super helpful. Even if I had to think a little to make sense of the 'breaking up the components and adding them up', it helped me a lot when I got it how algebraically we're doing the same thing as adding up the figures
Although some algorithms in the software does the calculation in a straight forward way, the decomposition of x[n] into a sum of its discrete components is a great intuition, which clarifies this concept so much. Algorithms can also be developed based on this method, I think it may not be any slower than the former method, if not faster. Thank you so much for sharing it!
@@iain_explains The steps we were taught by my professor was to first convert x(n) to x(k) and do the same for h(n). Then,we do a time reversal on either x(k) or h(k),afterward, we perform a left hand shift until we get to the zero values. We then perform a right hand shift until we get continuous zero values. This sequence produced gives the convolution sum.
It's so much easier to understand convolution in the discrete case imo. I think it should be taught discretely first, then continuously - especially if you start with the example of a system's unit impulse response :)
That's interesting. Indeed, I think it's often easier to think in terms of (discrete) summations rather than (continuous) integrals, but on the other hand, the "natural" response of a real (continuous) system can be more motivating (easier to visualise), compared to the "response" of a (non-natural-world) digital system.
I'm embarrassed I didn't realize convolution was a sum till you spelt it out. I see the capital Sigma but somehow it didn't register. Until you. Great work Professor I'm really grateful
I'm so glad I was able to help clear that up for you. Thanks for your comment. It's always nice to hear from people who are finding the videos helpful.
The result of discrete convolution between two signal looks like the coefficient obtain by multiplying two polynomials together. what would be an appropriate to explain this issue ?
One confusion here: In the continuous time, we flip h(\tau), shift it by t , multiply with x(\tau), and then find the area to determine different values of y(t). In discrete time, it looks like we shift h(k) by n and then multiply with x(k), then add to get y(n). So, this flipping operation is not done in discrete time. Is my understanding correct?
Many people (basically everyone) talks about "flipping" and "shifting", but I never use these terms. I believe these terms are the cause of most students finding convolution difficult. In these videos on continuous time convolution I use the same approach each time (and it's the same approach as for the discrete time case): "Convolution of Square with Rectangle" ruclips.net/video/xVbONyYEipU/видео.html and "Convolution Square with Exponential" ruclips.net/video/lsHkWFBm3so/видео.html
You got my like about 2 seconds in, for your langauge / pronunciation alone; At some point (I passed long long ago) hearing Indian English becomes tiring, but hearing actual British English (or what sounds to me like that)...
Hi Iain, I have been working on a PAM-3 differential code that I have not been able to find any similar code. I would like someone who has worked on Digital Comms research to analyze my ideas. Can I send you a small word document that shows the encoding? I am a senior design engineer who has been working for many years.
One of the best intuitive explanations of convolution so far. Thank you!
Glad it was helpful!
the last part explained the first part better. thanks for this
super helpful. Even if I had to think a little to make sense of the 'breaking up the components and adding them up', it helped me a lot when I got it how algebraically we're doing the same thing as adding up the figures
Great. I'm glad it helped. Perhaps this video may provide more intuition: "How to Understand Convolution" ruclips.net/video/x3Fdd6V_Hok/видео.html
Although some algorithms in the software does the calculation in a straight forward way, the decomposition of x[n] into a sum of its discrete components is a great intuition, which clarifies this concept so much. Algorithms can also be developed based on this method, I think it may not be any slower than the former method, if not faster. Thank you so much for sharing it!
I'm glad you found the video interesting and useful.
@@iain_explains The steps we were taught by my professor was to first convert x(n) to x(k) and do the same for h(n). Then,we do a time reversal on either x(k) or h(k),afterward, we perform a left hand shift until we get to the zero values. We then perform a right hand shift until we get continuous zero values. This sequence produced gives the convolution sum.
You helped me figure out graphing convolution 15 mins before my exam! Thank you!
I'm glad you found it useful. And I hope your exam went well.
Watched ~5 vids and this is the best approach
I'm so glad you think so. Thanks for letting me know.
