Sometimes a Laplace transform ,H(s), can be identified as the product of two other transforms, F(s) and G(s). The transforms F(s) and G(s) correspond to the functions f(t) and g(t) respectively. Although ordinary multiplication is not applicable to the Laplace transform, a "generalized product" can be used. $$ Given: $$ $$ \mathcal{L}[f(t)] = F(s) \;, \;\;\;\;\;\mathcal{L}[g(t)] = G(s) $$ $$ Then: $$ $$ H(s) = F(s)G(s) = \mathcal{L}[h(t)] $$
$$ ...where: $$ $$ \begin{align} h(t) &= \int_{0}^{t}f(t-\tau)g(\tau) \;d\tau = \int_{0}^{t}f(\tau) g(t-\tau) \;d\tau \\\\ &= (f*g)(t) = (g*f)(t) \\\\ &= the \; convolution \; integral \end{align} $$
The asterix symbol used above is a shorthand way of thinking of the convolution integral as a "generalized product" and represents the convolution of the terms to the left and right of it. (f*g)(t) is referred to as the "convolution of f and g" and is a factor of "t". The integrals expressed above are known as the "Convolution integral". $$ Additionally: $$ $$ (f*g)(t) = \mathcal{L^{-1}}[F(s)G(s)] $$ The convolution integral can be used to find the response to a signal, h(t), given and excitation, f(t), and knowing the impulse response, g(t), of the system.
Essentially, the convolution of two signals consists of time-reversing one of the signals, shifting it and multiplying it point by point with the second signal and then integrating the product. The convolution integral also allows you to write the solution of an initial value problem in terms of an integral.
Convolution integrals arise in various applications in which the behavior of a system at time "t" depends not only on its state at time "t" but also on its past history
Properties of the convolution integral:
$$ 1) \;\; f*g = g*f \;\;\;,(commutative \; law) $$ $$ 2) \;\; f*(g_1 + g_2) = f*g_1 + f*g_2 \;\;\;, (distributive \; law) $$ $$ 3) \;\; (f*g)*h = f*(g*h) \;\;\;, (associative \; law) $$ $$ 4) \;\; f*0 = 0*f = 0 $$ For property #4, zeros do not denote the number "0", but the function that has the value "0" for each value of "t". $$ 5) \;\; f(t)*\delta(t) = \int_{-\infty}^{\infty}f(\tau) \; \delta(t-\tau) d\tau = f(t) $$ $$ \quad...where\;\delta(t) = unit\;impulse\;function $$ $$ 6) \;\; f(t)*\delta(t-t_0) = f(t-t_0) $$ $$ 7) \;\; f(t)*\delta '(t) = \int_{-\infty}^{\infty}f(\tau) \; \delta ' (t-\tau) d\tau = f'(t) $$ $$ 8) \;\; f(t)*u(t) = \int_{-\infty}^{\infty}f(\tau) \; u(t-\tau) d\tau = \int_{-\infty}^{t} f(\tau) d\tau $$ $$ \quad...where\;u(t) = unit\;step\;function $$ $$ 9) \;\; F_1(s)F_2(s) = \mathcal{L} [f_1(t)*f_2(t)] $$
Example:
Given: $$ f(t) = \sin(t) \;\;, \qquad g(t) = \cos(t) $$ Determine: $$ (f*g)(t) $$ Utilizing the definition of the Convolution Integral, we proceed as follows:
$$ \begin{align} (f*g)(t) &= h(t) \\\\ &= \int_{0}^{t}f(t-\tau)g(\tau) \;d\tau \\\\ &= \int_{0}^{t} \sin(t-\tau)\cos(\tau) d\tau \qquad,\;(Eqn\;1) \\\\ \end{align} $$
Recall the following trig identity: $$ \sin(A\pm B) = \sin(A) \cos(B) \pm \cos(A) \sin(B) $$ ...and apply it to the sine term in equation #1:
$$ \begin{align} (f*g)(t) &= \int_{0}^{t}\big[(\sin(t) \cos(\tau) - \cos(t) \sin(\tau)) \cos(\tau)\big] d\tau \\\\ &= \int_{0}^{t}\big[ \sin(t) \cos^2(\tau) - \cos(t)\sin(\tau)\cos(\tau) \big] d\tau \\\\ &= \int_{0}^{t} \sin(t) \cos^2(\tau) d\tau - \int_{0}^{t} \cos(t)\sin(\tau)\cos(\tau) d\tau \\\\ &= \sin(t) \int_{0}^{t} \cos^2(\tau) d\tau - \cos(t) \int_{0}^{t} \sin(\tau)\cos(\tau) d\tau \\\\ \end{align} $$
