Solving Impulse Response of RL Circuit

[riyaad_ali]

Hi all,

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

Homework Statement

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.

Homework 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)

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

 

[rude man]

Have you had the Laplace transform?

 

[riyaad_ali]

No, we haven't studied Laplace transforms yet, so I was hoping there would be a way to solve without it.

 

[rude man]

You know those ads on TV that say, "don't try this at home"? I say, "don't try to solve differential equations with Dirac delta (aka impulse) function excitations without a transform". The logical one here is the Laplace.

Maybe some math wiz will show us how to do it the classical, or another, way.

EDIT: I'm surprised that a 4th yr course would not have covered the Laplace or similar (e.g. heaviside) xfrm.

FYI my answer is v_L = k{δ(t) - (1/τ)e^-t/τ)} for a unit impulse input kδ(t) where k = 1.0 V-sec and τ = L/R.

(The units of δ(t) are sec^-1.)

 

[DeeNos]

Dear Riyaad,

Thank you for sharing your question with the community. It is great that you are taking a time-series analysis course for Physics students and exploring the impulse response of an RL circuit. Here is a step-by-step solution to help you understand the derivation of the impulse response:

1. Start with the basic circuit equation: V(t) = R*I + L*dI/dt

2. Substitute in the input potential V(t) = δ(t) to get: δ(t) = R*I + L*dI/dt

3. Since we are interested in the potential across the inductor, we can rearrange the equation to get: V_(t) = L*dI/dt = δ(t) - R*I

4. Next, we can use the definition of the delta function: δ(t) = dH(t)/dt to rewrite the equation as: V_(t) = L*dI/dt = dH(t)/dt - R*I

5. Integrate both sides with respect to time to get: ∫V_(t)dt = ∫dH(t)/dt dt - R∫I dt

6. The left side becomes: V_(t) = ∫V_(t)dt = V(t)

7. The first integral on the right side becomes: ∫dH(t)/dt dt = H(t) + C, where C is a constant of integration.

8. The second integral on the right side becomes: ∫I dt = I(t) + C, where C is another constant of integration.

9. Substituting these values back into the equation, we get: V(t) = H(t) + C - R*(I(t) + C)

10. Since we are assuming no initial current (I=0 at t=0), we can set C=0 and simplify the equation to: V_(t) = H(t) - R*I(t)

11. Finally, substitute in the expression for current I(t) = (1/R)*[V(t) - V_(t)] to get the final form of the impulse response: V_(t) = δ(t) - (R/L)*e^{(-R/L)*t}*H(t)

I hope this helps you understand the derivation of the impulse response and its relationship to the step response. Keep up the good work in your course and good luck with your future studies in time-series analysis