Imaginary Axis
Figure WE6.3: Root locus plot with PI control, KI = 4.5.
7 6 5 4 3 2 1 0 1
20
15
10
5
0
5
10
15
20
Root Locus
Real Axis
Imaginary Axis
Figure WE6.4: Root locus plot with PI control, KI = 6.
14 WORKED EXAMPLE SOLUTIONS 161
Problem 7
1. The t
If n 0, 0
K
GsK
s
the steady-state errors to various standard inputs, obtained from
Equations (5), (6), (7) and (8) are
00
0202
111
111
11
11
0
lim l
Position
im
li
Velocity
Acceleration m lim
ss
p
ss
ss
ss
ss
GK
K
e
K
e
sG s s
e
sGssK
(11)
ME 413 Systems
de t
u t K e t K e t dt K
d
t
i
p
d
t
pi
()
()
1
()
()
()()()
0
0
i
d
d
i
p
K
K
T
K
i
K
where T ,
proportional gain integral gain
derivative gain
integral time constant derivative
time constant
MCEN 467 Control Systems
Controller Effects
A proportional
14 WORKED EXAMPLE SOLUTIONS 149
The two conditions are
4.6
_!n 0.1
_ 0.5
Hence we require
4.6
(25 + 50KT )/2 0.1
25 + 50KT
2
p
50Kp 0.5
The first gives
KT 1.34
If KT = 1.34 then the second is satisfied with
Kp 170
14 WORKED EXAMPLE SOLUTIONS 150
Problem 2
0.5 1
s
n
T
s
ME 413 Systems Dynamics & Control Section 10-4: Stability
1/20
Chapter 10
Time Domain Analysis and Design of
Control System
9
A. Bazoune
10.7 STABILITY ANALYSIS
Stability Conditions Using Rolling Ball
The base of the hollow is in equilibrium
MCEN 467 Control Systems
PID Controller
In the time domain:
The signal u(t) will be sent to the
plant, and a new output y(t)
will be obtained. This new output y(t)
will be sent back to
the sensor again to find the new error
signal e(t). The
controllers
e
ss
ss
v
A
e
K
r(t) At
0
ss
ss
e
ss
e
e
ss
a
A
e
K
21
2
r(t) At
1 ss
p
A
e
K
r(t)A
0
0
e
ss e
ss
where 0p K G is defined as the position error constant.
where
0
lim
v
s
K sG s
is defined as the velocity error constant.
where
2
0
lim
a
s
KsGs
is defi
9/11
/ Example 5
Use block diagram algebra to solve the previous example.
/ Solution
Multiple-Inputs cases
In feedback control system, we often encounter multiple inputs (or even multiple output
cases). For a linear system, we can apply the superposition
2
/| |
kL( j
)
1
S
Figure 92: Infinity norm of the sensitivity as a stability margin.
21.510.500.511.52
2
1.5
1
0.5
0
0.5
1
1.5
2
1/gm
kL( j
)
Figure 93: Gain margin.
Gain margin: If kL(j!) crosses the negative real axis between 0 and 1 at
1/gm, then gm i
10
1
10
0
10
1
10
2
10
3
270
225
180
135
90
Phase (deg)
Bode Diagram
Frequency (rad/sec)
Figure T2.3: Bode plot of a plant.
TUTORIAL 3 168
Tutorial 3
Question 3.1
Consider the control system depicted in figure T3.1. The plant has transfer
function
G(s) =
x
in the original equation.
The above characteristic equation is:
Zero element in the first column,
we cant continue as usual.
ME 413 Systems Dynamics & Control Section 10-4: Stability
11/20
s
s 4 s 3 2s 2 2s 5 0
Substitute
1
s
x
in the above
4 3 2
B x 1
The control system does not need to
know why the measured
value is not currently what is required,
only that is so.
There are two possible causes of
such a disparity:
The system has been disturbed.
The set point has changed. In the
absence of external
1 + (T2/_)s
_
=k
_
s + 1/T1
s + 1/(_T1)
_
s + 1/T2
s + _/T2
_
The advantage of such a restriction is that the compensator (without the gain
also set to k = 1) can be implemented as a passive circuit with only two resistors
and two capacitors. Both lead an
op-amp has open-loop gain A then
vo(t) = A(vs(t) vf (t)
The feedback mechanism is a voltage divider so the feedback voltage is given as
vf (t) =
Rf
Rf + Rg
vo(t)
It follows that in closed-loop vo(t) can be expressed as
vo(t) =
A
1 + _A
vs(t)
1
_
vs(t) wi
(12)
Thus a system with n 1, or with one integration in Gshas
a zero position error,
a constant velocity error and
infinite acceleration error
3. Type-2 System.
If n 1, 2
K
Gs
s
, the steady-state errors to various standard inputs, obtained from
Equat
resonance, if not be unstable. As an extreme example, suppose the op-amp can
be characterised by the second order transfer function
A(s) =
M
(_s + 1)2
For large M this has phase margin PM 0o, with cross- over frequency !c M/_. The closedloop behaviour is
Figure 98: Nichols chart for Example 2 with k = 1.
