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Control Engineering (AEEC345)1. Introduction to Control Systems

Dr C. Themistos

1. Introduction to Control Systems

Control engineering is based on the foundations of feedback theory and linear system analysis, and it integrates the concepts of network theory and communication theory. Therefore control engineering is not limited to any engineering discipline but is equally applicable to aeronautical, chemical, mechanical, environmental, civil, and electrical engineering. For example, a control system often includes electrical, mechanical, and chemical components.A control system is an interconnection of components forming a system configuration that will provide a desired system response. Therefore, a component or process to be controlled, can be represented by a block, as shown in Figure 1.1.

Fig.1.1

1.1 Open- and closed-loop control systemsAn open-loop control system utilizes a controller or control actuator to obtain the desired response, as shown in Figure 1.2. An open-loop system is a system without feedback

Fig.1.2

In contrast to an open-loop control system, a closed-loop control system utilizes an additional measure of the actual output to compare the actual output with the desired output response. The measure of the output is called the feedback signal. A simple closed-loop feedback control system is shown in Figure 1.3.

Fig.1.3

A closed-loop control system uses a measurement of the output and feedback ofthis signal to compare it with the desired output (reference or command).


Dr C. ThemistosA common example of an open-loop control system is an electric toaster in the kitchen. An example of a closed-loop control system is a person steering an automobile (assuming his or her eyes are open) by looking at the auto’s location on the road and making the appropriate adjustments. The introduction of feedback enables us to control a desired output and can improve accuracy, but it requires attention to the issue of stability of response.

Fig.1.4

The system shown in Figure 1.4 is a negative feedback control system, because the output is subtracted from the input and the difference is used as the input signal to the power amplifier. With an accurate sensor the measured output is equal to the actual output of the system. The difference between the desired output and the actual output is the error, which is adjusted by the control device (usually an amplifier). The output of the control device causes the actuator to modulate the process in order to reduce the error.

1.2 Differential equations of Physical SystemsThe differential equations describing the dynamic performance of a physical system are obtained by utilizing the physical laws of the process and this approach applies equally well to mechanical, electrical, fluid and thermodynamic systems For example in mechanical systems, one utilizes Newton’s Laws, while in Electrical Systems, one utilizes Kirchoff’s laws. An electrical circuit (shown in Fig.1.5), can be described by Kirchoff’s current law as follows:

Fig.1.6

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1.3 Review of the Laplace and the Inverse Laplace TransformA great majority of physical systems are linear within some range of the variables. A system is defined as linear in terms of the excitation and response. The Laplace Transform is an operational method ideal for transforming linear differential equations, characterizing a system, into algebraic functions of a complex variable s. Operations, such as differentiation and integration can be replaced by algebraic operations in the complex plane.

A complex variable is a complex number that has a variable real and a variable imaginary part. In the Laplace transformation the notation s is used as a complex variable, where:

s = + j

and and are the real and imaginary part, respectively.

Complex variables can be plotted on the complex s-plane as shown in Fig.1.7, for s1.

Fig.1.7

Where the magnitude, and angle, , of s1, are:

A complex function, G(s) is a function of s with a real and an imaginary part. For example by considering:

then

G(s)= G( + j)=

By considering a function of time f(t), where f(t)=0 for t<0,

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Dr C. Themistosthe Laplace transform of f(t), F(S) is given by:

where, s is a complex variable.

By considering the step functionf(t)=0 for t<0

=A for t>0where A is a constant, its Laplace Transform is given by:

Some typical Laplace Transforms of commonly used functions are given in Table 1.1 below:

Functiondescription

Time function f(t),for t>0

Laplace Transform F(s)

Step A

Exponential Ae-at

Ramp At

Sinusoidal

Cosine

Table 1.1

The reverse process of finding the time function f(t) from the Laplace transform is called the Inverse Laplace transform, and therefore, Table 1.1 can be used vice-versa.

