Difficulty understanding group delay concept
$begingroup$
I’m having difficulty understanding the concept of group delay. The mathematical definition is not difficult, since it says it is the negative derivative of Bode plot’s phase curve wrt frequency. Most often qualitative definitions are easy but the math is difficult. In this case seems the opposite or it is due to my ignorance.
I have read some similar questions but still did not get the reason for such a concept. Is it possible to illustrate this concept with an example in elementary level. I know the meaning of Fourrier transform, frequency domain representation, and basic filter theory. Also a bit of modulation. What would the derivative of a phase frequency plot tell us regarding a low pass filter for instance? Im completely lost on the meaning of it so I cant even pose the question well.
delay group
asked Mar 18 at 17:27
user1999user1999
434315
$endgroup$
add a comment |
$begingroup$
I’m having difficulty understanding the concept of group delay. The mathematical definition is not difficult, since it says it is the negative derivative of Bode plot’s phase curve wrt frequency. Most often qualitative definitions are easy but the math is difficult. In this case seems the opposite or it is due to my ignorance.
I have read some similar questions but still did not get the reason for such a concept. Is it possible to illustrate this concept with an example in elementary level. I know the meaning of Fourrier transform, frequency domain representation, and basic filter theory. Also a bit of modulation. What would the derivative of a phase frequency plot tell us regarding a low pass filter for instance? Im completely lost on the meaning of it so I cant even pose the question well.
delay group
asked Mar 18 at 17:27
user1999user1999
434315
$endgroup$
$begingroup$
Group Delay is an indicator of how a waveform will be distorted, or preserved.
$endgroup$
– analogsystemsrf
Mar 19 at 2:45
add a comment |
$begingroup$
I’m having difficulty understanding the concept of group delay. The mathematical definition is not difficult, since it says it is the negative derivative of Bode plot’s phase curve wrt frequency. Most often qualitative definitions are easy but the math is difficult. In this case seems the opposite or it is due to my ignorance.
I have read some similar questions but still did not get the reason for such a concept. Is it possible to illustrate this concept with an example in elementary level. I know the meaning of Fourrier transform, frequency domain representation, and basic filter theory. Also a bit of modulation. What would the derivative of a phase frequency plot tell us regarding a low pass filter for instance? Im completely lost on the meaning of it so I cant even pose the question well.
delay group
asked Mar 18 at 17:27
user1999user1999
434315
$endgroup$
I’m having difficulty understanding the concept of group delay. The mathematical definition is not difficult, since it says it is the negative derivative of Bode plot’s phase curve wrt frequency. Most often qualitative definitions are easy but the math is difficult. In this case seems the opposite or it is due to my ignorance.
I have read some similar questions but still did not get the reason for such a concept. Is it possible to illustrate this concept with an example in elementary level. I know the meaning of Fourrier transform, frequency domain representation, and basic filter theory. Also a bit of modulation. What would the derivative of a phase frequency plot tell us regarding a low pass filter for instance? Im completely lost on the meaning of it so I cant even pose the question well.
delay group
delay group
asked Mar 18 at 17:27
user1999user1999
434315
asked Mar 18 at 17:27
user1999user1999
434315
asked Mar 18 at 17:27
user1999user1999
434315
asked Mar 18 at 17:27
user1999user1999
434315
asked Mar 18 at 17:27
user1999user1999
434315
434315
$begingroup$
Group Delay is an indicator of how a waveform will be distorted, or preserved.
$endgroup$
– analogsystemsrf
Mar 19 at 2:45
add a comment |
$begingroup$
Group Delay is an indicator of how a waveform will be distorted, or preserved.
$endgroup$
– analogsystemsrf
Mar 19 at 2:45
$begingroup$
Group Delay is an indicator of how a waveform will be distorted, or preserved.
$endgroup$
– analogsystemsrf
Mar 19 at 2:45
$begingroup$
Group Delay is an indicator of how a waveform will be distorted, or preserved.
$endgroup$
– analogsystemsrf
Mar 19 at 2:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
(1) Let us start with the PHASE DELAY: The response of a linear two-port to a sinusoidal excitation is an output signal with the same frequency w but with a phase delay $ phi $:
$ V_{out}=V_{max} times sin(wt+phi)=V_{max}
times sin[w(t+$ $phiover w$ $)] = V_{max} times sin[w(t-t_p)] $
Here, the expression $t_p=-$ $ phi over w$ is a delay time (phase delay) between input and output.
(2) For communication purposes of arbitrary waveforms we need the superposition of several sinusoidal waves with different frequencies. Of course we do not want that the various sinusoidal waves suffer from DIFFERENT delay figures.
