[Fourier transform and its application study notes] 21. The matrix definition of discrete Fourier transform, some properties

DFT at 0 points

$\underline{\mathcal{f}}\underline{f} (0) = \displaystyle{\sum_{n=0}^{n-1}\underline{f}[n]e^{-2\pi I\frac{n0}{N}} = \sum_{n=0}^{n-1}\underline{f}[n]}$

Remember, the Fourier transform has a similar formula at 0.

$\mathcal{f}f (0) = \displaystyle{\int_{-\infty}^{\infty}f (t) e^{-2\pi I0t}dt = \int_{-\infty}^{\infty}f (t) DT}$

In this view, the discrete Fourier transform is very similar to the Fourier transform.

Typical discrete signals and their DFT1. A typical discrete signal

The sampled values of discrete signal $\underline{1}$ at each sample point are $1$

$\underline{1} = (,..., 1) $

Discrete signal $\underline{\delta_0}$ only at 0 o ' clock is the $1$ pulse.

$\UNDERLINE{\DELTA_0} = (1,0,0,..., 0) $

Discrete signal $\underline{\delta_k}$ only at section $k$, is $1$ pulse.

$\underline{\delta_k} = (\underbrace{0,0,..., 0,1}_{k\ entries},0,..., 0) $

2. DFT for typical discrete signals

DFT for the $\underline{\delta_0}$

$\UNDERLINE{\MATHCAL{F}}\UNDERLINE{\DELTA_0} = \displaystyle{\sum_{n=0}^{n-1}\underline{\delta_0}[n] \underline{\ Omega}^{-n} = \underline{\delta_0}[0]\underline{\omega}^0 = 1\cdot \underline{\omega}^0 = (,..., 1) = \underline{1}}$

So

$\underline{\mathcal{f}}\underline{\delta_0}=\underline{1}$

DFT for the $\underline{\delta_k}$

$\underline{\mathcal{f}}\underline{\delta_k} = \displaystyle{\sum_{n=0}^{n-1}\underline{\delta_k}[n] \underline{\ Omega}^{-n} = \underline{\delta_k}[k]\underline{\omega}^{-k} = 1\cdot \underline{\omega}^{-k} = \underline{\omega}^{- k}}$

So

$\underline{\mathcal{f}}\underline{\delta_k} = \underline{\omega}^{-k}$

DFT for the $\underline{\omega}^{k}$

$\begin{align*}
\UNDERLINE{\MATHCAL{F}}\UNDERLINE{\OMEGA}^{K}[M]
&=\sum_{n=0}^{n-1}\underline{\omega}^k[n]\underline{\omega}^{-n}[m]\\
&=\sum_{n=0}^{n-1}e^{2\pi I\frac{kn}{n}}\cdot E^{-2\pi i\frac{mn}{n}}\\
&=\underline{\omega}^k\cdot\underline{\omega}^m\\
&=\begin{cases}
0 & \text{,} k\neq m \ \
N & \text{,} k= m
\end{cases}
\end{align*}$

So

$\underline{\mathcal{f}}\underline{\omega}^k = (\underbrace{0,0,..., 0,n}_{k\ entries},0,..., 0) = N\underline{\delta _k}$

DFT Matrix 1. Definition of DFT Matrix

Make $\omega = E^{2\pi i\frac{1}{n}}$, note that the $\omega$ here are not underlined and are not discrete signals. Then there are

$\OMEGA^{-MN} = E^{-2\pi i\frac{mn}{n}}$

The transformation of the DfT can be written

$\UNDERLINE{\MATHCAL{F}}\UNDERLINE{F}[M] = \displaystyle{\sum_{n=0}^{n-1}\omega^{-mn} \underline{f}[n]}$

In this way, a new expression is presented, that is, the DFT for discrete signal $\underline{f}$ is equivalent to multiplying by $\underline{f}$ with the Matrix $\left (\omega^{-} \right) ^{mn}$, $mn $ for the $m of the Matrix $ row $n$ Column

