naver-site-verification: naver47238a1b6bfbb19a2fd4b619734fa9a6.html Bode diagram program page 1 | Ti84-plus programs
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The Bode diagram with the Ti-84 Plus (CE) page 1

 Explanation of the Bode diagram program : P-control

The Bode diagram is useful for determining the stability of a control system  (phase margin  and  gain margin). It is an important subject in Electrical and mechanical Engineering. With this Ti-84 Basic program, the Bode diagram is plotted to determine for example the proportional gain for a desired phase margin or gain margin. As a case, a speed control system is analyzed, of which the proportional gain for a phase margin of 60 degrees has to be determined. This phase and the gain margin are determined of an open loop system, See the figure below. The example is related to a real existing drive system at Avans University. 

Control system to be analyzed with the Bode Program of the Ti-84
Open loop transfer function

For Kp=1 the Bode diagram of Hopen is plotted, Kasyn=2.071Nm/A, J=0.018887kgm^2,
tauuni=10msec , tautacho=1msec.

The Bode Program menu has 5 options.

1: New transfer function open loop= Hs

2: The open loop function * Controller = H(s)*Hr 

3: existing model is used for adapting the scaling of  the axes

4: Closed loop =(Hs*Hr/(1+Hs*Hr)

5: Stop​

Start of the Bode program on your Ti for open loop
Transfer input function with open loop Bode results

Open loop transfer function. Green= amplitude, Red=phase/10 (to fit in the same screen as the amplitude). At 0 dB the phase=-13.48*10=-134.8 degrees. Which is a phase margin of 180-134.8=45.2 degrees

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Improved phase margin by adjusting the proportional gain
New Bode diagram with adjusted proportional gain to achieve the right phase marginfor

To achieve 0dB of total gain, the gain Kp should be -5.65 dB =10^(-5.65/20)=0.52. The corresponding frequency equals 51.12 rad/sec. 
If a P-control is added by inserting Hr(s)=0.52, and we run the program Bode again, then 0dB gives a phase margin of 180-120-= 60deg phase margin. See figures below. Remember that the real phase has to be multiplied by 10 (12*10=120deg). The corresponding frequency  =  51.12 rad/sec, which = 10^1.71  

Bode plot results
Check wit Simulink of the Ti results

A check with Simulink (Matlab) gives the same result, for Kp=0.52 there is a phase margin of 60.1 degrees at a corresponding radian frequency of 51 rad/sec

Do you want to adapt this Ti-Basic program ? 

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