A damped control system for aiming a hydrophonic array on a minesweeper vessel has the following open-loop transfer function from the driveshaft to the array.
G
(
s
)
=
K
J
s
2
+
K
d
s
{\displaystyle G(s)={\frac {K}{Js^{2}+K_{d}s}}}
The gain parameter K can be varied. The moment of inertia, J , of the array and the force due to viscous drag of the water, Kd are known constants and given as:
J
=
9
N
m
s
2
r
a
d
−
1
{\displaystyle J=9\,N\,m\,s^{2}\,rad^{-1}}
K
d
=
2
N
m
s
r
a
d
−
1
{\displaystyle K_{d}=2\,N\,m\,s\,rad^{-1}}
The system is arranged as a closed loop system with unity feedback. Find the value of K such that, when the input is a unit step, the closed loop response has at most a 50% overshoot (approximately). You may use standard response curves. Should K be greater or less than this value for less overshoot?
Find the corresponding time-domain response of the system.
The system is now given an input of constant angular velocity, V . For the limiting value of K found above, calculate the maximum value of V such that the array follows the input with at most 5° error. First, let us draw the block diagram of the system. We know the open-loop transfer function, and that there is unit feedback. Therefore, we have:
The closed-loop gain is given by:
H
(
s
)
{\displaystyle H(s)}
=
G
(
s
)
1
+
G
(
s
)
{\displaystyle ={\frac {G(s)}{1+G(s)}}}
=
K
J
s
2
+
K
d
s
J
s
2
+
K
d
s
+
K
J
s
2
+
K
d
s
{\displaystyle ={\frac {\dfrac {K}{Js^{2}+K_{d}s}}{\dfrac {Js^{2}+K_{d}s+K}{Js^{2}+K_{d}s}}}}
=
K
J
s
2
+
K
d
s
+
K
{\displaystyle ={\frac {K}{Js^{2}+K_{d}s+K}}}
We now need to express the closed-loop transfer function in the standard second order form.
ω
n
2
s
2
+
2
ζ
ω
n
s
+
ω
n
2
{\displaystyle {\frac {\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}}}
=
K
J
s
2
+
K
d
s
+
K
{\displaystyle ={\frac {K}{Js^{2}+K_{d}s+K}}}
=
K
J
s
2
+
K
d
J
s
+
K
J
{\displaystyle ={\frac {\frac {K}{J}}{s^{2}+{\frac {K_{d}}{J}}s+{\frac {K}{J}}}}}
We can now express the natural frequency ωn and damping ratio, ζ :
ω
n
2
=
K
J
=
K
9
{\displaystyle \omega _{n}^{2}={\frac {K}{J}}={\frac {K}{9}}}
2
ζ
ω
n
=
K
d
J
{\displaystyle 2\zeta \omega _{n}={\frac {K_{d}}{J}}}
ζ
=
K
d
2
J
J
K
=
K
d
2
1
K
J
=
1
3
K
{\displaystyle \zeta ={\frac {K_{d}}{2J}}{\sqrt {\frac {J}{K}}}={\frac {K_{d}}{2}}{\sqrt {\frac {1}{KJ}}}={\frac {1}{3{\sqrt {K}}}}}
We now look at the standard response curves for second order systems.
We see that for 50% overshoot, we need ζ =0.2 or more.
ζ
=
1
3
K
≥
1
5
{\displaystyle \zeta ={\frac {1}{3{\sqrt {K}}}}\geq {\frac {1}{5}}}
K
≤
25
9
{\displaystyle K\leq {\frac {25}{9}}}
This is the maximum permissible value, thus K should be less than this value for less overshoot. We can now evaluate the natural frequency fully:
ω
n
=
5
9
{\displaystyle \omega _{n}={\frac {5}{9}}}
The output of the second order system is given by the following equation:
y
(
t
)
{\displaystyle y(t)}
=
1
−
1
1
−
ζ
2
e
−
ζ
ω
n
t
sin
(
1
−
ζ
2
ω
n
t
+
sin
−
1
1
−
ζ
2
)
{\displaystyle =1-{\frac {1}{\sqrt {1-\zeta ^{2}}}}e^{-\zeta \omega _{n}t}\sin \left({\sqrt {1-\zeta ^{2}}}\omega _{n}t+\sin ^{-1}{\sqrt {1-\zeta ^{2}}}\right)}
=
1
−
1
0.96
e
−
t
/
9
sin
(
5
0.96
9
t
+
sin
−
1
0.96
)
{\displaystyle =1-{\frac {1}{\sqrt {0.96}}}e^{-t/9}\sin \left({\frac {5{\sqrt {0.96}}}{9}}t+\sin ^{-1}{\sqrt {0.96}}\right)}
=
1
−
1.02
e
−
t
/
9
sin
(
0.544
t
+
0.436
π
)
{\displaystyle =1-1.02e^{-t/9}\sin \left(0.544t+0.436\pi \right)}
We can plot the output of this system:
The tracking error signal, E(s) , is equal to the output's deviation from the input.
E
(
s
)
=
R
(
s
)
−
Y
(
s
)
{\displaystyle E(s)=R(s)-Y(s)\,}
Now, we can find the gain from the reference input, R(s) to the error tracking signal:
E
(
s
)
R
(
s
)
=
R
(
s
)
−
Y
(
s
)
R
(
s
)
{\displaystyle {\frac {E(s)}{R(s)}}={\frac {R(s)-Y(s)}{R(s)}}}
The gain from the input to the error tracking signal of a unity feedback system like this is simply
1
1
+
G
(
s
)
{\displaystyle {\frac {1}{1+G(s)}}}
E
(
s
)
R
(
s
)
{\displaystyle {\frac {E(s)}{R(s)}}}
=
1
1
+
K
9
s
2
+
2
s
{\displaystyle ={\frac {1}{1+{\frac {K}{9s^{2}+2s}}}}}
=
9
s
2
+
2
s
9
s
2
+
2
s
+
2.78
{\displaystyle ={\frac {9s^{2}+2s}{9s^{2}+2s+2.78}}}
Now, R(s) is given by the Laplace transform of a ramp of slope V :
R
(
s
)
=
v
s
2
{\displaystyle R(s)={\frac {v}{s^{2}}}}
We now use the final value theorem to find the value of E(s) in the steady state:
lim
t
→
∞
e
(
t
)
{\displaystyle \lim _{t\to \infty }{e(t)}}
=
lim
s
→
0
s
E
(
s
)
{\displaystyle =\lim _{s\to 0}{sE(s)}}
=
lim
s
→
0
s
9
s
2
+
2
s
9
s
2
+
2
s
+
2.78
V
s
2
{\displaystyle =\lim _{s\to 0}{s{\frac {9s^{2}+2s}{9s^{2}+2s+2.78}}{\frac {V}{s^{2}}}}}
=
2
V
2.78
{\displaystyle ={\frac {2V}{2.78}}}
We require this to be less than
5
∘
=
5
π
180
rad
{\displaystyle 5^{\circ }={\frac {5\pi }{180}}{\mbox{rad}}}
V
≤
2.78
2
×
5
π
180
=
0.12
rad/s
{\displaystyle V\leq {\frac {2.78}{2}}\times {\frac {5\pi }{180}}=0.12\,{\mbox{rad/s}}}