$\begin{pmatrix}
1 &1 &1 &, ... &1 \
1 &\omega^{-1} &\omega^{-2} &, .... &\omega^{-(N-1)} \ \
1 &\omega^{-2} &\omega^{-4} &, .... &\omega^{-2 (N-1)} \ \
\vdots &\vdots &\vdots & ... & \vdots\\
1 &\omega^{-(N-1)} &\omega^{-2 (N-1)} &.. &\omega^{-(N-1) ^2}
\end{pmatrix}
\begin{pmatrix}
\underline{f}[0]\\
\underline{f}[1]\\
\underline{f}[2]\\
\vdots\\
\UNDERLINE{F}[N-1]
\end{pmatrix}
=
\begin{pmatrix}
\underline{\mathcal{f}}\underline{f}[0]\\
\underline{\mathcal{f}}\underline{f}[1]\\
\underline{\mathcal{f}}\underline{f}[2]\\
\vdots\\
\UNDERLINE{\MATHCAL{F}}\UNDERLINE{F}[N-1]
\end{pmatrix}$

The matrix to the left of the above equation is the DfT matrix, which is a matrix of $n\times n$ that can be expressed

$ (\mathcal{f}) _{nm} = \omega^{-nm}$

2. Conjugate transpose of the DFT matrix

Conjugate transpose of the DFT matrix $\mathcal{f}$, i.e.

$ (\mathcal{f}^{*}) _{nm} = \overline{(\underline{\mathcal{f}}) _{mn}} = \overline{\omega^{-mn}} = \omega^{mn}$

The DfT matrix is multiplied by its conjugate transpose matrix, which has

$\underline{\mathcal{f}}\underline{\mathcal{f}}^{*} =
\begin{pmatrix}
1 &1 &1 &, ... &1 \
1 &\omega^{-1} &\omega^{-2} &, .... &\omega^{-(N-1)} \ \
1 &\omega^{-2} &\omega^{-4} &, .... &\omega^{-2 (N-1)} \ \
\vdots &\vdots &\vdots & ... & \vdots\\
1 &\omega^{-(N-1)} &\omega^{-2 (N-1)} &.. &\omega^{-(N-1) ^2}
\end{pmatrix}
\begin{pmatrix}
1 &1 &1 &, ... &1 \
1 &\omega^{1} &\omega^{2} &, .... &\omega^{(N-1)} \ \
1 &\omega^{2} &\omega^{4} &, .... &\omega^{2 (N-1)} \ \
\vdots &\vdots &\vdots & ... & \vdots\\
1 &\omega^{(N-1)} &\omega^{2 (N-1)} &.. &\omega^{(N-1) ^2}
\end{pmatrix}$

$=
\begin{pmatrix}
\underline{\omega}^0\cdot\underline{\omega}^0 &\underline{\omega}^1\cdot\underline{\omega}^0 &\underline {\omega}^2\cdot\underline{\omega}^0 & &\underline{\omega}^{n-1}\cdot\underline{\omega}^0 \ \
\underline{\omega}^0\cdot\underline{\omega}^1 &\underline{\omega}^1\cdot\underline{\omega}^1 &\underline {\omega}^2\cdot\underline{\omega}^1 & &\underline{\omega}^{n-1}\cdot\underline{\omega}^1 \ \
\underline{\omega}^0\cdot\underline{\omega}^2 &\underline{\omega}^1\cdot\underline{\omega}^2 &\underline {\omega}^2\cdot\underline{\omega}^2 & &\underline{\omega}^{n-1}\cdot\underline{\omega}^2 \ \
\vdots &\vdots &\vdots &amp ... & \vdots\\
\underline{\omega}^0\cdot\underline{\omega}^{n-1} &\underline{\omega}^1\cdot\underline{\omega}^{n-1} &\ Underline{\omega}^2\cdot\underline{\omega}^{n-1} &.. &\underline{\omega}^{n-1}\cdot\underline{\omega}^{ N-1} \ \
\end{pmatrix}\qquad\qquad\qquad\qquad$