Some basic Laplace Transform theorems are the following:

(i) Real Differentiation theorem.The Laplace Transform of the derivative of a function is given by:

Laplace transform of

Where f(0) is the initial value of f(t) at t=0.

(ii) Real Integration theorem.If f(t) is of exponential order the Laplace Transform of the definite integral of f(t) is given by:

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Laplace Transform of

(iii) Final-value theoremThe final value theorem relates the steady state behaviour of f(t) to the behaviour of sF(s) in the neighborhood of s=0 and is stated as follows:

If f(t) and df(t)/dt are Laplace transformable, if F(s) is the Laplace transform of f(t), and if exists, then

1.4 The Transfer function of linear systemsThe transfer function of linear systems is defined as the ratio of the Laplace transform of the output variable to Laplace transform of the input variable, with all initial conditions assumed to be zero.By considering the voltage divider network shown in Fig.1.8

Fig.1.8

The Kirchoff’s voltage law in Laplace form yields:

For the input: V1(s)=I(s)(Z1(s) + Z2(s))

For the output V2(s)=I(s)Z2(s)

Therefore the transfer function G(S) can be defined as:

Note: In the preceding sections, as a convention, the lower case variable names will denote time dependent variables, where upper case variable names will denote frequency dependent variables. For example, v denotes v(t), while V denotes V(s).

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2. Mathematical modelling of dynamical Control Systems

The mathematical model of a dynamical (time-dependent) system can be described in different forms such as the transfer function or the state-space representation. Once a mathematical model of a system is obtained in differential form and linearized using the Laplace transform, there are several numerical tools used for analysis and synthesis problems.A control system may consist of a number of components, where the function of each component and the flow of signals can be demonstrated using block diagrams or signal flow diagrams that will be examined in this section.

2.1 Transfer function of dynamical systemsAs it has been mentioned earlier the differential equations describing the dynamical behaviour of dynamical systems, are obtained by utilizing physical laws of a process and apply equally well to electrical, mechanical, fluid and thermodynamic systems.

2.1.1 Transfer function of electrical systemsBy considering the basic elements of an electrical circuit, the resistor, R, the inductor, L, and the capacitor, C, the basic differential equations describing their behaviour based on Ohm’s law and the respective Laplace transforms and element impedance (in Laplace form) are given below:

Differential equation:

Laplace transform:

Element Impedance:

Consider the circuit diagram of a potential divider described earlier in section 1.4, shown in Fig.2.1:

Fig.2.1

The transfer function of the above circuit has been derived to be:

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By assuming the impedances Z1 and Z2 to be Z1=R and Z2=1/sC the circuit yields a passive low-pass filter as shown in Fig.2.2.

Fig.2.2The transfer function is then given by:

2.1.2 Transfer function of operational amplifiersThe operational amplifier, particularly in the inverting configuration, is widely used in control systems.

Fig.2.3

By considering the above circuit in terms of impedances, as shown in Fig.2.3 the transfer function can be given by:

ExampleDetermine the transfer function Vo/Vi of the op-amp configuration shown in Fig.2.4

Fig.2.4

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Solution:

Zf=R

2.1.3 Transfer function of cascaded elementsConsider the network of cascaded elements, shown in Fig.2.5.

Fig.2.5

The transfer function, G(s)= Vo/V1 of the above system can be obtained using the node method.

Node ABy applying by applying Kirchoff’s current law the following equation is obtained:

I1=I2+ I3 (1)

Where:

By substituting in (1):

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(2)

Node BI3=I4 (3)

Where:

By substituting in (3):

(4)

By substituting (4) in (2):

Therefore, the transfer function of the system, G(s) becomes:

(5)

ExampleDerive the transfer function of the following network of cascaded RC elements

By considering:

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where Z1=R and Z2=1/sC

Therefore

By assuming R=1 M and C=1F, then RC=1 and the transfer function yields to:

2.2 Mechanical SystemsThe fundamental law governing mechanical systems is the Newton’s second law:

where m is the mass, the acceleration and F the applied Force of a body.

