Saturday, October 10, 2009

Multiplication and convolution

Using the tool, review the transforms of the unit pulse function and the cosine function. For the moment it is best to view these using the magnitude and phase representation of the frequency domain.
Now switch to one of the 8ms segment of a cosine or sine waveforms. You should observe that the frequency domain plot is some form of combination of the two types of signal. Strictly speaking, the time domain signal is the multiplication of a unit pulse of 8ms duration delayed by 4ms, and a cosinusoid or sinusoid waveform of the selected frequency. The frequency domain transform is then the addition of two sa functions which have been shifted in frequency. Notice where the highest peaks are and you should observe that these correspond with the frequency of the sine or cosine signal. What has happened is that in the frequency domain the sa function from the unit pulse and the two impulses from the sine or cosine function have been convolved together. This is an example of the general rule that multiplication in the time domain equates to convolution in the frequency domain.
You can reconstruct the two constituent waveforms by shifting the frequency response of the 8ms unit pulse to 500Hz, and to -500Hz.You should find that the real component of the two shifted signals are the same, but that the quadrature components are the complement of each other. Thus when they are summed together, the result is a signal with a real component and a zero quadrature component.
In fact an equivalent rule also holds that convolution in the time domain equates to multiplication in the frequency domain. Thus, for example, a complex phasor in the frequency domain multiplied by a given signal's transform produces a time domain function where an impulse is convolved with signal. This is precisely what is happening when the delay value is being altered.

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