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Frequency response method

Frequency response method. Given: G ( j ω ) as a function of ω is called the freq. resp. For each ω , G ( jω ) = x ( ω ) + jy ( ω ) is a point in the complex plane As ω varies from 0 to ∞, G ( jω ) = x ( ω ) + jy ( ω ) defines a curve in the x-y plane or the s=x+jy complex plane.

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Frequency response method

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  1. Frequency response method Given: • G(jω) as a function of ω is called the freq. resp. • For each ω, G(jω) = x(ω) + jy(ω) is a point in the complex plane • As ω varies from 0 to ∞, G(jω) = x(ω) + jy(ω) defines a curve in the x-y plane or the s=x+jy complex plane. • This curve in the complex plane is called the Nyquist plot of G(jw) G(s) nyquist(sys)

  2. Can rewrite in Polar Form: • |G(jω)| as a function of ω is called the amplitude resp. • as a function of ω is called the phase resp. • The two plots: with log scale-ω, are Bode plot bode(sys)

  3. To obtain freq. Resp from G(s): • Select • Evaluate G(jω) at those to get • Plot Imag(G) vs Real(G): Nyquist • or plot with log scale ω

  4. To obtain freq. resp. experimentally: • Select • Give input to system as: • Adjust A1 so that the output is not saturated or distorted. • Measure amp B1 and phase φ1 ofoutput: System

  5. Then is the freq. resp. of the system at freq ω1 • Repeat for all ωK • Either plot or plot

  6. G1(s) G2(s) Product of T.F.

  7. System type & Bode plot C(s) Gp(s)

  8. System type is for steady state,i.e. t →∞ i.e. s → 0 i.e. ω → 0 As ω → 0 i.e. slope = N(–20) dB/dec.

  9. N = 0, type zero: at low freq: mag. is flat at 20 log K phase plot is flat at 0°

  10. If Bode gain plot is flat at low freq, system is type zero (confirm by phase plot flat at 0°) Then: Kv = 0, Ka = 0 Kp = Bode gain as ω→0 = DC gain (convert dB to values)

  11. Example

  12. N = 1, type = 1 Bode mag. plot has –20 dB/dec at low freq. (ω→0) (straight line with slope = –20) Bode phase plot becomes flat at –90° when ω→0 Kp= DC gain → ∞ Kv = K = value of asymptotic straight line at ω = 1 =ws0dB =asymptotic straight line 0 dB crossing frequency Ka = 0

  13. Example

  14. The matching phase plot at lowfreq. must be → –90° type = 1 Kp= ∞ ← position error const. Kv = value of low freq. straight line at ω = 1 = 23 dB ≈ 14 ← velocity error const. Ka = 0 ← acc. error const.

  15. N = 2, type = 2 Bode gain plot has –40 dB/dec slope at low freq. Bode phase plot becomes flat at –180° at low freq. Kp= DC gain → ∞ Kv = ∞ also Ka = value of straight line at ω = 1 = ws0dB^2

  16. Example Ka Sqrt(Ka)

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