What is the period of this signal?
$begingroup$
Below is the signal :
$y[n] = j ^ n$
Someone told me that the period is 4 ,but he didn't explain me why. Can anyone help me ?
signal-processing periodic-functions
asked Jan 13 at 11:16
zil derzil der
31
$endgroup$
add a comment |
$begingroup$
Below is the signal :
$y[n] = j ^ n$
Someone told me that the period is 4 ,but he didn't explain me why. Can anyone help me ?
signal-processing periodic-functions
asked Jan 13 at 11:16
zil derzil der
31
$endgroup$
add a comment |
$begingroup$
Below is the signal :
$y[n] = j ^ n$
Someone told me that the period is 4 ,but he didn't explain me why. Can anyone help me ?
signal-processing periodic-functions
asked Jan 13 at 11:16
zil derzil der
31
$endgroup$
Below is the signal :
$y[n] = j ^ n$
Someone told me that the period is 4 ,but he didn't explain me why. Can anyone help me ?
signal-processing periodic-functions
signal-processing periodic-functions
asked Jan 13 at 11:16
zil derzil der
31
asked Jan 13 at 11:16
zil derzil der
31
edited Jan 13 at 11:32
zil der
asked Jan 13 at 11:16
zil derzil der
31
asked Jan 13 at 11:16
zil derzil der
31
asked Jan 13 at 11:16
zil derzil der
31
31
add a comment |
add a comment |
1 Answer
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I presume that "j" here is the "imaginary unit", that I would call "i", the principal square root of -1 in the complex number system. By definition then $j^2= -1$. Then $j^3= j(j^2)= j(-1)= -j$ and $j^4= (j^2)(j^2)= (-1)(-1)= 1$. If we continue the same way we get $j^5= j$, $j^6= -1$, $j^7= -j$, $j^8= 1$, etc. Every power of j that is a multiple of 4 gets us back to 1 so the sequence of powers repeats every 4 places. It is in that sense that multiplication by j is "periodic with period 4".
answered Jan 13 at 12:26
user247327user247327
11.7k1516
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$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
I presume that "j" here is the "imaginary unit", that I would call "i", the principal square root of -1 in the complex number system. By definition then $j^2= -1$. Then $j^3= j(j^2)= j(-1)= -j$ and $j^4= (j^2)(j^2)= (-1)(-1)= 1$. If we continue the same way we get $j^5= j$, $j^6= -1$, $j^7= -j$, $j^8= 1$, etc. Every power of j that is a multiple of 4 gets us back to 1 so the sequence of powers repeats every 4 places. It is in that sense that multiplication by j is "periodic with period 4".
answered Jan 13 at 12:26
user247327user247327
11.7k1516
$endgroup$
$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
add a comment |
$begingroup$
I presume that "j" here is the "imaginary unit", that I would call "i", the principal square root of -1 in the complex number system. By definition then $j^2= -1$. Then $j^3= j(j^2)= j(-1)= -j$ and $j^4= (j^2)(j^2)= (-1)(-1)= 1$. If we continue the same way we get $j^5= j$, $j^6= -1$, $j^7= -j$, $j^8= 1$, etc. Every power of j that is a multiple of 4 gets us back to 1 so the sequence of powers repeats every 4 places. It is in that sense that multiplication by j is "periodic with period 4".
answered Jan 13 at 12:26
user247327user247327
11.7k1516
$endgroup$
$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
add a comment |
$begingroup$
I presume that "j" here is the "imaginary unit", that I would call "i", the principal square root of -1 in the complex number system. By definition then $j^2= -1$. Then $j^3= j(j^2)= j(-1)= -j$ and $j^4= (j^2)(j^2)= (-1)(-1)= 1$. If we continue the same way we get $j^5= j$, $j^6= -1$, $j^7= -j$, $j^8= 1$, etc. Every power of j that is a multiple of 4 gets us back to 1 so the sequence of powers repeats every 4 places. It is in that sense that multiplication by j is "periodic with period 4".
answered Jan 13 at 12:26
user247327user247327
11.7k1516
$endgroup$
I presume that "j" here is the "imaginary unit", that I would call "i", the principal square root of -1 in the complex number system. By definition then $j^2= -1$. Then $j^3= j(j^2)= j(-1)= -j$ and $j^4= (j^2)(j^2)= (-1)(-1)= 1$. If we continue the same way we get $j^5= j$, $j^6= -1$, $j^7= -j$, $j^8= 1$, etc. Every power of j that is a multiple of 4 gets us back to 1 so the sequence of powers repeats every 4 places. It is in that sense that multiplication by j is "periodic with period 4".
answered Jan 13 at 12:26
user247327user247327
11.7k1516
answered Jan 13 at 12:26
user247327user247327
11.7k1516
answered Jan 13 at 12:26
user247327user247327
11.7k1516
answered Jan 13 at 12:26
user247327user247327
11.7k1516
11.7k1516
$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
add a comment |
$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
$begingroup$
Thank you ,very much !!!
$endgroup$
– zil der
Jan 14 at 22:31
add a comment |
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