单边指数时域信号的傅里叶反变换

再一次学习傅里叶变换的时候，遇到了一个之前没有考虑过的问题：如何计算单边/双边指数信号的傅里叶反变换？

二者所面临的问题实际上是一样的，双边指数信号的傅里叶变换无非是两个分式的和，分别对其进行傅里叶反变换即等效于对两个关于y轴对称的单边指数信号对应的频域方程进行傅里叶反变换。

想了很久，用常规方法似乎无解。上网查询才发现已有完善的解答，特在此借鉴一下。

方法一最“常规”，即通过构造一阶常微分方程并求解的方式得到结论；

方法二利用了留数定理，数学上最为直观，但如果不熟悉相关的理论比较难以想到这种解法，且有点“循环证”的感觉；

方法三就是傅里叶变换的定义，如果能够接受则无需绞尽脑汁想办法去算这个积分。

其根本的难点在于阶跃信号在零点造成的一系列非线性效应，在连续信号中处理起来并不方便。然而对于离散信号而言单位阶跃响应则简单许多，在z域也更好理解与计算一些。

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QUESTION:

\[\] Fourier transform is defined as \[H(jw)=\int_{-\infty}^{\infty}h(t)e^{-jwt}dt\] Inverse Fourier transform is defined as \[h(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}H(jw)e^{jwt}dw\]

Let \(h(t)=e^{-at}u(t),a>0\), where \(u(t)\) is heaviside function and a is real constant.

Fourier transform of this function is \[H(jw)=\int_{0}^{\infty}h(t)e^{-jwt}dt=\int_{0}^{\infty}e^{-at}e^{-jwt}dt=\frac{1}{a+jw}\] How can I calculate inverse Fourier transform of \(\frac{1}{a+jw}\)? \[h(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{a+jw}e^{jwt}dw\]

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METHOD 1:

We can proceed to evaluate the integral if we invoke Generalized Functions. To that end, we write the inverse Fourier Transform representation for f as

\[f(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 1\] Then, note that the derivative f’ is given by \[f’(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{j\omega\,e^{j\omega t}}{a+j\omega}\,d\omega \tag 2\]

Adding (2) and a times (1) reveals that \[f’(t) +af(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}e^{j\omega t}\,d\omega =\delta(t) \tag 3\] where \(\delta(t)\) is the Dirac Delta Distribution. Solution to the ODE expressed in (3) is \(f(t)=e^{-at}h(t)\) as was to be shown!

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METHOD 2:

We begin by writing the integral representation for f(t) as the Cauchy Principal Value

\[f(t)=\frac{1}{2\pi}\lim_{R\to \infty}\int_{-R}^{ R}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 4\] Note that we need not evaluate the integral in terms of its Cauchy Principal Value. Inasmuch as the improper integral converges, its Cauchy Principal Value converges to the same value.

Now, we define a new function \(f_R(t)\) as \[f_R(t)=\oint_C \frac{e^{jzt}}{a+jz}\,dz\] where the closed contour C is comprised of the real line segment from z=-R to z=+R and for t>0 (t<0) the semicircle \(C_R\) of radius R in the upper (lower) half plane.

Jordan’s Lemma guarantees that as \(R\to \infty\), the contribution of the integral from the integration over \(C_R\) goes to zero. Therefore, we have \(f(t)=\lim_{R\to \infty}f_R(t)\)

Now, we note that since a>0 that the only singularity of \[\frac{e^{jzt}}{a+jz}\] is a \(z=ja\). Thus, from the Residue Theorem we have

\[f(t)=2\pi i\text{Res}\left(\frac{e^{jzt}}{a+jz},z=ja\right)=e^{-at}, t>0　 OR　 0,t<0\]

as expected!

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METHOD 3:

In general, we have \[{\cal F}(t \mapsto f(-t))(\omega) = ({\cal F}f) (-\omega)\]

If \(f(t) = e^{-at} u(t)\), with a>0, we have \[({\cal F}f)(\omega) = { 1 \over i \omega +a}\]

If we let \(g(t) = -f(-t)\), we have \[({\cal F}g)(\omega) = -({\cal F}f)(-\omega) = {1 \over i \omega -a}\]

Hence the inverse is \[t \mapsto - e^{at} u(-t)\]