ch13 - 13-1 13-2 For a first-order Pade approximation...

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Unformatted text preview: 13-1 13-2 For a first-order Pade approximation 2121sses+--from which we obtain 1=PadeAR -=-2tan21PadeBoth approximations represent the original function well in the low frequency region. At higher frequencies, the Pad approximation matches the amplitude ratio of the time delay element exactly (ARPade = 1), while the two-term approximation introduces amplification (ARtwo term>1). For the phase angle, the high-frequency representations are: 90-termtwo180-PadeSince the angle of -jeis negative and becomes unbounded as , we see that the Pade representation also provides the better approximation to the time delay element's phase angle, matching of the pure time delay element to a higher frequency than the two-term representation. 13.3 Nominal temperature F1232F119F127=+=TF4)F119F127(21=-=A.,sec5.4=rad/s189.sec)60/8.1(2==Using Eq. 13-2 with K=1, ( 29F25.51)5.4()189.(412222=+=+=AAActual maximum air temperature = F25.128=+ATActual minimum air temperature = F75.117=-AT13-3 13.4 12.1)()(+=ssTsTm)()12.()(sTssTm+=amplitude of T=3.464 467.31)2.(2=+phase angle of T= + tan-1(0.2) = + 0.04 Since only the maximum error is required, set = 0 for the comparison of Tand mT. Then Error = mT-T=3.464 sin (0.2t) 3.467sin(0.2t+ 0.04) = 3.464 sin(0.2t) 3.467[sin(0.2t) cos 0.04 + cos(0.2t)sin 0.04] = 0.000 sin(0.2t) -0.1386 cos(0.2t) Since the maximum absolute value of cos(0.2t) is 1, maximum absolute error = 0.1386 13-4 13.5 a) Bode DiagramFrequency (rad/sec)Phase (deg)Magnitude (abs)10-210-11010-210-110101-180-135-90-45AR (absolute) 0.1 4.44 -32.41 0.69 -12410 0.005 -173b) Bode DiagramFrequency (rad/sec)Phase (deg)Magnitude (abs)10-21010-210-110101-270-225-180-135-90-45AR (absolute) 0.1 4.42 -38.21 0.49 -16910 0.001 -25713-5 c) Bode DiagramFrequency (rad/sec)Phase (deg)Magnitude (abs)10-11010-210-110101102-90-45AR (absolute) 0.1 4.48 -22.11 2.14 -44.910 0.003 -87.6d) Bode DiagramFrequency (rad/sec)Phase (deg)Magnitude (abs)10-11010-210-110101102-270-225-180-135-90-45AR (absolute) 0.1 4.48 -33.61 1.36 -13610 0.04 -26613-6 e) Bode DiagramFrequency (rad/sec)Phase (deg)Magnitude (abs)10-110101-180-150-120-9010-210-110101102AR (absolute) 0.1 44.6 -1171 0.97 -16910 0.01 -179f) Bode DiagramFrequency (rad/sec )Phase (deg)Magnitude (abs)10-110101-180-150-120-9010-110101102AR (absolute) 0.1 44.8 -1121 1.36 -13510 0.04 -15813-7 13.6 a)Multiply the AR in Eq. 13-41a by 122+a. Add to the value of in Eq. 13-41b the term + )(tan1a-. KjG=)(222222)4.()1(1+-+a--=-22114....
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ch13 - 13-1 13-2 For a first-order Pade approximation...

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