필터 지우기
필터 지우기

Integration of a function which has limits in terms of parameters

조회 수: 2 (최근 30일)
aakash dewangan
aakash dewangan 2021년 5월 12일
댓글: Walter Roberson 2021년 5월 16일
Hi,
I have a function f3(t), defined as
clc; clear all;
syms t L rho m L n T k G v
N(t) = (T*L/2)*(n*pi/L)^2+k*sin(n*pi*v*t/L);
D(t) = (rho*L/2+m*(sin(n*pi*v*t/L))^2);
alpha(t) = N/D;
i = 6; j = 1
f3(t) = alpha(t)*cos(2*pi*i*t/L)*sin(2*pi*j*t/L);
I3 = int(f3,t,0,L)
when i am integrating the function f3(t) from limits 0 to L, the code is not giving me the output in the form of some expression. The output i am getting is in the form of int(f3,t,0,L), which I dont want. I want output in the form of expression (which will contain t L rho m L n T k G v).
Please help me with this!! Any help will be appreciated..
  댓글 수: 5
aakash dewangan
aakash dewangan 2021년 5월 15일
편집: aakash dewangan 2021년 5월 15일
Walter Roberson, thanks again for your reply and help.
Yes, you are right. I got integration output for a few combinations of parameter values (not for all values). Thanks for help.
"Do you have any suggestion for me to do this (integration) using any other approach / method / technique??"
Thanks,
Walter Roberson
Walter Roberson 2021년 5월 16일
I have no suggestions.
There just might be a change of variables available to make sin(n*pi*v*t/L) linear.
Is n*v known to be integer? If so then that would make a difference in the integration, as sin(n*pi*v*t/L) at t=L would be sin(n*pi*v) and if n*v were integer that would be 0 .

댓글을 달려면 로그인하십시오.

채택된 답변

Jan
Jan 2021년 5월 14일
syms t L rho m L n T k G v
N(t) = (T*L/2)*(n*pi/L)^2+k*sin(n*pi*v*t/L);
D(t) = (rho*L/2+m*(sin(n*pi*v*t/L))^2);
alpha(t) = N/D;
i = 6; j = 1;
f3(t) = alpha(t)*cos(2*pi*i*t/L)*sin(2*pi*j*t/L);
I3 = int(f3, t, [0,L])
I3 = 
As Walter has said already: There is no closed form solution of this integral. You can simplify it only, if you provide specific values for the parameters.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Logical에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by