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Fourier and Inverse Fourier Transforms |
The Fourier transform of a function f(x) is defined as
and the inverse Fourier transform (IFT) as
This documentation refers to this formulation as the Fourier transform of f with respect to x as a function of w. Or, more concisely, the Fourier transform of f with respect to x at w. Mathematicians often use the notation F[f] to indicate the Fourier transform of f. In this setting, the transform is taken with respect to the independent variable of f (if f = f(t), then t is the independent variable; f = f(x) implies that x is the independent variable, etc.) at the default variable w. This documentation refers to F[f] as the Fourier transform of f at w and F–1[f] is the IFT of f at x. See fourier and ifourier in the reference pages for tables that show the Symbolic Math Toolbox commands equivalent to various mathematical representations of the Fourier and inverse Fourier transforms.
For example, consider the Fourier transform of the Cauchy density function, (π(1 + x2))–1:
syms x cauchy = 1/(pi*(1+x^2)); fcauchy = fourier(cauchy)
fcauchy = (pi*exp(-w)*heaviside(w) + pi*heaviside(-w)*exp(w))/pi
fcauchy = expand(fcauchy)
fcauchy = exp(-w)*heaviside(w) + heaviside(-w)*exp(w)
ezplot(fcauchy)
The Fourier transform is symmetric, since the original Cauchy density function is symmetric.
To recover the Cauchy density function from the Fourier transform, call ifourier:
finvfcauchy = ifourier(fcauchy)
finvfcauchy = -(1/(x*i - 1) - 1/(x*i + 1))/(2*pi)
simplify(finvfcauchy)
ans = 1/(pi*(x^2 + 1))
An application of the Fourier transform is the solution of ordinary and partial differential equations over the real line. Consider the deformation of an infinitely long beam resting on an elastic foundation with a shock applied to it at a point. A "real world" analogy to this phenomenon is a set of railroad tracks atop a road bed.
The shock could be induced by a pneumatic hammer blow.
The differential equation idealizing this physical setting is
Here, E represents elasticity of the beam (railroad track), I is the "beam constant," and k is the spring (road bed) stiffness. The shock force on the right side of the differential equation is modeled by the Dirac Delta function δ(x). The Dirac Delta function has the following important property:
A definition of the Dirac Delta function is
where
Let Y(w) = F[y(x)](w) and Δ(w) = F[δ(x)](w). Indeed, try the command fourier(dirac(x), x, w). The Fourier transform turns differentiation into exponentiation, and, in particular,
To see a demonstration of this property, try this
syms w y(x) fourier(diff(y(x), x, 4), x, w)
which returns
ans = w^4*fourier(y(x), x, w)
Note that you can call the fourier command with one, two, or three inputs (see the reference pages for fourier). With a single input argument, fourier(f) returns a function of the default variable w. If the input argument is a function of w, fourier(f) returns a function of t. All inputs to fourier must be symbolic objects.
Applying the Fourier transform to the differential equation above yields the algebraic equation
or
Y(w) = Δ(w)G(w),
where
for some function g(x). That is, g is the inverse Fourier transform of G:
g(x) = F–1[G(w)](x)
The Symbolic Math Toolbox counterpart to the IFT is ifourier. This behavior of ifourier parallels fourier with one, two, or three input arguments (see the reference pages for ifourier).
Continuing with the solution of the differential equation, observe that the ratio
is a relatively "large" number since the road bed has a high stiffness constant k and a railroad track has a low elasticity E and beam constant I. Make the simplifying assumption that
This is done to ease the computation of F –1[G(w)](x). Now type
G = 1/(w^4 + 1024); g = ifourier(G, w, x); g = simplify(g); pretty(g)
and see
1/2 / pi \
2 sin| -- + 4 x | exp(-4 x) heaviside(x)
\ 4 /
------------------------------------------- -
512
1/2 / pi \
2 heaviside(-x) sin| 4 x - -- | exp(4 x)
\ 4 /
-------------------------------------------
512Notice that g contains the Heaviside distribution
Since Y is the product of Fourier transforms, y is the convolution of the transformed functions. That is, F[y] = Y(w) = Δ(w) G(w) = F[δ] F[g] implies
by the special property of the Dirac Delta function. To plot this function, substitute the domain of x into y(x), using the subs command.
XX = -3:0.05:3;
YY = double(subs(g, x, XX));
plot(XX, YY)
title('Beam Deflection for a Point Shock')
xlabel('x'); ylabel('y(x)');The resulting graph
shows that the impact of a blow on a beam is highly localized; the greatest deflection occurs at the point of impact and falls off sharply from there.
The Laplace transform of a function f(t) is defined as
while the inverse Laplace transform (ILT) of f(s) is
where c is a real number selected so that all singularities of f(s) are to the left of the line s = c. The notation L[f] indicates the Laplace transform of f at s. Similarly, L–1[f] is the ILT of f at t.
