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Group delay群延时


Group Delay

Ron Hranac

Group Delay

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1

The Basics
Bandpass f

ilter equivalent
? Consider the 6 MHz spectrum occupied by an analog TV channel or digitally modulated signal, the 5-42 MHz upstream spectrum, or any specified bandwidth or passband as the equivalent of a bandpass filter.

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2

The Basics
Transit time and velocity of propagation
? A signal takes a certain amount of time to pass through a filter
The transit time through the filter is a function of the filter’s velocity of propagation (also called velocity factor) Velocity of propagation is the speed that an electromagnetic signal travels through some medium, usually expressed as a percentage of the speed of light in a vacuum

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3

The Basics
Velocity of propagation versus frequency
? In many instances the velocity of propagation through a filter varies with frequency
The velocity of propagation may be greater in the center of the filter’s passband, but slower near the band edges
Passband center

-3 dB point

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4

The Basics
Delay and absolute delay
? The finite time required for a signal to pass through a filter—or any device for that matter— is called delay
Absolute delay is the delay a signal experiences passing through the device at some reference frequency

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The Basics
Delay versus frequency
? If delay through a filter is plotted on a graph of frequency (x-axis) versus time delay (y-axis), the plot often has a parabola- or bathtub-like shape

Time delay

5 4 3 2 1 0

Frequency

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The Basics
Network analyzer plot of Ch. T8 bandpass filter
? The upper trace shows magnitude versus frequency: the filter’s bandpass characteristics. The x-axis is frequency, the y-axis is amplitude. ? The lower trace shows group delay versus frequency. The x-axis is frequency, the y-axis is time. Note the bathtublike shape of the curve.

Graphic courtesy of Holtzman, Inc.
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7

The Basics
Phase versus frequency
? If propagation or transit time through a device is the same at all frequencies, phase is said to be linear with respect to frequency
If phase changes uniformly with frequency, an output signal will be identical to the input signal—except that it will have a time shift because of the uniform delay through the device

? If propagation or transit time through a device is different at different frequencies, the result is delay shift or non-linear phase shift
If phase changes non-linearly with frequency, the output signal will be distorted
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8

The Basics
Delay and phase distortion
? Delay distortion—also known as phase distortion— is usually expressed in units of time: milliseconds (ms), microseconds (?s) or nanoseconds (ns) relative to a reference frequency ? Phase distortion is related to phase delay ? Phase distortion is measured using a parameter called envelope delay distortion, or group delay distortion

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9

The Formal Definition
Group delay is “the derivative of radian phase with respect to radian frequency. It is equal to the phase delay for an ideal non-dispersive delay device, but may differ greatly in actual devices where there is a ripple in the phase versus frequency characteristic.”
IEEE Standard Dictionary of Electrical and Electronics Terms

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10

The Math (Yikes!)
? In its simplest mathematical representation…

d? GD = dω

where GD is group delay variation, φ is phase in radians, and ω is frequency in radians per second ? For the purists, group delay τ also is defined

?? (ω ) τ (ω ) = ? ?ω

? And yet another definition is

d d D (ω )? ? Θ(ω )? ? ∠H (e jωΤ ) dω dω
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11

The Translation (Whew!)

? If phase versus frequency is non-linear, group delay exists. ? In a system, network, device or component with no group delay, all frequencies are transmitted through the system, network, device or component in the same amount of time—that is, with equal time delay. ? If group delay exists, signals at some frequencies travel faster than signals at other frequencies.

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12

The Analogy
Imagine a group of runners with identical athletic abilities on a smooth, flat track …

Example courtesy of Holtzman, Inc.
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All of the athletes arrive at the finish line at exactly the same time and with equal time delay from one end of the track to the other!
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13

Group Delay: An Analogy
Now let’s substitute a group of RF signals for the athletes. Here, the “track” is the equivalent of a filter’s passband.
Frequency Time All of the frequencies arrive at the destination at exactly the same time and with equal time delay through the filter passband!
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14

Group Delay: An Analogy
Back to athletes, but now there are some that have to run in the ditches next to the track.

Example courtesy of Holtzman, Inc.
Group Delay ? 2005 Cisco Systems, Inc. All rights reserved.

