A Discrete Time system is that in which the input and output are discrete time. It is illustrated in the Figure below.

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UNIT I – Classification and Components of Control Systems

The Control System is that mechanism by which any quantity of interest in a machine, mechanism or other equipment is maintained or altered in accordance with the desired manner. For example, the driving system of an automobile. The speed of the automobile is a function of the position of its accelerator. The desired speed can be maintained by controlling the pressure on the accelerator pedal. This automobile driving system constitutes a control system.

For the automobile driving system the input (command) signal is the force on the accelerator pedal which through linkages causes the carburettor valve to open (close) so as to increase or decrease fuel flow to the engine bringing the engine – vehicle speed to the desired value. It may be noted that the signal flow through

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Closed – Loop Control

Let us reconsider the automobile system. The route, speed and acceleration of the automobile are determined and controlled by the driver by observing the traffic and road conditions and by properly manipulating the accelerator, clutch, gear-lever, brakes and steering wheel, etc. Suppose the driver wants to maintain the speed of the vehicle constant. He accelerates the automobile to this speed by the help of the accelerator and then maintains it by holding the accelerator steady. These operations can be represented in a diagrammatic form as shown in the figure below. The events in the control sequence of this figure follow a closed-loop. i.e. the information about the instantaneous state of the output is feedback to the input and is used to modify it in such a manner as to achieve the desired output. This system is hence called a closed-loop system.

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Let us investigate another control aspect of the above example of an auto mobile, say its steering mechanism which is illustrated in the Figure above. The driver senses visually and by tactile means the error between actual and desired directions of the automobile. Systems of the type represented in Figures above involve manual control by a human operator. These are classified as manually controlled systems. Human involvement in the system may introduce human error which is most undesirable. Some of the systems such as missiles etc. which are self-destructive and in such systems human element must be excluded.

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Automatic Control Systems

Thus in most of the situations the use of some equipment which performs the same intended function as a continuously employed human operator is preferred. A system incorporating such an equipment is known as automatic control system.

In fact, in most situations an automatic control system could be made to perform intended functions better than a human operator. The general block diagram of such a system is given below.

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Open-Loop Control

Any physical system which does not automatically correct for variations in its output, is called an open-loop system. Such a system can be represented by a block diagram below.

In these systems the output remains constant for a constant input signal provided the external conditions remain unaltered. The output may be changed to any desired value by appropriately changing the input signal. It is important to note that that the fundamental difference between an open and closed-loop control system is that of the feedback action.

E.g. consider a traffic control system for regulating the flow of traffic at the crossing of two roads. The system will be termed open-loop if red and green lights are put on and off by timer set for predetermined fixed intervals of time. If on the other hand a scheme is introduced in which the rates of traffic flow along both the directions are measured and are compared and the difference is used to control the timings of the red and green lights, a closed loop system results.

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Linear Control Systems

In order to understand the linear control system, we should know the principle of superposition.

The principle of superposition theorem includes two important properties and they are explained below:

Homogeneity: A system is said to be homogeneous, if we multiply input with some constant A then output will also be multiplied by the same value of constant.

Additivity: Suppose we have a system S and we are giving the input to this system as a₁ for the first time and we are getting output as b₁ corresponding to input a₁. On second time we are giving input a₂ and correspond to this we are getting output as b₂. Now suppose this time we are giving input as summation of the previous inputs (i.e. a₁ + a₂) and corresponding to this input suppose we are getting output as (b₁ + b₂) then we can say that system S is following the property of additivity. Now we are able to define the linear control systems as those types of control systems which follow the principle of homogeneity and additivity. Consider a purely resistive network with a constant DC source. This circuit follows the principle of homogeneity and additivity. All the undesired effects are neglected and assuming ideal behaviour of each element in the network, we say that we will get near voltage and current characteristic. This is the example of linear control system.

Linear Control System

a1 b1

A*a1 A*b1

(a1+a2 ) (b1 + b2)

A*(a1+a2 ) A*(b1 + b2)

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Non-Linear Control Systems

We can simply define non linear control system as all those system which do not follow the principle of homogeneity. In practical life all the systems are non-linear system. Examples of Non-linear System: A well known example of non-linear system is magnetization curve or no load curve of a DC machine. We will discuss briefly no load curve of DC machines here: No load curve gives us the relationship between the air gap flux and the field winding mmf. It is very clear from the curve given below that in the beginning there is a linear relationship between winding mmf and the air gap flux but after this, saturation has come which shows the non linear behaviour of the curve or characteristics of the non linear control system.

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Continuous and Discrete Time Systems

A system is Continuous Time when its input and output are continuous time. It is illustrated in the Figure below.

Continuous Time System

X(t) Y(t)

A Discrete Time system is that in which the input and output are discrete time. It is illustrated in the Figure below.

Discrete Time System

X(n) Y(n)

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Deterministic and Stochastic Systems

Any system where the input and output can be uniquely described by an explicit mathematical expression is called the deterministic system. The example of such a system can be with a sinusoidal input and output signals. The deterministic system can be given by the block diagrams below.

