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% Now loading library
>load "user\control.e"
% Draw the tribute of (s+2)
>pbode([2,1],1);
% And inv(s+2)
>pbode(1,[5,1]);
% if we want to draw W(s)=(s+2)/(s+5) we simply must add
% the two tribute
>pbode([2,1],[5,1]);
% Into a term of the type (s+a) we can find the break
% point:
%
% a(s/a+1) 1/a is called break point
>pbode([1,1],1];
% in (s+1) break point is 1, we can see that a decade before
% and after bode's diagrams of the magnitude is linear.
% On a log-log paper of course.
>pbode([1,1],1,expi=-4,expf=4);
% In the diagrams we can look this better. For the phase
% diagram we can to notice that a decade before phase
% = 0 and after=90
>n=polycons([1,1]);
>pbode(n,1,expi=-4,expf=4);
% This is basically the method for plot asintotic Bode's
% diagrams. We simply add the slope of the single tribute
% for the magnitude and the tribute to the phase of each
% terms.
>n2=polymult(n,polycons([1000,1000,1000,1000]));
>pbode(n2,1,expi=-3,expf=6);
% There are 4 different tribute:
%
% (s+a) we have alredy noticed it
%
% (s-a) is like s+a in magnitude but not in phase
>pbode([-1,1],1);
>pbode([1,1],1);
% and they reciprocal
>pbode(1,[1,1]);
>pbode(1,[-1,1]);
% Now we can to examine the root's place. From a given
% W(s)=N(s)/D(s) we can exctract the poly:
%
% P(s,k)=N(s)+kD(s)
%
% And when the k change its value the roots change places.
% The root's place is the plot of the roots.
%
% For EXIT click out of the graph. If you click ON the
% graph you can get information about the value of k that
% permits the configuration.
>place(polycons([1,2,8]),polycons([6,3,7]));
%
%
% Some examples from: Esercizi di controlli automatici,
% ed sidera.
%
%
%
% III.4
>pbode(1,polycons([0,-2,-4]),-1,2);
>place(1,polycons([0,-2,-4]));
%
%
% III.5
>pbode([-0.5, 1],polycons([0,-1,-5]));
>place([-0.5, 1],polycons([0,-1,-5]));
%
%
% III.7
>pbode([2,1],polycons([-1,1,pbode([2,1],polycons([-1,1,1]));
>place([2,1],polycons([-1,1,pbode([2,1],polycons([-1,1,1]));
%
%
% III.8
>pbode([1, 0, 1],polycons([2,-2,4]),pt=300); .. 300 campioni!
>place([1, 0, 1],polycons([2,-2,4]));
%
%
% III.9
>pbode([1,0,1],polycons([0,-2,-4]),pt=600); .. 600 campioni!
>place([1,0,1],polycons([0,-2,-4]));
%
%
% III.10
>pbode([-2,1],polymult([2,1],[1,0,1]));
>place([-2,1],polymult([2,1],[1,0,1]));
%
%
% III.11
>pbode(1,polycons([0,2,2]));
>place(1,polycons([0,2,2]));
%
%
% III.12
>pbode(1,polymult([1,0,1],[2,1]));
>place(1,polymult([1,0,1],[2,1]));
%
%
% III.13
>pbode([-1,1],polymult(polycons(0),[1,0,1]));
>place([-1,1],polymult(polycons(0),[1,0,1]));
%
%
% III.14
>pbode([1,1],polymult([2,1],[5,4,1]));
>place([1,1],polymult([2,1],[5,4,1]));
%
%
% III.15
>pbode(polymult([3,1],[2,2,1]),polycons([0,1,-2,-4]));
>place(polymult([3,1],[2,2,1]),polycons([0,1,-2,-4]));
%
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