Figure (1): A Control System

Examples of Control Systems

Different examples of control systems are,

A Switch

One among the examples of a control system is a switch in which the input is mechanically pressing or applying force on button with the fingers and output is flow or non-flow of current. The diagrammatic representation is shown below

Figure (2): A switch

A Driving System

Here the input is the Acceleration which is given by a human being to the vehicle which controls or regulates the output i.e., the speed of the vehicle. The desired speed can be obtained by controlling the Acceleration. The diagrammatic representation is shown below

Figure (3): Driving System of an Automobile

Biological Control System

In this system a person with his finger points towards a particular object and the output is desired pointed direction. The control signal here is the position of the object. The diagrammatic representation is shown below

Figure (4): Biological Control System.

Examples of control system include temperature measurements of thermometer, refrigerator, washing machine, electric frying pans, household devices with thermoset like iron etc.

Elements of a Control System

In a system where, the variations in the output quantity are continuously measured through feedback and compared by the input quantity, then such a system is called ‘control system’. The reference input is the excitation signal given to the system. It is also called as command signal (or) excitation. The output quantity is the response obtained after the processing of the signal. It is also called as response (or) controlled variable. The basic components of control system are,

1. Plant
2. Feedback path elements
3. Error detector
4. Controller.

Figure: Block Diagram of Components Connected

The functioning of the component of the system shown in figure is,

Plant

It is a unit where actual processing is performed. The input to the plant is the control signal generated by the controller. The plant performs the necessary action on this signal and generates the desired output which is called as controlled signal.

Feedback

Feedback is a controlled action in which the sampled output is given to the input for automatic correction of output due to any changes or disturbances occurring in the system. Generally negative feedback is employed for controlling of systems as it provides accuracy, better stability and reject disturbances. The feedback signal is fed to the error detector.

Error Detector

The function of error detector is to generate error signal. There are two inputs to the error detector i.e., the feedback signal and the reference signal. The error detector generates an output which is the difference of these two signals. This error signal is fed to the controller for the necessary control action.

Controller

The controller modifies and amplifies the error signal so that, the signal is a bit modified. Now a modified signal is obtained by the controlled action of controller. The modified signal is then passed to the plant for output rectification so as to reduce the error signal.

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Fig. 1: Representation of an taken out of the block.

Input X(s) is given to the block by a straight line with an arrow head and output Y(s) by another is taken out of the block by another st. line with arrow head as indicated in the figure. The output Y(s) is the product of input X(s) and transfer function G(s) of the block and is given by equation

Y(s) = G(s) X(s)

Similarly block for each element of control system are developed and they are interconnected to obtain the block diagram of the control system. For the connection of blocks, the following constituents are used.

Summing Point

It is represented by a circle with a cross inside it. It is used when two signals are to be added or subtracted (+) sign near arrow head indicates addition of signal and (-) sign indicates subtraction of signal are shown in Fig. 2.

Fig. 2. Summing point.

Take Off Point

If the input signal goes to more than one point in the block diagram, then a take off point is used as shown in Fig. 3.

Fig. 3. Take off point.

Branch Point

If the input from a block goes to more than one point, then a branch point is used as shown in Fig. 4.

Fig. 4. Branch point.

Thus, overall transfer function is product of individual transfer function.

Cascade Connection

When a number of blocks are connected in series so that output of one block goes to the input of next block, then this connection is called cascade connect. It is shown in Fig. 5.

Fig. 5. Cascade connection.

R₁(s) = G₁(s) . R₁(s)

R₂(s) = G₁(s) . R₁(s)

= G₁(s) G₂(s) R(s)

C(s) = G₃(s) R₂(s)

= G₁(s) G₂(s) G₃(s) R(s)

= [G₁(s) G₂(s) G₃(s)] R(s)

Thus, over all transfer function is product of individual transfer function.

Parallel Connection

If input is given to different blocks and output is the sum of outputs then parallel connection is used as shown in the Fig. 6.

Fig. 6. Parallel connection.

C₁ = R . G₁(s)

C₂ = R . G₂(s)

C₃ = R . G₃(s)

C(s) = C₁ + C₂ + C₃

= [G₁(s) + G₂(s) + G₃(s)] R(s)

Thus in parallel connection, the over all transfer function is sum of individual transfer function of blocks.

Colonial Connection

When blocks are connected to form a closed loop then it is called colonial connection as shown in Fig. 7.

The block diagram does not represent the physical nature of a control system. Two different physical system may have same block diagram. Also, a physical system may not have a unique block diagram. It can be represented by different block diagram. However, the overall transfer function for the system is same.

Fig. 7. Colonial connection.

