23_Freq_Resp_t_domain

23_Freq_Resp_t_domain - freq_resp 1/2 Frequency Responses: t

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Unformatted text preview: freq_resp 1/2 Frequency Responses: t "domain" 1 1/(2*3.14159)s+1 Sine Wave1 Transfer Fcn1 CHE 361 + 7 + Constant1 Sum1 + 9 + Constant2 Sum2 1 1/(2*3.14159)s+1 Sine Wave2 Transfer Fcn2 + 1 + Constant3 Sum3 Mux + 3 + Constant4 Sum4 1 1/(2*3.14159)s+1 Sine Wave3 Transfer Fcn3 yout + -5 + Sum5 Constant5 + -3 + Constant6 Sum6 Mux For G1st order ( s ) K1 = τ1 s 1 K1 AR1 = from Eqns (5 26), (5 27) and (13 3a) we have and 2 order ( s ) = 2 τ2 ζ s 1 K2 2 τ2 s 2 from Eqns (5 61), (5 62) and (13 41a,b) we have K2 AR2 = 1 ( ω τ2 )2 For ζ < AR 2, max = tan 1 ( ω τ1 ) 1 ω2τ1 For G2nd φ1 = 2 2 and 2 2 ζ ω τ2 φ2 = tan 1 2 2 ζ ω τ2 1 ( ω τ )2 2 from Eqns (5 65,66) or (13 44,45) we have K2 > K2 occurs at ωresonant = 1 2ζ 2 τ2 2 ζ 1 ζ2 See page 2 for Simulink plots of 3 frequencies into G1 and Bode plot freq_resp 2/2 τprocess = 1 = 0.16 min , ∆ t i = t input 2π ∆t i t output < 0 at t = 6 min, φ i = P 360 deg. (Fig13.1,pg335) i Low, Medium, or High Frequency: Input u'(t) on Top, Output y'(t) on Bottom 10 sin ( π t / 2) u A (t ) = 9 y', output = lower curve : u', input = upper curve 8 P A = 4 min π rad = 1.57 ωA = 2 min 6 4 uB ( t ) = 3 sin ( 2 π t ) 2 P B = 1 min ω B = 2 π = 6.28 0 rad min -2 uC ( t ) = -4 3 sin ( 4 π t ) P C = 0.5 min -6 0 Amplitude Ratio 10 10 10 1 2 3 4 Time 5 6 7 8 AR A = 0.970 ∆ t A = 0.156 min φ A = 14.04 degrees -1 -2 10 0 1 10 Frequency (rad/min) 10 2 0 Phase (deg) rad min Data points circled: Bode Plot: made by che_bode.m using freqg.dat 0 ω C = 4 π = 12.57 -45 -90 0 10 AR B = 0.707 ∆ t B = 0.125 min φ B = 45.00 degrees AR C = 0.447 ∆ t C = 0.088 min φ C = 63.4 degrees 1 10 Frequency (rad/min) 10 2 ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.

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