h ( r ) e-"'dr in .NET framework

Generation QR Code in .NET framework h ( r ) e-"'dr

h ( r ) e-"'dr
Reading QR Code In VS .NET
Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET framework applications.
Draw Denso QR Bar Code In .NET
Using Barcode maker for VS .NET Control to generate, create QR Code image in .NET applications.
(2.88)
QR Code JIS X 0510 Reader In .NET Framework
Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in .NET applications.
Bar Code Creation In VS .NET
Using Barcode printer for .NET framework Control to generate, create barcode image in VS .NET applications.
Since the system is stable, that is,
Barcode Recognizer In .NET Framework
Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications.
Paint Denso QR Bar Code In C#.NET
Using Barcode printer for Visual Studio .NET Control to generate, create QR Code image in .NET applications.
since le-j"'J
Make Quick Response Code In Visual Studio .NET
Using Barcode creation for ASP.NET Control to generate, create QR Code ISO/IEC18004 image in ASP.NET applications.
Quick Response Code Drawer In VB.NET
Using Barcode encoder for .NET framework Control to generate, create QR Code ISO/IEC18004 image in VS .NET applications.
1. Thus, H( jw) converges for any w. Using Euler's formula, we have
1D Barcode Maker In .NET
Using Barcode creation for .NET framework Control to generate, create Linear 1D Barcode image in Visual Studio .NET applications.
Create Bar Code In Visual Studio .NET
Using Barcode maker for Visual Studio .NET Control to generate, create bar code image in .NET applications.
H(jw)
Make Bar Code In .NET Framework
Using Barcode printer for .NET framework Control to generate, create barcode image in VS .NET applications.
Drawing 2 Of 5 Standard In Visual Studio .NET
Using Barcode generation for .NET Control to generate, create Industrial 2 of 5 image in VS .NET applications.
/- (r)
Make Code 39 Extended In Java
Using Barcode maker for Java Control to generate, create Code 3/9 image in Java applications.
Matrix 2D Barcode Maker In VS .NET
Using Barcode drawer for ASP.NET Control to generate, create Matrix Barcode image in ASP.NET applications.
e'~"'di
Creating DataMatrix In None
Using Barcode drawer for Software Control to generate, create Data Matrix 2d barcode image in Software applications.
Recognize Barcode In Visual Studio .NET
Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications.
jWh ( r ) ( c o so r - j sin w r ) dr x
Generating GS1 - 12 In Objective-C
Using Barcode maker for iPhone Control to generate, create UPC A image in iPhone applications.
Bar Code Printer In VB.NET
Using Barcode maker for .NET Control to generate, create bar code image in VS .NET applications.
= j::(r)
Code-128 Encoder In Objective-C
Using Barcode generation for iPhone Control to generate, create ANSI/AIM Code 128 image in iPhone applications.
EAN-13 Recognizer In VB.NET
Using Barcode recognizer for .NET framework Control to read, scan read, scan image in Visual Studio .NET applications.
cos wr dr
I-. h ( r )sin o r d r
(2.90)
Since cos o r is an even function of r and sin o r is an odd function of 7 , and if h ( t ) is real and ) even, then h(r)coso r is even and h ( r )sin wr is odd. Then by Eqs. ( 1 . 7 5 ~ and (1.77), Eq. (2.89 becomes
H( jo)
= 2/
h ( r )cos o r d r
CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
Since cos w r is an even function of w, changing w to - w in Eq. (2.90) and changing j to - j in Eq. (2.891, we have
H ( -jw)
= H ( jw)* = 2 /
h ( r ) COS( o r ) d r r dr
=2jrnh(r) cos o 0
Thus, we see that the eigenvalue H ( j w ) corresponding to the eigenfunction el"' is real. Let the system be represented by T . Then by Eqs. (2.231, (2.241, and (2.91) we have
T ( e j U ' }= H ( j w ) e'"'
T [ ~ - J " '= H ( - j w ) e-J"' }
= H ( j o ) e-I"'
(2.92~)
( 2.92b)
Now, since T is linear, we get
o T[COS t } = T{;(ej"'
+ e-'"I
) } = ; T ( ~ J ~ ': } ( ~ - J "1 + T '
(2.93~)
( j ~ ) { ; ( ~ je -"J w r ) } = ~ ( j w ) c o wt + ' s
Thus, from Eqs. ( 2 . 9 3 ~and (2.93b) we see that cos wt and sin wt are the eigenfunctions of the ) system with the same real eigenvalue H ( j o ) given by Eq. (2.88) or (2.90).
