How to use sampling and summation instead of integration to calculate FS coefficients?

Question

I have to calculate the fourier series coefficients of the sawtooth function but instead of integrating i have to sample the function and the carrier then summate. when i do that the results arent correct. i dont know what i'm doing wrong, any ideas?

Answer 1

Ok so they're asking you to perform a Discrete Fourier Transform.
(http://en.wikipedia.org/wiki/Discrete_Fourier_transform )
on a real-valued series
(http://en.wikipedia.org/wiki/Discrete_Fourier_transform#The_real-input_DFT )

Additionally since you know that the series is either real-valued and odd (most likely for sawtooth), or real-valued and even, this tells you half the FS coefficients, and gives you this relation on both the samples x_k and the coefficients X_k:
x_(N-k) = +/- x_k
X_(N-k) = +/- X_k ; (use + sign for even; - for odd)
for k =0,...,(N/2 -1) (this halves the range needed for k!)

Then this helps you with the coefficient:
X_k = Sum_(k=0)_(N-1) x_n * e^((-2pi i/N)kn)
for k=0,...,N-1

So fold the sum in half by applying
X_(N-k) = +/- X_k ; (use + sign for even; - for odd)
for k =0,...,(N/2 -1)
(Let us assume + for even, if it is formulated odd just insert the -)
(Also let us assume N is even, it is trivial to rewrite if N is odd).

Then we are going to separate the upper and lower halves of the summation and apply the odd/even relation on samples x_k:
X_k = Sum_(k=0)_(N-1) x_n * e^((-2pi i/N)kn)
= Sum_(k=0)_(N/2-1) x_n * e^((-2pi i/N)kn)
+ Sum_(k=N/2)_(N) x_n * e^((-2pi i/N)kn)

= Sum_(k=0)_(N/2-1) x_n * e^((-2pi i/N)kn)
+ Sum_(k=0)_(N/2) (+/-)x_n * e^((-2pi i/N)(N-k)n) ; (setting k'=N-k)
(now we assume without loss of generality the even case: + for the +/- sign)

X_k = Sum_(k=0)_(N/2-1) x_n * e^((-2pi i/N)kn)
+ Sum_(k=0)_(N/2) (+/-)x_n * e^((-2pi i/N)(N-k)n)
(and noting that e^((-2pi i/N)(N)n) = 1)

X_k = Sum_(k=0)_(N/2-1) x_n * e^((-2pi i/N)kn)
+ Sum_(k=0)_(N/2) x_n * e^((-2pi i/N)(-k)n)
= Sum_(k=0)_(N/2-1) x_n * [ e^((-2pi i/N)kn) + e^((-2pi i/N)(-k)n)]

e^(i theta) = cos theta + i sin theta, then
e^(i theta) + e^(-i theta) = 2 cos theta

X_k = Sum_(k=0)_(N/2-1) x_n * 2 cos ((-2pi/N)kn)

So just plug in the numbers for X_k; k =0,...,(N/2 -1)
Suggestion: X_0 should be easy (the DC component), so make sure you get that right. Next check X_1

This should help?

Answer 2

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