Download Lecture 24: LTI Differential Systems and Rational

Lecture 25: Block Diagram Representation of LTI Differential Systems
8.3
Analysis of LTI Differential Systems Using Block Diagrams
Block diagrams are useful to analyze LTI differential systems composed of subsystems. They are also
used to represent a realization of an LTI differential system as a combination of three basic elements: the
integrator, the gain, and the summing junction.
8.3.1
System Interconnections
We have already studied system interconnections using block diagrams in the context of general systems
(not necessarily linear) and also in the context of LTI systems with the convolution and Fourier
transforms. However, this approach to analyze complex systems becomes quite powerful when one uses
transfer functions to describe the interconnected LTI systems. The reason is that the space of all transfer
functions forms an algebra with the usual arithmetic operations (+,-,x,/). That is, the sum, difference,
multiplication, or division of two transfer functions yield another transfer function.
Example: Find the step response of the car cruise control system depicted below to a step in the desired
velocity input vdes from 0 km/h to 100 km/h, where
1
, Re{s}  3
s3
10
K ( s)  , Re{s}  0
s
F ( s)  2, s  C
P ( s) 
(8.14)
and the time unit is a minute.
vdes (t )  100u(t )
+
e( t )
K ( s)
u( t )
P ( s)
vm ( t )
v (t )
F ( s)
This feedback system represents a car cruise control system where
P( s) is the car's dynamics from the
F ( s) is a scaling factor on the tachometer speed
measurement to get, say, km/h units, and K ( s) is the controller, here a pure integrator that integrates
the error between the desired speed and the actual car speed to obtain a throttle command u( t ) for the
engine throttle input to the speed output,
engine.
It is interesting to note that the theory of Laplace transforms and transfer functions allows us to analyze
within a single framework a mechanical system (the car), an electromechanical sensor (the tachometer),
and a control circuit or a control algorithm residing in a microcontroller chip.
The first task of a control engineer would be to check that this system is stable (you don't want the car to
accelerate out of control when the driver switches on the cruise control system). Thus, we need to find the
1
transfer function relating the input vdes ( t ) to the scaled measured speed of the car
its poles. Let's denote the Laplace transforms of the signals with "hats". We have
v(t ) and compute
v( s)  F ( s) P( s) K ( s)e( s)
e( s)  vdes ( s)  v( s)
(8.15)
thus the error can be expressed as,
e( s)  vdes ( s)  F ( s) P( s) K ( s)e( s)

1
vdes ( s)
1  F ( s) P( s) K ( s)
(8.16)
and substituting this expression back into the first equation of (8.15), we obtain
v( s) 
F ( s) P( s) K ( s)
vdes ( s)  H ( s)vdes ( s) .
1
F (
s)
P(
s) K
(
s)
(8.17)
H ( s ):
This is the transfer function relating the desired car velocity to its measurement. Using the actual transfer
functions provided above, we get
H ( s) 
2 s 1 3 10s
20
 2
.
1 10
1  2 s 3 s s  3s  20
This is a second-order transfer function with undamped natural frequency
damping ratio

(8.18)
 n  20
radians/min and
3
 0.34 . Hence the poles are in the open left-half plane
2 20
p1   n   n  2  1  15
.  j 4.2131
p2   n   n  2  1  15
.  j 4.2131
(8.19)
H ( s) is Re{s}  15
. . The Laplace transform
of the car speed response to a step in the desired velocity input vdes from 0 km/h to 100 km/h is
which means that the system is stable and the ROC of
v( s)  H ( s)vdes ( s)
20
100
, Re{s}  0
s  3s  20 s
A( s  15
. )  B4.2131
C


2
2
( s 
 15
.) 
 (
4.2131
)
s



2
(8.20)
Re{s} 0
Re{s}1.5
where the second-order term of the partial fraction expansion was written to match the form of the
Laplace transform of the sum of a damped sine and cosine (see Table 9.2 in the book). We find
v( s) 
100( s  15
. )  35.603(4.2131) 100

