This page shows the workflow for Fourier and inverse Fourier transforms in Symbolic Math
Toolbox™. For simple examples, see
ifourier. Here, the workflow for Fourier
transforms is demonstrated by calculating the deflection of a beam due to a force. The
associated differential equation is solved by the Fourier transform.
The Fourier transform of f(x) with respect to x at w is
The inverse Fourier transform is
Symbolic workflows keep calculations in the natural symbolic form instead of numeric form. This approach helps you understand the properties of your solution and use exact symbolic values. You substitute numbers in place of symbolic variables only when you require a numeric result or you cannot continue symbolically. For details, see Choose Numeric or Symbolic Arithmetic. Typically, the steps are:
Fourier transform can be used to solve ordinary and partial differential equations. For example, you can model the deflection of an infinitely long beam resting on an elastic foundation under a point force. A corresponding real-world example is railway tracks on a foundation. The railway tracks are the infinitely long beam while the foundation is elastic.
E be the elasticity of the beam (or railway track).
I be the second moment of area of the cross-section of the beam.
k be the spring stiffness of the foundation.
The differential equation is
Define the function
y(x) and the variables. Assume
k are positive.
syms Y(x) w E I k f assume([E I k] > 0)
Assign units to the variables by using
u = symunit; Eu = E*u.Pa; % Pascal Iu = I*u.m^4; % meter^4 ku = k*u.N/u.m^2; % Newton/meter^2 X = x*u.m; F = f*u.N/u.m;
Define the differential equation.
eqn = diff(Y,X,4) + ku/(Eu*Iu)*Y == F/(Eu*Iu)
eqn(x) = diff(Y(x), x, x, x, x)*(1/[m]^4) + ((k*Y(x))/(E*I))*([N]/([Pa]*[m]^6)) == ... (f/(E*I))*([N]/([Pa]*[m]^5))
Represent the force
f by the Dirac delta function δ(x).
eqn = subs(eqn,f,dirac(x))
eqn(x) = diff(Y(x), x, x, x, x)*(1/[m]^4) + ((k*Y(x))/(E*I))*([N]/([Pa]*[m]^6)) == ... (dirac(x)/(E*I))*([N]/([Pa]*[m]^5))
Calculate the Fourier transform of
eqn by using
fourier on both sides of
eqn. The Fourier
transform converts differentiation into exponents of
eqnFT = fourier(lhs(eqn)) == fourier(rhs(eqn))
eqnFT = w^4*fourier(Y(x), x, w)*(1/[m]^4) + ((k*fourier(Y(x), x, w))/(E*I))*([N]/([Pa]*[m]^6)) ... == (1/(E*I))*([N]/([Pa]*[m]^5))
fourier(Y(x),x,w) in the equation.
eqnFT = isolate(eqnFT, fourier(Y(x),x,w))
eqnFT = fourier(Y(x), x, w) == (1/(E*I*w^4*[Pa]*[m]^2 + k*[N]))*[N]*[m]
Y(x) by calculating the inverse Fourier transform of the
right side. Simplify the result.
YSol = ifourier(rhs(eqnFT)); YSol = simplify(YSol)
YSol = ((exp(-(2^(1/2)*k^(1/4)*abs(x))/(2*E^(1/4)*I^(1/4)))*sin((2*2^(1/2)*k^(1/4)*abs(x) + ... pi*E^(1/4)*I^(1/4))/(4*E^(1/4)*I^(1/4))))/(2*E^(1/4)*I^(1/4)*k^(3/4)))*[m]
YSol has the correct dimensions by substituting
eqn and using the
checkUnits returns logical
eqn now has
compatible units of the same physical dimensions.
ans = struct with fields: Consistent: 1 Compatible: 1
Separate the expression from the units by using
YSol = separateUnits(YSol)
YSol = (exp(-(2^(1/2)*k^(1/4)*abs(x))/(2*E^(1/4)*I^(1/4)))*sin((2*2^(1/2)*k^(1/4)*abs(x) + ... pi*E^(1/4)*I^(1/4))/(4*E^(1/4)*I^(1/4))))/(2*E^(1/4)*I^(1/4)*k^(3/4))
Use the values E = 106 Pa, I = 10-3
m4, and k = 106
N/m2. Substitute these values into
YSol and convert to
floating point by using
vpa with 16 digits of accuracy.
values = [1e6 1e-3 1e5]; YSol = subs(YSol,[E I k],values); YSol = vpa(YSol,16)
YSol = 0.0000158113883008419*exp(-2.23606797749979*abs(x))*sin(2.23606797749979*abs(x) + ... 0.7853981633974483)
Plot the result by using
fplot(YSol) xlabel('x') ylabel('Deflection y(x)')
The plot shows that the deflection of a beam due to a point force is highly localized. The deflection is greatest at the point of impact and then decreases quickly. The symbolic result enables you to analyze the properties of the result, which is not possible with numeric results.
YSol is a product of terms. The term with
sin shows that the response is vibrating oscillatory behavior. The
exp shows that the oscillatory behavior is quickly damped by
the exponential decay as the distance from the point of impact increases.