Convolution integral

The convolution of two functions $f(t)$ and $g(t)$ is

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

(For functions defined on $[0,\infty )$. The general definition integrates over $\mathbb{R}$ if the functions are defined there.)

The convolution combines two functions into a third by sliding one over the other and integrating the product. Despite looking complicated, it has a simple meaning in the s-domain: products of transforms.

Bilateral (LTI) form

For continuous-time LTI systems on the whole real line, the convolution integrates over all $\tau$:

$y(t)=(x\ast h)(t)={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau .$

This is the form used in Fourier-transform analysis and in the Sampling derivations. For causal signals (zero for $t<0$), the integrand is nonzero only for $0<\tau <t$, recovering the one-sided form above.

The integral picks out the structure of LTI: the impulse response $h$ is flipped and shifted, multiplied by the input $x$, and the area is the output at time $t$. The flip-and-slide method is the picture of this operation.

This is the second formula you have to memorize in Continuous-Time Signals and Systems (after the impulse-response definition). It is not on the formula sheet, so make sure it sticks.

Convolution theorem

The fundamental property:

$\mathcal{L}\{f\ast g\}(s)=F(s)G(s)$

Convolutions in time correspond to products in the s-domain. Equivalently:

${\mathcal{L}}^{-1}\{F(s)G(s)\}(t)=(f\ast g)(t)$

This lets you invert products of Laplace transforms by computing convolutions in time. Very useful when neither factor is on a standard table.

Note: ${\mathcal{L}}^{-1}\{F(s)G(s)\}\ne f(t)\cdot g(t)$. Convolution is not pointwise multiplication.

Why it’s useful for ODEs

Consider the IVP $y^{\prime \prime }+y=g(t)$, $y(0)=0$, $y^{\prime }(0)=0$. Take Laplace:

$(s^2+1)Y(s)=G(s)\Rightarrow Y(s)=\frac{1}{s^2+1}\cdot G(s)$

The first factor inverts to $\sin t$. The second is whatever $G(s)$ is. By the convolution theorem:

$y(t)={\mathcal{L}}^{-1}\{Y(s)\}=\sin t\ast g(t)={\int }_0^t\sin (t-\tau )g(\tau )d\tau$

So you can write the solution explicitly without knowing $\mathcal{L}\{g\}$ in closed form. Plug in any $g$ and integrate. Particularly useful when $g$ is given numerically or by a complicated formula.

Worked example: convolution

Compute $f\ast g$ where $f(t)=t$ and $g(t)=t^2$.

$(f\ast g)(t)={\int }_0^t(t-\tau ){\tau }^{2}d\tau ={\int }_0^t(t{\tau }^2-{\tau }^3)d\tau =\frac{t\cdot t^3}{3}-\frac{t^4}{4}=\frac{t^4}{12}$

Note: $f\cdot g=t\cdot t^2=t^3$, a different function. Convolution and multiplication are not the same.

Properties

The convolution operation is:

1. Commutative: $f\ast g=g\ast f$.
2. Distributive: $f\ast (g+h)=f\ast g+f\ast h$.
3. Associative: $(f\ast g)\ast h=f\ast (g\ast h)$.
4. Has identity: $f\ast \delta =f$, where $\delta$ is the Dirac delta function — the delta is the convolution identity. The cleanest statement uses the two-sided convolution: ${\int }_{-\infty }^{\infty }f(t-\tau )\delta (\tau )d\tau =f(t)$ by the sifting property. For the one-sided form ${\int }_0^{t}f(t-\tau )\delta (\tau )d\tau$ used here, the result depends on whether the integration lower limit is taken as $0^{-}$ or $0^{+}$. The standard engineering convention (matching the Laplace transform) uses $0^{-}$, so the delta at $\tau =0$ is fully inside the interval and $f\ast \delta =f$ unambiguously.
5. Zero: $f\ast 0=0$.

The first three mirror the s-domain product: $FG=GF$, $F(G+H)=FG+FH$, $(FG)H=F(GH)$. Algebra in the s-domain is much cleaner than in time.

Worked example: inverse via convolution

Find ${\mathcal{L}}^{-1}\left\{\frac{1}{(s-a)(s-b)}\right\}$.

Decompose: ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$ and ${\mathcal{L}}^{-1}\{1/(s-b)\}=e^{bt}$. By the convolution theorem:

${\mathcal{L}}^{-1}\left\{\frac{1}{(s-a)(s-b)}\right\}=e^{at}\ast e^{bt}={\int }_0^t{e}^{a(t-\tau )}{e}^{b\tau }d\tau =e^{at}{\int }_0^t{e}^{(b-a)\tau }d\tau$

For $a\ne b$:

$=e^{at}\cdot \frac{e^{(b-a)t}-1}{b-a}=\frac{e^{bt}-e^{at}}{b-a}$

You could also do this by partial fractions and reach the same answer.

Worked example 2: $\frac{1}{s^2(s^2+a^2)}$

Decompose: $\frac{1}{s^2(s^2+a^2)}=\frac{1}{a}\cdot \frac{a}{s^2(s^2+a^2)}=\frac{1}{a}\cdot \frac{1}{s^2}\cdot \frac{a}{s^2+a^2}$.

So ${\mathcal{L}}^{-1}\{1/s^2\}=t$ and ${\mathcal{L}}^{-1}\{a/(s^2+a^2)\}=\sin (at)$. Convolution:

${\mathcal{L}}^{-1}\left\{\frac{1}{s^2(s^2+a^2)}\right\}=\frac{1}{a}(t\ast \sin (at))=\frac{1}{a}{\int }_0^t(t-\tau )\sin (a\tau )d\tau$

Computing this integral by parts and simplifying yields $\frac{at-\sin (at)}{a^3}$.

Convolution as system response

For a linear time-invariant system with impulse response $h(t)$, the output for any input $g(t)$ is

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

The impulse response $h(t)$ is exactly the inverse Laplace of the system’s Transfer function $H(s)=1/(a{s}^2+bs+c)$.

So convolution is the time-domain way to compute system response. Equivalent to: take Laplace, multiply by $H(s)$, inverse-Laplace. But sometimes it’s easier to evaluate the convolution integral directly.

See Convolution properties for the full algebra.