# Sketching Fourier Transform

satchmo05

## Homework Statement

Let h(t) be impulse response of unity-gain ideal lowpass filter with bandwidth of 50[Hz] and a time delay of 5[ms]. Sketch magnitude and phase of Fourier transform of h(t).

## The Attempt at a Solution

I know that the magnitude2 of H(f) is total power gain, so perhaps by taking the square of this expression might get me the magnitude of simply H(f). However, I am not exactly sure how to take the information given to convert into an expression I can actually deal with.

Any help would be most appreciated. Thanks!

## Answers and Replies

rootX
How does low pass filter frequency response looks like?

Now, you are making a shift in time domain, how would this impact in frequency domain?

In impulse response input is δ(t), so output will be: y(t) = δ(t) conv h(t). What would this equal in frequency domain? Y(s) = ?

satchmo05
rootX,

Thank you for your response! I believe I understand the magnitude of the response. The magnitude is simply a function of the lowpass filter itself, drawing the bandwidth as given where the function's bandwidth breaks at the corner frequency (-3db). I do not believe the time delay affects a low pass filter. However, the phase graph may and that is where I am having trouble now. I believe the phase plot will look like a downward slope from -90 degrees to 0 degrees, where -45 degrees and 20[Hz] intersect.

If I introduce a time delay in the continuous time domain (h(t)), that would add a complex exponential factor in front of the H(f). Does this only change the order of magnitude of the two magnitude and phase plots?

If you convolve in the time domain, you multiply in the frequency domain. What did you mean by that statement? Thanks again for your help.