Is compressive sensing a type of interpolation? [closed]
$begingroup$
In what ways is compressive sensing different from traditional numerical interpolation of sampled points from a given signal?
numerical-methods signal-processing numerical-optimization
asked Dec 30 '18 at 21:23
TerrellTerrell
132
$endgroup$
closed as off-topic by Brian Borchers, Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz Dec 31 '18 at 8:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In what ways is compressive sensing different from traditional numerical interpolation of sampled points from a given signal?
numerical-methods signal-processing numerical-optimization
asked Dec 30 '18 at 21:23
TerrellTerrell
132
$endgroup$
closed as off-topic by Brian Borchers, Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz Dec 31 '18 at 8:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
In compressed sensing the measurements are random projections of the signal that we want to recover. In traditional numerical interpolation, we are given measurements of the signal itself. In compressed sensing, we know that the signal we want to recover is sparse, and the strategy is to find a sparse signal which is consistent with the given measurements. In traditional interpolation, sparsity plays no role.
$endgroup$
– littleO
Dec 30 '18 at 22:43
add a comment |
$begingroup$
In what ways is compressive sensing different from traditional numerical interpolation of sampled points from a given signal?
numerical-methods signal-processing numerical-optimization
asked Dec 30 '18 at 21:23
TerrellTerrell
132
$endgroup$
In what ways is compressive sensing different from traditional numerical interpolation of sampled points from a given signal?
numerical-methods signal-processing numerical-optimization
numerical-methods signal-processing numerical-optimization
asked Dec 30 '18 at 21:23
TerrellTerrell
132
asked Dec 30 '18 at 21:23
TerrellTerrell
132
asked Dec 30 '18 at 21:23
TerrellTerrell
132
asked Dec 30 '18 at 21:23
TerrellTerrell
132
asked Dec 30 '18 at 21:23
TerrellTerrell
132
132
closed as off-topic by Brian Borchers, Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz Dec 31 '18 at 8:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brian Borchers, Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz Dec 31 '18 at 8:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lee David Chung Lin, Andrew, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
In compressed sensing the measurements are random projections of the signal that we want to recover. In traditional numerical interpolation, we are given measurements of the signal itself. In compressed sensing, we know that the signal we want to recover is sparse, and the strategy is to find a sparse signal which is consistent with the given measurements. In traditional interpolation, sparsity plays no role.
$endgroup$
– littleO
Dec 30 '18 at 22:43
add a comment |
2
$begingroup$
In compressed sensing the measurements are random projections of the signal that we want to recover. In traditional numerical interpolation, we are given measurements of the signal itself. In compressed sensing, we know that the signal we want to recover is sparse, and the strategy is to find a sparse signal which is consistent with the given measurements. In traditional interpolation, sparsity plays no role.
$endgroup$
– littleO
Dec 30 '18 at 22:43
2
2
$begingroup$
In compressed sensing the measurements are random projections of the signal that we want to recover. In traditional numerical interpolation, we are given measurements of the signal itself. In compressed sensing, we know that the signal we want to recover is sparse, and the strategy is to find a sparse signal which is consistent with the given measurements. In traditional interpolation, sparsity plays no role.
$endgroup$
– littleO
Dec 30 '18 at 22:43
$begingroup$
In compressed sensing the measurements are random projections of the signal that we want to recover. In traditional numerical interpolation, we are given measurements of the signal itself. In compressed sensing, we know that the signal we want to recover is sparse, and the strategy is to find a sparse signal which is consistent with the given measurements. In traditional interpolation, sparsity plays no role.
$endgroup$
– littleO
Dec 30 '18 at 22:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The "traditional" compressive sensing (CS) considers the problem of recovering exactly an unknown vector $mathbf{x} in mathbb{R}^N$ from $m<N$ linear observations, i.e., from a vector $mathbf{y} = mathbf{A} mathbf{x}$, where $mathbf{A} in mathbb{R}^{mtimes N}$ is a, so called, sensing matrix (which is known). As the resulting system of linear equations is underdetermined, CS makes the additional assumption that $mathbf{x}$ has only $s ll N$ non-zero elements when expressed with respect to a certain (known) basis, i.e., $mathbf{x}$ is sparse. CS theory investigates what properties $mathbf{A}$ should have in order to recover $mathbf{x}$ with the smallest possible number of observations $m$, as well as designing practical (numerical) algorithms to do that.
