# From transfer function to frequency response

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?

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.

• 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 8:03
• 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 8:04
• 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 8:16
• 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 8:37
• @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