Hence, we want a constant delay time tp for all these frequencies and we require that the equation $|phi|=t_ptimes w$ results in a LINEAR rising function between $phi$ and $w$ (for $t_p$=const).
From system theory we know that such a requirement (linearity between $phi$ and $w$) can be realized within a relatively small frequency band only. Hence, we define this requirement to be valid only within a frequency band that is realtively small if compared with the mean value of these frequencies:
We express this linearity requirement in form of the slope of the function and arrive at the so-called group delay
$t_g=-$ $dphiover dw$ $=const$ .
In practice, this requirement can be fulfilled with some errors only. Therefore, the constancy of the value for the group delay tg is a good measure for the quality of a communication channel (low distortion).
For example, a constant group delay is very important for a "good" pulse transmission.
edited Mar 18 at 21:15
Toor
1,277210
answered Mar 18 at 18:18
LvWLvW
14.7k21230
$endgroup$
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
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Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
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Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
add a comment |
$begingroup$
Group delay is the delay, in seconds, to a signal.
Imagine a cable that's (say) 1uS long, electrically. If you put a step into one end, the step will come out 1uS later. If you plot the phase response of the cable, the phase at DC will be zero, and at 1MHz will be 2pi. The slope of the phase, d(phase)/d(frequency) is therefore 1uS.
The same goes for a filter. The delay to a signal through the filter is dp/df. As this measure will be frequency dependant, it only applies to signals which are bandlimited to lie within the range for which dp/df is fairly constant.
As a step in voltage is wideband, if this is passed through a filter with group delay that varies with frequency, the different frequencies of the step will be passed with different delays, and the result is dispersion, or smearing out of the step at the filter output.
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
$endgroup$
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
add a comment |
$begingroup$
If the group delay for signals between $f_1$ and $f_2$ is $tau_{1,2}$, then a signal that's band-limited to lie between $f_1$ and $f_2$ will be delayed by $tau_{1,2}$.
If the group delay across the entire frequency band is the same number, then any signal will be delayed by that -- and it'll just be "delay", because the term "group delay" applies technically, but you don't need to get that specific.
If you have one signal (e.g. a pulse) that encounters a filter with a group delay that varies across the spectrum of your signal, then your signal will get spread out, because various components will arrive at different times. This is why a pulse that's run through a simple lumped-component low-pass filter will be smeared out, but the same pulse run through a constant group-delay FIR filter will just be rounded.
(Because you asked about phase delay in the comments): https://en.wikipedia.org/wiki/Group_delay_and_phase_delay
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
$endgroup$
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
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– user1999
Mar 18 at 18:02
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They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(1) Let us start with the PHASE DELAY: The response of a linear two-port to a sinusoidal excitation is an output signal with the same frequency w but with a phase delay $ phi $:
$ V_{out}=V_{max} times sin(wt+phi)=V_{max}
times sin[w(t+$ $phiover w$ $)] = V_{max} times sin[w(t-t_p)] $
Here, the expression $t_p=-$ $ phi over w$ is a delay time (phase delay) between input and output.
(2) For communication purposes of arbitrary waveforms we need the superposition of several sinusoidal waves with different frequencies. Of course we do not want that the various sinusoidal waves suffer from DIFFERENT delay figures.
Hence, we want a constant delay time tp for all these frequencies and we require that the equation $|phi|=t_ptimes w$ results in a LINEAR rising function between $phi$ and $w$ (for $t_p$=const).
From system theory we know that such a requirement (linearity between $phi$ and $w$) can be realized within a relatively small frequency band only. Hence, we define this requirement to be valid only within a frequency band that is realtively small if compared with the mean value of these frequencies:
We express this linearity requirement in form of the slope of the function and arrive at the so-called group delay
$t_g=-$ $dphiover dw$ $=const$ .
In practice, this requirement can be fulfilled with some errors only. Therefore, the constancy of the value for the group delay tg is a good measure for the quality of a communication channel (low distortion).
For example, a constant group delay is very important for a "good" pulse transmission.
edited Mar 18 at 21:15
Toor
1,277210
answered Mar 18 at 18:18
LvWLvW
14.7k21230
$endgroup$
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
$begingroup$
Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
$begingroup$
Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
add a comment |
$begingroup$
(1) Let us start with the PHASE DELAY: The response of a linear two-port to a sinusoidal excitation is an output signal with the same frequency w but with a phase delay $ phi $:
$ V_{out}=V_{max} times sin(wt+phi)=V_{max}
times sin[w(t+$ $phiover w$ $)] = V_{max} times sin[w(t-t_p)] $
Here, the expression $t_p=-$ $ phi over w$ is a delay time (phase delay) between input and output.