( s  15
. ) 2  (4.2131) 2
s
(8.21)
which for the ROC's of (8.20) and using Table 9.2 in the textbook, corresponds to
v (t )  100  100e 1.5t cos 4.2131t  35.603e 1.5t sin 4.2131t u(t ) .
2
(8.22)
This speed response is shown below.
140
( km / h)
120
v (t )
100
80
60
40
20
0
0
1
2
3
4
5
t (min)
6
It can be seen on this plot that there is too much overshoot in the speed response. By choosing a smaller
integrator gain in K ( s) , we would obtain a smoother response with less overshoot.
8.3.2
Realization of a Transfer Function
It is possible to obtain a realization of the transfer function of an LTI differential system as a combination
of three basic elements:
the integrator,
1
s
the gain,
k

and the summing junction.

8.3.2.1 Simple First-order Transfer Function
Consider the transfer function H ( s) 
1
. It can be realized with a feedback interconnection of the
sa
three basic elements:
x (t )
+
-
y (t )
1
s
3
a
We have
1
1
Y ( s)  a Y ( s)  X ( s)
s
s
1
1
 s 1 X ( s) 
X ( s)  H ( s) X ( s)
1 a s
sa
(8.23)
With this block diagram, we can basically realize any transfer function of any order after it is written as a
partial fraction expansion. This leads to the parallel form introduced below.
8.3.2.2 Simple Second-Order Transfer Function (constant numerator)
Consider the transfer function H ( s) 
1
. It can be realized with a feedback interconnection
s  a1s  a0
2
of the three basic elements in a number of ways. One way is to expand the transfer function as a sum of
two first-order transfer functions (partial fraction expansion). The resulting form is called the parallel form
and it is discussed below. Another way is to break up the transfer function as a cascade (multiplication) of
two first-order transfer functions. This cascade form is also discussed below. Yet another way to realize
the second-order transfer function is the so-called direct form or controllable canonical form. To develop
this form, consider the system equation
s2Y ( s)  a1sY ( s)  a0Y ( s)  X ( s) .
(8.24)
This equation can be realized as follows (the idea is that the variable at the input of an integrator is
thought as the derivative of its output.)
2
X ( s)
+ s Y ( s)
-
sY ( s)
1
s
1
s
Y ( s)
a1
a0
8.3.2.3 Parallel Realization
A parallel realization can be obtained by expanding the transfer function into partial first-order fractions
with real coefficients or complex coefficients for complex poles.
4
H ( s) 
Example: Consider the system
2
1
s 1
 3  3 . Its parallel realization is shown
( s  1)( s  2) s  1 s  2
below.
+
x (t )
1
s
23
y (t )
+
1
+
+
1
s
-
13
2
8.3.2.4 Cascade Realization
A cascade realization can be obtained by expressing the numerator and denominator of the transfer
function as a product of zeros of the form ( s  zi ) and of poles of the form ( s  pi ) respectively.
Example: Consider the system
H ( s) 
s 1
. Its cascade form is shown below. The zero in the
( s  1)( s  2)
first first-order block is implemented using a feedthrough term. This first block is actually in direct form,
which is explained below.
1
x (t )
+
-
1
s
1
+ +
+
1
s
-
2
y (t )
1
1
8.3.2.5 Direct Form (Controllable Canonical Form)
A direct form can be obtained by breaking up a general transfer function into two subsystems as follows
X ( s)
1
n
n 1
s  an 1s a1s  a0
Y ( s)
W ( s)
5
m1
bm s  bm1s b1s  b0
m
The input-output system equation of the first subsystem is
snW ( s)  an1sn1W ( s) a1sW ( s)  a0W ( s)  X ( s) ,
(8.25)
and for the second subsystem we have
Y ( s)  bm smW ( s)  bm1sm1W ( s)b1sW ( s)  b0W ( s) .
(8.26)
The direct form realization is then (for a second-order system):
b2
b1
s 2W ( s)
X ( s)
+
-
-
sW ( s)
1
s
1
s
a1
a0
6
W ( s)
+ +
b0
+
Y ( s)