Question

When using the transfer function, when is it necessary to use H(j*omega) instead of H(s), and why?

(If I don't receive a response, I'll assume that this is the wrong subject area in which to post this question. :D)

Answer 1

We get \(H(s)\) when we find the Laplace transform of a function. And it is called the transfer function. Here \(s=\sigma+\mathbf j \omega\). And when \(\sigma \to 0\) we will get the Fourier transform of the function where we call it as the 'frequency response' of the function which can be given an actual meaning. Now \(H(s)\to H(\mathbf j \omega)\). So if you're asked for transfer function it is the laplace transform, and if it is the frequency response you need it is the Fourier transform.

Answer 2

Great explanation, thanks!

Answer 3

Tell me though, is omega frequency? And if it is, what is sigma?

Answer 4

I understand that sigma is the real part of something, and omega is the imaginary part. But how can frequency have an imaginary component?

Answer 5

yes omega is the frequency. Before going further I have to say that both 'Exponential' Fourier transform and Laplace transforms are mathematical concepts. It is the sinusoidal fourier transform which has some kind of a meaning (frequency component concept).

From \(e^{j\omega t}=\cos(\omega t)+\mathbf j\sin(\omega t)\) we derive the exponential fourier transform from the sinusoidal fourier transform. And that's how we get a imaginary part. I believe that we prefer exponential fourier transform over the other since it is easier to handle mathematically.

The \(\sigma\) here does not have a direct meaning. it is used because so that we can find the transform similar to fourier transform for the signals which we cannot find the fourier transform.

As I've learned the sigma comes to play like this,

say we need to find the fourier transform for a signal that is not a energy signal (say \(\sin(\omega t)\)). What we do is multiply it with \(e^{-\sigma t}\). Now if sigma is positive (only in this case), \(\sin(\omega t)e^{-\sigma t}\) will convert which will enable us to find the fourier transform, so from definition,

\[\mathrm{F}\{\sin(\omega t)e^{-\sigma t}\}=\int \limits_{-\infty}^\infty \sin(\omega t)e^{-\sigma t}e^{-j\omega t}dt=\int \limits_{-\infty}^\infty \sin(\omega t)e^{-(\sigma+j\omega )t}dt\]

So no we call multiplying with \(e^{-\sigma t}\) and taking the fourier transform - Laplace tranform

Answer 6

Thanks! Wish I could give you another medal!