Fourier series FS, continuous time Fourier transform CTFT, discrete time Fourier transform DTFT, discrete Fourier transform DFT, derivation and connection (1)

This article mainly discusses in detail the differences and connections between Fourier series FS, continuous time Fourier transform CTFT, discrete time Fourier transform DTFT, and discrete Fourier transform DFT, focusing mainly on the formula Formal derivation, relevant image examples are omitted, rigorous mathematical proofs such as Fourier series, and the properties of various transformations are not the focus of this article. This article mainly refers to the following textbooks.

"Digital Signal Processing – Computer-Based Methods (Fourth Edition)" Sanjit K. Mitra, Electronic Industry Press.

Since the formula in this article takes up a lot of characters, it cannot be published completely in one blog, so it is divided into two blogs. This article mainly introduces Fourier series FS, continuous time Fourier transform CTFT, and discrete time Fourier transform DTFT. For the content of Discrete Fourier Transform DFT and related summary, please go to the following link.

Fourier series FS, continuous time Fourier transform CTFT, discrete time Fourier transform DTFT, discrete Fourier transform DFT, derivation and connection (2)
https://blog.csdn.net/qq_33552519/article/details/130260657

The discrete cosine transform and discrete sine transform derived from the discrete Fourier transform are introduced in another article of mine. You are welcome to read and provide relevant modification comments.

• DFT：https://blog.csdn.net/qq_33552519/article/details/124917126
• DCT：https://blog.csdn.net/qq_33552519/article/details/124917473
• DST：https://blog.csdn.net/qq_33552519/article/details/124962042

1 Fourier series of periodic functions

Based on advanced mathematical knowledge, let $f\left(t\right)$ This is one or more $T$ is a function of period, and when$[-T/2,T/2]$ Above bounded, my name is number skewer

$$\begin{array}{c}{a}_n=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}f\left(t\right)\cos \left(n{\omega }_{0}t\right)dt,\\ b_n=\frac{2}{T}\underset{-T/2}{\overset{T/2}{\int }}f\left(t\right)\sin \left(n{\omega }_{0}t\right)dt,\\ where{oh}_0=2\pi /T.\end{array}\text{\hspace{1em}(1.1)}$$

为函数 $fFourier\; coefficients\; of\; (t)$. Using Fourier coefficients as coefficients, the trigonometric series

$$f\left(t\right)\sim \frac{a_0}{2}+\sum _{n=1}^{\infty }\left(a_n\cos \left(n{\omega }_{0}t\right)+b_n\sin \left(n{\omega }_{0}t\right)\right),\text{\hspace{1em}(1.2)}$$

is called a function $f\left(t\right)$ 的傅里叶级数 (Fourier Series, FS)，其中 $\cos \left(n{\omega }_{0}t\right)$ 与 $\sin \left(n{\omega }_{0}t\right)$ is called the basic trigonometric function system. Considering Euler's formula, we have

$$\begin{array}{c}\cos \left(n{\omega }_{0}t\right)=\frac{{It\; is}^{jn{\omega }_{0}t}+{It\; is}^{-jn{\omega }_{0}t}}{2},\\ \sin \left(n{\omega }_{0}t\right)=\frac{{It\; is}^{jn{\omega }_{0}t}-{It\; is}^{-jn{\omega }_{0}t}}{2j}.\end{array}\text{\hspace{1em}(1.3)}$$

And according to equation (1.1), we have

$$\begin{array}{c}{a}_{-n}=a_n,\\ b_{-n}=-b_n.\end{array}\text{\hspace{1em}(1.4)}$$

then

$$\begin{array}{rl}f\left(t\right)& \sim \frac{a_0}{2}+\sum _{n=1}^{\infty }\left[a_n\cos \left(n{\omega }_{0}t\right)+b_n\sin \left(n{\omega }_{0}t\right)\right]\\ & =\frac{a_0}{2}+\sum _{n=1}^{\infty }\left(\frac{a_n-j{b}_n}{2}{It\; is}^{jn{\omega }_{0}t}+\frac{a_n+j{b}_n}{2}{It\; is}^{-jn{\omega }_{0}t}\right)\\ & =\frac{a_0-j{b}_0}{2}{It\; is}^{j0{\omega }_{0}t}+\sum _{n=1}^{\infty }\left(\frac{a_n-j{b}_n}{2}{It\; is}^{jn{\omega }_{0}t}+\frac{a_{-n}-j{b}_{-n}}{2}{It\; is}^{-jn{\omega }_{0}t}\right)\\ & =\sum _{n=-\infty }^{\infty }\frac{a_n-j{b}_n}{2}{It\; is}^{jn{\omega }_{0}t}=\sum _{n=-\infty }^{\infty }{F}_n{It\; is}^{jn{\omega }_{0}t}.\end{array}\text{\hspace{1em}(1.5)}$$

