# 0.3 Modelling corruption  (Page 6/11)

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The discrete time counterpart of $\delta \left(t\right)$ is the (discrete) delta function

$\delta \left[k\right]=\left\{\begin{array}{cc}1\hfill & k=0\hfill \\ 0\hfill & k\ne 0\hfill \end{array}\right).$

While there are a few subtleties (i.e., differences) between $\delta \left(t\right)$ and $\delta \left[k\right]$ , 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 $\delta \left(t-{t}_{0}\right)$ 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

$\delta \left(f-{f}_{0}\right)⇔{e}^{j2\pi {f}_{0}t}.$

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?

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
$\sum _{k=-\infty }^{\infty }\delta \left(t-k{T}_{s}\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}⇔\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\frac{1}{{T}_{s}}\sum _{n=-\infty }^{\infty }\delta \left(f-n{f}_{s}\right),$
where ${f}_{s}=1/{T}_{s}$ . Hint: Let $w\left(t\right)=1$ 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 $w\left(t\right)=A\mathrm{sin}\left(2\pi {f}_{0}t\right)$ , and use Euler's identity [link] to rewrite $w\left(t\right)$ as

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