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So there's this problem I am trying to solve:


Consider a series RL circuit corrected to a DC voltage source and an on-off switch. Assume R = 2 ohms, and L = 1H, and applied voltage V = 10 volts. Find the transfer function of the circuit if the current I is the measured output. Suppose the switch is closed at t = 0, find the current in the circuit as a function of time.


Based on the problem, I believe the circuit should look like this:

circuit

Obviously, when the switch is closed we will have to deactivate the inductor, so the current would be 5A. Now, when the switch is open, the current would normally be zero, so after doing the Laplace Transform of the inductor, I cannot find a way to find the transfer function, and then the current as a function of time.

I would appreciate your help here guys. Thank you.

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  • \$\begingroup\$ Your problem statement says you want the behavior after the switch closes, but your diagram shows the switch being opened at t=0. Could you edit to clarify? \$\endgroup\$
    – The Photon
    Jan 24, 2019 at 19:24

3 Answers 3

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I think you are mixing up the ideas of the transient behavior of the circuit, and the steady state behavior in the frequency domain (by using Laplace methods).

Transient response would have to do with closing the switch, and seeing the current/ voltage in the inductor as time goes on from t=0.

The steady state behavior in the frequency domain would have to do with replacing the inductor with its frequency domain impedance of L*s. In that case, you would solve for the current through the network, and divide by your input voltage to get the transfer function from input voltage to output current as a function of s.

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The time domain step response of a series R-L system is simple and well documented. If current \$i(t)\$ at time \$t=0\$ is zero (\$i(0)=0\$), then:

$$ i(t) = i_{t\rightarrow \infty} \left(1-e^{-t \over \tau}\right) $$

where time constant \$\tau = \frac{L}{R} \$. Intuitively, current \$i_{t\rightarrow \infty}\$ is when the system has long settled into steady state, and the rate of change of current is zero. Then the inductor is effectively a short circuit, the voltage across it being zero, and the voltage across R is at a maximum (10V) Therefore, by ohm's law:

$$ i_{t\rightarrow \infty} = \frac{v_{t\rightarrow \infty}}{R} = \frac{10V}{2\Omega} = 5A $$

This yields the following unambiguous expression for current, as a plain old function of time:

$$ i(t) = 5 \left(1-e^{-2t }\right) $$

If you are being asked to derive this expression for yourself, you have at least two approaches. You can do everything in the time domain, using the time-domain expressions for current vs. voltage in the two components R and L, or (see below) you could use an inverse Laplace transform, using frequency (s) domain terms for those relationships.

Time domain analysis

Since the current is the same in all components around the loop, the algebra is quite simple. The voltage across R is an application of Ohm's law:

$$ v_R(t) = i(t)\times R $$

The relationship between current through inductance L and the voltage across it is:

$$ v_L(t) = L \frac{di(t)}{dt} $$

After time \$t=0\$, by Kirchhoff's voltage law we can state that the sum of the voltages across each element must be equal to the voltage source of 10V:

$$ \begin{aligned} v_R + v_L &= 10 \\ \\ iR + L\frac{di}{dt} &= 10 \\ \\ 2i + \frac{di}{dt} &= 10 \end{aligned} $$

If you solve this for \$i\$, you will find the exact same expression from above:

$$ i(t) = 5 \left(1-e^{-2t }\right) $$

This is not a transfer function, it's just a function, telling you what \$i\$ will be at time \$t\$ assuming that \$i=0\$ when \$t=0\$. A transfer function must relate some input to an output, so we still need a function describing an input signal.

I think your problem assumes that the input shall be the voltage across the R-L pair. If so, then we can say that this input is 0V for \$t<0\$ (the switch is open), and 10V for \$t \ge 0\$ (following switch closure). That's a step function:

$$ v(t) = \begin{cases} 0V & \text{if } t < 0 \\ 10V & \text{if } t \ge 0 \end{cases} $$

The exponential output we arrived at above is correct for this input, but only for this input. It isn't a general expression that works for any \$v(t)\$. A transfer function would also need to be valid for whatever input is described by \$v(t)\$, not just this step function. The differential equation above came close though, we just need to keep it general, by replacing the constant 10V with the variable input voltage \$v(t)\$:

$$ \begin{aligned} iR + L\frac{di}{dt} &= v \end{aligned} $$

That still obeys Ohm's and Kirchhoff's laws, and accurately describes the state of the circuit at any time \$t\$, and as such is very close to becoming a transfer function. The only thing wrong with it is that the subject of the formula is \$v\$ instead of \$i\$. The last step would be to make \$i\$ the subject, equal to some expression in terms of \$v\$, which is usually difficult.

Frequency domain analysis

To simplify that task, we use Laplace transforms. Just for completeness, I will repeat what we just did in the time domain, to derive \$i(t)\$, and then I'll work on a proper transfer function.