$=\begin{pmatrix}
N &0 &0 &, ... &0 \
0 &n &0 &, ... &0 \
0 &0 &n &, ... &0 \
\vdots &\vdots &\vdots & ... &\vdots \
0 &0 &0 &, ... &n
\end{pmatrix} \qquad \left (
\underline{\omega}^l\cdot\underline{\omega}^k = \begin{cases}
0 & \text{,} k\neq l \ \
N & \text{,} k= L
\end{cases}
\right) \qquad\qquad\qquad\qquad$

$= NI \qquad (i\ is\ n\times n\ identity\ matrix) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$

So

$\underline{\mathcal{f}}\underline{\mathcal{f}}^{*} = ni$

The same can be done

$\UNDERLINE{\MATHCAL{F}}^{*}\UNDERLINE{\MATHCAL{F}} = ni$

3. Idft Matrix

By the result of the result of the operation of the transpose matrix above, the IDFT matrix can be obtained

$\underline{\mathcal{f}}^{-1} = \frac{1}{n}\underline{\mathcal{f}}^{*}$

That

$\left (\underline{\mathcal{f}}^{-1} \right) _{mn} = \frac{1}{n}\omega^{mn}$

The result can also be derived from the process of deriving a DFT matrix.

4. DFT Matrix Operation Complexity

Operation Complexity here is the number of multiplication operations in the value matrix operation, each of the $\underline{\mathcal{f}}\underline{f}$ operations will cost $n$ multiplication, in which there are $n$ entries, that is, the operation complexity is $n\times n$, with $o (N ^2) $ representation.

The purpose of FFT algorithm is to reduce the computational complexity of DFT, and the complexity of FFT is $o (NLOGN) $.

DFT characteristics

We left this last lesson on the discussion of DFT characteristics: Both the input and the output are periodic discrete functions with a period of $n$. This conclusion is due to the periodicity of discrete complex indices.

1. Proof of the periodicity of input and output discrete signals

The certification process is as follows:

$\underline{\omega}[m] = E^{2\pi I \frac{m}{n}}$

$\underline{\omega}[m+n] = E^{2\pi i\frac{m+n}{n}} = E^{2\pi I\frac{m}{n}}\cdot e^{2\pi i}=e^{2\pi I\frac{m}{N}} = \under line{\omega}[m]$

Similarly

$\UNDERLINE{\OMEGA}^{N}[M] = \underline{\omega}^n[m+n] \qquad \underline{\omega}^{-n}[m] = \underline{\omega}^{-n}[m +n]$

That is, $\underline{\omega},\underline{\omega}^n,\underline{\omega}^{-n}$ are discrete complex exponential signals with periodic $n$.

The above conclusions are extended to the DFT

$\underline{\mathcal{f}}\underline{f}[m+n] = \displaystyle{\sum_{n=0}^{n-1}\underline{f}[n]\underline{\omega}^{- n}[m+n]= \sum_{n=0}^{n-1}\underline{f}[n]\underline{\omega}^{-n}[m]} = \underline{\mathcal{f}}\underline{f}[m]$

therefore $\underline{\mathcal{f}}\underline{f}[m+n] = \underline{\mathcal{f}}\underline{f}[m]$, indicating the discrete signal $\underline{f}$ A discrete signal with a period of $n$ is obtained after DFT.

In the same vein, extended to IDFT

$\underline{\mathcal{f}}^{-1}\underline{f}[m+n] = \frac{1}{n}\displaystyle{\sum_{n=0}^{n-1}\underline{f}[n]\ underline{\omega}^{n}[m+n]= \frac{1}{n}\sum_{n=0}^{n-1}\underline{f}[n]\underline{\omega}^{-n}[m]} = \underline{\ mathcal{f}}^{-1}\underline{f}[m]$

therefore $\underline{\mathcal{f}}^{-1}\underline{f}[m+n] = \underline{\mathcal{f}}^{-1}\underline{f}[m]$, indicating the discrete signal $\ underline{f}$ the IDFT will get discrete signals with a period of $n$.

