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Proof of First Difference Property for Fourier Series

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1

$begingroup$

I am having trouble with deriving a proof for the first difference property for the Fourier Series.

Here is my attempt at the derivation:

$$
y[n] = x[n] - x[n-1]
$$

Fourier Series Representation:

$$
a_k - a_ke^-jkomega_0
$$

Fourier Series:

$$
y[n] = sum_k=<N>(a_k-a_ke^-jkomega_0)e^jkomega_0n
$$

I have set up the summation for the Fourier Series, however I have been having difficulty to compute the summation.

The summation should equal this:

$$
a_k(1-e^-jkomega_0)
$$

How would I evaluate the summation?

discrete-signals fourier-series

share|improve this question

edited 6 hours ago

Matt L.

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$endgroup$

add a comment | 

1

$begingroup$

I am having trouble with deriving a proof for the first difference property for the Fourier Series.

Here is my attempt at the derivation:

$$
y[n] = x[n] - x[n-1]
$$

Fourier Series Representation:

$$
a_k - a_ke^-jkomega_0
$$

Fourier Series:

$$
y[n] = sum_k=<N>(a_k-a_ke^-jkomega_0)e^jkomega_0n
$$

I have set up the summation for the Fourier Series, however I have been having difficulty to compute the summation.

The summation should equal this:

$$
a_k(1-e^-jkomega_0)
$$

How would I evaluate the summation?

discrete-signals fourier-series

share|improve this question

edited 6 hours ago

Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

asked 8 hours ago

user2443636user2443636

611 bronze badge

New contributor

user2443636 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I am having trouble with deriving a proof for the first difference property for the Fourier Series.

Here is my attempt at the derivation:

$$
y[n] = x[n] - x[n-1]
$$

Fourier Series Representation:

$$
a_k - a_ke^-jkomega_0
$$

Fourier Series:

$$
y[n] = sum_k=<N>(a_k-a_ke^-jkomega_0)e^jkomega_0n
$$

I have set up the summation for the Fourier Series, however I have been having difficulty to compute the summation.

The summation should equal this:

$$
a_k(1-e^-jkomega_0)
$$

How would I evaluate the summation?

discrete-signals fourier-series

share|improve this question

edited 6 hours ago

Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

asked 8 hours ago

user2443636user2443636

611 bronze badge

New contributor

user2443636 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 6 hours ago

Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

asked 8 hours ago

user2443636user2443636

611 bronze badge

New contributor

user2443636 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 6 hours ago

Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

asked 8 hours ago

user2443636user2443636

611 bronze badge

New contributor

user2443636 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 6 hours ago

Matt L.

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edited 6 hours ago

Matt L.

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asked 8 hours ago

user2443636user2443636

611 bronze badge

New contributor

user2443636 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

user2443636user2443636

611 bronze badge

1 Answer
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3

$begingroup$

You have the result already written down in your question. If $a_k$ are the Fourier coefficients of $x[n]$, then your third formula is the Fourier series of $y[n]=x[n]-x[n-1]$. So the Fourier series coefficients of $y[n]$ are

$$b_k=a_k-a_ke^-jkomega_0=a_k(1-e^-jkomega_0)$$

meaning that

$$y[n]=sum_k=0^N-1b_ke^jomega_0 nk,qquadomega_0=2pi/N$$

share|improve this answer

answered 6 hours ago

Matt L.Matt L.

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$endgroup$

add a comment | 

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3

$begingroup$

You have the result already written down in your question. If $a_k$ are the Fourier coefficients of $x[n]$, then your third formula is the Fourier series of $y[n]=x[n]-x[n-1]$. So the Fourier series coefficients of $y[n]$ are

$$b_k=a_k-a_ke^-jkomega_0=a_k(1-e^-jkomega_0)$$

meaning that

$$y[n]=sum_k=0^N-1b_ke^jomega_0 nk,qquadomega_0=2pi/N$$

share|improve this answer

answered 6 hours ago

Matt L.Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

You have the result already written down in your question. If $a_k$ are the Fourier coefficients of $x[n]$, then your third formula is the Fourier series of $y[n]=x[n]-x[n-1]$. So the Fourier series coefficients of $y[n]$ are

$$b_k=a_k-a_ke^-jkomega_0=a_k(1-e^-jkomega_0)$$

meaning that

$$y[n]=sum_k=0^N-1b_ke^jomega_0 nk,qquadomega_0=2pi/N$$

share|improve this answer

answered 6 hours ago

Matt L.Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

add a comment | 

$begingroup$

You have the result already written down in your question. If $a_k$ are the Fourier coefficients of $x[n]$, then your third formula is the Fourier series of $y[n]=x[n]-x[n-1]$. So the Fourier series coefficients of $y[n]$ are

$$b_k=a_k-a_ke^-jkomega_0=a_k(1-e^-jkomega_0)$$

meaning that

$$y[n]=sum_k=0^N-1b_ke^jomega_0 nk,qquadomega_0=2pi/N$$

share|improve this answer

answered 6 hours ago

Matt L.Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

share|improve this answer

answered 6 hours ago

Matt L.Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

answered 6 hours ago

Matt L.Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges

answered 6 hours ago

Matt L.Matt L.

52.9k22 gold badges3939 silver badges9999 bronze badges