Fourier coefficients of $sin + cos$Determining Fourier series coefficientsFourier series for $cos( frac x2)$Coefficients of Fourier Series of (Cos(t))^3reciprocal of a Fourier Cosine seriesHow to calculate the Fourier coefficientsDoubt regarding Fourier series coefficients.Find the Fourier series coefficients of $x_1[n]r[n]$.Is just the product of their coefficients?Fourier coefficients of $sin(2 pi f_0 t)$Fourier coefficients of a sumFourier coefficients of $|cos x|$
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Fourier coefficients of $sin + cos$
Determining Fourier series coefficientsFourier series for $cos( frac x2)$Coefficients of Fourier Series of (Cos(t))^3reciprocal of a Fourier Cosine seriesHow to calculate the Fourier coefficientsDoubt regarding Fourier series coefficients.Find the Fourier series coefficients of $x_1[n]r[n]$.Is just the product of their coefficients?Fourier coefficients of $sin(2 pi f_0 t)$Fourier coefficients of a sumFourier coefficients of $|cos x|$
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Fourier coefficients $x(t) = A_1 cos (2 pi f_0 t + E ) + A_3 cos (6 pi f_0 t + F )$
I applied the Fourier series definition with the integer from $0$ to $t * e^2if_0ktpi $ and I found that the series is $ 0 $ except for $k=1,-1,3,-3 $ but now I don’t know how to find the Fourier coefficients for this values of $k$.
fourier-series
$endgroup$
add a comment |
$begingroup$
Fourier coefficients $x(t) = A_1 cos (2 pi f_0 t + E ) + A_3 cos (6 pi f_0 t + F )$
I applied the Fourier series definition with the integer from $0$ to $t * e^2if_0ktpi $ and I found that the series is $ 0 $ except for $k=1,-1,3,-3 $ but now I don’t know how to find the Fourier coefficients for this values of $k$.
fourier-series
$endgroup$
$begingroup$
Thanks for trying to format your question! You just need dollar signs around the parts you want to format: $x(t)$ becomes $x(t)$.
$endgroup$
– Toby Mak
Mar 13 at 11:09
add a comment |
$begingroup$
Fourier coefficients $x(t) = A_1 cos (2 pi f_0 t + E ) + A_3 cos (6 pi f_0 t + F )$
I applied the Fourier series definition with the integer from $0$ to $t * e^2if_0ktpi $ and I found that the series is $ 0 $ except for $k=1,-1,3,-3 $ but now I don’t know how to find the Fourier coefficients for this values of $k$.
fourier-series
$endgroup$
Fourier coefficients $x(t) = A_1 cos (2 pi f_0 t + E ) + A_3 cos (6 pi f_0 t + F )$
I applied the Fourier series definition with the integer from $0$ to $t * e^2if_0ktpi $ and I found that the series is $ 0 $ except for $k=1,-1,3,-3 $ but now I don’t know how to find the Fourier coefficients for this values of $k$.
fourier-series
fourier-series
edited Mar 17 at 11:55
Rócherz
3,0013821
3,0013821
asked Mar 13 at 10:48
Elena MartiniElena Martini
113
113
$begingroup$
Thanks for trying to format your question! You just need dollar signs around the parts you want to format: $x(t)$ becomes $x(t)$.
$endgroup$
– Toby Mak
Mar 13 at 11:09
add a comment |
$begingroup$
Thanks for trying to format your question! You just need dollar signs around the parts you want to format: $x(t)$ becomes $x(t)$.
$endgroup$
– Toby Mak
Mar 13 at 11:09
$begingroup$
Thanks for trying to format your question! You just need dollar signs around the parts you want to format: $x(t)$ becomes $x(t)$.
$endgroup$
– Toby Mak
Mar 13 at 11:09
$begingroup$
Thanks for trying to format your question! You just need dollar signs around the parts you want to format: $x(t)$ becomes $x(t)$.
$endgroup$
– Toby Mak
Mar 13 at 11:09
add a comment |
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$begingroup$
Thanks for trying to format your question! You just need dollar signs around the parts you want to format: $x(t)$ becomes $x(t)$.
$endgroup$
– Toby Mak
Mar 13 at 11:09