The characteristics of this $n$ are determined by the periodicity of $\underline{\omega}$. For a discrete signal $\underline{f}$, a discrete signal $\underline{\mathcal{f}}\underline{f}$ with a period of $n$ is obtained, and then the $\underline{\mathcal{f}}\ underline{f}$ idft, or $\underline{\mathcal{f}}^{-1}\underline{\mathcal{f}}\underline{f}$, will also get discrete signals with $n$ cycles $\ underline{f}$, this shows that the original signal $\underline{f}$ is also a discrete signal with a period of $n$.

Therefore, the conclusion is as follows:

• The input discrete signal $\underline{f}$ and the output discrete signal $\underline{\mathcal{f}}\underline{f}$ of the DfT are periodic signals with a period of $n$ ($N $); idft input discrete signal $\ underline{f}$ and output discrete signal $\underline{\mathcal{f}}^{-1}\underline{f}$ are periodic signals with a period of $n$ ($N $).

In the practical application, the discrete signal we collect usually is not the signal that the period is $n$, even does not have the periodicity, but in the DfT analysis, all is taken from the acquisition signal to take part, this part will be regarded as a period in the periodic signal, it will produce the periodic signal after the DFT.

Independence of the Index

Periodicity can be said to be a limitation, because we need to use periodic signals to see the discrete signals that will be DFT, IDFT, but this feature also brings a simple and beneficial conclusion: the independence of the index (independence of indexing).

Since both $\underline{f}$ and $\underline{\omega}$ are cyclical, any continuous $n$ sampling can be used when calculating a DFT, i.e.

$\underline{\mathcal{f}}\underline{f} = \displaystyle{\sum_{n=0}^{n-1}\underline{f}[n]\underline{\omega}^{-n} = \ Sum_{n=1}^{n}\underline{f}[n]\underline{\omega}^{-n}}$

We can also place 0 points in the middle of the sampled values, i.e.

When the $n$ is odd

$\underline{\mathcal{f}}\underline{f} = \displaystyle{\sum_{n=-\frac{n-1}{2}}^{\frac{n-1}{2}}\underline{f}[n]\ Underline{\omega}^{-n}}$

When $n$ is an even number

$\underline{\mathcal{f}}\underline{f} = \displaystyle{\sum_{n=-\frac{n}{2}+1}^{\frac{n}{2}}\underline{f}[n]\ Underline{\omega}^{-n}}$

Different software implementations may have different expressions for subscripts (indexes), so we need to understand the independence of subscripts (indexes) through this section.

The inversion of discrete signal and the duality of DFT 1. Definition of discrete signal inversion

When discussing the reversal of a continuous signal, the continuous signal $f (x) $ is reversed to $f^-(x) = f (-X) $, where the reversal of the discrete signal is defined as $\underline{f}^{-}[m] = \underline{f}[-m]$, that is, if there is a discrete signal

$\underline{f} = \left (\underline{f}[0],\underline{f}[1],\underline{f}[2],..., \underline{f}[n-1] \right) $

It's counter-switched to

$\underline{f}^-= \left (\underline{f}[0],\underline{f}[-1],\underline{f}[-2],..., \underline{f}[-(N-1)] \right) $

Inverse conversion of discrete complex exponential $\underline{\omega}$

$\underline{\omega}^-= \left (1,e^{-2\pi i\frac{1}{n}},e^{-2\pi I\frac{2}{n}},..., e^{-2\pi I\frac{N-1}{N}} \right) = \ underline{\omega}^{-1}$

Similarly, there

$\left (\underline{\omega}^n \right) ^-= \underline{\omega}^{-n} \qquad \left (\underline{\omega}^{-n} \right) ^-= \ underline{\omega}^{n}$

