# Electronic Signal and System problem: Ratio of signal energy

1. Aug 27, 2011

### nidhalc

1. The problem statement, all variables and given/known data
x(t) is input into a perfect lowpass filter with frequency response H(ω), having a bandwidth of BHz and a passband gain of 1. For B = (2πT)-1 Hz, calculate the ratio of the output signal energy to the input signal energy.

2. Relevant equations
x(t) = Ae-|t|/T

3. The attempt at a solution
I got |x(jω)| = A/(1+jω)
Using input energy Wi = (1/pi)$\int$$\infty$0|x(jω)|2 I got Wi to be A2/2

The output energy Wo = (1/pi)$\int$$\infty$0|xo(jω)|2, where xo(jω) = X(jω)H(jω)

but I'm unable to find any equation which links the frequency response to the bandwidth in order to get the output energy. I have that |H(ω)| = gain which in this instance is = 1. Is that all I need? I've a feeling I'm missing something very basic here. Any hints or clues appreciated!!

2. Aug 27, 2011

### nidhalc

As it refers to it being a perfect low pass filter could you take it as being ideal and so xo(jω) = x(jω) and then use the bandwidth as the interval over which you're integrating, so you'd be integrating according to 0<|ω| < B? Sound right to anyone?

3. Aug 27, 2011

### uart

Yes that's what is expected here.

One question though. Is your signal $A e^{-|t|/T}$ for all t, or is it $A\,u(t)\,e^{-t/T}$?

Your Fourier transform seems to be that of the latter (but with T=1).

4. Aug 30, 2011

### nidhalc

Thank you so much for getting back to me. It never specified so I took t>0. Would the Fourier transform still be correct were that the case? In the Fourier transform examples we were given we were always given t>0.

5. Sep 1, 2011

### rude man

There is a difference between the Fourier integral and the Fourier transform. For transient inputs/outputs (t>= 0) it's the Fourier integral - used when the input is a single pulse, like yours. The Fourier transform covers all t and is analogous to the two-sided Laplace transform.

6. Sep 2, 2011

### uart

Why would they bother specifying |t| if they were only taking t>0?

The answers are quite different for the two cases. For a start the FT for the u(t)exp(-t/T) case is A/(1/T+jω) whereas for the other case it's 2T/(1 + (ωT)^2)

The case with the step response has more high frequency components than the other case so you lose more of the signal when you LPF.

Have you found any results for the first case yet?

Last edited: Sep 2, 2011