Small-Scale Fading II
(and basics about random processes)
PROF. MICHAEL TSAI 2014/3/31
Random processes
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t
t 𝑥2(𝑡)
𝑥1(𝑡)
t 𝑥3(𝑡)
𝑋(𝑡)
One realization of X(t)
………
𝑋(𝜏) is a random variable:
𝑋(𝜏) 𝑓𝑋 𝜏 𝑋
Joint CDF for a random process
• If we sample X(t) at times 𝒕𝟎, … 𝒕𝒏, we can have a joint cdf of samples at those times:
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𝑃𝑋 𝑡0 ,…,𝑋 𝑡𝑛 𝑥0, … , 𝑥𝑛 = 𝑝 𝑋 𝑡0 ≤ 𝑥0, 𝑋 𝑡1 ≤ 𝑥1, … , 𝑋 𝑡𝑛 ≤ 𝑥𝑛
t
X(t) 𝑋 𝑡0 𝑋 𝑡1 … 𝑋 𝑡𝑛
Stationary Random
Process (Strict-sense)
• A random process X(t) is stationary if for all T, all n, and all sets of sample times 𝒕𝟎, 𝒕𝟏, … , 𝒕𝒏 we have:
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𝑝 𝑋 𝑡0 ≤ 𝑥0, 𝑋 𝑡1 ≤ 𝑥1, … , 𝑋 𝑡𝑛 ≤ 𝑥𝑛 =
𝑝 𝑋 𝑡0 + 𝑇 ≤ 𝑥0, 𝑋 𝑡1 + 𝑇 ≤ 𝑥1, … , 𝑋 𝑡𝑛 + 𝑇 ≤ 𝑥𝑛
If time shifts does not matter, then it is stationary
Mean (First Moment)
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t
t 𝑥2(𝑡)
𝑥1(𝑡)
t 𝑥3(𝑡)
𝐸[𝑋 𝑡 ]
……… Averaging over all realizations
𝐸[𝑋 𝑡 ]
𝐸 𝑋 𝜏
Autocorrelation (Second Moment)
• “How similar a random process and a shifted version of itself is”
• Autocorrelation of a random process is defined as:
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𝐴𝑋 𝑡, 𝑡 + 𝜏 ≜ 𝐸 𝑋 𝑡 𝑋 𝑡 + 𝜏
t
t 𝑥𝑗(𝑡 + 𝜏)
𝑥𝑖(𝑡)
Shifted by 𝜏
×
All possible combinations of realizations
𝜏 𝐴 𝑡, 𝑡 + 𝜏
For a particular t
=
For stationary random processes…
• Mean
• Autocorrelation
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𝐸 𝑋 𝑡 = 𝐸 𝑋 𝑡 − 𝑡 = 𝐸 𝑋 0 = 𝜇𝑋
Constant. Does not change with t.
𝐴𝑋 𝑡, 𝑡 + 𝜏 = 𝐸 𝑋 𝑡 − 𝑡 𝑋 𝑡 + 𝜏 − 𝑡 = 𝐸[𝑋 0 𝑋 𝜏 ] ≜ 𝐴𝑋(𝜏)
Two random processes
• Two random processes X(t) and Y(t) defined on the same underlying probability space have a joint cdf:
for all possible sets of sample times 𝒕𝟎, … , 𝒕𝒏 and 𝒕𝟎′ , … , 𝒕𝒎′ .
• Two random processes are independent if we have
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𝑝𝑋 𝑡
0 ,…,𝑋 𝑡𝑛 𝑌 𝑡0′ ,…,𝑌 𝑡𝑚′ 𝑥0, … , 𝑥𝑛, 𝑦0, … , 𝑦𝑚 =
𝑝 𝑋 𝑡0 ≤ 𝑥0, … , 𝑋 𝑡𝑛 ≤ 𝑥𝑛, 𝑌 𝑡0′ ≤ 𝑦0, … , 𝑌 𝑡𝑚′ ≤ 𝑦𝑚
Similar to how you can define a joint cdf for two random variables
𝑝𝑋 𝑡
0 ,…,𝑋 𝑡𝑛 𝑌 𝑡0′ ,…,𝑌 𝑡𝑚′ 𝑥0, … , 𝑥𝑛, 𝑦0, … , 𝑦𝑚 =
𝑝 𝑋 𝑡0 ≤ 𝑥0, 𝑋 𝑡1 ≤ 𝑥1, … , 𝑋 𝑡𝑛 ≤ 𝑥𝑛 𝑝 𝑌 𝑡0′ ≤ 𝑦0, 𝑌 𝑡1′ ≤ 𝑦1, … , 𝑌 𝑡𝑛′ ≤ 𝑦𝑛
Cross-correlation
• The cross-correlation between two random processes X(t) and Y(t) is defined as
• Two random processes are uncorrelated if
for all t and 𝝉.