Substituting into equation (1.1), we can get

$$\begin{array}{rl}{F}_n& =\frac{a_n-j{b}_n}{2}\\ & =\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}f\left(t\right)\left[\cos \left(n{\omega }_{0}t\right)-j\sin \left(n{\omega }_{0}t\right)\right]dt\\ & =\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}f\left(t\right)\left(\frac{{It\; is}^{jn{\omega }_{0}t}+{It\; is}^{-jn{\omega }_{0}t}}{2}-j\frac{{It\; is}^{jn{\omega }_{0}t}-{It\; is}^{-jn{\omega }_{0}t}}{2j}\right)dt\\ & =\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}f\left(t\right){It\; is}^{-jn{\omega }_{0}t}dt.\end{array}\text{\hspace{1em}(1.6)}$$

In this way, we get a more unified and concise complex form of the Fourier series formula, which is also called the Fourier transform pair of periodic functions in signals and systems. It is customary to call the process of solving Fourier coefficients forward transformation, while the Fourier series itself is called inverse transformation, that is

F [ n ] = F { f ( t ) } = 1 T ∫ − T / 2 T / 2 f ( t ) e − j n ω 0 t d t , f ( t ) = F − 1 { F [ n ] } = ∑ n = − ∞ ∞ F n e j n ω 0 t . (1.7) \begin{gathered} F\left[ n \right] = \mathcal{F}\left\{ {f\left( t \right)} \right\} = \frac{1}{T}\int\limits_{ - T/2}^{T/2} {f\left( t \right){e^{ - jn{\omega _0}t}}dt} , \\ f\left( t \right) = {\mathcal{F}^{ - 1}}\left\{ {F\left[ n \right]} \right\} = \sum\limits_{n = - \infty }^\infty { {F_n}{e^{jn{\omega _0}t}}} . \\ \end{gathered} \tag{1.7} F[n]=F{ f(t)}=T1​−T/2∫T/2​f(t)It is−jnω0​tdt,f(t)=F−1{ F[n]}=n=−∞∑∞​Fn​It isjnω0​t.​(1.7)

Note that in advanced mathematics, due to the rigor of mathematics, the Fourier series uses wavy lines $\sim$ symbol, indicating that the Fourier series may not necessarily converge to $f\left(t\right)$, but in signals and systems, time domain and frequency are more considered. The domain energies are equal, which is Parseval's theorem, and the properties of most of the functions involved are also relatively good. To avoid trouble, they are generally represented by an equal sign. In the following discussion, in addition to some special functions such as the Dirac function and the series represented by the Dirac function, after ignoring the finite discontinuity points in the cycle, it may be assumed that the function $f\left(t\right)$ The city is connected.

Although periodic function $f\left(t\right)$'s own change $t$ itself is continuous, but the resulting Fourier coefficients$F\left[n\right]$ It's just a dispersive efficiency point $n{\omega }_0$Defined on . This property is similar to the spectrum of electronic energy level transitions, which is composed of electromagnetic wave components of several discrete frequencies. Therefore, the Fourier coefficient $F\left[n\right]$ Always symmetrical periodic signal $f\left(t\right)$ 的频谱，$FThe\; space\; where\; [n]$ is located is also called the frequency domain. In addition, because ${It\; is}^{jn{\omega }_{0}t}$ corresponds to a unit circle on the complex plane, rather than the infinite axis in the real number domain. This makes signal analysis and processing in the complex number domain worse than in the real number domain. More convenient, this property is particularly important in the later Laplace transform and z transform.

2 Fourier transform of non-periodic functions

If a non-periodic function is considered $f\left(t\right)$, I can allow that period $T\to \infty$，于は具き频${oh}_0\to 0$. Pull-in amount $\omega$, and ${oh}_n=n{\omega }_0$,So

$$\Delta {\omega }_n={oh}_n-{oh}_{n-1}={oh}_0\to 0.$$

You are there $\Delta {\omega }_n\to d\omega$, this time apart ${oh}_n$ 由为连续的 $\omega$ without any further distinction. At this time

$$\begin{array}{rl}\underset{T\to \infty }{\lim }f\left(t\right)& =\underset{T\to \infty }{\lim }\sum _{n=-\infty }^{\infty }\frac{F_n}{\Delta {\omega }_n}\Delta {\omega }_n{It\; is}^{j{\omega }_{n}t}\\ & =\underset{T\to \infty }{\lim }\sum _{n=-\infty }^{\infty }\left(\frac{1}{T\cdot \Delta {\omega }_n}\underset{-T/2}{\overset{T/2}{\int }}f\left(\tau \right){It\; is}^{-j{\omega }_n\tau }d\tau \right){It\; is}^{j{\omega }_{n}t}\Delta {\omega }_n\\ & =\underset{T\to \infty }{\lim }\sum _{n=-\infty }^{\infty }\left(\frac{1}{2\pi }\underset{-T/2}{\overset{T/2}{\int }}f\left(\tau \right){It\; is}^{-j{\omega }_n\tau }d\tau \right){It\; is}^{j{\omega }_{n}t}\Delta {\omega }_n\\ & =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\left(\underset{-\infty }{\overset{\infty }{\int }}f\left(\tau \right){It\; is}^{-j\omega \tau }d\tau \right){It\; is}^{j\omega t}d\omega .\end{array}\text{\hspace{1em}(2.1)}$$

