Mixing
Artifacts and low frequencies FFT.
$begingroup$
I am working on analyzing a time signal and want to preform a FFT. However I run in to some artifacts at low frequencies. I have managed to reproduce the behavior in a test signal. Given by
$S(t) = theta (0.3-t)[sin(2 pi 30 t)+sin(2 pi 100 t) + 3.0]$
In other words a signal that is zero up to t = 0.3 and then a combination of two sine waves with frequency 30 and 100 Hz.
This gives the following DFT, amplitudes.
I have tried removing the mean, windowing and combinations of the two. The only thing that seems to help is either to remove the constant or to make the time interval, where the function is zero, shorter.Also need to window it.
So why not just cut it off and be done with it? Well for this test signal that would be an ideal solution, but for my real signal I do not really want to do that and it do not seem to help either.
Do anyone know what causes this behavior and how to fix it?
EDIT: The plot do not include the DC term.
fourier-analysis signal-processing
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
$endgroup$
|
show 1 more comment
$begingroup$
I am working on analyzing a time signal and want to preform a FFT. However I run in to some artifacts at low frequencies. I have managed to reproduce the behavior in a test signal. Given by
$S(t) = theta (0.3-t)[sin(2 pi 30 t)+sin(2 pi 100 t) + 3.0]$
In other words a signal that is zero up to t = 0.3 and then a combination of two sine waves with frequency 30 and 100 Hz.
This gives the following DFT, amplitudes.
I have tried removing the mean, windowing and combinations of the two. The only thing that seems to help is either to remove the constant or to make the time interval, where the function is zero, shorter.Also need to window it.
So why not just cut it off and be done with it? Well for this test signal that would be an ideal solution, but for my real signal I do not really want to do that and it do not seem to help either.
Do anyone know what causes this behavior and how to fix it?
EDIT: The plot do not include the DC term.
fourier-analysis signal-processing
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
$endgroup$
$begingroup$
Can you elaborate on what part of the plot is an artifact? The DC component is there, so the value at zero frequency is certainly not an artifact. Are you referring to the small peaks between DC and your first frequency component?
$endgroup$
– rajb245
Jan 21 '14 at 17:06
1
$begingroup$
You have two step discontinuities and a DC component. Basically you are multiplying a signal composed of a DC term and two sinusoids by a rectangular signal. In the frequency domain this is convolution, hence you would expect to see $operatorname{sinc}$ 'smearing' of the basic components, which is exactly what you have.
$endgroup$
– copper.hat
Jan 21 '14 at 17:09
1
$begingroup$
If this is what you mean, then it is no artifact. Your signal "turns on" in an infinitesimal amount of time. Resolving this sharp transition requires infinite bandwidth, so you are seeing the effect of the sharp transition "spraying" frequencies out of band. In your real signal, the transition will not be a step transition; it will occur over a finite amount of time. You can replace your step function with a finite-rise time, smooth function like atan or erfc, appropriately shifted and scaled. This will suppress some of what you call "artifact".
$endgroup$
– rajb245
Jan 21 '14 at 17:14
$begingroup$
Ah, I forgot to mention that this is not including the DC component. The zero frequency amplitude is not in this plot.
$endgroup$
– HaakonA
Jan 21 '14 at 17:16
$begingroup$
You still have the $operatorname{sinc}$ 'smearing' of the DC component, so you will likely have some large low frequency terms.
$endgroup$
– copper.hat
Jan 22 '14 at 20:30
|
show 1 more comment
$begingroup$
I am working on analyzing a time signal and want to preform a FFT. However I run in to some artifacts at low frequencies. I have managed to reproduce the behavior in a test signal. Given by
$S(t) = theta (0.3-t)[sin(2 pi 30 t)+sin(2 pi 100 t) + 3.0]$
In other words a signal that is zero up to t = 0.3 and then a combination of two sine waves with frequency 30 and 100 Hz.
This gives the following DFT, amplitudes.
I have tried removing the mean, windowing and combinations of the two. The only thing that seems to help is either to remove the constant or to make the time interval, where the function is zero, shorter.Also need to window it.
So why not just cut it off and be done with it? Well for this test signal that would be an ideal solution, but for my real signal I do not really want to do that and it do not seem to help either.
Do anyone know what causes this behavior and how to fix it?
EDIT: The plot do not include the DC term.
fourier-analysis signal-processing
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
$endgroup$
I am working on analyzing a time signal and want to preform a FFT. However I run in to some artifacts at low frequencies. I have managed to reproduce the behavior in a test signal. Given by
$S(t) = theta (0.3-t)[sin(2 pi 30 t)+sin(2 pi 100 t) + 3.0]$
In other words a signal that is zero up to t = 0.3 and then a combination of two sine waves with frequency 30 and 100 Hz.
This gives the following DFT, amplitudes.
I have tried removing the mean, windowing and combinations of the two. The only thing that seems to help is either to remove the constant or to make the time interval, where the function is zero, shorter.Also need to window it.
So why not just cut it off and be done with it? Well for this test signal that would be an ideal solution, but for my real signal I do not really want to do that and it do not seem to help either.
