Mixing
Coefficients of Fourier series
$begingroup$
i’d like to calculate Fourier coefficients of $cos 2 pi f_0 t$.
This is what I did :
$$ c_k = frac{1}{T_0}int_{0}^{T} cos 2 pi f_0 t cdot e^{-2ipi f_0 t}. $$
From Euler formulas:
$$ frac{1}{T_0}int_{0}^{T} frac{ e^{2ipi f_0 t} + e^{-2ipi f_0 t} }{2} cdot e^{-2ipi f_0 t}. $$
$ frac{1}{2T_0}int_{0}^{T} 1 + e^{-2pi f_0 t(1+i) } $
Solving i obtained $c_k = frac{1}{2T_0}[T_0 + frac{e^{-2pi (1+i)} - 1 }{-2pi f_0 - 2 i pi f_0 }] $ (because $f_0 = 1/T_0 $).
But $e^{-2ipi}e^{-2pi} = e^{-2pi}[cos2pi + i sin(-2pi)]= e^{-2pi}$.
So $frac{1}{2T_0}[ T_0 + frac{e^{-2pi } - 1 }{-2pi f_0}(1+i)]$.
On my book the result is $1/2$ If $k = pm 1$ And $0$ If $k$ Is different from $1$. Can someone can tell me where is the error ? I’d really like to learn how to correct this exercise . Thank you so much
fourier-series signal-processing
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
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add a comment |
$begingroup$
i’d like to calculate Fourier coefficients of $cos 2 pi f_0 t$.
This is what I did :
$$ c_k = frac{1}{T_0}int_{0}^{T} cos 2 pi f_0 t cdot e^{-2ipi f_0 t}. $$
From Euler formulas:
$$ frac{1}{T_0}int_{0}^{T} frac{ e^{2ipi f_0 t} + e^{-2ipi f_0 t} }{2} cdot e^{-2ipi f_0 t}. $$
$ frac{1}{2T_0}int_{0}^{T} 1 + e^{-2pi f_0 t(1+i) } $
Solving i obtained $c_k = frac{1}{2T_0}[T_0 + frac{e^{-2pi (1+i)} - 1 }{-2pi f_0 - 2 i pi f_0 }] $ (because $f_0 = 1/T_0 $).
But $e^{-2ipi}e^{-2pi} = e^{-2pi}[cos2pi + i sin(-2pi)]= e^{-2pi}$.
So $frac{1}{2T_0}[ T_0 + frac{e^{-2pi } - 1 }{-2pi f_0}(1+i)]$.
On my book the result is $1/2$ If $k = pm 1$ And $0$ If $k$ Is different from $1$. Can someone can tell me where is the error ? I’d really like to learn how to correct this exercise . Thank you so much
fourier-series signal-processing
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
$endgroup$
$begingroup$
You know that $T=2pi$, yes?
$endgroup$
– uniquesolution
Jan 5 at 14:32
add a comment |
$begingroup$
i’d like to calculate Fourier coefficients of $cos 2 pi f_0 t$.
This is what I did :
$$ c_k = frac{1}{T_0}int_{0}^{T} cos 2 pi f_0 t cdot e^{-2ipi f_0 t}. $$
From Euler formulas:
$$ frac{1}{T_0}int_{0}^{T} frac{ e^{2ipi f_0 t} + e^{-2ipi f_0 t} }{2} cdot e^{-2ipi f_0 t}. $$
$ frac{1}{2T_0}int_{0}^{T} 1 + e^{-2pi f_0 t(1+i) } $
Solving i obtained $c_k = frac{1}{2T_0}[T_0 + frac{e^{-2pi (1+i)} - 1 }{-2pi f_0 - 2 i pi f_0 }] $ (because $f_0 = 1/T_0 $).
But $e^{-2ipi}e^{-2pi} = e^{-2pi}[cos2pi + i sin(-2pi)]= e^{-2pi}$.
So $frac{1}{2T_0}[ T_0 + frac{e^{-2pi } - 1 }{-2pi f_0}(1+i)]$.
