# Impulse response of an RL circuit

I've got homework in the Signals & Systems class I'm taking, and I'm trying to find the impulse response to this circuit whose transfer function I've found. In the class we've done practice problems, but we've had numerical values for the resistors, capacitors and inductors.

I don't know how to approach the impulse response without having values for the components. I've tried to give common factor, but I didn't reach anything. Can you give me any advice on how to get the impulse response?

• Are you aware that the TF is the laplace of the impulse response? Maybe I'm misinterpreting your question? Please use lower case formal s instead of the handwriting form of s you have used. I'm assuming you mean s as in the complex operator? Commented Dec 23, 2023 at 12:09

Well, as you stated the transfer function of your circuit is given by:

\begin{alignat*}{1} \mathscr{H}\left(\text{s}\right)&=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}\\ \\ &=\frac{\displaystyle\text{R}_2\space\text{||}\space\text{sL}}{\displaystyle\text{R}_1+\left(\text{R}_2\space\text{||}\space\text{sL}\right)}\\ \\ &=\frac{\displaystyle\frac{\displaystyle\text{sLR}_2}{\displaystyle\text{sL}+\text{R}_2}}{\displaystyle\text{R}_1+\frac{\displaystyle\text{sLR}_2}{\displaystyle\text{sL}+\text{R}_2}}\\ \\ &=\frac{\displaystyle\text{sLR}_2}{\displaystyle\text{R}_1\left(\text{sL}+\text{R}_2\right)+\text{sLR}_2}\\ \\ &=\frac{\displaystyle\text{sLR}_2}{\displaystyle\text{sLR}_1+\text{R}_1\text{R}_2+\text{sLR}_2}\\ \\ &=\frac{\displaystyle\text{sLR}_2}{\displaystyle\text{R}_1\text{R}_2+\text{sL}\left(\text{R}_1+\text{R}_2\right)} \end{alignat*} \tag2

Where $$\\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$$.

Now, when we aplly the inverse Laplace transform, we can see:

\begin{alignat*}{1} \text{y}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\mathscr{H}\left(\text{s}\right)\right]_{\left(t\right)}\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{sLR}_2}{\displaystyle\text{R}_1\text{R}_2+\text{sL}\left(\text{R}_1+\text{R}_2\right)}\right]_{\left(t\right)}\\ \\ &=\text{LR}_2\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{s}}{\displaystyle\text{R}_1\text{R}_2+\text{sL}\left(\text{R}_1+\text{R}_2\right)}\right]_{\left(t\right)} \end{alignat*} \tag2

Now, we can use the derivative property of the Laplace transform:

$$\text{y}\left(t\right)=\text{LR}_2\cdot\frac{\partial}{\partial t}\left(\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle1}{\displaystyle\text{R}_1\text{R}_2+\text{sL}\left(\text{R}_1+\text{R}_2\right)}\right]_{\left(t\right)}\right)\tag3$$

Now, let's recall that:

$$\mathscr{L}_t\left[\frac{\displaystyle\exp\left(-\frac{\displaystyle\text{a}t}{\displaystyle\text{b}}\right)}{\displaystyle\text{b}}\right]_{\left(\text{s}\right)}=\frac{\displaystyle1}{\displaystyle\text{a}+\text{bs}}\tag4$$

Edit, so we get:

\begin{alignat*}{1} \text{y}\left(t\right)&=\text{LR}_2\cdot\frac{\partial}{\partial t}\left(\frac{\displaystyle\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\\ \\ &=\text{LR}_2\cdot\frac{\partial}{\partial t}\left(\frac{\displaystyle\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\\ \\ &=\frac{\displaystyle\text{LR}_2}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\cdot\frac{\partial}{\partial t}\left(\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\right)\\ \\ &=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\frac{\partial}{\partial t}\left(\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\right)\\ \\ &=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\cdot\frac{\partial}{\partial t}\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\\ \\ &=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1+\text{R}_2}\cdot\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\cdot\left(-\frac{\displaystyle\text{R}_1\text{R}_2}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right)\\ \\ &=-\frac{\displaystyle\text{R}_1\text{R}_2^2}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)^2}\cdot\exp\left(-\frac{\displaystyle\text{R}_1\text{R}_2t}{\displaystyle\text{L}\left(\text{R}_1+\text{R}_2\right)}\right) \end{alignat*} \tag5

• You are missing a scaled impulse ($b_n \delta(t)$) in the impulse response.
– Carl
Commented Dec 23, 2023 at 20:30

You don't have to get bogged down in so much algebra if you know a little circuit theory. If $$\e(t)=\delta(t)\$$, then during the initial voltage spike, the inductor is essentially open: inductors oppose sudden changes in current, so the inductor current cannot spike like a delta function. Consequently, initially we have a simple resistive voltage divider and we can write $$h(t)=v_\text{out}(t)=\frac{R_2}{R_1+R_2}\delta(t)+\text{bounded terms}.$$ The "bounded terms" here contain no delta functions. We now need to determine what these are.

The inductor current immediately after the initial input voltage spike is $$i_L(0^+)=\frac{1}{L}\int\limits_{0^-}^{0^+}h(t)dt=\frac{1}{L}\int\limits_{0^-}^{0^+}\frac{R_2}{R_1+R_2}\delta(t)dt=\frac{R_2}{L(R_1+R_2)}.$$

Here we have simply ignored the "bounded terms" because they cannot integrate to a non-zero value during an infinitesimal time interval. After $$\t=0^+\$$, the circuit is simply an inductor discharging through a resistance $$\R_1\parallel R_2\$$, with time constant $$\tau=\frac{L}{R_1\parallel R_2}=\frac{L(R_1+R_2)}{R_1R_2}.$$

We thus have $$i_L(t)=i_L(0^+)e^{-t/\tau}u(t)=\frac{R_2}{L(R_1+R_2)}e^{-t/\tau}u(t),$$ where $$\u(t)\$$ is the unit step function, and $$h(t)=L\frac{di_L}{dt}=\frac{R_2}{R_1+R_2}\delta(t)-\frac{R_1R_2^2}{L(R_1+R_2)^2}e^{-t/\tau}u(t).$$