Fall 2014 2014 Wireless Wireless Comm Communicat unications ions Solution of Homework Assignment #1
Problem 1 (20 points) (a) Consider a two-path channel with impulse response h(t) = α1 δ (τ ) τ ) + α2 δ (τ − − 0.022 × 10 6 ). Find the distance separating the transmitter and receiver, as well as α as α 1 and and α α 2 , assuming free space path loss on each path with a reflection coefficient of −1. Assume the transmitter and receiver are located 8 meters above the ground and the carrier frequency is 900 MHz. Sol: Since we know h(t) = α1 δ (t − τ ) τ ) + α 2 δ (t − (τ + 0.22 × 10 6)), ht = hr = 8, f c = 900MHz, R = −1 and use Gr = G l = 1, we have λ = 13 and the delay spread is −
−
x + x + x delay spread = c
− l = 0.022 × 10
6
−
.
So we can get
2 82 + ( d2 )2 3
× 10
8
−d
= 0.022
6
−
16.1(m) × 10 ⇒ d = 16.
(b) Consider a system with a transmitter, receiver, and scatterer as shown in Figure 1. Assume the transmitter and receiver are both at heights ht = hr = 4m and are separated by distance d, with the scatterer at distance 0. 0.5d along both dimensions in a two-dimensional grid of the ground, i.e. on such a grid the transmitter is located at (0, 0), the receiver is located at (0, (0, d) and the scatterer is located at (0. (0.5d, 0.5d). Assume 2 a radar cross section of 20 dBm dBm . Find the path loss of the scattered signal for d for d = = 1, 10, 100, and 1000 meters. Compare with the path loss at these distances if the signal is just reflected with reflection coefficient R = −1.
Figure 1: System with scatting
Sol: λ = 13 m since f c = 900M 900MHz Hz.. Let G = 1 and the radar cross section A = 20 √ dBm dB m2 =100 m2 . Also, x = x = (0. (0.5d)2 + (0. (0.5d)2 = 0.5d. Thus, the path loss due to scattering is given by
P r = P t
√
λ GA
√
4π
3/2
1
xx
2
=
0.0224 . d2
r Hence, for d = 1, 10, 100 and 1000 meters, P = 2.244 10 2 , 2.244 10 4 , 2.244 10 6 , P t 2.244 10 8 , respectively. The path loss due to reflection for 2 ray model is −
×
−
×
P r = P t
√ 2
R G x + x
λ 4π
2
=
3.52
× 10 d2
×
−
×
−
4
−
.
r For for d = 1, 10, 100 and 1000 meters, P = 3.52 10 4 , 3.52 10 6 , 3.52 10 8 , P t 3.52 10 10 , respectively. So the path loss due to reflection is much smaller than that due to only scattering.
×
−
×
−
×
−
×
−
Problem 2 (20 points) This problem shows how different propagation models can lead to very different SNRs (and therefore different link performance) for a given system design. Consider a linear cellular system using frequency division, as might operate along a highway or rural road, as shown in Figure 2 below. Each cell is allocated a certain band of frequencies, and these frequencies are reused in cells spaced a distance d away. Assume the system has square cells which are two kilometers per side, and that all mobiles transmit at the same power P. For the following propagation models, determine the minimum distance that the cells operating in the same frequency band must be spaced so that uplink SNR (the ratio of the minimum received signal-to-interference power (S/I) from mobiles to the base station) is greater than 20 dB. You can ignore all interferers except from the two nearest cells operating at the same frequency.
Figure 2: Linear Cells (a) Propagation for both signal and interference follow a free-space model. Sol: Let d be the distance between cells with reuse frequency and p be the transmit power of all the mobiles. Also, we know S 20dB. Minimum S/I is required I uplink when main user is at A and interferers are at B. The distance between A and base station 1 is dA = 2km, and the distance between B and base station 1 is dB = 2km. Thus,
√
S I
P = min
2P
Gλ 4πdB
√
2
Gλ 4πdA
≥
2
d2B (dmin 1)2 = 2 = = 100, dA 4
−
which gives dmin = 21 km. Since integer number of cells should be accommodated in distance d, dmin = 22 km. 2
(b) Propagation for both signal and interference follow the simplified path loss model(i.e. (2.39) in the note) with d0 =100m, K = 1, and γ = 3. Sol: γ d γ P k dA P r d0 S 1 dB γ = k = γ = P u d I min 2P k d 2 dA dB
0
⇒ − √ ⇒
0
3
1 dmin 1 2 2 Therefore, we use dmin = 10km.
= 100
⇒d
min
= 9.27
(c) Propagation for the signal follows the simplified path loss model with d0 = 100m, K = 1, and γ = 2, while propagation of the interference follows the same model but with γ = 4. Sol: γ d k dA S (dmin 1)4 = = 100. γ = I min 2k d 0.04 dB
0
−
0
So dmin = 2.41 km and thus we should take dmin = 4 km.
