Why I'm I getting results in terms of t for an and bn even though it is an integration with specific limits?

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Stefanos 2022-12-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1877337-why-i-m-i-getting-results-in-terms-of-t-for-an-and-bn-even-though-it-is-an-integration-with-specific

编辑： Matt J 2023-1-9

%Excercise 4

syms t
syms N
%prompt = "Please enter a function rules for variable segments of the function f: ";
%'Enter 2D array with [] around it and commas between the columns and semicolon at end of each row '
f = input('Please enter function rules for variable segments of the function f:')
flim= input('Please enter real numbers to define the limits of the segments of f:')
N= input('Please enter an intiger to specifie the max order of Fourier coefficients of f:')
%flimFirst= flim(1);
%filmEnd= flim(end);
%a0= 2/(flim(end)-flim(1))*int(f(1),flim(1),flim(2));
%a0= sprintf('%.4f', a0)
%a02= 2/(flim(end)-flim(1))*int(f(2),flim(2),flim(3));
%a02= sprintf('%.4f', a02)
T = flim(end)-flim(1);
w0 = 2*pi/T;
a0= (2/T)*int(f(length(f)),t,flim(length(flim)-1),flim(length(flim)));
sprintf('%.4f', a0)
a02= (2/T)*int(f(1),t,flim(1),flim(2));
sprintf('%.4f', a02)
a03= (2/T)*int(f(2),t,flim(2),flim(3));
sprintf('%.4f', a02)
for i = 1:N
    for k= 2:length(flim)
        an= (2/T)*int((f(k-1))*cos(N*w0*t)),t,flim(k-1),flim(k);
        an(i)=an;
        bn= (2/T)*int((f(k-1))*sin(N*w0*t)),t,flim(k-1),flim(k);
        bn(i)=bn;
        fs= (a0/2)+(an*cos(w0*N*t))+(bn*sin(w0*N*t));
    end
end
an
%sprintf('%.4f', an);
bn
%sprintf('%.4f', bn);
fs
%sprintf('%.4f', fs);

3 个评论
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Stephen23 2022-12-14

编辑：Stephen23 2022-12-14

在 MATLAB Online 中打开

Original question by Stefanos retrieved from Google Cache:

https://web.archive.org/web/20221214200211/https://webcache.googleusercontent.com/search?q=cache:FV1lEzkAZhgJ:https://www.mathworks.com/matlabcentral/answers/1877337-why-i-m-i-getting-results-in-terms-of-t-for-an-and-bn-even-though-it-is-an-integration-with-specific

https://web.archive.org/web/20221214200534/https://www.mathworks.com/matlabcentral/answers/1877337-why-i-m-i-getting-results

"Why I'm I getting results in terms of t for an and bn even though it is an integration with specific limits?"

%Excercise 4
Theme
syms t
syms N
%prompt = "Please enter a function rules for variable segments of the function f: ";
%'Enter 2D array with [] around it and commas between the columns and semicolon at end of each row '
f = input('Please enter function rules for variable segments of the function f:')
flim= input('Please enter real numbers to define the limits of the segments of f:')
N= input('Please enter an intiger to specifie the max order of Fourier coefficients of f:')
%flimFirst= flim(1);
%filmEnd= flim(end);
%a0= 2/(flim(end)-flim(1))*int(f(1),flim(1),flim(2));
%a0= sprintf('%.4f', a0)
%a02= 2/(flim(end)-flim(1))*int(f(2),flim(2),flim(3));
%a02= sprintf('%.4f', a02)
T = flim(end)-flim(1);
w0 = 2*pi/T;
a0= (2/T)*int(f(length(f)),t,flim(length(flim)-1),flim(length(flim)));
sprintf('%.4f', a0)
a02= (2/T)*int(f(1),t,flim(1),flim(2));
sprintf('%.4f', a02)
a03= (2/T)*int(f(2),t,flim(2),flim(3));
sprintf('%.4f', a02)
for i = 1:N
    for k= 2:length(flim)
        an= (2/T)*int((f(k-1))*cos(N*w0*t)),t,flim(k-1),flim(k);
        an(i)=an;
        bn= (2/T)*int((f(k-1))*sin(N*w0*t)),t,flim(k-1),flim(k);
        bn(i)=bn;
        fs= (a0/2)+(an*cos(w0*N*t))+(bn*sin(w0*N*t));
    end
end
an
%sprintf('%.4f', an);
bn
%sprintf('%.4f', bn);
fs
%sprintf('%.4f', fs);

Rena Berman 2022-12-27

(Answers Dev) Restored edit

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回答（1 个）

Torsten 2022-12-13

在 MATLAB Online 中打开

syms t

f = [pi+t,pi-t];

flim= [-pi,0,pi];

N = 5;

T = flim(end)-flim(1);

w0 = 2*pi/T;

fs = 0.0;

for i = 0:N

for k= 2:length(flim)

an= (2/T)*int(f(k-1)*cos(i*w0*t),t,flim(k-1),flim(k));

bn= (2/T)*int(f(k-1)*sin(i*w0*t),t,flim(k-1),flim(k));

fs= fs + an*cos(w0*i*t)+bn*sin(w0*i*t);

end

if i==0

fs = fs/2;

end

fs = matlabFunction(fs);

t = -pi:0.01:pi;

plot(t,fs(t))

15 个评论
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Walter Roberson 2022-12-14

When you have a for loop with : operator, then after the loop, the loop control variable is assigned one of the following:

[] (empty double) if no increment is specified and the terminal value (right side of : operator) is less than the left side of the colon operator
[] (empty double) if positive increment is specified and the terminal value (far right side of : operator) is less than the left side of the colon operator
[] (empty double) if negative increment is specified and the terminal value (far right side of : operator) is greater than the left side of the colon operator
last value assigned by the colon operator if none of the operations inside the loop assign to the loop control variable or if the loo control variable is not assigned to in the last iteration of the loop
last value assigned to the loop control variable if the loop control variable is assigned to in the last iteration of the loop

Or to state more compactly: [] if the loop body was never entered, and otherwise the last value assigned to the loop control variable.

So, with your for i=1:N provided that N >= 1, the value would be the last value assigned to the loop control variable. In that loop, it would be floor(N) .

You then test i==0 after the loop. Is that ever a possibility? Assuming that the disp() calls do not somehow assign anything to i then after the first loop, i is going to be empty (which is not 0) if the loop was never executed at all, and i is going to be a minimum of 1 if N>=1. So i will never be 0 after that first loop.

Exception: if you overwrote disp as a function that does an assignin('caller', 'i', 0) then Yes, in theory i could be 0 after the loop. That would be rather poor programming practice though.