为什么狄拉克梳子的傅立叶变换就是狄拉克梳子？

16

这是没有意义的，我的，因为海森堡不平等指出 $\Delta t\Delta \omega$ 〜1。

因此，当您在时间上进行了完全本地化时，您会得到频率上完全分布的东西。因此，基本关系 $\mathfrak{F}\{\delta(t)\} = 1$ ，其中 $\mathfrak{F}$ 是傅立叶变换算符。

但是对于狄拉克梳子，应用傅立叶变换，您会收到另一个狄拉克梳子。直观地，您还应该获得另一行。

为什么这种直觉会失败？

fourier-transform

13

我相信谬论是要相信狄拉克梳子在时间上是本地化的。这不是因为它是周期函数，因此它只能在其基本频率的倍数（即离散的频率点）上具有频率分量。它不能有连续的频谱，否则就不会是周期性的。就像任何其他周期函数一样，狄拉克梳子可以用傅立叶级数表示，即复指数的无穷大。每个复指数对应于不同频率的频域中的狄拉克脉冲。将这些狄拉克脉冲求和得到频域的狄拉克梳。

— 马特·L。
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是的，两个周期梳都没有定位在其各自的独立变量（时间/频率）中。

— 彼得·克

11

您的直觉失败是因为您从错误的假设开始。海森堡的不确定性并未说明您的想法。正如您在问题中已经说过的那样，这是一个不平等。确切地说，是

Δ t \cdot Δ f \geq \frac{1}{4 π}

$\Delta t \cdot \Delta f \geq\frac{1}{4\pi}$

没有理由为什么不确定性乘积必须接近所有信号的下限。实际上，唯一达到此最低限度的信号就是Gabor原子。对于所有其他信号，期望它更大，甚至可能无限大。

— 爵士乐
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1

是的，但是主要的谬误是认为狄拉克梳子在时间上是局部的。不是因为它是周期性的。因此，不确定性定理对狄拉克梳没有任何帮助。

— 马特·L。

@MattL。，那不是我对原始问题的理解。我认为他实际上是在争论狄拉克火车是在其本机领域完全被本地化的，因此，傅里叶应该转变成非常本地化的东西。

— Jazzmaniac，2015年

1

好的，看起来OP对“另一条线”的含义有误解。我认为这是指平坦频谱（就像他之前提到的狄拉克脉冲的频谱一样）。但是您认为这是指一条频谱线，即一个单一频率。至少现在我了解您的答案将如何回答OP的问题。

— 马特·L.

1

@MattL。，我实际上以为他写“线”时，他是指狄拉克分布的通常图形表示。无论如何，他将必须澄清，因为至少可以通过两种不同的方式来真正理解该问题。

— Jazzmaniac，2015年

1

好的，“标准”定义是一种有关动量和位置不确定性（特别是标准偏差）的物理陈述，并且在其中带有

。即便如此，在这种情况下，你必须定义什么是“的意思

”和“

”。该常数（您指定为

ℏ

$\hbar$

Δ t

$\Delta t$

Δ f

$\Delta f$

）不能太远统一（在对数标度），但它不一定是

\frac{1}{4 π}

$\frac{1}{4 \pi}$

除外由于对“的一个具体定义

”和“

”。

\frac{1}{4 π}

$\frac{1}{4 \pi}$

Δ t

$\Delta t$

Δ f

$\Delta f$

— 罗伯特·布里斯托

6

电气工程师使用Dirac delta函数有点快而松散，数学家认为这不是函数（或者至少不是“常规”函数，而是“分布”）。数学事实是，如果 $f(t)=g(t)$ “几乎无处不在”（其以每值的装置 $t$ 除了离散值的可数），则

\int f (t) d t = \int g (t) d t

$\int f(t)dt = \int g(t)dt$ 。

well the functions $f(t)=0$ and $g(t)=\delta(t)$ are equal everywhere except at $t=0$ , yet we electrical engineers insist that their integrals are different. but if you set aside this little (and, in my opinion, non-practical) difference, the answer to your question is:

狄拉克梳功能
${I I I}_{T} (t) ≜ \sum_{k = - \infty}^{+ \infty} δ (t - k T)$ $\mathrm{III}_T(t) \triangleq \sum\limits_{k=-\infty}^{+\infty} \delta(t - kT)$ 是周期的周期函数 $T$ 并因此具有傅立叶级数： ${I I I}_{T} (t) = \sum_{n = - \infty}^{+ \infty} c_{n} e^{j 2 π n t / T}$ $\mathrm{III}_T(t) = \sum\limits_{n=-\infty}^{+\infty} c_n \ e^{j 2 \pi n t/T}$
如果将傅立叶级数的系数 $c_n$ 爆破，则会得到：

\begin{aligned} c_{n} & = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} {I I I}_{T} (t) e^{- j 2 π n t / T} d t \\ = \frac{1}{T} \int_{- T / 2}^{T / 2} δ (t) e^{- j 2 π n t / T} d t (k = 0) \\ = \frac{1}{T} \int_{- T / 2}^{T / 2} δ (t) e^{- j 2 π n 0 / T} d t \\ = \frac{1}{T} \forall n \end{aligned}

