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Definite integral with complex number

Asked by Pilar Jiménez on 17 Jan 2017
Latest activity Commented on by Star Strider
on 18 Jan 2017
I need to solve an integral of the function f=(wn.*t).*exp(1i.*2.*pi.*t) where wn=1 and have complex numbers, it is a definite integral of 0 to 0.3 and I have to obtain a numeric answer. I tried to do with the comands int and integral but doesn`t help me because this comands uses symbolics variables and I need numeric answers, and I tried to obtain the sum under the wave but I couldn`t solve it. Can someone help me?

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2 Answers

Answer by Star Strider
on 18 Jan 2017

Use the vpa function:
syms wn t
wn = sym(1);
f = (wn.*t).*exp(1i.*2.*pi.*t);
f_int = int(f, t, 0, 0.3)
f_int_num = vpa(f_int)
f_int =
- 1/(4*pi^2) - (((pi*3i)/5 - 1)*(1/4 + (2^(1/2)*(5^(1/2) + 5)^(1/2)*1i)/4 - 5^(1/2)/4))/(4*pi^2)
f_int_num =
0.012251815898938149373515863015179 + 0.038845017631697804582142824751429i

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I checked with my professor and I have a mistake, the function is incorrect. Wn is not multiplied by t, it is in function of t. I mean, the correct expression is wn(t) not wn.*t, do you know how I can register that in the code?
My pleasure.
Please let me know.
To define ‘wn(t)’ as uniformly equal to 1 definitely changes the result:
syms wn t u_lim wn(t)
wn(t) = sym(1);
f = wn*exp(1i*2.*pi*t);
upper_limit = vpasolve(abs(int(f, t, 0, u_lim)) == 0.3, u_lim)
abs_upper_limit = abs(upper_limit)
upper_limit =
61.162453359143770665259861917249 - 0.1257699093208763500021861131513i
abs_upper_limit =
61.162582670939654308556507428042
Is the rest correct? Are you solving for the upper limit of integration that will make the integral equal to 0.3? If so, this works.

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Answer by David Goodmanson on 18 Jan 2017
Edited by David Goodmanson on 18 Jan 2017

Hello Diana, symbolic variables are a great thing, but if you are looking for a numerical result and are happy with 15 or so sigfigs, it isn't like they have to be invoked. You can just do
ff = @(t,wn) (wn.*t).*exp(1i.*2.*pi.*t) % or you could define this in an mfile
integral(@(t) ff(t,1),0,.3) % pass in wn =1
format long
ans = 0.012251815898938 + 0.038845017631698i
Now that symbolic variables are much better integrated into Matlab, sometimes I wonder if they are getting overused.

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Hi Diana, I think it's time to talk to your prof because it seems like there must be some kind of mistake. You and two of us on this site have all gotten the same answer, given the integrand and the integration limits that were specified.
Yes, I will share with him to check it. Thanks for you support.
I checked with my professor and I have a mistake, the function is incorrect. Wn is not multiplied by t, it is in function of t. I mean, the correct expression is wn(t) not wn.*t, do you know how I can register that in the code?

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