360 315 270 225 180 135 90 45 0
100
80
60
40
20
0
20
40
6 dB
3 dB
1 dB
0.5 dB
0.25 dB
0 dB
1 dB
3 dB
6 dB
12 dB
20 dB
40 dB
60 dB
80 dB
100 dB
Nichols Chart
OpenLoop Phase (deg)
OpenLoop Gain (dB)
Figure
0
10
20
Complementary sensitivity
Gain dB
Frequency rad/s
Figure 109: Closed-loop sensitivities with k = 1.
Example: G(s) =
1
s(s + 1)
.
If C(s) = 1 so that kL(s) = G(s), the gain margin is infinite and the phase
margin is 51.8o (Fig 108). The resulting s
_ _ _
_
_
_
In order to compute the output, we resort the above equation into partial fraction
expansion
1212
12
. .
n i i ij
m i ii ik
KKKKKK
Cs
spspspspspsp
_ _ _ _
_ _ _ _
_ _ _ _
The first bracket contains time that originates from the system itself
CsEsGs(2)
Substitution of Equation (2) into (1) yields
1
1
EsRs
Gs
(3)
The steady-state error
follows:
ss
e may be found by use of the Final Value Theorem (FVT) as
0 0 1
lim lim lim
ss
tss
sR s
e e t SE s
G s
(4)
Equation (4) shows that the steady state
S(j!) 0 and T (j!) 1 at low frequencies
S(j!) 1 and T (j!) 0 at high frequencies
Typical (good) values of S and T are shown in Fig 105. The closed-loop
bandwidth !b is shown, usually defined as the frequency at which the complementary
sensitivity drops be
=
k(8 3!2)
(1 + !2)(64 + !4)
+j
k(4! + !3)
(1 + !2)(64 + !4)
The real part is always negative for ! > 0, but tends to zero as ! .
The imaginary part is zero when ! = 0, ! = 2 and as ! . Hence we
can tabulate the following values:
! Re [kL(j!)] Im[kL(j!)]
in particular for this example that for small gain the closed-loop damping ratio
increases as we increase the gain. This effect is reversed for higher gains.
4 3 2 1 0 1 2
3
2
1
0
1
2
3
0.76 0.64 0.5 0.34 0.16
0.86
0.94
0.985
0.76 0.64 0.5 0.34 0.16
0.86
KA
1 + TAs
This change in structure adds one additional pole, moves the two zeros and adds
an additional two zeros. With the choice KF = 0.04 and TF = 1 the zeros are
at 0.40 j10.69 and 1.20 j0.83 while the additional pole is at 1. The
root locus is sho
0
a
A
A
E
AC
BD
C
0
0
0
a
AC
a
C
0
00
AC
C
0
F
CE
Da
E
00
0
0
0
0
0
0
0
a
EF
a
F
000
_ Example 1 Given
2
5432
2 2 25
3 9 16 10
ss
Hs
sssss
Check whether this system is stable or not.
_ Solution
The characteristic equation is:
Construct the Routh array
543
k
(s + 1)3
The forward loop transfer function kL(s) has three stable poles and no
unstable poles, so P = 0. From Fig 84 we see that the Nyquist curve does
not encircle 1 for k positive and k < 8. Hence N = 0 for 0 < k < 8. In
this case the Nyquist criteri
A phase lag compensator has the same structure as a phase lead compensator
C(s) = k
1 + TLs
1 + _TLs
but with _ > 0. Its effect is opposite to that of a phase lead compensator; it
buys extra gain at low frequencies, at the expense of increased phase. In
t
its final value the very first time.
Rise Time. The rise time r T is the time required for the response to rise from 10%
to 90%, 5% to 95%, or 0% to 100% of its final value. For underdamped second order
systems,
the 0% to 100% rise time is normally used.
1 + G(s)C(s)
=
Kp(1 + Td)
16s2 + (10 + KpTd)s + 1 + Kp
We still require Kp = 51. A peak overshoot of 10% corresponds to _ 0.6.
So we require
0.6 =
(10 + 51Td)/16
2 1.8
which gives
Td 0.48
Note that this analysis disregards the effect of the zero that the