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 :


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

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


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 :


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

  • \$\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\$
    – user16222
    Commented Oct 28, 2018 at 17:11
  • \$\begingroup\$ @JonRB still how would I eliminate \$\phi\$ and \$\theta\$? \$\endgroup\$
    – paulplusx
    Commented Oct 28, 2018 at 17:14
  • \$\begingroup\$ you don't :) thats what you are after. \$\endgroup\$
    – user16222
    Commented Oct 28, 2018 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
    Commented Oct 28, 2018 at 17:16
  • \$\begingroup\$ by determinig them \$\endgroup\$
    – user16222
    Commented Oct 28, 2018 at 17:17

1 Answer 1


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

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

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