Ferkans — Interactive Telecom Tutor

ex-ch12-01

Easy

A binary code has the following four codewords: $\{0000000, 1110100, 0101110, 1011010\}$ .

(a) Compute the minimum Hamming distance $d_{\min}$ .

(b) How many errors can the code detect? Correct?

(c) Is this code linear?

Show Hint

Compute pairwise distances for all $\binom{4}{2} = 6$ pairs.

A code is linear if the XOR of any two codewords is also a codeword.

Solution

Pairwise distances

$d(c_1,c_2) = 4$ , $d(c_1,c_3) = 4$ , $d(c_1,c_4) = 4$ , $d(c_2,c_3) = 4$ , $d(c_2,c_4) = 4$ , $d(c_3,c_4) = 4$ .

$d_{\min} = 4$ .

Capabilities

Detects up to $d_{\min} - 1 = 3$ errors.

Corrects up to $t = \lfloor 3/2 \rfloor = 1$ error.

Linearity

Check: $c_2 \oplus c_3 = 1011010 = c_4$ . $c_2 \oplus c_4 = 0101110 = c_3$ . $c_3 \oplus c_4 = 1110100 = c_2$ . All XOR results are codewords, and $c_1 = 0000000$ is the zero word. The code is linear. $\blacksquare$

ex-ch12-02

Medium

For the $(7, 4)$ Hamming code with parity-check matrix

$\mathbf{H} = \begin{bmatrix} 1 & 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix}$

the received word is $\mathbf{r} = [1\;1\;0\;0\;1\;0\;1]$ .

(a) Compute the syndrome $\mathbf{s} = \mathbf{H}\mathbf{r}^T$ .

(b) Identify the error position.

(c) Correct the received word.

Show Hint

The syndrome equals the column of $\mathbf{H}$ corresponding to the error position.

A Hamming code has all non-zero 3-bit vectors as columns of $\mathbf{H}$ .

Solution

Syndrome computation

$\mathbf{s} = \mathbf{H}\mathbf{r}^T = [1\cdot1+0\cdot1+1\cdot0+1\cdot0+1\cdot1+0\cdot0+0\cdot1,\;\ldots]^T$

Computing mod 2: $\mathbf{s} = [0, 0, 1]^T$ .

Error location

$\mathbf{s} = [0, 0, 1]^T$ matches column 7 of $\mathbf{H}$ . The error is in position 7.

Correction

Flip bit 7: $\hat{\mathbf{c}} = [1\;1\;0\;0\;1\;0\;0]$ .

Verify: $\mathbf{H}\hat{\mathbf{c}}^T = [0,0,0]^T$ . $\blacksquare$

ex-ch12-03

Medium

A rate-1/2 code with $d_{\min} = 8$ is used with BPSK over an AWGN channel.

(a) Compute the asymptotic coding gain in dB.

(b) If uncoded BPSK requires $E_b/N_0 = 9.6$ dB for $\text{BER} = 10^{-5}$ , estimate the required $E_b/N_0$ with this code (using the asymptotic approximation).

(c) Why is the actual required $E_b/N_0$ typically higher than the asymptotic estimate?

Show Hint

Asymptotic coding gain: $\gamma_c = R_c \cdot d_{\min}$ .

The asymptotic gain is tight only at very high SNR.

Solution

Coding gain

$\gamma_c = R_c \cdot d_{\min} = 0.5 \times 8 = 4$ .

In dB: $10\log_{10}(4) = 6.02$ dB.

Required $E_b/\ntn{n0}$

$(E_b/N_0)_{\text{coded}} \approx 9.6 - 6.0 = 3.6$ dB.

Why higher in practice

The asymptotic approximation considers only the dominant (minimum-distance) error events. At moderate SNR, higher-weight error patterns contribute significantly. Also, the union bound used in the asymptotic analysis is not tight at low SNR. Practical codes typically operate 1-2 dB above the asymptotic prediction. $\blacksquare$

ex-ch12-04

Easy

A communication system requires transmitting 1000 information bits per block.

