What does this notation mean? Double arrow with $z$ above
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I am reading a paper about digital filtering (for the very first time) and I found this notation (double arrow with $z$ above) which I do not quite understand.
Could you please give me some hint?
Defining the inverse convolution operator $(b_1^n)^{-1}(k) stackrel{z}{longleftrightarrow} 1 / B_1^n(z)$, the solution is found by inverse filtering (cf. [97])
convolution spline
edited Nov 26 at 10:15
user302797
19.4k92252
asked Nov 26 at 9:56
Ramiro Scorolli
64513
add a comment |
up vote
1
down vote
favorite
I am reading a paper about digital filtering (for the very first time) and I found this notation (double arrow with $z$ above) which I do not quite understand.
Could you please give me some hint?
Defining the inverse convolution operator $(b_1^n)^{-1}(k) stackrel{z}{longleftrightarrow} 1 / B_1^n(z)$, the solution is found by inverse filtering (cf. [97])
convolution spline
edited Nov 26 at 10:15
user302797
19.4k92252
asked Nov 26 at 9:56
Ramiro Scorolli
64513
2
Does it make sense if it is a Z-transform? en.wikipedia.org/wiki/Z-transform
– Kusma
Nov 26 at 10:00
indeed, it makes sense! I wasn't aware of the existence of such a transform, I thought that they were dealing with Discrete Time Fourier Transform. You can post it as an answer.
– Ramiro Scorolli
Nov 26 at 10:04
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading a paper about digital filtering (for the very first time) and I found this notation (double arrow with $z$ above) which I do not quite understand.
Could you please give me some hint?
Defining the inverse convolution operator $(b_1^n)^{-1}(k) stackrel{z}{longleftrightarrow} 1 / B_1^n(z)$, the solution is found by inverse filtering (cf. [97])
convolution spline
edited Nov 26 at 10:15
user302797
19.4k92252
asked Nov 26 at 9:56
Ramiro Scorolli
64513
I am reading a paper about digital filtering (for the very first time) and I found this notation (double arrow with $z$ above) which I do not quite understand.
Could you please give me some hint?
Defining the inverse convolution operator $(b_1^n)^{-1}(k) stackrel{z}{longleftrightarrow} 1 / B_1^n(z)$, the solution is found by inverse filtering (cf. [97])
convolution spline
convolution spline
edited Nov 26 at 10:15
user302797
19.4k92252
asked Nov 26 at 9:56
Ramiro Scorolli
64513
edited Nov 26 at 10:15
user302797
19.4k92252
asked Nov 26 at 9:56
Ramiro Scorolli
64513
edited Nov 26 at 10:15
user302797
19.4k92252
edited Nov 26 at 10:15
user302797
19.4k92252
edited Nov 26 at 10:15
user302797
19.4k92252
19.4k92252
asked Nov 26 at 9:56
Ramiro Scorolli
64513
asked Nov 26 at 9:56
Ramiro Scorolli
64513
asked Nov 26 at 9:56
Ramiro Scorolli
64513
64513
2
Does it make sense if it is a Z-transform? en.wikipedia.org/wiki/Z-transform
– Kusma
Nov 26 at 10:00
indeed, it makes sense! I wasn't aware of the existence of such a transform, I thought that they were dealing with Discrete Time Fourier Transform. You can post it as an answer.
– Ramiro Scorolli
Nov 26 at 10:04
add a comment |
2
Does it make sense if it is a Z-transform? en.wikipedia.org/wiki/Z-transform
– Kusma
Nov 26 at 10:00
indeed, it makes sense! I wasn't aware of the existence of such a transform, I thought that they were dealing with Discrete Time Fourier Transform. You can post it as an answer.
– Ramiro Scorolli
Nov 26 at 10:04
2
2
Does it make sense if it is a Z-transform? en.wikipedia.org/wiki/Z-transform
– Kusma
Nov 26 at 10:00
Does it make sense if it is a Z-transform? en.wikipedia.org/wiki/Z-transform
– Kusma
Nov 26 at 10:00
indeed, it makes sense! I wasn't aware of the existence of such a transform, I thought that they were dealing with Discrete Time Fourier Transform. You can post it as an answer.
– Ramiro Scorolli
Nov 26 at 10:04
indeed, it makes sense! I wasn't aware of the existence of such a transform, I thought that they were dealing with Discrete Time Fourier Transform. You can post it as an answer.
– Ramiro Scorolli
Nov 26 at 10:04
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
It is a Z transform. Just like Fourier transform, it turns convolutions into multiplications, so it can be used to undo convolutions.
answered Nov 26 at 10:33
Kusma
3,7331219
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It is a Z transform. Just like Fourier transform, it turns convolutions into multiplications, so it can be used to undo convolutions.
answered Nov 26 at 10:33
Kusma
3,7331219
add a comment |
up vote
1
down vote
accepted
It is a Z transform. Just like Fourier transform, it turns convolutions into multiplications, so it can be used to undo convolutions.
answered Nov 26 at 10:33
Kusma
3,7331219
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It is a Z transform. Just like Fourier transform, it turns convolutions into multiplications, so it can be used to undo convolutions.
answered Nov 26 at 10:33
Kusma
3,7331219
It is a Z transform. Just like Fourier transform, it turns convolutions into multiplications, so it can be used to undo convolutions.
answered Nov 26 at 10:33
Kusma
3,7331219
answered Nov 26 at 10:33
Kusma
3,7331219
answered Nov 26 at 10:33
Kusma
3,7331219
answered Nov 26 at 10:33
Kusma
3,7331219
3,7331219
add a comment |
add a comment |
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2
Does it make sense if it is a Z-transform? en.wikipedia.org/wiki/Z-transform
– Kusma
Nov 26 at 10:00
indeed, it makes sense! I wasn't aware of the existence of such a transform, I thought that they were dealing with Discrete Time Fourier Transform. You can post it as an answer.
– Ramiro Scorolli
Nov 26 at 10:04