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3ece 16 Q5/A1 G (s) = Kp = lim G(s) = 530 Ku Ка P 10 S(0.15+1) E(s) R(S) lim SG(s) 530 = lim S²G(S) = 5-0 = then > K₁ = ∞ K₂ = K3 = ㅎ Ku = lo 0(0.1(0)+1 1 I+G(s) 0.15² +5 10 +0.15² +5 0.1 = 10 18 -0.001 0.1(0)+1 ok lo 0.1*0+1 1+ = 10 =-1000 દરે 10 Sco-15+1) +1 1 10 0.15² +5 =o.ls+os² -0.00 15³. (4 Explain to me how the step in the red box occurred. And the step in the blue box also ■

3ece 16 Q5/A1 G (s) = Kp = lim G(s) = 530 Ku Ка P 10 S(0.15+1) E(s) R(S) lim SG(s) 530 = lim S²G(S) = 5-0 = then > K₁ = ∞ K₂ = K3 = ㅎ Ku = lo 0(0.1(0)+1 1 I+G(s) 0.15² +5 10 +0.15² +5 0.1 = 10 18 -0.001 0.1(0)+1 ok lo 0.1*0+1 1+ = 10 =-1000 દરે 10 Sco-15+1) +1 1 10 0.15² +5 =o.ls+os² -0.00 15³. (4 Explain to me how the step in the red box occurred. And the step in the blue box also ■

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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3ece 16 Q5/A/ G (s) = Kp = lim G(s) = So Ku = Ka 10 S(0.15+1) E(s) R(3) = lim S²G(S) = = 5-0 K3 = Ku = lim SG(S) = So then > K₁ = K₂ = ols² +s 10 +0.15² +5 lo 0(0.1(0)+1 1 I+G(s) = ㅎ = 10 18 -0.001 0.1(0)+1 0x 10 0.1*0 +1 = 50 = 10 =-1000 =O 1+√10 S(0.15+1) +1 1 10 0.15² +5 =o.ls+os² -0.0015³~ Explain to me how the step in the red box occurred. And the step in the blue box also .
Transcribed Image Text:3ece 16 Q5/A/ G (s) = Kp = lim G(s) = So Ku = Ka 10 S(0.15+1) E(s) R(3) = lim S²G(S) = = 5-0 K3 = Ku = lim SG(S) = So then > K₁ = K₂ = ols² +s 10 +0.15² +5 lo 0(0.1(0)+1 1 I+G(s) = ㅎ = 10 18 -0.001 0.1(0)+1 0x 10 0.1*0 +1 = 50 = 10 =-1000 =O 1+√10 S(0.15+1) +1 1 10 0.15² +5 =o.ls+os² -0.0015³~ Explain to me how the step in the red box occurred. And the step in the blue box also .
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