Some athletes take a little longer than others to arrive at the finish line. Their time delay from one end of the track to the other is unequal.
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15

Group Delay: An Analogy
Substitute RF signals for the athletes again. The “track” is a filter’s passband, the “ditches” are the filter’s band edges.
Frequency Time Time difference (unequal time delay through the filter)
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16

Group Delay: An Analogy

? Group delay exists, because some frequencies— the ones near the band edges—took longer than others to travel through the filter! ? Now take the dotted line connecting the frequencies and flip it on its side. The result is the classic bathtub-shaped group delay curve.
Time difference (unequal time delay)
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Frequency
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17

Group Delay

The Classic Group Delay “Bathtub Curve”
? In this example, the group delay between 25 and 40 MHz is about 300 ns (3 vertical divisions at 100 ns each) ? That is, it takes the 40 MHz signal 300 ns longer to reach the headend than the 25 MHz signal
Time difference in nanoseconds

Frequency
Graphic courtesy of Holtzman, Inc.
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18

Adaptive Equalization
? Adaptive equalization is a tool to combat linear distortions such as group delay ? All DOCSIS cable modems support downstream adaptive equalization (always on) ? DOCSIS 1.1 and 2.0 cable modems support upstream adaptive equalization (pre-equalization in transmitted signal, may be turned on/off by cable operator)
DOCSIS 1.1 modems: 8-tap adaptive equalization DOCSIS 2.0 modems: 24-tap adaptive equalization Adaptive equalization is not supported in DOCSIS 1.0 modems

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19

Adaptive Equalization

? Recall the group delay analogy using athletes on a track:
Adaptive equalization can be thought of as analogous to delaying the runners in the middle of the track, allowing the slower runners in the “ditches” to catch up. This allows all runners to arrive at the finish line at the same time, with equal time delay.

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20

Phase Versus Frequency

? OK, group delay exists if phase versus frequency is non-linear ? But just what does that mean? ? Let’s look at an example, using a 100 ft piece of .500 feeder cable

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21

Phase Versus Frequency
Back to Basics: Velocity of Propagation
? Hardline coax used for feeder applications has a velocity of propagation of around 87%
The speed of light in free space or a vacuum is 299,792,458 meters per second, or 983,571,056.43 feet per second—1 foot in about 1.02 ns In coaxial cable with a velocity of propagation of 87%, electromagnetic signals travel at a velocity equal to 87% of the free space value of the speed of light. That works out to 260,819,438.46 meters per second, or 855,706,819.09 feet per second—1 foot in about 1.17 ns So, electromagnetic signals will travel 100 ft in a vacuum in 101.67 ns, and through a 100 ft piece of coax in 116.86 ns
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22

Phase Versus Frequency
Back to Basics: Wavelength and Period
? Wavelength (λ) is the speed of propagation of an electromagnetic signal divided by its frequency (f) in hertz (Hz). It is further defined as the distance a wave travels through some medium in one period.
Period (T) of a cycle (in seconds) = 1/f, where f is frequency in Hz
One cycle + Amplitude

0

Time or phase

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23

Phase Versus Frequency
Back to Basics: Wavelength Formulas
? In a vacuum, wavelength in feet (λft) = 983,571,056.43/fHz, which is the same as λft = 983.57/fMHz ? In coaxial cable with 87% VP, λft = 855,706,819.09 /fHz or λft = 855.71/fMHz

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24

Phase Versus Frequency
Back to Basics: An Example
? For instance, the period of a 10 MHz sine wave is 1/10,000,000 Hz = 1x10-7 second, or 0.1 microsecond. That means a 10 MHz signal takes 0.1 ?s to complete one cycle, or 1 second to complete 10,000,000 cycles. ? In a vacuum, the 10 MHz signal travels 98.36 ft in 0.1 ?s. This distance is one wavelength in a vacuum. ? In 87% velocity of propagation coax, the 10 MHz signal travels 85.57 ft in 0.1 ?s. This distance is one wavelength in coax.
10 MHz signal completes one cycle in 0.1 ?s + Amplitude

0

Time or phase

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25

Phase Versus Frequency

? If we can calculate a given frequency’s wavelength in feet, we can say that a 100 ft piece of .500 coax is equivalent to a certain number wavelengths at that frequency! ? From the previous example, it stands to reason that a 100 ft piece of coax is equivalent to just over one wavelength at 10 MHz. That is, the 10 MHz signal’s 85.57 ft wavelength in coax is just shy of the 100 ft overall length of the piece of coax.

