20 2S 1Ω ww Io(s) Using ladder method Let Vo(s) = Vo 12 = Vo 2 1 V r4s + 1/13 (142) = 16 [48 + 1] V₁ = ½ + ½ = V +; 2s V₁ Vo I = S+4 = 2s Vo 4s [4s + 4s s+ 4 =V₂ 4s vo [48 + 1] ≤ 14 = Vo 2s² + 12s+1 => H(s) = 4s(s + 4) 10 = 1₁ + 12 = V Often we set the output equal to 1. 4Ω 2s V₁ 12(s) ↓11(s) 202 4s + 1 [4s (s + 4) 4s (s + 4) 2s212s1 Solve for h(t)as before. ww +5°1 Vo(s) 4s + 1 V₁ 4s + 1 1 Let Vo(s) = 1 => 12 = 1/1/2 I₁ S+4 4s S+4 H(s) = = 4s(s + 4) V₁ = 1 + 1 [11 調 2s 2 = 1+ 4s 1 4s +1 4s + 1 4s 10 = 11 + 12 = 1 + 4s(s + 4) 2s212s+1 Solve for h(t)as before. 4s(s + 4) 2
20 2S 1Ω ww Io(s) Using ladder method Let Vo(s) = Vo 12 = Vo 2 1 V r4s + 1/13 (142) = 16 [48 + 1] V₁ = ½ + ½ = V +; 2s V₁ Vo I = S+4 = 2s Vo 4s [4s + 4s s+ 4 =V₂ 4s vo [48 + 1] ≤ 14 = Vo 2s² + 12s+1 => H(s) = 4s(s + 4) 10 = 1₁ + 12 = V Often we set the output equal to 1. 4Ω 2s V₁ 12(s) ↓11(s) 202 4s + 1 [4s (s + 4) 4s (s + 4) 2s212s1 Solve for h(t)as before. ww +5°1 Vo(s) 4s + 1 V₁ 4s + 1 1 Let Vo(s) = 1 => 12 = 1/1/2 I₁ S+4 4s S+4 H(s) = = 4s(s + 4) V₁ = 1 + 1 [11 調 2s 2 = 1+ 4s 1 4s +1 4s + 1 4s 10 = 11 + 12 = 1 + 4s(s + 4) 2s212s+1 Solve for h(t)as before. 4s(s + 4) 2
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
Related questions
Question
Explain this method to be plz step by step if possible
![20
2S
1Ω
ww
Io(s)
Using ladder method
Let
Vo(s) = Vo
12
=
Vo
2
1
V
r4s
+ 1/13 (142) = 16 [48 + 1]
V₁ = ½ + ½ = V +;
2s
V₁ Vo
I
=
S+4
=
2s
Vo
4s
[4s +
4s
s+ 4
=V₂
4s
vo [48 + 1] ≤ 14
= Vo
2s² + 12s+1
=>
H(s)
=
4s(s + 4)
10 = 1₁ + 12 = V
Often we set the output equal to 1.
4Ω
2s
V₁
12(s)
↓11(s)
202
4s + 1
[4s (s + 4)
4s (s + 4)
2s212s1
Solve for h(t)as before.
ww
+5°1
Vo(s)
4s + 1
V₁
4s + 1
1
Let Vo(s)
= 1
=>
12 = 1/1/2
I₁
S+4
4s
S+4
H(s)
=
=
4s(s + 4)
V₁ = 1 +
1 [11
調
2s 2
= 1+
4s
1
4s +1
4s + 1
4s
10 = 11 + 12
=
1
+
4s(s + 4)
2s212s+1
Solve for h(t)as before.
4s(s + 4)
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5b75143-9ad0-4b41-a9ea-1d4e86bf4b3a%2F0e42e8b8-ee66-4c7e-b326-ce5ba091476b%2Fojb4k2u_processed.png&w=3840&q=75)
Transcribed Image Text:20
2S
1Ω
ww
Io(s)
Using ladder method
Let
Vo(s) = Vo
12
=
Vo
2
1
V
r4s
+ 1/13 (142) = 16 [48 + 1]
V₁ = ½ + ½ = V +;
2s
V₁ Vo
I
=
S+4
=
2s
Vo
4s
[4s +
4s
s+ 4
=V₂
4s
vo [48 + 1] ≤ 14
= Vo
2s² + 12s+1
=>
H(s)
=
4s(s + 4)
10 = 1₁ + 12 = V
Often we set the output equal to 1.
4Ω
2s
V₁
12(s)
↓11(s)
202
4s + 1
[4s (s + 4)
4s (s + 4)
2s212s1
Solve for h(t)as before.
ww
+5°1
Vo(s)
4s + 1
V₁
4s + 1
1
Let Vo(s)
= 1
=>
12 = 1/1/2
I₁
S+4
4s
S+4
H(s)
=
=
4s(s + 4)
V₁ = 1 +
1 [11
調
2s 2
= 1+
4s
1
4s +1
4s + 1
4s
10 = 11 + 12
=
1
+
4s(s + 4)
2s212s+1
Solve for h(t)as before.
4s(s + 4)
2
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