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Let us say that we are interested in finding the voltage (potential difference) $y$ across the resistor.

circuit

The circuit consists of a battery, a resistor and an inductor. The problem can be solved by following the Laplace transform "recipe".

$$ -U + L\left(s\left(\frac{Y}{R}\right) -i(0)\right) + Y = 0 \implies \\ \implies Y = \frac{U + Li(0)}{1+ \frac{Ls}{R}} \implies \\ \implies y(t) = \frac{R}{L}\int_0^te^{-\tau/\frac{R}{L}}u(t-\tau)d\tau + Ri(0)e^{-t/\frac{R}{L}} $$

Having gotten this answer I am not sure of how to interpret the convolution. I don't mean that I wouldn't know how to evaluate the integral but rather that I don't understand why it takes on the form it does. Is it adding up contributions from earlier emf:s from the battery?

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  • $\begingroup$ Yes, what you said about adding past influences is exactly right. $\endgroup$
    – DanielSank
    Sep 6, 2015 at 22:02
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    $\begingroup$ It's quite enlightening to work out the link between this convolution business, which is also called "Green's functions", and matrices. $\endgroup$
    – DanielSank
    Sep 6, 2015 at 23:26

3 Answers 3

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I'll try to rearrange the problem: we want to calculate the difference of potential $y(t)$ using a Fourier transform and the convolution theorem. We can now write the Kirchoff equation for the circuit:

$$y(t)=U(t)-L\frac{dI}{dt}(t)$$

and then we can apply the Fourier transform on the two sides of the equation obtaining this (using the tilde symbol for the transform):

$$\tilde{Y}(\omega)=\tilde{U}(\omega)-i\omega L \tilde{I}(\omega)$$

Here we have applied the Fourier rule for the derivative of a function. Note that now we have a simple equation in $\tilde{I}$ since we can write $\tilde{Y}(\omega)=R\tilde{I}(\omega)$. Solving the equation for $\tilde{I}$ leads to a simple expression for $\tilde{Y}(\omega)$:

$$\tilde{Y}(\omega)=\frac{R\tilde{U}(\omega)}{R+i\omega L }$$

We can now use the convolution theorem and split out the previous equation in two terms $h(t)$ and $g(t)$ in order to have $\tilde{Y}(\omega)=F[h(t)*g(t)]=\sqrt{2\pi}F[h(t)]F[g(t)]=\sqrt{2\pi}\tilde{H}(\omega)\tilde{G}(\omega)$:

$$\tilde{Y}(\omega)=\sqrt{2\pi}\left[\frac{1}{\sqrt{2\pi}}\frac{1}{R+i\omega L}\right][R\tilde{U}(\omega)]$$

If we calculate the antitrasform of the two functions in the brackets, then we have a convolution integral for $y(t)$. We know that the antitrasform of $R\tilde{U}(\omega)$ will be $RU(t)$, so we can only calculate one antitrasform:

$$F^{-1}[\tilde{H}(\omega)]=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{R+i\omega L}$$

Now we must choose a suitable closed path of integration on which we can apply the Cauchy theorem. We can for example use the half circumference in the Argand-Gauss plan that lies on the positive part of the ordinates axis and then complete the path with a straight line on the abscissa's axis. Inside this integration path we have a pole of order 1 (namely $\omega=i\frac{R}{L}$). Using the Jordan's lemma we can say that the integral over the circumference is 0 and so we only have to calculate the residue in $\omega=i\frac{R}{L}$. Finally we conclude that

$$y(t)=\frac{1}{\tau}\int_{-\infty}^{+\infty} e^{-(t-t')/\tau}U(t')dt'$$

where $\tau=L/R$. This is the convolution of a given input signal $U(t)$ and the function $G(t)=\frac{1}{\tau}e^{-t/\tau}$. In particular $G(t)$ is referred to as the system's transfer function or gain. What's the meaning of $G(t)$? It represents the impulsive response of the system because we have $y(t)=G(t)$ for a delta-like input $U(t)=\delta(t)$. What's the meaning of the convolution? If we have many inputs that follow one another, the overall output must be the sum of the responses of the systems, in fact, linear systems like RL circuits obey to the superposition principle. For example, if we have two delta-like inputs that follow one another, we must sum the two exponential responses to obtain $y(t)$. Note that in this case, if the delta inputs are near in time (the time interval must be $<\tau$), the maximum value for $y(t)$ is reached when the second delta impulse is applied to the system.

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The way to understand the convolution operation intuitively is to look at the differential form. $dy(u) = f(u)g(t-u)du$ (we will add up $dy(u)$ from $0$ to $t$ to get $y(t)$). At any moment $u\ (0\le u\le t)$, an amount of some quantity $f(u)$ is produced, then it undergoes a change according to function $g$ from that moment until $t$ ($u$ reaches $t$). Attempt to visualize this:

enter image description here

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To add to the mathematical explanations given in the other answers, physically speaking, an inductor is a device that resists changes in current, but which gradually allows current to change as time goes by if needed. That is, if a sudden change is applied, the current going through the inductor will gradually change towards the desired change. If a potential is increased across the inductor, the current through the inductor will increase slowly, and if the potential is reduced, the current will decrease slowly. Therefore, as time goes by, the influence of a past changes will diminish (hence the exponential term multiplied by time shifted external forcing). Note that if additional changes happen in time at successive moments, if any, such changes will be added together (as the inductor is a linear device and obeys the principle of superposition). Therefore you are right, convolution is used as you add contributions from earlier voltage forcing (here your battery) to see what the current current is (which gives you potential across the resistor), with a time-dependent exponential reduction.

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