Discrete-time systems review
導入済み| English | 意味 |
|---|---|
| impulse response | impulse response |
| convolution | 畳み込み |
| causality | 因果律 |
| BIBO stability | BIBO stability |
| frequency response | 周波数応答 |
A program on a list of numbers
- Every filter, every spectrum and every speedup in this course is a program running on a list of numbers.
- One list decides everything that program does: the impulse response.
Convolution
- A linear time-invariant system is fully described by its impulse response 单位脉冲响应 $h[n]$.
- The output is the convolution 卷积 $y[n] = \sum_{k} h[k]\,x[n-k]$: slide the taps along the input, multiply and add.
h = [1, 2, 1] and x = [1, 1]. What is y?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
y[0] = 1; y[1] = 2 + 1 = 3; y[2] = 1 + 2 = 3; y[3] = 1. The output is len(h) + len(x) − 1 = 4 samples long.
Causal and stable
- Causality 因果性: $h[n] = 0$ for $n < 0$, so no output comes before its input.
- BIBO stability BIBO 稳定: $\sum |h[n]|$ is finite, so every bounded input gives a bounded output.
A system with h[n] = 1.1ⁿ for n ≥ 0 is BIBO stable.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The terms grow, so their sum diverges. A step input would drive the output without limit.
Match each property to its test.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
All three are read straight off the impulse response.
Frequency response
- Feed in $e^{j\omega n}$ and the same exponential comes out, multiplied by $H(e^{j\omega}) = \sum h[n]e^{-j\omega n}$.
- That multiplier is the frequency response 频率响应. At $\omega = 0$ it is the tap sum, the DC gain.
What is the DC gain of h = [0.25, 0.5, 0.25]?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
The tap sum is 1, so a constant input passes unchanged.
Worked example: the sliding table
- $h = [1, 2, 1]$ on a step: $y = 1, 3, 4, 4, \ldots$ It settles at the tap sum, 4, after three samples.
- A pulse $[1, 1, 1]$ gives $1, 3, 4, 3, 1$.
h = [1, 3, 1] filters a unit step. What value does the output settle at?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
A step through any FIR settles at the tap sum, 1 + 3 + 1 = 5, after three samples.
Worked example: the recursive averager
- $y[n] = \dfrac{y[n-1] + x[n]}{2}$ has $h[n] = 0.5^{n+1}$, which never ends. Its step response is $0.5, 0.75, 0.875, 0.9375, \ldots = 1 - 2^{-(n+1)}$.
- Its single pole is at $z = 0.5$, inside the unit circle, so it is stable. It approaches 1 but never arrives.
The averager y[n] = (y[n−1] + x[n])/2 starts at rest. What is its step response at n = 4? Give 5 decimal places.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$1 - 2^{-5} = 0.96875$: it halves the remaining distance each step.
Three common slips
- Inconsistent flipping. In a hand table, flip one signal and keep to it.
- "$\sum h$ is the gain." It is the gain at DC only.
- Expecting a recursive filter to settle. It gets closer each step, but never exactly arrives.
Which recursive filter is stable?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The pole sits at the feedback coefficient. Only 0.8 is inside the unit circle; a coefficient of 1 is an accumulator, which is only marginal.
Recap
- Output = input convolved with $h[n]$.
- Causal: $h[n] = 0$ for $n < 0$. Stable: $\sum |h| < \infty$, or poles inside the unit circle.
- $H(e^{j\omega})$ is the frequency response; the tap sum is the DC gain.