So it can be defined, assuming the non-periodic function $f\left(t\right)$ 在 $\left(-\infty ,+\infty \right)$ 上绝对对可积，则 $fThe\; Fourier\; Transform\; (FT)\; and\; inverse\; transform\; of\; (t)$ are respectively for

F ( j ω ) = F { f ( t ) } = ∫ − ∞ ∞ f ( t ) e − j ω t d t , f ( t ) = F − 1 { F ( j ω ) } = 1 2 π ∫ − ∞ ∞ F ( j ω ) e j ω t d ω . (2.2) \begin{gathered} F\left( {j\omega } \right) = \mathcal{F}\left\{ {f\left( t \right)} \right\} = \int\limits_{ - \infty }^\infty {f\left( t \right){e^{ - j\omega t}}dt} , \\ f\left( t \right) = {\mathcal{F}^{ - 1}}\left\{ {F\left( {j\omega } \right)} \right\} = \frac{1}{ {2\pi }}\int\limits_{ - \infty }^\infty {F\left( {j\omega } \right){e^{j\omega t}}d\omega } . \\ \end{gathered} \tag{2.2} F(jω)=F{ f(t)}=−∞∫∞​f(t)It is−jωtdt,f(t)=F−1{ F(jω)}=2π1​−∞∫∞​F(jω)It isjωtdω.​(2.2)

In order to distinguish it from the subsequent discrete time Fourier transform, it is sometimes called the continuous time Fourier transform (Continuous Time FT, CTFT). Note that the above derivation is not strict. Here we focus on its form and omit the relevant proof. In addition $F\left(j\omega \right)$ 也可Direct copying $F\left(\omega \right)$, the imaginary number symbol is added mainly to express the result of Fourier transform is a complex number, and is also distinguished from the subsequent discrete-time Fourier transform.

Compared with the discrete coefficients obtained by Fourier series $F\left[n\right]$，傅りば变卢 $F\left(j\omega \right)$ The rate of change of the fixed rate $\omega$upper, and $\omega$given $n$ 的Quantity纲也owned区别，$\omega$ 是由 $\underset{{oh}_0\to 0}{\lim }n{\omega }_0$ is defined by this limit. Consider it defined in the finite interval $\left[-T/2,T/2\right]$ 的函数 $f\left(t\right)$，对其以 $T$ Extend the period to obtain the periodic function$\tilde{f}\left(t\right)$，那么可得 $f\left(t\right)$ 的傅利变换谱$F\left(j\omega \right)$ 与 $\tilde{f}\left(t\right)$ 的傅りｺ级数组$\tilde{F}\left[n\right]$ Division

$$\begin{array}{c}F\left(j\omega \right)=\underset{-T/2}{\overset{T/2}{\int }}f\left(t\right){It\; is}^{-j\omega t}dt,\\ \tilde{F}\left[n\right]=\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}\tilde{f}\left(t\right){It\; is}^{-jn{\Omega }_{T}t}dt=\frac{1}{T}{F\left(j\omega \right)|}_{\omega =n{\Omega }_T},\\ where{Oh}_T=2\pi /T.\end{array}\text{\hspace{1em}(2.3)}$$

In other words, the Fourier coefficient of the periodic function $\tilde{F}\left[n\right]$ is obtained by Fourier transforming the periodic function within one period $F\left(j\omega \right)$ 以 ${Oh}_T$ is evenly spaced sampling and $T$ The result of amplitude scaling. This involves the concept of sampling, which leads to the discrete-time Fourier transform we are talking about below. We will also see that the properties of the time domain and the frequency domain are always mirror symmetries.

