离散时间系统
引入| English | 释义 | 拼音 |
|---|---|---|
| difference equation | 差分方程 | chā fēn fāng chéng |
| DC gain | 直流增益 | zhí liú zēng yì |
| unit circle/ˈjuːnɪt ˈsɜːkl/ | 单位圆 | dān wèi yuán |
An equaliser in a loop
- Once sound is sampled, an equaliser is a few lines of code: each new output is built from the new input and the outputs just before it.
- Running that loop by hand, one sample at a time, is the clearest way to see what it does. This lesson does exactly that.
Difference equations
- A discrete-time system is often a difference equation 差分方程, such as $y[n] = 0.5\,y[n-1] + x[n]$: half the last output plus the new input.
- Discrete convolution plays the same role as in continuous time: $y[n] = \displaystyle\sum_{k} x[k]\,h[n - k]$.
y[n] = 0.5y[n−1] + x[n], starting at rest. What is the impulse response h[n]?
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
An impulse gives $y[0] = 1$, then each sample halves: $1, 0.5, 0.25, \dots = 0.5^{n}$.
Iterate, one sample at a time
- State the starting condition, here $y[-1] = 0$, then compute forwards. Each line uses only numbers already found.
- With a unit step, $y[0] = 1$, $y[1] = 1.5$, $y[2] = 1.75$, $y[3] = 1.875$.
![Iterating y[n] = 0.5·y[n−1] + x[n] on a step: each step halves the gap to the limit 2.](/handout-media/university_electrical_engineering/assets/06-iteration.png?v=1793069544)
y[n] = 0.5y[n−1] + x[n] with a unit step and y[−1] = 0. What is y[4]?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
$1,\ 1.5,\ 1.75,\ 1.875,\ 1.9375$: each step halves the gap to 2.
The closed form
- For $y[n] = a\,y[n-1] + x[n]$ with a step, the outputs $1,\ 1 + a,\ 1 + a + a^{2}, \dots$ are geometric partial sums: $y[n] = \dfrac{1 - a^{n+1}}{1 - a}$.
- If $|a| < 1$, the limit is $\dfrac{1}{1 - a}$, the DC gain 直流增益. For $a = 0.5$ it is 2, and the gap to it halves every sample.
y[n] = 0.75y[n−1] + x[n] has a DC gain of
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
$\dfrac{1}{1 - 0.75} = 4$. A pole near 1 means a large DC gain and a slow approach.
For y[n] = 0.5y[n−1] + x[n] with a step, the gap to the final value halves every sample.
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
The gap is $2 - y[n] = 2 \times 0.5^{n+1}$, which halves with each $n$.
Worked example: a negative root
- $a = -0.8$: $y[n] = \dfrac{1 - (-0.8)^{n+1}}{1.8}$. The samples go $1,\ 0.2,\ 0.84,\ 0.328, \dots$, swinging either side of the limit 0.556.
- The sign flip every sample is the signature of a negative root. Since $|a| = 0.8 < 1$, the swings still shrink: stable.
y[n] = −0.8y[n−1] + x[n] on a step. The output
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
The negative root flips the sign of the gap each sample: $1, 0.2, 0.84, 0.328, \dots$ The swings shrink because $|-0.8| < 1$.
Stability: inside the unit circle
- The mode of $y[n] = a\,y[n-1]$ is $a^{n}$. It decays if $|a| < 1$ and grows if $|a| > 1$.
- So a discrete system is stable when all its characteristic roots lie inside the unit circle 单位圆, $|a| < 1$.
Which recursion is unstable?
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
$|1.1| > 1$: the mode $1.1^{n}$ grows. The others lie inside the unit circle.
Three common slips
- An unstated starting value. Write $y[-1]$ before the first line.
- Judging stability from a few samples. Only $|a| < 1$ decides.
- Reading the DC gain from $y[0]$. It is the limit $\dfrac{1}{1 - a}$, reached as the samples settle.
Order the steps for iterating a difference equation by hand.
难度 3/5 · 今日首次完成可获 200 金币及少量额外奖励
Each line uses only values already found; the closed form then checks where the samples are heading.
Recap
- Iterate a difference equation one sample at a time from a stated start.
- $y[n] = a\,y[n-1] + x[n]$ has DC gain $\dfrac{1}{1 - a}$ when $|a| < 1$.
- Roots inside the unit circle: stable; negative roots alternate in sign.