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26

Phase Versus Frequency
? OK, let’s figure out the wavelength in feet for several frequencies in a vacuum and in our 100 ft piece of coax, using the previous formulas. Because of the cable’s velocity of propagation, each frequency’s wavelength in the cable will be a little less than it is in a vacuum.
Frequency
1 MHz 5 MHz 10 MHz 30 MHz 42 MHz 50 MHz 65 MHz 100 MHz

λft in a vacuum
983.57 feet 196.71 feet 98.36 feet 32.97 feet 23.42 feet 19.67 feet 15.13 feet 9.84 feet

λft in coax
855.71 feet 171.14 feet 85.57 feet 28.52 feet 20.37 feet 17.11 feet 13.16 feet 8.56 feet
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Group Delay

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Phase Versus Frequency
? Next let’s figure out the number of wavelengths for each frequency in the 100 ft piece of coax

Frequency

λft in a vacuum

λft in coax

Number of λ in 100 ft of coax
0.12 λ 0.58 λ 1.17 λ 3.51 λ 4.91 λ 5.84 λ 7.60 λ 11.69 λ

1 MHz 5 MHz 10 MHz 30 MHz 42 MHz 50 MHz 65 MHz 100 MHz

983.57 feet 196.71 feet 98.36 feet 32.97 feet 23.42 feet 19.67 feet 15.13 feet 9.84 feet

855.71 feet 171.14 feet 85.57 feet 28.52 feet 20.37 feet 17.11 feet 13.16 feet 8.56 feet

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10 MHz Sine Wave: A Closer Look
One cycle

Amplitude

Time or phase

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10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

180° °

0° °

225° ° Amplitude: 0 270° °

315° °

0° °

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10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

180° ° Amplitude: 0.707

0° °

45° °
225° ° 270° ° 315° °

0° °
0.0125 ?s

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31

10 MHz Sine Wave: A Closer Look
90° ° 135° ° Amplitude: 1 45° °

90° °
180° ° 0° °

45° °
225° ° 270° ° 315° °

0° °
0.0250 ?s

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32

10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

90° °
180° ° Amplitude: 0.707 0° °

45° °

135° °
225° ° 270° ° 315° °

0° °
0.0375 ?s

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33

10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

90° °
180° ° 0° °

45° °

135° °
225° ° 315° ° 270° °

Amplitude: 0

0° °

180° °
0.0500 ?s

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10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

90° °
180° ° 0° °

45° °

135° °
225° ° 315° ° 270° °

0° °

180° °
0.0625 ?s

Amplitude: -0.707

225° °

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10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

90° °
180° ° 0° °

45° °

135° °
225° ° 315° ° 270° °

0° °

180° °
0.0750 ?s

225° °
Amplitude: -1

270° °
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10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

90° °
180° ° 0° °

45° °

135° °
225° ° 315° ° 270° °

0° °

180° °
0.0875 ?s

Amplitude: -0.707

225° °

315° °

270° °
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37

10 MHz Sine Wave: A Closer Look
90° ° 135° ° 45° °

90° °
180° ° 360° °

45° °

135° °
225° ° 315° ° 270° °

Amplitude: 0

0° °

180° °

360° °
0.1000 ?s Period of one cycle of a 10 MHz sine wave is 0.1 ?s

225° °

315° °

270° °
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38

Phase Versus Frequency
? Knowing that one wavelength (cycle) of a sine wave equals 360 degrees of phase, we can now figure out the total number of degrees of phase the 100 ft piece of cable represents at each frequency
One cycle

90° °

Amplitude

0° °

180° °

360° Time or phase °

270° °

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39

Phase Versus Frequency

Frequency

λft in a vacuum

λft in coax

Number of λ in 100 ft of coax
0.12 λ 0.58 λ 1.17 λ 3.51 λ 4.91 λ 5.84 λ 7.60 λ 11.69 λ

Total phase in degrees
42.07° ° 210.35° ° 420.71° ° 1262.12° ° 1766.96° ° 2103.53° ° 2734.58° ° 4207.05° °

1 MHz 5 MHz 10 MHz 30 MHz 42 MHz 50 MHz 65 MHz 100 MHz

983.57 feet 196.71 feet 98.36 feet 32.97 feet 23.42 feet 19.67 feet 15.13 feet 9.84 feet

855.71 feet 171.14 feet 85.57 feet 28.52 feet 20.37 feet 17.11 feet 13.16 feet 8.56 feet

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Phase Versus Frequency
? Next, we can plot the 100 ft piece of cable’s phase versus frequency on a graph. In this example, the line is straight— that is, phase versus frequency is linear.
4000° °
x

Phase (total degrees)

3000° °
x

2000° °
x

x

x

1000° °
x x

0° ° 0 10 20 30 40 50 60 70 80 90 100 110

Frequency (MHz)
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Phase Versus Frequency
? Finally, we can plot the time delay for each frequency through the 100 ft piece of cable. This line is the derivative of radian phase with respect to radian frequency. It’s flat, because phase versus frequency is linear—there is no group delay variation!
400

Time delay (nanoseconds)

300

200

100

0° ° 0 10 20 30 40 50 60 70 80 90 100 110

Frequency (MHz)
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42

Phase Versus Frequency

? Another way of looking at this is to say that the cable’s velocity of propagation is the same at all frequencies! ? In other words, every frequency takes 116.86 ns to travel from one end of the 100 ft piece of cable to the other end. ? But what happens if something in the signal path causes some frequencies to travel a little slower than other frequencies?

