Chapter 2 - Solution Controle Digital

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Chapter 2 Problem 1P
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Chapter 2 Problem 2P
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Chapter 2 Problem 3P
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Chapter 2 Problem 4P
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Chapter 2 Problem 5P
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Chapter 2 Problem 6P
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Using long division method,
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The power series form of formula is:
 …… (4)
The coefficient of quaint expressed in Equation (4):
Step 3 of 4
a)
From the real translation property of z – transform,
 …… (5)
Substitute 0 for , 1 for , –1.1 for ,–1.79 for , and –0.73 for and 2T for n in
Equation (5):
Thus, the z transform of is .
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b)
To find the z – transform of using property of z – transform
From the real translation property of z – transform
 …… (6)
By adding and subtracting terms and factoring 
Substitute 2T for n, 0 for , 1 for , –1.1 for , –1.79 for , and –0.73 for 
Thus, the z transform of is .
c)
To find the z – transform of using property of z – transform
From the time shifting property of z – transform
 …… (7)
Substitute 0 for , 1 for , –1.1 for , –1.79 for , and –0.73 for and T for n in
Equation (7):
Thus, the z transform of is .
Chapter 2 Problem 7P
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Chapter 2 Problem 8P
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Chapter 2 Problem 9P
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Consider 
Substitute 0 for a in Equation (1):
 …… (2)
The above transfer function is in first order.
Hence, for first order transfer function, the condition for parameter is .
Thus, the parameter a is obtained for first-order transfer function for the given equation.
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(b)
From Equation (2), write the first-order transfer function:
Thus, the first-order transfer function, obtained is .
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(c)
Consider the following data:
 …… (3)
Where,
 is unit-step response.
For unit-step response
Cancelling Z transform both sides in Equation (3):
Substitute 1 for in the above equation.
Thus, the value parameter, a is .
Chapter 2 Problem 10P
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Find for
Substitute 2 for k in Equation (2):
Substitute 1 for and 4 for in the above equation:
Given that for the value of is 0
Substitute 0 for in the above value:
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Find for 
Substitute 3 for k in Equation (2):
Substitute 4 for and 10 for and 0 for in the above equation:
Thus, the value for different values of k are .
(b)
To find using Z transform
Take z transform for Equation (2):
 …… (3)
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The formula for z transform of is:
Substitute the given values 1 for , 1 for , and 0 for in the above value:
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Substitute the value in Equation (3).
 …… (4)
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Use partial fraction technique and find 
 …… (5)
 …… (6)
Equate constant term on both sides:
 …… (7)
Equate z coefficient on both sides in Equation (6):
 …… (8)
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Add Equation (7) and Equation (8):
Multiply Equation (8) by 2:
Add Equation (7) with the above value:
Substitute –2 for A and 3 for B in Equation (5):
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Substitute above value in Equation (4):
 …… (9)
Take inverse z transform:
Thus, the value of is .
(c)
Final value theorem:
Substitute Equation (9) in the above equation:
Thus, from final value theorem the value of is .
Chapter 2 Problem 11P
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Chapter 2 Problem 12P
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Chapter 2 Problem 13P
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Chapter 2 Problem 14P
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Chapter 2 Problem 15P
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From Figure 1, write the state equation using left-side rule.
 …… (1)
Therefore, the difference equation is .
(b)
Take z transform for Equation (1).
Therefore, the transfer function is .
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(c)
The curve for right-side rule is shown in Figure 2.
Figure 2
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From the Figure 2, write the state equation using right-side rule.
 …… (2)
Therefore, the difference equation is .
(d)
Take z transform for Equation (2).
Therefore, the transfer function is .
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(e)
Substitute 0 for k in Equation (1):
 …… (3)
Substitute 1 for k in Equation (1):
 …… (4)
Substitute Equation (3) in Equation (4).
 …… (5)
Substitute 2 for k in Equation (1):
 …… (6)
Substitute Equation (5) in Equation (6).
From the above equation the value of :
Therefore, the value of is .
