考研851 自动控制原理
题海 · pdf-page · p.15
\[ f_1(t) * f_2(t) = \int_0^t f_1(t-\tau) f_2(\tau) \mathrm{d}\tau \]

由于 \(\tau > t\) 时,有 \(f_1(t-\tau)1(t-\tau)=0\),因此

\[ \int_0^t f_1(t-\tau) f_2(\tau) \mathrm{d}\tau = \int_0^\infty f_1(t-\tau)1(t-\tau) f_2(\tau) \mathrm{d}\tau \]

于是

\[ \mathscr{L}\left[\int_0^t f_1(t-\tau) f_2(\tau) \mathrm{d}\tau\right] = \mathscr{L}\left[\int_0^\infty f_1(t-\tau)1(t-\tau) f_2(\tau) \mathrm{d}\tau\right] \]
\[ = \int_0^\infty \mathrm{e}^{-st}\left[\int_0^\infty f_1(t-\tau)1(t-\tau) f_2(\tau) \mathrm{d}\tau\right]\mathrm{d}t \]

\(t-\tau=\lambda\),并改变积分次序,可得

\[ \mathscr{L}\left[\int_0^t f_1(t-\tau) f_2(\tau) \mathrm{d}\tau\right] = \int_0^\infty f_1(t-\tau)1(t-\tau)\mathrm{e}^{-st}\mathrm{d}t \int_0^\infty f_2(\tau)\mathrm{d}\tau \]
\[ = \int_0^\infty f_1(\lambda)\mathrm{e}^{-s(\lambda+\tau)}\mathrm{d}\lambda \int_0^\infty f_2(\tau)\mathrm{d}\tau \]
\[ = \int_0^\infty f_1(\lambda)\mathrm{e}^{-s\lambda}\mathrm{d}\lambda \int_0^\infty f_2(\tau)\mathrm{e}^{-s\tau}\mathrm{d}\tau \]
\[ = F_1(s) F_2(s) \]

式中

\[ F_1(s) = \int_0^\infty f_1(t)\mathrm{e}^{-st}\mathrm{d}t = \mathscr{L}[f_1(t)], \qquad F_2(s) = \int_0^\infty f_2(t)\mathrm{e}^{-st}\mathrm{d}t = \mathscr{L}[f_2(t)] \]

拉普拉斯变换的基本性质,如表 1-2 所示。

表 1-2 拉普拉斯变换的基本性质

序号 基本运算 \(f(t)\) \(F(s)=\mathscr{L}[f(t)]\)
1 拉普拉斯变换定义 \(f(t)\) \(F(s)=\displaystyle\int_0^\infty f(t)\mathrm{e}^{-st}\mathrm{d}t\)
2 位移(时间域) \(f(t-\tau_0)1(t-\tau_0)\) \(\mathrm{e}^{-\tau_0 s}F(s), \ \tau_0>0\)
3 相似性 \(f(at)\) \(\dfrac{1}{a}F\left(\dfrac{s}{a}\right), \ a>0\)
4 一阶导数 \(\dfrac{\mathrm{d}f(t)}{\mathrm{d}t}\) \(sF(s)-f(0)\)
5 \(n\) 阶导数 \(\dfrac{\mathrm{d}^n f(t)}{\mathrm{d}t^n}\) \(s^nF(s)-s^{n-1}f(0)-s^{n-2}\dot f(0)-\cdots-f^{(n-1)}(0)\)
6 不定积分 \(\displaystyle\int f(t)\mathrm{d}t\) \(\dfrac{1}{s}\left[F(s)+f^{-1}(0)\right]\)
7 定积分 \(\displaystyle\int_0^t f(t)\mathrm{d}t\) \(\dfrac{1}{s}F(s)\)
8 函数乘以 \(t\) \(tf(t)\) \(-\dfrac{\mathrm{d}}{\mathrm{d}s}F(s)\)
9 函数除以 \(t\) \(\dfrac{1}{t}f(t)\) \(\displaystyle\int_t^\infty F(s)\mathrm{d}s\)
10 位移(\(s\) 域) \(\mathrm{e}^{-at}f(t)\) \(F(s+a)\)
11 初始值 \(\displaystyle\lim_{t\to 0_+} f(t)\) \(\displaystyle\lim_{s\to\infty} sF(s)\)
12 终值 \(\displaystyle\lim_{t\to\infty} f(t)\) \(\displaystyle\lim_{s\to 0} sF(s)\)
13 卷积 \(f_1(t)*f_2(t)=\displaystyle\int_0^t f_1(\tau)f_2(t-\tau)\mathrm{d}\tau\) \(F_1(s)F_2(s)\)

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