The Laplace transform has many applications including the solution of ordinary differential equations/initial value problems. Consider the resistance-inductor-capacitor (RLC) circuit below.
Let Rj and Ij, j = 1, 2, 3 be resistances (measured in ohms) and currents (amperes), respectively; L be inductance (henrys), and C be capacitance (farads); E(t) be the electromotive force, and Q(t) be the charge.
By applying Kirchhoff's voltage and current laws, Ohm's Law, and Faraday's Law, you can arrive at the following system of simultaneous ordinary differential equations.
Solve this system of differential equations using laplace. First treat the Rj, L, and C as (unknown) real constants and then supply values later on in the computation.
syms R1 R2 R3 L C real syms I1(t) Q(t) s; dI1(t) = diff(I1(t), t); dQ(t) = diff(Q(t),t); E(t) = sin(t); % Voltage eq1(t) = dI1(t) + R2*dQ(t)/L - (R2 - R1)*I1(t)/L; eq2(t) = dQ(t) - (E(t) - Q/C)/(R2 + R3) - R2*I1(t)/(R2 + R3);
At this point, you have constructed the equations in the MATLAB workspace. An approach to solving the differential equations is to apply the Laplace transform, which you will apply to eq1(t) and eq2(t). Transforming eq1(t) and eq2(t)
L1(t) = laplace(eq1,t,s) L2(t) = laplace(eq2,t,s)
returns
L1(t) = s*laplace(I1(t), t, s) - I1(0) + ((R1 - R2)*laplace(I1(t), t, s))/L - (R2*(Q(0) - s*laplace(Q(t), t, s)))/L L2(t) = s*laplace(Q(t), t, s) - Q(0) - (R2*laplace(I1(t), t, s))/(R2 + R3) - (C/(s^2 + 1) - laplace(Q(t), t, s))/(C*(R2 + R3))
Now you need to solve the system of equations L1 = 0, L2 = 0 for laplace(I1(t),t,s) and laplace(Q(t),t,s), the Laplace transforms of I1 and Q, respectively. To do this, make a series of substitutions. For the purposes of this example, use the quantities R1 = 4 Ω (ohms), R2 = 2 Ω, R3 = 3 Ω, C = 1/4 farads, L = 1.6 H (henrys), I1(0) = 15 A (amperes), and Q(0) = 2 A/sec. Substituting these values in L1
syms LI1 LQ
NI1 = subs(L1(t),{R1,R2,R3,L,C,I1(0),Q(0)}, ...
{4,2,3,1.6,1/4,15,2})returns
NI1 = s*laplace(I1(t), t, s) + (5*s*laplace(Q(t), t, s))/4 + (5*laplace(I1(t), t, s))/4 - 35/2
The substitution
NQ = subs(L2,{R1,R2,R3,L,C,I1(0),Q(0)},{4,2,3,1.6,1/4,15,2})returns
NQ(t) = s*laplace(Q(t), t, s) - 1/(5*(s^2 + 1)) + (4*laplace(Q(t), t, s))/5 - (2*laplace(I1(t), t, s))/5 - 2
To solve for laplace(I1(t),t,s) and laplace(Q(t),t,s), make a final pair of substitutions. First, replace the strings laplace(I1(t),t,s) and laplace(Q(t),t,s) by the sym objects LI1 and LQ, using
NI1 =...
subs(NI1,{laplace(I1(t),t,s),laplace(Q(t),t,s)},{LI1,LQ})to obtain
NI1 = (5*LI1)/4 + LI1*s + (5*LQ*s)/4 - 35/2
Collecting terms
NI1 = collect(NI1,LI1)
gives
NI1 = (s + 5/4)*LI1 + (5*LQ*s)/4 - 35/2
A similar string substitution
NQ = ...
