The circuit contains two independent state variables but the output of the circuit simulated in LT spice shows a first-order response.

Where am I wrong?

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  • 1
    \$\begingroup\$ Second order system (not saying it is one though) can have an over-damped response which looks like this one. \$\endgroup\$
    – Eugene Sh.
    Jul 13, 2022 at 14:50
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    \$\begingroup\$ It's not clear what your voltage nodes are from your graph -- it's good to label your nodes in LTSpice, then the traces are labeled, too. Just from context, I'm assuming that v001 is the cold end of R3, and that v002 is the junction of the two other resistors and the two caps. \$\endgroup\$
    – TimWescott
    Jul 13, 2022 at 15:00
  • \$\begingroup\$ I commented on your previous question, but what you see is not a single exponential. Why not calculate the transfer function? Things will be immediately become apparent. \$\endgroup\$ Jul 13, 2022 at 15:02
  • 2
    \$\begingroup\$ Why has your previous question (to which I answered) been deleted? There is no valid reason for doing this as far as I can see @voltageSpike \$\endgroup\$
    – Andy aka
    Jul 13, 2022 at 16:10

2 Answers 2


I'm not sure if you've been presented with an interesting problem in pole-zero cancellation and have gotten it wrong*, or if you really are trying to analyze that circuit.

For such simple circuits, you should only use SPICE to validate your hand calculations.

You're driving a current, so all the interesting dynamics are in that network composed of R1, R2, C1, and C2.

The impedance of that network between the cold end of R3 and ground, is

$$Z_l = \frac{1}{C_1 s + \frac{1}{R_1}} + \frac{1}{C_2 s + \frac{1}{R_2}}. \tag 1 $$

Do some simple algebra, and you get:

$$\begin{align} Z_l &= \frac{R_1}{R_1 C_1 s + 1} + \frac{R_2}{R_2 C_2 s + 1} \\ &= \frac{R_2\left(R_1 C_1 s + 1\right) + R_1\left(R_2 C_2 s + 1\right)} {\left(R_1 C_1 s + 1\right)\left(R_2 C_2 s + 1\right)} \end{align} \tag 2 $$

Because that network is driven by a current, the voltage response is just the current times the impedance -- so the poles & zeros of the impedance are the poles & zeros of the gain at that point (and R3 just adds a constant -- which is why I'm leaving it out).

If \$R_1 C_1 = R_2 C_2 \$, then the zero in the numerator of (2) would be cancelled by one of the poles (work out the math -- you'll see it). That is not the case.

Instead, you have a very over-damped system (because \$C_2 R_2 = 4C_1 R_1\$), and you cannot pick out the 2nd-order response visually -- which is why simulation is a bad idea when you can easily just do the math.

Going back to the \$R_1 C_1 = R_2 C_2 \$ case. If that were true (i.e. if you take your schematic above and either exchange the values of R1 and R2, or C1 and C2 but not both) then you could see intuitively that the wire between all four components would only carry current in the event that one of the capacitors happened to have some suitable initial value at startup. In all other cases, the C1-C2 voltage divider and the R1-R2 voltage divider would always have the same voltage, and there would be nothing to drive a current in that wire.

This would be reflected in the math. If \$R_1 C_1 = R_2 C_2 \$, then the denominator of (2) would be \$\left(R C s + 1\right)^2\$ (where \$R_1 = R_2 = R\$ and \$C_1 = C_2 = C\$). The numerator would be \$R^2 \left(R C s + 1\right)\$. There would be a pole-zero cancellation, and you'd have a simple example of a rather important result in systems theory -- that you can have a system with unobservable or uncontrollable** modes. Such modes can cause trouble, because they can misbehave and you can't fix that with feedback.

* Or your prof copied it into their notes wrong, or out of their notes wrong. It has happened before.

** I'm 99.44% sure that this circuit has one uncontrollable mode, but I'm too lazy to verify -- if you get that far along in your studies, you can revisit it.


With 2 independent reactances (L or C), it is always a 2nd order circuit but here , overdamped with 2 different time constants that affect each other so they become longer than just RC=T but two separate combinations of 4 parts.

In the frequency domain, without R3, the lower frequency response starts flat at 0 dB for nodes "A" and B with some attenuation on B at some breakpoint. So it might go from 0 to -10 dB then flat yet overdamped in a region like the dominant 1st order response. But with R3 you have a 2nd order LPF but 2 still distinct breakpoints now reacting with R3 to attenuate -40 dB/decade after the 2nd breakpoint.

Either way in the time domain asymptote or the frequency domain, you ought to be able to measure the distinct 2 breakpoints which from a distance look like one RC filter since the values are close to each other but are not exactly the same as 1 pair, but close.


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