By considering the basic elements of the Translational and the Rotational mechanical systems, such as the mass the spring and the damper, the differential equations describing their behaviour based on Newton’s laws and their respective Laplace transforms are given below.

2.2.1 Translational Motion

Element: Spring MassDamper

Differential Equation:

Laplace Transform:

where F is the Force, v is the translational velocity, 1/k is the reciprocal translational stiffness, M is the mass and b is the viscous friction

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Dr C. ThemistosAlternatively in translational displacement:



Laplace Transform:

Consider the simple spring-mass damper system and the free body diagram of the mass shown in Fig.2.6a and Fig.2.6b, respectively. The system could represent , for example an automobile shock absorber and the wall friction is described as viscous damper (linear friction force)..

Fig.2.6a Fig.2.6b

By summing the forces acting on the mass from the free-body diagram Newton’s second law yields:

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Dr C. ThemistosThe Laplace Transform of the above equation is:

By assuming the following initial condition: F(t)=0, y(0)=y0 and y’(0)=0 the equation becomes:

Therefore solving for Y, the transfer function of the system is described by:

The denominator of the equation, when is set to zero is called the characteristic equation of the system, because the roots of the equation (called poles) determine the character of the time response

2.2.2 Rotational Motion


Differential Equation:

Laplace Transform:

where T is the Torque, is the angular velocity, 1/k is the reciprocal rotational stiffness, J is the moment of inertia and b is the viscous friction

Alternatively in translational displacement:


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Laplace Transform:

Example: Modelling a DC MotorConsider the armature controlled DC motor, where the electric circuit of the armature and the free body diagram of the rotor is shown in Fig.2.7a and Fig.2.7b, respectively. Assume the rotor has inertia Jm and viscous friction coefficient b.

Fig.2.7a Fig.2.7b

From the electric circuit of the armature, by applying Kirchoff’s law:

(1)

Ra is the armature resistance and inductance. Respectively, ia is the armature current and e is the back emf that generates the electromotive force, defined as:

where Ke is the electric constant and m is the angular displacement of the shaft. Equation (1) then becomes:

(2)

In Laplace form the above equation is transformed in:

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(3)

From the free body diagram of the shaft shown in Fig.2.7b, Newton’s law

yields:

(4)

where T=Ktia and Kt, is the torque constant, and therefore,

and by applying the Laplace transform and assuming zero

initial conditions equation (4) yields:

(5)

By substituting equation (3) in (5)

(6)

The transfer function of equation (6) m/Ua would then be:

3. Control System RepresentationThe importance of the transfer function to represent the relationship between the system variables can be viewed by diagrammatic means. The Block diagram and Signal flow graph models relationships are the most important means of pictorial system representation in Control Engineering.

3.1 Block diagram modelsThe simplest form of the Block diagram is the functional block, shown in Fig.3.1 consisting of a unidirectional block that represent the transfer function, G(s), of a system and the input/ output arrows that indicate the direction of the flow of signals, as shown in Fig.3.1.

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Dr C. ThemistosFig.3.1

The arrowhead pointing towards the block indicates the input signal, A(s) and the arrowhead leading away from the block represents the output signal, B(s). The signal can pass only in the direction of the arrows. The dimension of the output signal from the block is the dimension of the input signal multiplied by the dimension of the transfer function in the block.

e.g. B(s)=A(s)*G(s)

Interconnecting the blocks of the components according to the signal flow, the block diagram of the entire system can be formed. Apart from the basic functional block, there are two other basic symbols used in block diagrams, such as the summing point and the branch point (or pick off point), shown in Fig.3.2a and Fig.3.2b, respectively.