On my book the result is $1/2$ If $k = pm 1$ And $0$ If $k$ Is different from $1$. Can someone can tell me where is the error ? I’d really like to learn how to correct this exercise . Thank you so much
fourier-series signal-processing
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
$endgroup$
i’d like to calculate Fourier coefficients of $cos 2 pi f_0 t$.
This is what I did :
$$ c_k = frac{1}{T_0}int_{0}^{T} cos 2 pi f_0 t cdot e^{-2ipi f_0 t}. $$
From Euler formulas:
$$ frac{1}{T_0}int_{0}^{T} frac{ e^{2ipi f_0 t} + e^{-2ipi f_0 t} }{2} cdot e^{-2ipi f_0 t}. $$
$ frac{1}{2T_0}int_{0}^{T} 1 + e^{-2pi f_0 t(1+i) } $
Solving i obtained $c_k = frac{1}{2T_0}[T_0 + frac{e^{-2pi (1+i)} - 1 }{-2pi f_0 - 2 i pi f_0 }] $ (because $f_0 = 1/T_0 $).
But $e^{-2ipi}e^{-2pi} = e^{-2pi}[cos2pi + i sin(-2pi)]= e^{-2pi}$.
So $frac{1}{2T_0}[ T_0 + frac{e^{-2pi } - 1 }{-2pi f_0}(1+i)]$.
On my book the result is $1/2$ If $k = pm 1$ And $0$ If $k$ Is different from $1$. Can someone can tell me where is the error ? I’d really like to learn how to correct this exercise . Thank you so much
fourier-series signal-processing
fourier-series signal-processing
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
asked Jan 5 at 14:24
Elena MartiniElena Martini
1
1
$begingroup$
You know that $T=2pi$, yes?
$endgroup$
– uniquesolution
Jan 5 at 14:32
add a comment |
$begingroup$
You know that $T=2pi$, yes?
$endgroup$
– uniquesolution
Jan 5 at 14:32
$begingroup$
You know that $T=2pi$, yes?
$endgroup$
– uniquesolution
Jan 5 at 14:32
$begingroup$
You know that $T=2pi$, yes?
$endgroup$
– uniquesolution
Jan 5 at 14:32
add a comment |
2 Answers
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$begingroup$
The problem start in you very first expression
$$
c_k = frac{1}{T}int_0^T cos 2pi f_0 t cdot e^{-2pi i color{red}{k}t / T}~{rm d}t
$$
and now do the same trick you did
begin{eqnarray}
c_k &=& f_0int_0^{1/f_0} frac{e^{2pi i f_0 t} + e^{-2pi i f_0 t}}{2} e^{-2pi i f_0 color{red}{k}t} ~{rm d}t = frac{f_0}{2} int_0^{1/f_0}left[ e^{2pi i f_0(1 - k)t} + e^{-2pi i f_0(1 + k)t}right]{rm d}t
end{eqnarray}
Now consider three cases
$k = 1$
$$
c_1 = frac{f_0}{2} int_0^{1/f_0}left[1 + e^{-4pi i f_0t}right]{rm d}t = frac{1}{2}
$$
$k = -1$
Same idea
$$
c_{-1} = frac{1}{2}
$$
$k not= 1$ and $k not= -1$
$$
c_k = frac{f_0}{2} left[frac{e^{2pi i f_0(1 - k)t}}{2pi i f_0 (1 - k)} - frac{-e^{2pi i f_0(1 + k)t}}{2pi i f_0 (1 + k)} right]_0^{1/f_0} = 0
$$
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
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add a comment |
$begingroup$
First off, you didn't set up the initial integral correctly. Assuming $T_0=frac{1}{f_0}$ is the period of $cos(2pi f_0t)$, it should be:
$$c_k=frac 1 {T_0}int_0^{T_0}cos(2pi f_0t)e^{-2ipi kf_0t}dt$$
Critically, notice how $k$ is now in the exponent of the complex exponential where as in your integral, $k$ is not present in the formula at all. Now, we can use Euler's formula:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0t}+e^{-2ipi f_0t}}{2}e^{-2ipi kf_0t}dt$$
Distribute the $e^{-2ipi kf_0t}$:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t}}{2}dt$$
