I have a simple RL circuit such as the one shown below

RL circuit

and I want to derive the differential equation relating the input and output voltages. I want to take the output voltage as the one across the inductor. So far I have done the following but I'm not sure whether I'm making a mistake or not, since I haven't been able to find a similar process in a couple books and internet. I want to do this as an exercise for finding Fourier Transform of output.

I know the differential equation starts out like this:

$$ V_{in}(t) = L\frac{di}{dt} + Ri $$

From which I know that the current \$i\$ is

$$ \frac{V_{out}(t) - V_{in}(t)}{R} $$

I replace this, and get

$$ V_{in}(t) = \frac{L}{R}\left(\frac{dV_{out}}{dt} - \frac{dV_{in}}{dt}\right) + V_{out}(t) - V_{in}(t) $$

Is this correct? All that would remain would be to replace $$V_{in} = e^{jwt}$$ and $$V_{out} = H(w)e^{jwt}$$ and simplify, getting

$$ H(w) = \frac{2+\frac{L}{R}jw}{1+\frac{L}{R}jw} $$

I believe I made a mistake somewhere since this result looks kind of weird.

Also: is this a high pass or low pass filter? How do I figure this out from looking at the equation? And how do I find the cut off frequency?


You got the second relation wrong:

\$ i(t) = \dfrac{v_{in}(t)-v_{out}(t)}{R} \$

substituting in the first equation you get:

\$ v_{in}(t) = \dfrac{L}{R}\left(\dfrac{dv_{in}}{dt} - \dfrac{dv_{out}}{dt}\right) + v_{in}(t) - v_{out}(t) \$

Thus \$v_{in}\$ cancels out in both members. Rearranging you get:

\$ \dfrac{L}{R}\dfrac{dv_{out}}{dt} + v_{out} = \dfrac{L}{R} \dfrac{dv_{in}}{dt} \$


\$ \dfrac{L}{R} \cdot j\omega V_{out} + V_{out} = \dfrac{L}{R} \cdot j\omega V_{in} \$

\$ H(\omega) = \dfrac{V_{out}}{V_{in}} = \dfrac{ \dfrac{L}{R} j\omega } { 1 + \dfrac{L}{R} j\omega } \$

This is coeherent with the standard s-domain approach for determining the transfer function of the system \$W(s)\$. In fact if you replace the inductor with its s-domain equivalent \$sL\$ and apply the voltage divider formula in the s-domain you get:

\$ V_{out}(s) = V_{in}(s) \dfrac{Ls}{R+Ls} \qquad \Leftrightarrow \qquad W(s) = \dfrac{V_{out}(s)}{V_{in}(s)} = \dfrac{Ls}{R+Ls} = \dfrac{\dfrac{L}{R}s}{1+\dfrac{L}{R}s} \$

and since

\$ H(\omega)=W(j\omega)=\dfrac{ \dfrac{L}{R} j\omega } { 1 + \dfrac{L}{R} j\omega } \$

you can see that the two methods give the same result.

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  • \$\begingroup\$ Aww jeez I can't believe I messed up KCL. Thank you! \$\endgroup\$ – Emilio Botero May 13 '15 at 0:25

The standard time-domain approach is given in this tutorial: RL Circuit Analysis example

An interesting frequency-domain approach (maybe closer to what you want) is given in RL frequency domain

As to your question, an easy way to see if a circuit is highpass is to imagine it at DC and at infinite frequency (and maybe an intermediate frequency) to see if the transmission is large or zero. In this case, at DC the inductor is a short circuit so Vout would be shorted to ground. At infinite frequency the inductor is an open circuit so the transmission would be 1. Therefore this must be a high-pass filter. You find the cutoff of a first-order circuit as one over the time-constant (L/R) in radians. In other words is is R/L radians.

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  • \$\begingroup\$ Hey, thank you for your answer. I'm actually trying to do this through the differential equations. The RL frequency domain approach does indeed solve the problem, but using voltage dividers. I was just wondering how to get the same result using the differential equation and the simple fact that output is scaled version of input. Besides, where did the complex factor go in those derivations? \$\endgroup\$ – Emilio Botero May 12 '15 at 21:26

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