Why does decimation make a system time variant?
On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."
Why does decimation makes system time variant?
wavelet linear-systems
edited 2 days ago
Matt L.
49.3k13684
asked 2 days ago
Sweeper
1548
add a comment |
On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."
Why does decimation makes system time variant?
wavelet linear-systems
edited 2 days ago
Matt L.
49.3k13684
asked 2 days ago
Sweeper
1548
1
On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
2 days ago
That link doesnt work.
– Sweeper
2 days ago
1
try again please.
– Olli Niemitalo
2 days ago
add a comment |
On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."
Why does decimation makes system time variant?
wavelet linear-systems
edited 2 days ago
Matt L.
49.3k13684
asked 2 days ago
Sweeper
1548
On Wikipedia I read this : "The Discrete Wavelet Transform, often used in modern signal processing, is time variant because it makes use of the decimation operation."
Why does decimation makes system time variant?
wavelet linear-systems
wavelet linear-systems
edited 2 days ago
Matt L.
49.3k13684
asked 2 days ago
Sweeper
1548
edited 2 days ago
Matt L.
49.3k13684
asked 2 days ago
Sweeper
1548
edited 2 days ago
Matt L.
49.3k13684
edited 2 days ago
Matt L.
49.3k13684
edited 2 days ago
Matt L.
49.3k13684
49.3k13684
asked 2 days ago
Sweeper
1548
asked 2 days ago
Sweeper
1548
asked 2 days ago
Sweeper
1548
1548
1
On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
2 days ago
That link doesnt work.
– Sweeper
2 days ago
1
try again please.
– Olli Niemitalo
2 days ago
add a comment |
1
On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
2 days ago
That link doesnt work.
– Sweeper
2 days ago
1
try again please.
– Olli Niemitalo
2 days ago
1
1
On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
2 days ago
On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
2 days ago
That link doesnt work.
– Sweeper
2 days ago
That link doesnt work.
– Sweeper
2 days ago
1
1
try again please.
– Olli Niemitalo
2 days ago
try again please.
– Olli Niemitalo
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
HINT:
Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.
answered 2 days ago
Matt L.
49.3k13684
I dont understand.
– Sweeper
2 days ago
2
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
1
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
add a comment |
The continuous wavelet transform is aimed at being shift-invariant (or equivariant, if you do care about terms, see Difference between “equivariant to translation” and “invariant to translation”). While some redundant versions of their discrete implementations can be made invariant (stationary, undecimated or shift/time-invariant discrete wavelets, see Shift invariant in wavelet), the classical critically sampled discrete wavelet is not.
For an $M$-band (classically $M=2$) discrete wavelet transform, over $L$ levels, the transform is equivariant to multiples of $M^L$ shifts only, and generally not in-between, unless the deterministic signal, or the stochastic noise properties, allows for more.
answered yesterday
Laurent Duval
16.4k32058
add a comment |
Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse discrete wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation (usually with quadrature mirror filters) that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.
However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance as in user Matt L's example.
Instead of aliasing cancellation, it would also work to have perfect anti-alias filters, but this is typically not practical, because of the infinite length required of the impulse responses of such filters, or because they would introduce Gibbs phenomenon ringing.
answered 2 days ago
Olli Niemitalo
7,9381234
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT:
Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.
answered 2 days ago
Matt L.
49.3k13684
I dont understand.
– Sweeper
2 days ago
2
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
1
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
add a comment |
HINT:
Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.
answered 2 days ago
Matt L.
49.3k13684
I dont understand.
– Sweeper
2 days ago
2
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
1
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
add a comment |
HINT:
Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.
answered 2 days ago
Matt L.
49.3k13684
HINT:
Think about the following situation: you have an input signal with alternating ones and zeros, and you decimate by a factor of $2$. Now shift the input signal by one sample and compute a new output signal by decimating by a factor of $2$. Is the output signal a shifted version of the previous output signal? If the system is time-invariant, shifting the input signal causes the output signal to shift by the same amount.
answered 2 days ago
Matt L.
49.3k13684
answered 2 days ago
Matt L.
49.3k13684
answered 2 days ago
Matt L.
49.3k13684
answered 2 days ago
Matt L.
49.3k13684
49.3k13684
I dont understand.
– Sweeper
2 days ago
2
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
1
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
add a comment |
I dont understand.
– Sweeper
2 days ago
2
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
1
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
I dont understand.
– Sweeper
2 days ago
I dont understand.
– Sweeper
2 days ago
2
2
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
@Sweeper: Imagine the signal I gave as an example. If you take, say, all even samples you end up with only ones at the output. If you take all odd samples you end up with all zeros at the output. So shifting the input by 1 sample totally changes the output, hence the system cannot be time-invariant.
– Matt L.
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
Ah yes, I get it now. My problem was that I thought decimation of 1 - 0 - 1 - 0 signal would give 0.5 - 0.5 signal.
– Sweeper
2 days ago
1
1
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
Depending on the source, decimation or downsampling may include a prior filtering step, see for instance dsp.stackexchange.com/questions/45276/…
– Laurent Duval
yesterday
add a comment |
The continuous wavelet transform is aimed at being shift-invariant (or equivariant, if you do care about terms, see Difference between “equivariant to translation” and “invariant to translation”). While some redundant versions of their discrete implementations can be made invariant (stationary, undecimated or shift/time-invariant discrete wavelets, see Shift invariant in wavelet), the classical critically sampled discrete wavelet is not.