Problem 3 (20 points) Consider a cellular system where the received signal power is distributed according to a log-normal distribution with mean µ dBm and standard deviation σψ dBm. Assume the received signal power must be above 10 dBm for acceptable performance. (a) What is the outage probability when the log-normal distribution has µ ψ = 15 dBm and σψ = 8 dBm? Sol: For µψ = 15 dBm and σψ = 8 dBm, the outage probability for P min = 10 dBm is P[P r
< P min] = P =1
P r
− µψ < P − µψ min
σψ σψ Q( 0.625) = 0.266
− −
=
P
P r
− µψ < 10 − 15
σψ
8
(b) For σψ = 4 dBm, what value of µψ is required such that the outage probability is less than 1%, a typical value for high-quality PCS systems? Sol: P r µψ 10 µψ P[P r < P min ] = P < = 0.01 σψ 4 10 µψ =1 Q = 0.01, 4
which gives µψ
−
−
−
− µψ < 10 − µψ
−
≥ 19.3 dBm.
(c) Repeat (b) for σψ = 12 dBm. Sol: P[P r
< P min ] = P =1
P r
−Q
σψ 10
3
12
− µψ
12
= 0.01
= 0.01,
which gives µψ
≥ 37.9 dBm.
Problem 4 (20 points) The following table lists a set of empirical path loss measurements: Distance from transmitter 5m 25m 65m 110m 400m 1000m
P r /P t -60 dB -80 dB -105 dB -115 dB -135 dB -150 dB
(a) Find the parameters of a simplified path loss model plus log normal shadowing that best fit this data. Sol: The ratio of received power to transmitted power in dB is given by P r = P t
−10γ log
10
d + 10 log10 K d0
− ψdB ,
where ψ dB is a Gauss-distributed random variable with mean zero and variance σ 2 . The values of P r /P t in the above table is the means of P r /P t . Thus, assuming d 0 = 1 and we have the following:
−6.989γ + 10log −13.98γ + 10log −18.13γ + 10log −20.41γ + 10log −26.02γ + 10log −30.00γ + 10log
K = 10 K = 10 K = 10 K = 10 K = 10 K = 10
−60 −80 −105 −115 −135 −150.
Using lease square error fitting on the above 6 linear equations yields the following
−
6 115.529
−115.529 2566.59
10log10 K = γ
−645
13801.24
,
which gives γ = 4.04 and 10 log10 K = 29.72. (For the least square error fitting, please refer to the website http://mathworld.wolfram.com/LeastSquaresFitting.html).
−
(b) Find the path loss at 2 Km based on this model. Sol: For d = 2 Km, we have P r = P t
−10 · 4.04 log
10
(2000)
− 29.72 = −1163.08 dB.
(c) Find the outage probability at a distance d assuming the received power at d due to path loss alone is 10 dB above the required power for no outage. Sol: The outage probability is p out = P[P r < P min ], and for path loss only we know the 4
pl received power P r,dB = explicitly written as
−10γ log
pout =
P
10
d + 10 log10 K + P t,dB = 10 + P min,dB . which can be
P r,dB
−
= P[ψdB <
P pl r,dB
< P min,dB
−10] = P
−
ψdB < σ
P pl r,dB
−
10 , σ
where σ is the standard deviation of random variable ψdB due to shadowing. The variance can be found by the data given in the table and γ and K found in (a), i.e. σ
2
≈
1 6
6
2
− − P r P t
i=1
10γ log10 d
i
− 29.72
= 9.246 σ 3.04 dB.
⇒ ≈ Therefore, pout
≈ P
ψdB < σ
−3.289
= Q(3.289) = 0.00050272.
Problem 5 (15 points) Consider a cellular system operating at 900 MHz where propagation follows free space path loss with variations from log normal shadowing with σ = 6 dB. Suppose that for acceptable voice quality a signal-to-noise power ratio of 15 dB is required at the mobile. Assume the base station transmits at 1 W and its antenna has a 3 dB gain. There is no antenna gain at the mobile and the receiver noise in the bandwidth of interest is -10 dBm. Find the maximum cell size so that a mobile on the cell boundary will have acceptable voice quality 95% of the time. Sol: We have the following free space path loss parameters: f c = 900 MHz, λ = 13 , σ = 6 dB, required SN R = 15 dB, P t = 1 W, g = 3 dB, P noise = 40 dB, and thus we know the minimum received signal power P r = 15 40 = 25 dB. Suppose we choose a cell of radius d and thus the received signal power due to the path loss in dB is
−
√ Gl λ 4πd
P pl = 10 log10 P t Also we know
P[P r P
P
−
Since r σ pl further have
= 10 log10
−25 dB] = P
P r
1.4
× 10
3
−
d2
=
− P pl > −25 − P pl σ
σ
−28.54 − 20log
10
d.
= 0.95.
X is a Gaussian random variable with zero mean and unit variance, we P
which gives
>
−
2
−
3.54 + 20 log10 d X> = Q 6 3.54 + 20 log10 d = 6
3.54 + 20 log10 d = 0.95, 6
−1.645 ⇒ d = 0.21355. 5