Differentiator

Sin(ωt) cos(ωt)

Integrator Sin(ωt)

cos(ωt)

In many practical applications there are systems that either cannot be described to any reasonable degree of accuracy by explicit mathematical formulas, or such a description is too complicated to be of any practical use. The input and output of such a system is random. The mathematical framework for the theoretical analysis of random signals is provided by the theory of probability and stochastic processes. This type of system can be given by the block diagram below. Both X(t) and Y(t) are random signals.

Stochastic System

X(t) Y(t)

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SISO and MIMO systems

Single Input – Single Output (SISO) Systems are those where the system takes only one input and there is only one output. One such system can be illustrated in the figure below.

X(t) SISO Y(t)

Multiple Input – Multiple Output (MIMO) Systems are those where the system takes multiple inputs and give out multiple outputs. The block diagram of such a system is given below. Here 𝑀 ≤ 𝑁 𝑜𝑟 𝑀 ≥ 𝑁.

MIMO X₁(t)

X₂(t)

XN(t)

Y₁(t) Y₂(t)

YM(t)

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Static Systems

A system is said to be static if its output y(t) depends only on the input u(t) at the present time t, mathematically described as,

𝑦 𝑡 = ℎ 𝑢 𝑡 = ℎ(𝑢₁ 𝑡 , 𝑢₂ 𝑡 , … … , 𝑢_𝑚 𝑡 )

where h(u(t)) is an algebraic function of the input variables 𝑢₁ 𝑡 , 𝑢₂ 𝑡 , … … … , 𝑢_𝑚 𝑡 . Figure given below gives an example of a static system, which is a resistive circuit excited by an input voltage u(t). Let the output be the voltage across R₃, and according to the circuit theory we have,

𝑦 𝑡 = ℎ 𝑢 𝑡 = ^𝑅²^𝑅³

𝑅₁ 𝑅₂+𝑅₃ +𝑅₂𝑅₃ 𝑢 𝑡 = 𝑘. 𝑢(𝑡)

Where k is a constant. Clearly, the output can be simply determined by the present input u(t), so it is a static system.

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Dynamic Systems

Unlike static systems, the output y(t) of a dynamic system will be affected by the input u(t) at the present time but also by its past u(t). Mathematically, a dynamic system can be expressed by the state space description, which describes a system by its state equation and output equation. The state equation is often given as a first order differential equation:

𝑥(𝑡) = 𝑓 𝑡, 𝑥 𝑡 , 𝑢 𝑡 , 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑡₀ = 𝑥₀ And the output equation is in the form of

𝑦 𝑡 = 𝑔 𝑡, 𝑥 𝑡 , 𝑢 𝑡 = ℎ(𝑡, 𝑢 𝑡 )

The figure below shows the block diagram of the dynamic system expressed in state-space description above.

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Causal and Non-Causal Systems

Causal systems are those whose output depends only on x(t) or input for 𝑡 ≤ 𝑡

₀

. That is present output depends only on the past and present inputs, not on future inputs. Any practical REAL TIME system must be causal.

Non-Causal systems are those whose output depends on the present, past and the future inputs. Non-causal systems are important because:

1.Realizable when the independent variable is something other than “time” (e.g. space)

2.Even for temporal systems, can prerecord the data (non-

real time), mimic a non-causal system.

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Time Varying and Time Invariant System

Time Invariant systems are those which follow the properties given below:

TI System Delay by T seconds

X(t) Y(t) Y(t-T)

Delay by T seconds TI System

X(t) X(t-T) Y(t-T)

Time varying systems are those which do not possess the above properties i.e.

output changes with time.

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Forced and Force-Free Systems

In order to explain stability of a control system, we start from the fact that total response of a system is the sum of the natural response and forced response.

Total Response = Natural Response + Forced Response

For a control system to be useful, the natural response must (1)

eventually approach zero, thus leaving only the force response, or (2)

oscillate. In some systems, however, the natural response grows

without bound rather than diminish to zero or oscillate. Eventually, the

natural response is so much greater than the forced response that the

system is no longer controlled. This condition, called instability, could

lead to self-destruction of the physical device if limit stops are not part

of the design. Such a system is forced-free system.

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Laplace Transform Review

The Laplace Transform is defined by the equation below:

𝐿 𝑓 𝑡 = 𝐹 𝑠 =

₀₋^∞

𝑓 𝑡 𝑒

^−𝑠𝑡

𝑑𝑡

Where s = 𝜎 + 𝑗𝜔, a complex variable. Thus knowing f(t), and the integral in the above equation exists, we can find the function F(s), that is called the Laplace Transform of f(t).

𝑠

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f(t) F(s)

1 𝛿(𝑡) 1

2 u(t) 1

𝑠

+ 𝜔

²

7 cos 𝜔𝑡 𝑢(𝑡) 𝑠

𝑠

²

+ 𝜔

²

Laplace Transform Table

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Laplace Transform Theorems

Theorem Name

1. 𝐿 𝑓 𝑡 = 𝐹 𝑠 =

0−

∞

𝑓(𝑡)𝑒^−𝑠𝑡𝑑𝑡 Definition

2. 𝐿 𝑘𝑓 𝑡 = 𝑘𝐹(𝑠) Linearity Theorem

3. 𝐿 𝑓₁ 𝑡 + 𝑓₂(𝑡) = 𝐹₁ 𝑠 + 𝐹₂(𝑠) Linearity Theorem 4. 𝐿 𝑒^−𝑎𝑡𝑓(𝑡) = 𝐹(𝑠 + 𝑎) Frequency Shift Theorem 5. 𝐿 𝑓 𝑡 − 𝑇 = 𝑒^−𝑠𝑇𝐹(𝑠) Time Shift Theorem

6. 𝐿 𝑓 𝑎𝑡 = 1

𝑎𝐹(𝑠

𝑎) Scaling Theorem

7. 𝐿 𝑑𝑓

𝑑𝑡 = 𝑠𝐹 𝑠 − 𝑓(0) Differentiation Theorem 8. 𝐿 𝑑²𝑓

𝑑𝑡² = 𝑠²𝐹 𝑠 − 𝑠𝑓 0 − 𝑓^′(0) Differentiation Theorem 9.

