Impedances for RLC circuits in the limit of zero frequency

If a voltage source is sinusoidal, then we can introduce a phasor map and come up with equations like$$V_0 e^{i \omega t} = I(R + i\omega L + \frac{1}{\omega C} i)$$where $I$ would also differ from $V$ by a complex phase.

But if you set $\omega = 0$, which would appear to correspond to the case where the source voltage is constant, you get nonsense results. If there is a capacitor then you get an infinity, and the effects of any inductor disappears completely (as if it were no longer in the circuit) because $i\omega L = 0$.

I wondered if anyone could explain what's going on here. Is this approach only valid for a sinusoidal source? Or if not, then how do we interpret these weird results? If we were to solve the above example with $\omega = 0$ in the normal way, we'd be dealing with the constant coefficients differential equation $\ddot{I} + \frac{R}{L}\dot{I} + \frac{1}{LC}I = 0$.

 

[vanhees71]

For $\omega=0$ you have the limiting case of time-independent fields and sources, i.e., magnetostatics. It's not so easy to take the limit $\omega \rightarrow 0$.

 

[A.T.]

If a voltage source is sinusoidal, then we can introduce a phasor map and come up with equations like$$V_0 e^{i \omega t} = I(R + i\omega L + \frac{1}{\omega C} i)$$where $I$ would differ from $V$ by a complex phase.

Is this for a RLC in series?

 

Is this for a RLC in series?

It should be, yes. Did I make a mistake?

 

[A.T.]

It should be, yes.

If the equation describes the long term behavior (ω has been constant for a long time), then for ω = 0 the source voltage has been constant for a long time, the capacitor is charged, and I = 0. So the RHS is indeterminate, which is not wrong, but not useful to determine the LHS.

But take this with a grain of salt. I'm no expert on circuits.

 

[anorlunda]

But if you set ω=0, which would appear to correspond to the case where the source voltage is constant, you get nonsense results.

That's a great question. Fortunately, it has a simple answer.

In AC analysis, we ignore the startup transient and ignore zero or nonzero initial conditions. Think of the source as V=0 when t<0 and V=sin(wt) when t>0. That's the startup transient. But we treat is as if the sin(wt) extended back to t= minus infinity.

But in your case, when w=0, the only conditions remaining are the initial conditions, so our sinusoidal equation reduces to a DC equation.

 

In AC analysis, we ignore the startup transient and ignore zero or nonzero initial conditions. Think of the source as V=0 when t<0 and V=sin(wt) when t>0. That's the startup transient. But we treat is as if the sin(wt) extended back to t= minus infinity.

But in your case, when w=0, the only conditions remaining are the initial conditions, so our sinusoidal equation reduces to a DC equation.

Interesting, so if I understand correctly the transient solution we'd get from solving $\ddot{I} + \frac{R}{L}\dot{I} + \frac{1}{LC}I = 0$ in the $\omega = 0$ case (which, even in underdamped conditions, would be enveloped by a negative exponential) would die away pretty quickly.

And if we were to instead use the AC analysis assumptions, we'd ignore that phase and go straight to our weird indeterminate solution, that @A.T. alluded to in #5.

 

[A.T.]

... our weird indeterminate solution, that @A.T. alluded to in #5.

It's not that weird, if you think about it. If ω = 0 then any value of V will result in I=0 after a while. So you cannot reverse the relation to get a specific V from the final I, because I is the same for all V.

 

[anorlunda]

Interesting, so if I understand correctly the transient solution we'd get from solving $\ddot{I} + \frac{R}{L}\dot{I} + \frac{1}{LC}I = 0$ in the $\omega = 0$ case (which, even in underdamped conditions, would be enveloped by a negative exponential) would die away pretty quickly.

And if we were to instead use the AC analysis assumptions, we'd ignore that phase and go straight to our weird indeterminate solution, that @A.T. alluded to in #5.

There's another way to look at it. $\ddot{I} + \frac{R}{L}\dot{I} + \frac{1}{LC}I = 0$ is not the complete equation, because it ignores initial conditions. With initial conditions, we would also get decaying expotential terms superimposed.

Yet another way to look at it: sin(wt) is valid for $-\infty < t < +\infty$. But adding an initial condition V=0 for t<0 makes it an aperiodic waveform. In your w=0 case, either V and I are zero for all time, or it is aperiodic.

Edit: I should have said, adding an initial condition V=0 and $\frac{dV}{dt}=0$ for t<0

 

[anorlunda]

I just thought of a fun theorem to prove. Is it possible to generate a periodic solution with an aperiodic forcing function? No doubt, someone did that before, but it might be fun to prove from scratch.

 

[Lord Jestocost]

If a voltage source is sinusoidal, then we can introduce a phasor map and come up with equations like$$V_0 e^{i \omega t} = I(R + i\omega L + \frac{1}{\omega C} i)$$where $I$ would also differ from $V$ by a complex phase.
But if you set $\omega = 0$, which would appear to correspond to the case where the source voltage is constant, you get nonsense results.

