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?


1 Answer 1


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.

  • \$\begingroup\$ Also, \$s\rightarrow j\omega\$ arises from the close relationship between the Laplace transform and unilateral Fourier transform for causal systems. \$\endgroup\$
    – Chu
    Mar 24, 2017 at 8:03
  • \$\begingroup\$ I cannot really get rid of the terms A and B. $$ X = \frac{A}{\omega \; cos \phi }$$ $$ B = \frac{A \;tan \phi}{\omega} $$ $$ \phi = arctan \frac{B \omega}{A} $$ \$\endgroup\$
    – luis
    Mar 24, 2017 at 8:04
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Chu
    Mar 24, 2017 at 8:16
  • \$\begingroup\$ 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. \$\endgroup\$
    – luis
    Mar 24, 2017 at 8:37
  • 1
    \$\begingroup\$ @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. \$\endgroup\$
    – The Photon
    Mar 25, 2017 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.