 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
2d barcode generator vb.net e  a f u (  r ) e  f l d t =  (Oe("")' d f in Visual Studio .NET
e  a f u (  r ) e  f l d t =  (Oe("")' d f Recognizing QR Code ISO/IEC18004 In VS .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET framework applications. QR Printer In .NET Using Barcode creation for Visual Studio .NET Control to generate, create QRCode image in .NET framework applications. Thus, we obtain
Decode QR Code JIS X 0510 In Visual Studio .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET applications. Barcode Creation In VS .NET Using Barcode creation for .NET framework Control to generate, create barcode image in .NET framework applications. e"'u(  I ) Bar Code Decoder In .NET Using Barcode recognizer for VS .NET Control to read, scan read, scan image in .NET applications. QR Code Maker In Visual C# Using Barcode printer for Visual Studio .NET Control to generate, create Quick Response Code image in .NET framework applications. ( b ) Similarly, QR Code 2d Barcode Drawer In .NET Using Barcode maker for ASP.NET Control to generate, create QR Code ISO/IEC18004 image in ASP.NET applications. QR Code 2d Barcode Creation In VB.NET Using Barcode generation for VS .NET Control to generate, create QR Code JIS X 0510 image in VS .NET applications. LAPLACE TRANSFORM AND CONTINUOUSTIME LTI SYSTEMS
Draw Code 3 Of 9 In VS .NET Using Barcode encoder for Visual Studio .NET Control to generate, create Code 39 Extended image in VS .NET applications. Print Barcode In Visual Studio .NET Using Barcode generator for .NET Control to generate, create barcode image in .NET framework applications. [CHAP. 3
Making USS128 In VS .NET Using Barcode creation for .NET Control to generate, create EAN 128 image in .NET applications. Generate Code 2/5 In .NET Framework Using Barcode creation for .NET Control to generate, create Industrial 2 of 5 image in .NET applications. Thus, we obtain
Barcode Creation In None Using Barcode maker for Online Control to generate, create barcode image in Online applications. Code 128 Scanner In Java Using Barcode reader for Java Control to read, scan read, scan image in Java applications. ealu(  t ) Code 128 Code Set C Scanner In C#.NET Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in .NET applications. Bar Code Encoder In .NET Framework Using Barcode creator for Reporting Service Control to generate, create bar code image in Reporting Service applications. Re(s) < a
Painting EAN13 In None Using Barcode printer for Software Control to generate, create European Article Number 13 image in Software applications. UCC  12 Generation In None Using Barcode drawer for Font Control to generate, create UPC A image in Font applications. A finiteduration signal x ( t ) is defined as
Make ECC200 In None Using Barcode drawer for Word Control to generate, create Data Matrix image in Microsoft Word applications. Creating Code 128B In None Using Barcode creator for Software Control to generate, create Code 128C image in Software applications. t,I t I t , otherwise
where I , and I, are finite values. Show that if X ( s ) converges for at least one value of s , then the ROC of X ( s ) is the entire splane. Assume that X(s) converges at s
= a,;then
by Eq. (3.3) Since (u, 0,) > 0, e(ul"~)l is a decaying exponential. Then over the interval where the maximum value of this exponential is e("l"o)'l, and we can write x(t) + 0, Thus, X(s) converges for Re(s) = a, > u,,.By a similar argument, if
a, < u,, then (3.57) dl < e
( w ~  ~ ~ ) l ~
l''l~(r)le~~'<m dt
and again X(s) converges for Re(s) = u, <u,. Thus, the ROC of X(s) includes the entire splane.
~ ( t=) O I t l T
otherwise
Find the Laplace transform of x ( t ) . By Eq. (3.3) s + a
[ 1 e(s+u)T~
( 3.58) Since x(f is a finiteduration signal, the ROC of X(s) is the entire splane. Note that from Eq. (3.58) it appears that X(s) does not converge at s = a. But this is not the case. Setting CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUSTIME LTI SYSTEMS
s = a in the integral in Eq. (3.581, we have
The same result can be obtained b applying L'Hospitalls rule to Eq. (3.58). y
Show that if x ( t ) is a rightsided signal and X(s)converges for some value of s, then the R O C of X ( s ) is of the form equals the maximum real part of any of the poles of X ( s ) . where amax
Consider a rightsided signal x(t) so that
and X(s) converges for Re(s) = a,. Then
Thus, X(s) converges for Re(s) = a , and the ROC of X(s) is of the form Re($) > a ( , . Since the ROC of X(s) cannot include any poles of X(s), we conclude that it is of the form Re( s > ~ m a x where a , ,equals the maximum real part of any of the poles of X(s). ,,, Find the Laplace transform X ( s ) and sketch the polezero plot with the ROC for the following signals x( t ): ( a ) x ( t )=e"u(t) +eP3'u(t) ( b ) x ( t ) =e"u(t) + e2'u(t) (c) x ( t ) = e 2 ' u ( t )+ e  3 ' u (  t ) (a) From Table 31 LAPLACE TRANSFORM AND CONTINUOUSTIME LTI SYSTEMS
[CHAP. 3
Fig. 311 We see that the ROCs in Eqs. (3.59) and (3.60) overlap, and thus, From Eq. (3.61) we see that X(s) has one zero at s =  and two poles at s s =  3 and that the ROC is Re(s) >  2, as sketched in Fig. 31 l(a). (b) From Table 31  2 and
We see that the ROCs in Eqs. (3.62) and (3.63) overlap, and thus, From Eq. (3.64) we see that X(s) has no zeros and two poles at s = 2 and s that the ROC is  3 < Re(s) < 2, as sketched in Fig. 31 l(b). From Table 31  3 and
Re(s) <  3 (3.66) s +3 We see that the ROCs in Eqs. (3.65) and (3.66) do not overlap and that there is no common ROC; thus, x(t) has no transform X(s). e  3 r ~ (t ) CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUSTIME LTI SYSTEMS
Find X(s) and sketch the zeropole plot and the ROC for a > 0 and a < 0.
The signal x ( t ) is sketched in Figs. 312(a) and ( b ) for both a > 0 and a < 0. Since x ( t ) is a twosided signal, we can express it as x ( t ) =e"u(t) +ea'u(r) (3.67) Note that x ( t ) is continuous at
t =0 and x(O) =x(O) = x ( O + ) = 1 . From Table 31 earu(  t ) Re(s) < a
(3.69) Fig. 312 LAPLACE TRANSFORM AND CONTINUOUSTIME LTI SYSTEMS
[CHAP. 3
If a > 0, we see that the ROCs in Eqs. (3.68)and (3.69) overlap, and thus, 1 1  2a x ( s )=  =  a < Re(s) < a s+a sa sZaZ From Eq. (3.70)we see that X ( s ) has no zeros and two poles at s = a and s =  a and that the ROC is  a < Re(s) < a , as sketched in Fig. 312(c).If a < 0, we see that the ROCs in Eqs. (3.68) and (3.69) do not overlap and that there is no common ROC; thus, X ( I ) has no transform X ( s ) . PROPERTIES OF THE LAPLACE TRANSFORM Verify the timeshifting property (3.161, that is, By definition ( 3 . 3 )