Do anyone know what causes this behavior and how to fix it?
EDIT: The plot do not include the DC term.
fourier-analysis signal-processing
fourier-analysis signal-processing
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
edited Jan 21 '14 at 17:17
HaakonA
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
asked Jan 21 '14 at 16:58
HaakonAHaakonA
513
513
$begingroup$
Can you elaborate on what part of the plot is an artifact? The DC component is there, so the value at zero frequency is certainly not an artifact. Are you referring to the small peaks between DC and your first frequency component?
$endgroup$
– rajb245
Jan 21 '14 at 17:06
1
$begingroup$
You have two step discontinuities and a DC component. Basically you are multiplying a signal composed of a DC term and two sinusoids by a rectangular signal. In the frequency domain this is convolution, hence you would expect to see $operatorname{sinc}$ 'smearing' of the basic components, which is exactly what you have.
$endgroup$
– copper.hat
Jan 21 '14 at 17:09
1
$begingroup$
If this is what you mean, then it is no artifact. Your signal "turns on" in an infinitesimal amount of time. Resolving this sharp transition requires infinite bandwidth, so you are seeing the effect of the sharp transition "spraying" frequencies out of band. In your real signal, the transition will not be a step transition; it will occur over a finite amount of time. You can replace your step function with a finite-rise time, smooth function like atan or erfc, appropriately shifted and scaled. This will suppress some of what you call "artifact".
$endgroup$
– rajb245
Jan 21 '14 at 17:14
$begingroup$
Ah, I forgot to mention that this is not including the DC component. The zero frequency amplitude is not in this plot.
$endgroup$
– HaakonA
Jan 21 '14 at 17:16
$begingroup$
You still have the $operatorname{sinc}$ 'smearing' of the DC component, so you will likely have some large low frequency terms.
$endgroup$
– copper.hat
Jan 22 '14 at 20:30
|
show 1 more comment
$begingroup$
Can you elaborate on what part of the plot is an artifact? The DC component is there, so the value at zero frequency is certainly not an artifact. Are you referring to the small peaks between DC and your first frequency component?
$endgroup$
– rajb245
Jan 21 '14 at 17:06
1
$begingroup$
You have two step discontinuities and a DC component. Basically you are multiplying a signal composed of a DC term and two sinusoids by a rectangular signal. In the frequency domain this is convolution, hence you would expect to see $operatorname{sinc}$ 'smearing' of the basic components, which is exactly what you have.
$endgroup$
– copper.hat
Jan 21 '14 at 17:09
1
$begingroup$
If this is what you mean, then it is no artifact. Your signal "turns on" in an infinitesimal amount of time. Resolving this sharp transition requires infinite bandwidth, so you are seeing the effect of the sharp transition "spraying" frequencies out of band. In your real signal, the transition will not be a step transition; it will occur over a finite amount of time. You can replace your step function with a finite-rise time, smooth function like atan or erfc, appropriately shifted and scaled. This will suppress some of what you call "artifact".
$endgroup$
– rajb245
Jan 21 '14 at 17:14
$begingroup$
Ah, I forgot to mention that this is not including the DC component. The zero frequency amplitude is not in this plot.
$endgroup$
– HaakonA
Jan 21 '14 at 17:16
$begingroup$
You still have the $operatorname{sinc}$ 'smearing' of the DC component, so you will likely have some large low frequency terms.
$endgroup$
– copper.hat
Jan 22 '14 at 20:30
$begingroup$
Can you elaborate on what part of the plot is an artifact? The DC component is there, so the value at zero frequency is certainly not an artifact. Are you referring to the small peaks between DC and your first frequency component?
$endgroup$
– rajb245
Jan 21 '14 at 17:06
$begingroup$
Can you elaborate on what part of the plot is an artifact? The DC component is there, so the value at zero frequency is certainly not an artifact. Are you referring to the small peaks between DC and your first frequency component?
$endgroup$
– rajb245
Jan 21 '14 at 17:06
1
1
$begingroup$
You have two step discontinuities and a DC component. Basically you are multiplying a signal composed of a DC term and two sinusoids by a rectangular signal. In the frequency domain this is convolution, hence you would expect to see $operatorname{sinc}$ 'smearing' of the basic components, which is exactly what you have.
$endgroup$
– copper.hat
Jan 21 '14 at 17:09
$begingroup$
You have two step discontinuities and a DC component. Basically you are multiplying a signal composed of a DC term and two sinusoids by a rectangular signal. In the frequency domain this is convolution, hence you would expect to see $operatorname{sinc}$ 'smearing' of the basic components, which is exactly what you have.
$endgroup$
– copper.hat
Jan 21 '14 at 17:09
1
1
$begingroup$
If this is what you mean, then it is no artifact. Your signal "turns on" in an infinitesimal amount of time. Resolving this sharp transition requires infinite bandwidth, so you are seeing the effect of the sharp transition "spraying" frequencies out of band. In your real signal, the transition will not be a step transition; it will occur over a finite amount of time. You can replace your step function with a finite-rise time, smooth function like atan or erfc, appropriately shifted and scaled. This will suppress some of what you call "artifact".
