Signal representations and system properties
Введено| English | Значение |
|---|---|
| signal/ˈsɪɡnl/ | Information or cues that communicate a condition and can influence a response or decision. |
| system/ˈsɪstəm/ | система |
| linearity/lɪˈnɪərɪti/ | линейность |
| time invariance | неизменность во времени |
| causality | причинность |
| BIBO stability | устойчивость BIBO |
Which tools are allowed?
- An audio channel runs a microphone through filters, an equaliser and a speed-up stage. Before analysing it, you must know which mathematical tools it permits.
- Two questions decide that: is it linear, and is it time-invariant? This lesson turns each property into a quick experiment.
Signals and systems
- A signal 信号 is a function of time: continuous $x(t)$ or discrete $x[n]$. A system 系统 maps an input signal to an output signal.
- Each property below is a test you run on chosen inputs: feed something in, compare two outputs. It is never decided by how the system looks.
Linearity
- Linearity 线性: if $x_{1} \to y_{1}$ and $x_{2} \to y_{2}$, then $ax_{1} + bx_{2} \to ay_{1} + by_{2}$. Superposition and scaling must both hold.
- $y = 3x$ passes. $y = x^{2}$ fails: doubling the input quadruples the output. So does $y = x + 1$, because zero input gives a non-zero output.
Which system is linear?
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
A scaled delay obeys superposition and scaling. Squaring and the absolute value fail scaling; adding 1 gives a non-zero output for zero input.
The system y = 2x + 5 is linear.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Zero input gives 5, but a linear system must map zero to zero. Its graph is a straight line, but "linear" here means superposition.
Time invariance
- Time invariance 时不变性: delay the input by $T$ and the output must be the same, delayed by exactly $T$.
- The test compares two routes: shift then apply the system, or apply the system then shift. If they differ, the system is time-varying.
$y(t) = x(2t)$ with $x = u(t)$. Shift by 1 first: $u(t-1) \to u(2t - 1) = u\left(t - \tfrac{1}{2}\right)$. System first, then shift: $u(2(t-1)) = u(t - 1)$. Half a shift against a full one: time-varying.
Is y(t) = x(2t) time-invariant?
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Shift then system gives $x(2t - 1)$; system then shift gives $x(2t - 2)$. They differ, so the system is time-varying, though it is linear.
Causality
- Causality 因果性: the output at time $t$ depends only on the input up to time $t$. Every real-time physical system is causal.
- $y(t) = x(t - 1)$ is causal, a delay. $y(t) = x(t + 1)$ is not: it needs the future. Systems working on recorded data can be non-causal.
Match each system to the property it fails.
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
Each fails one test: it needs the future, it squares, its gain changes with time, or its impulse response is not absolutely integrable.
BIBO stability
- BIBO stability BIBO 稳定性: every bounded input gives a bounded output. For an LTI system that is the same as $\displaystyle\int_{-\infty}^{\infty} |h(t)|\,dt < \infty$.
- $h(t) = e^{-t}u(t)$ integrates to 1: stable. The integrator $h(t) = u(t)$ integrates to infinity: a bounded step input gives an unbounded ramp.
The integrator h(t) = u(t) is
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
$\int_{0}^{\infty} 1\,dt = \infty$. A bounded impulse response is not enough; it must be absolutely integrable.
Compute ∫|h(t)| dt for h(t) = 4e^(−2t)u(t).
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
$\int_{0}^{\infty} 4e^{-2t}\,dt = \dfrac{4}{2} = 2 < \infty$, so the system is BIBO stable.
Three common slips
- Testing linearity with one input. You need two inputs, plus scaling.
- Judging time invariance from a diagram. Run the shift test.
- Confusing causality with stability. $h(t) = -e^{t}u(-t)$ is stable but not causal; the two are independent.
Every BIBO-stable system is causal.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
$h(t) = -e^{t}u(-t)$ decays into the past: $\int |h| = 1$, so it is stable, yet it responds before the impulse arrives.
Recap
- Linear: superposition and scaling. Time-invariant: shift in, same shift out.
- Causal: no future inputs needed. BIBO stable: $\int |h|\,dt < \infty$.
- Linear plus time-invariant unlocks convolution and the transforms.