Question

b) Given a second order system with the following open loop transfer function where damping ratio, } = 0.707 and natural frequency, Wn= 2.5. wn? G(S) = S2 + 23wns +wn? i. Determine the steady state error to an appropriate input via a calculation method using the transfer function. Compare your answer with the steady state error from the exact frequency response for this system given in Figure Q4(b). (5 marks) ii. Evaluate the difference of the exact frequency response in Figure Q4(b) if damping ratio, 3 = 0.3 and what are the effects to the closed-loop system's bandwidth and transient response? (5 marks) 20 + 0 Magnitude (dB) -20 -40 -60 0 -45 Phase (deg) -90 -135 -180 10-1 10° 10 102 Frequency (rad/s) Figure Q4(6): Exact frequency response of the second-order system

Answer 1

Given on2 G1s) 3 +2&ugs+with ISC W = 6.25 ; Ep = 0.707 (2.572 Gls) 8+2x0-807x2.55 +(2-5)2 6.25 GIS) = 3²+3.5355+ 6.25 Emor ECS) = R(S) 1+ GIS) HIS) Steady state Error It ES R(S) sto I+G(S) HIS) Hell appropriate input will be step input so RIS) & ㅎ S. ess 50 Ys wn2 3+28 wystwyz 1+ 17 1s S. 나 Sto + 6.25 s²+ 3.5355+ 6.25 S70 1 + 6.25 Ps2 +3.5355+ 6.25 Ja = 0.5 1+1 It lt 6.25 so 3 +35355+ 6.25

.. Steady state Error = 0.5 from bode plot; at www frequency G15) u kp Her kp = 1 M (dx)=0 at low tepune for Type o systems ess Hkp . - 0.5 1+1 Both are some a 20 (1) from bode plot Gain cross over frequency & wgc. radrans / see phose coose over frequency & upc = radrass /sec R Gain Margm Gm = if we chance = 0.3 phose Margin pm 6.25 G1s) - 3 + 2x0-3x2-55+ 6.25 GIS) 6.25 s² 1.55+ 6.25
frequencys response of modified system is given below

Bode Diagram Gm = Inf dB (at Inf rad/s), Pm = 50.2 deg (at 3.2 rad/s) 20 0 -20 Magnitude (dB) System: Frequency (rad/s): 3.23 Magnitude (dB): -0.199 -40 -60 -80 0 -45 Phase (deg) -90 -135 -180 10-1 100 102 101 Frequency (rad/s)

3.2 red/sec Gain cross over frequency age pm 50.2° puase crossover frequency not defined Gm not defined or infinite Band widty is given by un -2&2) + 1 & 4472 W gw = 0.707 Wow 1 - 240.10 7+ ((0.7077 48 42 2.5 2.5 1- 2x 0.707² +0.5 o.fo7 WBwa at &= 0.3 - 3.62 wow 3.62 Transtent wew= Response) settling time at 2% &wn &= 0.707 ts - 2.26 Sec = 0.3 ts 2 1.33 sec