• If both random processes are stationary, we have
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𝐴𝑋𝑌 𝑡, 𝑡 + 𝜏 ≜ 𝐸 𝑋 𝑡 𝑌 𝑡 + 𝜏
𝐸 𝑋 𝑡 𝑌 𝑡 + 𝜏 = 𝐸 𝑋 𝑡 𝐸 𝑌 𝑡 + 𝜏
𝐴𝑋𝑌 𝑡, 𝑡 + 𝜏 = 𝐸 𝑋 𝑡 𝑌 𝑡 + 𝜏 = 𝐸 𝑋 0 𝑌 𝜏 = 𝐴XY 𝜏
Wide-Sense Stationary (WSS)
• A process is wide-sense stationary if
and
• 𝑨𝑿(𝝉) has its maximum value at 𝝉 = 𝟎.
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𝐸 𝑋 𝑡 = 𝜇𝑋
𝐴𝑋 𝑡, 𝑡 + 𝜏 = 𝐸 𝑋 𝑡 𝑋 𝑡 + 𝜏 = 𝐴𝑋 𝜏
𝐴𝑋 𝜏 ≤ 𝐴𝑋 0 = 𝐸 𝑋2 𝑡
A random process is always “the most similar” to the version of itself without shifting.
Ergodicity
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t
t 𝑥2(𝑡)
𝑥1(𝑡)
t 𝑥3(𝑡)
𝑋(𝑡)
Expectation value over time is the same as expectation over all possible realizations 𝐸𝑡 .
𝐸𝑖 .
Power Spectral Density
• The power spectral density of a WSS process is
defined as the Fourier transform of its autocorrelation function with respect to 𝝉:
• PSD takes its name from the fact that the expected power of a random process X(t) is the integral of its PSD:
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𝑆𝑋 𝑓 =
−∞
∞
𝐴𝑋 𝜏 exp −𝑗2𝜋𝑓𝜏 𝑑𝜏
𝐸 𝑋2 𝑡 = 𝐴𝑋 0 =
−∞
∞
𝑆𝑋 𝑓 𝑑𝑓
Gaussian random processes
• A random process X(t) is a Gaussian process if, for all values of T and all functions g(t), the random variable
has a Gaussian distribution.
• We usually use this to model the noise for a communication receiver.
• Mean & variance:
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𝑋𝑔 =
0 𝑇
𝑔 𝑡 𝑋 𝑡 𝑑𝑡
Linear combination of samples
𝐸 𝑋𝑔 =
0 𝑇
𝑔 𝑡 𝐸 𝑋 𝑡 𝑑𝑡
𝑉𝑎𝑟 𝑋𝑔 =
0 𝑇
0 𝑇
𝑔 𝑡 𝑔 𝑠 𝐸 𝑋 𝑡 𝑋 𝑠 𝑑𝑡 𝑑𝑠 − 𝐸2[𝑋𝑔]
Gaussian random processes
• Samples of a random process, 𝑿 𝒕𝒊 , 𝒊 = 𝟎, … , 𝒏, are jointly Gaussian random variables, if we let 𝒈 𝒕 = 𝜹 𝒕 − 𝒕𝒊 .
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𝑋𝑔 =
0 𝑇
𝑔 𝑡 𝑋 𝑡 𝑑𝑡
𝑋𝑔 =
0 𝑇
𝛿 𝑡 − 𝑡𝑖 𝑋 𝑡 𝑑𝑡 = 𝑋 𝑡𝑖
Recap: 2 important aspects (of channel time variation)
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Example: white noise
• White noise is a zero-mean WSS random process with a PSD that is constant over all frequencies.
