Control CDA 24 Octubre 2016 PDF

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Control CDA 24 Octubre 2016...

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CDA CITTEL 24 d’octubre de 2016. Duració de l’examen 1:45 hores. Dept. Teoria del Senyal i Comunicacions No es pot fer ús de cap dispositiu electrònic excepte calculadores no programables.

We consider a communication channel affected by path loss, shadowing and fast fading . The path loss is LdB = 50dB and the shadowing SdB is assumed to be log-normal with a 0dB mean and standard deviation σ ψ dB = 15 dB such the mean received power can be expressed as a function of the transmitted power Pt as:

P(dBm) = Pt (dBm) − LdB − SdB We assume the channel to be narrowband Rayleigh within the signal bandwidth such that instantaneous P(t) 2 = w(t) = c(t) follows an exponential distribution, fw (w) = e− w ; w ≥ 0 . normalized channel power P The coherence time of the channel is Δt c = 100ms . a) We first ignore the fast fading and analyze only the effect of the path loss and shadowing in the mean received power P . For Pt = 10dBm and using the function Q(z) = ∫

∞ z

1 − y 2 /2 dy , compute the outage e 2π

probability of a receiver characterized by a minimum required mean power Pmin = −70dBm . From now on we assume there is a slow power control mechanism that can compensate the path loss and shadowing, but not the fast fading such that the mean received power is fixed at P(dBm) = −40dBm and the instantaneous power is P(t) = Pw(t) . b) Compute the outage probability considering the fast fading when using slow power control for

Pmin = −70dBm . c) Assuming channel knowledge at the receive side only, compute the outage capacity for the outage probability obtained in b), when this channel is used in a bandwidth Bs = 1KHz and with a receiver with spectral noise density of N 0 = 10−13 J . d) Provide an integral expression for the ergodic capacity for the same bandwidth and noise spectral density as in c). Compute the capacity of the AWGN channel with the same mean SNR and compare this results using Jensen inequality with the ergodic capacity. e) We want to evaluate the data rate bound of a communication system that transmits blocks of data of duration TB = 0.5ms in this channel. Justify whereas the channel can be regarded constant (block fading) or time-varying (i.i.d. fading) during the transmission of one block of data. Which is appropriate capacity expression for this communication system, the outage capacity or the ergodic capacity? Repetition in time We think of improving the performance of our communication system by repeating the transmission of each block of data after Δt1 sec . f) Give the minimum value of Δt1 for which the fast fading in the repeated block is uncorrelated with the fast fading of the original block.

g) The receiver decodes the block received with the highest power. Compute the outage capacity for a 2% outage probability when this channel is used in a bandwidthBs = 1KHz and with a receiver of spectral noise density of N 0 = 10−13 J . Repetition in frequency Another option to improve the performance of our communication system is by repeating the transmission of each block at a different frequency. In order to do so we look into the wideband autocorrelation function of the channel, which is given by:

Ac (τ1 , τ 1 + τ;t,t + Δt) = E[c∗ ( τ1;t)c(τ 1 + τ ;t + Δt)] = Ac (τ ; Δt) = Bc Π(Bcτ )sinc( fD Δt) Bc = 10KHz ; f D = 10Hz ⎧⎪ 1 − 12 ≤ x ≤ 21 Π(x) = ⎨ else ⎩⎪ 0 This function corresponds to the 2-D autocorrelation of the normalized time-varying impulse response c(τ ;t) in which the variable τ models the temporal dispersion introduced by the channel and t the time varying nature of the channel caused by the Doppler effect. We want to analyze the time-varying channel behavior in the frequency domain by looking into the Fourier Transform of the 2-D autocorrelation Ac (τ ; Δt) with respect to τ : ∞

AC (Δf ; Δt) = ∫ Ac (τ; Δt)e − j 2π Δf τ d τ −∞

h) Obtain AC (Δf ; Δt) . i) Give the minimum value of carrier frequency separation Δf1 for which the fading in the repeated block is uncorrelated with the fading of the original block. We transmit every block two times using two carrier frequencies separated Δf1 Hz . In order to make a fair comparison with the previous system we assume the signal bandwidth has been reduced to B P B′s = s = 0.5KHz and the signal power to P′ = so that the total bandwidth and power are the same as 2 2 before. The spectral noise density of the receiver is also the same, N 0 = 10−13 J . j) The receiver decodes the block received with the highest power. Compute the system outage capacity for a 2% outage probability. Repetition in time and frequency Finally, we transmit every block four times: the original block, a block repetition using the same carrier frequency after Δt1 sec , another block repetition at the same time but using a carrier frequency separated Δf1 Hz , and a repetition after Δt1 sec and at a carrier frequency separated Δf1 Hz . Again, in order to make a fair comparison we assume the signal bandwidth has been reduced to B′s =

Bs = 0.5KHz and the signal 2

P so that the total bandwidth and power are the same as before. The spectral noise density 2 of the receiver is also the same, N 0 = 10−13 J . k) The receiver decodes the block received with the highest power. Compute the system outage capacity for power to P′ =

a 2% outage probability.

