## C = W log 1 + P ) W = log 1 + E ) E b = W C

A.2 Spectral Efficiency as a Function of Eb /N0 • Spectral efficiency: bits/Hz – For a CT bandlimited AWGN channel,  C = W log 1 + P N0 W  . – T...
A.2 Spectral Efficiency as a Function of Eb /N0

• Spectral efficiency: bits/Hz – For a CT bandlimited AWGN channel,  C = W log 1 +

P N0 W

 .

– Thus,   P C = log 1 + W N0W – However, the right side is a function of W . – Since P = CEb,

  Eb C C = log 1 + . W N0 W

– Thus, C

Eb W C 2W − 1 W = (2 − 1) = , C N0 C W which is the minimum SNR to achieve the spectral efficiency C/W [bps/Hz]. ∗ C/W → 0 =⇒ Eb/N0 & ln 2 ≈ −1.59 [dB], called the Shannon limit. ∗ This means that, even if we extremely reduce the spectral efficiency, zero-error transmission requires Eb/N0 ≥ −1.59 dB. 4

– The inverse function C/W as a function of Eb/N0 is mostly used: bandwidthefficiency plane ∗ For large enough Eb/N0, 10 log(Eb/N0) ∝ C/W ∗ This means that to increase the spectral efficiency n times, we need the SNR n times in dB scale. For example, if you use 10 dB power then to triple the spectral efficiency you must use 30 dB power. which is in practice impossible. ∗ In MIMO systems, the slope of C/W vs. Eb/N0 is dramatically changed. This is why the idea of using multiple antennas is revolutionary. • The next topic is to derive the relation between Pb and Eb/N0. For this, we need the rate-distortion theory.

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A.5 Channel Coding with Non-Zero Pb

• Given a DMC with capacity C [bits/channel use], what is the maximum throughput with Pb 6= 0? – First, we perform lossy source coding: R(Pb) = 1 − hb(Pb) < 1 – Then, we channel code the source encoder output. Thus, C(Pb) =

C C = > C. R(Pb) 1 − hb(Pb)

• What is the capacity of a DT AWGN channel with nonzero Pb? – Since

we have

  1 P = log 1 + 2 , C 2 σ    1   P log 1 + P P 2 σ2 C , P = > C b σ2 1 − hb(Pb) σ2 

P σ2



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– Let C˜ = C

P , Pb σ2



˜ b, we have [bits/channel use] and σ 2 = N0/2. Since P = CE   E 1 ˜ b 2 log 1 + 2C N0 C˜ = 1 − hb(Pb) ˜

Eb 2(1−hb(Pb))(2C) − 1 = ⇐⇒ , N0 2C˜ which is the minimum SNR to achieve the spectral efficiency C˜ [bits/chanel use] with BER Pb. ∗ Fix Pb: · C˜ → 0 =⇒ Eb/N0 & (1 − hb(Pb)) ln 2 ≈ 10 log10(1 − hb(Pb)) − 1.59 < −1.59 [dB]. – Pb vs. Eb/N0 ˜ ∗ Fix C:  1 − Pb = h−1 b for Eb/N0
C(P, N0, W ) 1 − hb(Pb) • What is the spectral efficiency of a bandlimited AWGN channel with nonzero Pb? C(P, N0, W, Pb) =

– Let C 0 , C(P, N0, W, Pb). Then, 

C 0 log 1 + = W 1 − hb(Pb) – Since P = C 0Eb, C0 = W

  C 0 Eb log 1 + W N0 1 − hb(Pb)

– Thus,

0

Eb 2 = N0

(1−hb (Pb )) C W

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C0 W

−1

∗ Fix Pb: · C 0/W → 0 =⇒ Eb/N0 & (1−hb(Pb)) ln 2 ≈ 10 log10(1−hb(Pb))−1.59 < −1.59 [dB]. · C 0/W does not improve much up to Pb ≈ 10−4. See the figures in the next 3 pages. · Thus, the bandwidth-efficiency plane with Pb 6= 0 is the same as that with Pb = 0 up to Pb ≈ 10−4. ∗ Fix C 0/W : · Now, we can plot Pb vs. Eb/N0 of the optimal channel coding schemes.    C 0 Eb log 1 + W N0 −1  , Pb = hb 1− C0 W

where Eb/N0