6.1 Laplace Transform of a Signal

信号的拉普拉斯变换

P1 · Damping rescues divergent signals 收敛因子救场

The Fourier transform requires — many useful signals fail this. Multiplying by rescues them. The Laplace transform is exactly the Fourier transform of .
傅里叶变换要求信号绝对可积，许多有用的信号不满足。乘以一个收敛因子 e^−σt 可以救场——拉氏变换正是 x(t)e^−σt 的傅里叶变换。

LT:

u(t) e^t·u(t) cos(2t)·u(t) e^−tcos(2t)·u(t)

σ = 0.50

— —

Time domain — original vs damped

Running integral

Key idea: the set of σ for which the integral converges is the region of convergence (ROC) — . The σ-threshold is determined by the rightmost pole of .
使积分收敛的 σ 取值范围就是收敛域，σ 的下界由 X(s) 最右侧的极点决定。

P2 · The 3D Laplace surface — FT is the σ=0 slice 三维拉氏曲面：FT 是 σ=0 的切片

Closed-form evaluated on the complex plane. Rotate the surface; the orange slice curve is at the chosen σ. Slide σ to 0 — if 0 is inside the ROC, the slice is exactly the Fourier transform.
三维曲面是 |X(s)|；橙色切片为当前 σ 处的 |X(σ+jω)|。当 σ=0 且 0 落在收敛域内时，该切片就是傅里叶变换。

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u(t) → 1/s e^t·u(t) → 1/(s−1) cos(2t)·u(t) → s/(s²+4) e^−tcos(2t)·u(t) → (s+1)/((s+1)²+4)

σ = 0.00

— —

Slice |X(σ + jω)| at current σ — Fourier transform when σ ∈ ROC

Reading the surface. Red dots = poles. Yellow line = jω-axis (the FT path). Orange line = current σ slice. Surface plots clipped to a finite range so poles read as tall spikes rather than ∞.
红点是极点；黄线是 jω 轴（傅里叶变换的积分路径）；橙线是当前 σ 切片。曲面绘制的是经过截断的 log₁₀|X(s)|。

Think About It 想一想

Q1: Why does the Fourier transform of e^tu(t) not exist, but its Laplace transform does?为什么 e^tu(t) 没有傅里叶变换，却有拉普拉斯变换？

The pole of sits at , so the ROC is . The Fourier integration path (i.e. σ=0) is outside the ROC — the integrand blows up. The Laplace transform exists because we are free to pick any σ in the ROC.

X(s)=1/(s−1) 的极点在 s=+1，收敛域是 σ>1。傅里叶变换走的是 σ=0 的虚轴，落在收敛域之外，积分发散。但拉氏变换允许我们选任意 σ 进入收敛域，因而存在。

Q2: How do you read a pole's position on the s-plane?如何读懂 s 平面上的极点位置？

The real part of a pole tells you the decay/growth rate of the corresponding time-domain mode : negative ⇒ decay, positive ⇒ growth. The imaginary part tells you the oscillation frequency. Complex-conjugate pairs give damped sinusoids; pure imaginary poles give undamped oscillation (FT lives on the boundary).

极点的实部决定时域分量 e^pt 的衰减/增长率：负值衰减，正值增长。虚部决定振荡频率。共轭复极点对应衰减正弦；纯虚极点对应无衰减振荡（FT 存在于边界）。

Q3: When σ=0 lies in the ROC, what does the slice |X(jω)| equal?当 σ=0 在收敛域内时，切片 |X(jω)| 等于什么？

It equals the Fourier transform magnitude exactly: . The Laplace and Fourier transforms agree on the imaginary axis when the FT integral converges — which is precisely when σ=0 is in the ROC.

它就是傅里叶变换的幅度：X(ω)=X(s)|_s=jω。当 σ=0 在收敛域内时，拉氏变换在虚轴上的值等于傅里叶变换。