Interpolation is usually considered as the problem of recovering an unknown vector $mathbf{x} in mathbb{R}^N$ by directly observing only $m<N$ of its elements. Since, as stated, this problem is ill posed, common interpolation methods such as linear, cubic or spline make the additional assumption that the values of consecutive elements of $mathbf{x}$ are not significantly different, i.e., $mathbf{x}$ is "smooth". In general, conventional interpolation does not guarantee exact recovery of the unknown vector. A notable exception are bandlimited signals, which can indeed be recovered exactly under a sampling scheme with periodic samples, taken with sufficiently large (Nyquist) frequency.
Therefore, the major differneces of CS and conventional interpolation are
Observation model: CS considers a more general observation model than conventional interpolation
Unknown vector assumptions: CS considers the recovery of sparse vectors (w.r.t. some basis) whereas interpolation considers "smooth"/"bandlimited" vectors. Note that, in general, a sparse vector is not "smooth"/"bandlimited" and vice versa.
Theoretical analysis: CS provides rigorous guarantees for the exact recovery of sparse vectors, whereas few things can be said on the exact recovery of signals by numerical interpolation.
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
$endgroup$
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The "traditional" compressive sensing (CS) considers the problem of recovering exactly an unknown vector $mathbf{x} in mathbb{R}^N$ from $m<N$ linear observations, i.e., from a vector $mathbf{y} = mathbf{A} mathbf{x}$, where $mathbf{A} in mathbb{R}^{mtimes N}$ is a, so called, sensing matrix (which is known). As the resulting system of linear equations is underdetermined, CS makes the additional assumption that $mathbf{x}$ has only $s ll N$ non-zero elements when expressed with respect to a certain (known) basis, i.e., $mathbf{x}$ is sparse. CS theory investigates what properties $mathbf{A}$ should have in order to recover $mathbf{x}$ with the smallest possible number of observations $m$, as well as designing practical (numerical) algorithms to do that.
Interpolation is usually considered as the problem of recovering an unknown vector $mathbf{x} in mathbb{R}^N$ by directly observing only $m<N$ of its elements. Since, as stated, this problem is ill posed, common interpolation methods such as linear, cubic or spline make the additional assumption that the values of consecutive elements of $mathbf{x}$ are not significantly different, i.e., $mathbf{x}$ is "smooth". In general, conventional interpolation does not guarantee exact recovery of the unknown vector. A notable exception are bandlimited signals, which can indeed be recovered exactly under a sampling scheme with periodic samples, taken with sufficiently large (Nyquist) frequency.
Therefore, the major differneces of CS and conventional interpolation are
Observation model: CS considers a more general observation model than conventional interpolation
Unknown vector assumptions: CS considers the recovery of sparse vectors (w.r.t. some basis) whereas interpolation considers "smooth"/"bandlimited" vectors. Note that, in general, a sparse vector is not "smooth"/"bandlimited" and vice versa.
Theoretical analysis: CS provides rigorous guarantees for the exact recovery of sparse vectors, whereas few things can be said on the exact recovery of signals by numerical interpolation.
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
$endgroup$
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
add a comment |
$begingroup$
The "traditional" compressive sensing (CS) considers the problem of recovering exactly an unknown vector $mathbf{x} in mathbb{R}^N$ from $m<N$ linear observations, i.e., from a vector $mathbf{y} = mathbf{A} mathbf{x}$, where $mathbf{A} in mathbb{R}^{mtimes N}$ is a, so called, sensing matrix (which is known). As the resulting system of linear equations is underdetermined, CS makes the additional assumption that $mathbf{x}$ has only $s ll N$ non-zero elements when expressed with respect to a certain (known) basis, i.e., $mathbf{x}$ is sparse. CS theory investigates what properties $mathbf{A}$ should have in order to recover $mathbf{x}$ with the smallest possible number of observations $m$, as well as designing practical (numerical) algorithms to do that.
Interpolation is usually considered as the problem of recovering an unknown vector $mathbf{x} in mathbb{R}^N$ by directly observing only $m<N$ of its elements. Since, as stated, this problem is ill posed, common interpolation methods such as linear, cubic or spline make the additional assumption that the values of consecutive elements of $mathbf{x}$ are not significantly different, i.e., $mathbf{x}$ is "smooth". In general, conventional interpolation does not guarantee exact recovery of the unknown vector. A notable exception are bandlimited signals, which can indeed be recovered exactly under a sampling scheme with periodic samples, taken with sufficiently large (Nyquist) frequency.
Therefore, the major differneces of CS and conventional interpolation are
Observation model: CS considers a more general observation model than conventional interpolation
Unknown vector assumptions: CS considers the recovery of sparse vectors (w.r.t. some basis) whereas interpolation considers "smooth"/"bandlimited" vectors. Note that, in general, a sparse vector is not "smooth"/"bandlimited" and vice versa.
