# 信号与系统学习笔记（三）

## 信号与系统学习笔记（三）

Contents

### 卷积

$$f(t)=\int_{-\infty}^\infty f_1(\tau)f_2(t-\tau)\mathrm d\tau$$

$$f(t)=f_1(t)\otimes f_2(t) \qquad or \qquad f(t)=f_1(t)*f_2(t)$$

$$e(t)=\lim_{\Delta t_1\rightarrow 0}\sum_{t_1=-\infty}^\infty e(t_1)\delta(t-t_1)\Delta t_1\\ r(t)=\lim_{\Delta t_1\rightarrow 0}\sum_{t_1=-\infty}^\infty e(t_1)h(t-t_1)\Delta t_1\\$$

$$r(t)=\int_{-\infty}^\infty e(\tau)h(t-\tau)\mathrm d\tau\\ r_{zs}(t)=e(t)*h(t)$$

e.g.

$\text{RL}$ 串联电路，激励信号为电压源 $e(t)$ ,响应为电流 $i(t)$ ,求冲激响应 $h(t)$ ，并求 $e(t)=u(t)-u(t-t_0)$ 时的响应 $i(t)$ .

$$Li'(t)+Ri(t)=e(t)\\$$
$\delta(t)$ 配平得到 $i(0_+)=\dfrac{1}{L}$ ,特征根显然为 $\alpha=-\dfrac{R}{L}$,得到冲激响应
$$h(t)=\frac{1}{L}e^{-\frac{R}{L}t}u(t)$$

\begin{align*} i(t)&=(u(t)-u(t-t_0))*(\frac{1}{L}e^{-\frac{R}{L}t}u(t))\\ &=(u(t)-u(t-t_0))\int_0^t\frac{1}{L}e^{-\frac{R}{L}(t-\tau)}\mathrm d\tau+u(t-t_0)\int_0^{t_0}\frac{1}{L}e^{-\frac{R}{L}(t-\tau)}\mathrm d\tau\\ &=(u(t)-u(t-t_0))\frac{1}{R}(1-e^{-\frac{R}{L}t})+u(t-t_0)\frac{1}{R}(e^{-\frac{R}{L}(t-t_0)}-e^{-\frac{R}{L}t})\\ &=\frac{1}{R}(1-e^{-\frac{R}{L}t})u(t)-\frac{1}{R}(1-e^{-\frac{R}{L}(t-t_0)})u(t-t_0) \end{align*}

### 卷积的性质

#### 交换律

$$f_1(t)*f_2(t)=f_2(t)*f_1(t)$$

$$f_1(t)*f_2(t)=\int_{-\infty}^\infty f_1(\tau)f_2(t-\tau)\mathrm d\tau\\ =\int_{-\infty}^\infty f_1(t-\lambda)f_2(\lambda)\mathrm d\lambda\\ =f_2(t)*f_1(t)$$

#### 分配律

$$f(t)*[h_1(t)+h_2(t)]=f(t)*h_1(t)+f(t)*h_2(t)$$

#### 结合律

$$f(t)*h_1(t)*h_2(t)=f(t)*[h_1(t)*h_2(t)]$$

\begin{align*} f(t)*h_1(t)*h_2(t)=&\int_{-\infty}^\infty[\int_{-\infty}^\infty f(\lambda)h_1(\tau-\lambda)\mathrm d\lambda]h_2(t-\tau)\mathrm d\tau\\ =&\int_{-\infty}^\infty f(\lambda)[\int_{-\infty}^\infty h_1(\tau-\lambda)h_2(t-\tau)\mathrm d\tau]\mathrm d\lambda\\ =&\int_{-\infty}^\infty f(\lambda)[\int_{-\infty}^\infty h_1(\tau)h_2(t-\lambda-\tau)\mathrm d\tau]\mathrm d\lambda\\ =&f(t)*[h_1(t)*h_2(t)] \end{align*}

#### 微积分性质

$$g'(t)=f(t)*h'(t)=f'(t)*h(t)$$

$$\frac{\mathrm d}{\mathrm dt}[f(t)*h(t)]=\frac{\mathrm d}{\mathrm dt}\int_{-\infty}^\infty f(\tau)h(t-\tau)\mathrm d\tau\\ =\int_{-\infty}^{\infty}f(\tau)\frac{\mathrm dh(t-\tau)}{\mathrm dt}\mathrm d\tau\\ =f(t)*h'(t)$$

$$g^{(n-m)}(t)=f^{(n)}(t)*h^{(-m)}(t)$$
$n$ 为微分 ，$-m$ 为积分。

$$f(-\infty)=0\\ h(-\infty)=0$$

#### 与冲激函数卷积

$$f(t)*\delta(t)=\int_{-\infty}^\infty f(\tau)\delta(t-\tau)\mathrm d\tau\\ =f(t)$$

