How Do You Compute the DFT of Periodic Signals?

[ankh]

Homework Statement

Find the discrete Fourier transform X[k] = DFTn {x[n]} of the following
periodic sequences x[n] = x[n - N] with period N:

(a) For n = 0 . . .N - 1 we have x[n] =$$\delta$$[n].
(b) For n = 0 . . .N - 1 we have x[n] = $$\mu$$[n] -$$\mu$$[n - K] with K < N.
(c) x[n] = cos( (2*pi*M*n)/N ).

Homework Equations

We don't have a book for my digital processing class and i missed couple of classes so i have no idea how to start these problems. A little hint or a link to a good tutorial/source would be greatly appreciated.

The Attempt at a Solution

 

[Redbelly98]

Wikipedia has good info:

http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Definition

 

[Mene]

The discrete Fourier transform (DFT) is a mathematical operation that converts a discrete signal from its original domain (usually time or space) to a representation in the frequency domain. It is used extensively in digital signal processing and has many applications in various fields such as communications, image processing, and audio processing.

To find the DFT of a periodic sequence x[n] with period N, we can use the formula:

X[k] = \sum_{n=0}^{N-1} x[n] e^{-i2\pi kn/N}

where k is the frequency index and i is the imaginary unit. This formula essentially calculates the contribution of each sample in the time domain to the frequency domain at a particular frequency k.

(a) For the sequence x[n] = \delta[n], we can see that it is a single impulse at n=0 with all other samples being zero. Plugging this into the DFT formula, we get:

X[k] = \sum_{n=0}^{N-1} \delta[n] e^{-i2\pi kn/N}

Since \delta[n] is zero for all n except n=0, the only term that contributes to the sum is when n=0. Therefore, we get:

X[k] = e^{-i2\pi k(0)/N} = 1

This means that the DFT of the sequence x[n] = \delta[n] is a constant value of 1 for all frequency indices k.

(b) For the sequence x[n] = \mu[n] -\mu[n - K], we can see that it is a step function with a step of K at n=0. Plugging this into the DFT formula, we get:

X[k] = \sum_{n=0}^{N-1} (\mu[n] -\mu[n - K]) e^{-i2\pi kn/N}

We can split this into two separate sums:

X[k] = \sum_{n=0}^{K-1} (\mu[n] -\mu[n - K]) e^{-i2\pi kn/N} + \sum_{n=K}^{N-1} (\mu[n] -\mu[n - K]) e^{-i2\pi kn/N}

The first sum is equal to:

\sum_{n=0}^{K-1} (\mu[n] -\mu[n - K