How Are Step Functions Used in Calculating Fourier Transforms?

[mickonk]

Hi. Here is one example from my book.
Calculate Fourier transform of signal:

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

,

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

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

,
Fourier transform of our signal will be:

We will find Fourier transform of x1(n):

How they calculated that u(n+N)-u(n-N-1) equals 1 for -N<=n<=N and 0 for other values of N?

 

[milesyoung]

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.

 

[mickonk]

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?

 

[milesyoung]

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.

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?