Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2024 / Paper 3 / 7 / d

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- Solution d

Solution

u_{0} = 0

. For a nonzero integer

r

, integrate the two constant halves:

u_{r} = \frac{1 V}{4 s} [\int_{0}^{2 s} e^{- i r ω t} d t - \int_{2 s}^{4 s} e^{- i r ω t} d t] = \frac{1 - ( - 1 ) ^{r}}{iπ r} V .

(1)

Consequently

u_{0} = 0, u_{r} = {- 2 i / (π r) V, 0, r odd, r \neq = 0 even .

(2)

The complex Fourier series is

g (t) \sim \sum_{r \in Z} u_{r} e^{i r π t / (2 s)}

. The negative-index coefficients are the complex conjugates of the positive-index ones, as required for a real signal. Combining each pair gives the equivalent real expansion

g (t) \sim \frac{4 V}{π} \sum_{j = 0}^{\infty} \frac{s i n [( 2 j + 1 ) π t / ( 2 s )]}{2 j + 1} .

(3)

At a jump, pointwise convergence of a piecewise smooth Fourier series gives zero, the average of the two one-sided limits. The value assigned to

g

at isolated jumps does not affect its Fourier coefficients or mean-square signal power.

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Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2024 / Paper 3 / 7 / d

Solution

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