Are the period of signal and its fourier coefficient the same?
Are the period of signal and its fourier coefficient the same?i mean if the the period of $x[n]$ is $5$,will the period of its fourier coefficient ,$a_k$, also $5$? how to prove it?
Because the definition of a periodic is $x[n]=x[n+N]$,but if we do fourier series at the same time ,it will become $a_k=a_k times e^{jwkN}$,and in this formula,$w=frac{2 pi}{N}$,so i it will become $a_k=a_k times e^{j pi k}$
,i can't know the period of $a_k$ from here.
fourier-series fourier-transform
asked Nov 28 '18 at 2:39
electronic component
387
|
show 4 more comments
Are the period of signal and its fourier coefficient the same?i mean if the the period of $x[n]$ is $5$,will the period of its fourier coefficient ,$a_k$, also $5$? how to prove it?
Because the definition of a periodic is $x[n]=x[n+N]$,but if we do fourier series at the same time ,it will become $a_k=a_k times e^{jwkN}$,and in this formula,$w=frac{2 pi}{N}$,so i it will become $a_k=a_k times e^{j pi k}$
,i can't know the period of $a_k$ from here.
fourier-series fourier-transform
asked Nov 28 '18 at 2:39
electronic component
387
Did you try a single sine wave signal first?
– Somos
Nov 28 '18 at 2:54
@Somos but if i do that,that maybe the exception,can we just use the definition to prove that?
– electronic component
Nov 28 '18 at 3:03
Next try a finite linear combination of sine waves all with periods a sub-multiple (subharmonics) of a fundamental period.
– Somos
Nov 28 '18 at 3:05
@Somos I have proved it by definition,the answer is "yes",if $a_k$ is a periodic,its period must be equal to its inverse fourier coefficient $x[n]$
– electronic component
Nov 28 '18 at 4:29
I don't understand your question. Can you give a simple example of what you are asking? I don't understand what a coefficient being periodic means. It is just a number. What is $x[n]$? What is $a_k$?
– Somos
Nov 28 '18 at 4:35
|
show 4 more comments
Are the period of signal and its fourier coefficient the same?i mean if the the period of $x[n]$ is $5$,will the period of its fourier coefficient ,$a_k$, also $5$? how to prove it?
Because the definition of a periodic is $x[n]=x[n+N]$,but if we do fourier series at the same time ,it will become $a_k=a_k times e^{jwkN}$,and in this formula,$w=frac{2 pi}{N}$,so i it will become $a_k=a_k times e^{j pi k}$
,i can't know the period of $a_k$ from here.
fourier-series fourier-transform
asked Nov 28 '18 at 2:39
electronic component
387
Are the period of signal and its fourier coefficient the same?i mean if the the period of $x[n]$ is $5$,will the period of its fourier coefficient ,$a_k$, also $5$? how to prove it?
Because the definition of a periodic is $x[n]=x[n+N]$,but if we do fourier series at the same time ,it will become $a_k=a_k times e^{jwkN}$,and in this formula,$w=frac{2 pi}{N}$,so i it will become $a_k=a_k times e^{j pi k}$
,i can't know the period of $a_k$ from here.
fourier-series fourier-transform
fourier-series fourier-transform
asked Nov 28 '18 at 2:39
electronic component
387
asked Nov 28 '18 at 2:39
electronic component
387
asked Nov 28 '18 at 2:39
electronic component
387
asked Nov 28 '18 at 2:39
electronic component
387
asked Nov 28 '18 at 2:39
electronic component
387
387
Did you try a single sine wave signal first?
– Somos
Nov 28 '18 at 2:54
@Somos but if i do that,that maybe the exception,can we just use the definition to prove that?
– electronic component
Nov 28 '18 at 3:03
Next try a finite linear combination of sine waves all with periods a sub-multiple (subharmonics) of a fundamental period.
– Somos
Nov 28 '18 at 3:05
@Somos I have proved it by definition,the answer is "yes",if $a_k$ is a periodic,its period must be equal to its inverse fourier coefficient $x[n]$
– electronic component
Nov 28 '18 at 4:29
I don't understand your question. Can you give a simple example of what you are asking? I don't understand what a coefficient being periodic means. It is just a number. What is $x[n]$? What is $a_k$?
– Somos
Nov 28 '18 at 4:35
|
show 4 more comments
Did you try a single sine wave signal first?
– Somos
Nov 28 '18 at 2:54
@Somos but if i do that,that maybe the exception,can we just use the definition to prove that?
– electronic component
Nov 28 '18 at 3:03
Next try a finite linear combination of sine waves all with periods a sub-multiple (subharmonics) of a fundamental period.
– Somos
Nov 28 '18 at 3:05
@Somos I have proved it by definition,the answer is "yes",if $a_k$ is a periodic,its period must be equal to its inverse fourier coefficient $x[n]$
– electronic component
Nov 28 '18 at 4:29
I don't understand your question. Can you give a simple example of what you are asking? I don't understand what a coefficient being periodic means. It is just a number. What is $x[n]$? What is $a_k$?
– Somos
Nov 28 '18 at 4:35
Did you try a single sine wave signal first?
– Somos
Nov 28 '18 at 2:54
Did you try a single sine wave signal first?
– Somos
Nov 28 '18 at 2:54
@Somos but if i do that,that maybe the exception,can we just use the definition to prove that?
– electronic component
Nov 28 '18 at 3:03
@Somos but if i do that,that maybe the exception,can we just use the definition to prove that?