3 Discrete time Fourier transform

Before discussing sampling and discrete-time Fourier transform, we need to introduce the Dirac function, which is usually defined based on the limit, that is

$$d\left(t\right)=\underset{\epsilon \to 0}{\lim }\{\begin{array}{cc}1/\epsilon ,& t\in \left(-\epsilon /2,\epsilon /2\right)\\ 0,& others\end{array}\text{\hspace{1em}(3.1)}$$

Note that the definition of Dirac function is not unique. The above uses a relatively simple definition of rectangular function. Other definitions such as triangular function can also be used, etc. It only occurs when $t\to 0$ has infinite values ​​and is 0 elsewhere, so the Dirac function is also called the impulse function and has

$$\underset{-\infty }{\overset{\infty }{\int }}d\left(t\right)dt=\underset{\epsilon \to 0}{\lim }\underset{-\epsilon /2}{\overset{\epsilon /2}{\int }}\frac{1}{e}dt=1.\text{\hspace{1em}(3.2)}$$

Hypothetical function $f\left(t\right)$ 连续，因为 $\underset{t\to t_0}{\lim }f\left(t\right)=f\left(t_0\right)$, it is easy to derive the following properties

$$\begin{array}{rl}\underset{-\infty }{\overset{\infty }{\int }}f\left(t\right)d\left(t-t_0\right)dt& =\underset{\epsilon \to 0}{\lim }\underset{t_0-\epsilon /2}{\overset{t_0+\epsilon /2}{\int }}f\left(t\right)\frac{1}{e}dt\\ & =f\left(t_0\right)\underset{\epsilon \to 0}{\lim }\underset{t_0-\epsilon /2}{\overset{t_0+\epsilon /2}{\int }}\frac{1}{e}dt=f\left(t_0\right)\\ & =\underset{-\infty }{\overset{\infty }{\int }}f\left(t_0\right)d\left(t-t_0\right)dt.\end{array}\text{\hspace{1em}(3.3)}$$

This property is also called the sampling property of the Dirac function. In addition, the convolution property of the Dirac function is also very commonly used, namely

$$\begin{array}{rl}f\left(t\right)\ast d\left(t-t_0\right)& =\underset{-\infty }{\overset{\infty }{\int }}f\left(\tau \right)d\left(t-t_0-\tau \right)d\tau \\ & =\underset{\epsilon \to 0}{\lim }\underset{t-t_0-\epsilon /2}{\overset{t-t_0+\epsilon /2}{\int }}f\left(\tau \right)\frac{1}{e}d\tau \\ & =f\left(t-t_0\right)\underset{\epsilon \to 0}{\lim }\underset{t-t_0-\epsilon /2}{\overset{t-t_0+\epsilon /2}{\int }}\frac{1}{e}d\tau \\ & =f\left(t-t_0\right).\end{array}\text{\hspace{1em}(3.4)}$$

In other words, the Dirac function can make the time axis $The\; origin\; of\; t$ moves according to the position where the impulse is generated, so that the time shift of the function can be expressed in the form of convolution.

The Dirac function provides an analysis method from continuous time signals to discrete time signals. Realistic sampling usually involves a sampling period $T$ and sampling time $\epsilon$ Two concepts. The sampling period is the time distance between two adjacent discrete sampling points. Generally, we only discuss equally spaced sampling. The sampling time is mainly limited by the sampling technology. Existing technology usually cannot measure a relatively accurate value of the signal in an instant. For example, when measuring voltage, it is usually necessary to charge the capacitor, and charging takes a certain amount of time. This process can be expressed as an integral, that is:

$$\bar{f}\left(nT\right)=\frac{1}{e}\underset{nT-\epsilon /2}{\overset{nT+\epsilon /2}{\int }}f\left(t\right)dt.\text{\hspace{1em}(3.5)}$$

It can be found that the Dirac function is equivalent to an ideal instantaneous sampling process. Using the Dirac function can ignore the concept of sampling time, thereby eliminating many unnecessary operations in signal analysis.

Then, suppose there is a continuous time signal $f_a\left(t\right)$，以 $T$ 为采样周期，对 $f_a\left(t\right)$ 在 $t=nT$ is uniformly sampled, and a discrete sequence can be obtained$f\left[n\right]$, inside

$$f\left[n\right]=f_a\left(nT\right),-\infty <n<\infty \text{\hspace{1em}(3.6)}$$

The sampling process can be expressed as a weighted superposition of periodic Dirac functions, thereby obtaining a continuous time signal$f_p\left(t\right)$，即

$$f_p\left(t\right)=\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right)d\left(t-nT\right).\text{\hspace{1em}(3.7)}$$

当然，$f_p\left(t\right)$ 也等价于 $f_a\left(t\right)$ 与周期冲激串 $p\left(t\right)$ 的乘积，current

$$\begin{array}{c}{f}_p\left(t\right)=f_a\left(t\right)p\left(t\right),\\ wherep\left(t\right)=\sum _{n=-\infty }^{\infty }d\left(t-nT\right).\end{array}\text{\hspace{1em}(3.8)}$$

Pay attention to the difference between continuous time signals and continuous signals. Continuous time only refers to independent variables $t$ is continuous, and it does not mean that the signal itself is continuous. In addition, since $f_p\left(t\right)$ is represented by the weighted superposition of Dirac functions, which only It only makes sense under interval integrals, for example

$$\underset{nT-\tau }{\overset{nT+\tau }{\int }}{f}_p\left(t\right)dt=f_a\left(nT\right)=f[n],t<T/2.\text{\hspace{1em}(3.9)}$$

