# Control - response time

1. Sep 25, 2006

### LM741

hey guys...

Say i have a second order system. From this i can establish a peak time, a settling time, and overshoot percentage...
but what is my response time - is the settling time?

Also is there some sort of rule of thumb for decreseing the response time??
like say, for example, increasing the frequency of where my poles occur will decrease my response time??? this question may be bit general as there are probably many ways to decrease response time - but if you could tell what sort of realtionship the frequency of my poles/zeros have on the response time - that would be great!!

thanks

john

2. Sep 25, 2006

### SGT

I don't think there is some property named response time.
There is the settling time: the time the response takes for the oscillations stay below a certain margin (1% or 2%) of the steady state response. To reduce settling time the poles must be moved to the left, far from the imaginary axis.
There is also the rise time: the time the response takes to go from 10% to 90% of the steady state value (before the first overshoot, if it exists). To reduce it the frequency must increase. An overdamped system has a slow rise time. That's why we require the dominant poles to be complex.

3. Sep 26, 2006

### LM741

thanks SGT
something i don't get that you mentioned: "To reduce settling time the poles must be moved to the left, far from the imaginary axis."
the thing is i've got this textbook which portrays the peaks at higher values (on an amplitude vs time graph) as the poles move further to the left...surely this means it will take longer for the signal to settle (i.e. settling time INCREASES). I figure this, beacause the peak has shot up higher therfore it will take longer to settle down...
thanks again

4. Sep 26, 2006

### SGT

If your poles are:
$$p_{1,2} = -\sigma \pm j\omega_d$$
Your transient response will be:
$$Ke^{-\sigma t}cos(\omega_d t + \phi)$$
Your settling time (time for the transient to reach 1% of its initial value) is:
$$t_s =\frac{5}{\sigma}$$
So, the larger is $$\sigma$$, the smaller the settling time.

Last edited by a moderator: Sep 26, 2006
5. Sep 27, 2006

### LM741

do you aggree with me the the peak amplitude increases as the poles move further away from the origin??

thanks for reply

6. Sep 27, 2006

### SGT

Yes. But no matter how big is the initial overshoot, the oscillations will reduce fast.

7. Sep 28, 2006

### LM741

really?? are you trying to say that the settling time is independent of the size of overshoot the system experiences???
i find that very hard to believe??
take a very basic analogy: I perform two experiments with a bouncing ball:
The first experiment i throw the bouncing ball extrelemy hard against a hard flat surface and then wait for the ball to SETTLE down.
The second experiment, i merley drop the ball letting gravity do the work and once again - wait for the ball to SETTLE down.

Surely the first experiment (larger overshoot) will take a longer time for the bouncing ball to settle down??
Thhis is assuming the ball is the signal and the height of the first bounce is the overshoot.

this is why i can't easily except that the settling time is independent of overshoot.

thanks for reply
John

8. Sep 28, 2006

### SGT

Your example with the bouncing ball is not valid. The valid experiment would be:
1. Take a ball and drop it in a hard surface (concrete, for instance).
2. Put a layer of foam in the hard surface and throw the cball very hard against it, so that the first bounce is higher than when the ball fell in the hard surface.
Even if the first bounce is higher in the second case, it will settle faster, because of the damping caused by the foam.
For the second order system, the transient term, as I said is of the form:
$$Ke^{-\sigma t}cos(\omega_d t + \phi)$$
The first peak occurs at a time $$t_p$$, when the cosine equals 1.
The value of the peak is $$Ke^{-\sigma t_p}$$.
The amplitude of the peak depends of the values of K and $$t_p$$. High values of K and small values of $$t_p$$ increase the amplitude.
Both effects happen when the poles are displaced to the left.
The damping depends on $$\sigma$$. After a time $$t_s = \frac{5}{\sigma}$$ the amplitude is only 1% of K. Clearly, the greater is $$\sigma$$, the smaller is $$t_s$$.

9. Sep 28, 2006

### LM741

brilliant analogy with the foam!! thanks!!
so is it safe to say that the settling time is TOTALLY independent of the overshoot/undershoot amplitude?- r or is this definition to general!! it seems like at the end of the day - the settling time really depends on the damping factor???
thanks so much again.
John

10. Sep 28, 2006

### SGT

Yes, settling time is the time the response takes to fall below 1% of its final value. Since $$e^{-5} = 0.0067$$, we see that after 5 time constants the transient term is only 0.67% of the maximum amplitude, no matter what its value.

11. Sep 29, 2006

### LM741

thanks sgt!!!
really helps

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