Exercise 5.7Z: Application of the IDFT

Three sets $\rm A$, $\rm B$ and $\rm C$
for the spectral coefficients

In the Discrete Fourier Transform $\rm (DFT)$, the time sample values $d(ν)$ with the control variable $ν = 0$, ... , $N – 1$ the discrete spectral coefficients $D(μ)$ with $μ = 0$, ... , $N – 1$ is calculated as follows:

$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1} d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

Here, $w$ abbreviates the complex rotation factor, which is defined as follows:

$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N} = \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right) \hspace{0.05cm}.$$

Correspondingly, for the Inverse Discrete Fourier Transform $\rm (IDFT)$ practically as an "inverse function" of the DFT the following applies:

$$d(\nu) = \sum_{\mu = 0 }^{N-1} D(\mu) \cdot {w}^{-\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

In this exercise, the time coefficients $d(ν)$ are to be determined for different complex-valued example sequences $D(μ)$ – which are labeled $\rm A$, $\rm B$ and $\rm C$ in the table. Thus, $N = 8$ is always valid.

Notes:

The exercise belongs to the chapter Implementation of OFDM Systems.
Reference is also made to the chapter Discrete Fourier Transform in the book "Signal Representation".
You can check your results with the interactive applet Discrete Fourier Transform and Inverse.

Questions

${\rm Re}\big[d(1)\big] \ = \ $

${\rm Im}\big[d(1)\big] \ = \ $

${\rm Re}\big[d(1)\big] \ = \ $

${\rm Im}\big[d(1)\big] \ = \ $

${\rm Re}\big[d(1)\big] \ = \ $

${\rm Im}\big[d(1)\big] \ = \ $

Solution

(1) Because of $D(μ) = 0$ for $μ ≠ 0$ all time coefficients $d(ν) = D(0)= 1 - {\rm j}$. Thus also holds:

$${\rm Re}[d(1)] \hspace{0.15cm}\underline {=+ 1}, \hspace{0.3cm}{\rm Im}[d(1)] \hspace{0.15cm}\underline {= -1}.$$

(2) Here, all spectral coefficients are zero except $D_1 = 1 - {\rm j}$ and $D_7 = 1 + {\rm j}$. It follows for all time coefficients $(0 ≤ ν ≤ 7)$:

$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {7\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}.$$

However, due to periodicity, also holds:

$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ +{\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}= \left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} + {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right]+ {\rm{j}} \cdot\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} - {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right].$$

Using Euler's theorem, this expression can be transformed as follows:

$$d(\nu) = 2 \cdot \cos \left( {\pi}/{4}\cdot \nu \right)+ 2 \cdot \sin \left( {\pi}/{4}\cdot \nu \right).$$

This time function $d(ν)$ is purely real and characterizes a harmonic oscillation with amplitude $ 2 \cdot \sqrt{2}$ and phase $φ = 45^\circ$.
The time coefficient with index $ν = 1$ indicates the maximum:

$$ {\rm Re}[d(1)] = 2 \cdot \frac {\sqrt{2}}{2}+ 2 \cdot \frac {\sqrt{2}}{2} = 2 \cdot {\sqrt{2}} \hspace{0.15cm}\underline {\approx 2.828}, \hspace{0.5cm}{\rm Im}[d(1)] \hspace{0.15cm}\underline {= 0}.$$

(3) According to the general equation:

$$d(1) = \sum\limits_{\mu = 0}^{7} D(\mu)\cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\mu} = \left[ D(1) + D(7) \right]\cdot \cos \left( {\pi}/{4} \right) + \left[ D(3) + D(5) \right]\cdot \cos \left( {3\pi}/{4} \right)+ {\rm j} \cdot \left[ D(2) - D(6) \right]\cdot \sin \left( {\pi}/{2} \right) + D(4) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}}.$$

The first three terms give pure real results:

$${\rm Re}[d(1)] = (1+1) \cdot \frac{1}{\sqrt{2}}-(3+3) \cdot \frac{1}{\sqrt{2}}+ {\rm j} \cdot4{\rm j} \cdot 1 = -\frac{4}{\sqrt{2}}-4\hspace{0.15cm}\underline { \approx -6.829}.$$

For the imaginary part, we obtain:

$${\rm Im}[d(1)] = {\rm Im}\left[4 \cdot{\rm j} \cdot (-1) \right] \hspace{0.15cm}\underline {= -4}.$$