Then, it can be deduced that $f_p\left(t\right)$ to change the location

$$\begin{array}{rl}{F}_p\left(j\omega \right)& =\underset{-\infty }{\overset{\infty }{\int }}{f}_p\left(t\right){It\; is}^{-j\omega t}dt\\ & =\underset{-\infty }{\overset{\infty }{\int }}\left(\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right)d\left(t-nT\right)\right){It\; is}^{-j\omega t}dt\\ & =\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right)\underset{-\infty }{\overset{\infty }{\int }}d\left(t-nT\right){It\; is}^{-j\omega t}dt\\ & =\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right)\underset{-\infty }{\overset{\infty }{\int }}d\left(t-nT\right){It\; is}^{-j\omega \left(t-nT\right)-j\omega nT}d\left(t-nT\right)\\ & =\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right){It\; is}^{-j\omega nT}.\end{array}\text{\hspace{1em}(3.10)}$$

Because we are more concerned about the signal itself, when the continuous-time signal is sampled to obtain the discrete-time signal, the sampling period can be ignored $T$, thereby simplifying subsequent operations. In order to distinguish it from the Fourier transform of general continuous-time signals, we use $\Omega$ Alternative expression (3.10) Neutral $\omega$，Immediate

$$F_p\left(j\Omega \right)=\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right){It\; is}^{-j\Omega nT}.\text{\hspace{1em}(3.11)}$$

Then the sequence can be defined $fThe\; Discrete\; Time\; FT\; (DTFT)\; of\; [n]$ is

$$F\left(e^{j\omega }\right)=\sum _{n=-\infty }^{\infty }f\left[n\right]{It\; is}^{-j\omega n},whereoh=\Omega T.\text{\hspace{1em}(3.12)}$$

and have

$$oh=2\pi \iff Oh=\frac{\omega }{T}=\frac{2\pi }{T}={Oh}_T.\text{\hspace{1em}(3.13)}$$

This way we have a method to analyze discrete-time signals, which is especially important for digital circuits that can only process discrete data. However, although we intentionally ignore the sampling period $T$However, this does not mean that it does not require further discussion. Because sampling is a conversion process from continuous time to discrete time, intuitively this will lead to the loss of some information, so how to prove and measure this information distortion is the focus of our next discussion.

It can be found that, unlike the Fourier transform of general continuous-time signals, the continuous-time signal defined by the weighted superposition of periodic Dirac functions$f_p\left(t\right)$ 的傅里叶变换 $F_p\left(j\Omega \right)$，以及序列 $f\left[n\right]$ Dispersal time change $F\left(e^{j\omega }\right)$ are all clearly periodic, that is

$$\begin{array}{rl}{F}_p\left(j\Omega \right)& =\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right){It\; is}^{-j\Omega nT}=\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right){It\; is}^{-j\Omega n\frac{2\pi }{{Oh}_T}}\\ & =\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right){It\; is}^{-j\left(\Omega +k{\Omega }_T\right)n\frac{2\pi }{{Oh}_T}}=F_p\left(j\left(\Omega +k{\Omega }_T\right)\right),\\ F\left(e^{j\omega }\right)& =\sum _{n=-\infty }^{\infty }f\left[n\right]{It\; is}^{-j\omega n}=\sum _{n=-\infty }^{\infty }f\left[n\right]{It\; is}^{-j\left(\omega +2k\pi \right)n}\\ & =F\left(e^{j\left(\omega +2k\pi \right)}\right)=F\left(e^{j\left(\omega +k{\Omega }_{T}T\right)}\right),\\ where\omega & =\Omega T,{Oh}_T=2\pi /T.\end{array}\text{\hspace{1em}(3.14)}$$

即 $F_p\left(j\Omega \right)$ This is the ratio < /span>${Oh}_T$ is a function of period, $F\left(e^{j\omega }\right)$ $2\pi$ is the period. This periodicity is caused by the periodic impulse train used in the sampling process$pIntroduced\; by\; (t)$. In addition, although we pass $F_p\left(j\Omega \right)$ 与 $F\left(e^{j\omega }\right)$ defines the form of discrete-time Fourier transform, but does not compare it with the Fourier transform of general continuous-time signals. Transformation$F_a\left(j\Omega \right)$ Establish a connection. Since the continuous time function $f_a\left(t\right)$ の傅りば变卢 $F_a\left(j\Omega \right)$ Body submersion periodicity , immediately $F_a\left(j\Omega \right)$ possible complete number $Oh\in \left(-\infty ,+\infty \right)$ extends non-periodically, then if $F_p\left(j\Omega \right)$ 是由 $F_a\left(j\Omega \right)$ 推导而来的，$F_p\left(j\Omega \right)$ periodicity is impossible Guide to refuge $F_a\left(j\Omega \right)$ .