For simplicity, take the $frac 1 2$ out of the integral:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dt$$
Integrate:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi f_0(1-k)T_0}-e^{2ipi f_0(1-k)0}}{2ipi f_0(1-k)}+frac{e^{-2ipi f_0(1+k)T_0}-e^{-2ipi f_0(1+k)0}}{-2ipi f_0(1+k)}right)$$
Substitute $f_0T_0=1$ and $e^0=1$:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi (1-k)}-1}{2ipi f_0(1-k)}+frac{e^{-2ipi (1+k)}-1}{-2ipi f_0(1+k)}right)$$
Now, since $k$ is an integer, $1-k$ and $-(1+k)$ are also both integers, so $e^{2ipi (1-k)}=e^{-2ipi (1+k)}=1$:
$$c_k=frac{1}{2T_0}left(frac{1-1}{2ipi f_0(1-k)}+frac{1-1}{-2ipi f_0(1+k)}right)=0$$
Now, this seems like a very unintuitive answer, because it means $c_k=0$ for all $k$. However, if you look closely, you will see that we divided by $1-k$ and $1+k$ in our integration in order to get $c_k=0$. Clearly, this does not work for $k=1$ and $k=-1$ since it causes a division-by-zero error. Therefore, $c_k=0$ for all $kneq pm 1$ and we need to treat the cases $k=1$ and $k=-1$ separately.
First, let's do $k=1$. From a previous equation, we have:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dtrightarrow c_1=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_00cdot t}+e^{-2ipi f_0cdot 2cdot t})dt$$
Simplify and substitute $e^0=1$:
$$c_1=frac 1 {2T_0}int_0^{T_0}(1+e^{-4ipi f_0t})dt$$
Integrate:
$$c_1=frac 1 {2T_0}left(T_0+frac{e^{-4ipi f_0T_0}-e^{-4ipi f_0cdot 0}}{-4ipi f_0}right)$$
Substitute $e^{-4ipi f_0T_0}=e^{-4ipi}=1$ and $e^0=1$:
$$c_1=frac 1 {2T_0}left(T_0+frac{1-1}{-4ipi f_0}right)=frac{1}{2T_0}T_0=frac 1 2$$
I will leave it as an exercise to you to finish it off by showing $c_{-1}=frac 1 2$.
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
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2 Answers
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2 Answers
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$begingroup$
The problem start in you very first expression
$$
c_k = frac{1}{T}int_0^T cos 2pi f_0 t cdot e^{-2pi i color{red}{k}t / T}~{rm d}t
$$
and now do the same trick you did
begin{eqnarray}
c_k &=& f_0int_0^{1/f_0} frac{e^{2pi i f_0 t} + e^{-2pi i f_0 t}}{2} e^{-2pi i f_0 color{red}{k}t} ~{rm d}t = frac{f_0}{2} int_0^{1/f_0}left[ e^{2pi i f_0(1 - k)t} + e^{-2pi i f_0(1 + k)t}right]{rm d}t
end{eqnarray}
Now consider three cases
$k = 1$
$$
c_1 = frac{f_0}{2} int_0^{1/f_0}left[1 + e^{-4pi i f_0t}right]{rm d}t = frac{1}{2}
$$
$k = -1$
Same idea
$$
c_{-1} = frac{1}{2}
$$
$k not= 1$ and $k not= -1$
$$
c_k = frac{f_0}{2} left[frac{e^{2pi i f_0(1 - k)t}}{2pi i f_0 (1 - k)} - frac{-e^{2pi i f_0(1 + k)t}}{2pi i f_0 (1 + k)} right]_0^{1/f_0} = 0
$$
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
$endgroup$
add a comment |
$begingroup$
The problem start in you very first expression
$$
c_k = frac{1}{T}int_0^T cos 2pi f_0 t cdot e^{-2pi i color{red}{k}t / T}~{rm d}t
$$
and now do the same trick you did
begin{eqnarray}
c_k &=& f_0int_0^{1/f_0} frac{e^{2pi i f_0 t} + e^{-2pi i f_0 t}}{2} e^{-2pi i f_0 color{red}{k}t} ~{rm d}t = frac{f_0}{2} int_0^{1/f_0}left[ e^{2pi i f_0(1 - k)t} + e^{-2pi i f_0(1 + k)t}right]{rm d}t
