the point I don't understand about signal systems - Page 1
Forum section: electronics / general

author: circuit975
author_rank: Contributor
timestamp: July 05, 2025, 09:17:04 pm
content: i sincerely greet the entire communityi have difficulty understanding the output of this system, i.e. y(t), I have difficulty understandingy(t) = v(t) − h(t)*y(t)I know this is the right output. but I need to understand the logic of how this happened.actually, what I see is this: y(t) = v(t) - [v(t)*h(t)]but if it is true, as I said above, y(t) = v(t) − h(t)*y(t)if anyone knows what the logic of this is, I would be very glad if they would explain.every answer given will be referred to with gratitude.

author: TimFox
author_rank: Super Contributor
timestamp: July 05, 2025, 09:20:18 pm
content: That equation (withh yterm) must be a typo.y(t) is the output of the subtractor, and is not applied to the input thereof.

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2025, 01:58:04 am
content: The negative input to the summing junction is v(t) * h(t), i.e. v(t) convoluted with h(t). The positive input to the summing junction is v(t).Therefore the output is: y(t) = v(t) − v(t) * h(t)As Tim said, there is probably a typo somewhere.

author: circuit975
author_rank: Contributor
timestamp: July 06, 2025, 05:57:03 am
content: Quote from: IanB on July 06, 2025, 01:58:04 amwhat I saw at the beginning was true, wasn't it?however, many sources, including artificial intelligence, point out that the output is y (t) = v (t) − h (t) * y(t). My mind went blank.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 06, 2025, 06:44:19 am
content: Quote from: IanB on July 06, 2025, 01:58:04 amTo me, the formula given above looks correct by inspection of the circuit diagram in the first post.I have no idea what "convolution" means, but I take it that the box containingh(t)multiplies its input signal (v(t)) by that function (= gain?) and then sends it to the negative input of the summer. So @IanB's formula seems correct.Don't trust "AI" with this stuff ...

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2025, 06:44:55 am
content: Quote from: circuit975 on July 06, 2025, 05:57:03 amAI, or "Artificial Incorrectness", is not a source. You should ignore it unless you are able to find an expert human to confirm what it says.

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2025, 06:53:50 am
content: Quote from: Analog Kid on July 06, 2025, 06:44:19 amConvolution is the process of taking a time-varying signal, like v(t), and passing it through a dynamic processing element such as a filter, like h(t), to obtain the output in the time domain. The mathematical process of convolution is required because the output at any given instant depends on the history of the input over all preceding time.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 06, 2025, 07:11:14 am
content: So how do we know that that part of that circuit indicates convolution? Is it the shape of the rectangle? The use of the letterhto describe the function? The overall topology of the circuit?Since you say that "the output at any given instant depends on the history of the input over all preceding time", that sounds like integration, as one function I can think of. Does that qualify as convolution?

author: circuit975
author_rank: Contributor
timestamp: July 06, 2025, 08:30:06 am
content: Quote from: IanB on July 06, 2025, 06:53:50 amYes, I don’t fully trust AI either, so I wanted to verify it. So, in short, my initial assumption was correct, right?

author: TimFox
author_rank: Super Contributor
timestamp: July 06, 2025, 02:30:49 pm
content: Quote from: Analog Kid on July 06, 2025, 07:11:14 amConvolution is often denoted by “*”, which is confusing since that symbol is often used for simple multiplication.In careful usage, "convolution" is the process, but "convolve" is the verb."Convoluted" is another thing entirely.

author: circuit975
author_rank: Contributor
timestamp: July 06, 2025, 03:31:47 pm
content: Quote from: TimFox on July 06, 2025, 02:30:49 pmactually, as a result, this is the right thing right now, isn't it?y(t) = v(t) - [v(t) * h(t)]

author: Analog Kid
author_rank: Super Contributor
timestamp: July 06, 2025, 07:45:24 pm
content: Quote from: TimFox on July 06, 2025, 02:30:49 pmOK, that's interesting, but you didn't answer my question:How do we know (from the diagram, presumably) that the rectangle in the OP's circuit containingh(t)indicates convolution?

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2025, 08:00:52 pm
content: Quote from: Analog Kid on July 06, 2025, 07:45:24 pmWe know because of the "h(t)", which means the functions are in the time domain. In the time domain, combining functions on a block diagram involves convolution.If you transform to the Laplace domain, i.e. "h(s)", then you can replace convolution with simple multiplication.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 06, 2025, 08:08:31 pm
content: Quote from: IanB on July 06, 2025, 08:00:52 pmSo it's thetfor time that does it?Doeshhave any special meaning, or is it arbitrary?Still learning all these conventions ...

author: TimFox
author_rank: Super Contributor
timestamp: July 06, 2025, 08:21:55 pm
content: Quote from: Analog Kid on July 06, 2025, 07:45:24 pmWe know because of the symbol "*" in his text description.A bit opaque.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 06, 2025, 09:35:30 pm
content: Quote from: TimFox on July 06, 2025, 08:21:55 pmYou mean the*symbol as iny(t) = v(t)−v(t)*h(t)?But isn't that just the multiplication symbol? Which to me could mean thath(t)is just a simple gain function, not a convolution. (WhereA = h(t).)

author: TimFox
author_rank: Super Contributor
timestamp: July 06, 2025, 09:39:11 pm
content: The original question is ill-posed, but "*" is the standard symbol for convolution.If the source were cited, I might continue.

author: Analog Kid
author_rank: Super Contributor
timestamp: July 06, 2025, 09:44:41 pm
content: Quote from: TimFox on July 06, 2025, 09:39:11 pmOK, so what then is the multiplication symbol (assuming the function is a linear gain stage)?Just put the two variables next to each other, as in algebra?

author: TimFox
author_rank: Super Contributor
timestamp: July 06, 2025, 10:01:28 pm
content: Quote from: Analog Kid on July 06, 2025, 09:44:41 pmI use "x" or the two variables next to each other.Microsoft Word's equation editor uses a true St Andrew's cross instead of the miniscule x.One can also use "." with the dot centered vertically."*" for multiplication is a Hollerith-card convention.In other contexts, "*" can mean a footnote, or a birth date, or a wild-card character, or omission, or lots of other stuff.However, I have seen convolution use only the "*".

author: CatalinaWOW
author_rank: Super Contributor
timestamp: July 06, 2025, 10:46:52 pm
content: The only way you can know that convolution is involved is context. This is why dragging an equation out of a book or off the internet without fully understanding it is often confusing and sometimes dangerous.Different texts will use different conventions. Most but not all texts carefully describe how they will use the various symbols and maintain a consistent usage throughout. Over my lifetime I have seen some gradual moves toward standard notation, but it hasn't happened yet as far as I can tell. Part of the reason is that different fields use the same mathematics to solve very different problems and have long engrained practices. Also all these notations are to simplify and make the concepts easier to handle. Different problems and different writing/publishing tools make the different notations more applicable in some contexts than in others.This is one reason that AI is so often wrong. It searches reams of text and finds patterns and mashes them together in some manner. Sometimes taking a bit from here and a bit from there. Both sources were correct in their own context, but the combination is meaningless.

author: circuit975
author_rank: Contributor
timestamp: July 07, 2025, 07:04:10 am
content: Quote from: TimFox on July 06, 2025, 09:39:11 pmthe question was not asked badly, everything is quite clear and obvious. is it not quite clear that such a graph is a convolution? what part didn't you understand?

author: TimFox
author_rank: Super Contributor
timestamp: July 07, 2025, 02:02:25 pm
content: What graph? Did you mean drawing?