$endgroup$
– rajb245
Jan 21 '14 at 17:14
$begingroup$
If this is what you mean, then it is no artifact. Your signal "turns on" in an infinitesimal amount of time. Resolving this sharp transition requires infinite bandwidth, so you are seeing the effect of the sharp transition "spraying" frequencies out of band. In your real signal, the transition will not be a step transition; it will occur over a finite amount of time. You can replace your step function with a finite-rise time, smooth function like atan or erfc, appropriately shifted and scaled. This will suppress some of what you call "artifact".
$endgroup$
– rajb245
Jan 21 '14 at 17:14
$begingroup$
Ah, I forgot to mention that this is not including the DC component. The zero frequency amplitude is not in this plot.
$endgroup$
– HaakonA
Jan 21 '14 at 17:16
$begingroup$
Ah, I forgot to mention that this is not including the DC component. The zero frequency amplitude is not in this plot.
$endgroup$
– HaakonA
Jan 21 '14 at 17:16
$begingroup$
You still have the $operatorname{sinc}$ 'smearing' of the DC component, so you will likely have some large low frequency terms.
$endgroup$
– copper.hat
Jan 22 '14 at 20:30
$begingroup$
You still have the $operatorname{sinc}$ 'smearing' of the DC component, so you will likely have some large low frequency terms.
$endgroup$
– copper.hat
Jan 22 '14 at 20:30
|
show 1 more comment
1 Answer
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compare it with the FFT of $theta (0.3-t)$ and of $$H(t) = theta (0.3-t)[exp(2i pi 30 t)+exp(2 i pi 100 t) + 3.0]$$
use a window (hanning)
if you understand what's happening, you will understand nearly everything of the FFT.
answered Jul 20 '15 at 16:43
reunsreuns
28515
$endgroup$
add a comment |
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$begingroup$
compare it with the FFT of $theta (0.3-t)$ and of $$H(t) = theta (0.3-t)[exp(2i pi 30 t)+exp(2 i pi 100 t) + 3.0]$$
use a window (hanning)
if you understand what's happening, you will understand nearly everything of the FFT.
answered Jul 20 '15 at 16:43
reunsreuns
28515
$endgroup$
add a comment |
$begingroup$
compare it with the FFT of $theta (0.3-t)$ and of $$H(t) = theta (0.3-t)[exp(2i pi 30 t)+exp(2 i pi 100 t) + 3.0]$$
use a window (hanning)
if you understand what's happening, you will understand nearly everything of the FFT.
answered Jul 20 '15 at 16:43
reunsreuns
28515
$endgroup$
add a comment |
$begingroup$
compare it with the FFT of $theta (0.3-t)$ and of $$H(t) = theta (0.3-t)[exp(2i pi 30 t)+exp(2 i pi 100 t) + 3.0]$$
use a window (hanning)
if you understand what's happening, you will understand nearly everything of the FFT.
answered Jul 20 '15 at 16:43
reunsreuns
28515
$endgroup$
compare it with the FFT of $theta (0.3-t)$ and of $$H(t) = theta (0.3-t)[exp(2i pi 30 t)+exp(2 i pi 100 t) + 3.0]$$
use a window (hanning)
if you understand what's happening, you will understand nearly everything of the FFT.
answered Jul 20 '15 at 16:43
reunsreuns
28515
answered Jul 20 '15 at 16:43
reunsreuns
28515
answered Jul 20 '15 at 16:43
reunsreuns
28515
answered Jul 20 '15 at 16:43
reunsreuns
28515
28515
add a comment |
add a comment |
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$begingroup$
Can you elaborate on what part of the plot is an artifact? The DC component is there, so the value at zero frequency is certainly not an artifact. Are you referring to the small peaks between DC and your first frequency component?
$endgroup$
– rajb245
Jan 21 '14 at 17:06
1
$begingroup$
You have two step discontinuities and a DC component. Basically you are multiplying a signal composed of a DC term and two sinusoids by a rectangular signal. In the frequency domain this is convolution, hence you would expect to see $operatorname{sinc}$ 'smearing' of the basic components, which is exactly what you have.
$endgroup$
– copper.hat
Jan 21 '14 at 17:09
1
$begingroup$
If this is what you mean, then it is no artifact. Your signal "turns on" in an infinitesimal amount of time. Resolving this sharp transition requires infinite bandwidth, so you are seeing the effect of the sharp transition "spraying" frequencies out of band. In your real signal, the transition will not be a step transition; it will occur over a finite amount of time. You can replace your step function with a finite-rise time, smooth function like atan or erfc, appropriately shifted and scaled. This will suppress some of what you call "artifact".
$endgroup$
– rajb245
Jan 21 '14 at 17:14
$begingroup$
Ah, I forgot to mention that this is not including the DC component. The zero frequency amplitude is not in this plot.
$endgroup$
– HaakonA
Jan 21 '14 at 17:16
$begingroup$
You still have the $operatorname{sinc}$ 'smearing' of the DC component, so you will likely have some large low frequency terms.
$endgroup$
– copper.hat
Jan 22 '14 at 20:30