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In this question I asked about the difference between transfer function and frequency response. One user replied that "the frequency response is the transfer function where the transients are assumed to be completely dissipated". He showed an example, to prove his statement. It was like this:

Take, as an example, a sinusoid, \$\small \sin(\omega t) \rightarrow \dfrac{\omega}{s^2+\omega^2}\$, applied to a simple first order lag, \$\small G(s)=\dfrac{1}{1+s}\$. The response is: \$\small > R(s)=\dfrac{\omega}{(s^2+\omega^2)(1+s)}\$, and this can be expressed in partial fractions:

$$\small \frac{\omega}{(s^2+\omega^2)(1+s)}=\frac{A+Bs}{(s^2+\omega^2)}+\frac{C}{(1+s)}$$

Inverse LT gives:$$\small r(t)=\frac{A}{\omega}\sin(\omega t)+ B\cos(\omega t)+Ce^{-t/\tau}$$

The exponential term decays to zero, leaving the steady-state response as:

$$\small \frac{A}{\omega}\sin(\omega t)+B\cos(\omega t)= X\sin(\omega t+\phi)$$

Solving for \$\small X\$ and \$\small\phi\$ gives \$ \frac{1}{\sqrt{1+\omega^2}}\$, and \$\small \arctan{(-\omega)}\$, respectively, as is obtained using \$\small s\rightarrow j\omega\$ in the Laplace TF.

I don't quite understand the last part. How does he calculate \$\small X\$ and \$\small\phi\$ and what does he deduce by plugging \$\small s\rightarrow j\omega\$ into the transfer function? How is his original statement verified?

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luis
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This is just answering one part of your question:

I don't quite understand the last part. How does he calculate X and ϕ

This is just applying the trigonometric identity $$\sin\left(\alpha + \beta\right)=\sin\alpha\cos\beta + \cos\alpha\sin\beta$$ with \$\alpha=\omega{}t\$ and \$\beta=\phi\$.

Using this identity on the r.h.s. gives $$X\sin\left(\omega{}t+\phi\right)=X\left(\sin{}\omega{}t\cos\phi+\cos\omega{}t\sin\phi\right)$$ so our equation becomes $$\frac{A}{\omega}\sin\left(\omega{}t\right)+B\cos\left(\omega{}t\right)=X\left(\sin{}\omega{}t\cos\phi+\cos\omega{}t\sin\phi\right)$$

Which we can break into two parts, $$\frac{A}{\omega}\sin\left(\omega{}t\right)=X\cos\phi\sin{}\omega{}t$$ and $$B\cos\left(\omega{}t\right)=X\sin\phi\cos\omega{}t$$

So, $$\frac{A}{\omega} = X\cos\phi$$ and $$B=X\sin\phi$$

From there you should be able to get the conclusions from your source.

The Photon
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  • Also, \$s\rightarrow j\omega\$ arises from the close relationship between the Laplace transform and unilateral Fourier transform for causal systems. – Chu Mar 24 '17 at 08:03
  • I cannot really get rid of the terms A and B. $$ X = \frac{A}{\omega \; cos \phi }$$ $$ B = \frac{A \;tan \phi}{\omega} $$ $$ \phi = arctan \frac{B \omega}{A} $$ – luis Mar 24 '17 at 08:04
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    From the partial fraction equation: \$(A+Bs)(1+s)+C(s^2+\omega ^2)=\omega\$. Comparing coefficients gives: \$A+C\omega ^2=\omega\$; \$A+B=0\$; \$B+C=0\$; and hence \$A=\frac{\omega}{1+\omega ^2}\$ and \$B=-\frac{\omega}{1+\omega ^2}\$. Then plug these back into the partial fraction equation. – Chu Mar 24 '17 at 08:16
  • Ok thanks, that i got now. We've shown that the steady state part of the response can be expressed as a single sinusoid with specific amplitude and phase. Now how do we relate that to the fourier transform. \$ R(j \omega) \$ blows up as far as I can see. – luis Mar 24 '17 at 08:37
  • \$s\rightarrow j\omega\$ in \$\large \frac{1}{1+s}\$ gives \$\large \frac{1}{1+j\omega}\$ – Chu Mar 24 '17 at 09:29
  • So you are saying I should look at the inverse of \$\small R(s)=\dfrac{\omega}{(s^2+\omega^2)(1+j \omega)}\$ ? – luis Mar 24 '17 at 09:32
  • The TF is \$\large \frac{1}{1+s}\$ and its Fourier transform or spectrum is \$\large \frac{1}{1+j\omega}\$ – Chu Mar 24 '17 at 10:12
  • I know, but somehow I need to show that by FT I obtain only the steady state response in order to prove your claim. How do I continue from this substitution? – luis Mar 24 '17 at 10:19
  • Do the unilateral Fourier integral on \$e^{-t}\$, which Laplace transforms to \$\frac{1}{1+s}\$ – Chu Mar 24 '17 at 11:02
  • \$F(e^{-t}u(t)) = \frac{1}{1+j\omega}\$ . Sorry but I still don't get how it proves your point. Why did we go through the trouble of calculating \$X\$ and \$\phi\$ ? – luis Mar 24 '17 at 12:33
  • You quoted from my answer to a separate, older question. You asked how I calculated X and phi, and what info I got from s->jw. Well I think I've answered that in my comments here. – Chu Mar 25 '17 at 01:09
  • Sure the calculation was one part of the question. Thanks for your explanation! But in the end I would like to have a proof that (at least for our example) "the frequency response is the transfer function where the transients are assumed to be completely dissipated". That was the last part of the question in this post. – luis Mar 25 '17 at 08:49
  • There's nothing to prove, it's the definition of "frequency response". Or rather, "frequency response" is usually just a sloppy way to say "transfer function restricted to steady-state ac signals". As my earlier answer says, "frequency response" might also refer to other responses besides the transfer response. – The Photon Mar 25 '17 at 15:27
  • This means there is no way to show via Fourier transform that the output signal in the time domain is the same without the decay-term? I thought that was what he was going for when he said: "as is obtained using \$s \rightarrow j\omega\$ in the Laplace TF". – luis Mar 25 '17 at 20:29
  • You asked for "proof that 'the frequency response is the transfer function where the transients are assumed to be completely dissipated'". This is the part I'm saying is simply the definition of what is meant by "frequency response" (in some particular context). You can't prove this or disprove it. It's just an explanation of what you mean when you say "frequency response". – The Photon Mar 25 '17 at 22:12
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    @luis, my point is, you start with this definition and try to express it mathematically. You can't prove it with mathematics because it's not a mathematical statement. – The Photon Mar 25 '17 at 22:13