Procedure for Drawing Block Diagram

The steps to be followed to draw a block diagram of a system are given below:

1. Write down the differential equations representing the system.
2. Find Laplace transform of these equation assuming zero initial conditions.
3. Represent each equation in the form of a block.
4. Interconnect these blocks to get the final block diagram.

In this step use summing point when two signals are added/subtracted. Use cascade connection when output of one block is input to other block. Similarly, parallel connection is to be followed accordingly to the situation.

Properties of Block Diagram

1. Block diagram is a pictorial representation of a control system.
2. It does not gives any idea about physical nature of system.
3. Different analogous systems may have same block diagram.
4. Block diagram of a system is not unique.
5. It shows flow of signal thought out the system.

Significance of Block Diagram

1. Any system like electrical mechanical, thermal, chemical, plant process control, etc., can be represented by block diagram.
2. A complicated system can be represented by block diagram which can be further reduce to make it a simple diagram.
3. Transfer function of the system can be easily determined from the block diagram.
4. A system element can be easily added or deleted in the block diagram so as to study the behaviour of a given system in the presence or absence of the new element.

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Let us consider the Field controlled DC servomotor as shown in Figure 1.

The parameters are taken as

${R_f}$ = Field winding resistance.
${L_f}$ = Field winding inductance.
${i_a}$ = Armature current.
${i_f}$ = Field current.
${e_f}$ = applied field (control) voltage.
${e_b}$ = Motor back emf.
${T_m}$ = torque developed by motor.
$\theta $ = Angular displacement of motor shaft.
$J$ = Equivalent moment of inertia (of load and motor) referred to motor shaft.
$f_o$ = Equivalent viscous friction coefficient of motor and load referred to motor shaft.

In field controlled DC motors, the input field volatge ${e_a}$ controls the motor shaft output while the armature current ${i_a}$ remains constant. The DC motor operates in linear region for servo motor application. Hence, the air-gap flux $\phi $ is proportional of the field current $i_f$ as
\[\phi \propto {i_f}\]

\[\phi = {K_f}\hspace{0.1cm}{i_f}\]

The torque $T_m$ developed by the motor is proportional to the armature current i.e.

\[{T_m} \propto {i_a}\]

Also,

\[{T_m} \propto {K_f}\hspace{0.1cm}{i_f}\hspace{0.1cm}{i_a}\]

\[{T_m} = ({K_1}\hspace{0.1cm}{K_f}\hspace{0.1cm}{i_a})\hspace{0.1cm}{i_f}\]

In field controlled DC motors, the armature current $i_a$ remains constant. Therefore,

\[{T_m} = {{K’}_T}\hspace{0.1cm}{i_f}\]

where, ${{K’}_T}$ is the motor torque constant.

Using KVL on field circuit shown in Figure 1, the differential equation of the field circuit

\[{L_f}\left( {\frac{{d{i_f}}}{{dt}}} \right) + {R_f}\hspace{0.1cm}{i_f} = {e_f}…(1)\]

The torque equation is

\[J\left( {\frac{{{d^2}\theta }}{{d{t^2}}}} \right) + {f_o}\left( {\frac{{d\theta }}{{dt}}} \right) = {T_m} = {{K’}_T}\hspace{0.1cm}{i_f}….(2)\]

Transfer function calculation of the system

Taking the Laplace transforms of Equations 1 and 2 with assuming zero initial conditions, we get

\[{E_f}(s) = (s{L_f} + {R_f}){I_f}(s)…(3)\]
\[({s^2}J + sf)\theta (s) = {T_m}(s) = {{K’}_T}\hspace{0.1cm}{I_f}(s)….(4)\]

The transfer function of the system is obtained From Equations 3 and 4 as,

\[G(s) = \frac{{\theta (s)}}{{{E_f}(s)}} = \frac{{{K’}_T}}{{s[(s{L_f} + {R_f})(sJ + {f_o})]}}….(5)\]

Block diagram of the system

The complete block diagram is obtained From Equations 3 and 4 as shown below in Figure 2.

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Let us consider the Armature controlled DC servomotor as shown in Figure 1.

The parameters are taken as

${R_a}$ = Armature winding resistance.
${L_a}$ = Armature winding inductance.
${i_a}$ = Armature current.
${i_f}$ = Field current.
${e_a}$ = applied armature (control) voltage.
${e_b}$ = Motor back emf.
${T_m}$ = torque developed by motor.
$\theta $ = Angular displacement of motor shaft.
$J$ = Equivalent moment of inertia of load and motor referred to motor shaft.
$f_o$ = Equivalent viscous friction coefficient of motor and load referred to motor shaft.

