Problem with a function

조회 수: 1 (최근 30일)
Paul Rogers
Paul Rogers 2020년 10월 21일
댓글: Paul Rogers 2020년 10월 24일
Hello everyone, I wrote this function:
function dz = nocontrol(v,z,parameters)
gammaT=??????????;
phi_0 =0.6;
psi_0 =0.6857;
psi_c0 = 0.3;
B=1.8;
Lc = 3; %m
W = 0.25;
H = 0.18;
C = 0;
dz = zeros(2,1);
dz(1)=(1/(4*B*B*Lc))*(z(2)-gammaT*(z(1))^0.5);
psi_c=psi_c0+H*(1+1.5*(z(2)/W-1)-0.5*(z(2)/W-1).^3);
dz(2) =(1/Lc)*(psi_c-z(1));
end
my problem is that I don't know how to write properly gammaT. I'd like to express it as a function, something like this:
gammaT_max = 0.8;
gammaT_min = 0.7;
A = (gammaT_max - gammaT_min)/2; %amplitude
b = gammaT_max - ((gammaT_max - gammaT_min)/2);
gammaT = A*sin(w*t)+b
but I don't know how.
  댓글 수: 2
madhan ravi
madhan ravi 2020년 10월 21일
what's your question? Are you asking if the way you defined the gamma function is correct?
Paul Rogers
Paul Rogers 2020년 10월 21일
편집: Paul Rogers 2020년 10월 21일
I'd like to define gammaT as a function of the time, but I don't knwo how.

댓글을 달려면 로그인하십시오.

채택된 답변

Walter Roberson
Walter Roberson 2020년 10월 21일
You have defined gammaT as a variable that you multiply by, but it needs to be a function of time. You will likely need something like
gammaT = @(t) A*sin(w*t)+b
dz(1)=(1/(4*B*B*Lc))*(z(2)-gammaT(SOMETHING)*(z(1))^0.5);
The SOMETHING would have to be replaced with your time variable, but it is not obvious that you have a time variable. Perhaps v is your time variable.
  댓글 수: 3
Walter Roberson
Walter Roberson 2020년 10월 22일
You do not have a time variable, so I had to substitute the closest I could find.
function dz = nocontrol(v,z,parameters)
gammaT=parameters(1);
phi_0=parameters(2);
psi_0=parameters(3);
psi_c0=parameters(4);
B=parameters(5);
Lc=parameters(6);
W=parameters(7);
H=parameters(8);
C = 0;
gammaT_max = 0.8;
gammaT_min = 0.7;
A = (gammaT_max - gammaT_min)/2; %amplitude
b = gammaT_max - ((gammaT_max - gammaT_min)/2);
gammaT = @(t) A*sin(W*t)+b
dz = zeros(2,1);
dz(1)=(1/(4*B*B*Lc))*(z(2)-gammaT(v)*(z(1))^0.5);
psi_c=psi_c0+H*(1+1.5*(z(2)/W-1)-0.5*(z(2)/W-1).^3);
dz(2) =(1/Lc)*(psi_c-z(1));
end
Paul Rogers
Paul Rogers 2020년 10월 24일
thanks a lot, you really solved me something I couldn't figured out.
just the argument in the sin was't W but another variable w in radians.
It woorks now

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

제품


릴리스

R2014b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by