Canete J.F. System Engineering and Automation: An Interactive Educational Approach
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58 3 System Description
In this way we can rewrite as the partial fraction expansion
1
69
3
3
Using (3.24) and (3.26) we obtain the residues
lim
1
9
1
21!
lim
3
lim
1
lim
1
1
9
1
22!
lim
3
lim
1
1
3
We can check the results obtained by applying the MATLAB code
>> num=[1];den=[1 6 9 0];
>> [r,p]=residue(num,den)
r =
-0.1111
-0.3333
0.1111
p =
-3
-3
0
In this way, the partial fraction expansion becomes
1
69
1
9
1
9 3
1
3 3
so that the time response can be calculated adding the corresponding inverse
transform of each fraction in (3.47), yielding the solution
1
9
1
3
In fig. 3.10 it is shown that the step response after applying the MATLAB code
>> t = 0:0.1:5; y= 0.1111*(1-exp(-3*t)-3*t.*exp(-3*t));
>> plot(t,y)
3.4 Laplace Transfor
m
59
Fig. 3.10 Step response of the mechanical system of example 3.3
Example 3.5
Obtain the time response
to input signal
0.5
for the electric
circuit shown in fig. 3.11 whose dynamic equation is given by
for values of R
1
= 0.2, R
2
= 1, C = 0.01 and L = 0.1, starting from initial conditions
0
0 and
0
0.
Fig. 3.11 Diagram of the electric circuit for example 3.5
60 3 System Description
Solution:
Applying the Laplace transform properties to the differential equation of the
cirtcuit we get
so that the system response is given by
1
In case of step input
0.5
,
.
, so that the system response
particularized for R
1
, R
2
, C and L is given by
0.2 0.5
0.001
0.012 1
Now we proceed to calculate the roots of the denominator
0.001
0.012 10
by applying the MATLAB code
>> den= [0.001 0.012 1 0];
>> roots(den)
ans =
0
-6.0000 +31.05i
-6.0000 -31.05i
Expanding
into partial fractions, we have
0.1
0.001
0.012 1
100
12 1000
100
12 1000
6 31.05
6 31.05
Using (3.24) we obtain the residues
lim
100
1000
0.1
3.4 Laplace Transfor
m
61
lim
.
6 31.05
lim
.
100
6 31.05
0.05 0.0097
lim
.
6 31.05
lim
.
100
6 31.05
0.05 0.0097
We can check the results obtained by applying the MATLAB code
>> num=100;den=[1 12 1000 0];
>> [r,p]=residue(num,den)
r =
-0.0500 + 0.0097i
-0.0500 - 0.0097i
0.1000
p =
-6.0000 +31.05i
-6.0000 -31.05i
0
In this way, the partial fraction expansion becomes
100
12 1000
0.1
0.05 0.0097
6 31.05
0.05 0.0097
6 31.05
By grouping the two complex fractions we have
0.1
0.1 6
6
31.05
0.0194 · 31.05
6
31.05
so that the time response can be calculated adding the corresponding inverse
transform of each fraction, yielding the solution
0.1 0.1
cos 31.05 0.6
sin 31.05
In fig. 3.12 the step response after applying the MATLAB code is shown
>> t = 0:0.2:2; y= 0.1*(1-exp(-6*t).cos(31.05*t)-6*exp(-6*t).*sin(31.05*t));
>> plot(t,y)
62 3 System Description
Fig. 3.12 Step response of the electric circuit of example 3.5
3.5 Transfer Function
The transfer function of an invariant linear system is defined as the ratio between
the Laplace transform of system’s output and the transform of the system’s
input under the assumption of zero initial conditions. This function enables
the separation of the input, system, and output into three separate and distinct
parts, unlike the differential equation which define its dynamics.
Considering the single-input single-output continuous-time system defined by a
linear and invariant differential equation in the form
(3.27)
and taking the Laplace transform of both sides, we have
(3.28)
by assuming that the initial conditions are null. Taking the ratio between and
, we obtain the transfer function of this system as defined by
(3.29)
The transfer function is a property of the system itself, since it is not depend on
the input, and can be represented as a block diagram (fig. 3.13).
3.5 Transfer Function 63
Fig. 3.13 Block diagram of the transfer function .
In fact, the output response for an input signal can be obtained by
using the relation
(3.30)
assuming there are no initial conditions.
The highest power of the denominator of (characteristic equation) is
called the system order. The roots of the characteristic equation are called the
system poles, while the roots of the numerator are called the system zeros
(fig. 3.14)
Fig. 3.14 Pole-zero diagram in the s plane for the transfer function
.
This form of representation corresponds to the external description of the
system, which provides no information on its internal structure. Moreover, the
transfer function of different systems may be the same, which is the case of analog
systems.
64 3 System Description
Example 3.6
Obtain the transfer function of the hydraulic system of fig. 3.15 by considering
and
as input and output signal respectively, with a dynamic equation
given by
and indicate the poles and zeros locations inside the s plane for C = 2, R = 3, and I
= 0.5. Then, derive and plot the step response using the Laplace transform. Create
the transfer function and obtain the step response by using the MATLAB
appropriate commands, comparing both responses.
Fig. 3.15 Hydraulics system of example 3.6
Solution
Applying the Laplace transform, the differential equation assuming zero initial
conditions we have
Then, taking the ratio between
and
we obtain the transfer function
1
1
Substituting for C, R and I we have
1
61
The system’s poles are located in the solutions of the characteristic equation
610
3.5 Transfer Function 65
For this calculation we use the MATLAB command
>> roots([1 6 1])
ans =
-5.8284
-0.1716
while there are no zeros for this system.
In order to calculate the step response, we obtain
, so that
1
61
1
61
The partial fraction expansion becomes
1
61
0.1716
5.8284
and the residuals are obtained using the MATLAB code
>> num=[1];den=[1 6 1 0];
>> [r,p]=residue(num,den)
r =
0.0303
-1.0303
1.0000
p =
-5.8284
-0.1716
0
In this way, the partial fraction expansion becomes
1
1.0303
0.1716
0.0303
5.8284
so that the time response can be calculated adding the corresponding inverse
transform of each fraction in (3.47), yielding the solution
1 1.0303
.
0.0303
.
In fig. 3.16 the step response after applying the MATLAB code is shown
>> t = 0:0.1:5; y= (1-1.0303*exp(-0.1716*t)+0.0303*exp(-5.8284*t));
>> plot(t,y)
66 3 System Description
Fig. 3.16 Step response of the hydraulic system of example 3.6
To create the transfer function directly we use the following MATLAB code
>> s=tf('s');
>> G=1/(s^2+6*s+1)
Transfer function:
1
-------------
s^2 + 6 s + 1
>> step(G,50)
In fig. 3.17 the step response of the system running under MATLAB is shown.
Fig. 3.17 MATLAB step response of the hydraulic system of example 3.6.
3.6 The State Space Approach 67
In the case of multi-input multi-output systems, we define the transfer matrix
, whose elements
relate each output
to each input
,
assuming zero both the remaining entries
, and the initial conditions
(fig. 3.18).
Fig. 3.18 Continuous multi-input, multi-output system.
Each individual transfer function element
is given by the relation
(3.31)
and assuming no initial conditions, we derive each of the system output, following
the expressions
(3.32)
which can be transformed to a vector notation
(3.33)
assuming there are no initial conditions, with
…
and
…
.
3.6 The State Space Approach
The state of a dynamic system is defined by the minimum set of variables
,
,…,
such that the knowledge of the value of
together with the knowledge of the input
,
, completely determines the