In armature controlled DC motors, the armature input volatge ${e_a}$ controls the motor shaft output while the field current ${i_f}$ remains constant. The DC motor operates in linear region for servo motor application. Hence, the air-gap flux $\phi $ is proportional of the field current $i_f$ as
\[\phi \propto {i_f}\]

\[\phi = {K_f}{i_f}\]

The torque $T_m$ developed by the motor is proportional to the product of armature current and air gap flux i.e.

\[{T_m} \propto \phi {i_a}\]

Also,

\[{T_m} \propto {K_f}{i_f}{i_a}\]

\[{T_m} = ({K_1}{K_f}{i_f}){i_a}\]

In armature controlled DC motors, the field current $i_f$ remains constant. Therefore,

\[{T_m} = {K_T}{i_a}\]

where, $K_T$ is the motor torque constant.

The motor back emf being proportional to speed is given by

\[{e_b} = {K_b}\left( {\frac{{d\theta }}{{dt}}} \right)….(1)\]

where, $K_b$ is the back emf constant.

Using KVL on armature circuit shown in Figure 1, the differential equation of the armature circuit is

\[{L_a}\left( {\frac{{d{i_a}}}{{dt}}} \right) + {R_a}{i_a} + {e_b} = {e_a}…(2)\]

The torque equation is

\[J\left( {\frac{{{d^2}\theta }}{{d{t^2}}}} \right) + {f_o}\left( {\frac{{d\theta }}{{dt}}} \right) = {T_m} = {K_T}{I_a}….(3)\]

Transfer function calculation of the system

Taking the Laplace transforms of Equations 1, 2 and 3 with assuming zero initial conditions, we get

\[{E_b} = s{K_b}\theta (s)…(4)\]
\[(s{L_a} + {R_a}){I_a}(s) = {E_a}(s) – {E_b}(s)…(5)\]
\[({s^2}J + s{f_o})\theta (s) = {T_m}(s) = {K_T}{I_a}(s)….(6)\]

The transfer function of the system is obtained From Equations 4, 5 and 6 as,

\[G(s) = \frac{{\theta (s)}}{{{E_a}(s)}}\]

or

\[G(s) = \frac{{{K_T}}}{{s[(s{L_a} + {R_a})(sJ + {f_o}) + {K_T}{K_b}]}}….(7)\]

Block diagram of the system

The complete block diagram is obtained From Equations 4, 5 and 6 as shown below in Figure 2.

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Open loop control systems

It is a control system in which the control action independent of the output signal and depends only on its input signal, also the signal flows through the block is unidirectional. It is shown below,

Closed-loop control systems

It is a control system in which the control action dependents on both the output signal and the input signal. It is shown below,

As compared with the open-loop control system, an additional feedback feature in the closed-loop control system results in the relationship between the output and the input.

Comparison between Open loop and Closed loop control systems

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It is a control system in which the control action dependents on both the output signal and the input signal. It is shown below in Figure 1.

As compared with the open-loop control system, an additional feedback feature in the closed-loop control system results in the relationship between the output and the input.

Examples

Some of the most common examples in our surrounding as

• Automatic electric iron.
• A missile launcher system.
• DC motor speed control.

Advantages

• Accurate and reliable
• External noise and disturbance effects are insensitive due to the presence of feedback.

Disadvantages

• The system is complex and costly
• Complicate for maintenance
• They are less stable under certain conditions.

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It is a control system in which the control action independent of the output signal and depends only on its input signal, also the signal flows through the block is unidirectional. It is shown below in Figure 1.

Examples

Some of the most common examples in our surrounding as

• traffic light controller.
• washing machine.
• bread toaster.

Advantages

• Simple design and easy to construct.
• Cost is lesser.
• Ease of maintenance and simple in construction.
• There is no stability problem.

Disadvantages

• Inaccurate
• Unrealisable
• External noise and disturbance result in a change in output from the desired value.

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A block diagram of the second order closed-loop control system with unity negative feedback is shown below in Figure 1,

The general expression for the time response of a second order control system or underdamped case is

\[c(t) = 1 – \frac{{{e^{ – \xi {\omega _n}t}}}}{{\sqrt {1 – {\xi ^2}} }}\sin \left[ {({\omega _n}\sqrt {1 – {\xi ^2}} )t + \theta } \right]…(1)\]

Also Equation 1, is plotted in Figure 2 as shown below

Settling time $(t_s)$

In second order underdamped control system when unity step input applied, oscillation in the response occurs initially in the output time response and the magnitude of the oscillations decay exponentially with time constant $ 1/(\xi {\omega _n})$ . Settling time is the time taken by the response to settle down oscillations and stay within 2

For 2

\[{t_s} = \frac{4}{{\xi {\omega _n}}}\]

For 5

\[{t_s} = \frac{3}{{\xi {\omega _n}}}\]

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A block diagram of the second order closed-loop control system with unity negative feedback is shown below in Figure 1,

