Communication Systems · 96 h · Topic 9 of 11

Noise in digital modulations

For one error per million bits, a QPSK needs 10.5 dB of Eb/N0 and a 64-QAM, 18.8 dB. Tripling the bits per symbol costs almost seven times more energy per bit.

Matched filter Q function Eb/N0 Error rate M-QAM

01The optimum receiver

In a digital system noise does not degrade quality gradually: it produces decision errors. The performance measure is the bit error probability, or BER. The receiver that minimizes it with white Gaussian noise has two parts: a matched filter for the pulse, which maximizes the signal-to-noise ratio at the sampling instant, and a comparator that chooses the nearest symbol. At the output of the matched filter, that ratio depends only on the energy of the pulse, not on its shape:

\[ h(t) = p(T - t) \] \[ \mathrm{SNR_{max}} = \frac{2E}{N_0} \] \( E \) is the pulse energy and \( N_0/2 \) the noise spectral density. The matched filter is the time-reversed version of the pulse: it is equivalent to correlating the received signal with the expected pulse.

The output noise is Gaussian, as seen in probability distributions. The probability that it exceeds half the distance between two symbols is written with the \( Q \) function, the tail area of the standard Gaussian:

\[ Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^{\infty} e^{-u^2/2}\, du \] Q(3) ≈ 1.35·10⁻³ and Q(4.75) ≈ 10⁻⁶. The function falls so fast that a couple of dB of signal change the error rate by orders of magnitude.

02Binary modulations

To compare modulations we use the energy per bit \( E_b \) divided by \( N_0 \): this way they are compared with the same power and the same bit rate, independently of bandwidth. In BPSK, the two symbols are opposite and are separated by \( 2\sqrt{E_b} \):

\[ P_b^{\mathrm{BPSK}} = Q\!\left(\sqrt{\frac{2E_b}{N_0}}\right) \] For an error rate of 10⁻⁶, 10.5 dB is needed. QPSK is two BPSK in quadrature: it has the same bit error rate and occupies half the bandwidth.

Modulation	Bit error probability	E_b/N₀ for 10⁻⁶
Coherent BPSK and QPSK	\( Q(\sqrt{2E_b/N_0}) \)	10.5 dB
DPSK (differential)	\( \tfrac12 e^{-E_b/N_0} \)	11.2 dB
Coherent FSK	\( Q(\sqrt{E_b/N_0}) \)	13.5 dB
Noncoherent FSK	\( \tfrac12 e^{-E_b/2N_0} \)	14.2 dB

Coherent FSK loses 3 dB relative to BPSK because its two signals are orthogonal rather than opposite: they are closer together. Noncoherent receivers, which do not need to recover the carrier phase, are simpler and lose a little less than 1 dB in DPSK.

03Multilevel modulations

With more points in the constellation, at the same average energy the points are closer together and the error rate rises. For a square M-QAM with Gray code, with an approximation valid for low error rates:

\[ P_b \approx \frac{4}{k}\left(1 - \frac{1}{\sqrt M}\right) Q\!\left(\sqrt{\frac{3k}{M - 1}\, \frac{E_b}{N_0}}\right) \] \[ k = \log_2 M \] For 10⁻⁶, 16-QAM needs 14.4 dB and 64-QAM, 18.8 dB. Every two additional bits per symbol cost about 4 dB more.

Lab · error rate and noisy constellation

20,000 symbols are transmitted with Gaussian noise of the power corresponding to the chosen E_b/N₀. On the left, the first 1500 received points; on the right, the theoretical curves of the four modulations and the rate measured in the simulation.

E_b/N₀ =

04Eb/N0, signal-to-noise ratio and the link

At the receiver, the carrier-to-noise ratio is measured in the bandwidth \( B \). The relation to \( E_b/N_0 \) goes through the bit rate:

\[ \frac{S}{N} = \frac{E_b}{N_0} \cdot \frac{R_b}{B} \] A 10 Mbit/s QPSK in 6 MHz needs a signal-to-noise ratio 2.2 dB greater than its E_b/N₀. With the sensitivity and noise figure from the introduction, the link budget is closed.

Error-correcting codes add redundant bits that allow errors to be corrected at the receiver: in exchange for bit rate or bandwidth, they lower the required \( E_b/N_0 \) by several dB. This coding gain is what brings present-day systems closer to the limit studied in the unit on information theory.

05In the lab

Activity 1 · BER curve

With a pseudorandom sequence generator, a modulator, a noise generator and an error counter, measure the error rate of a BPSK for several E_b/N₀ values and compare it with theory.

Activity 2 · Measurement time

Calculate how many bits must be transmitted to measure a rate of 10⁻⁹ with confidence at 1 Gbit/s. Rule of thumb: count at least 100 errors.

Activity 3 · Matched filter

Program a correlator for rectangular pulses on a microcontroller and compare the error rate with that of simple sampling at the center of the pulse.

06Common mistakes

Comparing modulations at the same S/N instead of the same E_b/N₀. It hides the difference in bit rate and bandwidth.
Confusing symbol error rate and bit error rate. With Gray code, a symbol error generally costs a single bit out of the \( \log_2 M \).
Measuring too few bits. With 10⁶ bits you cannot claim a rate of 10⁻⁶: many more are needed.
Forgetting the losses of the real receiver. Synchronization, equalization and phase noise add one or two dB to the theoretical curves.

07Self-assessment

What error rate does a BPSK with E_b/N₀ = 9.6 dB have?

\( E_b/N_0 = 9.12 \) and \( Q(\sqrt{18.2}) = Q(4.27) \approx 10^{-5} \).

Why does QPSK have the same BER as BPSK with half the bandwidth?

Because they are two independent BPSK signals on quadrature carriers, each with the same energy per bit.

How many more dB than QPSK does 16-QAM need for 10⁻⁶?

About 4 dB: 14.4 versus 10.5.

A 16-QAM at 40 Mbit/s occupies 12 MHz. If E_b/N₀ = 15 dB, what S/N does it have?

\( 15 + 10\log_{10}(40/12) \approx 20.2 \) dB.

08Further reading

John G. Proakis and Masoud Salehi. Digital Communications. 5th ed., McGraw-Hill, 2008. The optimum receiver and the error probabilities of all the modulations.
Bernard Sklar. Digital Communications: Fundamentals and Applications. 2nd ed., Prentice Hall, 2001. The matched filter, BER curves and the link budget.
Simon Haykin. Digital Communication Systems. Wiley, 2014. Detection in Gaussian noise with a signal-space approach.