Sketching Fourier Transform of Lowpass Filter with Time Delay and Bandwidth

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SUMMARY

The discussion focuses on the Fourier Transform of a unity-gain ideal lowpass filter with a bandwidth of 50 Hz and a time delay of 5 ms. The magnitude of the Fourier Transform, H(f), is determined primarily by the filter's bandwidth, with the corner frequency marking the -3 dB point. The time delay introduces a complex exponential factor in the frequency domain, affecting the phase response but not the magnitude. The phase plot is expected to slope from -90 degrees to 0 degrees, intersecting at -45 degrees and 20 Hz.

PREREQUISITES
  • Understanding of Fourier Transform principles
  • Knowledge of lowpass filter characteristics
  • Familiarity with time-domain and frequency-domain relationships
  • Basic concepts of convolution and its effects in signal processing
NEXT STEPS
  • Study the properties of the Fourier Transform for linear time-invariant systems
  • Learn about the impact of time delays on frequency response
  • Explore the concept of convolution in the time domain and its multiplication in the frequency domain
  • Investigate the phase response of filters and its implications in signal processing
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Electrical engineers, signal processing students, and anyone involved in the design and analysis of filters and their frequency responses.

satchmo05
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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!
 
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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) = ?
 
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.
 

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