Step Response Of 2nd Order Instruments

Step Response of Second-order Systems and Damping Ratio H. R. Pota {[email protected]} June 12, 2005 Transfer function for a second-order system can be written as, ωn 2 (1) s2 + 2ζωn s + ωn 2 where ζ, ωn > 0. The transfer function is parameterised in terms of ζ and ωn . The value of ωn doesn’t qualitatively change the system response but there are three important cases—with qualitatively different system behaviour—as ζ varies. The three cases are discussed below. G(s) =

(a) ζ > 1 This is known as an overdamped system. To see why, let’s look at the step-response for this case. 1 ωn 2 Y (s) = s (s + p1 )(s + p2 ) p2 1 p1 1 1 − + (2) = s p2 − p1 s + p 1 p2 − p1 s + p 2 √ √ where p2 > p1 > 0, p1 = ωn (ζ − ζ 2 − 1), and p2 = ωn (ζ + ζ 2 − 1). This gives:  √  √   ! ! −ωn ζ− ζ 2 −1 t −ωn ζ+ ζ 2 −1 t 1 1 ζ ζ √ 2 √ 2 +1 e −1 e y(t) = 1 − + 2 2 ζ −1 ζ −1  q    q ζ e−ωn t sinh ωn ζ 2 − 1 t (3) = 1 − e−ωn t cosh ωn ζ 2 − 1 t − √ 2 ζ −1 In the above equation (3) it can be seen that 

cosh ωn

q

ζ2





− 1 t > 0 and sinh ωn

q

ζ2



− 1 t > 0,

∀t > 0.

This means that two positive numbers are always subtracted in equation (3) from the steadystate value of 1. Hence the step-response y(t) in equation (3) never goes above 1, hence the classification of overdamped. (b) ζ < 1 This is known as an underdamped system. To see why, let’s look at the step-response for this case. 1 ωn 2 Y (s) = s s2 + 2ζωn s + ωn 2 1 s + 2ζωn = − 2 s s + 2ζωns + ωn 2 1 ζωn s + ζωn = − − (4) 2 2 2 s (s + ζωn ) + ωn (1 − ζ ) (s + ζωn )2 + ωn 2 (1 − ζ 2 )

This gives: ζe−ζωnt sin ωd t − e−ζωn t cos ωd t y(t) = 1 − √ 1 − ζ2 e−ζωn t = 1− √ sin(ωd t + ψ) (5) 1 − ζ2 √ √ 1−ζ 2 −1 2 where ωd = ωn 1 − ζ and ψ = tan . ζ It can be clearly seen that the step-response in equation (5) will overshoot the steady-state value of 1 as the sinusoidal function takes negative values, hence the classification as underdamped system. (c) ζ = 1 This is the critically damped case. To see why let’s look at the step-response. Step-response: 1 ωn 2 s (s + ωn )2 1 1 ωn − − = s s + ωn (s + ωn )2

Y (s) =

(6)

This gives, y(t) = 1 − e−ωn t − ωn te−ωn t .

(7)

Comparing the above expression (7) for the step-response with the expression for overdamped case in expression (3), we see that since (for ζ > 1): 

cosh ωn

q

ζ2



− 1 t > 1,

∀t > 0 and √

ζ ζ2 − 1



sinh ωn

q

ζ2



− 1 t > ωn t,

∀t > 0,

the step-response of a critically damped system will always be higher than the overdamped system step-response (ζ > 1). In other words, of all the second order damped systems with real poles (parameterised by ζ as in expression (1)) this will reach the steady-state in the shortest time. That’s why it’s called critically damped. This can be clearly seen in Figure 1. Second order system step−response 1.4 ζ=2 ζ=1 ζ=0.3 1.2

Step Response

1

0.8

0.6

0.4

0.2

0

0

0.5

1

1.5

2

2.5

Time (s)

Figure 1: Step-response of Second-order System (ωn = 10)

2