Group Delay

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43

Phase Versus Frequency
? Take a look at the phase versus frequency plot on this screen shot ? Where phase is not linear versus frequency— that is, where the sloped line is not straight— group delay exists
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Graphic courtesy of Holtzman, Inc.
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44

Phase Versus Frequency
? As before, the group delay between 25 and 40 MHz is about 300 ns (3 vertical divisions at 100 ns each) ? It takes the 40 MHz signal 300 ns longer to reach the headend than the 25 MHz signal

Time difference in nanoseconds

Graphic courtesy of Holtzman, Inc.
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Why Is Group Delay Important?

? Group delay, a linear distortion, causes intersymbol interference to digitally modulated signals ? This in turn degrades modulation error ratio (MER)—the constellation symbol points get “fuzzy” Q
Average error power Average symbol power

I
MER = 10log(average symbol power/average error power)
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Why Is Group Delay Important?
? This test equipment screen shot is from a cable network’s upstream spectrum in which in-channel group delay was negligible— about 60 to 75 ns peakto-peak. ? Unequalized MER is 28.3 dB, well above the 17~20 dB MER failure threshold for 16-QAM.
Graphic courtesy of Sunrise Telecom
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Why Is Group Delay Important?

? In this example, inchannel group delay was around 270 ns peak-to-peak. ? Unequalized MER is about 21 dB, very close to the 16-QAM failure threshold. 16QAM would not work on this upstream!

Graphic courtesy of Sunrise Telecom
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48

Why Is Group Delay Important?
MATLAB? simulation for 64-QAM—no group delay

Group Delay

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Why Is Group Delay Important?
MATLAB? simulation for 64-QAM—with group delay

Group Delay

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50

Wrapping Up

? Common sources of group delay in a cable network
AC power coils/chokes (affects 5~10 MHz in the upstream) Node and amplifier diplex filters (affect frequencies near the diplex filter cutoff region in the upstream and downstream) Band edges and rolloff areas High-pass filters, data-only filters, step attenuators, taps or in-line equalizers with filters Group delay ripple caused by impedance mismatch-related micro-reflections and amplitude ripple (poor frequency response)
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Group Delay

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Wrapping Up

? The Fix?
Use adaptive equalization available in DOCSIS 1.1 and 2.0 modems (not supported in DOCSIS 1.0 modems) Avoid frequencies where diplex filter group delay is common Sweep the forward and reverse to ensure frequency response is flat (set equipment to highest resolution available; use resistive test points or probe seizure screws to see amplitude ripple) Identify and repair impedance mismatches that cause microreflections Use specialized test equipment to characterize and troubleshoot group delay (group delay can exist even when frequency response is flat)

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References

? Farmer, J., D. Large, W. Ciciora and M. Adams. Modern Cable Television Technology: Video, Voice and Data Communications, 2nd Ed., Morgan Kaufmann Publishers; 2004 ? Freeman, R. Telecommunications Transmission Handbook, 4th Ed., John Wiley & Sons; 1998 ? Hranac, R. “Group delay” Communications Technology, January 1999 www.ct-magazine.com/archives/ct/0199/ct0199e.htm ? Williams, T. “Tackling Upstream Data Impairments, Part 1” Communications Technology, November 2003 www.ct-magazine.com/archives/ct/1103/1103_upstreamdata.html

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References

? Williams, T. “Tackling Upstream Data Impairments, Part 2” Communications Technology, December 2003 www.ct-magazine.com/archives/ct/1203/1203_upstreamdata2.html ? Hranac, R. “Microreflections and 16-QAM” Communications Technology, March 2004 www.ct-magazine.com/archives/ct/0304/0304_broadband.html ? Hranac, R. “Linear Distortions, Part 1” Communications Technology, July 2005 www.ct-magazine.com/archives/ct/0705/0705_lineardistortions.htm ? Hranac, R. “Linear Distortions, Part 2” Communications Technology, August 2005 www.ct-magazine.com/archives/ct/0805/0805_lineardistortions.htm
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Q and A

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