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I'm taking a fourth year time-series analysis course for Physics students.

1. The problem statement, all variables and given/known data

Let us consider a simple physical system consisting of a resistor (with resistance R) and an inductor (with inductance L) in series. We apply an input voltage a(t) across the pair in series, and measure the output voltage b(t) across the inductor alone.

We are asked to show analytically what the step and impulse responses would be.

2. Relevant equations

Kirchhoff's equations

H(t) = step or Heaveside function

δ(t) = delta function

[itex]\frac{dH(t)}{dt}[/itex] = δ(t)

Let

V(t) = source potential

V_(t) = potential across inductor

Assume

I = 0 at t = 0 (no initial current)

3. The attempt at a solution

I was able to use the basic circuit equation:

V(t) = R*I + L*[itex]\frac{dI}{dt}[/itex]

To solve for the current across the inductor:

I(t) = [itex]\frac{V(t)}{R}[/itex]-[itex]\frac{V(t)}{R}[/itex]*e[itex]^{(-R/L)*t}[/itex]

Now setting V(t) = H(t) to find the step response, and knowing that:

L*[itex]\frac{dI(t)}{dt}[/itex] = V_(t)

I found that:

V_(t) = H(t)*e[itex]^{(-R/L)*t}[/itex]

...which is the correct step response. However, my problem is in deriving the impulse response. When I try use the current I(t) and substitute in δ(t) for the input potential V(t), my derivation falls apart. I'm not used to working with the δ(t) function, so this is a bit tricky for me. I've since found that the impulse response is:

V_(t) = δ(t) - (R/L)*e[itex]^{(-R/L)*t}[/itex]*H(t)

...but have no idea how to get to this point. Clearly, the impulse response is the time derivative of the step response, but is this just coincidental?

Any help would be much appreciated. Thanks!

Riyaad

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# Solving Impulse Response of RL Circuit

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