0
\$\begingroup\$

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)

enter image description here

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?

\$\endgroup\$
  • \$\begingroup\$ 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 \$\endgroup\$ – JonRB Oct 28 '18 at 17:11
  • \$\begingroup\$ @JonRB still how would I eliminate \$\phi\$ and \$\theta\$? \$\endgroup\$ – paulplusx Oct 28 '18 at 17:14
  • \$\begingroup\$ you don't :) thats what you are after. \$\endgroup\$ – JonRB Oct 28 '18 at 17:14
  • \$\begingroup\$ @JonRB I don't understand. How can I get the ratio \$\displaystyle \left|\frac{\alpha}{\beta}\right|\$ without eliminating them? \$\endgroup\$ – paulplusx Oct 28 '18 at 17:16
  • \$\begingroup\$ by determinig them \$\endgroup\$ – JonRB Oct 28 '18 at 17:17
1
\$\begingroup\$

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: $$\displaystyle \left|\frac{H(w=1)}{H(w=100)}\right|$$ $$= \displaystyle \left|\frac{10}{10-i}*\frac{10-100i}{10}\right|$$ $$= \displaystyle \left|\frac{10-100i}{10-i}\right|$$ $$= \displaystyle \left|10*\frac{1-10i}{10-i}\right|$$ $$= 10$$

\$\endgroup\$
  • \$\begingroup\$ Thank you for the verification. I did it in the same way (if you look in the chat). \$\endgroup\$ – paulplusx Oct 30 '18 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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