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Step Response 1.4

1.2

Amplitude

1

0.8

0.6

0.4

0.2

0

0

0.5

1

1.5

Time (sec)

The main issue guiding the design of the above is to avoid the resonant peak at 2 rad/sec. Consequently, the frequency crossover must be > 2 rad/sec. As shown in the bode plot above the bandwidth > 6.61 rad/sec, thus avoiding instability due to oscillation (see the step response above). In part b) we are to add an integrator, which will increase the slope of the system. So, the controller has 2 objectives: 1) Avoid resonant peak at 2 rad/sec by crossing over > 2 rad/sec. 2) Stabilize an increased slope. Since we must get more ‘lift’ out of our lead the simplest approach is to provide 2 sections of the original lead as below. s +a s +b K  s + a  s + a    Lead in part b): D ( s ) =  s  s + b  s + b 

Lead in part a): D ( s ) = K

The matlab code is: function prob_2_8b() % w = 7; a = w/(10);

Solution for Take Home Exam 1  
Solution for Take Home Exam 1  

, a) Design feedback assuming you have access to all the state elements. Place the closed- loop system poles at s = [-1 +- 1j, -0.5 +- 5j]....

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