Two main types of control

Open-Loop Control Systems

utilize a controller or control
actuator to obtain the desired
response.

Closed-Loop Control
Systems utilizes feedback to
compare the actual output to
the desired output response.

Multivariable Control System

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Control System Design

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Transient and steady state response

3
 Three terms needed to know

• Transient response
• Steady state response
• (transient response has decayed to zero)
remainin part resembles the input: output
• Stability: most important thing in analysis and
design objectives

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 Analysis and Design Objectives
• Transient Response must meet certain criteria.
– Mainly: Speed and max. overshoot.
• Steady-State Response must meet certain criteria.
– Mainly: small or zero error
• The system must have Stability.
– Total Response = Natural Response +
Forced Response
• Natural response describes the way the system
dissipates or gain energy. It is dependent only on
the system not the input
• Forced response depends on the input.
• Natural response must go to zero leaving only the
forced response or oscillate
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 Modeling in the Frequency Domain
• The first step in developing a mathematical model is
to apply the fundamental physical laws.
– Kirchoff’s Voltage Law - The sum of voltages around a closed path is
zero.
– Kirchoff’s Current Law - The sum of currents flowing from a node is
zero.
– Newton’s Laws - The sum of forces on a body is zero (considering
mass times acceleration as a force).
- The sum of moments on a body is zero.

• The model describes the relationship between

the input and the output of the dynamic
system.
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• Applying the laws of physics generates a differential
equation which can be represented in general form.

• The form and the coefficients of the differential

equation are a formulation or description of the
system.
• Although the differential equation relates the system
to its input and output, it is not a satisfying
representation from a system perspective.

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• We would prefer a mathematical representation such
as

in which the input, output and the system are

distinct and separate parts.
• We would also like to represent cascaded
interconnections, such as
 Laplace Transformation Review
• The Laplace transform is defined as,

where is a complex variable.

• The inverse Laplace transform, which allows us to
find f(t) given F(s), is

where

• Multiplication of f(t) by u(t) yields a time function

that is zero for t < 0.
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Description
Impulse (Dirac-delta)
Step
Ramp
Parabolic
 Example – 1 (Laplace transform)
• Find the Laplace transform of
Since the time function does not contain an impulse
function, we can replace the lower limit of the
integral in the Laplace transform definition with 0.
Hence,

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Example – 2 (Inverse Laplace transform)

• Find the inverse Laplace transform of

For this example we make use of the frequency shift
theorem and the Laplace transform of
If the inverse transform of is
then the inverse transform of
is .
Hence,

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 Partial-Fraction Expansion
• In order to find the inverse Laplace transform of a complicated function,
we can convert the function to a sum of simpler terms for which we know
the Laplace transform of each term. The result is called a partial-fraction
expansion.
• If where the order of N(s) is less than the order of D(s), then
a partial-fraction expansion can be made.
• If the order of N(s) is greater than or equal to the order of D(s), then N(s)
must be divided by D(s) successively until the result has a remainder
whose numerator is of order less than its denominator.

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 Case 1. Roots of the Denominator of F(s) Are Real and Distinct
An example of an F(s) with real and distinct roots in the denominator is

We can write the 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.

f(t) is the sum of the inverse Laplace transform of each term, or

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 Example 2.3 (Laplace Transform Solution of a
Differential Equation)

Solving for the response, Y(s), yields

Since each of the three component parts

is represented as an F(s) in Table 2.1, y(t)
is the sum of the inverse Laplace
transforms of each term.

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Case 2. Roots of Denominator of F(s) are
Real and Repeated

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Case 3. Roots of the Denominator of F(s) Are Complex or
Imaginary

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 The Transfer Function
• We are now ready to formulate the system representation by
establishing a viable definition for a function that algebraically
relates a system’s output to its input.
• This function will allow separation of the input, system and
output into three separate and distinct parts, unlike the
differential equation.
• The function will also allow us to algebraically combine
mathematical representations of subsystems to yield a total
system representation.

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 The Transfer Function

Notice that Eq. (2.53) separates the output, C(s), the input, R(s), and the system,
the ratio of polynomials in s on the right. We call this ratio, G(s), the transfer
function and evaluate it with zero initial conditions.

The transfer function can be represented as a block diagram, as shown in Figure

2.2, with the input on the left, the output on the right, and the system transfer
function inside the block. Notice that the denominator of the transfer function is
identical to the characteristic polynomial of the differential equation. Also, we can
find the output, C(s) by using

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Example 2.4: Transfer Function for a Differential Equation
• Find the transfer function represented by

Taking the Laplace transform of both sides, assuming zero initial conditions,
we have

Use the result of Example 2.4 to find the response, c(t) to an input,
a unit step, assuming zero initial conditions.

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Electrical Network Transfer Functions

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 Electrical Network Transfer Functions
• Equivalent circuits for the electric networks that we work with
first consist of three passive linear components: resistors,
capacitors, and inductors.
• We now combine electrical components into circuits, decide
on the input and output, and find the transfer function. Our
guiding principles are Kirchhoff’s laws.
• We sum voltages around loops or sum currents at nodes,
depending on which technique involves the least effort in
algebraic manipulation, and then equate the result to zero.
From these relationships we can write the differential
equations for the circuit.
• Then we can take the Laplace transforms of the differential
equations and finally solve for the transfer function.

