Frequency Resp. method Given: G(s) 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 y(s) u(s) G(s)

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

Relationship between bode and nyquist length vector

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: only if system is stable Select Give input to system: Adjust A1 so that the output is not saturated or distorted. Measure amp B1 and phase φ1 of output: u(s) y(s) System

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

y(s) u(s) G1(s) G2(s) Product of T.F. G(s)

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 evaluated 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)  -∞

System type on Nyquist plot

System relative order

Examples System type = Relative order = System type = Relative order =

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