EE150 信号与系统学习笔记

Coinred 的手稿们 / 2024-11-13 / 原文

Chapter 1: An Overview

Energy and power

Total energy & average power

\[\begin{aligned} & E = \int_{t_1}^{t_2} |x(t)|^2 dt \\ & P=\dfrac{E}{t_2-t_1} & \qquad\text{Continuous-time signal} \\ & E = \sum_{n=n_1}^{n_2} |x[n]|^2 \\ & P=\dfrac{E}{n_2-n_1+1} & \qquad\text{Discrete-time signal} \end{aligned} \]

Infinite time:

\[\begin{aligned} & E = \lim_{T\to\infty} \int_{-T}^{T} |x(t)|^2 dt = \int_{-\infty}^{\infty} |x(t)|^2 dt \\ & P= \lim_{T\to\infty} \dfrac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt & \qquad\text{Continuous-time signal} \\ & E = \lim_{N\to\infty} \sum_{n=-N}^{N} |x[n]|^2 = \sum_{n=-\infty}^{\infty} |x[n]|^2 \\ & P= \lim_{N\to\infty} \dfrac{1}{2N+1} \sum_{n=-N}^{N} |x[n]|^2 & \qquad\text{Discrete-time signal} \end{aligned} \]

Infinite/Finite-energy/power signal.

Exponential and Sinusoidal Signals

$x(t) = e^{j\omega_0 t}$ 的基波周期 (fundamental frequency) $T_0 = \dfrac{2\pi}{\omega_0}$ ， $\omega_0$ 是基波频率。

对于离散信号 $x[n] = e^{j\omega_0 n}$ ：

$e^{j(\omega_0 + 2\pi)n} = e^{j\omega_0 n}$ 是相同信号。
- 只考虑频率为 $0 \leq \omega_0 < 2\pi$ or $-\pi < \omega_0 \leq \pi$
$0 \leq \omega_0 < 2\pi$，振荡速率 (oscillation rate) 先快后慢，$\omega_0 = \pi$ 时最大，信号为 $(-1)^n$ 。
周期性：需要满足 $\dfrac{\omega_0}{2\pi} = \dfrac{m}{N}$ 为有理数，若 $\gcd(m,N) = 1$，则 $N = m\dfrac{2\pi}{\omega_0}$ 为该信号的基波周期，$\dfrac{\omega_0}{m}$ 为基波频率。

一组信号 $\phi_k[n] = e^{j\frac{2\pi}{N}k\cdot n}$ 只有 $N$ 个不同的信号（离散傅里叶变换时用得到）。

The Unit Impulse and Unit Step Functions

$\delta(t)$ 单位脉冲信号 Unit impulse，$u(t)$ 单位阶跃信号 Unit step 。

Relationship

\[\begin{aligned} & \delta[n] = u[n]-u[n-1] \\ & u[n] = \sum_{m = -\infty}^{n} \delta[m] = \sum_{k = 0}^{\infty} \delta[n-k] & \qquad\text{Discrete-time signal} \\ & \delta(t) = \dfrac{\mathrm{d} u(t)}{\mathrm{d} t} \\ & u(t) = \int_{-\infty}^{t} \delta(\tau) d\tau = \int_{0}^{\infty} \delta[t-\sigma] d\sigma & \qquad\text{Continuous-time signal} \end{aligned} \]

Sampling property

\[\begin{aligned} x[n]\delta[n-n_0] = x[n_0]\delta[n-n_0] & \qquad\text{Discrete-time signal} \\ x(t)\delta(t-t_0) = x(t_0)\delta(t-t_0) & \qquad\text{Continuous-time signal} \end{aligned} \]