• 𝑁0 is often called as one-sided white noise PSD.
• By inverse Fourier transform, the autocorrelation can be obtained:
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𝐸 𝑋 𝑡 = 0 𝑆𝑋 𝑓 = 𝑁0
2 for some constant 𝑁0
𝐴𝑋 𝜏 = 𝑁0
2 𝛿 𝜏
White noise is not correlated with any shifted version of itself.
(Not similar at all after ANY time period)
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t
t0
0 1 2 3 4 5 6 (t
0)
(t1) t1
t2
(t2) t3
(t3)
hb(t,)
Two main aspects
of the wireless
channel
Doppler Effect
• Difference in path lengths 𝚫𝐥 = 𝒅 𝒄𝒐𝒔𝜽 = 𝒗𝚫𝐭 𝐜𝐨𝐬𝜽
• Phase change 𝚫𝝓 = 𝟐𝝅𝚫𝐥
𝝀 = 𝟐𝝅𝒗𝚫𝐭
𝝀 𝐜𝐨𝐬𝜽
• Frequency change, or Doppler shift,
𝒇𝒅 = 𝟏 𝟐𝝅
𝚫𝝓
𝚫𝐭 = 𝒗
𝝀 𝒄𝒐𝒔𝜽
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Example
• Consider a transmitter which radiates a sinusoidal carrier frequency of 1850 MHz. For a vehicle moving 60 mph, compute the received carrier frequency if the mobile is moving
1. directly toward the transmitter.
2. directly away from the transmitter
3. in a direction which is perpendicular to the direction of arrival of the transmitted signal.
• Ans:
• Wavelength=𝜆 = 𝑐
𝑓𝑐 = 3×108
1850×106 = 0.162 (𝑚)
• Vehicle speed 𝑣 = 60𝑚𝑝ℎ = 26.82 𝑚
𝑠
1. 𝑓𝑑 = 26.82
0.162cos 0 = 160 𝐻𝑧 2. 𝑓𝑑 = 26.82
0.162cos 𝜋 = −160 (𝐻𝑧) 3. Since cos 𝜋
2 = 0, there is no Doppler shift!
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𝒇𝒅 = 𝟏 𝟐𝝅
𝚫𝝓
𝚫𝐭 = 𝒗
𝝀𝒄𝒐𝒔𝜽
Doppler Effect
• If the car (mobile) is moving toward the direction of the arriving wave, the Doppler shift is positive
• Different Doppler shifts if different 𝜽 (incoming angle)
• Multi-path: all different angles
• Many Doppler shifts Doppler spread
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Narrow-band Fading Model
• Sending an unmodulated carrier wave with random phase offset 𝝓𝟎:
• Received signal becomes
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𝑠 𝑡 = 𝑅𝑒{exp 𝑗 2𝜋𝑓𝑐𝑡 + 𝜙0 } = cos 2𝜋𝑓𝑐𝑡 + 𝜙0
𝑟 𝑡 = 𝑅𝑒
𝑛=1 𝑁 𝑡
𝛼𝑛 𝑡 exp −𝑗𝜙𝑛 𝑡 exp 𝑗2𝜋𝑓𝑐𝑡
= 𝑟𝐼 𝑡 cos(2𝜋𝑓𝑐𝑡) − 𝑟𝑄 𝑡 sin(2𝜋𝑓𝑐𝑡)
Sum of many MPC Carrier with frequency 𝑓𝑐
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𝑟𝐼 𝑡 =
𝑛=1 𝑁 𝑡
𝛼𝑛 𝑡 cos 𝜙𝑛 𝑡 𝑟𝑄 𝑡 =
𝑛=1 𝑁 𝑡
𝛼𝑛 𝑡 sin 𝜙𝑛 𝑡
𝜙𝑛 𝑡 = 2𝜋𝑓𝑐𝜏𝑛 𝑡 − 𝜙𝐷𝑛 − 𝜙0
= 𝑟𝐼 𝑡 cos(2𝜋𝑓𝑐𝑡) − 𝑟𝑄 𝑡 sin(2𝜋𝑓𝑐𝑡) 𝑟 𝑡 = 𝑅𝑒
𝑛=1 𝑁 𝑡
𝛼𝑛 𝑡 exp −𝑗𝜙𝑛 𝑡 exp 𝑗2𝜋𝑓𝑐𝑡
Doppler Shift Carrier phase shift (same for all MPC) Phase shift due to delay
Since N(t) is large & we assume 𝛼𝑛(𝑡) and 𝜙𝑛(𝑡) are independent for different MPC, we can approximate 𝑟𝐼(𝑡) and 𝑟𝑄(𝑡) as jointly Gaussian random processes.