Fig.3.2a Fig.3.2bThe summing point, is the symbol that indicates a summing operation. The plus ir minus symbol at each arrowhead determines whether the signal is to be added or subtracted. The quantities being added or subtracted should have the same dimensions and units.

The branch point is a point from which the signal from a block goes concurrently to other blocks or summing points.

3.2 Block diagram reductionThe block diagram representation of a system with no loading effects (the output of a block is not affected by the next following block), often can be reduced by block diagram reduction techniques, to a simplified block diagram with fewer blocks. The basic rules for block diagram algebra are given below:

3.2.1 Combining blocks in Cascade

Original block diagram: Equivalent diagram:

3.2.2 Moving a summing point behind a block


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3.2.3 Moving a branching point ahead of a block


3.2.4 Moving a branching point behind a block


3.2.5 Moving a summing point ahead of a block


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Dr C. Themistos3.2.6 Negative feedback control systemConsider the negative feedback control system shown in Fig.3.3.

Fig.3.3

G and H are the gain and feedback transfer functions, respectively, and E the error signal, i.e. the difference between the required output and the measured output.From the above figure the actual output, C, is given by:

C= GE = G(R - HC)

C= GR – GHC

C + GHC = GR

Therefore the closed loop transfer function Y/R would become:

For a unity gain feedback H=1, shown in Fig.3.4,

Fig.3.4

the closed loop transfer function would then be:

Example: Block diagram reductionConsider the block diagram of the multiple-loop feedback control system shown in Fig.3.5. The block diagram reduction is based on the utilization of the rule that eliminates feedback loops feedback loop.

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Fig.3.5

Step1: Initially to eliminate feedback loop G3G4H1, we H2 behind block G4 using rule in section 3.2.4

Step 2: Combine G3G4

Eliminate feedback loop G3G4H1 using rule in section 3.2.6.

Step3: Combine G2 with with the resultant transfer function in step 2 and eliminate upper feedback loop by using rule in section 3.2.6, as follows:

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Dr C. ThemistosThe block diagram would then be reduced to:

Step4: Combine G1 and eliminate the resultant feedback loop as follows

The final block diagram would then become:

4. Transient Response AnalysisThe transient response is the response of a system as a function of time. In practice the input signal to a control system is random and cannot be expressed analytically. Therefore, we must have a basis of comparison of performance of various control systems, by specifying particular test input signals and by comparing the responses of various systems to these input signals.

4.1 Test input signalsThe most common test input signals used in control systems are the unit step, the ramp and the parabolic, as shown in Fig.4.1

(a) (b) (c)Fig.4.1 Test input signals: (a) step, (b) parabolic and (c) impulse

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Dr C. ThemistosAll systems having the same transfer function will exhibit the same output in response to the same input.

4.2. Unit step response of first order systemA first order system (a system where order of the Laplace operator, s, in the denominator of the transfer function has order 1) can be represented in a block diagram form as shown in Fig.4.2.

Fig.4.2By using the feedback loop block reduction rule:

where G= , H=1 and E=R-C

the Closed Loop Transfer Function (CLTF) of the system can be expressed as:

The step input signal shown in Fig.4.1(a) can be expressed in Laplace form as .

For a unit step input (A=1) the Laplace transform would then be

From the CLTF of the first order system, the output can be expressed as:

Therefore, for a unit step input the output would be:

By expanding the above product into partial fractions:

So,

From the numerator:

let s=0 A =1 let s=-1/T 1 = -B/T B=-T

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Therefore,

By applying the Inverse Laplace transform:(4.1)

where c is the unit step response of the first order system and T is the time constant of the system.

The response curve of the first order system (T=1) is then shown in Fig.4.3.

Fig.4.3The time constant, T, defines the initial speed of the response, thus the smaller the time constant the faster the response.

By substituting t=T the response becomes:

Therefore, the time constant also defines the time when the response reaches 63% of the peak value.

