Question

f(t) A t (ms) 0 -4 -3 -2 -1 1 2 3 4 -A- What are the cosine Fourier series coefficients, $a_n$? a) $\frac{6}{n^2 \pi^2} [1 - \cos(n\pi)]$ b) $| \frac{20}{n\pi} \cos(\frac{2n\pi}{3}) - \frac{30}{n^2 \pi^2} \sin(\frac{2n\pi}{3}) |$ c) $\frac{20}{n^2 \pi^2} [\cos(n\pi) - 1]$ d) 0 e) $\frac{-30}{n^2 \pi^2} \sin(\frac{2n\pi}{3}) + \frac{20}{n\pi} \cos(\frac{2n\pi}{3})$ f) $\frac{40}{n^2 \pi^2}$

          f(t)
A
t (ms)
0
-4 -3 -2 -1 1 2 3 4
-A-
What are the cosine Fourier series coefficients, $a_n$?
a) $\frac{6}{n^2 \pi^2} [1 - \cos(n\pi)]$
b) $| \frac{20}{n\pi} \cos(\frac{2n\pi}{3}) - \frac{30}{n^2 \pi^2} \sin(\frac{2n\pi}{3}) |$
c) $\frac{20}{n^2 \pi^2} [\cos(n\pi) - 1]$
d) 0
e) $\frac{-30}{n^2 \pi^2} \sin(\frac{2n\pi}{3}) + \frac{20}{n\pi} \cos(\frac{2n\pi}{3})$
f) $\frac{40}{n^2 \pi^2}$

f(t)
A
t (ms)
0
-4 -3 -2 -1 1 2 3 4
-A-
What are the cosine Fourier series coefficients, an?
a) (6)/(n^2 π^2) [1 - cos(nπ)]
b) | (20)/(nπ)cos((2nπ)/(3)) - (30)/(n^2 π^2)sin((2nπ)/(3)) |
c) (20)/(n^2 π^2) [cos(nπ) - 1]
d) 0
e) (-30)/(n^2 π^2)sin((2nπ)/(3)) + (20)/(nπ)cos((2nπ)/(3))
f) (40)/(n^2 π^2)

Added by Sharon L.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert James Kiss

Step 1

The waveform is antisymmetric about the origin: for every t in one period the value at -t is the negative of the value at t, i.e. f(-t) = -f(t). (For example the triangular lobe at +1 has equal magnitude and opposite sign to the lobe at -1.) Show more…

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What are the cosine Fourier series coefficients, an? a) [1 - cos(nT)] / n^2 b) 20 - cos(nT) c) n^2 sin(2Ï€n) / 30 d) [cos(nT) - 1] / 30 e) 2Ï€n^2 / (30 * 2T) f) 20 / (2T) * n^2