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The function w(t)=δ(t+10)-2δ(t+1)+3δ(t-5) consisting of three weighted δ functions is represented graphically as three weighted arrows at t= -10, -1, 5, weighted by the appropriate constants.
The function w ( t ) = δ ( t + 10 ) - 2 δ ( t + 1 ) + 3 δ ( t - 5 ) consisting of three weighted δ functions is represented graphically as three weighted arrows at t = -10, -1, 5, weighted by the appropriate constants.

The discrete time counterpart of δ ( t ) is the (discrete) delta function

δ [ k ] = 1 k = 0 0 k 0 .

While there are a few subtleties (i.e., differences) between δ ( t ) and δ [ k ] , for the most part they act analogously. For example, the program specdelta.m calculates the spectrum of the (discrete) delta function.

time=2; % length of timeTs=1/100;                   % time interval between samples t=Ts:Ts:time;               % create time vectorx=zeros(size(t));           % create signal of all zeros x(1)=1;% delta function plotspec(x,Ts)              % draw waveform and spectrum
specdelta.m plots the spectrum of a delta function (download file)

The output of specdelta.m is shown in [link] . As expected from [link] , the magnitude spectrum of the delta function is equal to 1 at all frequencies.

Calculate the Fourier transform of δ ( t - t 0 ) from the definition. Now calculate it using the time shift property [link] . Are they the same?Hint: They had better be.

Use the definition of the IFT [link] to show that

δ ( f - f 0 ) e j 2 π f 0 t .

Mimic the code in specdelta.m to find the spectrum of the discrete delta function when:

  1. The delta does not occur at the start of x . Try x[10]=1 , x[100]=1 , and x[110]=1 . How do the spectra differ? Can you use the time shift property [link] to explain what you see?
  2. The delta changes magnitude x . Try x[1]=10 , x[10]=3 , and x[110]= 0.1 . How do the spectra differ? Can you use the linearity property [link] to explain what you see?
A (discrete) delta function at time 0 has a magnitude spectrum equal to 1 for all frequencies.
A (discrete) delta function at time 0 has a magnitude spectrum equal to 1 for all frequencies.

Mimic the code in specdelta.m to find the magnitude spectrum of the discrete delta function when:

  1. The delta does not occur at the start of x . Try x[10]=1 , x[100]=1 , and x[110]=1 . How do the spectra differ? Can you use the time shift property [link] to explain what you see?
  2. The delta changes magnitude x . Try x[1]=10 , x[10]=3 , and x[110]= 0.1 . How do the spectra differ? Can you use the linearity property [link] to explain what you see?

Modify the code in specdelta.m to find the phase spectrum of a signal that consists of a delta function when the nonzero termis located at the start ( x(1)=1 ), the middle ( x(100)=1 ) and at the end ( x(200)=1 ).

Mimic the code in specdelta.m to find the spectrum of a train of equally spaced pulses. For instance, x(1:20:end)=1 spaces the pulses 20 samples apart, and x(1:25:end)=1 places the pulses 25 samples apart.

  1. Can you predict how far apart the resulting pulses in the spectrum will be?
  2. Show that
    k = - δ ( t - k T s ) 1 T s n = - δ ( f - n f s ) ,
    where f s = 1 / T s . Hint: Let w ( t ) = 1 in [link] .
  3. Now can you predict how far apart the pulses in the spectrum are? Your answer should be in terms of how farapart the pulses are in the time signal.

In [link] , the spectrum of a sinusoid was shown to consist of two symmetrical spikes in the frequency domain,(recall [link] ). The next example shows why this is true by explicitly taking the Fouriertransform.

Let w ( t ) = A sin ( 2 π f 0 t ) , and use Euler's identity [link] to rewrite w ( t ) as

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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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