4.3 Steady state responseThe steady state response, css, is the response that exists for a long time following an input signal. It can be defined by applying the final value theorem to the response in the time- or in the s-domain as follows:

(4.2)Therefore in the time domain:

css=1In the s-domain:

css=1

4.3 Steady state error

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Dr C. ThemistosThe steady state error is the error after the transient response has decayed, leaving only the continuous response. From the block diagram in Fig.4.1 the error signal, E, is defined as:

E=R-CBy utilising the final value theorem in the time domain:

ess=0In the s-domain:

ess=0

ExampleDrive the unit step response of the system shown below and calculate the steady state response and the steady state error.

The CLTF of the above system is:

The response in Laplace form would be:

By partial fractions:

So,

From the numerator:

let s=0 A =1 let s=-10/3 10 = -(10B)/3 B=-3

Therefore,

By Inverse Laplace transform:

The steady state response will become: css=1

The steady state error then is:

ess=0

4.4 Unit step response of second order systems

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Dr C. ThemistosConsider the block diagram of a second order system shown in Fig.4.4

Fig.4.4The CLTF of the system is:

where, n is the natural frequency and is the damping ratio.

By using similar approach as for the first order system the transient response can be determined as:

(4.3)

where d is the damped natural frequency, and 0 < < 1

The unit step response of the second-order systems are classified with respect to the damping ratio as follows:

(a) Underdamped case (0 < < 1)

Equation (4.3) determines the unit step response of the underdamped second order system, shown in Fig.4.5

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Fig.4.5The basic features of the above response are the following:

(i) Peak time, Tp: The time required for the response to reach the first peak of the overshoot.

(ii) Maximum overshoot, Mp: The maximum peak value of the response curve measured from unity.. If the final steady state of the response differs from unity it is common then to use the maximum percentage overshoot, P.O., defined as:

The maximum overshoot can be calculated by using: . The above

derivative will then be evaluated at t=Tp (the peak time). By solving the above equation the peak time Tp is found to be:

The maximum overshoot can then be evaluated as:

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(iii) Rise time, Tr: The time required for the response to rise from 10% to 90%.

(iv) Settling time, Ts: The time required for the response to reach and stay within a range 2% or 5% about the final value. For the 2% range the settling time can be evaluated as:

(v) Delay time, Td: the time required for the response to reach half the final value the very first time.

ExampleCalculate the natural frequency, n, the damping ratio, the peak time, Tp the percentage overshoot, P.O. and the settling time, Ts, of the unit step response for the system shown below:

The closed loop transfer function of the system will be:

Therefore, by comparing with the standard form:

n2

=64 n=8 rad/sec and 2n=8 =

Then,

(b) Overdamped case ( >1) Consider the closed loop transfer function of the system to be:

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The two poles of the CLTF are negative real and unequal and the step response will be as shown in Fig.4.6

Fig.4.6The response is similar to the first order step response

(c) Undamped case ( =0) Consider the closed loop transfer function of the system to be:

In this case, the transient response shown in Fig.4.7 has a sinusoidal shape and does not die out.

Fig.4.7

5, Frequency Response Plots (Bode Plots)

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Dr C. ThemistosThe frequency response of a system is defined as the steady state response of the system to sinusoidal input.

The transfer function of a system, G(s), can be described in the frequency domain by the relation:

where and

The transfer function can also be expressed by a magnitude and a phase as:

where

and

A simplified graphical representation of the frequency response of a system G(j) can be obtained utilizing Logarithmic plots often called Bode plots, where the logarithmic Magnitude of the transfer function can be expressed as:

Logarithmic gain = where the units are decibels.

The logarithmic gain in dB and the phase can be plotted versus the frequency by using several different arrangements. For a Bode diagram, the plot of the logarithmic gain in dB versus is normally plotted in one set of axes and the phase versus on another set of axes.

