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 — 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.

For the related concepts of system response and transfer functions, see Transfer function and Impulse response. For the bilateral version used in signals-and-systems analysis, see the LTI form above and Convolution properties. For the graphical method, see Graphical convolution. For the frequency-domain dual (convolution becomes multiplication), see Convolution theorem.