# Problem related to the transfer function of RC low pass filter

I have the following question:

For the passive RC low pass filter shown below: $$V_S(t)=\cos(t)+\cos(100t)$$ $$V_0(t)=\alpha\cos(t+\theta)+\beta\cos(100t+\phi)$$

(where $$\\alpha\$$, $$\\beta\$$, $$\\theta\$$ and $$\\phi\$$ are constants)

The value of $$\\displaystyle \left|\frac{\alpha}{\beta}\right|\$$ is?

The answer is $$\10\$$

I have tried using the transfer function of a RC low pass filter :

$$H(s)=\frac{1}{sRC+1}=\frac{10}{s+10}$$

$$\implies h(t)=10e^{-10t}u(t)$$

Then tried using the two given equation with $$\h(t)\$$ as below:

$$h(t)=\frac{V_0(t)}{V_S(t)}$$

After that I am stuck. Even after trying with Initial value theorem as $$\h(0)=\lim_{s \to \infty}sH(s)\$$, I end up with :

$$\alpha\cos(\theta)+\beta\cos(\phi)=20$$

Can someone please tell if I am missing something or if the question doesn't provide enough information?

• why not superposition. determine the gain and phase change per sinus stimulus. Hint. Replace s with jw and replace w with the freq of interest
– user16222
Commented Oct 28, 2018 at 17:11
• @JonRB still how would I eliminate $\phi$ and $\theta$? Commented Oct 28, 2018 at 17:14
• you don't :) thats what you are after.
– user16222
Commented Oct 28, 2018 at 17:14
• @JonRB I don't understand. How can I get the ratio $\displaystyle \left|\frac{\alpha}{\beta}\right|$ without eliminating them? Commented Oct 28, 2018 at 17:16
• by determinig them
– user16222
Commented Oct 28, 2018 at 17:17

## 1 Answer

The transfer function of this circuit is: $$H(i\omega)=\frac{10}{10-i\omega}$$

Insert w=1 and w=100 into the transfer function: $$H(\omega=1)=\frac{10}{10-i}$$ $$H(\omega=100)=\frac{10}{10-100i}$$

Build the absolute ratio of both transfer functions: $$\displaystyle \left|\frac{H(w=1)}{H(w=100)}\right|$$ $$= \displaystyle \left|\frac{10}{10-i}*\frac{10-100i}{10}\right|$$ $$= \displaystyle \left|\frac{10-100i}{10-i}\right|$$ $$= \displaystyle \left|10*\frac{1-10i}{10-i}\right|$$ $$= 10$$

• Thank you for the verification. I did it in the same way (if you look in the chat). Commented Oct 30, 2018 at 4:18