Example 5.1 Bode diagram of an RC filterConsider an RC filter shown in Fig.5.1

Fig.5.1

By voltage divider rule the output Voltage in Laplace form is:

therefore the Transfer function, G(s), will be:

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and the sinusoidal steady-state transfer function is:

By setting

where, is the time constant of the network, then ,

the magnitude of the gain is:

Therefore the logarithmic gain of the system would be:

Therefore,

For small frequencies, such as 1/ then:

For example: =0.01/

=0. 1/ =1/

The above frequency =1/ is also known as the break frequency or corner frequency

For =10/ For =100/

An interval of two frequencies with a ratio equal to 10 is called a decade. For frequencies above the corner frequency 1/ there there is a decrease in the gain, where the slope of the asymptotic line for the 1st order system is –20 dB/decade. By plotting the gain with respect to the frequency variation in terms of the corner frequency the gain response is obtained as shown in Fig.5.2

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Fig.5.2The phase response, , of the above circuit can be obtained from

Where R() and X() are the real and the imaginary part of the transfer function, respectively.

For the transfer function: the phase response can be found to be:

For small frequencies, such as 1/ then For example:=0.01/ =0.1/ At the corner frequency =1/ For frequencies, above the corner frequency, >>1/ then For =10/ For =100/ Therefore the phase response in terms of the corner frequency will be as shown in Fig.5.3

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Dr C. ThemistosFig.5.3

The advantage of the logarithmic plots (Bode plots) is the conversion of products of transfer functions into a sum of standard factors. By considering a generalised transfer function of a system, consisting of sub-systems to be:

In terms of the logarithmic gain, the generalized transfer function can be expressed as:

Therefore, a bode diagram of a generalized transfer function can be expressed in terms of the addition of simpler bode plots of known simple transfer function factors.Similarly, the phase of the above system can be evaluated in terms of the sum of the individual phases of the sub-systems as follows:

G = G1 + G2 - G3 - G4

The different types of factors that may occur in a transfer functions are as follows:

1. Constant gain, Kb

2. Poles at the origin (j) Differentiator and Integrator (G=s or )

3. Poles or zeros on the real axis First order systems ( or )

4. Complex conjugate poles or zeros Second order systems ( )

Any generalized transfer function can be plotted using asymptotic approximations for each factor and adding them graphically.The curves for each standard form are determined as follows:

1. Constant gain, G=K b

The logarithmic gain is defined as:

and the phase as:

Graphically the gain and the phase plot for the constant gain is a straight horizontal line at and 0, respectively.

2. Differentiator or Integrator (G=s or )

The gain plots for G=s or are given by: and

, respectively. The gain plot for G=s and is a straight line

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Dr C. Themistoswith a positive slope of +20dB/decade and a negative slope of –20dB/decade, respectively.

The phase plots for G=s and are: and

The bode diagrams for the gain and the phase of the above functions are shown in Fig.5.4

Fig.5.4

3. First order system ( and )

The gain plots for ( and ) can be evaluated by:

and , respectively, where 1/ is the

break frequency. Below the break frequency the gain plots exhibit zero gain. Above

the break frequency the gain plot for and has increasing slope of

+20dB/decade and a decreasing slope of –20 dB/decade, respectively.

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The phase plots for and ) are:

and , respectively.

Both the phase plots for and originate from 0 and terminate at

+90 and -90, respectively. At the corner frequency the plots for and

have a phase of +45 and -45, respectively.

The bode diagrams for the gain and the phase of the above functions are shown in Fig.5.5

Fig.5.5

4. Second order system

The gain and the phase plot of the above functions can be calculated from:

and ,

respectively,

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Dr C. Themistoswhere n is the natural frequency of the system and corresponds to the corner frequency of the plot. The gain plot above the break frequency exhibits a slope of –40 dB/decade. The phase plots originates from 0 and terminates asymptotically to -180 having a phase of 90 at the break frequency, as shown in Fig.5.6.

