Standard Form for second order systems.

Click For Summary
In a second order system represented by T(s) = (ks + c) / (s^2 + as + b), achieving specific performance metrics like rise time, maximum overshoot, and settling time requires adjusting uncertain parameters. The standard form of a second order system is expressed as ωn^2 / (s^2 + 2ζωn s + ωn^2), where ωn is the natural frequency and ζ is the damping ratio. The numerator's constant does not influence the damping ratio or natural frequency, as these parameters are determined by ratios. Instead, it acts as a scaling factor for system response plots. Understanding this relationship is crucial for effectively tuning the system to meet desired performance criteria.
zoom1
Messages
63
Reaction score
0
Suppose there's a system with given uncertain parameters. And I would like to obtain certain Rise time, max. over shoot, settling time by adjusting those parameters.

Let's say this is the second order system;

T(s) = (ks + c) / (s2 + as + b)

First of all; for a second order system there is a standard form which is;

Wn2 / s2 + 2ζωns + ωn2

As we have to have the Wn2 in the numerator, it's not that way always. Just like in the example. So, what am I suppose to do at that point ?

If the transfer function T(s) looks like exactly the standard form, I could get the desired values by changing parameters. I think.
 
Engineering news on Phys.org
zoom1 said:
First of all; for a second order system there is a standard form which is;

Wn2 / s2 + 2ζωns + ωn2
Parentheses, please!

ωn2 / ( s2 + 2ζωns + ωn2 )

The constant in the numerator doesn't affect ζ, nor ωn, nor parameters such as % overshoot, rate of gain fall-off, etc., since these are calculated as ratios. The numerator is just a scaling factor for the plots.

The general expression is: A.ωn2 / ( s2 + 2ζωns + ωn2 )

where A can be seen to be the low frequency gain of this system.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

Replies
1
Views
4K
Replies
3
Views
7K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
3K
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 14 ·
Replies
14
Views
509
Replies
13
Views
2K