# Advanced Signals and Systems - Discrete Fourier Transform

### 15. Discrete Fourier transform.

Determine the DFT of the following sequences $$v(n)$$ with length $$M$$: \begin{align} \text{(a)} \;\;\;\;\;\;\;\; &v(n) = \gamma_0 (n-\kappa) & \kappa \in \{ 0,1,... M-1\} \nonumber \\ \text{(b)} \;\;\;\;\;\;\;\; &v(n) = \cos(\Omega_0n) & \Omega_0 = \frac{2\pi}{M}\mu_0, \, \mu_o \in \{ 0,1,... M-1\} \nonumber \end{align}

## Amount and difficulty

• Working time: approx. xx minutes
• Difficulty: xx

## Solution

\begin{align*} \text{(a)} \ \ V_M(\mu) &=&& \text{DFT} \left\{ v_1(n) \right\} = \sum \limits_{n=0}^{M-1} \gamma_0 (n-\kappa) e^{-j\mu \frac{ \pi}{M}n} = e^{-j\mu \frac{2\pi}{M}\kappa} \ \ \rightarrow |V_M(\mu)| = 1 \\ \text{(b)} \ \ V_M(\mu) &=&& \text{DFT} \left\{ \cos \left( \frac{2\pi}{M} \mu_0 n \right) \right\} = \cdots = \\ &&&\frac{M}{2} \left[ \sum \limits_{\lambda= -\infty}^\infty \gamma_0 \left(\mu -( \lambda M + \mu_0)\right) + \sum \limits_{ \lambda = -\infty}^\infty \gamma _0\left(\mu -(\lambda M -\mu_0)\right) \right] \end{align*}

### 16. Relation between DFT and Fourier transform.

The Fourier transform $$R(e^{j\Omega})$$ of the following sequence \begin{equation} r(n) = \begin{cases} 1 & |n|\leq N \\ 0 & \text{else} \nonumber \end{cases}\text{ ,} \end{equation} is given by \begin{equation} R(e^{j\Omega}) = \frac{\sin \left(\frac{\Omega}{2}(2N+1) \right)}{\sin \left(\frac{\Omega}{2} \right)} \nonumber \text{ .} \end{equation} Determine the DFT $$R_M(\mu)$$ of the given sequence $$r(n)$$. How are these two transformations linked with each other?

## Amount and difficulty

• Working time: approx. xx minutes
• Difficulty: xx  ## Solution

Applying the known definition of the DFT for deriving $$R_M(\mu)$$ we find:

\begin{align*} R_M(\mu) &= \sum \limits_{n= 0}^{M-1} r(n-N) \cdot e^{-j\frac{2\pi}{M} \mu n} = \sum \limits_{n= 0}^{2N} e^{-j\frac{2\pi}{M} \mu n} \end{align*}

The shift of $$N$$ has been carried out, in order to enable the calculation of the DFT. It can easily be reversed by using the shift theorem. Two cases should be considered:

1. $$\mu \neq 0$$ \begin{align*} R_M(\mu) &= e^{-j\frac{2\pi}{M} \mu N} \cdot \frac{\sin \left( \mu \frac{2\pi}{M} \frac{1}{2} (2N+1) \right) }{\sin \left( \mu\frac{2\pi}{M} \frac{1}{2} \right) } \end{align*}
2. $$\mu = 0$$ \begin{align*} R_M(\mu) &= \cdots = 2N+1 \end{align*}

Comparing this result to the result of the Fourier transform it is obviously that DFT is given by sampling the Fourier transform $$R(e^{j\Omega})$$ at $$\Omega = \mu \frac{2\pi}{M}$$.