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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