System intercomparison
No matter how much bandwidth is used, no system can transmit with less than −1.59 dB of energy per bit over the noise density. An uncoded QPSK needs 9.6 dB for one error in 100,000 bits.
01Two resources to share out
Every communication system spends two scarce resources: power and bandwidth. Comparing systems means seeing how much of each they need to deliver a given quality, which in analog is the output signal-to-noise ratio and in digital the error rate. Depending on which one is lacking, systems divide into two groups:
| System | Scarce resource | What is chosen | Examples |
|---|---|---|---|
| Power-limited | The energy of the received signal | Robust modulations and powerful codes, even if they take up bandwidth | Space probes, GPS, battery-powered sensors |
| Bandwidth-limited | The allocated spectrum | Constellations with many points and more power | Microwave radio links, cable modem, Wi-Fi |
Cellular and Wi-Fi systems move from one group to the other depending on distance: they choose the modulation and the code at each instant from the signal-to-noise ratio that the receiver measures.
02Analog systems, compared
With the results of noise in analog modulations, for a tone and the same received power:
| System | Bandwidth | Output SNR | Threshold |
|---|---|---|---|
| SSB | \( W \) | \( \gamma \) | No |
| DSB-SC | \( 2W \) | \( \gamma \) | No |
| AM, m = 1 | \( 2W \) | \( \gamma/3 \) | With an envelope detector |
| FM, β = 5 | \( 12W \) | \( 37.5\,\gamma \) | Yes, near γ = 20 dB |
SSB is the most efficient in spectrum and wideband FM the one that makes the best use of power. PCM adds another possibility: above its own error threshold, quality does not depend on the channel but on the quantization, and the signal can be regenerated at every repeater without accumulating noise. That is the reason networks became digital.
03The Shannon plane
In digital, the two measures are the spectral efficiency \( \eta = R_b/B \) and the required \( E_b/N_0 \). The Shannon–Hartley formula, with \( S = E_b R_b \) and \( N = N_0 B \), transmitting at capacity, gives the minimum possible for each efficiency:
Each modulation occupies a point in the plane, and the horizontal distance to the curve is the gain that a code could recover. M-QAM moves upward, gaining efficiency at the expense of power; orthogonal FSK moves to the left, spending bandwidth to save power.
The curve is the Shannon limit; below it no system is possible. The points are the uncoded modulations, at the \( E_b/N_0 \) that gives the chosen error rate, with ideal Nyquist pulses (α = 0). For coherent orthogonal FSK a tone spacing of \( 1/2T \) is assumed.
04Choosing a system
At an error rate of 10⁻⁵, uncoded modulations lie between 7 and 10 dB to the right of the limit. Error-correcting codes recover a good part of that distance: the convolutional codes of the 1970s gained about 5 dB, and present-day turbo and LDPC codes come within about 1 dB of the curve. The design of a link combines the pieces of the whole subject:
- The link budget, with the receiver noise figure, gives the available \( E_b/N_0 \).
- The allocated bandwidth sets the minimum spectral efficiency.
- The Shannon plane shows whether the pair is possible and what coding gain is needed.
- The modulation is chosen for its efficiency and for its envelope, depending on the available amplifiers.
05In the lab
Calculate the budget of a real microwave radio link from its datasheets and locate its operating point on the Shannon plane.
With a Wi-Fi analyzer, record the modulation and rate that a device uses at different distances from the access point, together with the reported signal-to-noise ratio.
Simulate on the computer a BPSK with and without a Hamming (7,4) code, and measure the coding gain at an error rate of 10⁻⁵, taking into account the energy of the redundancy bits.
06Common mistakes
- Comparing modulations by their error rate at the same S/N. The correct comparison is at the same \( E_b/N_0 \) and taking bandwidth into account.
- Believing that transmission is possible at any \( E_b/N_0 \) if a lot of bandwidth is used. Below −1.59 dB it is impossible.
- Forgetting the energy of the redundancy when measuring the gain of a code. Parity bits also cost power.
- Always choosing the most efficient modulation. In a power-limited system the opposite is better.
07Self-assessment
What is the minimum Eb/N0 for transmitting at 2 bit/s/Hz?
\( (2^2 - 1)/2 = 1.5 \): 1.76 dB.
An uncoded QPSK needs 9.6 dB for 10⁻⁵. How much could an ideal code gain at the same efficiency?
The limit at 2 bit/s/Hz is 1.76 dB: about 7.8 dB.
What resource does a space probe have little of, and what modulation suits it?
Power: a binary or few-point modulation, with a powerful code, even if it takes up more bandwidth.
What bandwidth does an FM with β = 5 occupy and how much does it improve the SNR over DSB?
12W, six times DSB, and it improves it by \( 1.5 \cdot 25 = 37.5 \) times: 15.7 dB.
08Further reading
- Bernard Sklar. Digital Communications: Fundamentals and Applications. 2nd ed., Prentice Hall, 2001. The spectral-efficiency plane and the design trade-offs of a link.
- John G. Proakis and Masoud Salehi. Communication Systems Engineering. 2nd ed., Prentice Hall, 2002. The comparison of analog and digital systems in the same framework.
- A. Bruce Carlson and Paul B. Crilly. Communication Systems. 5th ed., McGraw-Hill, 2010. The trade-off between bandwidth and power in analog and digital systems.