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SYSC-4607 —- Problem Set #8 Solutions
(a) ρ = 10dB = 10
Pe = ρ−d
So to have Pe ≤ 10 −^3 , we should have d ≤ 3, or at least d = 3. Solving the equation that relates diversity gain d to multiplexing gain r at high SNRs we get d = (Mr − r)(Mt − r) ⇒ 3 = (8 − r)(4 − r)
Solving for r we get
r = 3.35 or 8.64 We have that r ≤ min(Mr, Mt), so r ≤ 4 and so r = 3.35. But we know that r has to be an integer. So, we take the nearest integer which is smaller than the calculated value of r, which gives us r = 3. If we are allowed to assume that equations 10.23 and 10.24 hold at finite SNRs too and we are given that we can use base 2 for logarithms, we can find the data rate as R = rlog2(ρ) = 9. 96 bits/s/Hz
(b) With, r = 3, we can find d as d = (Mr − r)(Mt − r) = (8 − 3)(4 − 3) = 5 For this value of d, Pe = ρ−d^ = 10−^5
(a)
H =
[
]
[
] [
]
[
]
P = 10mW N 0 = 10−^9 W/Hz B = 100KHz
(b) When H is known both at the transmitter and at the receiver, the transmitter will use the optimal precoding filter and the receiver will use the optimal shaping filter to decompose the MIMO channel into 2 parallel channels. We can then do water-filling over the two parallel channels available to get capacity.
Finding the γi’s
γ 1 = γ 12 P N 0 B
γ 2 =
γ 22 P N 0 B
Finding γ 0 Now, we have to find the cutoff value γ 0. First assume that γ is less than both γ 1 and γ 2. Then
( 1 γ 0
γ 1
γ 0
γ 2
γ 0
γ 1
γ 2 ⇒ γ 0 =
1 + (^) γ^11 + (^) γ^12
which is less than both γ 1 and γ 2 values so our assumption was correct.
Finding the capacity Now we can use the capacity expression as
C =
∑^2
i=
Blog 2
γi γ 0
= 800Kbps
(c) Essentially we have two parallel channels after the precoding filter and the shaping filter are used at the transmitter and receiver respectively.
Once this is done the SNR at the combiner output is simply λmaxρ where λmax is the maxi- mum eigen value of the Wishart Matrix HHH^ and ρ is P/(N 0 B).
Finding γ As given in the question, λmax is 0.7592 and ρ was calculated to be 100. So we get that ρ = 75.92.
Finding Pb
When using BPSK,γs = γb. Now we can use the expression for Pb for BPSK
Pb = Q(
2 γb) = Q(
2 × 75 .92) ≈ 0;
Finding Rate R Since we are using BPSK and are given that B = 1/Tb, we get the rate using BPSK to be R = 100Kbps.
12-1 (a) ψi = cos(2πj/TN t + φj ) To form a set of orthonormal basis on [0, TN ] We need,
∫ TN
0 ψj^ ψkdt^ = 0, where the phases are uniformly distributed between [0,^2 π].
∫ (^) TN
0
ψj ψkdt =
∫ TN
0
cos(2πj/TN t + φj )cos(2πk/TN t + φk)dt
∫ TN
0
- 5 cos(2π(j + k)/TN t + φj + φk) + 0. 5 cos(2π(jk)/TN t + φj − φk)dt
= 0. 5
TN
2 π(j + k)
[sin(2π(j + k)t + φj + φk) − sin(φj + φk)]
TN
2 π(j − k)
[sin(2π(j − k)/t + φj − φk) − sin(φj − φk)]
= 0;
⇒j and k are integers. ⇒The minimum separation for sub-carriers is 1/TN for any φj. (b) If φj = 0∀j
∫ TN
0
ψj ψkdt = 0. 5
TN
2 π(j + k) sin(2π(j + k)t) + 0. 5
TN
2 π(j − k) sin(2π(j − k)/t) = 0
⇒ 2 π(j + k) = l 1 π, 2 π(j − k) = l 2 π l 1 , l 2 ∈ Z ⇒ j and k are multiples of 1/ The minimum separation:1/ 2 TN
12-2 (a) TN = 1/BN = 10Tm = 10/Bc = 10/ 10 KHz = 1ms (b) B = N^ (1+ TNβ +)= 128(1+1 1 ms.5+0 .1)= 333KHz (c) B = N^ +TNβ+ = 128+1 1 ms.5+0.^1 = 129KHz
Therefore, the total bandwidth using overlapping carriers is less than half of the non overlap- ping bandwidth.
12-4 (a) For FDM, the number of subchannels = (^) (ΔBf )c = 5
Ts = 10μsec, R = (^10) μsec^1 = 0. 1 M bps (b) P (^) bn = 0. 5
[
√ (^) γ bn γbn+
]
n γsn Pb 1 1000 2. 5 × 10 −^4 2 500 5 × 10 −^4 3 333 7. 5 × 10 −^4 4 250 1 × 10 −^4 5 200 1. 25 × 10 −^3
BER after decoding =
i=3 Pr[i channels in error] = 3.^5 ×^10 − 9 The total date rate of the system is the same as the data rate of any of the subcarriers (since they all have the same bits transmitted over them)⇒ R = 0. 1 M bps (c) Since it is not specified which equation to use for calculation of SNR, all answers based on any correct equation in the reader are being given full credit. What is given below is just one way to do the problem. Your answer can be totally different but we still give credit for it.