Theoretical analysis: CS provides rigorous guarantees for the exact recovery of sparse vectors, whereas few things can be said on the exact recovery of signals by numerical interpolation.
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
$endgroup$
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
add a comment |
$begingroup$
The "traditional" compressive sensing (CS) considers the problem of recovering exactly an unknown vector $mathbf{x} in mathbb{R}^N$ from $m<N$ linear observations, i.e., from a vector $mathbf{y} = mathbf{A} mathbf{x}$, where $mathbf{A} in mathbb{R}^{mtimes N}$ is a, so called, sensing matrix (which is known). As the resulting system of linear equations is underdetermined, CS makes the additional assumption that $mathbf{x}$ has only $s ll N$ non-zero elements when expressed with respect to a certain (known) basis, i.e., $mathbf{x}$ is sparse. CS theory investigates what properties $mathbf{A}$ should have in order to recover $mathbf{x}$ with the smallest possible number of observations $m$, as well as designing practical (numerical) algorithms to do that.
Interpolation is usually considered as the problem of recovering an unknown vector $mathbf{x} in mathbb{R}^N$ by directly observing only $m<N$ of its elements. Since, as stated, this problem is ill posed, common interpolation methods such as linear, cubic or spline make the additional assumption that the values of consecutive elements of $mathbf{x}$ are not significantly different, i.e., $mathbf{x}$ is "smooth". In general, conventional interpolation does not guarantee exact recovery of the unknown vector. A notable exception are bandlimited signals, which can indeed be recovered exactly under a sampling scheme with periodic samples, taken with sufficiently large (Nyquist) frequency.
Therefore, the major differneces of CS and conventional interpolation are
Observation model: CS considers a more general observation model than conventional interpolation
Unknown vector assumptions: CS considers the recovery of sparse vectors (w.r.t. some basis) whereas interpolation considers "smooth"/"bandlimited" vectors. Note that, in general, a sparse vector is not "smooth"/"bandlimited" and vice versa.
Theoretical analysis: CS provides rigorous guarantees for the exact recovery of sparse vectors, whereas few things can be said on the exact recovery of signals by numerical interpolation.
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
$endgroup$
The "traditional" compressive sensing (CS) considers the problem of recovering exactly an unknown vector $mathbf{x} in mathbb{R}^N$ from $m<N$ linear observations, i.e., from a vector $mathbf{y} = mathbf{A} mathbf{x}$, where $mathbf{A} in mathbb{R}^{mtimes N}$ is a, so called, sensing matrix (which is known). As the resulting system of linear equations is underdetermined, CS makes the additional assumption that $mathbf{x}$ has only $s ll N$ non-zero elements when expressed with respect to a certain (known) basis, i.e., $mathbf{x}$ is sparse. CS theory investigates what properties $mathbf{A}$ should have in order to recover $mathbf{x}$ with the smallest possible number of observations $m$, as well as designing practical (numerical) algorithms to do that.
Interpolation is usually considered as the problem of recovering an unknown vector $mathbf{x} in mathbb{R}^N$ by directly observing only $m<N$ of its elements. Since, as stated, this problem is ill posed, common interpolation methods such as linear, cubic or spline make the additional assumption that the values of consecutive elements of $mathbf{x}$ are not significantly different, i.e., $mathbf{x}$ is "smooth". In general, conventional interpolation does not guarantee exact recovery of the unknown vector. A notable exception are bandlimited signals, which can indeed be recovered exactly under a sampling scheme with periodic samples, taken with sufficiently large (Nyquist) frequency.
Therefore, the major differneces of CS and conventional interpolation are
Observation model: CS considers a more general observation model than conventional interpolation
Unknown vector assumptions: CS considers the recovery of sparse vectors (w.r.t. some basis) whereas interpolation considers "smooth"/"bandlimited" vectors. Note that, in general, a sparse vector is not "smooth"/"bandlimited" and vice versa.
Theoretical analysis: CS provides rigorous guarantees for the exact recovery of sparse vectors, whereas few things can be said on the exact recovery of signals by numerical interpolation.
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
answered Dec 30 '18 at 23:13
SteliosStelios
2,161279
2,161279
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
add a comment |
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
$begingroup$
Awesome explanation thanks for the comment
$endgroup$
– Terrell
Jan 6 at 23:26
add a comment |
2
$begingroup$
In compressed sensing the measurements are random projections of the signal that we want to recover. In traditional numerical interpolation, we are given measurements of the signal itself. In compressed sensing, we know that the signal we want to recover is sparse, and the strategy is to find a sparse signal which is consistent with the given measurements. In traditional interpolation, sparsity plays no role.
$endgroup$
– littleO
Dec 30 '18 at 22:43