Bode Plots. Hamid Roozbahani

Transcription

1 Bode Plots Hamid Roozbahani A Bode plot is a graph of the transfer function of a linear, time-invariant system versus frequency, plotted with a logfrequency axis, to show the system's frequency response. It is usually a combination of a Bode magnitude plot, expressing the magnitude of the frequency response gain, and a Bode phase plot, expressing the frequency response phase shift. [1]

2 Bode diagram Let us assume that transfer function can be divided into parts Thus the frequency response of the whole G can be calculated as a sum of the frequency responses of G 1 G n Magnitude Phase The idea in Bode s method is to plot magnitude curves using a logarithmic scale and phase curves using a linear scale. This strategy allows us to plot a high order G (jw) by simply adding the separate terms graphically Advantages of Bode plots 1. Dynamic compensator design can be based entirely on bode plots 2. Bode plots can be determined experimentally 3. Bode plots of system in series simply add which is quite convenient 4. The use of a log scale permits a much wider range of frequencies to be displayed on the single plot than is possible with linear scales

3 Bode diagram of constant G(s)=K,G(s)=1/s, G(s)=1/s 2 or s

4 Open-loop bode diagram K m =gain margin Φ m =phase margin If the both are positive the system is stable Recommended values Closed loop Bode-diagram M p =resonance peak < db (recommended for servo systems) Relation between resonance peak and overshoot The bandwidth of the system is a measure of speed of response. For control system it is defined as the frequency corresponding to (3db) in closed-loop magnitude Bode plot.

5 Calculation of responses by Matlab The open-loop transfer function of a servo is The bode diagram is calculated by typing NUM= [30] (nominator polynomial coefficients) DEN= [2.3e-4 1.2e-2 1 0] (denominator coefficients) Sys=tf (NUM, DEN) (calculates transfer function) Margin (sys) (calculates and plots Bode-diagram with gain and phase margins)

6 With direct feedback H(s)=1, the closed loop transfer function is This in Matlab is NUM=[30] DEN=[2.3e-4 1.2e ] bode(num,den) sys=tf(num,den) (calculates Bode-diagram without margins) Closed-loop frequency response

7 -If in the open-loop Bode-diagram the gain margin K m is bigger than 6dB and phase margin Φ m bigger than 45 o increase the gain until one reaches the limit -If K m or Φm or both are smaller than required the stability of servo is insufficient => increase the cylinder and valve or apply better control method -If 3dB bandwidth in closed-loop Bode-diagram is smaller than required ω -3dB the gain must be increased. If stability becomes insufficient => increase the cylinder and valve or apply better control method [3] Hand Sketching: Step-by-step approach 1- Put transfer function in ZPK form (factored zeros and poles, with a constant multiplier K out front). 2- Identify breakpoints: distance of poles and zeros from the origin. Mark those on the frequency axis of the plot. Remember to convert from rad/sec to Hz. 3- Determine low or high frequency constant asymptote of gain by taking the limit of H(s) as s 0 or infinity, respectively. Convert to db. 4- Start at one of the asymptotes that is constant. (see below if neither is constant). Move along in frequency until you get to a breakpoint. Each breakpoint is associated with a change in slope of +/-20 db/decade (+/-6 db/octave). From left to right, a zero produces an increase in slope (The increase could be from negative to less negative, or from positive to more positive, etc.) Each pole produces a decrease in slope. Work through all the breakpoints, and check that the final asymptote is correct. 5- Sketch in a smoother curve, 3 db below or above each breakpoint (unless it is a double pole or zero in which case it is 6 db below or above).

8 6- Now fill in the phase by the same procedure: Find the phase on the asymptotes by looking at the limit of H(jω) as ω 0 or ω infinity. If the limit is real, the phase is 180 degrees or zero. If the limit is imaginary, the phase is +/-90 degress (-90 for a negative imaginary limit). If the limit is zero, you have to look at what direction it came in to zero from. If it came from a positive imaginary number, the phase is 90 degree; a negative imaginary number -90 degrees. If it came from a positive real number, the phase is 0, and if it came from a negative real number, the phase is -90 Then from left to right, each pole causes a 90 transition in phase, and every zero causes a +90 transition in phase. You can use the rules for the slope of the transition, but it s usually not worth the trouble to get that exact. 7- OK, what if neither asymptote is flat? For example H(s) = s/[(s+1)(s+100)]: HF limit: s/100 -> 0 LF limit: s/(s2) -> 0 Solution: Consider the shape it will look like: With breakpoints at 1/(2π) and 100/(2π) Instead of starting at HF or LF asymptotes, you can start with the flat center part. Consider the approximate value there, where s>>1 and s<<100. Then we can write the transfer function, putting the negligible stuff in a small font: H(s) = s/[(s+1)(s+100)] Then. Thus, we can approximate H(s) as s/[100s] = 1/100 or 40 db in the flat middle part. Then we can start there and sketch the rest. (It slopes down with +/- 20 db/decade slope on either side of the plateau.) [4]

9 Bode Plots by hand 1. Bode Plots by Hand

10 Bode Phase Plots

11 [2]

12 Also you can find helpful information regarding Bode plots in below web address too: References: [1] Wikipedia- Bode plot- [2] Original EE 105 Discussion Notes from Meghdad Hajimorad ( Amin )- Last Modified by: Bill Hung- Date: 5 August 2006 [3] Course material by Prof. Huapeng Wu [4] Engineering Sciences 22 Systems Summer Bode Plots Page 1-BODE PLOTS