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): 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
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 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)
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)
H.w 2: The open loop response, that is, the speed of the motor to a voltage input of 20V, assuming a system without damping is dw 20 = (0.02) + (0.06)w. dt If the initial speed is zero (w(0) = 0) ,and using the Runge-Kutta 4th order method, what is the speed at t = 0.8s? Assume a step size of h = 0.4s.
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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