## A little segue into engineering mode

A homework problem in a physics course I’m taking this term contained the following question, paraphrased:

A resistor with resistance $R$ and an inductor with inductance $L$ are connected in series. Suppose that we apply a voltage $a(t)$ to this circuit. Find the voltage across the inductor $b(t)$ in the following cases:

1. the input voltage is given by the Heaviside step function $H(t)$ (in this case, the output voltage is called the step response of the circuit);
2. the input voltage is given by the Dirac delta function $\delta(t)$ (in this case, the output voltage is called the impulse response of the circuit).

First of all (and I know this is soft), let’s try to imagine what the step response looks like, without solving any equations. We know that for all $t < 0$, the voltage across the inductor must be zero because the voltage source has been off for an infinite amount of time (loosely speaking). At time $t = 0$, we switch on the voltage to some fixed value. (The problem as stated in its original form is dimensionally incorrect, but I'll just gloss over that for now.) If the circuit contained just a resistive element, then the current would jump to $V/R$ and stay that way as $t \to \infty$. However, because the circuit contains an inductive element opposing the change in current, the current must instead build up to this value continuously. It is further reasonable to imagine that the current increases quickly initially (because it is much less than the equilibrium value $V/R$) but then its increase slows. So a guess (which turns out to be right) is that the value $V/R - i(t)$ decays exponentially for $t \geq 0$. We can then obtain the inductor voltage as $Li'$.

The form of the impulse response is even simpler to guess. The current must again be zero for all $t < 0$. At $t = 0$ a very high voltage is suddenly applied for a very short period of time, which causes current to start flowing in the circuit. Immediately the voltage is switched off, which means that the resistor starts to dissipate the energy in the circuit. Ultimately the current must die down to zero because of this. However, the inductor tries to keep the current flowing, opposing the resistor. Initially the resistor is quite powerful, since the rate at which it dissipates energy is greater when the current is greater. Later, when the current is smaller, the decrease in current starts to slow. It is reasonable to imagine that the current in the circuit actually decays exponentially for $t > 0$, that is, $i(t) = V_{t=0} H(t) e^{-kt}$ for some positive constant $k$ with units of reciprocal time. This, too, is the correct answer.

Now for the math. Applying Kirchhoff's voltage law to the circuit yields
$Ri + Li' = VH(t)$
where $V$ is a constant with units of voltage added to make the equation dimensionally correct.

If you go online and search for "RL step response" or something like that, you'll see a solution that proceeds something like this. We rewrite the equation first as:
$i' = \frac{V}{L}H(t) - \frac{R}{L}i$
Then we "separate variables":
$\displaystyle \frac{di}{(V/L)H(t) - (R/L)i} = dt$
Then integrate both sides:

Wait, what?

Okay, well, I know this works in this case, because they wouldn’t be publishing it if it gave an answer known to be wrong. But this is bad math, because the variables have not been separated! $H(t)$ is a function of $t$, and yet we’ve got it on the $i$ side rather than the $t$ side. (Actually, I’ve noticed that instead of denoting the voltage $VH(t)$ they use $V_s$, so it’s easy to miss the fact that it’s a function of $t$.)

I decided to share what I wrote instead on my homework. I’m too lazy to transfer all the LaTeX over here, and besides, WordPress probably doesn’t support the environments I used. So just click here for extracted pages of the PDF in my solutions.

The cautious reader will notice that my solutions are not really that rigorous. The ODEs involved suffer from having discontinuous coefficients, so the Picard–Lindelöf theorem doesn’t apply and we are on shaky theoretical ground. Furthermore, even though I am a physics major I refuse to pretend that the Dirac delta really is a function. It’s not! But you can treat it as a distribution, which is a generalization of a function. The rigorously correct way to derive the step response, for example, is to solve the ODE separately in the regions $t < 0$ and $t \geq 0$ and then choose the constant to make the solution continuous at $t = 0$. The reason why I didn't write out fully rigorous solutions is that I wanted to show how to work with the Dirac delta and Heaviside step function, since they can be very confusing when you're seeing them for the first time.

Curiously, the initial spike in the impulse response has a magnitude independent of the resistance $R$. Intuitively, we can say that since the voltage is changing suddenly, the behaviour of the circuit is dominated by the inductive element.

Hi! I'm Brian Bi. As of November 2014 I live in Sunnyvale, California, USA and I'm a software engineer at Google. Besides code, I also like math, physics, chemistry, and some other miscellaneous things.
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### 2 Responses to A little segue into engineering mode

1. synagogue visitor says:

what’s wrong with using laplace transforms

2. Brian says:

I haven’t learned Laplace transforms.