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In the context of control theory, I found a text stating:

"Higher the loop gain of the system, larger is the percent overshoot"

Can anyone show that analytically? Consider it to be a second order system.

Edit: Doing the Maths I figured out that for a Step input, percentage peak overshoot is independent of the magnitude of step size, and is related only to damping ratio, zeta:

enter image description here

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  • \$\begingroup\$ How about you start this off with developing the math of a generic Low pass 2nd order open loop block Then adding an error amplifier with a gain block. If you want an analysis then get cracking on describing the opening scenario so that multiple answers are using the same model. \$\endgroup\$
    – Andy aka
    Jan 12 '20 at 11:11
  • \$\begingroup\$ While you are at that little task you should also consider that not formally accepting any answer to your 22 previously raised questions might be seen by some as rude. I’m sure you weren’t aware of this politeness; but now you are. \$\endgroup\$
    – Andy aka
    Jan 12 '20 at 11:14
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    \$\begingroup\$ Oh I did not know much of that stuff and how that works... I read about it just now! I think I was a bit reluctant to read much about the website's working. I am sorry if that seemed to be rude on my part. I really appreciate all the experience people have here and their willingness to help people like me. I did upvote the answers but didn't accept any. I'll accept the answers now onwards! \$\endgroup\$
    – Bhuvnesh
    Jan 12 '20 at 11:28
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    \$\begingroup\$ That’s good. If there isn’t an answer that is acceptable then leave a comment asking for clarification or, if you have no answers to some questions then maybe explain your own solution or revamp the question in some way. Sounds like you’re doing the right thing so good on you. \$\endgroup\$
    – Andy aka
    Jan 12 '20 at 11:36
  • \$\begingroup\$ Wow, your penmanship drastically improved for 100 *! \$\endgroup\$
    – pipe
    Jan 12 '20 at 23:12
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"Higher the loop gain of the system, larger is the percent overshoot"

Can anyone show that analytically? Consider it to be a second order system.

My answer is based on assuming that a 2nd order system is modified with gain and put inside a control loop hence, the system can then be regarded as having “loop gain”. This then makes it a “control-system” as per the tag in the question and, the term “loop gain” then makes sense. Other answers may not have made this assumption: -

enter image description here

Starting with an open-loop, 2nd order, low pass system; it may have an output/input equation similar to this: -

$$H(s) = \dfrac{\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}$$

Where \$\omega_o\$ is the natural resonant frequency and \$\zeta\$ is the damping ratio.

This is modified to include a gain factor (\$G\$) and, inserted inside a unity gain closed loop with negative feedback. The new equation for H(s) becomes: -

$$H(s) = \dfrac{\dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}}{1 + \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}}$$

$$= \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2 + G\cdot\omega_o^2}$$

$$= \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2 (1 + G)}$$

Notable is the \$1+G\$ term in the denominator because it modifies the old natural resonant frequency from \$\omega_o\$ to a new resonant frequency (\$\omega_n\$) that is \$\sqrt{1+G}\$ times bigger.

And, the important thing here is that this has the effect of lowering the damping ratio (in other words the old damping ratio is greater than the new damping ratio i.e.\$\zeta_o > \zeta_n\$). Hence, the terms in front of s (namely \$2\zeta\omega_o\$) relative to the new natural resonant frequency (\$\omega_n\$), become: -

$$\dfrac{2\zeta\omega_o}{\omega_n} \Rightarrow\dfrac{2\zeta\omega_o}{\omega_o\sqrt{1+G}} = \dfrac{2\zeta}{\sqrt{1+G}}$$

This, in effect, lowers the old damping ratio and creates more overshoot.

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For a standard 2nd order TF: $$ G(s)=\frac{\omega_n^2}{s^2+2\zeta \omega_n s+\omega_n^2} $$

the unit step response for: \$ 0\le\zeta<1 \$, is: $$ c(t)= 1-\frac{1}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_n t}\:sin\left(\omega_n \sqrt{1-\zeta^2}\:t+\phi\right), \:\:\phi = cos^{-1}(\zeta) $$

Differentiating and setting to zero, \$\large\frac{dc(t)}{dt}=\small 0\$, gives the 'time to 1st peak':

$$t_p=\frac{\pi}{\omega_n\sqrt{1-\zeta^2}} $$

Substituting this value of time into the \$c(t)\$ expression:

$$ c(t_p)= 1-\frac{1}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_n t_p}\:sin\left(\omega_n \sqrt{1-\zeta^2}\:t_p+\phi\right), \:\:\phi = cos^{-1}(\zeta) $$

The sine term evaluates to:

$$sin\left(\omega_n \sqrt{1-\zeta^2}\:t_p+ cos^{-1}\zeta\right)=-\sqrt{1-\zeta^2} $$

giving the magnitude of the 1st peak (above zero):

$$c(t_p)= 1+e^{-\zeta\omega_n t_p} $$

Substituting for \$t_p\$:

$$c(t_p)= \large 1+e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}}$$

Hence the % overshoot is:

$$PO= 100\large\:e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}}$$

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For 2nd order systems, ζ=1 for 0% step overshoot.

For greater than 2nd order, ζ needs to greater than 1 for 0% step overshoot. These are called Bessel response filters. These have the lowest component sensitivity and lowest Q<1, ζ>1 filters of all filter types with orders >2.

Butterworth filter
Chebyshev filter
Elliptic (Cauer) filter
Bessel filter
Gaussian filter
Optimum "L" (Legendre) filter
Linkwitz–Riley filter

Bessel Filters compromise "sharpness of spectral edges" of the frequency response, as the tradeoff for 0% step overshoot. You are taught about Stephen Butterworth's 1930 filter where the frequency response is montonic and flat with no frequency overshoot for 2nd order but has time overshoot on the step response which increases greatly with order of filter.

The ideal BESSEL filter stays at 0% overshoot regardless of filter order.

enter image description here

For a 2nd order system, your formula indicates

\$\zeta=0,1~~for~~ \dfrac{\zeta \pi}{\sqrt{1-\zeta^2}}=0\$

For higher order filters/systems Bessel filters use different low Q<1, ζ>1 values at different poles to give best shape factor with no overshoot.

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Consider a closed loop transfer function of the form of a theoretical 2 pole op amp.

Transfer Function

Where w1 and w2 are the angular frequencies of the two poles.

w1 is the angular frequency of the dominant pole and w2 is the angular frequency of the higher frequency pole.

If we were to increase w1 (by reducing the size of the compensation capacitor) then we are effectively increasing the open loop gain and therefore the loop gain (without increasing the dc open loop gain) and because zeta, the damping ratio, is equal to the average of the two pole frequencies divided by the natural resonant angular frequency….

Equations

….it means that zeta has reduced reducing stability.

If instead we were theoretically able to push the higher frequency pole up in frequency then this also increases the open loop and loop gains (all be it at higher frequencies) but in this case zeta increases improving stability. Although the loop gain has increased at higher frequencies, the loop phase lag has reduced improving phase margin, and therefore zeta has increased and stability has been improved.

Having said that, increasing the loop gain by increasing the dc open loop gain increases the open loop gain at all frequencies in which case zeta is reduced as are stability margins.

Just as an aside. In a practical op amp it is very difficult to move the poles independently like this. As you move the dominant pole down in frequency, the higher frequency pole simultaneously moves up in frequency amplifying the stability improvement. This is an effect known as pole splitting.

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