Fig.5.6

Procedure for drawing the Bode diagramConsider the following transfer function of a system

1. Factorise the transfer function in terms of the standard factors.

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2. Define the corner frequencies of the standard factors and put the in ascending order.

The transfer function consists of:a) Constant gain factor, 5, (Plot 1)b) Integrator with a corner frequency, c=1 (Plot 2)

c) First order factors: (Plot 3) and (Plot 4) with corner

frequencies of c=2 and c=10, respectively.

d) Second order factor: with a corner frequency of c=10 (Plot

5)3. Plot the standard gain plot for each factor and add the slope contribution from

each curve to obtain the overall gain response (thick line), as shown in Fig.5.7.

Fig.5.7Similarly, the phase response for each known factor determined earlier is plotted and the overall phase response (thick line) is obtained from the sum of the phases at each frequency, as shown in Fig.5.8

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Fig.5.86. Stability of Linear Feedback SystemsA stable system should exhibit a bounded (limited) output if the corresponding input is bounded.The stability of a feedback system is related to the location of the roots of the characteristic equation of the system.Consider the general closed loop transfer function of a linear closed loop system:

where an…a0 and bn…b0 are constants and m<n

By setting the equation of the numerator n(s)=0, the roots of the equation are known as the zeros of the system transfer function.By setting the equation of the denominator q(s)=0, the roots of the equation are known as the poles of the system transfer function. The equation q(s) is also defined as the characteristic equation of the system.

A necessary and sufficient condition for a feedback system to be stable is: All poles of the system transfer function have negative real parts (i.e. are

located in the Left Hand Side (LHS) of the s-plane.)An example regarding the stability of a feedback system with respect to the location of the poles of its transfer function on the s-plane, is shown in Fig.6.1

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Fig.6.1

Two more necessary conditions must be satisfied in order to determine whether a system is stable or not.

1. All the coefficients of the polynomial of the characteristic equation must must have the same sign.

2. All the coefficients of the polynomial must be non-zero.Note: The above conditions are necessary but not sufficient

ExampleDetermine whether the conditions for stability are satisfied in the following characteristic equations1. Stable (The two necessary conditions are satisfied)2. Unstable (The s4 coefficient is missing)3. Unstabe (The s3 coefficient has negative sign)6.1 The Routh-Hurwitch Stability Criterion.In order to satisfy the necessary and sufficient condition for the stability of linear systems it is required to determine the poles of the characteristic equation. For higher order systems this can only be achieved using computer software like MATALB.The Routh-Hurwitch Criterion is a necessary and sufficient criterion for the stability of linear feedback systems. The method provides an answer to the question of stability, without solving the characteristic equation of the transfer function.

It is based on ordering the coefficients of the characteristic equation

into an array as follows:

The rows are then completed as

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where

The Routh-Hurwitch stability criterion states that:The number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array.

Therefore, for a stable system there should be no sign changes in the first column. Three different cases or configurations of the first column of the Routh Array will be considered.

Case 1. No element in the first column is zero.The characteristic polynomial of a third order sytstem is:

The Routh array would be:

where: and

The Routh array will then become:

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Two changes in sign appear in the first column, therefore, the system is unstable with two roots in the RHS of the s-plane.

Case 2. A zero appears in the first column while other elements in the row containing the zero are nonzero.

The characteristic polynomial of the closed loop transfer function of a system is:

The Routh array is:

where and

The zero element in the third row can be replaced by a small positive number, , that is allowed to approach zero after completing the array.

The array will then become:

since when

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since when

and

Therefore the final Routh array is:

There are two sign changes in the first column, therefore the system has two unstable poles.

Case 3. A zero appears in the first column and other elements in the row containing the zero are also zero.This indicates that there are roots of equal magnitude that are symmetrical about the origin of the s-plane. (i.e. two real roots with equal magnitudes and opposite sign and/or two conjugate imaginary roots)In this case the evaluation of the rest of the array can be continued by forming an auxiliary polynomial with the coefficients of the last row and by using the coefficients of the derivative of this polyniomial in the next row.