(2) For communication purposes of arbitrary waveforms we need the superposition of several sinusoidal waves with different frequencies. Of course we do not want that the various sinusoidal waves suffer from DIFFERENT delay figures.
Hence, we want a constant delay time tp for all these frequencies and we require that the equation $|phi|=t_ptimes w$ results in a LINEAR rising function between $phi$ and $w$ (for $t_p$=const).
From system theory we know that such a requirement (linearity between $phi$ and $w$) can be realized within a relatively small frequency band only. Hence, we define this requirement to be valid only within a frequency band that is realtively small if compared with the mean value of these frequencies:
We express this linearity requirement in form of the slope of the function and arrive at the so-called group delay
$t_g=-$ $dphiover dw$ $=const$ .
In practice, this requirement can be fulfilled with some errors only. Therefore, the constancy of the value for the group delay tg is a good measure for the quality of a communication channel (low distortion).
For example, a constant group delay is very important for a "good" pulse transmission.
edited Mar 18 at 21:15
Toor
1,277210
answered Mar 18 at 18:18
LvWLvW
14.7k21230
$endgroup$
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
$begingroup$
Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
$begingroup$
Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
add a comment |
$begingroup$
(1) Let us start with the PHASE DELAY: The response of a linear two-port to a sinusoidal excitation is an output signal with the same frequency w but with a phase delay $ phi $:
$ V_{out}=V_{max} times sin(wt+phi)=V_{max}
times sin[w(t+$ $phiover w$ $)] = V_{max} times sin[w(t-t_p)] $
Here, the expression $t_p=-$ $ phi over w$ is a delay time (phase delay) between input and output.
(2) For communication purposes of arbitrary waveforms we need the superposition of several sinusoidal waves with different frequencies. Of course we do not want that the various sinusoidal waves suffer from DIFFERENT delay figures.
Hence, we want a constant delay time tp for all these frequencies and we require that the equation $|phi|=t_ptimes w$ results in a LINEAR rising function between $phi$ and $w$ (for $t_p$=const).
From system theory we know that such a requirement (linearity between $phi$ and $w$) can be realized within a relatively small frequency band only. Hence, we define this requirement to be valid only within a frequency band that is realtively small if compared with the mean value of these frequencies:
We express this linearity requirement in form of the slope of the function and arrive at the so-called group delay
$t_g=-$ $dphiover dw$ $=const$ .
In practice, this requirement can be fulfilled with some errors only. Therefore, the constancy of the value for the group delay tg is a good measure for the quality of a communication channel (low distortion).
For example, a constant group delay is very important for a "good" pulse transmission.
edited Mar 18 at 21:15
Toor
1,277210
answered Mar 18 at 18:18
LvWLvW
14.7k21230
$endgroup$
(1) Let us start with the PHASE DELAY: The response of a linear two-port to a sinusoidal excitation is an output signal with the same frequency w but with a phase delay $ phi $:
$ V_{out}=V_{max} times sin(wt+phi)=V_{max}
times sin[w(t+$ $phiover w$ $)] = V_{max} times sin[w(t-t_p)] $
Here, the expression $t_p=-$ $ phi over w$ is a delay time (phase delay) between input and output.
(2) For communication purposes of arbitrary waveforms we need the superposition of several sinusoidal waves with different frequencies. Of course we do not want that the various sinusoidal waves suffer from DIFFERENT delay figures.
Hence, we want a constant delay time tp for all these frequencies and we require that the equation $|phi|=t_ptimes w$ results in a LINEAR rising function between $phi$ and $w$ (for $t_p$=const).
From system theory we know that such a requirement (linearity between $phi$ and $w$) can be realized within a relatively small frequency band only. Hence, we define this requirement to be valid only within a frequency band that is realtively small if compared with the mean value of these frequencies:
We express this linearity requirement in form of the slope of the function and arrive at the so-called group delay
$t_g=-$ $dphiover dw$ $=const$ .
In practice, this requirement can be fulfilled with some errors only. Therefore, the constancy of the value for the group delay tg is a good measure for the quality of a communication channel (low distortion).
For example, a constant group delay is very important for a "good" pulse transmission.
edited Mar 18 at 21:15
Toor
1,277210
answered Mar 18 at 18:18
LvWLvW
14.7k21230
edited Mar 18 at 21:15
Toor
1,277210
edited Mar 18 at 21:15
Toor
1,277210
edited Mar 18 at 21:15
Toor
1,277210
1,277210
answered Mar 18 at 18:18
LvWLvW
14.7k21230
answered Mar 18 at 18:18
LvWLvW
14.7k21230
answered Mar 18 at 18:18
LvWLvW
14.7k21230
14.7k21230
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
$begingroup$
Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
$begingroup$
Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
add a comment |
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
$begingroup$
Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
$begingroup$
Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
$begingroup$
Oh this was a very great explanation. I was wondering the linearity requirement in phase delay and its connection with group delay. Your example is very clear.
