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?

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    \$\begingroup\$ They are different \$\endgroup\$ Jan 26 '17 at 2:27
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    \$\begingroup\$ In my opinion if you got a negative answer you did something wrong. \$\endgroup\$ Jan 26 '17 at 2:42

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!

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    \$\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

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