Solving Differential Equations

The Continuous Solution

You might remember using Laplace transforms to solve differential equations; well, now we're going to do something similar with Fourier transforms to solve them.

If I have a differential equation of the form

a₁ y'' + a₂ y' + a₃y = b₁ x'' + b₂ x' + b₃x

then this will transform to

a₁ (jw)² Y(w) + a₂ (jw) Y(w) + a₃ Y(w) = b₁ (jw)² X(w) + b₂ (jw) X(w) + b₃ X(w)

which we can then factorise to

Y(w) / X(w) = [( b₁(jw)² + b₂(jw) + b₃) / (a₁(jw)² + a₂(jw) + a₃)]

Oh look. We now have a transfer function in terms of jw that is very similar to what we would get from Laplace transforms. We can now apply the standard techniques we used in the Electronics lectures last year to work out the gain and phase response of the system. This is nothing new - except that now we can apply it to functions other than simple sinusoidal waves.

An Example: Automotives in Africa

Due to certain geological conditions, a road in Africa develops an uneven surface depth governed by the equation:

i) Show that this can be represented by a Fourier series containing only odd harmonic sine terms, and find an expression for the coefficients.

Since the surface is periodic, and has odd half- and even quarter-wave symmetry, we can conclude that it can be represented by such a series.

To find the Fourier coefficients, we must perform the integral:

If w = 2p/L, this yields

where f = p/4.

Now, the Fourier analysis is done.

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