0
\$\begingroup\$

i am third year Electrical engineering student, i have some issues that i could not find a good explanation for it. lets say we have an RC circuit (low pass configuration). if we calculate the transfer function we get Vo/vi=1/1+RCS. so we have a single pole in s=-1/RC. my question is what will happened to transfer function we stimulate the system in frequency that is equal to the system pole. i would expect it to blow since the denominator is zero in this case, but if we look at the frequency of this system it never blow. so what am i missing here? is it possible at all to stimulate the system in it pole frequency? thanks!

\$\endgroup\$
3
  • \$\begingroup\$ And s-domain is just a mathematical abstraction that helps us in this case (we do not need to solve a differential equation). And s-domain does not exist in reality. And in s-domain complex plane approaches infinity at the pole frequency but, such a situation is not possible in reality. In reality, we have a corner frequency at this point (-3dB for a first-order circuit). So to get a real response set s=jw into the transfer function and take the magnitude. \$\endgroup\$
    – G36
    Apr 18, 2020 at 20:15
  • \$\begingroup\$ ok thanks,can you please say more why this situation is nit possible in reality? \$\endgroup\$
    – Eli Cohen
    Apr 18, 2020 at 20:47
  • \$\begingroup\$ Eli, check Andy's answer here: electronics.stackexchange.com/a/316924/43172 \$\endgroup\$
    – Mike
    Apr 18, 2020 at 22:44

1 Answer 1

0
\$\begingroup\$

If your stimulus is a sinewave at the corner frequency, you will see an output sinewave that's 3db lower than the input.

The transfer function doesn't "blow up" because you're not causing the denominator to become zero. You're looking at the special case where s=jw, (i.e. a pure sine on the jw axis) and w is the corner frequency.

Plug that (s=jw) into the transfer function and take the magnitude and you'll see that the output is just a sinewave with the 3dB attenuation.

\$\endgroup\$
2
  • \$\begingroup\$ thank you John. by the frequency response its clear that if i stimulus a sinewave at the corner frequency, i get a output sinewave that's 3db lower than the input. but i cant understand why i am not causing the denominator to become zero? can you please explain me more please \$\endgroup\$
    – Eli Cohen
    Apr 18, 2020 at 19:16
  • \$\begingroup\$ Remember, s is a complex number, not a frequency. s=sigma + jomega. But for frequency response you're only looking at the case of a constant amplitude sinewave at a particular frequency, so sigma is zero, and jomega is j*2*pi*f. So s=jw, and is an imaginary quantity- s does not equal -1/RC. The magnitude of (1+RCjw) is not zero. \$\endgroup\$
    – John D
    Apr 19, 2020 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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