Microwave Circuits Review Part I

2.3. Spectral Efficiency and Power Efficiency

A fundamental upper limit or bound on data rate (b/s) exists for sending digital data (error-free) over a base-band channel of bandwidth B(Hz) in the presence of noise.

The upper limit is given by Shannon’s formula:

C = B*log(1 + SNR), where

C(b/s) = capacity, SNR = signal-to-noise ratio over the noisy channel, and log is of base-2.

Ideal tradeoff of Spectral Efficiency vs. Energy Efficiency

Ideally, for a capacity C = Rb, the Shannon’s formula gives

Rb/B = log(1 + SNR),

SNR = Ps/PN = Eb*Rb/(N0*B) = (Eb/N0)*(Rb/B), Eb/N0 is also called SNR per bit.

Thus,

Rb/B = log[1 + (Eb/N0)*(Rb/B)],

Eb/N0 = (B/Rb)2^(Rb/B) – 1.

We can see a tradeoff between spectral efficiency Rb/B and energy efficiency (Eb/N0)-1.

Higher data speeds are achievable with orthogonal-frequency-division-multiplexing (OFDM) transmission scheme. In OFDM a fast data stream is divided into a large number (10-1000) of slow parallel data streams, each of which is used to modulate a carrier of an orthogonal frequency. All modulated carriers are transmitted simultaneously. In the Rx the parallel data streams are retrieved through demodulation and recombined to reproduce the original transmitted fast data stream. OFDM is capable of the highest spectral efficiency Rb/B.

Bit-Error Rate

Pe and BER are used to denote bit-error rate.

In ASK, B = 2Rs = 2Rb, Eb = A2Tb/4, SNR = Ps/PN = EbRb/(N0B) = Eb/(2N0),

BERASK = Pe = 0.5erfc√[Eb/(2N0)] = Q[√(Eb/N0)],

i.e., Pe = 0.5erfc√(SNR) = Q[√(2SNR)].

In FSK, B ≈ Δf + 2(Rb/2) = 2Rb, Eb = A2Tb/2, SNR = Eb/(2N0),

BERFSK = Pe = 0.5erfc√[Eb/(2N0)] = Q[√(Eb/N0)].

In BPSK, B = 2Rb, Eb = A2Tb/2, SNR = Eb/(2N0),

BERBPSK = Pe = 0.5erfc√(Eb/N0) = Q[√(2Eb/N0)].

Note the superior performance of BPSK relative to FSK, and FSK relative to ASK. For the same BER, BPSK requires 1/2 the bit energy Eb of FSK. The same applies for FSK relative to ASK.

Symbol-Error Rate

For high-order modulation formats, each symbol = k bits and total number of symbols possible = 2k = M. We have

Es = kEb, SER = kBER.