1. Mathematical Models of Control Systems
Two methods to develop mathematical models from schematics of physical systems:
Transfer Functions in the Frequency Domain
State Equations in the Time Domain
The first step to develop a mathematical model is by applying the fundamental physical laws of science and engineering. E.g. Kirchhoff’s laws, Newton laws.
Differential equation can describe relationship between the input and output of a system.

2. Mathematical Models of Control Systems
It is preferable to have a mathematical representation where the input, output, and
system are distinct and separated. At the same time, the model should be able to
represent interconnection of several subsystems.
a. Block diagram representation of a system; b. block diagram representation of an interconnection of subsystems
Control Systems Engineering, Fourth Edition by Norman S. Nise

4. Laplace Transform
A system represented by a differential equation is difficult to model as a block diagram.
Thus, Laplace Transform is used to represent the input, output, and the system as separate entities.
Laplace Transform is defined as
where s is a Laplace operator in a form of complex frequency given by
where  and j are the real and imaginary frequency components respectively

5. Laplace Transform Example:
Obtained a Laplace Transform for a unit step
Solution:
f(t) 1 t

6. Laplace Transform
Obtained a Laplace Transform for the ramp function,

7. Laplace Transform
Find the Laplace Transform of

8. Laplace Transform Table
Laplace Transform’s table for common functions

9. Inverse Laplace
Find inverse Laplace transform of
Solution:

10. Characteristic of Laplace Transform
(1) Linear If
and
are constant and
and
are Laplace Transform

11. (2) Differential Theorem
where
Let
and

12. Partial Fraction Expansion
To find inverse Laplace transform of a complicated function, we can convert the function into a sum of simpler terms using partial fraction expansion.
If , where the order of N(s) is less than D(s), then a partial fraction expansion can be made.
Example:
N(s) must be divided by D(s) successively until the result has a remainder whose numerator is of order less than its denominator.
Taking the inverse Laplace transform, we obtain

13. Partial Fraction Expansion (Cont’d)
Three cases of partial fraction expansion:
Roots of the Denominator of F(s) are Real and Distinct
Roots of the Denominator of F(s) are Real and Repeated
Roots of the Denominator of F(s) are Complex and Imaginary
Example:
We write partial fraction expansion as a sum of terms where each factor of the original denominator forms the denominator of each term, and constants, called residues form the numerators.
(2.8)
Multiply equation (2.8) by (s+1),

14. Partial Fraction Expansion (Cont’d)
Letting s approach -1, we get K1=2.
Find K2 with the similar approach, letting s approach -2, we get K2=-2.
Thus,
Taking inverse Laplace transform, we get
Given the following differential equation, solve y(t) if all initial conditions are zero.

15. Partial Fraction Expansion (Cont’d)
Case 2: Example,
Add additional term consisting of denominator factor of reduced multiplicity,
K1=2 can be obtained as previously described.
Isolate K2 by multiplying by (s+2)2,
(2.24)
Let s approach -2, K2=-2.
To find K3, differentiate equation (2.24) with respect to s,
Let s approach -2, K3=-2.

16. Partial Fraction Expansion (Cont’d)
Take inverse Laplace transform,

17. Partial Fraction Expansion (Cont’d)
Case 3: Example
The function can be expanded in
(2.31)
Using the previous way, can be found.
Multiply Eq (2.31) by the lowest common denominator,
Balancing coefficients,

18. Transfer Function
G(s) is the transfer function. It is evaluated with zero initial conditions.

19. Transfer Function Find the transfer function represented by
Taking the Laplace transform,
The transfer function G(s), is

20. Transfer Function
From the previous slide, find the response, c(t), to an input, r(t)=u(t), a unit step, assuming zero initial condition.
Since r(t)=u(t), R(s)=1/s
Using partial fraction expansion,
Taking inverse Laplace,

21. Transfer Function
In general, a physical system that can be represented by a linear, time invariant differential equation can be modeled as a transfer function.
Transfer function can be used to represent electrical networks, translational mechanical systems, rotational mechanical systems, and electromechanical systems.

22. Electrical Network Transfer Functions
Equivalent circuits for the electronic networks consists of 3 passive linear components: resistor, capacitor, inductor.
Table 2.3 Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductors
impedance
admittance

23. Simple Circuits via Mesh Analysis
Find transfer function relating the capacity voltage, Vc(s) to the input voltage, V(s)
in the figure.
Summing the voltages around the loop,
assuming zero initial conditions,
(2.61)
Changing variables from current to charge,
Using i(t)=dq(t)/dt
From the voltage-charge relationship for a capacitor,
Thus,

24. Simple Circuits via Mesh Analysis
Taking Laplace,
Solving the transfer function, Vc(s)/V(s)
Block diagram of series RLC electrical network

25. Laplace transform of the equations in the voltage-current column of Table 2.3
For capacitor,
For resistor,
For inductor,
From previous example, Laplace transform of Eq(2.61) is
Thus,

26. Using mesh analysis without writing differentiation,
Solving for I(s)/V(s),
For capacitor,
Thus,

27. Simple Circuits via Nodal Analysis
Transfer function using current law and summing currents flowing from nodes.
Assume that currents leaving the node are positive, and currents entering the node
are negative.
Using previous example,
Solving the above equation for transfer function Vc(s)/V(s), yields

28. Simple Circuits via Voltage Division
Transfer function using voltage division.
Using previous example,
Solving the above equation for transfer function Vc(s)/V(s), yields

29. Complex Circuits
Given network of Figure 2.6(a), using mesh analysis, find transfer function, I2/V(s).
Figure 2.6 a. Two-loop electrical network; b. transformed two-loop electrical network; c. block diagram

30. Complex Circuits
Using nodal analysis for previous example, find the transfer function, Vc(s)/V(s).
(1)
(2)
Refer to circuit in Figure 2.6(b),
Rearranging and expressing the resistance as conductances, G1=1/R1 and
G2=1/R2

31. Complex Circuits
Write the mesh equations for the following network.

32. Complex Circuits

33. Complex Circuits
Mesh equations:

34. Operational Amplifier
Operational amplifier is an electronic amplifier used as a basic building block to implement transfer functions.
It has the following characteristics:
(a) Differential input, v2(t)-v1(t)
(b) High input impedance, Zi= (ideal)
(c) Low output impedance, Z0=0 (ideal)
(d) High constant gain amplification,
A=  (ideal)
Figure 2.10 a. Operational amplifier; b. schematic for an inverting operational amplifier;
c. Inverting operational amplifier
configured for transfer function realization.
Typically, the amplifier gain, A, is omitted.

35. Operational Amplifier
If v2 is grounded, the amplifier is called Inverting operational amplifier, e.g. Fig. 2.10(b).
For the inverting operational amplifier,
Find the transfer function, Vo(s)/Vi(s) as shown in Fig (c).
Ia(s)=0, I1(s)=-I2(s) since the input impedance is high
v1(t)0 since the gain, A is large.
Thus, I1(s)=Vi(s)/Z1(s), and I2(s)=-Vo(s)/Z2(s).
Equating the two currents,

36. Operational Amplifier
Find the transfer function, Vo(s)/Vi(s) for the circuit given below:

37. Operational Amplifier
Since ,