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

10- To enter the transfer functions G(s)= a) G= th(2, [3,2]) b) c) d) G(s) TH (2, [2, 3]) G= tf (2, [3,2]) G(s)= TF(2, [2, 3]) 2 3s+2 c) A marginally stable response d) A marginally unstable response to Matlab command use: 3s+1 11- To enter the transfer function g(s)= $2+35+2 ,can be use a) num=[3, 1]; den= [1, 3, 2]; g-tf (den, num) num=[3, 1]; den=[1, 3, 2]; g=tf (num, den) b) c) num=[3; 1]; den=[1, 3, 2]; g=tf (den, num) d) num=(3, 1); den=(1, 3; 2); g-tf (num, den) 12- What is the output of the code (>>p=[0 1]; q=[ 1 -1 0]; step(p,q) ) a) An unstable response b) A stable response (s+1)

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

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