end{eqnarray}
Now consider three cases
$k = 1$
$$
c_1 = frac{f_0}{2} int_0^{1/f_0}left[1 + e^{-4pi i f_0t}right]{rm d}t = frac{1}{2}
$$
$k = -1$
Same idea
$$
c_{-1} = frac{1}{2}
$$
$k not= 1$ and $k not= -1$
$$
c_k = frac{f_0}{2} left[frac{e^{2pi i f_0(1 - k)t}}{2pi i f_0 (1 - k)} - frac{-e^{2pi i f_0(1 + k)t}}{2pi i f_0 (1 + k)} right]_0^{1/f_0} = 0
$$
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
$endgroup$
add a comment |
$begingroup$
The problem start in you very first expression
$$
c_k = frac{1}{T}int_0^T cos 2pi f_0 t cdot e^{-2pi i color{red}{k}t / T}~{rm d}t
$$
and now do the same trick you did
begin{eqnarray}
c_k &=& f_0int_0^{1/f_0} frac{e^{2pi i f_0 t} + e^{-2pi i f_0 t}}{2} e^{-2pi i f_0 color{red}{k}t} ~{rm d}t = frac{f_0}{2} int_0^{1/f_0}left[ e^{2pi i f_0(1 - k)t} + e^{-2pi i f_0(1 + k)t}right]{rm d}t
end{eqnarray}
Now consider three cases
$k = 1$
$$
c_1 = frac{f_0}{2} int_0^{1/f_0}left[1 + e^{-4pi i f_0t}right]{rm d}t = frac{1}{2}
$$
$k = -1$
Same idea
$$
c_{-1} = frac{1}{2}
$$
$k not= 1$ and $k not= -1$
$$
c_k = frac{f_0}{2} left[frac{e^{2pi i f_0(1 - k)t}}{2pi i f_0 (1 - k)} - frac{-e^{2pi i f_0(1 + k)t}}{2pi i f_0 (1 + k)} right]_0^{1/f_0} = 0
$$
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
$endgroup$
The problem start in you very first expression
$$
c_k = frac{1}{T}int_0^T cos 2pi f_0 t cdot e^{-2pi i color{red}{k}t / T}~{rm d}t
$$
and now do the same trick you did
begin{eqnarray}
c_k &=& f_0int_0^{1/f_0} frac{e^{2pi i f_0 t} + e^{-2pi i f_0 t}}{2} e^{-2pi i f_0 color{red}{k}t} ~{rm d}t = frac{f_0}{2} int_0^{1/f_0}left[ e^{2pi i f_0(1 - k)t} + e^{-2pi i f_0(1 + k)t}right]{rm d}t
end{eqnarray}
Now consider three cases
$k = 1$
$$
c_1 = frac{f_0}{2} int_0^{1/f_0}left[1 + e^{-4pi i f_0t}right]{rm d}t = frac{1}{2}
$$
$k = -1$
Same idea
$$
c_{-1} = frac{1}{2}
$$
$k not= 1$ and $k not= -1$
$$
c_k = frac{f_0}{2} left[frac{e^{2pi i f_0(1 - k)t}}{2pi i f_0 (1 - k)} - frac{-e^{2pi i f_0(1 + k)t}}{2pi i f_0 (1 + k)} right]_0^{1/f_0} = 0
$$
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
answered Jan 5 at 15:09
caveraccaverac
14.5k31130
14.5k31130
add a comment |
add a comment |
$begingroup$
First off, you didn't set up the initial integral correctly. Assuming $T_0=frac{1}{f_0}$ is the period of $cos(2pi f_0t)$, it should be:
$$c_k=frac 1 {T_0}int_0^{T_0}cos(2pi f_0t)e^{-2ipi kf_0t}dt$$
Critically, notice how $k$ is now in the exponent of the complex exponential where as in your integral, $k$ is not present in the formula at all. Now, we can use Euler's formula:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0t}+e^{-2ipi f_0t}}{2}e^{-2ipi kf_0t}dt$$
Distribute the $e^{-2ipi kf_0t}$:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t}}{2}dt$$
For simplicity, take the $frac 1 2$ out of the integral:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dt$$
Integrate:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi f_0(1-k)T_0}-e^{2ipi f_0(1-k)0}}{2ipi f_0(1-k)}+frac{e^{-2ipi f_0(1+k)T_0}-e^{-2ipi f_0(1+k)0}}{-2ipi f_0(1+k)}right)$$