SYSTEMS DESCRIBED BY DIFFERENTIAL EQUATIONS 2.18. The continuous-time system shown in Fig. 2-18 consists of one integrator and one scalar multiplier. Write a differential equation that relates the output y( t ) and the input x( t 1.
Fig. 2-18
Let the input of the integrator shown in Fig. 2-18 be denoted by e(t). Then the input-output relation of the integrator is given by
Differentiating both sides of Eq. (2.94) with respect to t , we obtain
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
Next, from Fig. 2-18 the input e(t) to the integrator is given by e(t) = x ( t ) -ay(t) Substituting Eq. (2.96) into Eq. (2.95), we get
which is the required first-order linear differential equation.
2.19. T h e continuous-time system shown in Fig. 2-19 consists of two integrators and two scalar multipliers. Write a differential equation that relates the output y ( t ) and the input x ( t ).
Fig. 2-19
Let e ( 0 and w(t) be the input and the output of the first integrator in Fig. 2-19, respectively. Using Eq. (2.951, the input to the first integrator is given by
Since w(t) is the input to the second integrator in Fig. 2-19, we have
Substituting Eq. (2.99) into Eq. (2.98), we get
which is the required second-order linear differential equation. Note that, in general, the order of a continuous-time LTI system consisting of the interconnection of integrators and scalar multipliers is equal to the number of integrators in the system.
CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
2.20. Consider a continuous-time system whose input x ( t ) and output y ( t ) are related by
where a is a constant.
( a ) Find y ( t ) with
the auxiliary condition y(0) = y, and
x ( t ) =~ e - ~ ' u ( t )
(2.102)
( b ) Express y ( t ) in terms of the zero-input and zero-state responses.
( a ) Let
where y p ( t ) is the particular solution satisfying Eq. (2.101) and solution which satisfies
y h ( t ) is
the homogeneous
Assume that
y p ( t )= ~ e - ~ ' t>O
Substituting Eq. (2.104) into Eq. (2.101), we obtain
- b ~ e - 6 '+ a ~ e - b = K ~ - ~ ' '
from which we obtain A
=K/(a
- b), and
To obtain
yh(t),
we assume
y h ( t )= BeS'
Substituting this into Eq. (2.103) gives
sBe"+aBe"=(s+a)Be"'=O
from which we have s = - a and
y h ( t ) = Be-"'
Combining yp(t)and yh(t),we get
y ( t ) =Be-"+
-- b ~ e
From Eq. (2.106) and the auxiliary condition y(O) =yo, we obtain
B=yo- a-b
Thus, Eq. (2.106) becomes
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
For t < 0, we have x(t) = 0, and Eq. (2.101 becomes Eq. (2.103). Hence, y(r)
= Bepa'
From the auxiliary condition y(0) = y,, we obtain ~ ( r =yoe-"' ) r<O
( 2.108)
( b ) Combining Eqs. (2.107) and (2.108), y(r) can be expressed in terms of y,,(t) (zero-input response) and y,,(t) (zero-state response) as
where
2.21. Consider the system in Prob. 2.20.
( a ) Show that the system is not linear if y(0) = y,, # 0. ( b ) Show that the system is linear if y(0) = 0.
( a ) Recall that a linear system has the property that zero input produces zero output (Sec. 1.5E). However, if we let K = 0 in Eq. (2.102), we have x(r) = 0, but from Eq. (2.109) we see that
y(t)
= y,,e-"
Copyright © OnBarcode.com . All rights reserved.