Some assumptions
• No dominant LOS component
• 𝜶𝒏 𝒕 , 𝒇𝑫𝒏 𝒕 , 𝒂𝒏𝒅 𝝉𝒏 𝒕 change slowly over time
• 𝟐𝝅𝒇𝒄𝝉𝒏 changes rapidly relative to all other phase terms
• 𝝓𝒏(𝒕) uniformly distributed on [−𝝅, 𝝅].
• 𝜶𝒏 and 𝝓𝒏 are independent of each other.
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𝜙𝑛 𝑡 = 2𝜋𝑓𝑐𝜏𝑛 𝑡 − 𝜙𝐷𝑛 − 𝜙0
Very large
Zero-mean
• Similarly,
• So, E[r(t)]=0, and it is a zero-mean Gaussian process.
• If there is a dominant LOS component, then this is no longer true.
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𝐸 𝑟𝐼 𝑡 = 𝐸
𝑛
𝛼𝑛 cos 𝜙𝑛 𝑡 =
𝑛
𝐸 𝛼𝑛 𝐸[cos 𝜙𝑛(𝑡)] = 0
𝐸 𝑟𝑄 𝑡 = 0
Un-correlated
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𝐸 𝑟𝐼 𝑡 𝑟𝑄 𝑡 = 𝐸
𝑛
𝛼𝑛𝑐𝑜𝑠𝜙𝑛 𝑡
𝑚
𝛼𝑚 sin 𝛼𝑚 𝑡
=
𝑛 𝑚
𝐸 𝛼𝑛𝛼𝑚 𝐸 cos 𝜙𝑛 𝑡 sin 𝜙𝑚 𝑡
=
𝑛
𝐸 𝛼𝑛2 𝐸 cos 𝜙𝑛 𝑡 sin 𝜙𝑛 𝑡 =
𝑛
𝐸 𝛼𝑛2 𝐸 sin 2𝜙𝑛 𝑡
2 = 0
𝛼𝑛 and 𝜙𝑛 are not correlated.
=
𝑛,𝑚 𝑛≠𝑚
𝐸 𝛼𝑛 𝐸 𝛼𝑚 𝐸 cos 𝜙𝑛 𝑡 𝐸 sin 𝜙𝑚 𝑡 +
𝑛
𝐸 𝛼𝑛2 𝐸 cos 𝜙𝑛 𝑡 sin 𝜙𝑛 𝑡 Different MPC’s 𝛼𝑛 and 𝜙𝑛 are independent
Uniformly distributed over −𝜋, 𝜋 , so =0.
𝑟𝐼(𝑡) and 𝑟𝑄 𝑡 are uncorrelated, and they are Gaussian processes
they are independent.
Autocorrelation
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𝐴𝑟𝐼 𝑡, 𝑡 + 𝜏 = 𝐸 𝑟𝐼 𝑡 𝑟𝐼 𝑡 + 𝜏 =
𝑛
𝐸 𝛼𝑛2 𝐸 cos 𝜙𝑛 𝑡 cos 𝜙𝑛 𝑡 + 𝜏
= .5𝐸[cos(2𝜋𝑓𝐷𝑛𝜏)] + .5𝐸 cos 4𝜋𝑓𝑐𝜏𝑛 − 4𝜋𝑓𝐷𝑛𝑡 − 2𝜋𝑓𝐷𝑛𝜏 − 2𝜙0 𝐸 cos 𝜙𝑛 𝑡 cos 𝜙𝑛 𝑡 + 𝜏 =
= 𝐸[.5 cos 𝜙𝑛 𝑡 + 𝜏 − 𝜙𝑛 𝑡 + .5 cos 𝜙𝑛 𝑡 + 𝜏 + 𝜙𝑛 𝑡 ]
𝜙𝑛 𝑡 + 𝜏 = 2𝜋𝑓𝑐𝜏𝑛 − 2𝜋𝑓𝐷𝑛 𝑡 + 𝜏 − 𝜙0 𝜙𝑛 𝑡 = 2𝜋𝑓𝑐𝜏𝑛 − 2𝜋𝑓𝐷𝑛𝑡 − 𝜙0
Large and uniformly distributed over [−𝜋, 𝜋]
0
𝐴𝑟𝐼 𝑡, 𝑡 + 𝜏 = .5
𝑛
𝐸 𝛼𝑛2 𝐸[cos(2𝜋𝑓𝐷𝑛𝜏)] = .5
𝑛
𝐸 𝛼𝑛2 2𝜋𝑣𝜏 cos 𝜃𝑛 𝜆
Only depends on 𝜏, so WSS!
