Nyquist plot disagreeing with the unity step response of the closed loop

Question

Let us assume that the open loop transfer function of my system is:

\$L(s)=\frac{10(s+1)(s+2)}{(s-3)(s-4)}\$

Also, the system has unity feedback, meaning that the closed loop transfer function is

\$T(s)=\frac{L(s)}{1+L(s)}\$ since \$L(s)=G(s)H(s)\$ and \$H(s)=1\$

Then the Nyquist plot is this:

There are 2 RHP poles present and the plot circles the (-1,0) point 2 times, anti-clockwise. \$Z=N-P\$ where \$N=-2\$ and \$P=2\$ so \$Z=0\$ => Stable System.

And the unity step response is this (for the closed loop):

It looks stable, but if I press on "more time", it looks pretty unstable:

So the Nyquist plot is disagreeing with the unity step response. What is happening?

The transfer function is theoretical in nature, it does not necessarily represent a real system's response. I have taken care to make sure that the unity step response is that of the closed loop transfer function (and not that of the open loop, which is clearly unstable). I used WolframAlpha for all my plots.

Answer 1

Please double check the simulation/plots. That behavior seems way too fast ("noisy") for the dynamics of your system. I tried simulating the closed loop system in Octave and could notice that there is an unstable pole cancellation, but that does not lead to the "unstable" behavior in Octave.

I would say that is a problem in WolframAlpha and not really a control theory thing.

s = tf('s')
l = 10(s+1)(s+2)/((s−3)(s−4))
pzmap(l/(1+l))
step(l/(1+l),40)

Answer 2

The closed-loop system has a RHP pole and zero. If they are not cancelled, the unstable dynamics can sometimes show up in the output.

This is the result you are seeing from Wolfram Alpha.

((s + 7)/(s - 3))/(1 + (s + 7)/(s - 3));
Grid[{{tfm = TransferFunctionModel[%, s], ssm = StateSpaceModel[tfm]}}]
or = OutputResponse[ssm, UnitStep[t], {t, 0, 20}];
Plot[or, {t, 0, 20}]

If you plot the entire range, you will see that the output actually blows up.

Plot[or, {t, 0, 20}, PlotRange -> All]

If the pole and zero are cancelled then the dynamics is stable.

Grid[{{tfm1 = TransferFunctionCancel[tfm], 
ssm1 = StateSpaceModel[tfm1]}}]
or1 = OutputResponse[ssm1, UnitStep[t], {t, 0, 20}];
Plot[or1, {t, 0, 20}, PlotRange -> All]

To get a clearer picture, do the analysis with the Jordan canonical realization.

ssm2 = Last@JordanModelDecomposition@ssm
or2 = OutputResponse[ssm2, UnitStep[t], {t, 0, 20}];
Plot[or2, {t, 0, 20}, PlotRange -> All]

You can see that the output is stable and well behaved because it depends on just the first state. The second state is unstable and blows up, but does not manifest itself in the output. You can plot the individual state responses and see that the second one does indeed blow up.

sr2 = StateResponse[ssm2, UnitStep[t], {t, 0, 20}];
GraphicsRow[Plot[#, {t, 0, 20}, PlotRange -> All] & /@ sr2]

In summary, it depends on how you realize this transfer function. If the unstable dynamics cannot be cancelled, they will show up somewhere. You may not see it at the output, but things are blowing up internally.