R4 A point s0 on the real axis, i.e. s0 R, is part of the root locus (for 0),
if and only if, it is located to the left of an odd number of poles and zeros (so
that (5.6.6) is satisfied).
R5 When is close to zero, then n of the roots are located at the po
With respect to the important characteristic of stability, a continuous time
system is
Section 4.14. Summary 107
stable if and only if the real parts of all poles are strictly negative
marginally stable if at least one pole is strictly imaginary and no
function C(s) is designed to give a particular reference signal response; i.e.
Y (s)
R(s)
= Go(s)C(s)
1 + Go(s)C(s)
(5.2.8)
then this induces a unique output disturbance response
Y (s)
Do(s)
=
1
1 + Go(s)C(s)
(5.2.9)
without any further design freedom.
Fr
When a system has a non ideal frequency response we say it introduces distortion.
To describe the different kinds of distortion that we commonly met in practice,
consider a signal f(t) given by
92 Continuous Time Signals and Systems Chapter 4
f(t) =
_nf
i
m.s.t.
Figure 6.6. Plant step response
The parameter setting rules proposed in Table 6.2 are applied to the model
(6.4.1), where we have again normalized time in delay units. Loop responses for a
unit step reference are shown in Figure 6.7 on the next pag
Remark 7.4. Canceled nominal model poles and zeros will still appear,a s poles or
zeros,in some closed loop transfer functions. The discussion above shows that any
cancellation will force the canceled factor to be present in the closed loop polynomial
Acl
= 1/1 without a significant change in gain. Thus if we have a simple feedback
loop which passes through the 1 point at, say, frequency 1, then inserting a
lead compensator such that 11 = 1 will give a 45 phase margin. Of course,
the disadvantage is an in
Table 4.5 gives the time and frequency responses of simple linear models. The
reader is encouraged to evaluate some of these responses to verify the results since
familiarity with these responses can be very helpful when diagnosing a control problem,
or f
s2 + 0.707s+ 0.250
(4.16.15)
H2(s) =
0.0625
s4 + 1.3066s3 + 0.8536s2 + 0.3266s+ 0.0625
(4.16.16)
4.16.1 Draw their Bode diagrams and verify that both exhibit a low-pass filtering
behavior. Compute the bandwidth for each filter.
4.16.2 Compute the unit ste
In subsection 8.3.1, we learned that control loop performance is limited by the maximal
available movement available from actuators. This is heuristically reasonable.
What is perhaps less obvious is that control systems are often also limited by minimal
a
144 Analysis of SISO control loops Chapter 5
is retained when the true plant is controlled by the same controller. We call this
property robust stability.
Sufficient conditions for a feedback loop to be robustly stable are stated in the
following:
Theorem
sign(t) +
1
3e
3t(t) (4.10.7)
One of the more attractive features of the Fourier transform is its connection
to frequency response, since (4.10.2) describes f(t) as a linear combination of exponentials
of the form ejt, where ranges continuously from to .
acting individually
a powerful body of frequency domain properties and results holds: poles,
zeros, transfer functions, Bode and Nyquist plots with their associated
properties and results
relative degree, inversion, stability and inverse stability are e
We first compute the closed loop characteristic polynomial,which is given by
p(s) = s3 + 2s2 + s + K,an d we then build the Rouths array
s3 1 1
s2 2 K
s1 1 0.5K
s0 K
We can now see that there are no closed
loop unstable poles if and only if 1
0.5K > 0 an
G21(s) =
34.7e9.2s
8.15s+ 1
(6.7.4)
G22(s) =
0.87(11.6s+ 1)es
(3.89s + 1)(18.8s+ 1)
(6.7.5)
Note that the units of time here are minutes.
Also, note that u1 not only affects y1 (via the transfer function G11) but also
y2 (via the transfer function G21). S
delay that should be added to achieve the same critical condition.
|a|
1
(a) (b)
1
Go(j)C(j) Go(j)C(j)
= p
= s
Figure 5.7. Stability margins and sensitivity peak
Part (b) in Figure 5.7 defines an alternative indicator for relative stability. We
first re
We will next consider some examples.
Example 5.3. Let p(s) = s4 +s3 +3s2 +2s+1. Inspection of the polynomial does
not allow us to say whether or not it is a Hurwitz polynomial. We proceed to use
Rouths array.
s4 1 3 1
s3 1 2
s3 1 1
s1 1
s0 1
From the arra
sin(j)
(4.12.13)
where
= o
2
and = o +
2
(4.12.14)
The corresponding magnitudes are
|G_(j)| =2
_
sin
_
o
2
_
1
o
_
|F(j)| (4.12.15)
|G(j)| =2
_
sin
_
o
2
_
1
o
_
(4.12.16)
These expressions, together with the assumption that |F(j)| vanishes for very
hi
CPID(s) = Kp + KI
s
+ KDs
Ds + 1
(7.3.2)
For future reference, we note the following alternative representation of a PID
controller.
Lemma 7.2. Any controller of the form
186 Synthesis of SISO controllers Chapter 7
C(s) = n2s2 + n1s + no
d2s2 + d1s
(7.3.3
illustrated in Figure 6.2, which shows that the filtered derivative approximates
the exact derivative well at frequencies up to 1
D
[rad/s], but that it has finite
gain at high frequencies, whereas an exact derivative has infinite gain.
Since D _= 0 was l
the MME.
Solution
We have that
G(s) = e
sF(s) Go(s) =
_s + 2k
s + 2k
_k
F(s) (4.12.4)
Hence
G(s) = e
s
_s + 2k
s + 2k
_k
1 (4.12.5)
with frequency response magnitude given by
|G(j)| =
_
e
j n_ _c_a_n_ _b e
j2k
_
= arctan
2k
(4.12.6)
The result is depic
. If we make the residue equal to zero we obtain c =
2 and
c = 2.
The above rules allow one to sketch a root locus diagram. This can also be
obtained using the MATLAB command rlocus; the result is shown in Figure 5.3.
A powerful MATLAB environment, rltool
Control system design makes use of two key enabling techniques: analysis and synthesis.
Analysis concerns itself with the impact that a given controller has on a given
system when they interact in feedback while synthesis asks how to construct controllers
integral in [0,) vanishes. In other words, (8.6.54) implies that y(t) must evolve
along the opposite direction to that of the desired value, during one (or more) non
zero duration interval. This leads to the terminology of inverse response for the
behavio
1
Tr s
_
(6.2.2)
CPD(s) = Kp
_
1 + Tds
Ds + 1
_
(6.2.3)
CPID(s) = Kp
_
1+
1
Tr s
+ Tds
Ds + 1
_
(6.2.4)
where Tr and Td are known as the reset time and derivative time, respectively.
As seen from (6.2.1) to (6.2.4), the members of this family include, in
.
.
.
a0 b0
(7.2.20)
181
In view of Sylvesters theorem (see Theorem 7.1),t he matrix S is nonsingular
if and only if the polynomials Bo and A0 are relatively prime. The result follows.
Two additional cases will now be considered:
Case 1, nc = 2n 1 + , wit
Problem 5.8. Consider a feedback control loop with nominal complementary sensitivity
To(s). Assume that,i n the true feedback control loop, the measurement
system is not perfect,an d that the measured output, Ym(s) is given by
Ym(s) = Y (s) + Gm(s)Y (s) (
Rivera, D., Morari, M., and Skogestad, S. (1986). Internal model control. PID
controller design. Ind. Eng. Chem. Process Des. Dev., 25:252265.
Smith predictor
Smith, O. (1958). Feedback control systems. McGraw Hill, New York
Section 7.7. Problems for the