Z x 0.632 Li6. Z + 3.16 L56.86 2x 0.632 L26. 8G' - 2 = 3.16 LSG. 86 Z[ 0.632 L26.86 - 1] 3. 16 L56.86
Z x 0.632 Li6. Z + 3.16 L56.86 2x 0.632 L26. 8G' - 2 = 3.16 LSG. 86 Z[ 0.632 L26.86 - 1] 3. 16 L56.86
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can someone elaborate on how Z(0.632∠26.86° -1) became Z(0.521∠146.78°) Please show all calculations in detail. Thank you!
![Z x 0.632 [26.%6
Z + 3.16 L56.86
2x 0.632 L26. 8g' - 2 = 3. 1G L56. 86
Z[ 0.632 L2G - 86' - 1
* 3. 16 L56.86
Z[ 0.521 L146.78]
= 3.16 (56.86
3. I6 L56.86
0.521 L146.78g
6.06 L-89.9e' r
Polar form
8.46xi0 - 6.05
car tesian form
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa1fb531-a671-4652-9745-dc6f9da9adcd%2F10822ce0-0bbe-4208-b69f-780019d86de0%2Fypbg02_processed.png&w=3840&q=75)
Transcribed Image Text:Z x 0.632 [26.%6
Z + 3.16 L56.86
2x 0.632 L26. 8g' - 2 = 3. 1G L56. 86
Z[ 0.632 L2G - 86' - 1
* 3. 16 L56.86
Z[ 0.521 L146.78]
= 3.16 (56.86
3. I6 L56.86
0.521 L146.78g
6.06 L-89.9e' r
Polar form
8.46xi0 - 6.05
car tesian form
%3D
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