hellww For the circuit in Fig. 14.6, calculate the gain IL(@)/I,(w) and its poles Example 14.2 and zeros. Phasor Domain (Frequency Domain) Time domain R R Solution: C By current division, j@C i,() =0.5 F jwL 2 H 4 +j2w 1, (@) 4 + j2w + 1/j0.5w I,(o) Figure 14.6 For Example 14.2. or j0.5w(4 + j2w) s(s + 2) S = = jo 1, (@ 1+ j2w + (jw)² s? + 2s + 1' The zeros are s(s + 2) = 0 21 = 0, z2 = -2 How solve tge step?! The poles are at s2 + 2s + 1 = (s + 1)² = 0 %3D Thus, there is a repeated pole (or double pole) at p = -1.
hellww For the circuit in Fig. 14.6, calculate the gain IL(@)/I,(w) and its poles Example 14.2 and zeros. Phasor Domain (Frequency Domain) Time domain R R Solution: C By current division, j@C i,() =0.5 F jwL 2 H 4 +j2w 1, (@) 4 + j2w + 1/j0.5w I,(o) Figure 14.6 For Example 14.2. or j0.5w(4 + j2w) s(s + 2) S = = jo 1, (@ 1+ j2w + (jw)² s? + 2s + 1' The zeros are s(s + 2) = 0 21 = 0, z2 = -2 How solve tge step?! The poles are at s2 + 2s + 1 = (s + 1)² = 0 %3D Thus, there is a repeated pole (or double pole) at p = -1.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Question
![For the circuit in Fig. 14.6, calculate the gain IL(@)/I,(w) and its poles
and zeros.
Example 14.2
Phasor Domain
(Frequency Domain)
Time domain
R
Solution:
4Ω
C
By current division,
i,()
:0.5 F
jwL
2 H
4 + j2w
1, (@)
4 + j2w + 1/j0.5w
%3D
Figure 14.6
For Example 14.2.
or
j0.5@(4 +j2w)
1+ j2w + (jw)² s² + 2s + 1'
s(s + 2)
s = jo
%3D
1,(@
The zeros are a
s(s + 2) = 0
21 = 0, z2 = -2
How solve tge step?!
The poles are at
s2 + 2s + 1 = (s + 1)² = 0
%3D
%3D
Thus, there is a repeated pole (or double pole) at p = -1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a570cb6-afd9-4401-ae47-19a419d7374f%2F3090086a-17cd-464a-abcf-4aac13d774ee%2F2fz0cyc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For the circuit in Fig. 14.6, calculate the gain IL(@)/I,(w) and its poles
and zeros.
Example 14.2
Phasor Domain
(Frequency Domain)
Time domain
R
Solution:
4Ω
C
By current division,
i,()
:0.5 F
jwL
2 H
4 + j2w
1, (@)
4 + j2w + 1/j0.5w
%3D
Figure 14.6
For Example 14.2.
or
j0.5@(4 +j2w)
1+ j2w + (jw)² s² + 2s + 1'
s(s + 2)
s = jo
%3D
1,(@
The zeros are a
s(s + 2) = 0
21 = 0, z2 = -2
How solve tge step?!
The poles are at
s2 + 2s + 1 = (s + 1)² = 0
%3D
%3D
Thus, there is a repeated pole (or double pole) at p = -1.
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