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?

## Best Answer

This is just answering one part of your question:

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.