linear sequential machines1 linear sequential machine and reduction of linear sequential machine
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Linear Sequential Machines 1
LINEAR SEQUENTIAL MACHINE AND REDUCTION
OF LINEAR SEQUENTIAL MACHINE
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(I) Time Domain(2) Frequency Domain
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Module 2 Operation(0 and 1)
and so on…..
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Note:
• A is always square matrix and gives number of delays•Number of columns in B gives number of inputs•Number of rows in C gives number of output•Order of A and C should be compatible with present state and order of B and D should be compatible with input
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Example 1:
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State Space Representation:
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Note:
Y’s represent Next Statey’s represent Present Statez’s represent Output State
Here, we have 3 delays, one input and two outputs
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Representation:
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Example 2:Determine Characterizing matrices A,B,C and D fromthe given linear circuit diagram
Hint: From fig.,A=2 X 2,B=2 X 1,C=1 X 2
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Y1 = y2 + xY2 = y1 + xz = y1 + x
From the figure,
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Solution:
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Reduction of Linear Sequential Machines
Reduction of linear sequential machines means reducing or minimizing the number of delays
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Example 3:Minimize (reduce) the linear machine given by following characterizing matrices A,B,C,D(Example Kohavi Page-580)
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Reduction:
The rank of the diagnostic matrix is 3, and hencethe dimenstion of given linear machine CANNOTbe reduced
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Example 3:Minimize the linear machine given by following characterizing matrices A,B,C,D(Example Kohavi Page-582)
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A* = TAR
B* = TB
C* = CR
D* = D
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Note:
In reduction number of Inputs and Outputs do not change and only the number of delays change
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Example 4:Minimize the linear machine given by following characterizing matrices A,B,C,D(Example Kohavi Page-584)
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A* = TAR
B* = TB
C* = CR
D* = D
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Example 5:Minimize the linear machine given by following characterizing matrices A,B,C,D(Example Kohavi 15 - 19)
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A* = TAR
B* = TB
C* = CR
D* = D
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Example 6:Minimize the linear machine given by following characterizing matrices A,B,C,D(Example Kohavi 15 - 20)
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A* = TAR
B* = TB
C* = CR
D* = D
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Example 7:Minimize the linear machine given by following characterizing matrices A,B,C,D(Example Kohavi 15 - 21)
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Determine R such that TR = I (Here R=T_1 )
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A* = TAR
B* = TB
C* = CR
D* = D
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Realisation:
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Realisation
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Note: Original and reduced system will have same transfer function after cancellation of common terms (if there) in the numerator and denominator of the Delay Transfer Function Matrix. Also, the number of columns in DTF matrix gives number of inputs and number of rows in DTF matrix gives the number of inputs
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Analogy of Control theory and SequentialCircuits but for this course, we restrict ourselves purely to sequential circuits
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Example 8: Find the DTF matrix of a linear machine given byfollowing characterizing matrices
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Here D is the delay operator
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Thus, the DTF matrix is given by ,
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Identification of Sequential Machines
There are two methods to solve these problems
• With Assumption
• With Out Assumption
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Example 1: Determine whether the machine given in
the following table is linear and thus find A,B,C,D
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Example 1: Determine whether the machine given in
the following table is linear and thus find A,B,C,D(Example Kohavi Page 591)
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,
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Example 2: Determine whether the machine given in
the following table is linear and thus find A,B,C,D(Example Kohavi Page 592)
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Example 2: Determine whether the machine given in
the following table is linear and if it is , show alinear realization (Problem Kohavi 15-25(b))
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Yh00 , Yg
00
A= 0 0 1 0 0 1 0 1 1
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C= 0 0 1 0 0 1
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Example 3: Determine whether the machine given in
the following table is linear and if it is , show alinear realization (Problem Kohavi 15-25(a))
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Example 3: Test the given machine for linearity
.In particular, determine if the state transistionsare linear and if the outputs are linear(Problem Kohavi 15-26)
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Output and State Equations:
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Verifying whether state transitions and Outputsare linear
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Note:Case (viii) is not true according to the giventable
Thus, THE GIVEN MACHINE IS STATE LINEAR BUT NOT OUTPUT LINEAR
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HUFFMAN’S MELEY PROCEDURE FOR REDUCTION OF STATES OF SEQUENTIAL
MACHINES
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• Group the states into sets where all members of each set have the same output combination
1 2
(a,d) (b,c,f,g)
• Applying the procedure again,(a(1,2),d(2,2)) (b(1,2),c(1,2),f(1,2),g(1,2))
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Here a(1,2), means, a goes to a when X = 0and goes to b when X = 1
from the table, and a is in group 1 and b is in group2, so, we write as a(1,2)
•Applying the procedure again
1 2 3a d (b,c,f,g)
(b(2,3),c(1,3),f(2,3),g(1,3))
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•Applying the procedure again,1 2 3 4a d b,f c,g
(b(2,4),f(2,4)) (c(1,3),g(1,3))
so, there is no need to apply the previous procedure again
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Finally, f can be replaced by b and g can be replaced by c