The transfer function and frequency response. - Page 1
Forum section: electronics / beginners

author: Tamu23
author_rank: Newbie
timestamp: June 08, 2022, 04:53:04 am
content: Hello, I’ve been having some trouble understanding this concept. I understand that the transfer function is Vo/Vin and in an amplifier that is the gain and that we use the laplace transform of the two voltages and make them functions of s.But I always thought that evaluating the laplace transform at a given frequency just told you how much of that frequency is present in the signal. So how can we take it to be a literal ratio like 6/3 = 2 and say that the output voltage is 2 times as high as the input voltage when mathematically it means the output voltage contains twice as much of that frequency?Also, I don’t understand why we make the substitution s = jw when computing this either, and I would love an intuitive explanation of this because I haven’t found one online yet. Thank you!

author: IanB
author_rank: Super Contributor
timestamp: June 08, 2022, 05:13:43 am
content: The transfer function does not simply represent a gain, it represents some kind of function or transformation applied to a signal.For example, there may be an input signal x(t) (a function of time) being processed by something like an amplifier, giving a transformed output signal y(t). This can be represented schematically something like this:x(t) => [ G(t) ] => y(t)Working with this system in the time domain can be complicated, so the Laplace transform is a way of making it easier to work with. In the Laplace domain the transformed system becomes:x(s) => [ G(s) ] => y(s)The advantage is that in the time domain, the function G(t) can involve differential and/or integral equations, but in the Laplace domain G(s) is transformed into algebraic equations.This is a topic that can be the subject of a whole course or many chapters of a textbook, so there is no way to condense it into one post in a forum like this.

author: IanB
author_rank: Super Contributor
timestamp: June 08, 2022, 05:21:19 am
content: Further to your question, if the input signal x(t) were a pure sine wave at some frequency f, then the transfer function G would produce an output y(t) which would also be a periodic function of some kind. Comparing the amplitude of y(t) with x(t) would give the system gain at that frequency f.This calculation becomes easier to do by applying the Laplace transform to the transfer function and applying x(s) as an input to give y(s) as the output.

author: gf
author_rank: Super Contributor
timestamp: June 08, 2022, 06:12:37 am
content: Quote from: IanB on June 08, 2022, 05:21:19 amPresuming a linear system, this "periodic function of some kind" is of course still a pure sine wave with frequency f (just with potentially different amplitude and phase).

author: emece67
author_rank: Frequent Contributor
timestamp: June 08, 2022, 09:15:36 am
content: .

author: rstofer
author_rank: Super Contributor
timestamp: June 08, 2022, 04:06:36 pm
content: There are a lot of videos on this topic. Google for 'frequency response from laplace transform' - no quotes.3blue1brown has also done some animations. Google for 'fourier 3blue1brown'

author: Benta
author_rank: Super Contributor
timestamp: June 08, 2022, 06:29:15 pm
content: This might help:https://www.eevblog.com/forum/projects/analog-filters-a-compilation-of-standard-transfer-functions/