Impulse response

The impulse response $h(t)$ of a linear time-invariant (LTI) system is the system’s output when the input is a Dirac delta $\delta (t)$ and the system is in its zero state (no stored energy at $t=-\infty$). Equivalently, it’s the inverse Laplace transform of the system’s Transfer function $H(s)$:

$h(t):={\mathcal{L}}^{-1}\{H(s)\}(t)$

For a second-order linear ODE $a{y}^{\prime \prime }+b{y}^{\prime }+cy=g(t)$ with zero initial conditions, the transfer function is $H(s)=1/(a{s}^2+bs+c)$, and the impulse response is

$h(t)={\mathcal{L}}^{-1}\left\{\frac{1}{a{s}^2+bs+c}\right\}(t)$

Why it matters

Once you know $h(t)$, you can compute the system’s response to any input via convolution:

$y(t)=(h\ast g)(t)={\int }_0^{t}h(t-\tau )g(\tau )d\tau$

This is the convolution form of the input-output relationship. The impulse response captures everything about how the system responds, the same way a transfer function does — they’re Laplace pairs.

Why “impulse”

If you input $g(t)=\delta (t)$ (a Dirac delta — an idealized instantaneous impulse), then $G(s)=\mathcal{L}\{\delta (t)\}=1$, so $Y(s)=H(s)\cdot 1=H(s)$, and $y(t)={\mathcal{L}}^{-1}\{H(s)\}=h(t)$.

So the impulse response is, literally, the system’s output when “kicked” with an idealized impulse.

Worked example

For the damped oscillator $y^{\prime \prime }+2y^{\prime }+5y=g(t)$:

Transfer function: $H(s)=\frac{1}{s^2+2s+5}$.

Complete the square: $s^2+2s+5=(s+1{)}^2+4$.

So $H(s)=\frac{1}{(s+1{)}^2+4}=\frac{1}{2}\cdot \frac{2}{(s+1{)}^2+4}$.

Inverse Laplace using the shifted-sine form: ${\mathcal{L}}^{-1}\{2/((s+1{)}^2+4)\}=e^{-t}\sin (2t)$.

So $h(t)=\frac{1}{2}{e}^{-t}\sin (2t)$.

This is the system’s response to an impulse at $t=0$. The factor $e^{-t}$ is the damping; the $\sin (2t)$ is the underdamped oscillation.

In context

Two equivalent ways to compute system response:

1. s-domain: $Y(s)=H(s)G(s)$, then inverse Laplace.
2. Time domain: $y(t)=(h\ast g)(t)$ via convolution.

Both give the same answer. Method 1 is usually faster algebraically; method 2 is useful when $G(s)$ is unwieldy or unknown in closed form.

The impulse response also makes physical sense: a complex system “smears” each input over time according to its impulse response. A short input gets stretched out (if the system is sluggish); a long input gets averaged (if the system has limited bandwidth).

Stability via impulse response

A system is bounded-input-bounded-output (BIBO) stable iff the impulse response is absolutely integrable. For a causal system (the standard case here, where $h(t)=0$ for $t<0$):

${\int }_0^{\infty }|h(t)|dt<\infty$

The general statement uses ${\int }_{-\infty }^{\infty }$, but for any causal LTI system the integrand is zero for negative $t$, so the bound only needs to hold on $[0,\infty )$. Use the lower limit that matches the system you’re analyzing.

For our LTI systems, this holds iff all poles of $H(s)$ have strictly negative real parts.

Nonzero initial conditions

The transfer function and impulse response are defined for zero initial conditions. For nonzero initial conditions, the full solution is

$y(t)=(h\ast g)(t)+y_k(t)$

where the second term is the homogeneous solution that handles the initial conditions:

$a{y}_k^{\prime \prime }+b{y}_k^{\prime }+c{y}_k=0,y_k(0)=Y_0,y_k^{\prime }(0)=Y_1$

Compute $y_k$ via the Characteristic equation separately, then add to the convolution result.

For example: $y^{\prime \prime }+y=t$, $y(0)=2$, $y^{\prime }(0)=-1$.

Particular (zero IC): Laplace gives $Y(s)=\frac{1}{s^2(s^2+1)}$. Inverse: $y_p(t)=t-\sin t$.

Homogeneous (nonzero IC): general $y_k=A\cos t+B\sin t$. Apply $y_k(0)=2$, $y_k^{\prime }(0)=-1$: $A=2$, $B=-1$. So $y_k=2\cos t-\sin t$.

Full: $y(t)=(t-\sin t)+(2\cos t-\sin t)=t+2\cos t-2\sin t$.

For the s-domain perspective on system characterization, see Transfer function. For the convolution operation that lets you apply $h$ to general inputs, see Convolution integral.

Finding h(t) from a differential equation

For an LTI system defined by a linear constant-coefficient ODE

$a_n{y}^{(n)}+\cdots +a_{0}y=b_m{x}^{(m)}+\cdots +b_{0}x,$

the impulse response is what comes out when $x(t)=\delta (t)$. The technique splits into three time regions:

• $t<0$: $h(t)=0$ (causal system, zero input before the impulse). Also enforces causality.
• $t>0$: $h$ satisfies the homogeneous equation. Find the characteristic roots $s_k$; write $h(t)=\sum K_k{e}^{s_{k}t}$ for $t>0$.
• $t=0$: integrate the ODE across the singularity (from $0^{-}$ to $0^{+}$). The integrals of bounded functions are zero; the integral of $\delta$ is 1; the integral of ${\delta }^{\prime }$ is the jump in $\delta$, etc. This gives equations for the unknown constants.

Whether $h(t)$ contains an impulse at $t=0$ depends on the relative orders $m$ vs $n$: $m<n$ means no impulse component (Case 1); $m=n$ means an impulse at $t=0$ with some strength to determine (Case 2); $m>n$ is rare in practice.

This is the time-domain way to find $h(t)$. The s-domain approach is usually faster: write the transfer function $H(s)$ algebraically and inverse-transform — see Transfer function and Partial fraction decomposition.

Causality and stability from h(t)

For an LTI system, two of its properties can be read directly off the impulse response:

• Causal: $h(t)=0$ for $t<0$.
• BIBO stable: ${\int }_{-\infty }^{\infty }|h(t)|dt<\infty$.

A causal signal with a $u(t)$ factor is automatically the impulse response of a causal system. A signal that decays fast enough at $\pm \infty$ is the impulse response of a stable system.