D-11 - undefined because the steady state error to a unit...

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where Kp is the position error constant. Kp = L(z)| z=1                  K * 0.0020061 (z+1.339) (z+0.01679) L(z)| z=1  =  ------------------------------------------    = ∞                   (z-1) (z-0.8763) (z-4.54e-005) e(∞) = 1  = 0          1+∞ The steady state error to a unit step in terms of gain K is zero. The steady state error to a unit ramp in terms of gain K is given as: e(∞) = 1            Kv where Kv is the velocity error constant. Kv = 1  (z-1) L(z)| z=1 , T=0.1s         T             K * 0.0020061 (z+1.339) (z+0.01679) Kv = ------------------------------------------       = 0.3856 K              (0.1) (z-0.8763) (z-4.54e-005) e(∞) = 1/0.356K = 2.593/K is the steady state error due to ramp input in terms of  gain K. The value of K which yields a 5% steady state error to a unit step input is 
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Unformatted text preview: undefined because the steady state error to a unit step is zero. The value of K which yields a 5% steady state error to a unit ramp is given as: 5% = 2.593/K K = 51.86 Closed Loop Controller The given time domain specifications are a 4% overshoot and 5 second settling time (assumed 2%) for a unit step input. Additionally the steady state error due to a ramp is equal to 1%. Using the percent overshoot we are able to determine the damping ration for the system. And ζ using the settling time we are able to calculate the natural frequency n ω . Then using n and we are able to ω ζ calculate the desired pole locations. PO=.04=exp((-pi* )/(1-ζ ^2)) => =.7156. ................................... 26 ζ ζ Ts2% = 5 = 4/( n* ω )=4/( n*.7156) => n ζ ω ω =1.1179. ..................... 27...
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D-11 - undefined because the steady state error to a unit...

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