# Settling time and overshoot

1. Dec 15, 2012

### MathematicalPhysicist

1. The problem statement, all variables and given/known data + the attempt at solution
I have the next transfer function:

$$H(s)=\frac{P(s)}{1+C(s)P(s)}$$

where $$P(s)=\frac{20}{(s^2-2s+9)(s+100)}$$

$$C(s)=m(s+40)$$

Now I want to find the settling time and overshoot of $$H(s)$$ as $$m\rightarrow \infty$$ to whithin 5 percent accuracy.

Now what I thought when $$m\rightarrow \infty$$ $$H(s)\approx \frac{1}{C(s)}$$.

So basically in this case the quadratic equation (in the notation of next webpage:
http://wikis.lib.ncsu.edu/index.php/Second_Order_Dynamics) is:

$$s^2+(-2s+20ms)+(9+800m)$$

Is this right or am I just rambling nonsense here, I am lost here.

2. Relevant equations
I know that when we have the Quadratic ploynomial in the denominator of $$G(s)=P(s)C(s)$$ s.t:
$$s^2+2\zeta \omega_n s +\omega_n ^2$$

then the settling time is given by:

$$t_s \approx \frac{3}{\zeta \omega_n}$$

and the overshoot is given by:

$$\sigma = exp(\frac{-\zeta \pi}{\sqrt{1-\zeta^2}})$$

Any hints or advice are welcomed and much appreciated.

Thanks.

2. Dec 16, 2012

### aralbrec

Are you familiar with the root locus method? It describes the path the roots take when a single parameter, like m, is varied. I would need to do a little review before giving a better answer but maybe this will get you started.

3. Dec 17, 2012

### MathematicalPhysicist

Yes I know about the RL.

I am not sure how to calculate the overshoot and settling time here, though I know that I need to look at the Bode graph of $$\frac{1}{m(s+40)}$$

And I know that the frequency that the asymptote to the the magnitude of the above expression intersects the $$\omega$$ axis is given by: $$\omega= 40$$.

Other than that, I am not sure how to procceed, any advice?

4. Dec 19, 2012

### rude man

My reaction is to agree with you that as m → ∞, H(s) = 1/C(S) = 1/m(s+40).

The impulse response to this H(s) is simply (1/m)exp(-40t). The step response is simply (1/m)[1 - exp(-40t)]. In neither case is there any overshoot. So I am as confused as you ...

5. Dec 20, 2012

### MathematicalPhysicist

And what of the settling time of 1/C(s)?

6. Dec 20, 2012

### rude man

Since the output is an exponential decay you can compute settling time depending on your definition of "settling time". For example, if the input is a step function, and "settling time" ts is defined as 95% of "final" value, then you solve for ts from 1 - exp(-40ts) = 0.95.

7. Dec 20, 2012

### MathematicalPhysicist

OK, thanks.

Then the overshoot is zero.

8. Dec 20, 2012

### aralbrec

The problem with m→∞ is that the response goes to zero. So you can't let m→∞. Maybe you can say m becomes large enough that it dominates the response. Then I do agree with what you have been saying. You can call the (s+100) pole insignificant as I think you did or you could do this, which is another way to deal with these sorts of boundary case problems:

$0=1+C(s)P(s)=s^3+98s^2+(20m-191)s+(900+800m)$

Making m large can simplify this by making most terms too small to be counted. Dividing by m yields this:

$0=\frac{1}{m}(s^3+98s^2-191s+900)+20s+800\approx s+40$

This means you have a single pole response as you found.

Your root equation can be written in the standard form with the parameter m in the right place for root locus:

$1+C(s)P(s)=1+m\frac{20(s+40)}{(s^2-2s+9)(s+100)}=1+mQ(s)$

The root locus method says as m begins at 0 and approaches ∞, the roots of this equation begin at the poles of Q(s) and end at the zeroes of Q(s). Q(s) has two zeroes at ∞ and one zero at -40. It has three poles, one at -100 and two complex conjugate ones. The conclusion from root locus is that with m→∞, there will be two roots at ∞ and one at -40. The finite root is (s+40) as found above.

Last edited: Dec 21, 2012