Why does $x[k] = k^k$ not have a Z-transform?
0
$begingroup$
The signal $x[k] = k^k, k=1,2,3...$ does not have a Z-transform. Why?
The definition of the Z-transform is:
$$X[z] = sum_{n=-infty}^{infty} x[n] z^{-n}$$
control-theory z-transform
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
$endgroup$
add a comment |
0
$begingroup$
The signal $x[k] = k^k, k=1,2,3...$ does not have a Z-transform. Why?
The definition of the Z-transform is:
$$X[z] = sum_{n=-infty}^{infty} x[n] z^{-n}$$
control-theory z-transform
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
$endgroup$
2
$begingroup$
It doesn't converge for any value of $z$ except $0$. That doesn't necessarily matter, but if you're expecting the end result to be a function there is no such function.
$endgroup$
– Matt Samuel
Dec 5 '18 at 23:42
add a comment |
0
0
0
$begingroup$
The signal $x[k] = k^k, k=1,2,3...$ does not have a Z-transform. Why?
The definition of the Z-transform is:
$$X[z] = sum_{n=-infty}^{infty} x[n] z^{-n}$$
control-theory z-transform
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
$endgroup$
The signal $x[k] = k^k, k=1,2,3...$ does not have a Z-transform. Why?
The definition of the Z-transform is:
$$X[z] = sum_{n=-infty}^{infty} x[n] z^{-n}$$
control-theory z-transform
control-theory z-transform
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
asked Dec 5 '18 at 23:18
David FerrisDavid Ferris
1136
1136
2
$begingroup$
It doesn't converge for any value of $z$ except $0$. That doesn't necessarily matter, but if you're expecting the end result to be a function there is no such function.
$endgroup$
– Matt Samuel
Dec 5 '18 at 23:42
add a comment |
2
$begingroup$
It doesn't converge for any value of $z$ except $0$. That doesn't necessarily matter, but if you're expecting the end result to be a function there is no such function.
$endgroup$
– Matt Samuel
Dec 5 '18 at 23:42
2
2
$begingroup$
It doesn't converge for any value of $z$ except $0$. That doesn't necessarily matter, but if you're expecting the end result to be a function there is no such function.
$endgroup$
– Matt Samuel
Dec 5 '18 at 23:42
$begingroup$
It doesn't converge for any value of $z$ except $0$. That doesn't necessarily matter, but if you're expecting the end result to be a function there is no such function.
$endgroup$
– Matt Samuel
Dec 5 '18 at 23:42
add a comment |
0
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$begingroup$
It doesn't converge for any value of $z$ except $0$. That doesn't necessarily matter, but if you're expecting the end result to be a function there is no such function.
$endgroup$
– Matt Samuel
Dec 5 '18 at 23:42