𝐿 𝑑^𝑛𝑓

𝑑𝑡^𝑛 = 𝑠^𝑛𝐹 𝑠 −

𝑘=1 𝑛

𝑠^𝑛−𝑘𝑓^𝑘−1(0)

Differentiation Theorem

10. 𝐿[

0 𝑡

𝑓 𝜏 𝑑𝜏] = 𝐹(𝑠) 𝑠

Integration Theorem

11. 𝑓 ∞ = lim

𝑠→0𝑠𝐹(𝑠) Final Value Theorem 12. 𝑓 0 = lim 𝑠𝐹(𝑠) Initial Value Theorem

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Partial Fraction Expansion

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 the partial – fraction expansion. If F₁(s) = N(s)/D(s), 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 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 the order less than its denominator. For example, if

𝐹₁ 𝑠 = ^𝑠³^+2𝑠²^+6𝑠+7

𝑠²+𝑠+5 we must perform the division and we get, 𝐹₁ 𝑠 = 𝑠 + 1 + ²

𝑠²+𝑠+5 of which the inverse Laplace transform can be obtained easily.

𝑓₁ 𝑡 = ^{𝑑𝛿(𝑡)}

𝑑𝑡 + 𝛿 𝑡 + 𝐿⁻¹ ²

𝑠²+𝑠+5 which can be found very easily by partial fraction expansion.

For example, 𝐹 𝑠 = ²

𝑠²+3𝑠+2 = ²

(𝑠+1)(𝑠+2) = ²

(𝑠+1) − ²

(𝑠+2)

Hence, 𝑓 𝑡 = 2𝑒^−𝑡 − 2𝑒^−2𝑡 𝑢(𝑡)

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Laplace Transform Solution of Differential Equation

Problem: Given the following differential equation, solve for y(t) if all the initial conditions are zero. Use the Laplace Transform.

𝑑²𝑦

𝑑𝑡² + 12 ^𝑑𝑦

𝑑𝑡 + 32𝑦 = 32 𝑢(𝑡)

Solution: The above equation can be solved as 𝑠²𝑌 𝑠 + 12𝑠𝑌 𝑠 + 32𝑌 𝑠 = ³²

𝑠

𝑌 𝑠 = ³²

𝑠(𝑠²+12𝑠+32) = ³²

𝑠(𝑠+4)(𝑠+8) = ^𝐾¹

𝑠 + ^𝐾²

(𝑠+4) + ^𝐾³

(𝑠+8)

Solving for 𝐾₁, 𝐾₂ , 𝐾₃ 𝐾₁ = ³²

(𝑠+4)(𝑠+8) 𝑠 → 0 = 1, 𝐾₂ = ³²

𝑠(𝑠+8) 𝑠 → −4 =

− 2, 𝐾₃ = ³²

𝑠 𝑠+4 𝑠 → −8 = 1 We get,

𝑌 𝑠 = ¹

𝑠 − ²

𝑠+4 + ¹

𝑠+8 𝑦 𝑡 = 1 − 2𝑒^−4𝑡 + 𝑒^−8𝑡 𝑢(𝑡)

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The Transfer Function 𝑎_𝑛 ^𝑑^𝑛^𝑐(𝑡)

𝑑𝑡^𝑛 + 𝑎_𝑛−1 ^𝑑^𝑛−1^𝑐(𝑡)

𝑑𝑡^𝑛−1 + ⋯ + 𝑎₀𝑐 𝑡 = 𝑏_𝑚 ^𝑑^𝑚^{𝑟 𝑡}

𝑑𝑡^𝑚 + 𝑏_𝑚−1 ^𝑑^𝑚−1^{𝑟 𝑡}

𝑑𝑡^𝑚−1 + ⋯ + 𝑏₀𝑟(𝑡)

Taking the Laplace Transform of the equation above assuming all initial conditions are zero,

𝐶(𝑠)

𝑅(𝑠) = 𝐺 𝑠 = ^(𝑏^𝑚^𝑠^𝑚^+𝑏^𝑚−1^𝑠^𝑚−1^+⋯+𝑏⁰⁾

(𝑎_𝑛𝑠^𝑛+𝑎_𝑛−1𝑠^𝑛−1+⋯+𝑎₀)

C(s) = R(s)G(s)

R(s)

G(s)

C(s)

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Problem: Find the transfer function G(s)=C(s)/R(s), corresponding to the differential equation