$endgroup$
– user1999
Mar 18 at 18:28
$begingroup$
Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
Further explanation/interpretation: For tg=const we require a constant slope for the phase function phi(w) in a certain frequency band only. That means: We do not require that function phi(w) crosses the origin, which would be the case for the phase function phi(w)=-tp*w (for w=0)
$endgroup$
– LvW
Mar 18 at 18:42
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
If for w=0 phi(w) does not cross the origin, does that mean the DC input and output are not in phase 100%? Btw DC input and step input confused me because if we apply DC at a time it is actually a step and includes all freq range. w=0 is DC but yet in real applying DC happpens from zero to a level. Or is the DC here is steady state DC? In that case it is the speed of electric current I guess at w=0.
$endgroup$
– user1999
Mar 18 at 18:48
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
Mar 18 at 19:07
$begingroup$
Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
$begingroup$
Toor - thank you for formatting my text.
$endgroup$
– LvW
Mar 18 at 22:07
add a comment |
$begingroup$
Group delay is the delay, in seconds, to a signal.
Imagine a cable that's (say) 1uS long, electrically. If you put a step into one end, the step will come out 1uS later. If you plot the phase response of the cable, the phase at DC will be zero, and at 1MHz will be 2pi. The slope of the phase, d(phase)/d(frequency) is therefore 1uS.
The same goes for a filter. The delay to a signal through the filter is dp/df. As this measure will be frequency dependant, it only applies to signals which are bandlimited to lie within the range for which dp/df is fairly constant.
As a step in voltage is wideband, if this is passed through a filter with group delay that varies with frequency, the different frequencies of the step will be passed with different delays, and the result is dispersion, or smearing out of the step at the filter output.
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
$endgroup$
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
add a comment |
$begingroup$
Group delay is the delay, in seconds, to a signal.
Imagine a cable that's (say) 1uS long, electrically. If you put a step into one end, the step will come out 1uS later. If you plot the phase response of the cable, the phase at DC will be zero, and at 1MHz will be 2pi. The slope of the phase, d(phase)/d(frequency) is therefore 1uS.
The same goes for a filter. The delay to a signal through the filter is dp/df. As this measure will be frequency dependant, it only applies to signals which are bandlimited to lie within the range for which dp/df is fairly constant.
As a step in voltage is wideband, if this is passed through a filter with group delay that varies with frequency, the different frequencies of the step will be passed with different delays, and the result is dispersion, or smearing out of the step at the filter output.
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
$endgroup$
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
add a comment |
$begingroup$
Group delay is the delay, in seconds, to a signal.
Imagine a cable that's (say) 1uS long, electrically. If you put a step into one end, the step will come out 1uS later. If you plot the phase response of the cable, the phase at DC will be zero, and at 1MHz will be 2pi. The slope of the phase, d(phase)/d(frequency) is therefore 1uS.
The same goes for a filter. The delay to a signal through the filter is dp/df. As this measure will be frequency dependant, it only applies to signals which are bandlimited to lie within the range for which dp/df is fairly constant.
As a step in voltage is wideband, if this is passed through a filter with group delay that varies with frequency, the different frequencies of the step will be passed with different delays, and the result is dispersion, or smearing out of the step at the filter output.
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
$endgroup$
Group delay is the delay, in seconds, to a signal.
Imagine a cable that's (say) 1uS long, electrically. If you put a step into one end, the step will come out 1uS later. If you plot the phase response of the cable, the phase at DC will be zero, and at 1MHz will be 2pi. The slope of the phase, d(phase)/d(frequency) is therefore 1uS.
The same goes for a filter. The delay to a signal through the filter is dp/df. As this measure will be frequency dependant, it only applies to signals which are bandlimited to lie within the range for which dp/df is fairly constant.
As a step in voltage is wideband, if this is passed through a filter with group delay that varies with frequency, the different frequencies of the step will be passed with different delays, and the result is dispersion, or smearing out of the step at the filter output.
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
answered Mar 18 at 18:03
Neil_UKNeil_UK
78.1k284181
78.1k284181
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
add a comment |
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
$begingroup$
Your first paragraph I guess summerizes but mind blowing I read it ten times still cannot get it. I dont get the difference between the phase and group delay there. There is speed of electric voltage phase time electrical length ect.