– electronic component
Nov 28 '18 at 3:03
Next try a finite linear combination of sine waves all with periods a sub-multiple (subharmonics) of a fundamental period.
– Somos
Nov 28 '18 at 3:05
Next try a finite linear combination of sine waves all with periods a sub-multiple (subharmonics) of a fundamental period.
– Somos
Nov 28 '18 at 3:05
@Somos I have proved it by definition,the answer is "yes",if $a_k$ is a periodic,its period must be equal to its inverse fourier coefficient $x[n]$
– electronic component
Nov 28 '18 at 4:29
@Somos I have proved it by definition,the answer is "yes",if $a_k$ is a periodic,its period must be equal to its inverse fourier coefficient $x[n]$
– electronic component
Nov 28 '18 at 4:29
I don't understand your question. Can you give a simple example of what you are asking? I don't understand what a coefficient being periodic means. It is just a number. What is $x[n]$? What is $a_k$?
– Somos
Nov 28 '18 at 4:35
I don't understand your question. Can you give a simple example of what you are asking? I don't understand what a coefficient being periodic means. It is just a number. What is $x[n]$? What is $a_k$?
– Somos
Nov 28 '18 at 4:35
|
show 4 more comments
1 Answer
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if $a_k$ is periodic,then $a_k=a_{k+p}$,$p$ is the period of $a_k$
by definition
$a_k=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}kn}$,and $a_{k+p}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}(z-p)n}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}zn}e^{jfrac{2 pi}{N}pn}$,if $a_{k+p}$ wants to be equal to $a_k$,then $p$ must be equal to $N$,so that $e^{jfrac{2 pi}{N}pn}=e^{j2 pi n} = 1$.
$p$ must be equal to $N$.that is the period of the $a_k = $ the period of the $x[n]$
answered Nov 28 '18 at 6:23
electronic component
387
add a comment |
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1 Answer
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1 Answer
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if $a_k$ is periodic,then $a_k=a_{k+p}$,$p$ is the period of $a_k$
by definition
$a_k=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}kn}$,and $a_{k+p}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}(z-p)n}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}zn}e^{jfrac{2 pi}{N}pn}$,if $a_{k+p}$ wants to be equal to $a_k$,then $p$ must be equal to $N$,so that $e^{jfrac{2 pi}{N}pn}=e^{j2 pi n} = 1$.
$p$ must be equal to $N$.that is the period of the $a_k = $ the period of the $x[n]$
answered Nov 28 '18 at 6:23
electronic component
387
add a comment |
if $a_k$ is periodic,then $a_k=a_{k+p}$,$p$ is the period of $a_k$
by definition
$a_k=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}kn}$,and $a_{k+p}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}(z-p)n}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}zn}e^{jfrac{2 pi}{N}pn}$,if $a_{k+p}$ wants to be equal to $a_k$,then $p$ must be equal to $N$,so that $e^{jfrac{2 pi}{N}pn}=e^{j2 pi n} = 1$.
$p$ must be equal to $N$.that is the period of the $a_k = $ the period of the $x[n]$
answered Nov 28 '18 at 6:23
electronic component
387
add a comment |
if $a_k$ is periodic,then $a_k=a_{k+p}$,$p$ is the period of $a_k$
by definition
$a_k=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}kn}$,and $a_{k+p}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}(z-p)n}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}zn}e^{jfrac{2 pi}{N}pn}$,if $a_{k+p}$ wants to be equal to $a_k$,then $p$ must be equal to $N$,so that $e^{jfrac{2 pi}{N}pn}=e^{j2 pi n} = 1$.
$p$ must be equal to $N$.that is the period of the $a_k = $ the period of the $x[n]$
answered Nov 28 '18 at 6:23
electronic component
387
if $a_k$ is periodic,then $a_k=a_{k+p}$,$p$ is the period of $a_k$
by definition
$a_k=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}kn}$,and $a_{k+p}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}(z-p)n}=frac{1}{N}sumlimits_{n=-infty}^{infty}e^{-jfrac{2 pi}{N}zn}e^{jfrac{2 pi}{N}pn}$,if $a_{k+p}$ wants to be equal to $a_k$,then $p$ must be equal to $N$,so that $e^{jfrac{2 pi}{N}pn}=e^{j2 pi n} = 1$.
$p$ must be equal to $N$.that is the period of the $a_k = $ the period of the $x[n]$
answered Nov 28 '18 at 6:23
electronic component
387
answered Nov 28 '18 at 6:23
electronic component
387
answered Nov 28 '18 at 6:23
electronic component
387
answered Nov 28 '18 at 6:23
electronic component
387
387
add a comment |
add a comment |
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Did you try a single sine wave signal first?
– Somos
Nov 28 '18 at 2:54
@Somos but if i do that,that maybe the exception,can we just use the definition to prove that?
– electronic component
Nov 28 '18 at 3:03
Next try a finite linear combination of sine waves all with periods a sub-multiple (subharmonics) of a fundamental period.
– Somos
Nov 28 '18 at 3:05
@Somos I have proved it by definition,the answer is "yes",if $a_k$ is a periodic,its period must be equal to its inverse fourier coefficient $x[n]$
– electronic component
Nov 28 '18 at 4:29
I don't understand your question. Can you give a simple example of what you are asking? I don't understand what a coefficient being periodic means. It is just a number. What is $x[n]$? What is $a_k$?
– Somos
Nov 28 '18 at 4:35