𝑑³𝑐

𝑑𝑡³

+ 3

^𝑑²^𝑐

𝑑𝑡²

+ 7

^𝑑𝑐

𝑑𝑡

+ 5𝑐 =

^𝑑²^𝑟

𝑑𝑡²

+ 4

^𝑑𝑟

𝑑𝑡

+ 3𝑟

Solution: Taking the Laplace Transform of both sides yields,

𝑠

³

𝐶 𝑠 + 3𝑠

²

𝐶 𝑠 + 7𝑠𝐶 𝑠 + 5𝐶 𝑠 = 𝑠

²

𝑅 𝑠 + 4𝑠𝑅 𝑠 + 3𝑅(𝑠) From which we get,

(𝑠

³

+3𝑠

²

+ 7𝑠 + 5)𝐶 𝑠 = 𝑠

²

+ 4𝑠 + 3 𝑅(𝑠) Hence,

𝐶(𝑠)

𝑅(𝑠)

=

^𝑠²^+4𝑠+3

(𝑠³+3𝑠²+7𝑠+5)

= 𝐺(𝑠)

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Problem: Find the ramp response for a system whose transfer function is 𝐺 𝑠 = ^𝑠

(𝑠+4)(𝑠+8)

Solution: 𝑟 𝑡 = 𝑡𝑢 𝑡 𝐻𝑒𝑛𝑐𝑒, 𝑅 𝑠 = ¹

𝑠²

C(s) = G(s)R(s)

Hence, 𝐶 𝑠 = ¹

𝑠(𝑠+4)(𝑠+8) = ^𝐾¹

𝑠 + ^𝐾²

(𝑠+4) + ^𝐾³

(𝑠+8) 𝐾₁ = ¹

(𝑠+4)(𝑠+8) 𝑠 → 0 =

1 32

𝐾₂ = ¹

𝑠(𝑠+8) 𝑠 → −4 = − ¹

16 , 𝐾₃ = ¹

𝑠(𝑠+4) 𝑠 → −8 = ¹

32

Hence, 𝐶 𝑠 = ¹

32𝑠 − ¹

16 𝑠+4 + ¹

32 𝑠+8 𝑐 𝑡 = ¹

32 − ¹

16 𝑒^−4𝑡 + ¹

32 𝑒^−8𝑡

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Electrical Network Transfer Functions

Component Current Voltage

Or

Voltage Current

Impedance Z(s) = V(s)/I(s)

Capacitor 𝑖 𝑡 = 𝐶

^{𝑑𝑣(𝑡)}

𝑑𝑡

𝐼 𝑠 = 𝐶𝑠𝑉(𝑠)

1 𝑠𝐶

Resister

𝑖 𝑡 =

¹

𝑅

𝑣(𝑡) 𝐼 𝑠 =

¹

𝑅

𝑉(𝑠)

𝑅

Inductor

𝑣 𝑡 = 𝐿

^{𝑑𝑖(𝑡)}

𝑑𝑡

𝑉 𝑠 = 𝐿𝑠𝐼(𝑠)

𝑠𝐿

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Transfer Function – Multiple Loops

Problem: Given the network of Figures (a) and (b) below, find the transfer function I₂(s)/V(s).

Solution: the first step in the solution is to convert the network into Laplace transforms for the impedances and circuit variables, assuming zero initial conditions.

The circuit with which we are dealing requires two simultaneous equations to solve 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.

𝑅₁𝐼₁ 𝑠 + 𝐿𝑠𝐼₁ 𝑠 − 𝐿𝑠𝐼₂ 𝑠 = 𝑉(𝑠) 𝐿𝑠𝐼₂ 𝑠 + 𝑅₂𝐼₂ 𝑠 + ¹

𝑠𝐶 𝐼₂ 𝑠 − 𝐿𝑠𝐼₁ 𝑠 = 0

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Combining terms in the above equations, we get simultaneous equations in I

₁

(s) and I

₂

(s):

𝑅

₁

Using the Cramer’s Rule to solve the simultaneous equations we get,

𝐼

₂

𝑠 =

(𝑅₁+𝐿𝑠) 𝑉(𝑠)

−𝐿𝑠 0

∆

=

^{𝐿𝑠𝑉(𝑠)}

∆

Where ∆= (𝑅

₁

+ 𝐿𝑠) −𝐿𝑠

−𝐿𝑠 (𝐿𝑠 + 𝑅

₂

+

¹

𝐶𝑠

) Forming the transfer function, G(s) yields 𝐺 𝑠 =

^𝐼²^(𝑠)

𝑉(𝑠)

=

^𝐿𝑠

∆

=

^𝐿𝐶𝑠²

𝑅 +𝑅 𝐿𝐶𝑠²+ 𝑅 𝑅 𝐶+𝐿 𝑠+𝑅

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Transfer Function – Inverting Operational Amplifier Circuit

Problem: find the transfer function V_o(s)/V_i(s) for the circuit given in the Figure below.