In order to prove this information distortion, we need to introduce the Poisson summation formula, that is, for a non-periodic continuous time function$\varphi \left(t\right)$，有

$$\begin{array}{c}\sum _{n=-\infty }^{\infty }\varphi \left(t+nT\right)=\frac{1}{T}\sum _{k=-\infty }^{\infty }Phi\left(jk{\Omega }_T\right){It\; is}^{jk{\Omega }_{T}t},\\ where{Oh}_T=2\pi /T.\end{array}\text{\hspace{1em}(3.15)}$$

其中 $Phi\left(j\Omega \right)$ 为 $\varphi Continuous-time\; Fourier\; transform\; of\; (t)$, $T$ is any non-zero length. To prove this, we need to comprehensively use the knowledge of the Fourier series of periodic functions and the Fourier transform of continuous time functions we mentioned earlier, as well as the properties of the Dirac function, that is

$$\begin{array}{rl}\sum _{n=-\infty }^{\infty }\varphi \left(t+nT\right)& =\sum _{n=-\infty }^{\infty }\varphi \left(t\right)\ast d\left(t+nT\right)\\ & =\varphi \left(t\right)\ast \sum _{n=-\infty }^{\infty }d\left(t+nT\right)\\ & =\varphi \left(t\right)\ast \left[\sum _{k=-\infty }^{-\infty }\left(\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}d\left(\tau \right){It\; is}^{-jk{\Omega }_T\tau }d\tau \right){It\; is}^{jk{\Omega }_{T}t}\right]\\ & =\varphi \left(t\right)\ast \frac{1}{T}\sum _{k=-\infty }^{\infty }{It\; is}^{jk{\Omega }_{T}t}\\ & =\frac{1}{T}\sum _{k=-\infty }^{\infty }\varphi \left(t\right)\ast {It\; is}^{jk{\Omega }_{T}t}\\ & =\frac{1}{T}\sum _{k=-\infty }^{\infty }\left(\underset{-\infty }{\overset{\infty }{\int }}\varphi \left(\tau \right){It\; is}^{jk{\Omega }_T\left(t-\tau \right)}d\tau \right)\\ & =\frac{1}{T}\sum _{k=-\infty }^{\infty }\left(\underset{-\infty }{\overset{\infty }{\int }}\varphi \left(\tau \right){It\; is}^{-jk{\Omega }_T\tau }d\tau \right){It\; is}^{jk{\Omega }_{T}t}\\ & =\frac{1}{T}\sum _{k=-\infty }^{\infty }Phi\left(jk{\Omega }_T\right){It\; is}^{jk{\Omega }_{T}t}.\end{array}\text{\hspace{1em}(3.16)}$$

当 $t=0$ time, existence

$$\sum _{n=-\infty }^{\infty }\varphi \left(nT\right)=\frac{1}{T}\sum _{k=-\infty }^{\infty }Phi\left(jk{\Omega }_T\right).\text{\hspace{1em}(3.17)}$$

According to the CTFT frequency shift properties, that is

$$\begin{array}{rl}& \frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}Phi\left(\Omega +\Psi \right){It\; is}^{j\Omega t}d\Omega \\ & =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}Phi\left(\Omega +\Psi \right){It\; is}^{j\left(\Omega +\Psi \right)t-j\Psi t}d\left(\Omega +\Psi \right)\\ & ={It\; is}^{-j\Psi t}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}Phi\left(\Omega \right){It\; is}^{j\Omega t}d\Omega \\ & =\varphi \left(t\right){It\; is}^{-j\Psi t}.\end{array}\text{\hspace{1em}(3.18)}$$

So there is

$$\sum _{n=-\infty }^{\infty }\varphi \left(nT\right){It\; is}^{-j\Psi nT}=\frac{1}{T}\sum _{k=-\infty }^{\infty }Phi\left(j\left(k{\Omega }_T+\Psi \right)\right).\text{\hspace{1em}(3.19)}$$

将 $f_a\left(t\right)$ 替代 $\varphi \left(t\right)$，$k{\Omega }_T$ replace $\Psi$，即有

$$\sum _{n=-\infty }^{\infty }{f}_a\left(nT\right){It\; is}^{-j\Omega nT}=F_p\left(j\Omega \right)=\frac{1}{T}\sum _{k=-\infty }^{\infty }{F}_a\left(j\left(\Omega +k{\Omega }_T\right)\right).\text{\hspace{1em}(3.20)}$$

