Frequency Resp. 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 ∞, the plot of G(jω) is called the Nyquist plot

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

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 Matlab command to explore: nyquist, bode

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

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

Product of T.F.

System type, steady state tracking, & Bode plot R(s) C(s) Gp(s) Y(s)

As ω → 0 Therefore: gain plot slope = –20N dB/dec. phase plot value = –90N deg

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

Example

Steady state tracking error Suppose the closed-loop system is stable: If the input signal is a step, ess would be = If the input signal is a ramp, If the input signal is a unit acceleration,

N = 1, type = 1 Bode mag. plot has –20 dB/dec slope at low freq. (ω→0) (straight line with slope = –20 as ω→0) 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’s 0 dB crossing frequency Ka = 0

Example Asymptotic straight line ws0dB ~14

The matching phase plot at low freq. 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.

Steady state tracking error Suppose the closed-loop system is stable: If the input signal is a step, ess would be = If the input signal is a ramp, If the input signal is a unit acceleration,

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

Example Ka ws0dB=Sqrt(Ka) How should the phase plot look like?

Steady state tracking error Suppose the closed-loop system is stable: If the input signal is a step, ess would be = If the input signal is a ramp, If the input signal is a unit acceleration,

System type, steady state tracking, & Nyquist plot C(s) Gp(s) As ω → 0

Type 0 system, N=0 Kp=lims0 G(s) =G(0)=K Kp w0+ G(jw)

Type 1 system, N=1 Kv=lims0 sG(s) cannot be determined easily from Nyquist plot winfinity w0+ G(jw)  -j∞

Type 2 system, N=2 Ka=lims0 s2G(s) cannot be determined easily from Nyquist plot winfinity w0+ G(jw)  -∞

In most cases, stability of this closed-loop Margins on Bode plots In most cases, stability of this closed-loop can be determined from the Bode plot of G: Phase margin > 0 Gain margin > 0 G(s)

If never cross 0 dB line (always below 0 dB line), then PM = ∞. If never cross –180° line (always above –180°), then GM = ∞. If cross –180° several times, then there are several GM’s. If cross 0 dB several times, then there are several PM’s.

Example: Bode plot on next page.

Example: Bode plot on next page.

Where does cross the –180° line Answer: __________ at ωpc, how much is Closed-loop stability: __________

crosses 0 dB at __________ at this freq, Does cross –180° line? ________ Closed-loop stability: __________

Margins on Nyquist plot Suppose: Draw Nyquist plot G(jω) & unit circle They intersect at point A Nyquist plot cross neg. real axis at –k