# 0.1 Discrete fourier transform

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## Discrete fourier transform

The Discrete Fourier Transform, from now on DFT, of a finite length sequence $\left({x}_{0},...,{x}_{K-1}\right)$ is defined as

${\stackrel{}{\mathrm{x̂}}}_{k}=\sum _{n=0}^{K-1}{x}_{n}{e}^{-2\pi jk\frac{n}{K}}\phantom{\rule{2.em}{0ex}}\left(k=0,...,K-1\right)$

To motivate this transform think of ${x}_{n}$ as equally spaced samples of a $T$ -periodic signal $x\left(t\right)$ over a period, e.g., ${x}_{n}=x\left(nT/K\right)$ . Then, using the Riemann Sum as an approximation of an integral, i.e.,

$\sum _{n=0}^{K-1}f\left(\frac{nT}{K}\right)\frac{T}{K}\simeq {\int }_{0}^{T}f\left(t\right)dt$

we find

${\stackrel{ˆ}{x}}_{k}=\sum _{n=0}^{K-1}x\left(\frac{nT}{K}\right){e}^{-2\pi j\frac{nT}{K}k/T}\simeq \frac{K}{T}{\int }_{0}^{T}x\left(t\right){e}^{-2\pi jtk/T}dt=K{X}_{k}$

Note that the approximation is better, the larger the sample size $K$ is.

Remark on why the factor $K$ in [link] : recall that ${X}_{k}$ is an average while ${\stackrel{^}{x}}_{k}$ is a sum. Take for instance $k=0$ : ${X}_{0}$ is the average of the signal while ${\stackrel{^}{x}}_{0}$ is the sum of the samples.

From the above we may hope that a development similar to the Fourier series [link] should also exist in the discrete case. To this end, we note first that the DFT is a linear transform and can berepresented by a matrix multiplication (the “exponent” $T$ means transpose):

${\left({\stackrel{ˆ}{x}}_{0},...,{\stackrel{ˆ}{x}}_{K-1}\right)}^{T}=DF{T}_{K}·{\left({x}_{0},...,{x}_{K-1}\right)}^{T}.$

The matrix ${DFT}_{K}$ possesses $K$ lines and $K$ rows; the entry in line $k$ row $n$ is ${e}^{-2\pi jkn/K}$ . A few examples are

${DFT}_{1}=\left(\begin{array}{c}1\end{array}\right)\phantom{\rule{1.em}{0ex}}{DFT}_{2}=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\phantom{\rule{1.em}{0ex}}{DFT}_{4}=\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& -j& -1& j\\ 1& -1& 1& -1\\ 1& j& -1& -j\end{array}\right)$

The rows are orthogonal The scalar product for complex vectors $x=\left({x}_{1},{x}_{2},...,{x}_{K}\right)$ and $y=\left({y}_{1},{y}_{2},...,{y}_{K}\right)$ is computed as

$x·y={x}_{1}{y}_{1}^{*}+{x}_{2}{y}_{2}^{*}+...+{x}_{K}{y}_{K}^{*},$
where ${a}^{*}$ is the conjugate complex of $a$ . Orthogonal means $x·y=0$ .
to each other. Also, all rows have length Length is computed as $||x||=\sqrt{x·x}=\sqrt{{x}_{1}{x}_{1}^{*}+{x}_{2}{x}_{2}^{*}+...+{x}_{K}{x}_{K}^{*}}=\sqrt{|{x}_{1}{|}^{2}+|{x}_{2}{|}^{2}+...+{|{x}_{K}|}^{2}}$ . $\sqrt{K}$ . Finally, the matrices are symmetric (exchanging lines for rows does not change the matrix). So, the multiplying DFT with its conjugate complex matrix ${\left(DF{T}_{K}\right)}^{*}$ we get $K$ times the unit matrix (diagonal matrix with all diagonal elements equal to $K$ ).

Inverse DFT

From all this we conclude that the inverse matrix of $DF{T}_{K}$ is $IDF{T}_{K}=\left(1/K\right)·{\left(DF{T}_{K}\right)}^{*}$ . Since ${\left({e}^{-\alpha }\right)}^{*}={e}^{\alpha }$ we find

${x}_{n}=\frac{1}{K}\sum _{k=0}^{K-1}{\stackrel{ˆ}{x}}_{k}{e}^{2\pi jk\frac{n}{K}}\phantom{\rule{2.em}{0ex}}\left(n=0,...,K-1\right)$

Spectral interpretation, symmetries, periodicity

Combining [link] and [link] we may now interpret ${\stackrel{ˆ}{x}}_{k}$ as the coefficient of the complex harmonic with frequency $k/T$ in a decomposition of the discrete signal ${x}_{n}$ ; its absolute value provides the amplitude of the harmonic and its argument the phase difference.

If $x$ is even, ${\stackrel{ˆ}{x}}_{k}$ is real for all $k$ and all harmonics are in phase.

Using the periodicity of ${e}^{2\pi jt}$ we obtain ${x}_{n}={x}_{n+K}$ when evaluating [link] for arbitrary $n$ . Short, we can consider ${x}_{n}$ as equally-spaced samples of the $T$ -periodic signal $x\left(t\right)$ over any interval of length $T$ :

${\stackrel{ˆ}{x}}_{k}=\sum _{n=-K/2}^{K/2-1}{x}_{n}{e}^{-2\pi jk\frac{n}{K}}.$

Similarly, ${\stackrel{ˆ}{x}}_{k}$ is periodic: ${\stackrel{ˆ}{x}}_{k}={\stackrel{ˆ}{x}}_{k+K}$ . Thus, it makes sense to evaluate ${\stackrel{ˆ}{x}}_{k}$ for any $k$ . For instance, we can rewrite [link] as

${x}_{n}=\frac{1}{K}\sum _{n=-K/2}^{K/2-1}{\stackrel{ˆ}{x}}_{k}{e}^{2\pi jk\frac{n}{K}}$

Since ${\stackrel{ˆ}{x}}_{k}$ corresponds to the frequency $k/T$ , the period $K$ of ${\stackrel{ˆ}{x}}_{k}$ corresponds to a period of $K/T$ in actual frequency. This is exactly the sampling frequency (or sampling rate) of the original signal( $K$ samples per $T$ time units). Compare to the spectral repetitions.

However, the period $T$ of the original signal $x$ is nowhere present in the formulas of the DFT (cpre. [link] and [link] ). Thus, if nothing is known about $T$ , it is assumed that the sampling rate is 1 (1 sample per time unit), meaning that $K=T$ .

FFT

The Fast Fourier Transform (FFT) is a clever algorithm which implements the DFT in only $Klog\left(K\right)$ operations. Note that the matrix multiplication would require ${K}^{2}$ operations.

Matlab implements the FFT with the command fft(x) where $x$ is the input vector. Note that in Matlab the indices start always with 1! This means that the first entry ofthe Matlab vector $x$ , i.e. $x\left(1\right)$ is the sample point ${x}_{0}=x\left(0\right)$ . Similar, the last entry of the Matlab vector $x$ is, i.e. $x\left(K\right)$ is the sample point ${x}_{K-1}=x\left(\left(K-1\right)T/K\right)=x\left(T-T/K\right)$ .

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