Match the transfer function with correct Bode amplitude plots.G(s)=Ss²+7s+10SG(s) =s² + 10s+ 100s+1G(s) =s²+10x+100S+ 10G(s) =G(s) =s²+s+10s+ 10s²+101s+ 100A.20 dB/dec-20 dB/decB.-20 dB/dec-20 dB/decC.20 dB/dec20 dB/decE.20 dB/dec-20 dB/dec-40 dB/dec-20 dB/dec

Step 1: explanation for iii) explanation for iv)

Answered: Match the transfer function with…

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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Match the transfer function with correct Bode amplitude plots. G(s) = s+ 10 s(+100) s+ 100 G(s) = s(s+10) S G(s)= (s+10) (s+100) (+10) G(s) = (+100) s(+100) G(s) = (+10) (+10)(+100) G(s) = S A. -20 dB/dec 20 dB/dec B. D. 20 dB/dec -20 dB/dec 20 dB/dec -40 dB/dec -20 dB/dec 20 dB/dec 40 dB/dec E. 20 dB/dec -20 dB/dec 20 dB/dec 20 dB/dec -20 dB/dec
Match the transfer function with correct Bode amplitude plots. ✓ G(s): 5 s+7 G(s)=s+5 G(s): s+ 10 S S G(s) = s+ 10 A. B. D. slope 20 dB/dec = slope = -20 dB/dec = slope 20 dB/dec = slope -20 dB/dec
Q1
Match the transfer function with correct Bode phase plots. s+10 G(s)=(s+100) s+ 100 G(s)= s(s+10) G(s) = (s+10) (s+100) s(s+10) G(s) = (s+100) s(s+100) G(s)= (s+10) G(s) = (+10) (+500) S A. 180 deg- 90 deg 0 deg 90 deg 180 deg- 90 deg 00 00 0 deg C. 90 deg+ 90 deg+ 0 deg -90 deg D. 180 deg F. 90 deg 0 deg 90 deg -180 deg 90 deg 0 deg 90 deg -180 deg 90 deg 0 deg -90dex+ -180 deg 00
H-W Find Time response for 2S+3 Yes)= s° (s+3)
Sketch frequency response of the system from part (a) in the form of Bode plot as accurately as you can.
Find the transfer function of the system
For the following open loop transfer functions, identify the correct Bode plot from the Bode plots given below: Bode Plots: 1) 3) Magnitude (dB) (Bap) eseyd 50 0 -100 -50 -150 -90 -135 -180 Magnitude (dB) -225 Phase (deg) -270 1012 -20 -40 -60 -80 -100 0 -90 -180 -270 G(s) = 10-1 Transfer function: 10 O a. 1 O b. 2 O c. 3 O d. 4 16 Bode Diagram TOP Frequency (rad/s) Bode Diagram (s+4)(s² +1.65+4) 101 Frequency (rad/s) 10 Њ 10² 10² N + Magnitude (dB) Phase (deg) Magnitude (dB) Phase (deg) 50 -50 -100 -45 -90 -135 -180 10:12 -20 -40 -60 -80 -100 -120 90 0 -90 -180 -270 1012 10 10 Bode Diagram 10² Frequency (rad/s) Bode Diagram 10² Frequency (rad/s) 10 10¹ 10² 10²
5 Problem Match the following two frequency response diagrams: -0 0.4 |G(i w)| 0.2 10 |G(i w)| 5 Response A 2 3 4 5 6 7 8 Frequency, (rad/s) Response B 5 6 7 8 Frequency, w(rad/s) 1 2 3 4 6 6 10 10 10 to two of the following ODES 1. +x+100x = u(t) 5. +100x = u(t) 9. +100x = u(t) 3. x+x+5x= u(t) 4. x+x+x= u(t) 2. x+x+25x = u(t) : 6. x + 25xu(t) 10. +25x = u(t) 7. x+5x= u(t) 8. u(t) 11. +5x= u(t) = 12. xu(t)
Match the transfer function with correct Bode phase plots. G(s) = 5 $+7 G(s)=s+5 G(s) = s+10 S S G(s) = S+ 10 QUESTION 10 90 deg 00 0 deg -90 deg 180 deg B. D 90 deg @ 0 deg -90 deg 180 deg 90 deg @ 0 deg -90 deg 180 deg 90 deg @ 0 deg -90 deg 180 deg
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Problem-3. Consider the closed loop transfer function from R(s) to C(s) as given below as given below. 85 H(s) = (s+5)(s? + 2s +17) MATLAB gives the partial fraction values of the Laplace transform C(s) to the unit step input as follows: Num: Denom: -0.5313 -5.0000 -0.2344 + 0.3906i -1.0000 + 4.0000i -0.2344 - 0.3906i -1.0000 - 4.0000i 1.0000 d) Find the overshoot and peak time by using the reduced-order transfer function Hint:Tp = e-(n{//1-§2 %OV = Wn V1- 2)

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Match the transfer function with correct Bode amplitude plots. G(s)= S s²+7s+10 S G(s) = s² + 10s+ 100 s+1 G(s) = s²+10x+100 S+ 10 G(s) = G(s) = s²+s+10 s+ 10 s²+101s+ 100 A. 20 dB/dec -20 dB/dec B. -20 dB/dec -20 dB/dec C. 20 dB/dec 20 dB/dec E. 20 dB/dec -20 dB/dec -40 dB/dec -20 dB/dec