Advanced Signals and Systems - Discrete Fourier Transform

 

15. Discrete Fourier transform.

Task

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.

Task

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

Additional material

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 \(\Omega = \mu \frac{2\pi}{M}\).