zen barcode ssrs and I0 V0 100 20 A 5 R in Software

Paint Quick Response Code in Software and I0 V0 100 20 A 5 R

and I0 V0 100 20 A 5 R
QR Code Recognizer In None
Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.
Quick Response Code Drawer In None
Using Barcode maker for Software Control to generate, create QR Code JIS X 0510 image in Software applications.
At ! 500 rad/s, Z1 5 j 500 20 10 3 5 j10 11:15 63:48
QR Code 2d Barcode Scanner In None
Using Barcode decoder for Software Control to read, scan read, scan image in Software applications.
Denso QR Bar Code Maker In C#.NET
Using Barcode generation for VS .NET Control to generate, create QR-Code image in .NET applications.
and i1 V1;max 50 sin !t 63:48 4:48 sin !t 63:48 sin !t 1 11:15 Z1 A
QR-Code Generator In VS .NET
Using Barcode generation for ASP.NET Control to generate, create QR Code image in ASP.NET applications.
Draw QR Code ISO/IEC18004 In .NET
Using Barcode drawer for .NET Control to generate, create QR Code image in .NET framework applications.
At 3! 1500 rad/s, Z3 5 j30 30:4 80:548
Denso QR Bar Code Creator In VB.NET
Using Barcode generator for VS .NET Control to generate, create QR Code JIS X 0510 image in .NET framework applications.
GS1 - 12 Creator In None
Using Barcode generation for Software Control to generate, create UPC-A Supplement 2 image in Software applications.
and i3 V3;max 25 sin 3!t 80:548 0:823 sin 3!t 80:548 sin 3!t 3 30:4 Z3 A
DataMatrix Maker In None
Using Barcode generation for Software Control to generate, create Data Matrix ECC200 image in Software applications.
EAN / UCC - 13 Creation In None
Using Barcode creation for Software Control to generate, create EAN / UCC - 13 image in Software applications.
The sum of the harmonic currents is the required total response; it is a Fourier series of the type (8). i 20 4:48 sin !t 63:48 0:823 sin 3!t 80:548 A
Paint Barcode In None
Using Barcode generation for Software Control to generate, create bar code image in Software applications.
Code 3 Of 9 Printer In None
Using Barcode printer for Software Control to generate, create Code-39 image in Software applications.
This current has the e ective value q p Ieff 202 4:482 =2 0:8232 =2 410:6 20:25 A which results in a power in the 5-
Draw USPS POSTNET Barcode In None
Using Barcode drawer for Software Control to generate, create USPS POSTal Numeric Encoding Technique Barcode image in Software applications.
Decoding Barcode In .NET Framework
Using Barcode scanner for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications.
resistor of
ANSI/AIM Code 39 Decoder In Java
Using Barcode reader for Java Control to read, scan read, scan image in Java applications.
USS Code 128 Decoder In None
Using Barcode scanner for Software Control to read, scan read, scan image in Software applications.
2 P Ieff R 410:6 5 2053 W
ECC200 Generator In VS .NET
Using Barcode generator for Reporting Service Control to generate, create ECC200 image in Reporting Service applications.
Reading Barcode In Visual Basic .NET
Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications.
As a check, we compute the total average power by calculating rst the power contributed by each harmonic and then adding the results. At ! 0: At ! 500 rad/s: At 3! 1500 rad/s: Then, P0 V0 I0 100 20 2000 W P1 1 V1 I1 cos 1 1 50 4:48 cos 63:48 50:1 W 2 2 P3 1 V3 I3 cos 3 1 25 0:823 cos 80:548 1:69 W 2 2 P 2000 50:1 1:69 2052 W
Encode EAN13 In None
Using Barcode printer for Font Control to generate, create EAN13 image in Font applications.
Data Matrix Creation In Java
Using Barcode drawer for Android Control to generate, create ECC200 image in Android applications.
CHAP. 17]
FOURIER METHOD OF WAVEFORM ANALYSIS
Fig. 17-10
Fig. 17-11
Another Method The Fourier series expression for the voltage across the resistor is vR Ri 100 22:4 sin !t 63:48 4:11 sin 3!t 80:548 V r p 1 1 VReff 1002 22:4 2 4:11 2 10 259 101:3 V 2 2
2 Then the power delivered by the source is P VReff =R 10 259 =5 2052 W.
In Example 17.5 the driving voltage was given as a trigonometric Fourier series in t, and the computations were in the time domain. (The complex impedance was used only as a shortcut; Zn and n could have been obtained directly from R, L, and n!). If, instead, the voltage is represented by an exponential Fourier series, v t
1 X 1
Vn e jn!t
then we have to do with a superposition of phasors Vn (rotating counterclockwise if n > 0, clockwise if n < 0), and so frequency-domain methods are called for. This is illustrated in Example 17.6.
EXAMPLE 17.6 A voltage represented by the triangular wave shown in Fig. 17-12 is applied to a pure capacitor C. Determine the resulting current.
Fig. 17-12 In the interval  < !t < 0 the voltage function is v Vmax 27Vmax = !t; and for 0 < !t < , v Vmax 2Vmax = !t. Then the coe cients of the exponential series are determined by the evaluation integral Vn 1 2 0
Vmax 2Vmax = !t e jn!t d !t
1 2
Vmax 2Vmax = !t e jn!t d !t
from which Vn 4Vmax =2 n2 for odd n, and Vn 0 for even n. The phasor current produced by Vn (n odd) is
FOURIER METHOD OF WAVEFORM ANALYSIS
[CHAP. 17
In with an implicit time factor e jn!t .
Vn 4Vmax =2 n2 4Vmax !C j Zn 1=jn!C 2 n
The resultant current is therefore i t
1 X 1
In e jn!t j
1 4Vmax !C X e jn!t n 2 1
where the summation is over odd n only. The series could be converted to the trigonometric form and then synthesized to show the current waveform. However, this series is of the same form as the result in Problem 17.8, where the coe cients are An j 2V=n for odd n only. The sign here is negative, indicating that our current wave is the negative of the square wave of Problem 17.8 and has a peak value 2Vmax !C=.
FOURIER TRANSFORM OF NONPERIODIC WAVEFORMS A nonperiod waveform x t is said to satisfy the Dirichlet conditions if 1 (a) x t is absolutely integrable, 1 jx t j dt < 1, and (b) the number of maxima and minima and the number of discontinuities of x t in every nite interval is nite.
For such a waveform, we can de ne the Fourier transform X f by 1 X f x t e j2ft dt
22a The time function x t is
where f is the frequency. The above integral is called the Fourier integral. called the inverse Fourier transform of X f and is obtained from it by 1 x t X f e j2ft df
22b
x t and X f form a Fourier transform pair. Instead of f , the angular velocity ! 2f may also be used, in which case, (22a) and (22b) become, respectively, 1 X ! x t e j!t dt 23a
Copyright © OnBarcode.com . All rights reserved.