Exercise 5.7Z: Application of the IDFT

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

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

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

Correspondingly,  for the  Inverse discrete Fourier transform  $\rm (IDFT)$  practically as an  "inverse function"  of the DFT 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

1

What are the time coefficients  $d(ν)$  for the spectral coefficients  $D(μ)$  according to  $\rm A$?
Enter the first coefficient  $d(1)$  with real and imaginary parts.

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

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

2

What are the time coefficients  $d(ν)$  for the spectral coefficients  $D(μ)$  according to  $\rm B$?
Enter the first coefficient  $d(1)$  with real and imaginary parts.

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

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

3

What are the time coefficients  $d(ν)$  for the spectral coefficients  $D(μ)$  according to  $\rm C$?
Enter the first coefficient  $d(1)$  with real and imaginary parts.

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

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

Solution

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:

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