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(f)
Substitute 0 for k in Equation (2):
 …… (7)
Substitute 1 for k in Equation (2):
 …… (8)
Substitute Equation (7) in Equation (8):
 …… (9)
Substitute 2 for k in Equation (2):
 …… (10)
Substitute Equation (9) in Equation (10):
From the above equation the value of :
Therefore, the value of is .
Chapter 2 Problem 16P
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5192-2-16P AID: 1825 | 31/10/2013
(a)
Write the state equation using trapezoidal rule.
 …… (1)
Therefore, the difference equation is 
(b)
Take z transform for Equation (1).
Therefore, the transfer function is .
Chapter 2 Problem 17P
Step 1 of 2
(b)
The difference equation for differentiatoris,
The right side rectangular rule diagram for the above equation is shown in Figure 1.
Figure 1
Therefore, the right side rectangular rule diagram for differentiator is drawn.
(c)
The given transfer function is,
The reciprocal value for the above transfer function is known as the appromiate differentiation model of the
system.
Rearrange the terms.
Take inverse z transform on both sides of the equation.
Therefore, the difference equation for differentiator is,
.
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(d)
The difference equation for differentiator is,
The right side rectangular rule diagram for the above equation is shown in Figure 2.
Figure 2
Therefore, the right side rectangular rule diagram for differentiator is drawn.
Chapter 2 Problem 18P
Step 1 of 5
(f)
The difference equation for 3D structure is,
The MATLAB program for 3D structure is,
ykminus2=0;
ykminus1=0;
ekminus2=0;
ekminus1=0;
ek=1;
for k=0:5
yk=ek*b2 +ekminus1*b1+ekminus2* b0-ykminus1* a1- ykminus2* a0;
[k, ek, yk]
ekminus2=ekminus1;
ekminus1=ek;
ykminus2=ykminus1;
ykminus1=yk;
end
Step 2 of 5
Example: Consider the value of b2 as 2, b1 as –2.4, b0 as 0.72, a1 as –1.4, and a0 as 0.98.
ykminus2=0;
ykminus1=0;
ekminus2=0;
ekminus1=0;
ek=1;
for k=0:5
yk=2*ek-2.4*ekminus1+0.72*ekminus2+1.4*ykminus1- 0.98*ykminus2;
[k, ek, yk]
ekminus2=ekminus1;
ekminus1=ek;
ykminus2=ykminus1;
ykminus1=yk;
end
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The MATLAB output is:
ans =
0 1 2
ans =
1.0000 1.0000 2.4000
ans =
2.0000 1.0000 1.7200
ans =
3.0000 1.0000 0.3760
ans =
4.0000 1.0000 -0.8392
ans =
5.0000 1.0000 -1.2234
From the MATLAB output, the value of is 1 and the value of is –1.2234.
Therefore, the MATLAB code for 3D structure is obtained.
(g)
The difference equations for 1D structure is:
The MATLAB program for 1D structure is:
fkminus2=0;
fkminus1=0;
ek=1;
for k=0:5
fk=ek-a1*fkminus1-a0*fkminus2;
yk=b2*fk+b1*fkminus1+b0*fkminus2;
[k, ek, yk]
fkminus2=fkminus1;
fkminus1=fk;
end
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Example: Consider the value for b2 as 2, b1 as –2.4, b0 as 0.72, a1 as –1.4, and a0 as 0.98.
fkminus2=0;
fkminus1=0;
ek=1;
for k=0:5
fk=ek+1.4 *fkminus1-0.98*fkminus2;
yk=2*fk-2.4*fkminus1+0.72*fkminus2;
[k, ek, yk]
fkminus2=fkminus1;
fkminus1=fk;
end
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The MATLAB output is:
ans =
0 1 2
ans =
1.0000 1.0000 2.4000
ans =
2.0000 1.0000 1.7200
ans =
3.0000 1.0000 0.3760
ans =
4.0000 1.0000 -0.8392
ans =
5.0000 1.0000 -1.2234
From the MATLAB output, the value of is 1 and the value of is –1.2234.
Therefore, the MATLAB code for 1D structure is obtained.
