Fourier coefficients of $sin{(2 pi f_0 t)}$
$begingroup$
I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.
I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:
$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.
Can you help me to understand how to solve this exercise ?
fourier-series
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
$endgroup$
add a comment |
$begingroup$
I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.
I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:
$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.
Can you help me to understand how to solve this exercise ?
fourier-series
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
$endgroup$
$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39
$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52
add a comment |
$begingroup$
I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.
I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:
$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.
Can you help me to understand how to solve this exercise ?
fourier-series
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
$endgroup$
I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.
I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:
$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.
Can you help me to understand how to solve this exercise ?
fourier-series
fourier-series
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
edited Dec 19 '18 at 15:40
quid♦
37.2k95193
37.2k95193
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
asked Dec 19 '18 at 15:33
Elena MartiniElena Martini
113
113
$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39
$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52
add a comment |
$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39
$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52
$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39
$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39
$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52
$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so
$c_k = 0$ at $kneqpm1 $.
Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:
$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$
but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$
The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$
The case for $k=-1$ is similar to the computation above.
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
$endgroup$
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so
$c_k = 0$ at $kneqpm1 $.
Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:
$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$
but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$
The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$
The case for $k=-1$ is similar to the computation above.
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
$endgroup$
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
add a comment |
$begingroup$
The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so
$c_k = 0$ at $kneqpm1 $.
Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:
$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$
but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$
The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$
The case for $k=-1$ is similar to the computation above.
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
$endgroup$
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
add a comment |
$begingroup$
The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so
$c_k = 0$ at $kneqpm1 $.
Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:
$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$
but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$
The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$
The case for $k=-1$ is similar to the computation above.
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
$endgroup$
The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so
$c_k = 0$ at $kneqpm1 $.
Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:
$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$
but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$
The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$
The case for $k=-1$ is similar to the computation above.
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
answered Dec 19 '18 at 17:13
PoypoyanPoypoyan
473411
473411
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
add a comment |
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15
add a comment |
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$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39
$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52