2d barcode generator vb.net s 2 ~ ( s )3 s Y ( s ) + 2 Y ( s ) = 4 s X ( s ) + X ( s ) + in VS .NET

Drawing QR in VS .NET s 2 ~ ( s )3 s Y ( s ) + 2 Y ( s ) = 4 s X ( s ) + X ( s ) +

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 QR-Code 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
QR Code ISO/IEC18004 Recognizer In VS .NET
Using Barcode reader for .NET framework Control to read, scan read, scan image in VS .NET applications.
Barcode Printer In .NET Framework
Using Barcode generator for .NET framework Control to generate, create bar code image in VS .NET applications.
Y ( s ) = - 3 Y 1 Y ( s ) - 2 s C 2 ~ ( s ) 4 s - ' ~ ( s+ s - ~ x ( s ) + )
Scan Bar Code In Visual Studio .NET
Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications.
Encoding QR Code In Visual C#.NET
Using Barcode printer for .NET Control to generate, create QR image in VS .NET applications.
corresponds to integration of k times) the simulation diagram in from which (noting that Fig. 7-13 can be drawn. Choosing the outputs of integrators as state variables as shown in
Quick Response Code Encoder In Visual Studio .NET
Using Barcode generator for ASP.NET Control to generate, create QR-Code image in ASP.NET applications.
Paint QR Code In VB.NET
Using Barcode generator for Visual Studio .NET Control to generate, create QR Code JIS X 0510 image in .NET applications.
In matrix form
Barcode Generation In VS .NET
Using Barcode creator for Visual Studio .NET Control to generate, create bar code image in .NET framework applications.
Create UCC - 12 In VS .NET
Using Barcode maker for VS .NET Control to generate, create EAN128 image in Visual Studio .NET applications.
Fig. 7-13
GS1 RSS Drawer In Visual Studio .NET
Using Barcode printer for Visual Studio .NET Control to generate, create DataBar image in Visual Studio .NET applications.
ISSN Generator In .NET Framework
Using Barcode maker for .NET framework Control to generate, create ISSN - 13 image in Visual Studio .NET applications.
7.16. Find state equations of a continuous-time LTI system with system function
USS-128 Drawer In Visual Basic .NET
Using Barcode maker for Visual Studio .NET Control to generate, create EAN / UCC - 13 image in .NET framework applications.
Scanning Bar Code In VS .NET
Using Barcode scanner for VS .NET Control to read, scan read, scan image in VS .NET applications.
STATE SPACE ANALYSIS
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 UPC-A Supplement 2 image in Online applications.
GTIN - 128 Maker In Java
Using Barcode drawer for Java Control to generate, create UCC-128 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. 7-14 can be drawn. Choosing the outputs of integrators as state variables as shown in Fig. 7-14, we get
As in the discrete-time case, the simulation of H(s) shown in Fig. 7-14 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. 7-14 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. 7-15, the simulation diagram in Fig. 7-16 can be
Fig. 7-16 Canonical simulation of the second form.
STATE SPACE ANALYSIS
[CHAP. 7
drawn. Choosing the outputs of integrators as state variables as shown in Fig. 7-16, we have
Cl(t) =u,(t)
In matrix form
As in the discrete-time case, the simulation of H ( s ) shown in Fig. 7-16 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 continuous-time 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 continuous-time 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. 7-17 can be drawn. Thus, H( S ) = H J s ) + H 2 ( s )+ H , ( s ) can be simulated by the diagram in Fig. 7-18 obtained by parallel connection of three systems. Choosing the outputs of integrators as state variables as shown in Fig. 7-18, 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. 7-17
STATE SPACE ANALYSIS
[CHAP. 7
Fig. 7-18
SOLUTIONS OF STATE EQUATIONS FOR DISCRETE-TIME LTI SYSTEMS
7.20. Find An for
by the Cayley-Hamilton 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 z-transform 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 Cayley-Hamilton 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"-'
Copyright © OnBarcode.com . All rights reserved.