LTI systems and convolution

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:

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.

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 $t = 1.5$, $y = 0.777$.

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.

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 $W_{1} + W_{2}$.

Three common slips

Recap