Solution: The transfer function of the operational amplifier circuit is given by the equation:

𝑍₁ 𝑠 = ¹

𝐶₁𝑠+ ¹

𝑅1

= ¹

5.6𝑥10⁻⁶𝑠+ ¹

360𝑥103

= ^360𝑥10³

2.016𝑠+1

𝑍₂ 𝑠 = 𝑅₂ + ¹

𝐶₂𝑠 = 220𝑥10³ + ¹⁰⁷

𝑠

𝑉_𝑜(𝑠)

𝑉_𝑖(𝑠) = − ^𝑍² ^𝑠

𝑍₁ 𝑠 = −1.232^𝑠²+45.95𝑠+22.55 𝑠

The resulting circuit is called PID controller and can be used to improve the performance of a control system.

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Transfer Function – Noninverting Operational Amplifier Circuit

Problem: Find the transfer function V

_o

(s)/V

_i

(s) for the circuit given.

Solution: We find each of the impedance functions Z

₁

(s) and Z

₂

(s). Thus,

𝑍₁ 𝑠 = 𝑅₁ + ¹

𝐶₁𝑠 𝑎𝑛𝑑 𝑍₂ 𝑠 =

𝑅₂(1/𝐶₂𝑠)

𝑅₂+(1/𝐶₂𝑠) 𝑎𝑛𝑑 ^𝑉^𝑜^(𝑠)

𝑉_𝑖 𝑠 = ^𝑍¹ ^{𝑠 +𝑍}²^(𝑠)

𝑍₁(𝑠)

And Hence,

𝑉_𝑜(𝑠)

𝑉_𝑖 𝑠

=

^𝐶²^𝐶¹^𝑅²^𝑅¹^𝑠²^{+ 𝐶}²^𝑅²^+𝐶¹^𝑅²^+𝐶¹^𝑅¹ ^𝑠+1

𝐶₂𝐶₁𝑅₂𝑅₁𝑠²+ 𝐶₂𝑅₂+𝐶₁𝑅₁ 𝑠+1

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Translational Mechanical System Transfer Functions

Mechanical systems parallel electrical networks to such an extent that there are analogies between electrical and mechanical components and variables. They are given as,

Component Force - Velocity Force - Displacement Impedance 𝒁_𝑴 𝒔 = 𝑭(𝒔)/𝑿(𝒔)

𝑓 𝑡 = 𝐾

0 𝑡

𝑣 𝜏 𝑑𝜏 𝑓 𝑡 = 𝐾𝑥(𝑡) 𝐾

𝑓 𝑡 = 𝑓_𝑣𝑣(𝑡)

𝑓 𝑡 = 𝑓_𝑣 𝑑𝑥(𝑡) 𝑑𝑡

𝑓_𝑣𝑠

𝑓 𝑡 = 𝑀𝑑𝑣(𝑡)

𝑑𝑡 𝑓 𝑡 = 𝑀𝑑²𝑥(𝑡) 𝑑𝑡²

𝑀𝑠²

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Transfer Function – One Equation of Motion

Problem: Find the transfer function, X(s)/F(s), for the system of Figures given below.

Solution: We now write the differential equation of the motion using Newton’s law to sum to zero all of the forces shown on the mass shown in Figure (a) below.

𝑀 ^𝑑²^𝑥(𝑡)

𝑑𝑡² + 𝑓_𝑣 ^{𝑑𝑥(𝑡)}

𝑑𝑡 + 𝐾𝑥 𝑡 = 𝑓 𝑡 𝑀𝑠²𝑋 𝑠 + 𝑓_𝑣𝑠𝑋 𝑠 + 𝐾𝑋 𝑠 = 𝐹 𝑠 𝐺 𝑠 = ^𝑋(𝑠)

𝐹(𝑠) = ¹

𝑀𝑠²+𝑓_𝑣𝑠+𝐾

(32)

Transfer Function – Two Degrees of Freedom

Problem: find the transfer function X

₂

(s)/F(s) for the system given below.

Solution: The system has two degrees of freedom, since each mass can be

moved in horizontal direction while the other is held still. Thus, two

simultaneous equations of motion will be required to describe the system.

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Forces on M1 and M2

(34)

The Laplace Transform of the equations of motion can now be written as, 𝑀₁𝑠² + 𝑓_𝑣1 + 𝑓_𝑣3 𝑠 + 𝐾₁ + 𝐾₂ 𝑋₁ 𝑠 − 𝑓_𝑣3𝑠 + 𝐾₂ 𝑋₂ 𝑠 = 𝐹 𝑠

− 𝑓_𝑣3𝑠 + 𝐾₂ 𝑋₁ 𝑠 + 𝑀₂𝑠² + 𝑓_𝑣2 + 𝑓_𝑣3 𝑠 + 𝐾₂ + 𝐾₃ 𝑋₂ 𝑠 = 0 From this, the transfer function 𝑋₂(𝑠) 𝐹(𝑠) can be calculated as,

𝑋₂(𝑠)

𝐹(𝑠) = 𝐺 𝑠 = ^(𝑓^𝑣3^𝑠+𝐾²⁾

∆

∆= [𝑀₁𝑠² + 𝑓_𝑣1 + 𝑓_𝑣3 𝑠 + (𝐾₁ + 𝐾₂)] −(𝑓_𝑣3𝑠 + 𝐾₂)

−(𝑓_𝑣3𝑠 + 𝐾₂) [𝑀₂𝑠² + 𝑓_𝑣2 + 𝑓_𝑣3 𝑠 + 𝐾₂ + 𝐾₃ ]

(35)

Potentiometers

These are simple reliable devices for measuring mechanical displacement (translatory linear or angular). A potentiometer is a simple voltage divider with three terminals – two fixed terminals and third a movable terminal attached to a jockey – symbolically represented in Figure below. The total resistance of the potentiometer (between fixed terminals) is spread out uniformly, linearly for translatory measurement or in helical form for the angular measurement.

