How to compute control system's performance parameters?

Hello everyone.

J = 0.2;
b = 0.1;
K = 0.2;
R = 10;
L = 5;
s = tf('s');
p = K/((J*s+b)*(L*s+R)+K^2);
step(p,200)
[y,t]=step(p,200);
stepinfo(y)

With the following code, I want to measure the rise time. Using the stepinfo command, this is what pops in my command window :

ans = 

  struct with fields:

        RiseTime: 94.5571 %%Rise time
    SettlingTime: 172.5924
     SettlingMin: 0.1735
     SettlingMax: 0.1923
       Overshoot: 0
      Undershoot: 0
            Peak: 0.1923
        PeakTime: 1378

But when I check the parameters using the step response graph, the answer is different.

So, why am I getting two different rise times? Any solution?

Thanks in advance!

[Muhammad]

%In the following program, peak is crossing the limit "2" where as it should be below "2".

clc

clear

N=zeros(1,2000);

K=zeros(1,2000);%redefined N & K to fix the error

Gr=[];

Ga=[];

e0=0.2;

gamass=0.05;

gamaL=0.05;

numda=.1;

numdas=.1;

delta=1;

em=0.1;

po=1;

fid=fopen('file.xls','a');

for n=1:2000

e=-0.5+n*1/2000;

N(1,n)=e;

A=[e-e0, 0,numdas/sqrt(2),numdas/sqrt(2);

0, e+e0, -numda/sqrt(2),-numda/sqrt(2);

numdas/sqrt(2), -numdas/sqrt(2), e-em, 0;

numdas/sqrt(2), -numdas/sqrt(2), 0, e+em;];

B=-(1i/2)*[gamaL,0,0,0;

0,gamaL,0,0;

0,0,0,0;

0,0,0,0;];

if abs(e)>delta;

ru=abs(e)/sqrt(delta*delta-e*e);

elseif abs(e)==delta

ru=1

else

ru=abs(e)/(i*sqrt(delta*delta-e*e));

end

ru=abs(e)/1i*sqrt(delta*delta-e*e);

C=-(1i/2)*[1,-delta/e,0,0;

-delta/e,1,0,0;

0,0,0,0;

0,0,0,0;];

D=gamass*ru*C;

F=B+D;

E=A-F;

Gr=(inv(E));

Ga=conj(Gr);%Gr';

G= Gr(1,2)*gamaL*Ga(2,1)*gamaL+...

Gr(1,4)*gamaL*Ga(4,1)*gamaL+ ...

Gr(3,2)*gamaL*Ga(2,3)*gamaL+ ...

Gr(3,4)*gamaL*Ga(4,3)*gamaL;

K(1,n)=G;

fprintf(fid,'%10.6f %10.6f\n',e ,G);

end

fclose(fid);

figure

plot(N,K);

hold on

axis([0.1,0.5,0.0,2.0]);

xlabel('e');

ylabel('G');

Answer 1

[Star Strider]

You are not calling stepinfo correctlly.

stepinfo(p) 

produces:

ans = 

  struct with fields:

        RiseTime: 4.4144
    SettlingTime: 8.0108
     SettlingMin: 0.1735
     SettlingMax: 0.1923
       Overshoot: 0
      Undershoot: 0
            Peak: 0.1923
        PeakTime: 23.2961

.

Answer 2

[Paul]

stepinfo(y,t)

If you want to use the output from step. Note that stepinfo using the transfer function gives a peculiar result for PeakTime:

>> sy=stepinfo(y,t)

sy = 

  struct with fields:

        RiseTime: 4.414431198715340e+00
    SettlingTime: 8.010849818943374e+00
     SettlingMin: 1.734586244553206e-01
     SettlingMax: 1.923076923076916e-01
       Overshoot: 0
      Undershoot: 0
            Peak: 1.923076923076916e-01
        PeakTime: 6.428571428571429e+01

>> sp=stepinfo(p)

sp = 

  struct with fields:

        RiseTime: 4.414430957167118e+00
    SettlingTime: 8.010849940068825e+00
     SettlingMin: 1.734589587177580e-01
     SettlingMax: 1.923064745779809e-01
       Overshoot: 0
      Undershoot: 0
            Peak: 1.923064745779809e-01
        PeakTime: 2.329614202821439e+01
        

PeakTime from stepinfo(y,t) is just taking the time at the first value that is the maximum of y. Maybe it should take the time at the last value that is the max?

>> [ymax,imax]=max(y);
>> t(imax)

ans =

     6.428571428571429e+01

>> all(y(imax:end)==ymax)

ans =

  logical

   1       
   

It appears that stepinfo(p) works by first generating a step response yy for some time vector tt and then essentially calling stepinfo(yy,tt). In this case, the time vector used by stepinfo(yy,tt) ended at the PeakTime, which also corresponds to max(yy).

I'm suprised by this result for this p. I would've thought that stepinfo would catch the case of all real, negative poles and then just compute the peak value as the final value (which can be determined without generating yy) and assigning the PeakTime = inf.

Answer 3

[Muhammad]

how to control conductance below "2" along yaxis in matlab program a normal-metal/quantum-dot/superconductor system with coupled Majorana bound states? I am working on it from six months but failed.

Answer 4

[Muhammad]

%In the following program, peak is crossing the limit "2" where as it should be below "2".

clc

clear

N=zeros(1,2000);

K=zeros(1,2000);%redefined N & K to fix the error

Gr=[];

Ga=[];

e0=0.2;

gamass=0.05;

gamaL=0.05;

numda=.1;

numdas=.1;

delta=1;

em=0.1;

po=1;

fid=fopen('file.xls','a');

for n=1:2000

e=-0.5+n*1/2000;

N(1,n)=e;

A=[e-e0, 0,numdas/sqrt(2),numdas/sqrt(2);

0, e+e0, -numda/sqrt(2),-numda/sqrt(2);

numdas/sqrt(2), -numdas/sqrt(2), e-em, 0;

numdas/sqrt(2), -numdas/sqrt(2), 0, e+em;];

B=-(1i/2)*[gamaL,0,0,0;

0,gamaL,0,0;

0,0,0,0;

0,0,0,0;];

if abs(e)>delta;

ru=abs(e)/sqrt(delta*delta-e*e);

elseif abs(e)==delta

ru=1

else

ru=abs(e)/(i*sqrt(delta*delta-e*e));

end

ru=abs(e)/1i*sqrt(delta*delta-e*e);

C=-(1i/2)*[1,-delta/e,0,0;

-delta/e,1,0,0;

0,0,0,0;

0,0,0,0;];

D=gamass*ru*C;

F=B+D;

E=A-F;

Gr=(inv(E));

Ga=conj(Gr);%Gr';

G= Gr(1,2)*gamaL*Ga(2,1)*gamaL+...

Gr(1,4)*gamaL*Ga(4,1)*gamaL+ ...

Gr(3,2)*gamaL*Ga(2,3)*gamaL+ ...

Gr(3,4)*gamaL*Ga(4,3)*gamaL;

K(1,n)=G;

fprintf(fid,'%10.6f %10.6f\n',e ,G);

end

fclose(fid);

figure

plot(N,K);

hold on

axis([0.1,0.5,0.0,2.0]);

xlabel('e');

ylabel('G');

why the peak is crossing 2 limit along y-axis? while matrics are correct?

So, why am I getting peak upto 4.5? Any solution, please guide?

Thanks in advance you guys