Modern Control Engineering

aa

Section 5–3 / Second-Order Systems 177

This value must be 0.2. Thus,

or

which yields

The peak time tpis specified as 1 sec; therefore, from Equation (5–20),

or

Sincezis 0.456,vnis

Since the natural frequency vnis equal to

Then Khis, from Equation (5–25),

Rise timetr: From Equation (5–19), the rise time tris

where

Thus,tris

Settling timets: For the 2%criterion,

For the 5%criterion,

ts=

3

s

=1.86 sec

ts=

4

s

=2.48 sec

tr=0.65 sec

b=tan-^1

vd

s

=tan-^1 1.95=1.10

tr=

p-b

vd

Kh=

21 KJz-B

K

=

21 Kz- 1

K

=0.178 sec

K=Jv^2 n=v^2 n=12.5 N-m

1 KJ,

vn=

vd

21 - z^2

=3.53

vd=3.14

tp=

p

vd

= 1

z=0.456

zp

21 - z^2

=1.61

e-Az^21 - z

(^2) Bp
=0.2