$$f(t)*\delta^{(k)}(t-t_0)=f^{(k)}(t-t_0)$$

#### e.g.

$$e(t)=u(t+\frac{1}{2})-u(t-1)\\ h(t)=\frac{1}{2}t(u(t)-u(t-2))$$

$$e(t)*h(t)=e'(t)*\int_{-\infty}^th(\lambda)\mathrm d\lambda\\ e'(t)=\delta(t+\frac{1}{2})-\delta(t-1)\\ \int_{-\infty}^th(\lambda)\mathrm d\lambda=\int_{-\infty}^t\frac{1}{2}\tau(u(\tau)-u(\tau-2))\mathrm d\tau\\ =\frac{1}{4}t^2u(t)-\frac{1}{4}(t^2-4)u(t-2)\\ =\frac{1}{4}t^2(u(t)-u(t-2))+u(t-2)\\ e'(t)*h^{(-1)}(t)=\frac{1}{4}(t+\frac{1}{2})^2(u(t+\frac{1}{2})-u(t-\frac{3}{2}))+u(t-\frac{3}{2})-\\ \{\frac{1}{4}(t-1)^2(u(t-1)-u(t-3))+u(t-3)\}$$

### 算子符号

#### 算子符号的基本规则

$$p=\frac{\mathrm d}{\mathrm dt}\\ \frac{1}{p}=\int_{-\infty}^t(\cdot)\mathrm d\tau\\ px=\frac{\mathrm d}{\mathrm dt}x\\ p^nx=\frac{\mathrm d^n}{\mathrm dt^n}x\\ \frac{1}{p}x=\int_{-\infty}^tx\mathrm d\tau$$

$$\frac{\mathrm d^2r(t)}{\mathrm dt^2}+5\frac{\mathrm dr(t)}{\mathrm dt}+6r(t)=\frac{\mathrm de(t)}{\mathrm dt}+3e(t)$$

$$p^2r+5pr+6r=pe+3e\\ (p^2+5p+6)r=(p+3)e$$

• $p$ 多项式可以进行因式分解或展开，如
\begin{align*} (p^2+5p+6)x&=(p+3)(p+2)x\\ &=(\frac{\mathrm d}{\mathrm dt}+3)(\frac{\mathrm d}{\mathrm dt}x+2x)\\ &=\frac{\mathrm d}{\mathrm dt}[\frac{\mathrm d}{\mathrm dt}x+2x]+3[\frac{\mathrm d}{\mathrm dt}x+2x]\\ &=\frac{\mathrm d^2}{\mathrm dt^2}x+5\frac{\mathrm d}{\mathrm dt}x+6x \end{align*}

• 等式两端的符号 $p$ 不可任意消去

$$\frac{\mathrm dx}{\mathrm dt}=\frac{\mathrm dy}{\mathrm dt}\\ x=y+C$$
可见
$$px=py\nLeftrightarrow x=y$$

• 微分和积分顺序不能倒换
$$p\cdot\frac{1}{p}x=\frac{\mathrm d}{\mathrm dt}\int_{-\infty}^tx\mathrm d\tau=x\\ \frac{1}{p}\cdot px=\int_{-\infty}^t\frac{\mathrm d}{\mathrm dt}x\mathrm d\tau=x(t)-x(-\infty)\ne x$$

#### 用算子符号建立微分方程

$$v_L(t)=L\frac{\mathrm d}{\mathrm dt}i(t)=Lpi_L(t)$$

$$v_C(t)=\frac{1}{C}\int_{-\infty}^ti_C(\tau)\mathrm d\tau=\frac{1}{Cp}i_C(t)$$

$$\begin{cases} (R_1+\dfrac{1}{Cp})i(t)-\dfrac{1}{Cp}i_L(t)=e(t)\\ -\dfrac{1}{Cp}i(t)+(Lp+R_2+\dfrac{1}{Cp})i_L(t)=0 \end{cases}$$

#### 传输算子

$$C_0p^nr(t)+C_1p^{n-1}r(t)+\cdots+C_{n-1}pr(t)+C_nr(t)=E_0p^me(t)+E_1p^{m-1}e(t)+\cdots+E_{m-1}pe(t)+E_me(t)\\ (C_0p^n+C_1p^{n-1}+\cdots+C_{n-1}p+C_n)r(t)=(E_0p^m+E_1p^{m-1}+\cdots+E_{m-1}p+E_m)e(t)$$

$$\begin{cases} D(p)=C_0p^n+C_1p^{n-1}+\cdots+C_{n-1}p+C_n\\ N(p)=E_0p^m+E_1p^{m-1}+\cdots+E_{m-1}p+E_m \end{cases}$$

$$D(p)[r(t)]=N(p)[e(t)]\\ r(t)=\frac{N(p)}{D(p)}e(t)$$

#### 牛刀小试

$$[H(p)\delta(t)]e^{-\alpha t}=H(p+\alpha)\delta(t)$$