The general expression for the time response of a second order control system or underdamped case is

\[c(t) = 1 – \frac{{{e^{ – \xi {\omega _n}t}}}}{{\sqrt {1 – {\xi ^2}} }}\sin \left[ {({\omega _n}\sqrt {1 – {\xi ^2}} )t + \theta } \right]…(1)\]

Also Equation 1, is plotted in Figure 2 as shown below

Delay time $(t_d)$

It is the time required by the response to reach 50

\[{t_d} = \frac{{1 + 0.7\xi }}{{{\omega _n}}}\]

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A block diagram of the second order closed-loop control system with unity negative feedback is shown below in Figure 1,

The general expression for the time response of a second order control system or underdamped case is

\[c(t) = 1 – \frac{{{e^{ – \xi {\omega _n}t}}}}{{\sqrt {1 – {\xi ^2}} }}\sin \left[ {({\omega _n}\sqrt {1 – {\xi ^2}} )t + \theta } \right]…(1)\]

Also Equation 1, is plotted in Figure 2 as shown below

Peak overshoot $(M_p)$

It is the difference between first peak of overshoot for output and the steady state output value, i.e.

\[{\text{Peak overshoot (}}{M_p}) = c({t_p}) – c(\infty )\]

where, $c({t_p})$ is the first peak of overshoot for output $c(t)$ and $c(\infty )$ is the steady state output value of $c(t)$.

Also,

\[{\text{Peak percent overshoot (}}{M_p}){\text{ = }}\frac{{c({t_p}) – c(\infty )}}{{c(\infty )}} \times 100% \]

In second order underdamped system with unity step input, the steady state output $c(\infty )$ is unity, therefore

\[{M_p} = c({t_p}) – 1….(2)\]

\[% {M_p} = \frac{{c({t_p}) – 1}}{1} \times 100\]

Also, Equation 1 at time $t = {t_p}$ can be written as,

\[c({t_p}) = 1 – \frac{{{e^{ – \xi {\omega _n}{t_p}}}}}{{\sqrt {1 – {\xi ^2}} }}\sin ({\omega _d}{t_p} + \theta )….(3)\]

Using Equation 2 and Equation 3 gives

\[{M_p} = 1 – \frac{{{e^{ – \xi {\omega _n}{t_p}}}}}{{\sqrt {1 – {\xi ^2}} }}\sin ({\omega _d}{t_p} + \theta ) – 1\]

\[{M_p} = \frac{{{e^{ – \xi {\omega _n}{t_p}}}}}{{\sqrt {1 – {\xi ^2}} }}\sin ({\omega _d}{t_p} + \theta )….(4)\]

As first peak of overshoot for output value of $c(t)$ i.e. n = 1 using in general equation of peak time $({t_p})$ gives

\[{t_p} = \frac{{1.\pi }}{{{\omega _d}}}….(5)\]

Put Equation 5 in Equation 4 gives

\[{M_p} = \frac{{{e^{ – \xi {\omega _n}.(\pi /{\omega _d})}}}}{{\sqrt {1 – {\xi ^2}} }}\sin ({\omega _d}.\frac{\pi }{{{\omega _d}}} + \theta )\]

or,

\[{M_p} = \frac{{{e^{ – \xi {\omega _n}.(\pi /{\omega _d})}}}}{{\sqrt {1 – {\xi ^2}} }}\sin (\pi + \theta )….(6)\]

Put ${\omega _d} = {\omega _n}\sqrt {1 – {\xi ^2}} $ in Equation 6 gives

\[{M_p} = \frac{{{e^{ – \xi {\omega _n}.(\pi /({\omega _n}\sqrt {1 – {\xi ^2}} ))}}}}{{\sqrt {1 – {\xi ^2}} }}\sin (\pi + \theta )\]

or simply,

\[{M_p} = \frac{{{e^{ – \xi \pi /(\sqrt {1 – {\xi ^2}} )}}}}{{\sqrt {1 – {\xi ^2}} }}\sin \theta ….(7)\]

Since,

\[\sin \theta = \sqrt {1 – {{\cos }^2}\theta } = \sqrt {1 – {\xi ^2}} ….(8)\]

Using Equation 7 and Equation 8 gives

\[{M_p} = \frac{{{e^{ – \xi \pi /(\sqrt {1 – {\xi ^2}} )}}}}{{\sqrt {1 – {\xi ^2}} }}\sqrt {1 – {\xi ^2}} \]

or simply,

\[{M_p} = {e^{ – \xi \pi /(\sqrt {1 – {\xi ^2}} )}}\]

Also, Peak percent overshoot will be

\[% {M_p} = \frac{{{e^{ – \xi \pi /(\sqrt {1 – {\xi ^2}} )}}}}{1} \times 100\]

or simply,

\[% {M_p} = {e^{ – \xi \pi /(\sqrt {1 – {\xi ^2}} )}} \times 100\]

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