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Lecture -4: Block diagrams -2 hours

By Mr S WijewardanaPhD student QMUL 16-04-2013

Learning Objectives:1. Why do we use block diagrams in control systems?2. Standard representation of blocks3. Simplification or reduction of block diagrams

1.Block diagrams have become more popular in linear control systems because of its simplicity and easy to manipulate.

2. Not only in control systems, block diagram representation is deployed in other areas like software engineering, process engineering, economics etc.

3. It is particularly important when we wanted to discuss the output or behaviour of a complete system as one unit while keeping the inside electronic undisclosed or insignificant.

4. Designers are happy to work with block diagrams when dealing with the analysis and simulations than the robust each and every components

vinV outiV outV inInputOutput

Block diagram representation of an LRC circuitIn lecture-2 we have derived the relationship between Output / Input =

Fig.1Fig.2Now we can completely ignore Fig.1 which is the RLC circuit and can instead concentrate on Fig.2 which is the same diagram with G(s), where G(s) is some function of s. No matter how complicated is the circuit diagram we are now dealing with a simple block diagram with a mathematical representation in s-domain. Similarly we can use many blocks to represent various circuit elements that we find in circuit diagrams.

G(s)OutputInputYout y(t)Block diagram representation in Laplace domain for a T.F, G(s)Since we defined that the T.F as the ration to Output to Input:g(t)OutputInput x(t) XinIf the system is in time domain:From the block diagram we can get the relationship: y(t)= g(t).x(t)1.2.Blocks connected in series(Cascade connection)G1(s)G2(s)G3(s) X(s) Y2(s) Y1(s) Y(s)

Combining all three equations:

G1(s).G2(s).G3(s) X(s) Y(s)When plants are cascaded we can get the products of functions to find the output. Addition and subtraction of two signal can be shown as below:- C(s)X(s)Y(s)+-y(t) = x(t) - c(t) - c(t)x(t)y(t)+-It is true in Laplace domain as well:Y(s)= X(s) C(s) X c(t)x(t)y(t)++Multiplication of two signals in time domainy(t) = x(t) . c(t)In Laplace domain we cannot multiply two functions: Instead we have to take the Convolution. * C(s)X(s)Y(s)Y(s) = X(s) * C(s)How to reduce a simple feedback control system with a, plant: G(s), summing junction, feedback plant: H(s) and a branch point?T.F= H(s))Here not the convolution but theDefinition of an input to output signal through a block Output signal C(s) R(s)E(s)+-T.F=G(s))B(s)B(s)=C(s).H(s)Feedback PathError detector(summing point)Forward blockBranch pointC(s)C(s)Error signal = E(s)Feedback block R(s)=Reference input signal in s-domainC(s)= Output signal(controlled variables)H(s)=Feedback transfer functionE(s) = Error signal= Actuating signal8Error detectors function is to find the difference between the Input signal and the feedback signal.(can be Opamp-comparator)

At a branch point same signal is divided into branches.

Error = Input signal Feedback signal

E(s) = R(s) C(s).H(s)

For the forward block we can write the Output/Input relationship

C(s)/E(s) = G(s)

G(s) = C(s)/[R(s)-C(s).H(s)]By cross multiplication: C(s) = G(s)R(s) G(s).C(s).H(s)Hence: C(s){ 1+ G(s).H(s)} = G(s).R(s)

The above equation can simply be represented by one block diagram as shown below: Note: simple feedback control system is reduced to an open-loop block diagramR(s)C(s)

Parallel ConnectionC(s)

OutputG1(s)G2(s)Gn(s)R(s)

Input

+-G2(s))H(s))B(s)C(s)G1(s) +- +-R(s)Example: Consider the block diagram as shown below. Simplify the blocks to get an OLTF.G2(s))H(s))B(s)C(s)G1(s) +- +-At first, control blocks marked within the dotted lines can be reduced to a single block)C(s)G1(s)

R(s) +-

Reference: http://en.wikibooks.org/wiki/Control_Systems/Block_Diagrams

Reference: http://en.wikibooks.org/wiki/Control_Systems/Block_DiagramsOutputHow to move plants or summing points to different locations:(very useful in reduction processes)G1(s)=G1(s)G1(s)G1(s)=1/G1(s)G1(s)Output=G1(s)1/G1(s)G1(s)AB-CG1(s)-1A, 1-BC{1-1/G1(s)}.G1(s)=G1(s)-1R(s)R(s) +Simplify the Block diagram given below to an OLTF.H1(s)H2(s)G2(s)G3(s)G1(s)C(s)-+-H2(s)G2(s)G3(s)G1(s)C(s)-+-

C(s)-

R(s)C(s)

R(s)R(s)C(s)

OLTF18Matlab can be used for the block diagram reductionG1(s)G2(s)++Summing junctionG1(s) +G2(s)R(s)C(s)Parallel:

Num1 = [1,0]Den1 = [1,0,2]

Num1 = 1Den1 = [3, 1]

Answer should be: num=[0,4,1,2]; den=[3,1,6,2]Matlab command:[num,den]=parallel(num1,den1,num2,den2)

When two blocks are in series: it will be G1(s).G2(s) = s/(3s3+s2+6s+2)Hence the num=[0,0,1,0]; den=[3,1,6,2]The Matlab command: [num,den]=series(num1,den1,num2,den2)

When G1(s) and G2(s) are in a feed back loop:G1G2-+

Answer= num[0,3,1,0]; den=[3,1,7,2]Matlab command for OLTF or feedback loop simplification:

[num,den]=feedback(num1,den1,num2,den2)

Manual Answer= num[0,3,1,0]; den=[3,1,7,2]G1G2G1G2R(s) +Use Matlab to find the OLTF and see the behaviour after unit step inputAssume G1= s/(s+3) and G2=1/sC(s)-Manual calculation gives the OLTF: (s3+s2+3s)/[s3+7s2+10s+3](s3+s2+3s)/s3+7s2+10s+3R(s) C(s)C(s)= R(s).(s3+s2+3s)/[s3+7s2+10s+3]

C(s)= For a unit step input:

numTF=[1,1,3,0]; denTF=[1,7,10,3];

There are two unwanted zeros in the num5, and den5....? why?When we manually simplify the numerator and denominator s on numerator and s in the denominator cancel off. But program doesnt know how to do it(it is not programmed to handle that situation)Hence we have to write a different programme.