0
\$\begingroup\$

I've been trying to calculate the frequency response of a series LR circuit with the output measured across the resistor.

I solved it in two ways, one by taking the Fourier Transform of the impulse function of this circuit, and the other method by simply using a voltage divider of the impedances. I've been getting different answers, for some reason. For one answer, I am getting the transfer function to be positive R/(R+jwL) and for the other answer, I am getting the negative of that.

Are they the same or did I do something wrong?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ They are different \$\endgroup\$ – Scott Seidman Jan 26 '17 at 2:27
  • 1
    \$\begingroup\$ In my opinion if you got a negative answer you did something wrong. \$\endgroup\$ – Claudio Avi Chami Jan 26 '17 at 2:42
1
\$\begingroup\$

I assume you are getting the negative sign with the Fourier transform, rather than with the voltage divider approach. So, just to perform the Fourier analysis for a series RL circuit with the output taken across the resistor, the impulse response is: $$\frac{1}{\tau}e^{-t/\tau}$$ and the resultant Fourier transform is: $$\int_{0}^{\infty}\frac{1}{\tau}e^{-t/\tau}e^{-j\omega t}dt$$

where the lower limit is zero since the impulse response is zero for \$t<0\$.

Performing the integral gives: $$\frac{1}{\tau}\left[\frac{-1}{\frac{1}{\tau}+j\omega}\large e^{-(\frac{1}{\tau}+j\omega)t}\right]_0^\infty$$

Taking the upper limit gives zero; and the lower limit gives: $$\frac{1}{1+j\omega \tau}$$

which is positive!

\$\endgroup\$
2
  • 1
    \$\begingroup\$ ...and the answer to the question is? \$\endgroup\$ – pipe Jan 26 '17 at 13:45
  • \$\begingroup\$ @pipe, fair comment, thank you! I've added some words. \$\endgroup\$ – Chu Jan 26 '17 at 13:59

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.