Recall the following power identity for the cosine function: $$ \cos^2(\theta) = \frac{1}{2}[1+\cos(2\theta)] $$ ...and apply it to the squared cosine term of the previous expression:
$$ (f*g)(t) = \frac{1}{2}\sin(t) \int_{0}^{t} [1+\cos(2\tau)] d\tau - \cos(t) \int_{0}^{t} \sin(\tau)\cos(\tau) d\tau \qquad,\;(Eqn\;2) $$
We will now use "u-substitution" to evaluate both integrals in equation #2:
$$ \int_{0}^{t} [1+\cos(2\tau)] d\tau = \int_{0}^{t} 1 d\tau + \int_{0}^{t} \cos(2\tau) d\tau $$ let: $$ u = 2\tau $$ $$ du = 2 d\tau $$ $$ d\tau = \frac{du}{2} $$ Now: $$ \begin{align} \int_{0}^{t} 1 d\tau + \int_{0}^{t} \cos(2\tau) d\tau &= \tau + \int_{0}^{t} \cos(u) \frac{du}{2} \\\\ &= \tau + \int_{0}^{t} \frac{1}{2} \cos(u) du \\\\ &= \tau + \frac{1}{2}\sin(u) \Big|^{t}_{0} \\\\ Now \; back \; substitute \\\\ \int_{0}^{t} 1 d\tau + \int_{0}^{t} \cos(2\tau) d\tau &= \big[ \tau + \frac{1}{2}\sin(2\tau) \big] \Big|^{t}_{0} \qquad,\;(Eqn\;3) \end{align} $$
$$ \int_{0}^{t} \sin(\tau)\cos(\tau) d\tau $$ let: $$ u = \sin(\tau) $$ $$ du = \cos(\tau) d\tau $$ $$ d\tau = \frac{du}{\cos(\tau)} $$ Now: $$ \begin{align} \int_{0}^{t} \sin(\tau)\cos(\tau) d\tau &= \int_{0}^{t} u \cancel{\cos(\tau)} \frac{du}{\cancel{\cos(\tau)}} \\\\ &= \int_{0}^{t} u \; du \\\\ &= \frac{1}{2} u^2 \Big|^{t}_{0} \\\\ Now \; back \; substitute \\\\ &= \big[ \frac{1}{2} \sin^2(\tau) \big] \Big|^{t}_{0} \qquad,\;(Eqn\;4) \end{align} $$
Now substitute equations #3 and #4 into equation #2:
$$ \begin{align} (f*g)(t) &= \frac{1}{2}\sin(t) \big[ \tau + \frac{1}{2}\sin(2\tau) \big] \Big|^{t}_{0} - \cos(t) \big[ \frac{1}{2} \sin^2(\tau) \big] \Big|^{t}_{0} \\\\ &= \frac{1}{2}\sin(t) \big[ t + \frac{1}{2}\sin(2t) \big] - \cos(t) \big[ \frac{1}{2} \sin^2(t) \big] \\\\ &= \frac{t}{2}\sin(t) + \frac{1}{4}\sin(t)\sin(2t) - \frac{1}{2}\cos(t)\sin^2(t) \\\\ Recall \; the \; following \; trig \; identity: \\\\ \sin(2t) = 2\sin(t)\cos(t) \\\\ (f*g)(t) &= \frac{t}{2}\sin(t) + \frac{1}{4}\sin(t)2\sin(t)\cos(t) - \frac{1}{2}\cos(t)\sin^2(t) \\\\ &= \frac{t}{2}\sin(t) + \cancel{\frac{1}{2}\sin^2(t)\cos(t)} - \cancel{\frac{1}{2}\cos(t)\sin^2(t)} \\\\ \end{align} $$
$$ (f*g)(t) = \frac{t}{2}\sin(t) $$
Graphical technique used to determine convolution integral
Given two signals, we can also use a graphical approach to find the convolution.
Steps to evaluate the convolution integral graphically
$$ \begin{align} 1) \; &Folding: \; Reflect \; f(\tau) \; about \; the \; vertical \; axis \; and \; obtain \; \\ &f(-\tau). \\\\ 2) \; &Displacement: \; Shift \; (or \; delay) \; f(\tau) \; by \; "t" \; to \; obtain \\ &f(t-\tau). \\\\ 3) \; &Multiplication: \; Determine \; the \; product \; of \; f(t-\tau) \; and \; g(\tau). \\\\ 4) \; &Integration: \; For \; a \; given \; time \; "t", \; calculate \; the \; area \\ &under \; the \; product \; \big[f(t-\tau)g(\tau)\big] \; for \; 0 \lt \tau \lt t \; to \\ &get \; h(t) \; at \; "t". \end{align} $$
In the following examples, these steps will hopefully become more clear when they are visually put into proactice.
Continue on to example problem #1 involving the graphical approach to solving the convolution integral...