Substitute $f_0T_0=1$ and $e^0=1$:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi (1-k)}-1}{2ipi f_0(1-k)}+frac{e^{-2ipi (1+k)}-1}{-2ipi f_0(1+k)}right)$$
Now, since $k$ is an integer, $1-k$ and $-(1+k)$ are also both integers, so $e^{2ipi (1-k)}=e^{-2ipi (1+k)}=1$:
$$c_k=frac{1}{2T_0}left(frac{1-1}{2ipi f_0(1-k)}+frac{1-1}{-2ipi f_0(1+k)}right)=0$$
Now, this seems like a very unintuitive answer, because it means $c_k=0$ for all $k$. However, if you look closely, you will see that we divided by $1-k$ and $1+k$ in our integration in order to get $c_k=0$. Clearly, this does not work for $k=1$ and $k=-1$ since it causes a division-by-zero error. Therefore, $c_k=0$ for all $kneq pm 1$ and we need to treat the cases $k=1$ and $k=-1$ separately.
First, let's do $k=1$. From a previous equation, we have:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dtrightarrow c_1=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_00cdot t}+e^{-2ipi f_0cdot 2cdot t})dt$$
Simplify and substitute $e^0=1$:
$$c_1=frac 1 {2T_0}int_0^{T_0}(1+e^{-4ipi f_0t})dt$$
Integrate:
$$c_1=frac 1 {2T_0}left(T_0+frac{e^{-4ipi f_0T_0}-e^{-4ipi f_0cdot 0}}{-4ipi f_0}right)$$
Substitute $e^{-4ipi f_0T_0}=e^{-4ipi}=1$ and $e^0=1$:
$$c_1=frac 1 {2T_0}left(T_0+frac{1-1}{-4ipi f_0}right)=frac{1}{2T_0}T_0=frac 1 2$$
I will leave it as an exercise to you to finish it off by showing $c_{-1}=frac 1 2$.
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
add a comment |
$begingroup$
First off, you didn't set up the initial integral correctly. Assuming $T_0=frac{1}{f_0}$ is the period of $cos(2pi f_0t)$, it should be:
$$c_k=frac 1 {T_0}int_0^{T_0}cos(2pi f_0t)e^{-2ipi kf_0t}dt$$
Critically, notice how $k$ is now in the exponent of the complex exponential where as in your integral, $k$ is not present in the formula at all. Now, we can use Euler's formula:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0t}+e^{-2ipi f_0t}}{2}e^{-2ipi kf_0t}dt$$
Distribute the $e^{-2ipi kf_0t}$:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t}}{2}dt$$
For simplicity, take the $frac 1 2$ out of the integral:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dt$$
Integrate:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi f_0(1-k)T_0}-e^{2ipi f_0(1-k)0}}{2ipi f_0(1-k)}+frac{e^{-2ipi f_0(1+k)T_0}-e^{-2ipi f_0(1+k)0}}{-2ipi f_0(1+k)}right)$$
Substitute $f_0T_0=1$ and $e^0=1$:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi (1-k)}-1}{2ipi f_0(1-k)}+frac{e^{-2ipi (1+k)}-1}{-2ipi f_0(1+k)}right)$$
Now, since $k$ is an integer, $1-k$ and $-(1+k)$ are also both integers, so $e^{2ipi (1-k)}=e^{-2ipi (1+k)}=1$:
$$c_k=frac{1}{2T_0}left(frac{1-1}{2ipi f_0(1-k)}+frac{1-1}{-2ipi f_0(1+k)}right)=0$$
Now, this seems like a very unintuitive answer, because it means $c_k=0$ for all $k$. However, if you look closely, you will see that we divided by $1-k$ and $1+k$ in our integration in order to get $c_k=0$. Clearly, this does not work for $k=1$ and $k=-1$ since it causes a division-by-zero error. Therefore, $c_k=0$ for all $kneq pm 1$ and we need to treat the cases $k=1$ and $k=-1$ separately.
First, let's do $k=1$. From a previous equation, we have:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dtrightarrow c_1=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_00cdot t}+e^{-2ipi f_0cdot 2cdot t})dt$$
Simplify and substitute $e^0=1$:
$$c_1=frac 1 {2T_0}int_0^{T_0}(1+e^{-4ipi f_0t})dt$$
Integrate:
$$c_1=frac 1 {2T_0}left(T_0+frac{e^{-4ipi f_0T_0}-e^{-4ipi f_0cdot 0}}{-4ipi f_0}right)$$
Substitute $e^{-4ipi f_0T_0}=e^{-4ipi}=1$ and $e^0=1$:
$$c_1=frac 1 {2T_0}left(T_0+frac{1-1}{-4ipi f_0}right)=frac{1}{2T_0}T_0=frac 1 2$$
I will leave it as an exercise to you to finish it off by showing $c_{-1}=frac 1 2$.
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
add a comment |
$begingroup$
First off, you didn't set up the initial integral correctly. Assuming $T_0=frac{1}{f_0}$ is the period of $cos(2pi f_0t)$, it should be:
$$c_k=frac 1 {T_0}int_0^{T_0}cos(2pi f_0t)e^{-2ipi kf_0t}dt$$
Critically, notice how $k$ is now in the exponent of the complex exponential where as in your integral, $k$ is not present in the formula at all. Now, we can use Euler's formula:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0t}+e^{-2ipi f_0t}}{2}e^{-2ipi kf_0t}dt$$
Distribute the $e^{-2ipi kf_0t}$:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t}}{2}dt$$
For simplicity, take the $frac 1 2$ out of the integral:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dt$$
Integrate:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi f_0(1-k)T_0}-e^{2ipi f_0(1-k)0}}{2ipi f_0(1-k)}+frac{e^{-2ipi f_0(1+k)T_0}-e^{-2ipi f_0(1+k)0}}{-2ipi f_0(1+k)}right)$$
Substitute $f_0T_0=1$ and $e^0=1$:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi (1-k)}-1}{2ipi f_0(1-k)}+frac{e^{-2ipi (1+k)}-1}{-2ipi f_0(1+k)}right)$$
Now, since $k$ is an integer, $1-k$ and $-(1+k)$ are also both integers, so $e^{2ipi (1-k)}=e^{-2ipi (1+k)}=1$:
$$c_k=frac{1}{2T_0}left(frac{1-1}{2ipi f_0(1-k)}+frac{1-1}{-2ipi f_0(1+k)}right)=0$$
Now, this seems like a very unintuitive answer, because it means $c_k=0$ for all $k$. However, if you look closely, you will see that we divided by $1-k$ and $1+k$ in our integration in order to get $c_k=0$. Clearly, this does not work for $k=1$ and $k=-1$ since it causes a division-by-zero error. Therefore, $c_k=0$ for all $kneq pm 1$ and we need to treat the cases $k=1$ and $k=-1$ separately.