$endgroup$
– user1999
Mar 18 at 18:18
add a comment |
$begingroup$
If the group delay for signals between $f_1$ and $f_2$ is $tau_{1,2}$, then a signal that's band-limited to lie between $f_1$ and $f_2$ will be delayed by $tau_{1,2}$.
If the group delay across the entire frequency band is the same number, then any signal will be delayed by that -- and it'll just be "delay", because the term "group delay" applies technically, but you don't need to get that specific.
If you have one signal (e.g. a pulse) that encounters a filter with a group delay that varies across the spectrum of your signal, then your signal will get spread out, because various components will arrive at different times. This is why a pulse that's run through a simple lumped-component low-pass filter will be smeared out, but the same pulse run through a constant group-delay FIR filter will just be rounded.
(Because you asked about phase delay in the comments): https://en.wikipedia.org/wiki/Group_delay_and_phase_delay
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
$endgroup$
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
Mar 18 at 18:02
$begingroup$
They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
add a comment |
$begingroup$
If the group delay for signals between $f_1$ and $f_2$ is $tau_{1,2}$, then a signal that's band-limited to lie between $f_1$ and $f_2$ will be delayed by $tau_{1,2}$.
If the group delay across the entire frequency band is the same number, then any signal will be delayed by that -- and it'll just be "delay", because the term "group delay" applies technically, but you don't need to get that specific.
If you have one signal (e.g. a pulse) that encounters a filter with a group delay that varies across the spectrum of your signal, then your signal will get spread out, because various components will arrive at different times. This is why a pulse that's run through a simple lumped-component low-pass filter will be smeared out, but the same pulse run through a constant group-delay FIR filter will just be rounded.
(Because you asked about phase delay in the comments): https://en.wikipedia.org/wiki/Group_delay_and_phase_delay
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
$endgroup$
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
Mar 18 at 18:02
$begingroup$
They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
add a comment |
$begingroup$
If the group delay for signals between $f_1$ and $f_2$ is $tau_{1,2}$, then a signal that's band-limited to lie between $f_1$ and $f_2$ will be delayed by $tau_{1,2}$.
If the group delay across the entire frequency band is the same number, then any signal will be delayed by that -- and it'll just be "delay", because the term "group delay" applies technically, but you don't need to get that specific.
If you have one signal (e.g. a pulse) that encounters a filter with a group delay that varies across the spectrum of your signal, then your signal will get spread out, because various components will arrive at different times. This is why a pulse that's run through a simple lumped-component low-pass filter will be smeared out, but the same pulse run through a constant group-delay FIR filter will just be rounded.
(Because you asked about phase delay in the comments): https://en.wikipedia.org/wiki/Group_delay_and_phase_delay
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
$endgroup$
If the group delay for signals between $f_1$ and $f_2$ is $tau_{1,2}$, then a signal that's band-limited to lie between $f_1$ and $f_2$ will be delayed by $tau_{1,2}$.
If the group delay across the entire frequency band is the same number, then any signal will be delayed by that -- and it'll just be "delay", because the term "group delay" applies technically, but you don't need to get that specific.
If you have one signal (e.g. a pulse) that encounters a filter with a group delay that varies across the spectrum of your signal, then your signal will get spread out, because various components will arrive at different times. This is why a pulse that's run through a simple lumped-component low-pass filter will be smeared out, but the same pulse run through a constant group-delay FIR filter will just be rounded.
(Because you asked about phase delay in the comments): https://en.wikipedia.org/wiki/Group_delay_and_phase_delay
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
edited Mar 18 at 18:35
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
answered Mar 18 at 17:55
TimWescottTimWescott
6,4531416
6,4531416
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
Mar 18 at 18:02
$begingroup$
They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
add a comment |
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
Mar 18 at 18:02
$begingroup$
They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
Mar 18 at 18:02
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
Mar 18 at 18:02
$begingroup$
They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
$begingroup$
They are different things. See en.wikipedia.org/wiki/Group_delay_and_phase_delay. Phase delay is the amount that the phase in a pure sine wave would be delayed; Group delay is the amount the information in the signal can be delayed. Just to really blow your mind, if you work out the equations for EM radiation in some media (specifically waveguides), the phase speed is actually faster than $c$ (as in the speed of light, $E = mc^2$, etc), so the phase delay is now phase advance -- but the group speed is less than $c$, so causality is not violated.
$endgroup$
– TimWescott
Mar 18 at 18:33
add a comment |
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$begingroup$
Group Delay is an indicator of how a waveform will be distorted, or preserved.
$endgroup$
– analogsystemsrf
Mar 19 at 2:45