Chapter 2 Problem 19P
Step 1 of 17
(a)
Refer to Figure P2-19 in the text book; the difference equations are,
 …… (1)
 …… (2)
 …… (3)
Therefore, the difference equations are,
.
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(b)
Take z transform for Equation (1).
 …… (4)
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Take z transform for Equation (2).
 …… (5)
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Take z transform for Equation (3).
 …… (6)
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Multiply the term in Equation (4).
 …… (7)
Multiply the term in Equation (5).
 …… (8)
Add Equation (7) from Equation (8).
 …… (9)
Substitute Equation (9) in Equation (6).
Therefore, the filter transfer function is,
 …… (10)
(c)
Refer to Figure P2-19 in the text book and draw the signal flow graph.
The signal flow graph for the 1X structure is shown in Figure 1.
Figure 1
Step 6 of 17
Find loop gain from Figure 1:
Loop gain:
The product of branch gains found by traversing a path that starts at a node and ends at the same node,
following the direction of the signal flow, without passing through any other node more than once.
Loop gain 1:
Figure 2
Step 7 of 17
From Figure 2, the loop gain 1 is . …… (11)
Loop gain 2:
Figure 3
Step 8 of 17
From Figure 3, the loop gain 2 is . …… (12)
Loop gain 3:
Figure 4
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From Figure 4, the loop gain 3 is . …… (13)
Find the forward-path gain from Figure 1.
Forward-path gain:
The product of gains found from Figure 1, the signal-flow graph demonstrates Mason's rule by traversing a
path from the input node to the output node of the signal-flow graph in the direction of signal flow.
From Figure 1, forward-path gain 1 is:
From Figure 1, forward-path gain 2 is:
From Figure 1, forward-path gain 3 is:
From Figure 1, forward-path gain 4 is:
Find the non-touching loops from Figure 1.
Non-touching loops:
Loop gain 1 is . …… (14)
Loop gain 2 is . …… (15)
Step 10 of 17
The Mason gain formula is:
 …… (16)
Where,
 is represented as path sign of the kth forward path.
 is represented as the value of for that part of the flow graph not touching the kth forward path.
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Find from Equations (11), (12), (13), (14), and (15):
 …… (17)
Find :
Where,
k is represented as the number of forward paths.
For forward path 1, 
For forward path 2, 
For forward path 3, 
For forward path 4, 
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Find : …… (18)
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Find : …… (19)
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Find : …… (20)
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Find : …… (21)
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Substitute 4 for p in Equation (16).
 …… (22)
Substitute Equations (17), (18), (19), (20), and (21) in Equation (22):
here,
The transfer function is,
 …… (23)
Equations (10) and (23) are equal.
Therefore, the transfer function is verified by using Mason’s gain formula.
(d)
The difference equations are,
The MATLAB program for 1X structure is,
f1kminus1=0;
f2kminus1=0;
ek=1;
for k=0:5
yk=b2*ek+f2kminus1;
[k, ek, yk]
f1k=g1*f1kminus1-g2*f2kminus1+g3*ek;
f2k=g1*f2kminus1-g2*f1kminus1+g4*ek;
f1kminus1=f1k;
f2kminus1=f2k;
end
Step 17 of 17
Consider 0.7 for , 0.7 for , –1.38 for , and 2 for ; and substitute these values in the
MATLAB program.
f1kminus1=0;
f2kminus1=0;
ek=1;
for k=0:5
yk=2*ek+f2kminus1;
[k, ek, yk]
f1k=0.7*f1kminus1-0.7*f2kminus1-1.38*ek;
f2k=0.7*f2kminus1-0.7*f1kminus1+0.4*ek;
f1kminus1=f1k;
f2kminus1=f2k;
end
The MATLAB output is,
ans =
 0 1 2
ans =
 1.0000 1.0000 2.4000
ans =
 2.0000 1.0000 3.6460
ans =
 3.0000 1.0000 5.3904
ans =
 4.0000 1.0000 7.8326
ans =
 5.0000 1.0000 11.2516
From the MATLAB output, the values of and are,
Therefore, the MATLAB code for the second-order transfer function’s 1X structure is obtained.
Chapter 2 Problem 20P
Step 1 of 12
Refer to Figure P2-19 in the textbook, the difference equations are,
 …… (11)
 …… (12)
 …… (13)
Take z transform for Equation (11).
 …… (14)
Step 2 of 12
Take z transform for Equation (12).
 …… (15)
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Take z transform for Equation (13).
 …… (16)
Multiply the term in Equation (14).
 …… (17)
Multiply the term in Equation (15).
 …… (18)
Add Equation (17) and Equation (18).
 …… (19)
Substitute Equation (19) in Equation (16).
 …… (20)
Take denominator term in Equation (1), which is equal to zero to find pole values.
The second order 1X digital filter structure is,
Substitute for p and for .
Substitute 2 for in the above equation:
 …… (21)
Equate Equation (1) with Equation (21).
Substitute for z.
The value of is,
Substitute A and values in Equation (21):
The value of is,
Substitute for p.
The value of is,
Substitute for p.
Step 4 of 12
The value of is,
Substitute for A.
The value of is,
Substitute for A.
Therefore, the coefficients values for 1X structure are,
. …… (22)
Step 5 of 12
(d)
The second order filter transfer functionis,
Enter the following code in MATLAB to find the residue value.
num=[2 –2.4 0.72];
den=[1 –1.4 0.98 ];
[r,p,k]=residue(num,den)
The MATLAB output is,
r =
0.2000 + 0.6857i
0.2000 - 0.6857i
p =
0.7000 + 0.7000i
0.7000 - 0.7000i
k =
2
From the MATLAB output, the value of is,
Substitute for p.
Step 6 of 12
The value of is,
Substitute for p.
The value of is,
Substitute for A in the above equation:
The value of is,
Substitute for A.
The coefficients values for 1X structure are,
 …… (23)
Equations (22) and (23) are equal.
Step 7 of 12
Therefore, the coefficients value for 1X structure is verified by using MATLAB.
Step 8 of 12
(e)
Refer to Figure P2-19 in the textbook and draw the signal flow graph.
The signal flow graph for the 1X structure is shown in Figure 1.
Figure 1
Step 9 of 12
Find the loop gain from Figure 1.
Three loop gains are present in Figure 1.
Loop gain 1:
Figure 2
From Figure 2, the loop gain 1 is . …… (24)
Loop gain 2:
Figure 3
From Figure 3, the loop gain 2 is . …… (25)
Step 10 of 12
Loop gain 3:
Figure 4
From Figure 4, the loop gain 3 is . …… (26)
Step 11 of 12
Find the Forward-path gain from the Figure 1.
Four forward paths are present in Figure 1.
From Figure 1, forward path gain 1 is,
From Figure 1, forward path gain 2 is,
From Figure 1, forward path gain 3 is,
From Figure 1, forward path gain 4 is,
Step 12 of 12
Find the non-touching loops from Figure 1.
Non-touching loops:
Loop gain 1 is …… (27)
Loop gain 2 is …… (28)
The Mason’s gain formula is,
 …… (29)
here,
 is represented as path sign of kth forward path.
 is represented as the value of for that part of the flow graph not touching the kth forward path.
Find from Equations (24), (25), (26), (27), and (28):
 …… (30)
Find :
here,
k is represented as number of forward paths.
For forward path 1, 
For forward path 2, 
For forward path 3, 
For forward path 4, 
Find :
 …… (31)
Find :
 …… (32)
Find :
 …… (33)
Find :
 …… (34)
Substitute 4 for p in Equation (29):
 …… (35)
Substitute Equations (30), (31), (32), (33), and (34) in Equation (35).
 …… (36)
here,
The transfer function for 1X structure is,
 …… (37)
Substitute 2 for , 0.7 for , 0.7 for , –1.38 for , and 0.4 for in Equation (37):
 …… (38)
From Equations (1) and (38), it concludes that both equations are equal.
Therefore, the transfer function gets verified by using Mason’s gain formula.
Chapter 2 Problem 21P
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Chapter 2 Problem 22P
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Chapter 2 Problem 23P
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Chapter 2 Problem 24P
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Chapter 2 Problem 25P
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Chapter 2 Problem 26P
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Chapter 2 Problem 27P
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Chapter 2 Problem 28P
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Chapter 2 Problem 29P
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Chapter 2 Problem 30P
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Chapter 2 Problem 31P
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Chapter 2 Problem 32P
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Chapter 2 Problem 33P
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Chapter 2 Problem 34P
Step 1 of 9
b)
Form the matrix for Equation (1):
Compare the coefficient in above matrices:
Step 2 of 9
Find the coefficient in terms of 
Compare the coefficient in above matrices:
Step 3 of 9
The simulation diagram for the state equation:
Figure (1): Simulation diagram for given state equation
(c)
Mason’s gain formula is 
Where,
 is path sign of the kth forward path.
 is value of for that of the flow graph not touching the kth forward path.
p is the number of forward paths.
Step 4 of 9
From Figure (1), the signal flow graph:
Step 5 of 9
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Figure (2): Signal flow graph
Step 7 of 9
From Figure (2), the signal flow graph has only one forward path. Hence, the transfer function is:
 …… (8)
From Figure (2),
Individual loop gain 1,
Individual loop gain 2, 
Forward path 
Substitute forand in the above value:
Step 8 of 9
Substitute the above value in Equation (8):
Substitute for and 1 for in the above equation:
The above Equation is compared with Equation (7) and the transfer function is verified.
Thus, the above equation is .
d)
To verify the results using MATLAB
The MATLAB program is:
A = [1 0 0; 1 1 0; 0 1 0];
B = [1; 0; 0];
C = [0 0 1];
D = 0;
[n,d]= ss2tf(A, B, C, D)
sys = tf(n,d)
Step 9 of 9
The MATLAB output is:
Thus, the numerator value and denominator value from MATLAB output are equal to Equation (7).
Thus, the transfer function is verified using MATLAB.
Chapter 2 Problem 35P
Step 1 of 4
The condition for the similarity transformation
 …… (1)
Take RHS term in Equation (1).
 …… (2)
Substitute the given values in Equation (2).
 …… (3)
Calculate the similarity transformation of .
Where,
 are the similar matrices.
Calculate the determinant of .
 …… (4)
Step 2 of 4
The determinant value of A is
 …… (5)
Compare Equations (4) and (5).
The determinant value for is equal to determinant value of A.
Hence, it satisfies the similarity transformation property
 …… (6)
Calculate the similarity transformation of .
Calculate the determinant of .
 …… (7)
Step 3 of 4
The determinant value of matrix B is:
 …… (8)
Compare Equations (7) and (8).
The determinant value for is equal to the determinant value of B.
Hence, it satisfies the similarity transformation property . …… (9)
Step 4 of 4
Substitute equations (6) and (9) in equation (3)
Therefore, the equation for similarity transformation is shown.
.
Chapter 2 Problem 36P
Step 1 of 3
From the MATLAB output, the state equation is,
 …… (1)
From the MATLAB output, the output equation is,
 …… (2)
Therefore, the sate equations are obtained using MATLAB.
.
Step 2 of 3
(b)
Modify Equation (1).
From the above matrix, the state equations are,
 …… (3)
 …… (4)
Modify Equation (2).
 …… (5)
Step 3 of 3
The simulation diagram using Equations (3), (4), and (5) is shown in Figure 1.
Figure 1
Therefore, the simulation diagram is drawn for computed state equations.
(c)
From Figure 1, the state equations are,
 …… (6)
 …… (7)
 …… (8)
Take z transform for Equation (6):
 …… (9)
Take z transform for Equation (7):
 …… (10)
Take z transform for Equation (8):
 …… (11)
Substitute Equation (10) in Equation (11):
 …… (12)
Substitute Equation (10) in Equation (9):
Substitute Equation (12) in the above equation:
 …… (13)
Consider .
Substitute the above value in Equation (13):
Therefore, the transfer function obtained from the simulation diagram is equal to the given standard
transfer function.
Hence, the simulation diagram from Figure 1 is one of the standard form.