find a and b x = A e-3+ J2-1 )@nt(-3- 132–1 )wnt+B eWith initial conditions x(0) and x(0), we found:= *(©)+(3+ Jz2-1 Jw,x(0)A =2wn./32–1- x(0)-(3- J32-1 )@nx(0)%3D2wn./32–1Where:3 = 1.16Wn = 5.08 rad/sec.

Answered: Image /qna-images/answer/6e810ec9-f9d2-4f81-9841-88a94f7007a3.jpg

(-3+ ]32–1 )wnt (-3- [32-1 )wnt X = A e +Be With initial conditions x(0) and x( 0), we found: = (0)+(3+ Jc2-1 Jwqx(0) )wnx(0) A = 2wn./32–1 - x(0)-(3- J32-1 )onx(0) B = 2wn. 32–1 Where: 3 = 1.16 Wn = 5.08 rad/sec.

(-3+ ]32–1 )wnt (-3- [32-1 )wnt X = A e +Be With initial conditions x(0) and x( 0), we found: = (0)+(3+ Jc2-1 Jwqx(0) )wnx(0) A = 2wn./32–1 - x(0)-(3- J32-1 )onx(0) B = 2wn. 32–1 Where: 3 = 1.16 Wn = 5.08 rad/sec.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

find a and b

x = A e-3+ J2-1 )@nt
(-3- 132–1 )wnt
+B e
With initial conditions x(0) and x
(0), we found:
= *(©)+(3+ Jz2-1 Jw,x(0)
A =
2wn./32–1
- x(0)-(3- J32-1 )@nx(0)
%3D
2wn./32–1
Where:
3 = 1.16
Wn = 5.08 rad/sec.

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