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2d barcode generator vb.net s 2 ~ ( s )3 s Y ( s ) + 2 Y ( s ) = 4 s X ( s ) + X ( s ) + in VS .NET
s 2 ~ ( s )3 s Y ( s ) + 2 Y ( s ) = 4 s X ( s ) + X ( s ) + Denso QR Bar Code Recognizer In Visual Studio .NET Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in .NET applications. Print QRCode In .NET Framework Using Barcode generation for .NET framework Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications. Dividing both sides of the above expression by s 2 and rearranging, we get
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Generating EAN 128 In None Using Barcode generation for Word Control to generate, create GS1 128 image in Office Word applications. EAN128 Generator In None Using Barcode generation for Online Control to generate, create GS1 128 image in Online applications. [CHAP. 7
UCC  12 Generation In None Using Barcode creation for Online Control to generate, create UPCA Supplement 2 image in Online applications. GTIN  128 Maker In Java Using Barcode drawer for Java Control to generate, create UCC128 image in Java applications. From the definition of the system function [Eq. (3.37)] Print Bar Code In None Using Barcode generation for Excel Control to generate, create barcode image in Microsoft Excel applications. Decoding GS1  13 In C# Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. we have
( s 3+ a l s 2+ a2s + a 3 ) Y ( s )= (bos3+ b l s Z+ b2s + b 3 ) x ( s ) Dividing both sides of the above expression by s3 and rearranging, we get
from which (noting that s  k corresponds to integration of k times) the simulation diagram in Fig. 714 can be drawn. Choosing the outputs of integrators as state variables as shown in Fig. 714, we get As in the discretetime case, the simulation of H(s) shown in Fig. 714 is known as the canonical simulation of the first form, and Eq. (7.106) is known as the canonical state representation of the first form. Fig. 714 Canonical simulation of the first form.
CHAP. 71
STATE SPACE ANALYSIS
7.17. Redo Prob. 7.16 by expressing H ( s ) as
H(s)=H,(s)H*(s) where
H 1 ( s )= s 3 + a , s 2 a 2 s + a , Then we have
Rearranging the above equations, we get
s 3 W ( s ) =  a 1 s 2 w ( s )  a 2 s W ( s ) a 3 W ( s ) + X ( s ) + Y ( s ) = b,s3W(s) + b l s 2 ~ ( s )b , s W ( s ) + b , W ( s ) from which, noting the relation shown in Fig. 715, the simulation diagram in Fig. 716 can be
Fig. 716 Canonical simulation of the second form.
STATE SPACE ANALYSIS
[CHAP. 7
drawn. Choosing the outputs of integrators as state variables as shown in Fig. 716, we have
Cl(t) =u,(t) In matrix form
As in the discretetime case, the simulation of H ( s ) shown in Fig. 716 is known as the canonical simulation of the second form, and Eq. (7.109) is known as the canonical state representation of the second form. 7.18. Consider a continuoustime LTI system with system function
Find a state representation of the system.
Rewrite H ( s ) as
Comparing Eq. (7.111) with Eq. (7.105) in Prob. 7.16, we see that
a, = 8 a , = 17 a , = 10 b , = b , =O b, Substituting these values into Eq. (7.106) in F'rob. 7.16, we get
y(t) = [1 7.19. Consider a continuoustime LTI system with system function
Find a state representation of the system such that its system matrix A is diagonal.
CHAP. 71
STATE SPACE ANALYSIS
First we expand H ( s ) in partial fractions as
where
1 H,(s) = s+l
H Z ( s )=  s+2
H 3 ( s )=  s+5
Let Then or
ffk yk(~) H k ( s ) = = s Pk X(s) ( S
pk)Yk(s) = % X ( S ) Y k ( s )= P ~ s  ~ Y ~+( ask)s  ' X ( s ) from which the simulation diagram in Fig. 717 can be drawn. Thus, H( S ) = H J s ) + H 2 ( s )+ H , ( s ) can be simulated by the diagram in Fig. 718 obtained by parallel connection of three systems. Choosing the outputs of integrators as state variables as shown in Fig. 718, we get 441) = s,(t) + x ( t ) &(t)  2 q ~ ( t ) ;x(t)  q3(t) = 5q3(t)  : x ( t ) ~ ( t= ) , ( t ) + q z ( t ) + q , ( t ) q
In matrix form
Note that the system matrix A is a diagonal matrix whose diagonal elements consist of the poles of H ( s ) . Fig. 717 STATE SPACE ANALYSIS
[CHAP. 7
Fig. 718 SOLUTIONS OF STATE EQUATIONS FOR DISCRETETIME LTI SYSTEMS
7.20. Find An for
by the CayleyHamilton theorem method.
First, we find the characteristic polynomial c(A) of A.
 A+
L)(A  L) Thus, the eigenvalues of A are A , i and A, + b,A =
a. Hence, by Eqs. (7.27) and (7.28) we have
b,, + 36, b1
An = b,I
and b, and b, are the solutions of
bo + b,(;) (4)" bo + b,(+)= ($)" CHAP. 7 1
STATE SPACE ANALYSIS
from which we get bo = Hence,  (;ln + 2(;ln
b , = 4($
 4($)" 7.21. Repeat Prob. 7.20 using the diagonalization method.
Let x be an eigenvector of A associated with A . Then
[ A l  A]x = 0
For A=Al = we have
The solutions of this system are given by x , = 2x2. Thus, the eigenvectors associated with A , are those vectors of the form For A = A 2 = $ we have
The solutions of this system are given by x, are those vectors of the form
= 4x2. Thus, the eigenvectors associated with A, Let a = p
= 1 in
the above expressions and let
Then
STATE SPACE ANALYSIS
(CHAP. 7
and by Eq. (7.29) we obtain
7.22. Repeat Prob. 7.20 using the spectral decomposition method.
Since all eigenvalues of A are distinct, by Eq. (7.33) we have
Then, by Eq. (7.34) we obtain
7.23. Repeat Prob. 7.20 using the ztransform method.
First, we must find ( z l  A)'. CHAP. 71
STATE SPACE ANALYSIS
Then by Eq. (7.35) we obtain
; , { ( z I A)'z) From the above results we note that when the eigenvalues of A are all distinct, the spectral decomposition method is computationally the most efficient method of evaluating An. 7.24. Find An for
The characteristic polynomial c(A) of A is
4 5A = (A
 l)(A
Thus, the eigenvalues of A are A , = 1 and A, = 4, and by Eq. (7.33) we have
E2= ( Thus, by Eq. (7.34) we obtain
STATE SPACE ANALYSIS
7.25. Find An for
The characteristic polynomial c(A) of A is
Thus, the eigenvalues of A are A, An. By Eq. (7.27) we have
= A, = 2. We use the CayleyHamilton theorem to evaluate
where b, and b, are determined by setting A Eqs. (A.59) and (A.60)I: 2 in the following equations [App. A, b, +b,A =An b, = n ~ "  ' Thus, + 2b, = 2" b, =n2"'

