0
\$\begingroup\$

As far as im aware Poles affect the stability of a transfer function primarily, and is defined whether they stay within the unit circuit & are on the left side of the pole-zero plot. Zeros have no role in stability? For the following transfer function which is in a feedback loop (Using a step input):

$$G(s) = \frac{K*50}{s^4+15s^3+100s^2+250s}$$

enter image description here

When I change the value of 'K', if the value is too low the system over damps, if the value is too high the system becomes unstable and oscillates into oblivion. Why?

For Example, when K = 1 enter image description here

When K = 10 enter image description here

When K = 100 enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ That would surprise me. The Laplace transform is linear, so is your system. All your \$K\$ does is scale the output. And if your system is stable for some input, then it's also stable for \$K\$ times that input. \$\endgroup\$ – Marcus Müller Mar 14 '18 at 12:18
  • 2
    \$\begingroup\$ Are you in fact applying G(s) inside a feedback loop? Did you forget to mention that? \$\endgroup\$ – Andy aka Mar 14 '18 at 12:20
  • 1
    \$\begingroup\$ Zeros do not affect stability of a system unless they precisely cancel a pole. \$\endgroup\$ – Parth Mar 14 '18 at 12:22
  • 3
    \$\begingroup\$ This is an \$s\$ domain transfer function, not a \$z\$ domain, and so the poles should be in the left half-plane and not within the unit circle. By the way, note that you have pole at the origin \$(s = 0)\$ so this is an integrator in cascade with a third-order low-pass filter. \$\endgroup\$ – Alfred Centauri Mar 14 '18 at 12:22
  • 2
    \$\begingroup\$ Your title is now wrong because there are now zeroes when you have G(s) in a feedback loop. Simple answer is that "this is what happens in control loops when you vary values" \$\endgroup\$ – Andy aka Mar 14 '18 at 12:36
1
\$\begingroup\$

As you've more or less drawn it, the overall transfer function is (assuming here that \$G(s)\$ is actually the transfer function of the plant, the transfer function enclosed by the box in your drawing)

$$H(s) = \frac{G(s)}{1 + G(s)}=\frac{K\cdot 50}{s^4 + 15s^3+ 100s^2+250^s + K\cdot 50}$$

So it's easy to see that the gain \$K\$ will affect the location of the poles.

Indeed, putting this into Wolfram Alpha with \$K = 100\$ shows two poles in the RHP.

You should try doing a root locus analysis to see how those poles move from the LHP to the RHP as \$K\$ is increased.

\$\endgroup\$
  • 1
    \$\begingroup\$ Ah! I did not realise in the feedback network, the numerator gets pulled down - obviously causing a root shift. Thanks \$\endgroup\$ – user160063 Mar 14 '18 at 18:45
  • \$\begingroup\$ i mentioned this a few minutes before this answer was posted in the comment section, but it was ignored. Great. :) \$\endgroup\$ – Abel Tom Mar 15 '18 at 6:26
1
\$\begingroup\$

Unfortunately, yes that transfer function will oscillate as you turn up gain factor \$k\$.

Think about the loop-gain around your feedback-loop. You are providing negative feedback so there is the first 180 deg phase shift. Your plant has an integrator (a pole at the origin), so, there is another 90 deg phase shift. Now as you increase the forward gain you give more opportunity for the next pole to add another 90 deg phase shift to the loop before the loop-gain drops below unity.

Consider a bode plot of \$G(s)\$ for the 3 cases of k,

enter image description here

Note the phase phase shift when \$k=100\$ is greater than 90 deg before the loop-gain declines below unity.

Hence when \$k=100\$, you have satisfied the conditions for oscillation.

\$\endgroup\$
0
\$\begingroup\$

Think about what happens with a simpler value for G(s) such as a 2nd order low pass filter formed from an inductor and capacitor. In open loop you have this circuit: -

enter image description here

And the phase and gain response is like this: -

enter image description here

Picture source

At DC there is no phase shift between output and input. At resonance there is 90 degrees of phase shift and at high frequencies you tend to get 180 degrees of phase shift but with very little output signal.

This circuit on its own cannot oscillate and, even if you put it within a feedback loop there will never be enough phase shift to produce oscillation. It could however produce over shoot like this: -

enter image description here

Picture source

But it won't be sustained i.e. it will die down.

However, your G(s) has an integrator cascaded with a 3rd order low pass filter so it could definitely oscillate if you don't control the feedback.

Consider the 2nd order RLC circuit above. If you added a phase shift of 90 degrees by cascading the RLC circuit with an integrator and "closed the loop" then it would likely oscillate close to resonance because at just past resonance the phase shift will be exactly 180 degrees and the gain (|G(s)|) could be greater than unity.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy