Mixing
Proof of Discrete Fourier Series Integration Property
$begingroup$
In my recent studies of the Fourier Series, I came along to proof the properties of the Fourier Series (just to avoid confusion, not the fourier transform but the series itself in discrete time domain).
I also came into the following property:
The question
For any Discrete Fourier Series whose average is zero (condition for convergence), the following property is met:
$$y[n] = sum_{r=-infty}^{n}x[r]$$
$$Longrightarrow Y_k = frac{1}{1-e^{-ifrac{2pi k}{N}}} X_k$$
so that:
$$y[n] = sum_{k=0}^{N-1} Big(frac{1}{1-e^{-ifrac{2pi k}{N}}} X_kBig) e^{ifrac{2pi k}{N} n}$$
Where $Y_k$ and $X_k$ are the respective Fourier Coefficients, and $N$ is the period.
The question is, ¿how can I prove that? or better, ¿what is wrong with what I have approached?
What I have been able to do
The first thing I did, is to define $y[n]$ as:
$$y[n] = sum_{r=0}^{n} x[r]$$
Since the average per period is zero.
Then, using the linearity property and the displacement in time property, I got into:
$$y[n] = sum_{r=0}^{n} sum_{k=0}^{N-1} X_ke^{-ifrac{2pi k}{N} r} e^{ifrac{2pi k}{N} n} = sum_{k=0}^{N-1} X_kBig( sum_{r=0}^{n}e^{-ifrac{2pi k}{N} r} Big) e^{ifrac{2pi k}{N} n}$$
Using the fact that, from hypothesis, $X_0$ is zero (since $X_0$ represents the average of one period), we can use the geometric progression into the second summatory:
$$y[n] = sum_{k=1}^{N-1}X_k frac{1-(e^{-ifrac{2pi k}{N} (n+1)})}{1-e^{-ifrac{2pi k}{N}}} e^{ifrac{2pi k}{N} n}$$
Which is, after splitting the summatory:
$$y[n] = sum_{k=1}^{N-1} underbrace{Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big)}_{text{Yay!}} e^{ifrac{2pi k}{N} n} - sum_{k=1}^{N-1} Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big) e^{-ifrac{2pi k}{N}}$$
The first expression is what I am looking for, but I haven't been able to work with the second expression/summatory, I suppose that there might be a way to prove that it is zero, but I am not very sure about it, so I don't know if either there is something wrong in the procedure or I am missing something.
Thanks for the help in advance.
discrete-mathematics proof-explanation fourier-series
asked Jan 26 at 19:08
ZumbockZumbock
135
$endgroup$
add a comment |
$begingroup$
In my recent studies of the Fourier Series, I came along to proof the properties of the Fourier Series (just to avoid confusion, not the fourier transform but the series itself in discrete time domain).
I also came into the following property:
The question
For any Discrete Fourier Series whose average is zero (condition for convergence), the following property is met:
$$y[n] = sum_{r=-infty}^{n}x[r]$$
$$Longrightarrow Y_k = frac{1}{1-e^{-ifrac{2pi k}{N}}} X_k$$
so that:
$$y[n] = sum_{k=0}^{N-1} Big(frac{1}{1-e^{-ifrac{2pi k}{N}}} X_kBig) e^{ifrac{2pi k}{N} n}$$
Where $Y_k$ and $X_k$ are the respective Fourier Coefficients, and $N$ is the period.
The question is, ¿how can I prove that? or better, ¿what is wrong with what I have approached?
What I have been able to do
The first thing I did, is to define $y[n]$ as:
$$y[n] = sum_{r=0}^{n} x[r]$$
Since the average per period is zero.
Then, using the linearity property and the displacement in time property, I got into:
$$y[n] = sum_{r=0}^{n} sum_{k=0}^{N-1} X_ke^{-ifrac{2pi k}{N} r} e^{ifrac{2pi k}{N} n} = sum_{k=0}^{N-1} X_kBig( sum_{r=0}^{n}e^{-ifrac{2pi k}{N} r} Big) e^{ifrac{2pi k}{N} n}$$
Using the fact that, from hypothesis, $X_0$ is zero (since $X_0$ represents the average of one period), we can use the geometric progression into the second summatory:
$$y[n] = sum_{k=1}^{N-1}X_k frac{1-(e^{-ifrac{2pi k}{N} (n+1)})}{1-e^{-ifrac{2pi k}{N}}} e^{ifrac{2pi k}{N} n}$$
Which is, after splitting the summatory:
$$y[n] = sum_{k=1}^{N-1} underbrace{Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big)}_{text{Yay!}} e^{ifrac{2pi k}{N} n} - sum_{k=1}^{N-1} Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big) e^{-ifrac{2pi k}{N}}$$
The first expression is what I am looking for, but I haven't been able to work with the second expression/summatory, I suppose that there might be a way to prove that it is zero, but I am not very sure about it, so I don't know if either there is something wrong in the procedure or I am missing something.
Thanks for the help in advance.
discrete-mathematics proof-explanation fourier-series
asked Jan 26 at 19:08
ZumbockZumbock
135
$endgroup$
add a comment |
$begingroup$
In my recent studies of the Fourier Series, I came along to proof the properties of the Fourier Series (just to avoid confusion, not the fourier transform but the series itself in discrete time domain).
I also came into the following property:
The question
For any Discrete Fourier Series whose average is zero (condition for convergence), the following property is met:
$$y[n] = sum_{r=-infty}^{n}x[r]$$
$$Longrightarrow Y_k = frac{1}{1-e^{-ifrac{2pi k}{N}}} X_k$$
so that:
$$y[n] = sum_{k=0}^{N-1} Big(frac{1}{1-e^{-ifrac{2pi k}{N}}} X_kBig) e^{ifrac{2pi k}{N} n}$$
Where $Y_k$ and $X_k$ are the respective Fourier Coefficients, and $N$ is the period.
The question is, ¿how can I prove that? or better, ¿what is wrong with what I have approached?
What I have been able to do
The first thing I did, is to define $y[n]$ as:
$$y[n] = sum_{r=0}^{n} x[r]$$
Since the average per period is zero.
Then, using the linearity property and the displacement in time property, I got into:
$$y[n] = sum_{r=0}^{n} sum_{k=0}^{N-1} X_ke^{-ifrac{2pi k}{N} r} e^{ifrac{2pi k}{N} n} = sum_{k=0}^{N-1} X_kBig( sum_{r=0}^{n}e^{-ifrac{2pi k}{N} r} Big) e^{ifrac{2pi k}{N} n}$$
Using the fact that, from hypothesis, $X_0$ is zero (since $X_0$ represents the average of one period), we can use the geometric progression into the second summatory:
$$y[n] = sum_{k=1}^{N-1}X_k frac{1-(e^{-ifrac{2pi k}{N} (n+1)})}{1-e^{-ifrac{2pi k}{N}}} e^{ifrac{2pi k}{N} n}$$
Which is, after splitting the summatory:
$$y[n] = sum_{k=1}^{N-1} underbrace{Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big)}_{text{Yay!}} e^{ifrac{2pi k}{N} n} - sum_{k=1}^{N-1} Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big) e^{-ifrac{2pi k}{N}}$$
The first expression is what I am looking for, but I haven't been able to work with the second expression/summatory, I suppose that there might be a way to prove that it is zero, but I am not very sure about it, so I don't know if either there is something wrong in the procedure or I am missing something.
Thanks for the help in advance.
discrete-mathematics proof-explanation fourier-series
asked Jan 26 at 19:08
ZumbockZumbock
135
$endgroup$
In my recent studies of the Fourier Series, I came along to proof the properties of the Fourier Series (just to avoid confusion, not the fourier transform but the series itself in discrete time domain).
I also came into the following property:
The question
For any Discrete Fourier Series whose average is zero (condition for convergence), the following property is met:
$$y[n] = sum_{r=-infty}^{n}x[r]$$
$$Longrightarrow Y_k = frac{1}{1-e^{-ifrac{2pi k}{N}}} X_k$$
so that:
$$y[n] = sum_{k=0}^{N-1} Big(frac{1}{1-e^{-ifrac{2pi k}{N}}} X_kBig) e^{ifrac{2pi k}{N} n}$$
Where $Y_k$ and $X_k$ are the respective Fourier Coefficients, and $N$ is the period.
The question is, ¿how can I prove that? or better, ¿what is wrong with what I have approached?
What I have been able to do
The first thing I did, is to define $y[n]$ as:
$$y[n] = sum_{r=0}^{n} x[r]$$
Since the average per period is zero.
Then, using the linearity property and the displacement in time property, I got into:
$$y[n] = sum_{r=0}^{n} sum_{k=0}^{N-1} X_ke^{-ifrac{2pi k}{N} r} e^{ifrac{2pi k}{N} n} = sum_{k=0}^{N-1} X_kBig( sum_{r=0}^{n}e^{-ifrac{2pi k}{N} r} Big) e^{ifrac{2pi k}{N} n}$$
Using the fact that, from hypothesis, $X_0$ is zero (since $X_0$ represents the average of one period), we can use the geometric progression into the second summatory:
$$y[n] = sum_{k=1}^{N-1}X_k frac{1-(e^{-ifrac{2pi k}{N} (n+1)})}{1-e^{-ifrac{2pi k}{N}}} e^{ifrac{2pi k}{N} n}$$
Which is, after splitting the summatory:
$$y[n] = sum_{k=1}^{N-1} underbrace{Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big)}_{text{Yay!}} e^{ifrac{2pi k}{N} n} - sum_{k=1}^{N-1} Big(frac{1}{1-e^{-frac{2pi k}{N}}}X_k Big) e^{-ifrac{2pi k}{N}}$$
The first expression is what I am looking for, but I haven't been able to work with the second expression/summatory, I suppose that there might be a way to prove that it is zero, but I am not very sure about it, so I don't know if either there is something wrong in the procedure or I am missing something.
Thanks for the help in advance.
discrete-mathematics proof-explanation fourier-series
discrete-mathematics proof-explanation fourier-series
asked Jan 26 at 19:08
ZumbockZumbock
135
asked Jan 26 at 19:08
ZumbockZumbock
135
edited Jan 26 at 19:55
Zumbock
asked Jan 26 at 19:08
ZumbockZumbock
135
asked Jan 26 at 19:08
ZumbockZumbock
135
asked Jan 26 at 19:08
ZumbockZumbock
135
135
add a comment |
add a comment |
1 Answer
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Let $alpha=e^{frac{-2pi ik}{N}}$ to ease the notation a little bit. Substituting into the definiton of DTFS yields:
$$begin{align}Y_k &=sum_{n=0}^{N-1}sum_{r=0}^nx[r]alpha^n=sum_{r=0}^{N-1}x[r]sum_{n=r}^{N-1}alpha^n=-sum_{r=0}^{N-1}x[r]sum_{n=0}^{r-1}alpha^n=sum_{r=0}^{N-1}x[r]frac{alpha^r-1}{1-alpha} \ &=frac{1}{1-alpha}left(sum_{r=0}^{N-1}x[r]alpha^r+sum_{r=0}^{N-1}x[r]right)=frac{X_k}{1-alpha}end{align}$$
since $sum_{r=0}^{N-1}x[r]=0$, and the other term is the definition of DTFS of $x[n]$.
answered Jan 26 at 20:09
gunesgunes
3747
$endgroup$
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $alpha=e^{frac{-2pi ik}{N}}$ to ease the notation a little bit. Substituting into the definiton of DTFS yields:
$$begin{align}Y_k &=sum_{n=0}^{N-1}sum_{r=0}^nx[r]alpha^n=sum_{r=0}^{N-1}x[r]sum_{n=r}^{N-1}alpha^n=-sum_{r=0}^{N-1}x[r]sum_{n=0}^{r-1}alpha^n=sum_{r=0}^{N-1}x[r]frac{alpha^r-1}{1-alpha} \ &=frac{1}{1-alpha}left(sum_{r=0}^{N-1}x[r]alpha^r+sum_{r=0}^{N-1}x[r]right)=frac{X_k}{1-alpha}end{align}$$
since $sum_{r=0}^{N-1}x[r]=0$, and the other term is the definition of DTFS of $x[n]$.
answered Jan 26 at 20:09
gunesgunes
3747
$endgroup$
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
add a comment |
$begingroup$
Let $alpha=e^{frac{-2pi ik}{N}}$ to ease the notation a little bit. Substituting into the definiton of DTFS yields:
$$begin{align}Y_k &=sum_{n=0}^{N-1}sum_{r=0}^nx[r]alpha^n=sum_{r=0}^{N-1}x[r]sum_{n=r}^{N-1}alpha^n=-sum_{r=0}^{N-1}x[r]sum_{n=0}^{r-1}alpha^n=sum_{r=0}^{N-1}x[r]frac{alpha^r-1}{1-alpha} \ &=frac{1}{1-alpha}left(sum_{r=0}^{N-1}x[r]alpha^r+sum_{r=0}^{N-1}x[r]right)=frac{X_k}{1-alpha}end{align}$$
since $sum_{r=0}^{N-1}x[r]=0$, and the other term is the definition of DTFS of $x[n]$.
answered Jan 26 at 20:09
gunesgunes
3747
$endgroup$
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
add a comment |
$begingroup$
Let $alpha=e^{frac{-2pi ik}{N}}$ to ease the notation a little bit. Substituting into the definiton of DTFS yields:
$$begin{align}Y_k &=sum_{n=0}^{N-1}sum_{r=0}^nx[r]alpha^n=sum_{r=0}^{N-1}x[r]sum_{n=r}^{N-1}alpha^n=-sum_{r=0}^{N-1}x[r]sum_{n=0}^{r-1}alpha^n=sum_{r=0}^{N-1}x[r]frac{alpha^r-1}{1-alpha} \ &=frac{1}{1-alpha}left(sum_{r=0}^{N-1}x[r]alpha^r+sum_{r=0}^{N-1}x[r]right)=frac{X_k}{1-alpha}end{align}$$
since $sum_{r=0}^{N-1}x[r]=0$, and the other term is the definition of DTFS of $x[n]$.
answered Jan 26 at 20:09
gunesgunes
3747
$endgroup$
Let $alpha=e^{frac{-2pi ik}{N}}$ to ease the notation a little bit. Substituting into the definiton of DTFS yields:
$$begin{align}Y_k &=sum_{n=0}^{N-1}sum_{r=0}^nx[r]alpha^n=sum_{r=0}^{N-1}x[r]sum_{n=r}^{N-1}alpha^n=-sum_{r=0}^{N-1}x[r]sum_{n=0}^{r-1}alpha^n=sum_{r=0}^{N-1}x[r]frac{alpha^r-1}{1-alpha} \ &=frac{1}{1-alpha}left(sum_{r=0}^{N-1}x[r]alpha^r+sum_{r=0}^{N-1}x[r]right)=frac{X_k}{1-alpha}end{align}$$
since $sum_{r=0}^{N-1}x[r]=0$, and the other term is the definition of DTFS of $x[n]$.
answered Jan 26 at 20:09
gunesgunes
3747
answered Jan 26 at 20:09
gunesgunes
3747
answered Jan 26 at 20:09
gunesgunes
3747
answered Jan 26 at 20:09
gunesgunes
3747
3747
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
add a comment |
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
$begingroup$
Thanks for the answer. Can't even imagine the answer was this easy.
$endgroup$
– Zumbock
Jan 26 at 20:29
add a comment |
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