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Real forward and inverse FFT

Jens Hee

https://jenshee.dk

March 2014

Change log

14. March 2014

1. Document started.

13. March 2020

1. Meta data added.

1 The Discrete Fourier Transform 1

1.1 Denition of the Discrete Fourier transform . . . . . . . . . . . . . . . . . . . . . 1

2 The Discrete Fourier Transform of real valued signals 2

2.1 Some denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.2 Forward DFT of a real values signal . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.3 Real DFT step by step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.4 Inverse DFT given a real values signal . . . . . . . . . . . . . . . . . . . . . . . . 3

2.5 Real IDFT step by step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Program examples 5

Chapter 1 The Discrete Fourier Transform

1.1 Denition of the Discrete Fourier transform

The denition of the Discrete Fourier transform (DFT) used in the following is given by:

X(k) =DFTNfx(n)g=N1X

n=0x(n)ej2N kn; k= 0;:::;N1 The Inverse Discrete Fourier transform (IDFT) is dened by: x(n) =IDFTNfX(k)g=1N N1X k=0X(k)ej2N nk

IDFT can be calculated from the DFT as follows:

x(n) =IDFTNfX(k)g=1N

DFTNfX(k)g

Chapter 2 The Discrete Fourier Transform of real

valued signals

2.1 Some denitions

In the next sections the following denitions are used: x

1(n) =x(2n)

2(n) =x(2n+ 1)

1(k) =DFTN=2fx1(n)g

2(k) =DFTN=2fx2(n)g

y(n) =x1(n) +jx2(n)

Y(k) =DFTN=2fy(n)g

2.2 Forward DFT of a real values signal

Given a real valued signalx(n),X(k) can be calculated as:

X(k) =N1X

n=0x(n)ej2N kn N=21X n=0x(2n)ej2N k2n+N=21X n=0x(2n+ 1)ej2N k(2n+1) N=21X n=0x

1(n)ej2N=2kn+N=21X

n=0x

2(n)ej2N=2knej2N

k =X1(k) +X2(k)ej2N k

Sincex1andx2are real valued:

Y(k) =X1+jX2

Y re=X1r Y ro=X2i Y ie=X2r Y io=X1i where: Y reis the even part of the real part ofY Y rois the odd part of the real part ofY 2 Y ieis the even part of the imaginary part ofY Y iois the odd part of the imaginary part ofY

X(k) is nally given by:

X(k) =X1r(k) +jX1i(k) + (X2r(k) +jX2i)ej2N

k =Yre(k) +jYio(k) + (Yie(k)jYro(k))ej2N k = 0:5(Yr(k) +Yr(N=2k) +j(Yi(k)Yi(N=2k)) +(Yi(k) +Yi(N=2k)j(Yr(k)Yr(N=2k)))ej2N k) = 0:5((Yr(k) +jYi(k))(1jej2N k) + (Yr(N=2k)jYi(k))(1 +jej2N k)) =Y(k)A(k) +Y(N=2k)B(k) where:

A(k) = 0:5(1jej2N

B(k) = 0:5(1 +jej2N

k) Note:

X(0) =Y(0)(1j) +Y(N=2)(1 +j) =Yr(0) +Yi(0)

X(N=2) =Y(N=2)(1 +j) +Y(0)(1j) =Yr(0)Yi(0)

2.3 Real DFT step by step

1. Form the complex signaly(n). This is straight forward since the complex input to an FFT is

often ordered as pairs of real and imaginary values, so no reordering of the input is necessary.

2. Do an N/2-point DFT using an N/2-point FFT.

3. Calculate the two arraysA(k) andB(k) for 0k < N=2.

4. Combine the spectral valuesY(k) withA(k) andB(k) to formX(k) for 0k < N=2.

UsualyX(k) is zero at half the samplig frequency andX(N=2) need not be calculated.

2.4 Inverse DFT given a real values signal

Given a spectrumX(k) of a real valued signalx(n),y(n) can be calculated as: y(n) =x(2n) +jx(2n+ 1) 1N N1X k=0X(k)ej2N

2nk+j1N

N1X k=0X(k)ej2N (2n+1)k 1N N1X k=0X(k)(1 +jej2N k)ej2N=2nk 1N N=21X k=0X(k)(1 +jej2N k)ej2N=2nk+1N N=21X k=0X(k+N=2)(1 +jej2N (k+N=2))ej2N=2n(k+N=2) 1N N=21X k=0X(k)(1 +jej2N k)ej2N=2nk+1N N=21X k=0X(k+N=2)(1jej2N k)ej2N=2nk 3

SinceX(k) =X(Nk) we have:

y(n) =1N=2N=21X k=00:5X(k)(1 +jej2N k)ej2N=2nk+1N=2N=1X k=00:5X(N=2k)(1jej2N k)ej2N=2nk

1N=2N=21X

k=0(X(k)A(k) +X(N=2k)B(k))ej2N=2nk

Y(k) =X(k)A(k) +X(N=2k)B(k)

whereAandBare dended above. Note:

Y(0) = 0:5(1 +j)X(0) + 0:5(1j)X(N=2)

2.5 Real IDFT step by step

1. Calculate the two arraysA(k) andB(k) for 0k < N=2.

2. Combine the spectral valuesX(k) withA(k) andB(k) to formY(k) for 0k < N=2.

UsualyX(k) is zero at half the samplig frequency and the last term inY(0) need not be taken into account.

3. Do an N/2-point IDFT ofY(k) using an N/2-point IFFT.

4. If the complex input and output to the FFT is ordered as pairs of real and imaginary values,

then no reordering of the output is necessary. 4

Chapter 3 Program examples

#include "math.h" static double pi = 4.0 * atan(1.0); void FFT(int M, double* X) const int N = (int)floor(pow(2.0, M) + 0.5); double wcos; double wsin; double ucos; double usin; int N2 = N >> 1; int L1 = N; int L2; for (int L = 0; L < M; L++)

L2 = L1 >> 1;

wcos = 1; wsin = 0; ucos = cos(pi / L2); usin = -sin(pi / L2); for (int J = 0; J < L2; J++) for (int I = J; I < N; I = I + L1) int Ir = 2 * I; int Ii = Ir + 1; int L22 = 2 * L2; double sumRe = X[Ir] + X[Ir + L22]; double sumIm = X[Ii] + X[Ii + L22]; double diffRe = X[Ir] - X[Ir + L22]; double diffIm = X[Ii] - X[Ii + L22];

X[Ir + L22] = diffRe * wcos - diffIm * wsin;

X[Ii + L22] = diffRe * wsin + diffIm * wcos;

X[Ir] = sumRe;

X[Ii] = sumIm;

double w = wcos * ucos - wsin * usin; wsin = wcos * usin + wsin * ucos; wcos = w;

L1 = L1 >> 1;

int K; int J = 0; int N1 = N - 1; 5 for (int I = 0; I < N1; I++) if (I < J) int Jr = 2 * J; int Ji = Jr + 1; int Ir = 2 * I; int Ii = Ir + 1; double TRe = X[Jr]; double TIm = X[Ji];

X[Jr] = X[Ir];

X[Ji] = X[Ii];

quotesdbs_dbs20.pdfusesText_26

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