calculate g(k)=integral(-infinite to +infinite)f(x)*(e^(-alxl))*e^(-ikx)
조회 수: 4 (최근 30일)
이전 댓글 표시
Assume f(x+1)=f(x),how to calculate g(k)=integral(-infinite to +infinite)f(x)*(e^(-alxl))*e^(-ikx) in matlab.
Then,input different a and k in matlab to try to find different integral values.
댓글 수: 2
Chunru
2021년 8월 9일
Are you sure the problem is well-defined? f(x+1)=f(x) indicate f(x) is an arbitrary periodic function with period of 1.
John D'Errico
2021년 8월 9일
PLEASE STOP ASKING THIS SAME QUESTION. This is the 9'th time you have asked it.
You are now becoming a spammer on the site.
답변 (1개)
Walter Roberson
2021년 8월 9일
syms x a k
syms f(x)
assume(a>0)
f(x) = sin(2*pi*x) + cos(6*pi*x)
inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)
g(k) = int(inner, x, -inf, inf)
Notice that the limit() portions involve sin(2*pi*x) as x goes to infinity. However, that is not defined, so the limit does not exist, so the integral is not defined.
What about constant f(x) ?
f(x) = 5
inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)
g(k) = int(inner, x, -inf, inf)
expand(rewrite(g,'sincos'))
... and we have the same problem, we have sin() and cos() of something involving x as x goes to infinity.
Your exp(-1i*k*x) is a phase term, and for constant k, as x goes to infinity you get infinite phase, which does not have a limit . The only way that you can get a definite integral in your f(x)*exp(-a*abs(x))*exp(-1i*k*x) expression as x goes to infinity, is if f(x) has a limit of 0. But we know that f(x+1) = f(x), so as x approaches infinity, for the limit f(x) to be 0, limit f(x) would have to be 0, so f(x) would have to be 0.
댓글 수: 13
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!