First, let's do $k=1$. From a previous equation, we have:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dtrightarrow c_1=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_00cdot t}+e^{-2ipi f_0cdot 2cdot t})dt$$
Simplify and substitute $e^0=1$:
$$c_1=frac 1 {2T_0}int_0^{T_0}(1+e^{-4ipi f_0t})dt$$
Integrate:
$$c_1=frac 1 {2T_0}left(T_0+frac{e^{-4ipi f_0T_0}-e^{-4ipi f_0cdot 0}}{-4ipi f_0}right)$$
Substitute $e^{-4ipi f_0T_0}=e^{-4ipi}=1$ and $e^0=1$:
$$c_1=frac 1 {2T_0}left(T_0+frac{1-1}{-4ipi f_0}right)=frac{1}{2T_0}T_0=frac 1 2$$
I will leave it as an exercise to you to finish it off by showing $c_{-1}=frac 1 2$.
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
First off, you didn't set up the initial integral correctly. Assuming $T_0=frac{1}{f_0}$ is the period of $cos(2pi f_0t)$, it should be:
$$c_k=frac 1 {T_0}int_0^{T_0}cos(2pi f_0t)e^{-2ipi kf_0t}dt$$
Critically, notice how $k$ is now in the exponent of the complex exponential where as in your integral, $k$ is not present in the formula at all. Now, we can use Euler's formula:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0t}+e^{-2ipi f_0t}}{2}e^{-2ipi kf_0t}dt$$
Distribute the $e^{-2ipi kf_0t}$:
$$c_k=frac 1 {T_0}int_0^{T_0}frac{e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t}}{2}dt$$
For simplicity, take the $frac 1 2$ out of the integral:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dt$$
Integrate:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi f_0(1-k)T_0}-e^{2ipi f_0(1-k)0}}{2ipi f_0(1-k)}+frac{e^{-2ipi f_0(1+k)T_0}-e^{-2ipi f_0(1+k)0}}{-2ipi f_0(1+k)}right)$$
Substitute $f_0T_0=1$ and $e^0=1$:
$$c_k=frac{1}{2T_0}left(frac{e^{2ipi (1-k)}-1}{2ipi f_0(1-k)}+frac{e^{-2ipi (1+k)}-1}{-2ipi f_0(1+k)}right)$$
Now, since $k$ is an integer, $1-k$ and $-(1+k)$ are also both integers, so $e^{2ipi (1-k)}=e^{-2ipi (1+k)}=1$:
$$c_k=frac{1}{2T_0}left(frac{1-1}{2ipi f_0(1-k)}+frac{1-1}{-2ipi f_0(1+k)}right)=0$$
Now, this seems like a very unintuitive answer, because it means $c_k=0$ for all $k$. However, if you look closely, you will see that we divided by $1-k$ and $1+k$ in our integration in order to get $c_k=0$. Clearly, this does not work for $k=1$ and $k=-1$ since it causes a division-by-zero error. Therefore, $c_k=0$ for all $kneq pm 1$ and we need to treat the cases $k=1$ and $k=-1$ separately.
First, let's do $k=1$. From a previous equation, we have:
$$c_k=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_0(1-k)t}+e^{-2ipi f_0(1+k)t})dtrightarrow c_1=frac 1 {2T_0}int_0^{T_0}(e^{2ipi f_00cdot t}+e^{-2ipi f_0cdot 2cdot t})dt$$
Simplify and substitute $e^0=1$:
$$c_1=frac 1 {2T_0}int_0^{T_0}(1+e^{-4ipi f_0t})dt$$
Integrate:
$$c_1=frac 1 {2T_0}left(T_0+frac{e^{-4ipi f_0T_0}-e^{-4ipi f_0cdot 0}}{-4ipi f_0}right)$$
Substitute $e^{-4ipi f_0T_0}=e^{-4ipi}=1$ and $e^0=1$:
$$c_1=frac 1 {2T_0}left(T_0+frac{1-1}{-4ipi f_0}right)=frac{1}{2T_0}T_0=frac 1 2$$
I will leave it as an exercise to you to finish it off by showing $c_{-1}=frac 1 2$.
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 5 at 15:11
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
add a comment |
add a comment |
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$begingroup$
You know that $T=2pi$, yes?
$endgroup$
– uniquesolution
Jan 5 at 14:32