Difficulty understanding group delay conceptPhysical significance of group delayCalculating Circuit Delayphase wrapping group delayUnderstanding max and min propagation delay in flip-flopsEstimating via propagation delayHow to introduce delay to a signalGroup delay and phase delay of a filterCalculating Phase/Time Delay induced by a 2-pole Bandpass filter? ie. Group Delay?Is group delay the same as the delay of a certain frequency?Physical significance of positive group delay with negative phase delayDelay pulse (trigger) for a variable delay respectively to 0-5V input
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Difficulty understanding group delay concept
Physical significance of group delayCalculating Circuit Delayphase wrapping group delayUnderstanding max and min propagation delay in flip-flopsEstimating via propagation delayHow to introduce delay to a signalGroup delay and phase delay of a filterCalculating Phase/Time Delay induced by a 2-pole Bandpass filter? ie. Group Delay?Is group delay the same as the delay of a certain frequency?Physical significance of positive group delay with negative phase delayDelay pulse (trigger) for a variable delay respectively to 0-5V input
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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 2 hours ago
user1999user1999
417313
417313
add a comment |
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:
Vout=vmax * sin(wt+phi)=vmax * sin[w(t+phi/w)] = vmax * sin[w(t-tp)]
Here, the expression tp=-phi/w is a delay time (phase delay) between input and output.
(2) For communication purposes of arbritrari 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|=tp*w results in a LINEAR rising function between phi and w (for tp=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
tg=-d(phi)/d(w)=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.
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
59 mins ago
$begingroup$
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
53 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
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.
$endgroup$
$begingroup$
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
1 hour ago
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
$endgroup$
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
1 hour ago
$begingroup$
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
1 hour ago
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:
Vout=vmax * sin(wt+phi)=vmax * sin[w(t+phi/w)] = vmax * sin[w(t-tp)]
Here, the expression tp=-phi/w is a delay time (phase delay) between input and output.
(2) For communication purposes of arbritrari 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|=tp*w results in a LINEAR rising function between phi and w (for tp=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
tg=-d(phi)/d(w)=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.
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
59 mins ago
$begingroup$
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
53 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
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:
Vout=vmax * sin(wt+phi)=vmax * sin[w(t+phi/w)] = vmax * sin[w(t-tp)]
Here, the expression tp=-phi/w is a delay time (phase delay) between input and output.
(2) For communication purposes of arbritrari 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|=tp*w results in a LINEAR rising function between phi and w (for tp=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
tg=-d(phi)/d(w)=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.
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
59 mins ago
$begingroup$
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
53 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
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:
Vout=vmax * sin(wt+phi)=vmax * sin[w(t+phi/w)] = vmax * sin[w(t-tp)]
Here, the expression tp=-phi/w is a delay time (phase delay) between input and output.
(2) For communication purposes of arbritrari 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|=tp*w results in a LINEAR rising function between phi and w (for tp=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
tg=-d(phi)/d(w)=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.
$endgroup$
(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:
Vout=vmax * sin(wt+phi)=vmax * sin[w(t+phi/w)] = vmax * sin[w(t-tp)]
Here, the expression tp=-phi/w is a delay time (phase delay) between input and output.
(2) For communication purposes of arbritrari 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|=tp*w results in a LINEAR rising function between phi and w (for tp=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
tg=-d(phi)/d(w)=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.
answered 1 hour ago
LvWLvW
14.6k21230
14.6k21230
$begingroup$
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
1 hour ago
$begingroup$
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
59 mins ago
$begingroup$
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
53 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
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
1 hour ago
$begingroup$
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
59 mins ago
$begingroup$
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
53 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
59 mins ago
$begingroup$
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
59 mins ago
$begingroup$
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
53 mins ago
$begingroup$
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
53 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
$begingroup$
@LvW Formatted your equations. Check for correctness when someone else approves my changes.
$endgroup$
– Toor
34 mins ago
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.
$endgroup$
$begingroup$
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
1 hour ago
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.
$endgroup$
$begingroup$
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
1 hour ago
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.
$endgroup$
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 1 hour ago
Neil_UKNeil_UK
77.7k284178
77.7k284178
$begingroup$
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
1 hour ago
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
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
$endgroup$
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
1 hour ago
$begingroup$
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
1 hour ago
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
$endgroup$
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
1 hour ago
$begingroup$
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
1 hour ago
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
$endgroup$
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
edited 1 hour ago
answered 1 hour ago
TimWescottTimWescott
5,9921415
5,9921415
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
1 hour ago
$begingroup$
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
1 hour ago
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
1 hour ago
$begingroup$
“then your signal will get spread out, because various components will arrive at different times” isnt this the result of “phase delay”?
$endgroup$
– user1999
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
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