Question 2

Find $F_r = {\em DFT}$ $\{f_k=$[1 0 0 1]$\}$ and plot the magnitude and phase of F_r. Find the ${\em IDFT}$ of F_r to verify it is equal to [1 0 0 1] as expected.

Solution.

The DFT of f_k is

$$\sum_{ k= 0}^{ N_0 - 1 }f_k (e^{j2\pi r/N_0})^{-k}$$ F_r = (17)

$$1\cdot e^0 + 0 + 0 + 1 \cdot (e^{j2\pi r/N_0})^{-3}$$ = (18)

$$1 + cos(3\cdot 2 \pi r/N_0) - jsin(3\cdot 2 \pi r/N_0)$$ = (19)

The magnitude and phase responses are

$$\sqrt{(1+cos(3\cdot 2 \pi r/N_0))^2 + sin^2(3\cdot 2 \pi r/N_0)}$$ |F_r| = (20)

$$arctan[\frac{-sin(3\cdot 2 \pi r/N_0)}{1+cos(3\cdot 2 \pi r/N_0)}]$$ arg(F_r) = (21)

  

Figure 1: Frequency responses of Fr

The IDFT of F_r is

$$\frac{1}{N_0}\sum_{ r= 0}^{ N_0 - 1 }F_r (e^{j2\pi r/N_0})^{k}$$ f_k = (22)

$$\frac{1}{N_0}\sum_{ r= 0}^{ N_0 - 1 }[e^{-j2\pi r/N_0 \cdot 0} + e^{-j2\pi r/N_0 \cdot 3}]e^{j2\pi rk/N_0}$$ = (23)

Using

$$\frac{1}{N_0}\sum_{ r= 0}^{ N_0 - 1 }e^{j2\pi rk/N_0}e^{-j2\pi rn/N_0} \delta_{kn}$$ = (24)

where

$$\delta_{kn} = \left \{ \begin{array}{ll} 1 & \mbox{for $k=n$ }\\ 0 & \mbox{otherwise} \end {array} \right.$$

f_k is

$$\delta_{k\cdot0} + \delta_{k\cdot3}$$ f_k = (25)

$$\delta[k] + \delta[k-3]$$ = (26)

or f_k = [1 0 0 1]