Solution ⎛ −70 − (10 − 50)⎞ a) Pout = 1− (Pr(P > Pmin ) = 1− Q⎜ ⎟⎠ = 1− Q(−2) = Q(2) ⎝ 15 10−3

(

b) Pout = Pr ( Pz < Pmin ) = Pr w < 10

−3

)= ∫ e

−w

dw = 1− e−0.001 = 0.001

0

⎛ 10 −710−3 ⎞ ⎛ P ⎞ c) Cout = (1− Pout )Bs log2 ⎜1+ min ⎟ = 10 3 (1 − 0.001)log2 ⎜ 1+ −13 3 ⎟ = 999 bps ⎝ 10 10 ⎠ N 0 Bs ⎠ ⎝ d) ∞

CE = E ⎡⎣Bs log2 (1+ γ )⎦⎤ = ∫ Bs log2 (1+ γ ) fγ (γ )d γ = γ = 0 ∞

∞

0

0

wP0 w10 −410−3 = 1000w = 10−1310 3 N 0 Bs

= Bs ∫ log2 (1+ 1000 w ) fw (w)dw = 10 3 ∫ log2 ( 1+ 1000 w)e− w dw [bps] ≤ Bs log2 (1+ E[ γ ]) = 10 3 log2 (1+ 1000 ) = 9.96Kbps=CAWGN e)

1 = 100ms ⇒ The channel can be regarded constant during one block fD Outage Capacity is the appropriate capacity measure 1 =100ms Ac (Δt) = sinc( fD Δt) = 0 ⇒ Δt1 = fD

0.5ms = TB ≪

f) g)

W

Pr(wm = max{w1 .w2 } < W ) = Pr(w1 < W ,w2 < W ) = ∫ e− w1 dw1 0

Pout = 0.02 = [1 − e

W

∫e

− w2

dw2 = [1 − e−W ]2

0

−W 2

] ⇒ Wout = 0.1525

⎛ W P⎞ ⎛ 0.1525 ⋅10 −410−3 ⎞ Cout = 12 (1− Pout )Bs log2 ⎜1+ out ⎟ = 12 10 3 (1 − 0.02)log2 ⎜ 1+ ⎟⎠ = 3.558 Kbps N 0 Bs ⎠ 10−1310 3 ⎝ ⎝ Note that there is a 50% penalty in the capacity caused by the block repetition in time h) i)

∞ ⎛ Δf ⎞ π τ AC (Δf ; Δt) = ∫ Bc Π(Bcτ )sinc( fD Δt)e− j 2 Δf dτ = sinc ⎜ ⎟ sinc( fD Δt) −∞ ⎝ Bc ⎠

⎛ Δf ⎞ AC (Δf1; Δt) = sinc ⎜ 1 ⎟ sinc( fD Δt) = 0 ⇒ Δf1 = Bc = 10KHz ⎝ Bc ⎠

j) 2

⎡W ⎤ Pr(wm = max{w1 ,w2 } < W ) = Pr(w1 < W ,w2 < W ) = ⎢ ∫e− w1 dw1 ⎥ = [1 − e−W ]2 ⎦ ⎣0 −W 2 Pout = 0.02 = [1 − e ] ⇒ Wout = 0.1525 ⎛ W P′⎞ ⎛ 0.1525 ⋅10 −4 ·0.5·10−3 ⎞ Cout = (1− Pout ) B′s log2 ⎜ 1+ out ⎟ = 0.5 ⋅10 3 (1 − 0.02)log2 ⎜ 1+ ⎟⎠ = 3.558 Kbps ⎝ N 0 Bs′ ⎠ 10−13 ⋅ 0.5·103 ⎝ k) 4

⎡W ⎤ Pr(wm = max{w1 ,w2 ,w3 ,w4 } < W ) = Pr(w1 < W ,w2 < W ,w3 < W ,w4 < W ) = ⎢ ∫ e− w1 dz1 ⎥ = [1 − e−W ]4 ⎦ ⎣0 Pout = 0.02 = [1 − e−W ]4 ⇒ Wout = 0.4717 ⎛ W P′⎞ ⎛ 0.4717 ⋅10 −410−3 ⎞ Cout = 12 (1− Pout ) B′s log2 ⎜ 1+ out ⎟ = 21 0.5 ⋅10 3 (1 − 0.02)log2 ⎜1+ ⎟⎠ = 2.17 Kbps N 0 B′s ⎠ 10−13 ⋅ 10 3 ⎝ ⎝...