Therefore, it can be seen that the function $f_{p}The\; continuous\; time\; Fourier\; transform\; of\; (t)$ is$f_{a}The\; continuous-time\; Fourier\; transform\; of\; (t)$ is taken at the sampling frequency${Oh}_T$ is the result of period translation and amplitude scaling. Then it is predictable that if $F_a\left(j\Omega \right)$ 在 $\left|\Omega \right|>{Oh}_{T}When\; /2$ has a non-zero value, it is very likely that $F_a\left(j\left(\Omega +k{\Omega }_T\right)\right)$ overlap, and the spectrum of the overlapping part will no longer be separated by filtering, that is, we cannot separate it from $F_p\left(j\Omega \right)$ 完整地恢复 $F_a\left(j\Omega \right)$，But the creation information is missing . However, the result is $F_a\left(j\Omega \right)$ 只在 $|\Omega |<{Oh}_{T}If\; there\; is\; a\; non-zero\; value\; in\; /2$, it will not match $F_a\left(j\left(\Omega +k{\Omega }_T\right)\right)$ There is an overlapping part. We only know $F_{p}In\; the\; case\; of\; (j\Omega )$, we It can be obtained by an ideal low-pass filter$F_p\left(j\Omega \right)$ 在 $Oh\in \left({\Omega }_T/2,{Oh}_T/2\right)$ $F_a\left(j\Omega \right)$, due to Fourier The transformation is reversible, which means that we pass the discrete sequence $f\left[n\right]$ Recover continuous-time signals without distortion$f_a\left(t\right)$。因为序列 $f\left[n\right]$ Dispersal time change $F\left(e^{j\omega }\right)$ 只是 $F_p\left(j\Omega \right)$ 将 ${Oh}_T$ 缩波到 $The\; result\; of\; 2\pi$ only needs to refer to the above inference for its properties.

Thus it can be found that although it is intuitive for continuous time signals $f_a\left(t\right)$ Perform sampling to obtain a discrete sequence$fThere\; will\; be\; information\; distortion\; in\; the\; process\; of\; [n]$, but this situation is not absolute. , which depends on the effective spectral range of the continuous-time signal and the sampling frequency used. Suppose $\left|\Omega \right|>{Oh}_m$ 时，有 $F_a\left(j\Omega \right)=0$, then the sampling frequency ${Oh}_T$As long as it satisfies

$${Oh}_m⩽{Oh}_T/2\iff {Oh}_T⩾2{\Omega }_m.\text{\hspace{1em}(3.21)}$$

can achieve the sequence $f\left[n\right]$ 到 $f_{a}Distortion-free\; recovery\; of\; (t)$. This conclusion is the sampling theorem of band-limited signals. Equation (3.21) is also called the Nyquist condition, $2{\Omega }_m$ is called the Nyquist rate, which determines complete recovery from the sampled form $f_{a}The\; minimum\; sampling\; frequency\; of\; (t)$. But it should be noted that $F_a\left(j\Omega \right)$ 在 $|\Omega |={Oh}_m$The impulse function (that is, the Dirac function) cannot be contained at because the impulse function is defined by the limit form, which means that $Oh\to {Oh}_m^{+}$ 时 $F_a\left(j\Omega \right)$ 并为 0、 $2{\Omega }_m$ can be restored $f_a\left(t\right)$. As an example ${Oh}_T=2{\Omega }_m$ 对 $f_a\left(t\right)=\sin \left({\Omega }_{m}t\right)$ When sampling, there are

$$f\left[n\right]=f_a\left(\frac{2\pi n}{{Oh}_T}\right)=\sin \left(\pi n\right)=0.$$

This is obviously impossible from $f\left[n\right]$ 恢复 $f_a\left(t\right)$ 的。

In summary, we analyzed the continuous-time signal $f_a\left(t\right)$ Obtain discrete sequence through uniform sampling$fThe\; process\; of\; [n]$ as well as the continuous time Fourier transform and the discrete time Fourier transform The inner connection of transformation. After this, we mainly focus on the discrete sequence $f\left[n\right]$ itself, without considering the problems caused by sampling frequency . As mentioned earlier, for the discrete sequence $f\left[n\right]$, assuming it is absolutely summable, define its discrete time Fourier transform for

$$F\left(e^{j\omega }\right)=\sum _{n=-\infty }^{\infty }f\left[n\right]{It\; is}^{-j\omega n}.\text{\hspace{1em}(3.22)}$$

In order to derive its inverse transformation form, because $F\left(e^{j\omega }\right)$ is here $2\pi$ is periodic, so we might as well imitate the continuous-time inverse Fourier transform form, but only when $\left(-\pi ,By\; integrating\; on\; \pi \right)$, we can get

$$\begin{array}{rl}& \frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}F\left(e^{j\omega }\right){It\; is}^{j\omega n}d\omega \\ =& \frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\left(\sum _{m=-\infty }^{\infty }f\left[m\right]{It\; is}^{-j\omega m}\right){It\; is}^{j\omega n}d\omega \\ =& \sum _{m=-\infty }^{\infty }f\left[m\right]\left(\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{It\; is}^{-j\omega \left(n-m\right)}d\omega \right).\end{array}\text{\hspace{1em}(3.23)}$$