The term in the brackets is the complex impedance $Z$ of a series RLC circuit. In case $\omega\rightarrow 0$, the magnitude of the impedance becomes infinite, $|Z| \rightarrow \infty$, viz. $I\rightarrow 0$. Nothing strange.

 

[vanhees71]

Let's see. I all starts from AC circuit theory. In the time domain the equation for an RLC circuit reads
$$L \ddot{Q} + R \dot{Q} + \frac{1}{C} Q=U,$$
where $Q$ is the charge on the capacitor. Now usually you are interested in AC with a given frequency $\omega$ and it's most convenient to use the complex approach with the understanding that the physical quantities are the real part of the corresponding complex functions.

In addition we are only interested in the stationary final state, i.e., we don't consider the damped transient states but look a long time after having switched on the circuit (long means long compared to the damping time of the homogeneous solution of the linear ODE).

Then we can write
$$Q=Q_0 \exp(\mathrm{i} \omega t), \quad U=U_0 \exp(\mathrm{i} \omega t),$$
and the ODE turns into a linear algebraic equation,
$$(-L\omega^2 + \mathrm{i} \omega R +\frac{1}{C}) Q_0=U_0.$$
Of course we are rather interested in the current, which is
$$I=\dot{Q}=\mathrm{i} \omega Q_0 \exp(\mathrm{i} \omega t)\; \Rightarrow \; Q_0=-\mathrm{i} I_0/\omega,$$
leading to
$$Z I_0=\left (R+\mathrm{i} \omega L-\frac{\mathrm{i}}{\omega C} \right) I_0=U_0.$$
This example leads to the usual AC circuit theory for the stationary state. It works exactly as DC circuit theory when instead of resistances you use impedances for the various elements, i.e., $Z_R=R$ for a resistor, $Z_L=\mathrm{i} \omega L$ for a coil, and $Z_C=-\mathrm{i}/(\omega C)$ for a capacitor. This of course only works for $\omega \neq 0$.

For $\omega=0$ you have $U=U_0=\text{const}$, and of course the stationary state is $Q=Q_0=\text{const}$ and thus $I=0$ and $\dot{I}=0$. Then the above ODE gets simply
$$\frac{Q_0}{C}=U_0=\text{const},$$
as it should be. There's of course no current because of the capacitor.

 

Then we can write
$$Q=Q_0 \exp(\mathrm{i} \omega t), \quad U=U_0 \exp(\mathrm{i} \omega t),$$
and the ODE turns into a linear algebraic equation,
$$(-L\omega^2 + \mathrm{i} \omega R +\frac{1}{C}) Q_0=U_0.$$

Maybe a silly final question... do we also need account for a phase here, e.g. putting $U = U_0 e^{i \omega t} e^{i \delta}$? I have the feeling that generally, excluding cases where we just have resistors, $\delta$ will often be non-zero.

 

[vanhees71]

What do you mean by phase? $I_0$ and $U_0$ are complex numbers. You can write them of course in "polar" form as $I_0=|I_0| \exp(\mathrm{i} \phi_Q)$ and $U_0=|U_0| \exp(\mathrm{i} \phi_U)$. The same holds for $Z=|Z| \exp(\mathrm{i} \phi_Z)$. Usually you are only interested in the relative phase shift between $U$ and $I$, and that's encoded in $\phi_Z$.

 

What do you mean by phase? $I_0$ and $U_0$ are complex numbers. You can write them of course in "polar" form as $I_0=|I_0| \exp(\mathrm{i} \phi_Q)$ and $U_0=|U_0| \exp(\mathrm{i} \phi_U)$. The same holds for $Z=|Z| \exp(\mathrm{i} \phi_Z)$. Usually you are only interested in the relative phase shift between $U$ and $I$, and that's encoded in $\phi_Z$.

Okay yes that makes sense if $U_0$, $I_0$ etc. are complex, I thought you had used them for the magnitude only. Thanks for clarifying!

 

[DaveE]

It should be, yes. Did I make a mistake?

Yes, the term i/(ωC) in you original equation should be 1/(iωC) = -i/(ωC).

 

Yes, the term i/(ωC) in you original equation should be 1/(iωC) = -i/(ωC).

Whoops, yes I meant to put it in the denominator.

 

[DaveE]

If you think of each of the terms R, iωL, and 1/(iωC) as impedances (i.e V/I), then the 0 and ∞ values you get when ω→0 make sense. In a DC circuit an inductor behaves as a short circuit (|Z|=0), and a capacitor behaves as an open circuit (|Z|=∞). If you apply a finite voltage across each of these impedances individually you will either get I→0 or I→∞.

 

[DaveE]

I just thought of a fun theorem to prove. Is it possible to generate a periodic solution with an aperiodic forcing function? No doubt, someone did that before, but it might be fun to prove from scratch.

Strictly speaking, no, not in linear circuits. The circuit would have to have no response at each of the frequencies of the forcing function but have a finite response at the periodic frequencies (the fundamental, natural response, plus perhaps harmonics). This, in the frequency domain, would look like a delta function (plus perhaps harmonics). Which is impossible to construct (it's not causal, I think). In practice, it's easy to approximate it, like a guitar string, or a high Q LC tank, for example.