We know the impedances of both L and R in the frequency domain:

$$ \begin{aligned} Z_R = R \\ Z_L = Ls \end{aligned} $$

The combined impedance of these elements in series is:

$$ \begin{aligned} Z_{TOTAL} &= Z_R + Z_L \\ \\ &= R + Ls \\ \\ &= 2 + s \end{aligned} $$

Knowing the voltage across this pair to be \$v(t)=10\$, for which the Laplace term is \$v(s)=\frac{10}{s}\$, apply Ohm's law to derive current \$i\$:

$$ \begin{aligned} i(s) &= \frac{v(s)}{Z_{TOTAL}} \\ \\ &= \frac{\frac{10}{s}}{2+s} \\ \\ &= \frac{10}{s(2+s)} \end{aligned} $$

Taking the inverse Laplace transform will get you back into the time domain, and yield the same expression for \$i(t)\$. I remind you again that this equation is still not a transfer function.

Frequency domain transfer function

Instead of focusing on the variation of \$i(s)\$ for some particular input, we are looking for something more general, relating \$i(s)\$ to any input \$v(s)\$. We got there already, when we said

$$ i(s) = \frac{v(s)}{Z_{TOTAL}} $$

Instead of replacing \$v\$ with some step function, we leave it as is, giving us:

$$ i(s) = \frac{v(s)}{2+s} $$

Now we have an equation relating \$i\$ and \$v\$, with \$i\$ as the subject, valid for any input \$v\$. It's in the frequency domain, but it still contains all the same information as the time domain differential equation above, and is a lot easier to solve.

The last step is to rearrange it to conform with the definition of a transfer function, which is:

$$ H(s) = \frac{y(s)}{x(s)} $$

where the \$y\$ is the output, and \$x\$ is the input. In other words, a transfer function is the ratio of output to input. Our relationship becomes:

$$ \begin{aligned} H(s) &= \frac{i(s)}{v(s)} \\ \\ &= \frac{1}{2+s} \end{aligned} $$

Notes

Rearranging that equation gives us:

$$ v(s) = i(s) \times (2 + s) \\ \\ $$

Notice that \$2 + s\$ is the combined impedance of R and L. This equation is simply a statement of Ohm's law, saying "voltage \$v\$ across the R-L series pair is equal to the product of current \$i\$ through them and their combined impedance \$R + Ls\$. There was no need for any complicated algebra.

This seems obvious when you consider that the output \$i\$ is the current that would flow through a path of impedance \$2+s\$ when an input voltage \$v\$ is applied across it.

Most transfer functions relate input and output of the same units, usually volts. You could, for example, be asked what is the relationship between the voltage across the inductor (output) and the voltage applied across the R-L series pair (input). In such a case, the transfer function would be dimensionless, \$\frac{out}{in}=\frac{[volts]}{[volts]}\$. In this case though, the units are \$\frac{[A]}{[V]}=[\Omega]^{-1}=[S]\$ or \$[mho]\$.

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The transfer function tells us how the output changes when the input changes.

In this case, the problem hasn't been stated clearly, but we can probably assume the input is the voltage provided by the voltage source. In a more general problem, this might not be a DC voltage, but might include a time-varying component.

It doesn't make much sense to find the transfer function with I as the output when the switch is open, because it would just be \$I=0\$. So we can probably also assume they want you to find the transfer function when the switch is closed.

I'd rather the problem was written more clearly rather than have to make these assumptions, but sometimes you gotta roll with it.

Now, if you know how to write the impedance of the resistor and the inductor in the Laplace domain, you can find the transfer function very easily.

Obviously, when the switch is closed we will have to deactivate the inductor,

This doesn't make any sense to me. When the switch is closed (in the time shortly after it closes, or if the source had a time-varying component) is exactly when we need to consider the effect of the inductor. You certainly can't just ignore it because the switch is closed.

I cannot find a way to find the transfer function, and then the current as a function of time.

You want to find the behavior after the switch closes, so the transfer function with the switch closed is what would be useful here.

note

If you were interested in how the circuit behaves if the switch is opened at t=0, then the circuit model is incomplete. You'd want to include the inter-winding capacitance of the inductor and probably the arcing behavior of the switch to get a realistic result. (The fact the model only works for switching the switch in one direction is likely to confuse learners, and therefore another reason the original problem statement is poorly written)

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  • \$\begingroup\$ By "deactivating" the inductor I mean replacing it with a current source connected in parallel with the inductor. The current source would be 5/s A, and we can write the impedance of the inductor as s ohms, based on the Laplace transform. Also, besides transfer function I would have to find i(t) \$\endgroup\$
    – snitchben
    Jan 24, 2019 at 19:33
  • \$\begingroup\$ @snitchben, consider what's the effect of closing the switch in terms of the voltage applied across the RL circuit. From there you should be able to solve it in the usual way in the Laplace domain. \$\endgroup\$
    – The Photon
    Jan 24, 2019 at 19:51

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