subs(NQ,{laplace(I1(t),t,s), laplace(Q(t),t,s)}, {LI1,LQ})yields
NQ(t) = (4*LQ)/5 - (2*LI1)/5 + LQ*s - 1/(5*(s^2 + 1)) - 2
which, after collecting terms,
NQ = collect(NQ,LQ)
gives
NQ(t) = (s + 4/5)*LQ - (2*LI1)/5 - 1/(5*(s^2 + 1)) - 2
Now, solving for LI1 and LQ
[LI1, LQ] = solve(NI1, NQ, LI1, LQ)
you obtain
LI1 = (5*(60*s^3 + 56*s^2 + 59*s + 56))/((s^2 + 1)*(20*s^2 + 51*s + 20)) LQ = (40*s^3 + 190*s^2 + 44*s + 195)/((s^2 + 1)*(20*s^2 + 51*s + 20))
To recover I1 and Q, compute the inverse Laplace transform of LI1 and LQ. Inverting LI1
I1 = ilaplace(LI1, s, t)
produces
I1 = 15*exp(-(51*t)/40)*(cosh((1001^(1/2)*t)/40) -... (293*1001^(1/2)*sinh((1001^(1/2)*t)/40))/21879) - (5*sin(t))/51
Inverting LQ
Q = ilaplace(LQ, s, t)
yields
Q = (4*sin(t))/51 - (5*cos(t))/51 +... (107*exp(-(51*t)/40)*(cosh((1001^(1/2)*t)/40) +... (2039*1001^(1/2)*sinh((1001^(1/2)*t)/40))/15301))/51
Now plot the current I1(t) and charge Q(t) in two different time domains, 0 ≤ t ≤ 10 and 5 ≤ t ≤ 25. The statements
subplot(2,2,1); ezplot(I1,[0,10]);
title('Current'); ylabel('I1(t)'); grid
subplot(2,2,2); ezplot(Q,[0,10]);
title('Charge'); ylabel('Q(t)'); grid
subplot(2,2,3); ezplot(I1,[5,25]);
title('Current'); ylabel('I1(t)'); grid
text(7,0.25,'Transient'); text(16,0.125,'Steady State');
subplot(2,2,4); ezplot(Q,[5,25]);
title('Charge'); ylabel('Q(t)'); grid
text(7,0.25,'Transient'); text(15,0.16,'Steady State'); generate the desired plots
Note that the circuit's behavior, which appears to be exponential decay in the short term, turns out to be oscillatory in the long term. The apparent discrepancy arises because the circuit's behavior actually has two components: an exponential part that decays rapidly (the "transient" component) and an oscillatory part that persists (the "steady-state" component).
The (one-sided) z-transform of a function f(n) is defined as
The notation Z[f] refers to the z-transform of f at z. Let R be a positive number so that the function g(z) is analytic on and outside the circle |z| = R. Then the inverse z-transform (IZT) of g at n is defined as
The notation Z–1[f] means the IZT of f at n. The Symbolic Math Toolbox commands ztrans and iztrans apply the z-transform and IZT to symbolic expressions, respectively. See ztrans and iztrans for tables showing various mathematical representations of the z-transform and inverse z-transform and their Symbolic Math Toolbox counterparts.
The z-transform is often used to solve difference equations. In particular, consider the famous "Rabbit Problem." That is, suppose that rabbits reproduce only on odd birthdays (1, 3, 5, 7, ...). If p(n) is the rabbit population at year n, then p obeys the difference equation
p(n+2) = p(n+1) + p(n), p(0) = 1, p(1) = 2.
You can use ztrans to find the population each year p(n). First, apply ztrans to the equations
syms p(n) z eq = p(n + 2) - p(n + 1) - p(n); Zeq = ztrans(eq, n, z)
to obtain
Zeq = z*p(0) - z*ztrans(p(n), n, z) - z*p(1) + z^2*ztrans(p(n), n, z) - z^2*p(0) - ztrans(p(n), n, z)
Next, replace ztrans(p(n), n, z) with Pz and insert the initial conditions for p(0) and p(1).
syms Pz
Zeq = subs(Zeq,{ztrans(p(n), n, z), p(0), p(1)}, {Pz, 1, 2})to obtain
Zeq = Pz*z^2 - z - Pz*z - Pz - z^2
Collecting terms
eq = collect(Zeq, Pz)
yields
eq = (z^2 - z - 1)*Pz - z^2 - z
Now solve for Pz
P = solve(eq, Pz)
to obtain
P = -(z^2 + z)/(- z^2 + z + 1)
To recover p(n), take the inverse z-transform of P.
p = iztrans(P, z, n); p = simplify(p)
The result is a bit complicated, but explicit:
p = (3*5^(1/2)*(1/2 - 5^(1/2)/2)^(n - 1))/5 -... (3*5^(1/2)*(5^(1/2)/2 + 1/2)^(n - 1))/5 +... (4*(-1)^n*cos(n*(pi/2 + asinh(1/2)*i)))/i^n
Finally, plot p:
m = 1:10;
y = double(subs(p,n,m));
plot(m, real(y),'rO')
title('Rabbit Population');
xlabel('years'); ylabel('p');
grid onto show the growth in rabbit population over time.
[1] Andrews, L.C., Shivamoggi, B.K., Integral Transforms for Engineers and Applied Mathematicians, Macmillan Publishing Company, New York, 1986
[2] Crandall, R.E., Projects in Scientific Computation, Springer-Verlag Publishers, New York, 1994
[3] Strang, G., Introduction to Applied Mathematics, Wellesley-Cambridge Press, Wellesley, MA, 1986
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