Fourier transform of symbolic function
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Hi, I want to conduct a frequenct analysis using the following code:
clear
syms t
syms y(t) z(t)
syms k m w J I
Dy = diff(y);
Dz = diff(z);
k = 4.5682;
m = 0.2;
w = 0.0014;
J = 0.0327;
I = 0.00362;
odes = [diff(y,2) == (-k/m)*y + (w/m)*z; diff(z,2) == (-J/I)*z + (w/I)*y];
cond1 = y(0) == 0.2;
cond2 = Dy(0) == 0;
cond3 = z(0) == 0;
cond4 = Dz(0) == 0;
conds = [cond1 cond2 cond3 cond4];
qwe = dsolve(odes, conds);
fplot(fft(qwe.y))
fplot(fft(qwe.z))
However, when I am using fft(), I need to use numerical data. Is there a way to conduct a symbolic fourier transform, or do I need to convert the symbolic function to numeric data?
If I do need to convert my symbolic function to numeric data, how do I do this?
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Accepted Answer
Star Strider
on 3 Nov 2021
syms t omega T
syms y(t) z(t)
syms k m w J I
Dy = diff(y);
Dz = diff(z);
k = 4.5682;
m = 0.2;
w = 0.0014;
J = 0.0327;
I = 0.00362;
odes = [diff(y,2) == (-k/m)*y + (w/m)*z; diff(z,2) == (-J/I)*z + (w/I)*y];
cond1 = y(0) == 0.2;
cond2 = Dy(0) == 0;
cond3 = z(0) == 0;
cond4 = Dz(0) == 0;
conds = [cond1 cond2 cond3 cond4];
qwe = dsolve(odes, conds);
wilberforce_y = simplify(qwe.y, 500)
Fy = simplify(fourier(wilberforce_y), 500)
wilberforce_z = simplify(qwe.z, 500)
Fz = simplify(fourier(wilberforce_z), 500)
figure
fplot(wilberforce_y, [0 25])
grid
figure
fplot(fourier(wilberforce_y), [-10*pi 10*pi])
figure
fplot(wilberforce_z, [0 25])
grid
figure
fplot(fourier(wilberforce_z), [-10*pi, 10*pi])
That they are plotted as straight lines at 0 are likely due to the δ funcitons.
Integrating them manually as:
F2y = int(wilberforce_y * exp(1j*omega*t), t, -T, T)
F2y = vpa(simplify(F2y, 500), 5)
F2y = subs(F2y, T, 1)
figure
fplot(F2y, [-10*pi 10*pi])
grid
F2z = int(wilberforce_z * exp(1j*omega*t), t, -T, T)
F2z = vpa(simplify(F2z, 500), 5)
F2z = subs(F2z, T, 1)
figure
fplot(F2z, [-10*pi 10*pi])
grid
produces a more tractible result, without the δ functions.
.
3 Comments
Star Strider
on 3 Nov 2021
As always, my pleasure!
(It’s necessary to add ‘omega’ and ‘T’ to the original syms call.)
More Answers (2)
Paul
on 3 Nov 2021
Replicating the code:
clear
syms t
syms y(t) z(t)
syms k m w J I
Dy = diff(y);
Dz = diff(z);
k = 4.5682;
m = 0.2;
w = 0.0014;
J = 0.0327;
I = 0.00362;
odes = [diff(y,2) == (-k/m)*y + (w/m)*z; diff(z,2) == (-J/I)*z + (w/I)*y];
cond1 = y(0) == 0.2;
cond2 = Dy(0) == 0;
cond3 = z(0) == 0;
cond4 = Dz(0) == 0;
conds = [cond1 cond2 cond3 cond4];
qwe = dsolve(odes, conds);
Check the solutions, using vpa to better see their form:
vpa(qwe.y,5)
vpa(qwe.z,5)
So both soutions are simple sums of two cos functions. The Fourier transform of a cos() is the sum of two Dirac delta functions. So we should expect the Fourier transform of each solution to be the sum of four Dirac delta functions
vpa(fourier(qwe.y),5)
vpa(fourier(qwe.z),5)
which is exactly what they are.
What additional analysis might you be interested in?
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Walter Roberson
on 3 Nov 2021
clear
syms t
syms y(t) z(t)
syms k m w J I
Dy = diff(y);
Dz = diff(z);
k = 4.5682;
m = 0.2;
w = 0.0014;
J = 0.0327;
I = 0.00362;
odes = [diff(y,2) == (-k/m)*y + (w/m)*z; diff(z,2) == (-J/I)*z + (w/I)*y];
cond1 = y(0) == 0.2;
cond2 = Dy(0) == 0;
cond3 = z(0) == 0;
cond4 = Dz(0) == 0;
conds = [cond1 cond2 cond3 cond4];
qwe = dsolve(odes, conds);
wilberforce_y = qwe.y
wyf = fourier(wilberforce_y)
wilberforce_z = qwe.z
wfz = fourier(wilberforce_z)
fplot(abs(wyf))
fplot(abs(wfz))
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