(36)

With reference to the figure above,

𝑅_𝑝 = 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑜𝑚𝑒𝑡𝑒𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑥_𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑜𝑚𝑒𝑡𝑒𝑟 𝑜𝑣𝑒𝑟 𝑤ℎ𝑖𝑐ℎ 𝑅_𝑝 𝑖𝑠 𝑠𝑝𝑟𝑒𝑎𝑑 𝑜𝑢𝑡 𝑢𝑛𝑖𝑓𝑜𝑟𝑚𝑙𝑦

𝑥_𝑖 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑗𝑜𝑐𝑘𝑒𝑦 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑔𝑟𝑜𝑢𝑛𝑑 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝐺 (𝑖𝑛𝑝𝑢𝑡) 𝑅_𝐿 = 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑉_𝑅𝐸𝐹 = 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑉_𝑜 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 (𝑉_𝑀𝐺)

As the resistance is spread out uniformly

𝑅_𝑖 = ( 𝑥_𝑖 𝑥_𝑡)𝑅_𝑝 𝑅_𝑒𝑞 = 𝑅_𝑖 𝑅_𝐿 = ₁ ¹

𝑅𝐿+ ^𝑥𝑡

𝑥𝑖 1 𝑅𝑃

Then by voltage dividing action 𝑉_𝑜 = 𝑉_𝑅𝐸𝐹 ^𝑅^𝑒𝑞

𝑅_𝑒𝑞+ 𝑅_𝑃−𝑅_𝑖 = ^𝑉^𝑅𝐸𝐹

𝑥_𝑡 𝑥_𝑖 + 𝑅_𝑃 𝑅_𝐿 1− 𝑥_𝑖 𝑥_𝑡

If R_L is open circuit,

(37)

Finite values of R

_L

causes the output voltage to be nonlinear function of (x

_i

/x

_t

). This relationship is plotted as V

_o

/V

_REF

vs x

_i

/x

_t

in the Figure below.

The figure also shows the actual relationship without loading which can

vary above or below the ideal case because of the manufacturing tolerances.

(38)

It is seen from the figure above that loading causes the potentiometer output to be a nonlinear function of (x

_i

), the mechanical displacement. This can be remedied by feeding the potentiometer output voltage to an analog buffer;

an OPAMP used as noninverting unity gain amplifier. This has an input impedance of around 1000 MΩ which eliminates loading.

Potentiometer specifications

1. Total Resistance (R

_P

)

2. Resolution: It is the smallest incremental change that is possible in a potentiometer.

3. Linearity: It is the variation of output vs input displacement from the linear relationship.

4. Precision: It is an indication of Linearity; say a linearity of 1%.

5. Equivalent noise resistance: It represents the electrical noise generated when moving the jockey of a wire wound potentiometer.

6. Power handling capacity: Power (P) in watts that the potentiometer can

dissipate continuously. Thus the maximum reference voltage that can be

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Potentiometer Classification

1. Wire wound potentiometers: These are made of resistance wire wound on an insulating former. These could be:

1. Linear (translatory) – available in shaft displacement 7.5, 10, 15 cm.

2. Rotary – single turn (360

^o

), three turn (1080

^o

), five turn (1800

^o

) and ten turns (3600

^o

).

2. Non-wire wound potentiometers: These are divided into three groups:

• Carbon composition – coated film and moulded type.

• Conductive plastic – resistance element is in the form of a cavity (coated with carbon resin mixture) in plastic base and the unit is hermetically sealed.

• Ceramic potentiometer – resistance material comprises of a

hybrid of ceramic and metal; immune to humidity.

(40)

Potentiometer as a Feedback Element

(41)

Potentiometers are often used for position feedback in

position control systems like robots. Wire wound

potentiometers are not used as feedback elements because

of noise and finite resolution. Instead moulded

composition potentiometers are widely used for this kind

of application as these have infinite resolution, low

internal noise and very robust. The figure of this system

shows that a signal from buffer is fed to sample and hold

circuit which holds constant value of the signal for the

sampling period. This is necessitated so that the analog to

digital converter (ADC) can convert the hold signal to

digital form. Both these blocks would need a finite amount

of time for the signal processing (unlike the analog

blocks).

(42)

Synchros

A synchro is an electromechanical transducer commonly used to convert angular position of a shaft into an electric signal. It is commercially known as selsyn or autosyn. The basic synchro unit is usually called synchro transmitter. Its construction is similar to that of a three-phase alternator. The stator (stationary member) is of laminated silicon steel and is slotted to accommodate a balanced three phase winding which is usually of concentric coil type (three identical coils are placed in the stator with their axis 120^o apart) and Y-connected. The rotor is of dumbbell construction and is wound with a concentric coil. An ac voltage is applied to the rotor winding through the slip rings. The constructional features and schematic diagram of a synchro are shown in the Figures below.

