Searching for the third pole on the RL at each of the
corresponding gains 7.36, 12.79 and 39.64.
Note that the third pole cannot be complex
as the CLTF is third order, ie. the third pole
must be on the real axis.
The poles are at s = -9.25, -8.6 and 1.8 respectively.
Root L ocus
10
8
6
4
System:System:
sys
sys
Gain: 7.4
G ain: 12.8
Pole: -9.25
Po le: -8.61
Dampin g:
Da1mping: 1
OvershoOotversh
(%): oot
0 (%): 0
Fre quenFreque
cy (ra d/se
ncyc):
(rad/sec):
9 .2 5
8.61
Imag Axis
2
System: sys
Ga in: 3 9.2
Pole: -1.8
Damping: 1
Ove rsho ot (%): 0
Frequen cy (ra d/sec): 1.8
0
-2
-4
-6
-8
-10
-1 0
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
Real A xis
Using Ts =
4
ςωn
; Tp =
π
ωd
and the real and imaginary parts of the dominant pole.
The velocity error constant is
sK ( s + 1.5)
K (1.5)
K v = lim sG (s ) = lim
=
s→ 0
s →0 s( s + 1)( s + 10)
(1)(10)
Subst. various values of K gives the K v values in the Table below.
1
Second- and third-order
responses
a. Case 2; b. Case 3
Cases 1 and 2 have the third
pole far away from the
complex pair. However,
there is no approx. pole zero
cancellation.
Case 3, the third pole is
closer to the zero, so a 2nd
order approx can be
considered valid.
The plots are relatively
close.
A step input is used to show the second order dynamics
and validity of the second order approximation.
We will now re-evaluate for the third pole analytically
knowing the gain at the corresponding dominant pair.
As an example, Case 3 will be used.
In case 3, K = 39.64. Using the magnitude criterion
assuming the third pole p exists somewhere between the
pole at − 10 and the zero at − 1.5. The point in question is p.
L1 = 10 − p
L2 = p − 1
L3 = p
L4 = p − 1.5
(10 − p )( p − 1) p
= 39.64
( p − 1.5)
Solving for p gives
K =
p = −1.795 which is the third pole.
The same can be done for the finding
the other third poles. Note the RL does NOT
exist on portions of the real axis where the sum
of poles and zeros to the left is even.
2
The closed loop TF is using K = 39.64
C ( s)
39.64 s + 59.64
=
R ( S ) s3 + 11s 2 + 49.64 s + 59.64
1
(discuss Matlab)
s2
The figure shows that there is a steady state error e( ∞)
of (4.5 − 4.33) = 0.17.
Since R ( s) =
NOTE THAT THIS IS =
1
Kv
For Case 3, ( K = 39.64 ) K v = 5.9
e (∞) =
1
= 0.17 which matches the Matlab plot.
5.9
Ste p Re spo n se
6
5
Syst em: sys
Time (se c): 4. 5
Amp lit ud e : 4 .3 3
Amp litude
4
3
2
1
0
0
1
2
3
4
5
6
Time (se c)
Example
Given a unity feedback system that has a forward TF
K
G (s ) =
, do the following
(s + 2)(s + 4)( s + 6)
(a) Obtain a RL using Matlab
(b) Using a 2nd order approx. determine the value of
K to give a 10% overshoot for a unit step input
(c) Determine settling and peak times
(d) The natural frequency
(e) The SSE for the value of K .
(f) Determine the validity of the 2nd order approx.
3
Repeat part (b) using the magnitude criterion.
This is to be done at the point − 2.028 + j 2.768
From the poles
L1 = (2.028 − 2) 2 + (2.768)2 = 2.76814
L2 = (2.028 − 4) 2 + (2.768) 2 = 3.399
L3 = (2.028 − 6) 2 + (2.768) 2 = 4.841
There are no zeros
K=
L1 L2 L3
= 45.55
1
(d) ωn = 2.0282 + 2.7682 = 3.432
(e) The system is Type 0, the position error constant is
K
45.55
=
= 0.949
2 *4 *6
48
1
Therefore, e(∞) =
= 0.51
1+ K p
lim G( s ) = K p =
s→ 0
(f)
In this case, K = 45.55. Using the magnitude criterion
assuming the third pole p exists somewhere to the left of the
pole at − 6. The point in question is p.
L1 = p − 6
L2 = p − 4
L3 = p − 2
4
( p − 6)( p − 4)( p − 2)
= 45.55
1
Solving for p gives
p = −7.943 which is the third pole.
K=
Since this third close loop pole is NOT 5 times or more
the magnitude of the real part of the closed loop dominant
pole, the second approx. NOT valid.
Roo t L o cu s
10
8
6
4
S yste m: sys
Ga in : 4 4. 6
P o le: -7 .9 2
Da mp in g: 1
Oversh oo t (%): 0
Fre qu e ncy (ra d/ se c): 7 .9 2
Ima g Axis
2
0
-2
-4
-6
-8
-1 0
-1 6
-1 4
-1 2
-1 0
-8
-6
-4
-2
0
2
Re al A xis
5