Exercises — Physical layer — encoding (NRZ, Manchester), bandwidth, Nyquist, Shannon-Hartley

WHAT: In NRZ-L high voltage $=1$. So a held-high line is a run of 1s. WHY the danger: with no voltage change, there is no edge for the receiver to lock its clock onto. The receiver's internal "tick-tock" drifts and it miscounts how many identical bits passed. The word: synchronization loss (loss of clocking).

• Noiseless → Nyquist: $C=2B{\log }_{2}M$.
• Noisy → Shannon: $C=B{\log }_2(1+S/N)$. The deeper reason (not just a name-match): on a noiseless line nothing limits how many levels $M$ you may use — so capacity is limited only by how fast you can push symbols, which is the $2B$ term. Adding $M$ always helps, so $M$ is a free knob and Nyquist has no $S/N$ in it. On a noisy line, noise smears levels into each other, so noise itself caps how many levels are distinguishable — that is why Shannon's formula contains $S/N$ (the noise) instead of $M$ (the levels). In short: Nyquist limits speed of change; Shannon limits distinguishability.

WHY ${\log }_2$: $n$ bits make $2^n$ distinct patterns. To label 8 levels we need $2^n=8$, so $n=3$. ${\log }_{2}8=3\text{ bits per symbol.}$

WHY Nyquist and not Shannon (the underlying principle): the channel is stated noiseless — there is no noise to smear levels, so the only thing capping us is how fast the signal can change, which the bandwidth $B$ fixes at $2B$ symbols/sec (the sampling picture in Figure 2). Shannon's $S/N$ term is meaningless here because $N=0$. So the right law is the one built purely from speed-of-change and levels: Nyquist. Step 1 (bits/symbol): ${\log }_{2}16=4$. Step 2 (Nyquist): $C=2B{\log }_{2}M=2(4000)(4)=8000\times 4=32000\text{ bits/s.}$

WHY Shannon here: the problem gives an $S/N$ — that is a fingerprint of noise. A noisy line's true ceiling is set by how many levels noise still lets you tell apart, which is exactly what Shannon captures. Step 1 (dB → linear): SNR and $S/N$ are the same ratio; Shannon needs it linear. $\frac{S}{N}=10^{20/10}=10^2=100.$ Step 2 (Shannon): $C=B{\log }_2(1+S/N)=3000{\log }_2(101).$ ${\log }_2(101)\approx 6.658$, so $C\approx 3000\times 6.658\approx 19975\text{ bits/s}$ (about $20$ kbit/s).

WHY not equal to baud: "2400 symbols per second" is the baud (symbol rate). Bit rate equals baud only when $M=2$. Here $M=4$. $C=\text{baud}\times {\log }_{2}M=2400\times {\log }_{2}4=2400\times 2=4800\text{ bits/s.}$

WHY Nyquist: noiseless again — no $S/N$, so the level count $M$ is the free knob and Nyquist is the governing law. Set up Nyquist and solve for $M$: $C=2B{\log }_{2}M\Rightarrow 12000=2(3000){\log }_{2}M=6000{\log }_{2}M.$ ${\log }_{2}M=\frac{12000}{6000}=2\Rightarrow M=2^2=4\text{ levels.}$ Here ${\log }_{2}M=2$ is already a whole number, so $M=4$ is an exact power of two — no rounding needed. Sanity check: 4 levels = 2 bits/symbol, $6000$ symbols/s $\times 2=12000$ bits/s. ✓

WHERE the $\frac{1}{2}$ comes from (traced to Nyquist): Nyquist's sampling result (Figure 2, right panel) says a bandwidth-$B$ channel supports at most $2B$ signal changes per second — the same $2$ that appears in $C=2B{\log }_{2}M$. So if you must produce $R$ transitions per second, the channel needs $2B=R\Rightarrow B_{\min }=\frac{1}{2}R.$ The $\frac{1}{2}$ is not magic — it is literally Nyquist's $2B$ solved for $B$.

• Manchester: up to $2$ transitions per bit at $10\times 10^6$ bits/s gives $R=2\times 10^7$ transitions/s, so $B_{\min }\approx \frac{1}{2}(2\times 10^7)=10^7=10$ MHz.
• NRZ: at most $1$ transition per bit gives $R=10^7$ transitions/s, so $B_{\min }\approx \frac{1}{2}(10^7)=5\times 10^6=5$ MHz. Conclusion: Manchester needs about double the bandwidth of NRZ for the same bit rate. That extra bandwidth is the price you pay to get a guaranteed clock edge every bit.

(a) Shannon: $1+S/N=256=2^8$, so ${\log }_{2}256=8$. $C=B{\log }_2(1+S/N)=10^6\times 8=8\times 10^6=8\text{ Mbit/s.}$ (b) Match with Nyquist: solve $2B{\log }_{2}M=8\times 10^6$ with $2B=2\times 10^6$: ${\log }_{2}M=\frac{8\times 10^6}{2\times 10^6}=4\Rightarrow M=2^4=16\text{ levels.}$ ${\log }_{2}M=4$ is a whole number, so $M=16$ is already a power of two. So $16$ levels just reach the Shannon limit here.

