Match the transfer function with correct Bode phase plots.s+10G(s)=(s+100)s+ 100G(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)SA.180 deg-90 deg0 deg90 deg180 deg-90 deg00000 degC.90 deg+90 deg+0 deg-90 degD.180 degF.90 deg0 deg90 deg-180 deg90 deg0 deg90 deg-180 deg90 deg0 deg-90dex+-180 deg00

Step 1: Step 2: Step 3: Step 4:

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 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
Construct the Bode plots of the following transfer functions. Indicate Gain Margin and Phase Margin of the systems. a) b) 8 s(1.25s + 1)(s+2) 1.6 (s+0.4)(s+0.8) (s + 1)
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
For the following open loop transfer functions, identify the correct Bode plot from the Bode plots given below: Bode Plots: Magnitude (dB) Phase (deg) O 50 -100 -50 -150 -90 0 -135 -180 Magnitude (dB) a 225 Phase (deg) -270 -20 10:2 -40 -60 -80 -100 0 -90 -180 G(s) = -270 Transfer function: a. 1 10-1 101 O b. 2 O c. 3 O d. 4 S Bode Diagram 10° Frequency (rad/s) Bode Diagram (₁² +s+2)(x² +58 + Frequency (rad/s) s+16) 101 10¹ 10² % 2) Magnitude (dB) Phase (deg) Magnitude (dB) Phase (deg) 50 -50 -100 -45 -90 -135 -180 10/2 -60 -80 -100 -120 90 0 -90 -180 -270 10:2 10¹ 10" Bode Diagram 10° Frequency (rad/s) Bode Diagram 10° Frequency (rad/s) 10¹ 101 10² 102
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)
(1) Consider the system represented by the block diagram. The closed loop transfer function T(s)-Y(s)/R(s) is (a) T(s)-50/(s+55 s+50). (b) T(s)=10/(s+50 s+55) (c) T(s)=10/(s+55 s+10). (d) None of the above. R(s)- 10 + s+5 5 Y(s)
The open loop transfer function of a humanoid's arm control system is given as: K G(s) = 2 s(s + 2s + 2) (a) Clearly locate all poles and zeros on a linear graph paper. Provide calculations for the following: asymptote angles, centroid for asymptotes, and departure angle from complex pole. (b) Plot the complete root locus, with the locus on the real axis is clearly shown. Use the scale of 4 cm : 1 unit for both axes and choose the longer side of the graph paper as the real axis.
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
össlg äbäi What is the the transfer function of the first-order system which has the impulse response curve as in * ?the following figure Impulse Response 1.8 1.6 1.4 1.2 System: sys Time (seconds): 0.333 Amplitude: 0.737 1 0.8 Slop 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 Time (seconds) 1/(s+3) O 2/(s+3) O 6/(s+3) O No one of the answers in a, b, and c ( Amplitude 2.
2- Using Matlab, what are the step response curves of the closed-loop system, as shown in fig.1. the feedback represents the second-order dynamic system. (fill in the following table) For=0.4 Wn 1 3 6 9 10 R(S) 0.1 0.3 0.6 0.9 1 For w 5 rad/sec 3 Settling time Peak response 2 Wn s(s+23wn) Settling time Peak response C(s) Discuss the follow Which parameters or w occur on the rise time of the response? Which parameter increases the speed of response? Which parameters can be decreases the response amplitude? Which parameter decreases the steady error state? fig.2
T(t) 02(1) 0(t) 18 N-m-s/rad 1 N-m-s/rad $ kg-m2 3 kg-m2 000 9 N-m/rad 3 N-m/rad (a) Figure P2.16a O John Wiley & Sons, Inc. All rights reserved. The matrix form of the system is 5s +9s +9 0,(s) s² +s +1)0,(s) (s +9) T(s) а. (s + 8) s*s+1)- 5s +9s +9 b. 0,(s) 3s +s+120,(s) -(s+9) -(s +9) T(s)
The Laplace transform of f t.cos 4t s2-16 f (s) = f(s) = (s2+16)2 (s2+16)2 Option 1 Option 2 s+4 f (s) s+1 f(s) : (s2-16)2 (s2+4)2 O Option 3

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