当 $n=m$, obviously

$$\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{It\; is}^{-j\omega 0}d\omega =1.\text{\hspace{1em}(3.24)}$$

当 $n\ne m$ time, existence

$$\begin{array}{rl}& \frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{It\; is}^{-j\omega \left(n-m\right)}d\omega \\ =& \frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\cos \left(\omega \left(n-m\right)\right)-j\sin \left(\omega \left(n-m\right)\right)d\omega \\ =& \frac{1}{2\pi \left(n-m\right)}{\left[\sin \left(\omega \left(n-m\right)\right)+j\cos \left(\omega \left(n-m\right)\right)\right]|}_{-\pi }^{\pi }\\ =& 0.\end{array}\text{\hspace{1em}(3.25)}$$

So there is

$$f\left[n\right]=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}F\left(e^{j\omega }\right){It\; is}^{j\omega n}d\omega .\text{\hspace{1em}((3.26)}$$

This can be summarized as the sequence $fThe\; discrete\; time\; Fourier\; transform\; of\; [n]$ is

F ( e j ω ) = F { f [ n ] } = ∑ n = − ∞ ∞ f [ n ] e − j ω n . f [ n ] = F − 1 { F ( e j ω ) } = 1 2 π ∫ − π π F ( e j ω ) e j ω n d ω . (3.27) \begin{gathered} F\left( { {e^{j\omega }}} \right) = \mathcal{F}\left\{ {f\left[ n \right]} \right\} = \sum\limits_{n = - \infty }^\infty {f\left[ n \right]{e^{ - j\omega n}}} . \\ f\left[ n \right] = {\mathcal{F}^{ - 1}}\left\{ {F\left( { {e^{j\omega }}} \right)} \right\} = \frac{1}{ {2\pi }}\int\limits_{ - \pi }^\pi {F\left( { {e^{j\omega }}} \right){e^{j\omega n}}d\omega } . \\ \end{gathered} \tag{3.27} F(ejω)=F{ f[n]}=n=−∞∑∞​f[n]It is−jωn.f[n]=F−1{ F(ejω)}=2π1​−π∫π​F(ejω)It isjωndω.​(3.27)

Similar to the Dirac function defined in continuous time, we can also define an impulse sequence in discrete time

$$d\left[n\right]=\{\begin{array}{cc}1,& n=0,\\ 0,& n\ne 0.\end{array}\text{\hspace{1em}(3.28)}$$

So there is

F { δ [ n − m ] } = ∑ n = − ∞ ∞ δ [ n − m ] e − j ω n = e − j ω m ∑ n = − ∞ ∞ δ [ n − m ] e − j ω ( n − m ) = e − j ω m . (3.29) \begin{gathered} \mathcal{F}\left\{ {\delta \left[ {n - m} \right]} \right\} = \sum\limits_{n = - \infty }^\infty {\delta \left[ {n - m} \right]{e^{ - j\omega n}}} \\ = {e^{ - j\omega m}}\sum\limits_{n = - \infty }^\infty {\delta \left[ {n - m} \right]{e^{ - j\omega \left( {n - m} \right)}}} = {e^{ - j\omega m}}. \\ \end{gathered} \tag{3.29} F{ δ[n−m]}=n=−∞∑∞​d[n−m]It is−jωn=It is−jωmn=−∞∑∞​d[n−m]It is−jω(n− m)=It is−jωm.​(3.29)

It just corresponds to the equation (3.23)

1 2 π ∫ − π π F { δ [ n − m ] } and ω and ω = 1 2 π ∫ − π π and − j ω and ω and ω = 1 2 π ∫ − π π e − j ω ( n − m ) d ω = δ [ n − m (3.30) \begin{aligned} &\frac{1}{ {2\pi }}\int\limits_{ - \pi }^\pi {\mathcal{F}\left\{ {\delta \left[ {n - m} \right ]} \right\}{e^{j\mega n}}d\mega } \\ = &\frac{1}{ {2\pi }}\int\limits_{ - \pi }^\pi { {e^{ - j\omega m}}{e^{j\omega n}}d\omega } \\ = &\frac{1}{ {2\pi }}\int\limits_{ - \pi }^\pi { {e^{ - j\omega \left( {n - m} \right)}}d\omega } = \delta \left[ {n - m} \right] \\ \end{aligned} \tag{3.30} ==​2π1​−π∫π​F{ δ[n−m]}It isjωndω2π1​−π∫π​It is−jωmejωndω2π1​−π∫π​It is−jω(n− m)dω=d[n−m].​(3.30)

Therefore, the discrete impulse sequence $d\left[n\right]$ is also very important in the analysis of discrete time signals, and its properties can be Provides considerable convenience when computing the discrete-time Fourier transform of a sequence.