(a) Convert then Shannon: $S/N=10^{30/10}=10^3=1000$. $C_{\text{Shannon}}=4\times 10^6\cdot {\log }_2(1001).$ ${\log }_2(1001)\approx 9.9672$, so $C_{\text{Shannon}}\approx 4\times 10^6\times 9.9672\approx 3.987\times 10^7\approx 39.87\text{ Mbit/s.}$ (b) Nyquist match: $2B{\log }_{2}M=C_{\text{Shannon}}$, $2B=8\times 10^6$: ${\log }_{2}M=\frac{3.987\times 10^7}{8\times 10^6}\approx 4.98\Rightarrow M\approx 2^{4.98}\approx \mathrm{31.5.}$ Rounding to a power of two: $M$ must be a whole number, and for clean bits/symbol we use $M=2^n$. The two candidate powers of two are $16$ ($n=4$) and $32$ ($n=5$). To reach the ceiling you round up to $M=32$ (which slightly exceeds the Shannon rate — noise would then cause errors, so you actually cap at Shannon). (c) Achievable rate = the smaller limit = Shannon ≈ 39.87 Mbit/s. Nyquist with an integer power-of-two $M$ tells you the design ($M=32$ to reach it, or $M=16$ to stay safely below); Shannon is the hard ceiling you cannot cross.

Line A (Nyquist — because noiseless, $M$ is the free knob): need $2B{\log }_{2}M\ge 10^6$ with $2B=5\times 10^5$: ${\log }_{2}M\ge \frac{10^6}{5\times 10^5}=2\Rightarrow M\ge 4.$ ${\log }_{2}M=2$ gives exactly $M=4$, already a power of two → achievable with $M=4$. ✓ Line B (Shannon — because $S/N$ given, noise is the limiter): $1+S/N=1024=2^{10}$, so $C=B{\log }_2(1024)=10^5\times 10=10^6=1\text{ Mbit/s exactly. Achievable, zero margin.}✓$ Verdict: both work. Line A has room to grow (add levels); Line B is at its absolute Shannon edge — any drop in $S/N$ breaks it.

(a) Current ceiling: $S/N=10^{24/10}=10^{2.4}\approx 251.2$. $C=B{\log }_2(1+251.2)=2\times 10^6\cdot {\log }_2(252.2).$ ${\log }_2(252.2)\approx 7.978$, so $C\approx 2\times 10^6\times 7.978\approx 1.596\times 10^7\approx 15.96\text{ Mbit/s.}$ 30 Mbit/s is NOT possible — the ceiling is only ~16 Mbit/s. (b) Solve for required $S/N$: need $B{\log }_2(1+S/N)=30\times 10^6$ with $B=2\times 10^6$: ${\log }_2(1+S/N)=\frac{30\times 10^6}{2\times 10^6}=15\Rightarrow 1+S/N=2^{15}=32768.$ $S/N=32767\Rightarrow (S/N{)}_{dB}=10{\log }_{10}(32767)\approx 45.15\text{ dB.}$ Reading: to nearly double the bit rate you must raise SNR from 24 dB to ~45 dB — a huge power jump, because capacity grows only logarithmically with $S/N$. This is why real designers widen $B$ or accept lower rates instead. See Signal-to-Noise Ratio and Bandwidth vs Bit Rate.

(a) NRZ, $M=2$: $C=2B{\log }_{2}2=2(5\times 10^6)(1)=10^7=10\text{ Mbit/s.}$ (b) Manchester: it consumes ~2× bandwidth per bit, so in the same 5 MHz it delivers half the bit rate of NRZ: $C_{\text{Manchester}}\approx \frac{1}{2}\times 10^7=5\times 10^6=5\text{ Mbit/s.}$ One-line trade: Manchester halves your throughput but buys a guaranteed clock edge every bit (self-clocking, no DC) — you pay speed for reliable synchronization.

(a) Shannon: $1+S/N=16=2^4$, so ${\log }_{2}16=4$. $C_{\text{Shannon}}=3\times 10^6\times 4=1.2\times 10^7=12\text{ Mbit/s.}$ (b) Nyquist levels to match: $2B{\log }_{2}M=12\times 10^6$, $2B=6\times 10^6$: ${\log }_{2}M=\frac{12\times 10^6}{6\times 10^6}=2\Rightarrow M=4.$ ${\log }_{2}M=2$ is a whole number, so $M=4$ is an exact power of two — it hits the ceiling with no rounding needed, so $M=4$ is the safe choice. (c) Actual rate with $M=4$: $C=2B{\log }_{2}4=6\times 10^6\times 2=1.2\times 10^7=12\text{ Mbit/s.}$ — right at the Shannon edge. ✓

• What is $B$? ::: The channel's bandwidth $=f_{\text{high}}-f_{\text{low}}$, the width of passable frequencies (Hz).
• What is $M$? ::: The number of distinct signal levels a symbol can take (carries ${\log }_{2}M$ bits).
• 8 levels = how many bits/symbol? ::: ${\log }_{2}8=3$ bits.
• Convert 30 dB to linear SNR. ::: $10^{30/10}=1000$.
• Nyquist at $B=3000$, $M=4$? ::: $2(3000)(2)=12000$ bit/s.
• Which rate do you actually deliver? ::: $\min (\text{Nyquist},\text{Shannon})$.
• Why does Manchester halve throughput vs NRZ? ::: it needs ~2× bandwidth per bit (self-clocking cost).
• Are "SNR" and "$S/N$" the same? ::: yes — the same signal-to-noise power ratio (convert from dB first).