2. The duality of DFT discussion

DFT for the inversion $\underline{f}^-$ of discrete signal $\underline{f}$

$\begin{align*}
\underline{\mathcal{f}}\underline{f}^-
&=\sum_{n=0}^{n-1}\underline{f}^-[n]\underline{\omega}^{-n}\\
&=\sum_{n=0}^{n-1}\underline{f}[-n]\underline{\omega}^{-n} \qquad (definition\ of\ reversed\ signal) \ \
&=\sum_{n=0}^{n-1}\underline{f}[n-n]\underline{\omega}^{-n} \qquad (\underline{f}\ is\ period\ of\ n) \ \
&=\sum_{l=n}^1\underline{f}[l]\underline{\omega}^{l-n} \qquad (letting\ l=n-n) \ \
&=\sum_{l=n}^1\underline{f}[l]\underline{\omega}^l \qquad (\underline{\omega}\ is\ period\ of\ N) \ \
&=\sum_{l=0}^{n-1}\underline{f}[l]\underline{\omega}^l \qquad (independence\ of\ indexing) \ \
&=\sum_{l=0}^{n-1}\underline{f}[l] (\underline{\omega}^{-l}) ^-\\
&=\left (\sum_{l=0}^{n-1}\underline{f}[l]\underline{\omega}^{-l} \right) ^-\qquad \left (\sum_{l=0}^{N-1}\ Underline{f}[l] (\underline{\omega}^{-l}) ^-[m] =\sum_{l=0}^{n-1}\underline{f}[l] (\underline{\omega}^{-l}) ^[-m]\ The\ definition\ of\ reversed\ signal\right) \ \
&=\left (\underline{\mathcal{f}}\underline{f} \right) ^-
\end{align*}$

So

$\underline{\mathcal{f}}\underline{f}^-= \left (\underline{\mathcal{f}}\underline{f} \right) ^-$

Two consecutive DFT for discrete signals

$\begin{align*}\underline{\mathcal{f}}\underline{\mathcal{f}}\underline{f}&=\sum_{k=0}^{n-1}\left (\sum_{n= 0}^{n-1}\underline{f}[n]\underline{\omega}^{-n}[k]\right) \underline{\omega}^{-k}\\
&=\sum_{n=0}^{n-1}\underline{f}[n]\left (\sum_{k=0}^{n-1}\underline{\omega}^{-n}[k]\omega^{-k}\right) \ \
&=\sum_{n=0}^{n-1}\underline{f}[n]\left (\underline{\mathcal{f}}\underline{\omega}^{-n}\right) \ \
&=\sum_{n=0}^{n-1}\underline{f}[n]\cdot n\underline{\delta_{-n}}\\
&=n\sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_{-n}}\\
&=n\sum_{n=0}^{n-1}\underline{f}[n]\left (\underline{\delta_n}\right) ^-\\
&=n\left (\sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}\right) ^-\qquad\left (\sum_{n=0}^{N-1}\ Underline{f}[n] (\underline{\delta_n}) ^-[m]=\sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}[-m]\right) \ \
&=n\left (\sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}[0],\sum_{n=0}^{n-1}\underline{f}[n]\ UNDERLINE{\DELTA_N}[1],..., \sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}[n-1]\right) ^-
\end{align*}$

Let's analyze $\displaystyle{\sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}[0]}$, where $\underline{\delta_n}[m]=\ Begin{cases} 1 & \text{,} m=n \ 0 & \text{,} m\neq n \end{cases}$, so

$\BEGIN{ALIGN*}\UNDERLINE{\MATHCAL{F}}\UNDERLINE{\MATHCAL{F}}\UNDERLINE{F}
&=n\left (\sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}[0],\sum_{n=0}^{n-1}\underline{f}[n]\ UNDERLINE{\DELTA_N}[1],..., \sum_{n=0}^{n-1}\underline{f}[n]\underline{\delta_n}[n-1]\right) ^-\\
&=n\left (\underline{f}[0],\underline{f}[1],..., \underline{f}[n-1] \right) ^-\\
&=n\underline{f}^-
\end{align*}$

Will eventually have to

$\underline{\mathcal{f}}\underline{\mathcal{f}}\underline{f} = n\underline{f}^-$

[Fourier transform and its application study notes] 21. The matrix definition of discrete Fourier transform, some properties