(43)

Let an ac voltage 𝑣_𝑟 𝑡 = 𝑉_𝑟𝑠𝑖𝑛𝜔_𝑐𝑡 be applied to the rotor of the synchro transmitter shown in the figure above. This voltage causes a flow of magnetizing current in the rotor coil produces a sinusoidally time varying flux directed along its axis and distributed nearly sinusoidally in the air gap along the stator periphery. Because of transformer action, voltages are induced in each of the stator coils. As the air gap flux is sinusoidally distributed, the flux linking any stator coil is proportional to the cosine of the angle between rotor and stator coil axes and so is the voltage induced in each stator coil. The stator coil voltages are in time phase with each other. Thus we see that the synchro transmitter acts like a single-phase transformer in which the rotor coil is the primary and the stator coils form the three secondaries.

Then, for the rotor position of the synchro transmitter shown in the Figure above, where the rotor axis makes an angle θ with the axis of the stator coils S₂

𝑣_𝑠₁_𝑛 = 𝐾𝑉_𝑟𝑠𝑖𝑛𝜔_𝑐𝑡𝑐𝑜𝑠(𝜃 + 120^𝑜) 𝑣_𝑠₂_𝑛 = 𝐾𝑉_𝑟𝑠𝑖𝑛𝜔_𝑐𝑡𝑐𝑜𝑠(𝜃)

𝑣_𝑠₃_𝑛 = 𝐾𝑉_𝑟𝑠𝑖𝑛𝜔_𝑐𝑡𝑐𝑜𝑠(𝜃 + 240^𝑜) The three voltages of the stator are

𝑣_𝑠₁_𝑠₂ = 𝑣_𝑠₁_𝑛 − 𝑣_𝑠₂_𝑛 = 3𝐾𝑉_𝑟 sin 𝜃 + 240^𝑜 𝑠𝑖𝑛𝜔_𝑐𝑡 𝑣_𝑠₂_𝑠₃ = 𝑣_𝑠₂_𝑛 − 𝑣_𝑠₃_𝑛 = 3𝐾𝑉_𝑟 sin 𝜃 + 120^𝑜 𝑠𝑖𝑛𝜔_𝑐𝑡

(44)

When the value of θ = 0, it is seen that maximum voltage is induced in the stator coil S2, while it appears from the terminal voltage 𝑣

_𝑠₃_𝑠₁

= 0 in this case. Thus it can be seen that the input to the synchro transmitter is the angular position of its rotor shaft and the output is a set of three single- phase voltages given by the above equations. The magnitudes of these voltages are functions of the shaft position.

The output of the synchro transmitter is applied to the stator winding of a

synchro - control transformer. The control transformer is similar in

construction to a synchro transmitter except for the fact that the rotor of the

control transformer is made cylindrical in shape so that the air gap is

practically uniform. The system (transmitter-control transformer pair) acts

as an error detector. Circulating currents of the same phase but of different

magnitudes flow through the two sets of the stator coils. The result is the

establishment of an identical flux pattern in the air gap of the control

transformer as the voltage drops in resistances and leakage reactances of

the two sets of the stator coils are usually small.

(45)

Synchro Error Detector

(46)

The control transformer flux axis thus being in the same position as that of the synchro transmitter rotor, the voltage induced in the control transformer rotor is proportional to the cosine of the angle between the two rotors and is given by

𝑒 𝑡 = 𝐾𝑉_𝑟𝑐𝑜𝑠∅𝑠𝑖𝑛𝜔_𝑐𝑡

Where Φ is the angular displacement between the two rotors. When Φ = 90^o, that is the two rotors at right angles, then the voltage induced in the control transformer rotor is zero. This position is known as the electrical zero position of the control transformer.

In the figure above the transmitter and control transformer rotors are shown in their respective electrical zero positions. Let the rotor of the transmitter rotate through an angle θ in the direction indicated and let the control transformer rotor rotate in the same direction through an angle α resulting in a net angular separation of 𝜑 = (90^𝑜 − 𝜃 + 𝛼) between the two rotors. The voltage at the rotor terminals of the control transformer is then

𝑒 𝑡 = 𝐾𝑉_𝑟 sin 𝜃 − 𝛼 𝑠𝑖𝑛𝜔_𝑐𝑡

For a small angle displacement between the two rotor positions,

(47)

The synchro transmitter – control transformer pair thus acts as an error detector giving a voltage signal at the rotor terminals of the control transformer proportional to the angular difference between the transmitter and control transformer shaft positions.

Equation above though derived for constant (θ - α), is valid for varying conditions as well, so long as the rate of the angle change is small enough for the speed voltages induced in the device to be negligible. This equation is represented graphically in the Figure given below for an arbitrary time variation of (θ - α).

(48)

We see from the curves that the output of the synchro error detector is a modulated signal, the modulating wave has the information regarding the lack of correspondence between the two rotor positions and the carrier wave is the ac input to the rotor of the synchro transmitter. This type of modulation is known as suppressed-carrier modulation. The equation for the modulating signal representing the discrepancy between the two shaft positions is,

𝑒

_𝑚

𝑡 = 𝐾

_𝑠

(𝜃 − 𝛼)

Where K

_s

is known as sensitivity of error detector and has the units

of volts (rms)/rad angular difference of the shafts of the synchro

pair.

