3_MAE 334Woodward - lab5

3_MAE 334Woodward - lab5 - Because this low pass filter can...

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MAE 334 – LAB 5 3 Because this low pass filter can be described by a well behaved continuous function and the input to the filter is a continuous function the Fourier transform can be found. The Fourier transform of a convolution is simply the product of the Fourier transform of the two functions involved in the convolution , that is: If H(f) = Fourier Transform of { h(t) } ) ( ˆ f y = Fourier Transform of { y(t) } ) ( ˆ f F = Fourier Transform of { F(t) } then the frequency response function of the filter can be expressed as ) ( ˆ ) ( ) ( ˆ f F f H f y = (6) The linear system simply takes each Fourier component (frequency) of the input signal and multiplies it by a gain which is frequency dependent. Note that since H(f) is generally complex, the phase of each Fourier component can also be changed. (This in effect scrambles the signal in time by delaying some frequency parts of the signal differently with respect to other parts. If you were to send your voice signal into such a filter it would sound lower and slightly distorted.) Obviously equation (6) represents a great simplification over equation (3) since it
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This document was uploaded on 01/03/2011.

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