = 15 < 5*5. Because you are using the same signal and shifting it over time f(x-t) at some point it must match with itself f(x)*f(x-t) for some t. In fact, it matches exactly when t = 0 and so one has f(x)^2 as an output which is the maximum possible at all points. But if the function is periodic or approximately then it will repeat itself for some non-zero t so you can find that t by looking at other maximum other than t = 0. Hence the period is simply the smallest nonzero t that. But the idea is simply that of squaring always gives the largest value in the sense that min(a,b)^2 = min(a,b)*min(a,b) <= a*b <= max(a,b)*max(a,b) = (max(a,b))^2 and when have a continuous function then we can ask about cross(e.g., at different points in time) values. You can, of course, use any binary function g(x,y) that has a maximum when x = y and the effect will be the same. E.g., x^n will also enable doing this and it can also make finding periods easier since it will, at least for normalized functions push the "noise" down to the floor. Also, the reason the shape decays is because the signal is being zero extended. If one makes them periodic then there is no decay. The problem with this though is then the signal has a natural period of the window size and this might cause some problems determining the correct period. Just to be clear, when one does f(x)*f(t - x) one is computing a pointwise product for at every x. One can then add up all the values to get the total relative "correlation". We know that at t = 0 this total is maximum because the product is f(x)*f(x) = f(x)^2 and this is the maximum value we could ever hope to get for each x. We can add up over the entire range of x to get the total "energy". At other values of t we would get other values we get other values but none can have more total power than at t = 0 or some period. All this comes out of the basic idea that a*b <= max(a,b)^2 and the only time a*b = max(a,b) is when a = b. This extends to entire functions by simply doing it pointwise and adding up to get an overall value. g(t) = sum(f(x)*f(t - x)). g(0) is maximum at t = k*period. This also extends to two different functions for the exact same reason so one can simply compare functions using this method knowing that it gives the maximum "comparison" at the maximum value of t. The reason for f(t - x) vs f(x - t) is just so our periods are positive rather than negative but it obviously makes no difference ultimately. Another way to think about it is that multiplying f(x) by a shifted copy will typically introduce a lot of destructive interference because the shifted copy will not match up and will introduce another set of zeros. E.g., f(x)*f(t - x) will have twice as many zeros, typically than f(x). If it is shifted by the period, if it has one, then those periods exactly align and one not have any excess. Generally if there are more zeros this means more crossings which means higher frequencies. Also note that when squaring one has no negative values. So when f(x) and f(t - x) align perfectly it must all be positive while there will otherwise be negative values. Hence when summing over the additional negative values will lower the overall sum value.
@ahmedali6519Ай бұрын
This is a really good approach for understanding Z transform!
@ahmedali6519Ай бұрын
This is a really good approach for understanding Z transform!
@ahmedali6519Ай бұрын
This is a really good approach for understanding Z transform!
@ahmedali6519Ай бұрын
This is a really good approach for understanding Z transform!
@ahmedali6519Ай бұрын
This is a really good approach for understanding Z transform
@ahmedali6519Ай бұрын
This is a really good approach for understanding Z transform!
@ahmedali6519Ай бұрын
JUST AMAIZING
@tjmozdzen2 ай бұрын
Wonderful details. The x-axis of the transform is easy to forget
@anandar11582 ай бұрын
Excellent
@NguyenHoa-er1ff3 ай бұрын
thank you David. Great demonstration. It is extremely helpful for me.
@tombouie3 ай бұрын
Thks for explaining the crux of correlation
@Nazhull3 ай бұрын
unipd? qualcuno?
@a-bm9ux4 ай бұрын
10 years later, this video still helps people out!
@danieldinh86905 ай бұрын
You are a beautiful man for uploading this treasure of a video.
@ddorran5 ай бұрын
Glad you enjoyed it!
@daa26225 ай бұрын
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?
@ddorran5 ай бұрын
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!
@daa26225 ай бұрын
okay, let me follow that then and ill try some experiments of my own. thank you!
@daa26224 ай бұрын
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!
