There is a nonsense signal about Convolution and I couldn't understand what to d - Page 1
Forum section: electronics / general

author: circuit975
author_rank: Contributor
timestamp: July 03, 2025, 01:13:04 pm
content: First of all, I wholeheartedly greet my colleagues and the entire community!I’ve set the signal asm(t) = −t [ u(t) − u(t−2) ]and the impulse response ash(t) = 2 [ u(t−3) − u(t−4) ].But while trying to find y(t), I ran into a small issue with the sketches.Because there is no overlap, and thus they don’t convolve.I just can’t figure out what I’m missing.There is a topic I can't get my head around.I have 2 signals. I’m going to convolve them, but even if I flip one of them, the other one lies horizontally on a different axis, so I can’t get them to overlap in any way — meaning I can’t convolve them.In problems like this, how are we supposed to perform the convolution operation?What’s the logic to solving these kinds of questions? What am I supposed to do?as I understand it, I drew my convolution, is there anyone competent to comment who has an opinion on whether this is true or not?

author: ejeffrey
author_rank: Super Contributor
timestamp: July 03, 2025, 01:36:38 pm
content: I can't make any sense of your graphs and you never defined u(t). I assume u(t) is a step function and what you have are two square "top hat" functions.Anyway, the output of convolution is another function, a function of a new variable tau. The fact that there is no overlap just means that the value of the convolution at tau=0 is zero...what you have to do for the convolution is to reflect one function, shift it by an amount tau, and then compute the integral of the product. So the first equation is, for what values of tau is there any overlap so that the convolution is non zero?

author: circuit975
author_rank: Contributor
timestamp: July 03, 2025, 01:54:31 pm
content: I’ve set the signal asm(t) = −t [ u(t) − u(t−2) ]and the impulse response ash(t) = 2 [ u(t−3) − u(t−4) ].I indicated the signs in this way at the beginning of my articlethe rectangular pulse on the left is h(t-tau), and the m(t) sign on the right.so the convolution I drew is wrong? honestly, what was supposed to happen?

author: Tation
author_rank: Regular Contributor
timestamp: July 03, 2025, 02:21:20 pm
content: Such sketches are not that good. The impulse response has width 1, the signal has width 2, but you have painted h wider than m.For each value of t, you must multiply h (displaced/reflected) by m. The sketches do not show the result of the multiplied signals. Obviously, there are values of t for which h (displaced/reflected) and m overlap, so the convolution will be != 0 for such t.Then, for each value of t, you need to integrate (the product of h —displaced/reflected— by m) from -∞ to ∞ to get the value of the convolution at such t.I think that, up to now, you have not yet grasped how to interpret «geometrically» the convolution. Sure there is some online, graphical, material on that.

author: circuit975
author_rank: Contributor
timestamp: July 03, 2025, 03:19:03 pm
content: Quote from: Tation on July 03, 2025, 02:21:20 pmI'm wondering which of the online materials is the most instructive and useful? i think if you share, we will learn as a community. thanks

author: ejeffrey
author_rank: Super Contributor
timestamp: July 03, 2025, 04:15:58 pm
content: You are mixing t and tau in your labels. If you can't avoid doing that use a completely different looking variable like q.Is your question how a positive and negative valued function can "overlap"? From your figures it seems maybe you are taking a geometric definition of overlap. That's the wrong way to think about overlap. "Overlap" just means non-zero at the same coordinate. What you do with convolution is multiplly the signals and integrate. Since x*0 is zero you only have to consider ranges where both functions (with the appropriate flip and shift) are non-zero.