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The rotor of the control transformer is made

cylindrical in shape so that the air gap is

practically uniform. This is essential for a

control transformer, since its rotor terminals are

connected to an amplifier, therefore the change

in the rotor output impedance with rotation of

the shaft must be minimized. Another

distinguishing feature is that the rotor winding of

the control transformer has a higher impedance

per phase. This feature permits several control

transformers to be fed from a single transmitter.

(50)

AC Tachometer

in space w.r.t.

the reference coil) so that the voltage at the quadrature coil terminals is zero.

Both fields induce currents of frequency 𝜔

_𝑐

in the short circuited rotor producing reaction flux.

Assume now that the rotor rotates at speed 𝜃 (say in the direction of 𝜑

_𝑓

). As

_𝑏

weakens. As a result of this unbalanced strengths of the fields

𝜑

_𝑓

𝑎𝑛𝑑 𝜑

_𝑏

, a net voltage is induced in the quadrature coil.

(52)

Because of the 90

^o

space displacement of this coil, its voltage is in phase quadrature to the voltage applied to the reference coil; the voltage magnitude being proportional to the rotor speed. Reverse phenomenon happens if the rotor rotates in opposite direction. Thus the tachometer voltage is approximately given as

𝑣

_𝑡

𝑡 = 𝐾

DC Servomotors

DC motors are constructed with permanent magnets (PM) which have high residual flux density and very high coercivity, dc servomotors are now constructed with PMs resulting in much higher torque/inertia ratio and also high operating efficiency as these motor have no field losses. The speed of a permanent magnet dc (PMDC) motor is nearly directly proportional to armature voltage at a given load torque. Also the speed-torque characteristic at a given voltage is more flat than in a wound field motor as the effect of armature reaction is less pronounced in PM motor.

Three types of constructions employed in PMDC servomotors are illustrated in

the Figure below in (a), (b), and (c). In Figure (a) the armature is slotted with dc

winding placed inside these slots (as in normal dc motor). In Figure (b) the

winding is placed on the armature surface. Because of the larger air gap,

stronger PMs are needed in this construction. In Figure (c) a much lower inertia

is achieved by placing the winding on a nonmagnetic cylinder which rotates in

annular space between the PM stator and stationary rotor. The air gap has to be

still larger with consequent need of much stronger PMs. The constructional

(54)

(55)

AC Servomotor

An AC servomotor is basically a two-phase induction motor except for certain special design features. A two phase induction motor consists of two stator windings oriented at 90ô. Figure given below shows the schematic diagram for balanced operation of the motor, i.e. voltages of equal rms magnitude and 90ô phase difference are applied to the two stator phases, thus making their respective fields 90ô apart in both time and space, resulting in magnetic field of constant magnitude rotating at synchronous speed. The direction of rotation depends upon phase relationship of the voltages V₁ and V₂. Due to this voltages are induced in the short circuited rotor due which currents are induced. The rotating magnetic field interacts with the these currents producing a torque on the rotor in the direction of the field rotation.

(56)

A two phase servomotor differs in two ways from a normal induction motor:

1. The rotor of the servomotor is built with high resistance so that its X/R ratio is small and the torque-speed characteristics as shown by the Figure above is linear in contrast to the highly non-linear characteristics when large X/R ratio is used for servo applications.

2. In servo applications, the voltages applied to the two stator windings are seldom balanced. As shown in the Figure below, one of the phases known as reference phase is excited by a constant voltage and the other phase known as control phase is energized by a voltage which is 90^o out of phase with respect to the voltage of the reference phase. The control phase voltage is supplied from a servo amplifier and it has variable magnitude and polarity (±90^𝑜 phase angle with respect to the reference phase).

(57)

It can be proved using symmetrical components that the starting torque of servomotor under unbalanced operation, is proportional to E, the rms value of the sinusoidal control voltage e(t). This can be depicted graphically in the Figure below. The torque generated by the motor is a function of both the speed 𝜃 and rms control voltage E, i.e. 𝑇_𝑀 = 𝑓( 𝜃, 𝐸).

(58)

DC Servo System

The most common structure of a high performance dc servo system is shown in Figure 1 below. There is virtually universal agreement that the cascaded control structure is the most effective approach to high performance servo systems. The cascade control structure includes an innermost current (or torque) regulator, a speed regulator around the current (or torque) regulator, and an outermost position regulator around the speed regulator. The sequence of position, speed, and current (torque) is natural as it matches the structure of the process to be controlled. Position is the integral of speed while speed is proportional to the integral of torque. The 4 quadrant power supply just means that the power converter can handle operation of the motor for all combinations of torque (current) and speed (voltage).

The cascade control structure will operate properly only if the bandwidth of the various regulators have the correct relationship. Bandwidth is the range of frequencies over which the controlled quantity tracks and responds to the command signal. In the cascade control, the current regulator has the highest bandwidth, then the speed regulator, and finally the position regulator has the lowest bandwidth. Therefore, the system is properly adjusted beginning with the innermost current regulator and working outward to the position regulator. The cascade control structure also has the benefit of easily limiting

(59)

(60)

(61)

(62)

(63)

AC Servo System

The purpose of this section is to review some of the theory behind the major components of an ac servo system. Figure 7 repeats the block diagram of the ac servo system. The digital ac servo system is typically available with three modes of operation:

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