Autocorrelation
• Finally,
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𝐴𝑟𝐼,𝑟𝑄 𝑡, 𝑡 + 𝜏 = 𝐴𝑟𝐼,𝑟𝑄 𝜏 = 𝐸 𝑟𝐼 𝑡 𝑟𝑄 𝑡 + 𝜏
= −.5
𝑛
𝐸 𝛼𝑛2 sin 2𝜋𝑣𝜏 cos𝜃𝑛
𝜆 = −𝐸 𝑟𝑄 𝑡 𝑟𝐼 𝑡 + 𝜏
𝑟 𝑡 = 𝑟𝐼 𝑡 cos 2𝜋𝑓𝑐𝑡 − 𝑟𝑄 𝑡 sin 2𝜋𝑓𝑐𝑡
𝐴𝑟 𝜏 = 𝐸 𝑟 𝑡 𝑟 𝑡 + 𝜏 = 𝐴𝑟𝐼 𝜏 cos 2𝜋𝑓𝑐𝜏 + 𝐴𝑟𝐼,𝑟𝑄 𝜏 sin(2𝜋𝑓𝑐𝜏) Also only depends on 𝜏, WSS!
The received signal, representing how the channel changes over time
Amplitude distribution - Rayleigh
• 𝒛 𝒕 = 𝒓 𝒕 = 𝒓𝑰𝟐 𝒕 + 𝒓𝑸𝟐 𝒕
• 𝒓𝑰(𝒕) and 𝒓𝑸(𝒕) are both zero-mean Gaussian random process (so at a given time, two Gaussian random
variables).
• z(t)’s distribution - the amplitude distribution of r(t):
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Channel path loss
t
𝑝𝑍 𝑧 = 2𝑧
𝑃𝑟 exp −𝑧2
𝑃𝑟 = 𝑧
𝜎2 exp − 𝑧2
2𝜎2 , 𝑧 ≥ 0
This is the famous Rayleigh distribution!
2-variable joint
Gaussian distribution
• PDF for 2-variable joint Gaussian distribution:
• 𝝆: X and Y’s correlation (in our case, 0)
• 𝝁𝑿 and 𝝁𝒀: X and Y’s mean
• 𝝈𝑿𝟐 and 𝝈𝒀𝟐: X and Y’s variance (in our case, both are 𝝈𝟐)
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𝑓 𝑋, 𝑌 = 1
2𝜋𝜎𝑋𝜎𝑌 1 − 𝜌2 exp − 1 2 1 − 𝜌2
𝑋 − 𝜇𝑋 2
𝜎𝑋2 − 2𝜌(𝑋 − 𝜇𝑋)(𝑌 − 𝜇𝑌)
𝜎𝑋𝜎𝑌 + 𝑌 − 𝜇𝑌 2 𝜎𝑌2
𝑓 𝑋, 𝑌 = 1
2𝜋𝜎2 exp −1 2
𝑋2 + 𝑌2 𝜎2 The rest of the derivation can be found here:
http://www.dsplog.com/2008/07/17/derive-pdf-rayleigh-random-variable/
Power distribution:
Rayleigh
• We can obtain the power distribution by making the change of variables 𝒛𝟐 𝒕 = 𝒓 𝒕 𝟐 to obtain
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𝑝𝑍2 𝑥 = 1
𝑃𝑟 exp − 𝑥
𝑃𝑟 = 1
2𝜎2 exp − 𝑥
2𝜎2 , 𝑥 ≥ 0
Example: Rayleigh fading
• Consider a channel with Rayleigh fading (no LOS!) and average received power 𝑷𝒓 = 𝟐𝟎 dBm. Find the probability that the received power is below 10 dBm.
