Does the sign matter on the frequency response?

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?

• They are different Jan 26 '17 at 2:27
• In my opinion if you got a negative answer you did something wrong. 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!

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