DT Convolution-Simple Example Part 1
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- Опубліковано 19 вер 2024
- Shows how to compute the discrete-time convolution of two simple waveforms.
This video was created to support EGR 433:Transforms & Systems Modeling at Arizona State University. Links to other videos can be found at www.engineering... and sites.google.co...
This video is made available under the Creative Commons BY-SA license.
Thanks, I had to do this for some test and could not understand the professor's explanation. This helped me a lot.
Your video was the only one with any clarity on how this addition was done during the convolution. Thank you
Isn't that exciting? We got our first value here. Made me smile sooo wide :DDDD This video was so helpful!
I recommend to watch this video, its so informative.
Well, actually that won't be the case. If you have a signal x[n] and a signal x[n-2], then the -2 would account for moving to the right (or delaying the signal). Here you are working with x[-n] and x[-n-2], however, and x[-n-2] is nothing but x[n-2] flipped around the y-axis (because you just swap n with -n). So in this case, you are reflecting a signal shifted two to the *right* of x[n] (i.e. x[n-2]) around the y-axis, which results in a signal shifted two the *left* of x[-n] (i.e. x[-n-2).
Thanks, i was thinking the same. But if convolution is defined from left to right shouldnt it be x[-t-tau] intead of x[-tau+t]?
Thank you. now I can clearly realize convolution
noveas la que te doy marialuisa
Ok so people here are wondering about the shifting. In this video he is right when he does shifting to the left for h(-4-k). Here he mirrors/flips it first and then does the shifting. For the h(-4-k), if I do the shifting first, then I would move it to the right for -4 and then just flip it, HOWEVER if you mirror first then do the shifting you cannot move it the right for -4, you have to go to the left. If you don't believe me, you can do it. So the important thing to note here is either you do the shifting or scaling first. h(-4-k)=h(-(4+k))
Won't deducting 4 move it to the right?
Aaahhhh...... thnx man! u jst saved my life! :)
denada jose manuel gutierrez
abnormal people.
look at it this way:
you invert the sequence x[n] which is centered at n = 0 and you get x[-n], then shift center to n = -3 or n+3 = 0. Then x[-n] becomes x[-(n+3)]. Hence, x[-n-3]. If you don't invert x[n] then you'll replace n by n+3 and get x[n+3]. So be careful about the - sign.
cheers darryl you have sorted me right out!
Great explanation, thanks!
what if in the question x[n] and h[n] are given? should we consider them as x[k] and h[k] and follow the rest of the steps u told??
uugh! I took a System Dynamics course and block diagrams jumped straight into the use of negative feedback and laplace transforms only. And that's natural after all, because control is all about constraint, and that's what negative feedback loops actually do. But then I started on my own doing MIT OCW 6.003 "Signals and Systems" to understand better how programs like MATALB might actually be doing the work and stuff and wow does that book (Oppenheim, et al 1996) waffle around in its examples. Impossible to follow. Diagrams nowhere near the text talking about them, going back and forth between the example and tangents.
There's a 2016 book, which is less problematic, but the structure is different, and just assumes students already know things like z-transform?
Your playlist here is a big help, especially this part where the convolution section is the worst section ever.
Good explanation. Thank you so much
Hi Darryl, in the 2nd term of the sum h(n-k) why does that not become h(n+k) when k is negative? For example when you calculate y(-2), here n = -2, then you would sum the products from k = -inf to k = +inf. There will be -k’s in that summation. For example x(-2)h(-2-(-2)) = x(-2)h(-2+2)? But that gives the wrong answer.
Great content! Trank you!
my prof wants it with 3 impulse responses.. any help? should i just do the first two, then the result with the 3rd?
Please, could you tell me, why are you taking n=-4? Is there some rule for that?
I think '-4' is more like an arbitrary starting point (number) because even if `n` is much lower than -4, y[n] would result in zero anyways.
At the very first place, why do we want to flip h[n] ? Is it not possible to make the convolution without fliping?
thnx brouston, u jst saved my life! :)
you didn't explain that why do you start from. -4
I think it would be 0 anyways if you shift it more to than -4 to the left. But theoretically you have to start at -inf. But all values are 0 until -4. So there is no point in calculating the previous values.
Thank you sir, u save my life :)
could you help me answer this please. if you have x[k] and y[k], how can one find h[k]? thank you.
Very good explanation...but can you explain why the hell do you need to flip h[n-k]...Actually I know why but the students need to understand the exact practical reason WHY?
There isn't an exact answer to why but since we are dealing with Linear and Time Invariant systems we can change the input to h[n-k] without changing the output and this helps us to visualize to convolution.
actually, x[-n-2] is x[-(n+2)] and is not equal to the flipped x(n-2),because in this case you shift to the right, because this is nothing but the signal x advanced (i.e shifted to the right ) and flipped about the 0 axis (or flipped and then shifted by 2 to the right if you want)
Thanks!!
can u explain x(n)+x(-n)=1+impulse function i.e del(n)
It seems that one of the elements of h is changed from 0 to -1, when flipped!
Thank YOu.
at 8:18 the product -1 * -1 is not -1 but 1 no?
i know this is real late but the value of h at -1 = 1 so 1*-1 = -1
sir you are an angel.
ADAMSIN KARDEŞİM
Exactly..this is what i wanna ask...
Actually, he's wrong. You can see my comment for the details.
Not sure how there are 31 dislikes. Did ya'll come here for a Taylor Swift video or something? Great job.
I know
Im sexy and I know it :p
Not so convoluted! Thanks for not being discrete on the teachings!
I thought the product of (-1) and (-1) is 1 not -1 !!! (8.20 min)