@ddorran4 ай бұрын
@@daa2622 Ah yes! That makes perfect sense. I did a video on linear phase filters that might be of interest kzbin.info/www/bejne/roG3lmqQjLGJnrMsi=WiqiA7-z8QENey7K
@이정호-f3u1s5 ай бұрын
The Best
@laurentthowai33595 ай бұрын
Simple and efficient Thanks !
@ddorran5 ай бұрын
Glad it was helpful!
@AbhinavRao-te9co5 ай бұрын
Why is the region of stability within the unit circle for the Z Tranform? Isnt that where the signal grows exponentially i.e. instability?
@ddorran5 ай бұрын
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 kzbin.info/www/bejne/mnatoWdsiKuajJY. Hopefully, one of these could provide some insight.
@AbhinavRao-te9co5 ай бұрын
@@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?
@ddorran5 ай бұрын
@@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).
@AbhinavRao-te9co5 ай бұрын
@@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?
@ddorran5 ай бұрын
@@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.
@MrKarnn6 ай бұрын
Thank you so much, why do people overcomplicate the explenation of this topic when you can explain it as simply as this
@Retronix217 ай бұрын
super nice demo thank you so much
@roderickmcleod48647 ай бұрын
is there a less laggy version that I can download?
@bookiefun4857 ай бұрын
Really appreciate your video and the detailed documentation.. Thanks a lot
@ddorran7 ай бұрын
Glad it helped!
@wai-fonglee11628 ай бұрын
Could you pls. tell me what is this 3D GUI tool?
@ddorran8 ай бұрын
It's MATLAB software available at dadorran.wordpress.com/2012/04/07/zpgui/
@wai-fonglee11628 ай бұрын
@@ddorran Thanks so much!
@oadka9 ай бұрын
Very nice comparison
@vasilisdimitriou66829 ай бұрын
σε αγαπώ!
@thucpham45989 ай бұрын
nice, thank you so much!
@muznamalik47989 ай бұрын
what about the effect function in Audacity? One can also apply church effect, isn't it?
@ddorran9 ай бұрын
Yes - absolutely. I can't be certain how that effect is applied in Audacity but I suspect it is using the convolution process.
@JeremyDismukes2259 ай бұрын
your accent is wild, but your explanation on the topic was very helpful review for my understanding for my FE review
@Edwinthebreadwin10 ай бұрын
I understand more now than I did in 3 months of control lectures, thank you for putting all the pieces together
@NineInchFailz11 ай бұрын
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!
@ddorran11 ай бұрын
F Well done. It's an achievement to persevere with the entire series. I'm glad it helped!
@NineInchFailz11 ай бұрын
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.
@ddorran11 ай бұрын
Thanks Dakotah! Good luck with your career as an Electrical engineer!
@nigelkundaigatsi20511 ай бұрын
Good video ,thank You Sir !
@Festus2022 Жыл бұрын
What about the frequency dependent phase shift response for V-in vs. V-out?
@melihcanyldz368 Жыл бұрын
finally , I found useful video to understand cross correlation , thank you David
@Drina00 Жыл бұрын
Thank you.
@cagedgandalf3472 Жыл бұрын
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.
@n4mmenam Жыл бұрын
🐐
@arash4232 Жыл бұрын
Funny😊, thank you. Please let me have the link for other applications, if there is. Thank you
@mujoetemi9689 Жыл бұрын
If I have this: H(z) = (-z^(-2)-2*z^(-1)+2*z+z^2)/8T, Which are the coefficiant a and b?
@teebee3881 Жыл бұрын
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)
@iamliam1241 Жыл бұрын
What a wonderful explanation!,much appreciated!
@apppurchaser2268 Жыл бұрын
Amazing explanation, thanks a lot
@bassrabbit9 Жыл бұрын
Best explanation on this I have ever seen. Thank you for the colors!!
@ddorran Жыл бұрын
No problem! Glad it helped!
@vbris Жыл бұрын
truly great material! no better video to explain the z transform than this one !
@mwerensteijn Жыл бұрын
Amazingly explained in simple words, thank you
@electropocalypse5877 Жыл бұрын
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.
@ddorran Жыл бұрын
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.