For an $M$-band (classically $M=2$) discrete wavelet transform, over $L$ levels, the transform is equivariant to multiples of $M^L$ shifts only, and generally not in-between, unless the deterministic signal, or the stochastic noise properties, allows for more.
answered yesterday
Laurent Duval
16.4k32058
add a comment |
The continuous wavelet transform is aimed at being shift-invariant (or equivariant, if you do care about terms, see Difference between “equivariant to translation” and “invariant to translation”). While some redundant versions of their discrete implementations can be made invariant (stationary, undecimated or shift/time-invariant discrete wavelets, see Shift invariant in wavelet), the classical critically sampled discrete wavelet is not.
For an $M$-band (classically $M=2$) discrete wavelet transform, over $L$ levels, the transform is equivariant to multiples of $M^L$ shifts only, and generally not in-between, unless the deterministic signal, or the stochastic noise properties, allows for more.
answered yesterday
Laurent Duval
16.4k32058
add a comment |
The continuous wavelet transform is aimed at being shift-invariant (or equivariant, if you do care about terms, see Difference between “equivariant to translation” and “invariant to translation”). While some redundant versions of their discrete implementations can be made invariant (stationary, undecimated or shift/time-invariant discrete wavelets, see Shift invariant in wavelet), the classical critically sampled discrete wavelet is not.
For an $M$-band (classically $M=2$) discrete wavelet transform, over $L$ levels, the transform is equivariant to multiples of $M^L$ shifts only, and generally not in-between, unless the deterministic signal, or the stochastic noise properties, allows for more.
answered yesterday
Laurent Duval
16.4k32058
The continuous wavelet transform is aimed at being shift-invariant (or equivariant, if you do care about terms, see Difference between “equivariant to translation” and “invariant to translation”). While some redundant versions of their discrete implementations can be made invariant (stationary, undecimated or shift/time-invariant discrete wavelets, see Shift invariant in wavelet), the classical critically sampled discrete wavelet is not.
For an $M$-band (classically $M=2$) discrete wavelet transform, over $L$ levels, the transform is equivariant to multiples of $M^L$ shifts only, and generally not in-between, unless the deterministic signal, or the stochastic noise properties, allows for more.
answered yesterday
Laurent Duval
16.4k32058
answered yesterday
Laurent Duval
16.4k32058
answered yesterday
Laurent Duval
16.4k32058
answered yesterday
Laurent Duval
16.4k32058
16.4k32058
add a comment |
add a comment |
Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse discrete wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation (usually with quadrature mirror filters) that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.
However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance as in user Matt L's example.
Instead of aliasing cancellation, it would also work to have perfect anti-alias filters, but this is typically not practical, because of the infinite length required of the impulse responses of such filters, or because they would introduce Gibbs phenomenon ringing.
answered 2 days ago
Olli Niemitalo
7,9381234
add a comment |
Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse discrete wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation (usually with quadrature mirror filters) that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.
However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance as in user Matt L's example.
Instead of aliasing cancellation, it would also work to have perfect anti-alias filters, but this is typically not practical, because of the infinite length required of the impulse responses of such filters, or because they would introduce Gibbs phenomenon ringing.
answered 2 days ago
Olli Niemitalo
7,9381234
add a comment |
Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse discrete wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation (usually with quadrature mirror filters) that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.
However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance as in user Matt L's example.
Instead of aliasing cancellation, it would also work to have perfect anti-alias filters, but this is typically not practical, because of the infinite length required of the impulse responses of such filters, or because they would introduce Gibbs phenomenon ringing.
answered 2 days ago
Olli Niemitalo
7,9381234
Decimation as part of the calculations does not necessarily lead to a time-invariant (shift-invariant) system. Consider the system: DWT → IDWT, where IDWT stands for inverse discrete wavelet transform. Typically, a DWT together with its inverse transform incorporates aliasing cancellation (usually with quadrature mirror filters) that enables to reconstruct the original signal even when decimation is used in DWT, making the full system shift-invariant.
However, typically the transformed signal would be somehow processed before IDWT, which might break the aliasing cancellation and lead to shift-invariance as in user Matt L's example.
Instead of aliasing cancellation, it would also work to have perfect anti-alias filters, but this is typically not practical, because of the infinite length required of the impulse responses of such filters, or because they would introduce Gibbs phenomenon ringing.
answered 2 days ago
Olli Niemitalo
7,9381234
edited yesterday
answered 2 days ago
Olli Niemitalo
7,9381234
answered 2 days ago
Olli Niemitalo
7,9381234
answered 2 days ago
Olli Niemitalo
7,9381234
7,9381234
add a comment |
add a comment |
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1
On the Wikipedia talk page someone complains about calling the discrete wavelet transform (DWT) a system. Its output is in a different domain than its input, so is it strictly a system? At least it's not a good example of one.
– Olli Niemitalo
2 days ago
That link doesnt work.
– Sweeper
2 days ago
1
try again please.
– Olli Niemitalo
2 days ago