• We want to find the probability that 𝒁𝟐 < 𝟏𝟎 𝒅𝑩𝒎 = 𝟏𝟎 𝒎𝑾.
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𝑝 𝑍2 < 10 =
0
10 1
100exp − 𝑥
100 𝑑𝑥 = 0.095
With a LOS component – Ricean (or Rician)
• If the channel has a fixed LOS component then 𝒓𝑰(𝒕) and 𝒓𝑸(𝒕) are no longer zero-mean variables.
• The received signal becomes the superposition of a complex Gaussian component and a LOS component.
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Rayleigh: lots of random nLOS components
Ricean: Rayleigh + one strong static component
(LOS or strong reflection nLOS)
Ricean distribution
• Amplitude distribution:
• 𝟐𝝈𝟐 = 𝒏,𝒏≠𝟎 𝑬[𝜶𝒏𝟐] is the average power in the nLOS MPCs.
• 𝒔𝟐 = 𝜶𝟎𝟐 is the power in the dominant strong component.
• 𝑰𝟎(𝒙): the modified Bessel function of zeroth order.
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𝑝𝑍 𝑧 = 𝑧
𝜎2 exp −𝑧2 + 𝑠2
2𝜎2 𝐼0 𝑧𝑠
𝜎2 , 𝑧 ≥ 0
Ricean distribution
• The average power in the Ricean fading is
• The Ricean distribution is often described in terms of a fading parameter K, defined by
• K is the ratio of the power in the dominant component to the power in the other random MPCs.
• K=0, then Ricean degenerates to Rayleigh
• K=∞, then Ricean becomes a non-fading LOS channel.
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𝑃𝑟 =
0
∞
𝑧2𝑝𝑍 𝑧 𝑑𝑧 = 𝑠2 + 2𝜎2
𝐾 = 𝑠2 2𝜎2
Ricean and Rayleigh
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Example: Intra-car Wireless Channel Measurements
Which distribution fit the empirical amplitude distribution function the best?
Lognormal, Nakagami, Rician, Rayleigh, and Weibull
Engine compartment Center
(Parked)
Engine compartment Under the engine
(Parked)
CDF CDF
Parked: Weibull
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Engine compartment Trunk
(Driving)
Under the engine Engine Compartment
(Driving)
CDF CDF
Driving: Rician/Nakagami
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Channel model
No Line-of-Sight component Rayleigh
Rician
Coherence Time
• Coherence Time:
Coherence time is a statistical measure of the range of time over which the channel can be considered “static”.
• 90% coherence time:
• We can define 50% coherence time in a similar way too.
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𝑻𝒄,𝟎.𝟗 = 𝒂𝒓𝒈𝒎𝒊𝒏𝝉 𝑨𝒓 𝝉
𝑨𝒓 𝟎 < 𝟎. 𝟗
The first time interval that normalized autocorrelation drops below the threshold.
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10-3 10-2 10-1 100 101 102 103 104
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time spacing (seconds)
Normalized Correlation
Scenario 1 (Parked) Scenario 2 (Driving)
50% correlation
~ 2.5 sec ~ 60 sec
Worst Channel
Coherence time is always on the order of seconds.
IE/UE Channel
Fast and slow fading channel
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𝜏
𝜏 𝐴𝑟 𝜏
𝐴𝑟 𝜏
𝑇𝑐
𝑇𝑐 𝑇𝑠 t
𝑇𝑠: symbol period
Slow fading
Fast fading
f 𝐵𝑠
𝐵𝑠: signal bandwidth 𝐵𝐷
𝐵𝐷: Doppler Spread
f
𝑇𝑆 > 𝑇𝑐
𝐵𝐷
𝐵𝐷: Doppler Spread
f 𝐵𝐷 > 𝐵𝑆
𝐵𝐷 ≪ 𝐵𝑆
𝑇𝐶 ≫ 𝑇𝑆