How Are Step Functions Used in Calculating Fourier Transforms?

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Step functions are utilized in calculating Fourier transforms by defining intervals where the function takes specific values. The discussion centers on the function x1(n) represented as u(n+N) - u(n-N-1), which equals 1 for the range -N ≤ n ≤ N and 0 elsewhere. Participants analyze how to derive this result using a number line to visualize the behavior of the step functions. There is confusion regarding the endpoints and the correct interpretation of the intervals, particularly concerning the values of N and n. Clarification is sought on the algebraic steps leading to the conclusion, emphasizing the importance of correctly identifying the intervals defined by the step functions.
mickonk
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Hi. Here is one example from my book.
Calculate Fourier transform of signal:

p1.png


Here is solution:
We can write x(n) as:

p2.png
,

where x1(n) is u(n+N)-u(n-N-1). We can write:

p3.png


(we used that cos(n)=(1/2)*(exp(j*n)+exp(-j*n)).
Using properties of Fourier transform of discrete signal:

p4.png
,
Fourier transform of our signal will be:

p5.png


We will find Fourier transform of x1(n):

p7.png
How they calculated that u(n+N)-u(n-N-1) equals 1 for -N<=n<=N and 0 for other values of N?
 
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mickonk said:
How they calculated that u(n+N)-u(n-N-1) equals 1 for -N<=n<=N and 0 for other values of N?
Try setting up a number line with -N and N + 1 on it, and then mark the intervals on that number line where u(n + N) and u(n - N - 1) are either 0 or 1.

You should be able to use that to say something about where u(n + N) - u(n - N - 1) is then 0 or 1.
 
Here is how I tried to solve this algebraically.
We have function u(n+N)-u(n-(N+1)). Let's say N is some positive number, for example N=1.
1. When n is less then -N, for example n=-2, we have u(n+N)-u(n-(N+1))=0-0=0.
2. When -N<n<N+1 , for example n=-0.5, we have u(n+N)-u(n-(N+1))=1-0=1.
3. When n>N+1, for example n=3, we have u(n+N)-u(n-(N+1))=1-1=0.
So our function equals 1 for n between -N and N+1 but solution from my book doesn't agree with it?
 
mickonk said:
Here is how I tried to solve this algebraically.
2. When -N<n<N+1 , for example n=-0.5, we have u(n+N)-u(n-(N+1))=1-0=1.
Take care to consider the endpoints of your intervals. What happened to -N?

Also, n and N are integers.

mickonk said:
3. When n>N+1, for example n=3, we have u(n+N)-u(n-(N+1))=1-1=0.
What happened to N + 1?
 

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