It's so much easier to understand convolution in the discrete case imo. I think it should be taught discretely first, then continuously - especially if you start with the example of a system's unit impulse response :)
That's interesting. Indeed, I think it's often easier to think in terms of (discrete) summations rather than (continuous) integrals, but on the other hand, the "natural" response of a real (continuous) system can be more motivating (easier to visualise), compared to the "response" of a (non-natural-world) digital system.
I'm embarrassed I didn't realize convolution was a sum till you spelt it out.
I see the capital Sigma but somehow it didn't register. Until you.
Great work Professor
I'm really grateful
I'm so glad I was able to help clear that up for you. Thanks for your comment. It's always nice to hear from people who are finding the videos helpful.
Best video on convolution, Thanks a lot !
Glad it was helpful!
wonderful and so simple. you are a remarkable doctor😍
I'm so glad you like the video.
This is a great help to understand convolution, thank you.
Glad it was helpful!
Great one.
The result of discrete convolution between two signal looks like the coefficient obtain by multiplying two polynomials together. what would be an appropriate to explain this issue ?
One confusion here: In the continuous time, we flip h(\tau), shift it by t , multiply with x(\tau), and then find the area to determine different values of y(t). In discrete time, it looks like we shift h(k) by n and then multiply with x(k), then add to get y(n). So, this flipping operation is not done in discrete time. Is my understanding correct?
Many people (basically everyone) talks about "flipping" and "shifting", but I never use these terms. I believe these terms are the cause of most students finding convolution difficult. In these videos on continuous time convolution I use the same approach each time (and it's the same approach as for the discrete time case): "Convolution of Square with Rectangle" ruclips.net/video/xVbONyYEipU/видео.html and "Convolution Square with Exponential" ruclips.net/video/lsHkWFBm3so/видео.html
thank you so much now i can understand how convolution work ,From Algeria
I'm so glad it was helpful.
Why at the beginnin you put n=-1 at y[n] you get x[0]?
I continued the maths.... it makes sense 🥺
thank you a lots❤
I'm glad you found it helpful!
Very good. Thank you so much. But what does convolution say. Any inference would be a value add
I''m not sure what you mean exactly. Perhaps this video will help: "How to Understand Convolution" ruclips.net/video/x3Fdd6V_Hok/видео.html
really good and understandable explanation!
Glad it was helpful!
You got my like about 2 seconds in,
for your langauge / pronunciation alone;
At some point (I passed long long ago) hearing Indian English becomes tiring,
but hearing actual British English (or what sounds to me like that)...
helpful but what if the response is negative ? will this still apply ?
Yes.
you are a lifesaver, thank you!
Glad it helped!
I better to share this video to my professor to show him how to teach... thank you
I'm glad you found it useful!
Great explanation 👍🏻👍🏻👍🏻
Glad you liked it!
great explanation
Thanks. Glad you liked it.
Finally, I think I am able to crack convolution quiz going to be happen tomorrow...
That's great to hear. Good luck with your quiz!
Hi Iain, I have been working on a PAM-3 differential code that I have not been able to find any similar code. I would like someone who has worked on Digital Comms research to analyze my ideas. Can I send you a small word document that shows the encoding? I am a senior design engineer who has been working for many years.
Sure, I'm happy to take a look. Feel free to email me.
Thank you, hi from Bolivia :D
Welcome!
Thank you sooooooo much!!!!
I'm glad it was helpful.
WHY IS K ZERO AND ONE- HOW DO I DETERMINE THE VALUE I USE FOR K
This video might help: "Discrete Time System Output Example" ruclips.net/video/Um564Ftq5s0/видео.html
Sir really, understand, thanks
Glad to hear that
That's a very Good method..I really Like It..Thanks:)
Glad you like it!
super!
Thankyou : )
❤
but what if x[n] is infinite, x[n]=u[n]
This is a discrete-time example. The values cannot be infinite. Perhaps you mean "infinite energy"? But that's OK, just use the same technique.
dr andrews if youre seeing this please dont make quiz 3 hard
wow amazing :) Türkiyeden Selamlar
Glad you liked it.