Consider the following characteristic equation:

The array of coefficients is

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Auxiliary polynomial U(s)

and

The coefficients of the s3 row are all zero, and therefore, the auxciliary polynomial, U(s), is formed from the coefficients of the s4 row as follows:

The derivative of P(s) would then be:

The above coefficients are then used for the row containing all the zero elements as follows and by completing all the rows the array will become::

Coefficients of dU(s)/ds

One sign change can be found in the first column of the new array, therefore the original equation has one root with positive real part.

6.2 Application of Routh’s Stability Criterion to control systemsConsider the system shown in Fig.6.2

Fig.6.2

The closed loop transfer function of the system will be:

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The characteristic equation will become:

The Routh Array will become:

For stability K must be positive (K > 0), and

6.3 Relative Stability in the frequency domain It is often required to determine the degree of stability of a stable system, known as the relative stability, that is measured by the relative real part (or other measure) of each root.

In the frequency domain the relative stability of a closed loop system can be utilized by the Nyquist stability criterion. The Nyquist criterion is defined in terms of the stability point 0dB and -180 on the Bode plot. The proximity of the magnitude and phase plot to the stability point is a measure of the relative stability of a system, that is determined in terms of the phase and gain margin.

The gain margin is defined as the reciprocal of the gain at the frequency at which the phase angle reaches 180, therefore

The phase margin is defined as the phase angle of the GH(j) plot at unity magnitude.

7. PID ControllersAn automatic controller compares the actual value of the plant output with the reference input (desired value), determines the deviation, and produces a control signal that will reduce the deviation to zero or to a small value. Consider the following unity feedback system shown in Fig.5.1.

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Fig.5.1Plant: A system to be controlledController: controls the overall system behavior

7.1 The three-term controllerThe transfer function of the PID controller looks like the following:

where: KP = Proportional gain KI = Integral gain KD = Derivative gain

Using the schematic diagram shown in Fig.5.1 the PID controller works in a closed-loop system as follows:The error signal (e) will be sent to the PID controller, and the controller computes both the derivative and the integral of this error signal.

The basic characteristics of P, I, and D controllers are the following: A proportional controller (KP) will have the effect of reducing the rise time and will

reduce, but never eliminate, the steady state error.. An integral control (KI) will have the effect of eliminating the steady-state error, but

it may make the transient response worse. A derivative control (KD) will have the effect of increasing the stability of the system,

reducing the overshoot, and improving the transient response.

5.2 Effects of control actionsConsider a second order system with the following closed loop transfer function,

To compare the above equation with the standard form, ,

the transfer function will become:

Therefore, the natural frequency, n, and the damping ratio, , can be calculated to be:

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From the damping factor, the system is found to be overdamped (>1), and therefore the rise time is quite slow.The steady state response can be calculated from:

Therefore the steady state error will be:

ess=1-Css= 0.95

which a quite high valueThe response obtained for the above system is shown in Fig.7.2

Fig.7.2

The DC gain of the plant transfer function is 1/20, so 0.05 is the final value of the output to an unit step input. This corresponds to the steady-state error of 0.95, quite large indeed. Furthermore, the rise time is about one second, and the settling time is about 1.5 seconds. Let's design a controller that will reduce the rise time, reduce the settling time, and eliminates the steady-state error.

7.3 Proportional controlThe proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces the steady-state error. The closed-loop transfer function of the above system with a proportional controller is:

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Let the proportional gain (Kp) equals 300 and the closed loop transfer function will become:

Therefore, the natural frequency, n, and the damping ratio, , can be calculated to be:

The system becomes underdamped with damped frequency :

The peak time is:

The Maximum overshoot

=

The above plot shows that the proportional controller reduced both the rise time and the steady-state error, increased the overshoot, and decreased the settling time by small amount.

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