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Example 2.6: Transfer Function—Single Loop via
the Differential Equation
• Find the transfer function relating the capacitor
voltage, , to the input voltage, V(s).

Summing the voltages around the loop, assuming

zero initial conditions, yields the integro-differential
equation for this network as
Changing variables from current to charge using
yields

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Example 2.6: Transfer Function—Single Loop via
the Differential Equation
• Substituting into Eq. (2.62) yields

• Taking the Laplace transform assuming zero initial

conditions, rearranging terms, and simplifying yields

• Solving for the transfer function, we obtain

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Example 2.6: Transfer Function—Single Loop via
the Differential Equation
• Let us now develop a technique for simplifying the
solution for future problems.
• For the capacitor,
• For the resistor,
• For the inductor,
• Now define the following transfer function:

• Notice that this function is similar to the definition of resistance, that is,
the ratio of voltage to current. But, unlike resistance, this function is
applicable to capacitors and inductors and carries information on the
dynamic behavior of the component, since it represents an equivalent
differential equation. We call this particular transfer function impedance.
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Example 2.6: Transfer Function—Single Loop via
the Differential Equation
• The Laplace transform of Eq. (2.61), assuming zero
initial conditions, is

• We summarize the steps as follows:

• Redraw the original network showing all time variables,
such as v(t), i(t), and vC(t), as Laplace transforms V(s), I(s),
and VC(s), respectively.
• Replace the component values with their impedance
values. This replacement is similar to the case of dc
circuits, where we represent resistors with their resistance
values. 27
Example 2.7: Transfer Function—Single Loop via
Transform Methods
• Repeat Example 2.6 using mesh analysis and transform
methods without writing a differential equation.

• Using Figure 2.5 and writing a mesh equation using the

impedances as we would use resistor values in a purely
resistive circuit, we obtain

• But the voltage across the capacitor, , is the product of the

current and the impedance of the capacitor. Thus,

• Solving Eq. (2.75) for I(s), substituting I(s) into Eq. (2.74), and
simplifying yields the same result as Eq. (2.66). 28
 Complex Circuits via Mesh Analysis
To solve complex electrical networks—those with multiple loops
and nodes—using mesh analysis, we can perform the following
steps:

• Replace passive element values with their impedances.

• Replace all sources and time variables with their Laplace
transform.
• Assume a transform current and a current direction in each
mesh.
• Write Kirchhoff’s voltage law around each mesh.
• Solve the simultaneous equations for the output.
• Form the transfer function.

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Example 2.10 Transfer Function—Multiple Loops

• Given the network of Figure 2.6(a), find the transfer function,

I2(s)/V(s).

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Example 2.10 Transfer Function—Multiple Loops

• The first step in the solution is to convert the network into

Laplace transforms for impedances and circuit variables,
assuming zero initial conditions.
• The result is shown in Figure 2.6(b). The circuit with which we
are dealing requires two simultaneous equations to solve for
the transfer function. These equations can be found by
summing voltages around each mesh through which the
assumed currents, I1(s) and I2(s), flow. Around Mesh 1, where
I1(s) flows,

Around Mesh 2, where I2(s) flows,

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 Example 2.10 Transfer Function—Multiple Loops

R1+Ls -Ls [I1

-Ls Ls+R2+1/Cs I2]
We can use Cramer’s rule (or any other
method for solving simultaneous
equations) to solve

Forming the transfer function, G(s), yields

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 Example 2.13 Mesh Equations via Inspection

• Write, but do not solve, the mesh equations for the network
shown

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 Operational Amplifiers

• An operational amplifier is an electronic amplifier used as a

basic building block to implement transfer functions. It has
the following characteristics:
• Differential input,
• High input impedance, (ideal)
• Low output impedance, (ideal)
• High constant gain amplification, (ideal)
The output is given by

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 Inverting Operational Amplifier

• If v2(t) is grounded, the amplifier is called an inverting

operational amplifier. For the inverting operational amplifier,
we have
• If the input impedance to the amplifier is high, then by
Kirchhoff’s current law, and .
• Also, since the gain A is large, v1(t)≈0.
• Thus, and . Equating the two
currents,
• The transfer function of the inverting operational amplifier
configured as shown in Figure 2.10(c) is

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Example 2.14 Transfer Function
Inverting Op-Amp Circuit

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 Non-inverting Operational Amplifier

• Another circuit that can be analyzed for its transfer function is

the non-inverting operational amplifier circuit shown in Figure
2.12.
• We now derive the transfer function. We see that

• But, using voltage division,

• Substituting Eq. (2.102) into Eq. (2.101), rearranging, and

simplifying, we obtain

• For large A, we disregard unity in the denominator and Eq.

(2.103) becomes

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Example 2.14 Transfer Function
Noninverting Op-Amp Circuit

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