$\begin{align} c_n & = \frac{1}{T}\int\limits_{t_0}^{t_0+T} \mathrm{III}_T(t) e^{-j 2 \pi n t/T} dt \\ & = \frac{1}{T}\int\limits_{-T/2}^{T/2} \delta(t) e^{-j 2 \pi n t/T} dt \quad \quad (k=0)\\ & = \frac{1}{T}\int\limits_{-T/2}^{T/2} \delta(t) e^{-j 2 \pi n 0/T} dt \\ & = \frac{1}{T} \quad \quad \forall n \\ \end{align}$

so the Fourier series for the Dirac comb is

{I I I}_{T} (t) = \sum_{n = - \infty}^{+ \infty} \frac{1}{T} e^{j 2 π n t / T}

$\mathrm{III}_T(t) = \sum\limits_{n=-\infty}^{+\infty} \frac{1}{T} \ e^{j 2 \pi n t/T}$

which means you're just summing up a bunch of sinusoids of equal amplitude.

the Fourier Transform of a single complex sinusoid is:

F {e^{j 2 π f_{0} t}} = δ (f - f_{0})

$\mathfrak{F} \left\{ e^{j 2 \pi f_0 t} \right\} = \delta(f-f_0)$

and there is this property of linearity regarding the Fourier Transform. the rest of the proof is an exercise left to the reader.

— robert bristow-johnson
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1

@Jazzmaniac, that's a falsehood. when have i ever been condescending toward mathematicians? (me thinks you're projecting a bit.) BTW, it's been 38 years since i have had 2 semesters of functional analysis at the graduate level. don't remember everything, but i sure do remember what a metric space is, a normed metric space (i think they were sometimes called "Banach spaces"), and inner product spaces (sometimes called "Hilbert spaces"), and what a functional is (maps from one of these to a number). and i know what linear spaces are. about

δ (t)

$\delta(t)$ , i don't mind them being naked.

— robert bristow-johnson

You go on with a wrong argument that suggests mathematicians don't get 1 when they integrate over a Dirac distribution. Well, you can't demonstrate any better that you haven't understood the Dirac distribution, even if you have taken a class on functional analysis. It doesn't need electrical engineers like you to "fix" mathematics. And I will keep pointing that out to you until you stop talking about mathematicians like that. It's entirely your choice.

— Jazzmaniac

that's a falsehood, too, @Jazzmaniac. i am saying that, consistent with what mathematicians tell us, the Dirac delta function is not really a function (even though we electrical engineers don't worry about that distinction and deal with it as if it were a function) because if it were a function that was zero almost everywhere, the integral would be zero. why do you keep misrepresenting me? what is the ax you're grinding?

— robert bristow-johnson

@robertbristow-johnson "electrical engineers play a little fast and loose with the Dirac delta function." Paul Dirac was an electrical engineer. Claude Shannon was also an electrical engineer. I admonish you from making such general and inaccurate statements. You claim to be an electrical engineer and clearly understand distribution theory.

— Mark Viola

nearly every undergraduate electrical engineering textbook on Linear System Theory or Signals and Systems or some similar name, will introduce and treat the Dirac Delta as a limiting case of a "nascent delta". e.g. :

δ (t) = lim_{a \to 0} \frac{1}{a \sqrt{π}} e^{- t^{2} / a^{2}}

$\delta(t) = \lim_{a \to 0}\frac{1}{a \sqrt{\pi}} \mathrm{e}^{-t^2/a^2}$ or some other unit area pulse function that you can make skinny. i would not be surprized that in published papers, folks like Shannon or Dirac (didn't know that) would stick with the conservative facts:

\int f (t) δ (t - τ) d t = f (τ)

$\int f(t) \delta(t-\tau) \ dt = f(\tau)$ and

δ (t) = 0 \forall t \neq 0

$\delta(t)=0 \quad \forall \ t \ne 0$ .

— robert bristow-johnson

1

I shall try to give an intuition. The way we could probably think is : "One Dirac delta gives us a 1 in frequency domain. Now I give infinite number of Dirac deltas. Shouldn't I get a higher DC?" Now let us see whether by adding all those frequency components mentioned in the Dirac comb in the frequency domain(FD), we get another Dirac comb in time domain(TD). We are adding continuous waveforms and getting deltas at discrete points. Sounds weird.

Coming back to the FD. We have a Dirac comb with spacing $\omega_0$ . To put it in words, we have deltas at $0,\pm\omega_0,\pm2\omega_0,\pm3\omega_0$ and so on. We thus have a DC and infinite number of cosines, namely $\cos(\omega_0 t), \cos(2\omega_0 t), \cos(3\omega_0 t)$ and so on.

Let's consider points in time domain corresponding to $t = \frac{2n\pi}{\omega_0}$ . All the above cosine waves will give us value 1. Hence they all add up and give us non zero value at those points. Now what about any other t? We need to get convinced that they will all add up to zero.

Now deviating slightly, let's consider a waveform $cos(kn) ; n = 0,1,2,3,4...\infty$ . We know that unless k can be expressed as a fraction multiplied by $\pi$ , it's aperiodic. What does that mean? There is not a single repeating sample. Each of the samples are unique. Looking it from another perspective, we have infinite number of samples which are unique and part of a cosine wave. This means taking all the infinite points, we will be able to construct a single CONTINUOUS cosine wave completely once. What if $cos(kn)$ is periodic? We already know that the sum of samples will be zero periodically based on value of k. Hence, sum of all the samples of $cos(kn)$ will give us zero for any value of k, except $k = 2\pi$ 's multiple.

Returning back to our original problem : We now take an arbitrary $t=t_0 \neq 2r\pi$ . Now we have $\cos(0\omega_0 t_0)[dc] + \cos(\omega_0 t_0) + \cos(2\omega _0 t_0) + \cos(3\omega_0 t_0)$ ....as the value at $t=t_0$ . But we have already proved this infinite sum =0 for any t except $t=\frac{2n\pi}{\omega _0}$ , where all these cosines add up to give dirac deltas.

— Subramanian T R
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