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.

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

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

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

Laplace Transform Obtained a Laplace Transform for the ramp function,

Laplace Transform Find the Laplace Transform of

Laplace Transform Table Laplace Transform’s table for common functions

Inverse Laplace Find inverse Laplace transform of Solution:

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

(2) Differential Theorem where Let and

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

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),

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.

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.

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

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,

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

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

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,

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.

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

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,

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

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,

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

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

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

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

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

Complex Circuits Write the mesh equations for the following network.

Complex Circuits

Complex Circuits Mesh equations:

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.

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. 2.10 (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,

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

Operational Amplifier Since ,