LTI systems and convolution
مُقدَّم| English | المعنى |
|---|---|
| impulse response | استجابة النبضة |
| linear time-invariant system | linear time-invariant system |
| convolution | The operation that gives an LTI system's output from its input and its impulse response. |
One recording predicts every sound
- Clap once in a concert hall and record the echo. That single recording predicts how the hall will colour any music played in it.
- The clap is an impulse; the recording is the impulse response. This lesson turns it into the output for any input.
The impulse response
- For a linear time-invariant system 线性时不变系统 (LTI), the impulse response 冲激响应 $h(t)$ is the output when the input is a unit impulse $\delta(t)$.
- It describes the system completely. Know $h(t)$ and you can find the output for any input.
Why one response is enough
- Any input is a sum of shifted, scaled impulses: $x(t) = \displaystyle\int x(\tau)\,\delta(t - \tau)\,d\tau$.
- Time invariance turns each $\delta(t - \tau)$ into $h(t - \tau)$; linearity adds up the scaled responses. The result is the convolution 卷积:
Convolution with h(t) gives the output of any system, linear or not.
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The derivation uses linearity (to add the responses) and time invariance (to shift one $h$). Without both, one impulse response does not describe the system.
Flip, slide, multiply, integrate
- Flip $h$ to get $h(-\tau)$, slide it to $t$, multiply by $x(\tau)$, and take the area of the overlap. That area is $y(t)$.
- Repeat for each $t$. When both signals start at 0, the overlap runs from $\tau = 0$ to $\tau = t$, and those are the limits.
Put the steps for computing y(t) at one value of t in order.
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Flip, slide, multiply, integrate. Repeat for each $t$ to build the whole output.
Worked example: a step into a decay
- $x(t) = u(t)$ into $h(t) = e^{-t}u(t)$. For $t \ge 0$: $y(t) = \displaystyle\int_{0}^{t} e^{-(t - \tau)}\,d\tau = 1 - e^{-t}$.
- Check the ends: $y(0) = 0$, since no input has arrived yet, and $y(\infty) = 1$, the DC gain $\int h\,dt$. At $t = 1.5$, $y = 0.777$.
A unit step into h(t) = e^(−t)u(t) gives, for t ≥ 0,
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$\int_{0}^{t} e^{-(t - \tau)}\,d\tau = 1 - e^{-t}$. It starts at 0 and settles at the DC gain 1.
For that step response, what is y(2)? Give 3 decimal places.
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$1 - e^{-2} = 1 - 0.135 = 0.865$.
Worked example: a 2-second pulse
- A pulse of height 1 from $t = 0$ to $t = 2$ is a step on at 0 minus a step on at 2. So the output is the step response $1 - e^{-t}$, minus the same response started at $t = 2$.
- At $t = 2$ the output peaks at $1 - e^{-2} = 0.865$. At $t = 3$ it has fallen to $e^{-1} - e^{-3} = 0.318$: the release tail.
A unit pulse lasting from t = 0 to t = 2 enters h(t) = e^(−t)u(t). What is y(3)?
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Step on at 0 minus step on at 2: $(1 - e^{-3}) - (1 - e^{-1}) = e^{-1} - e^{-3} = 0.318$, the release tail.
Properties worth knowing
- Convolution is commutative: $x * h = h * x$. Two systems in cascade have impulse response $h_{1} * h_{2}$.
- Widths add: a pulse of width $W_{1}$ convolved with one of width $W_{2}$ lasts $W_{1} + W_{2}$.
A 3 ms pulse is convolved with a 5 ms pulse. How long is the result?
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Widths add: the overlap is non-zero from when the pulses first touch until they fully separate, $3 + 5 = 8$ ms.
Three common slips
- Forgetting the flip. Convolution uses $h(t - \tau)$, not $h(\tau)$.
- Wrong limits. The overlap window, not $-\infty$ to $\infty$, sets the limits once the signals start at 0.
- Multiplying in time. $y \ne x \cdot h$; multiplication belongs to the frequency domain.
The output of an LTI system is the input multiplied by the impulse response, y(t) = x(t)h(t).
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In time the output is the convolution $x * h$. Multiplication, $Y = XH$, happens in the frequency domain.
Recap
- LTI systems are described completely by $h(t)$.
- $y = x * h$: flip, slide, multiply, integrate.
- Check the end points; cascades convolve; widths add.