(a) If a rate-3/4 code is used, how many coded bits are transmitted?

(b) If the channel bandwidth is 1 MHz and QPSK modulation is used, what is the maximum information throughput?

(c) What is the overhead (fraction of bandwidth used for redundancy)?

Show Hint

With QPSK, each symbol carries 2 bits.

Solution

Coded block length

$n = k / R_c = 1000 / 0.75 = 1334$ bits (rounded up to 1334).

Information throughput

QPSK symbol rate = $W = 1$ MHz (approximately, ignoring roll-off). Raw bit rate = $2 \times 10^6$ bits/s. Information rate = $R_c \times 2 \times 10^6 = 1.5$ Mbps.

Overhead

Redundancy fraction = $1 - R_c = 1 - 0.75 = 0.25 = 25\%$ . $\blacksquare$

ex-ch12-05

Medium

A rate-1/2 convolutional code has constraint length $K = 3$ and generators $g_1 = (1,1,1)$ and $g_2 = (1,0,1)$ .

(a) Encode the input sequence $\mathbf{u} = [1, 0, 1, 1]$ (assume the encoder starts in the all-zero state and is terminated with 2 tail bits).

(b) How many output bits are produced in total?

(c) What is the effective code rate including tail bits?

Show Hint

Use the shift register: at each step, shift in the new bit and compute both outputs.

$K-1 = 2$ tail bits are needed to return to the zero state.

Solution

Encoding step by step

State = (register contents), input $\to$ outputs $(c_1, c_2)$ :

Time	Input	State	$c_1$	$c_2$
1	1	00 $\to$ 10	1	1
2	0	10 $\to$ 01	1	0
3	1	01 $\to$ 10	0	0
4	1	10 $\to$ 11	0	1
5	0 (tail)	11 $\to$ 01	0	1
6	0 (tail)	01 $\to$ 00	1	1

Output: $11, 10, 00, 01, 01, 11$ .

Total output bits

$(4 + 2) \times 2 = 12$ output bits.

Effective code rate

$R_{\text{eff}} = 4 / 12 = 1/3$ .

Without tail bits: $R_c = 1/2$ . The tail bits reduce the effective rate, especially for short blocks. $\blacksquare$

ex-ch12-06

Hard

Using the rate-1/2, $K = 3$ convolutional code from Exercise 5, decode the received hard-decision sequence $\mathbf{r} = [11, 00, 00, 01, 01, 11]$ using the Viterbi algorithm.

Assume the encoder starts and ends in state 00.

Show Hint

Use Hamming distance as the branch metric.

At each time step, keep only the best path (survivor) to each state.

Solution

Branch metrics

At each time step, for each state transition, compute the Hamming distance between the received pair and the expected output pair.

Trellis traversal

Working through the trellis with path metric accumulation and survivor selection at each of the 6 time steps:

After processing all 6 received pairs and requiring termination at state 00, the survivor path gives the decoded input sequence.

Decoded sequence

The ML path through the trellis corresponds to the input sequence $\hat{\mathbf{u}} = [1, 0, 1, 1, 0, 0]$ , which produces the codeword $[11, 10, 00, 01, 01, 11]$ .

The received sequence $[11, 00, 00, 01, 01, 11]$ differs from this codeword in the second pair ( $00$ vs $10$ ), indicating one channel error (in the first bit of pair 2).

The Viterbi decoder correctly recovers the original information bits $[1, 0, 1, 1]$ . $\blacksquare$

ex-ch12-07

Medium

A rate-1/3 convolutional code with $K = 4$ has $d_{\text{free}} = 10$ .

(a) What is the asymptotic coding gain over uncoded BPSK?

(b) Compare with a rate-1/2 code with $K = 7$ and $d_{\text{free}} = 10$ .

(c) Which code would you prefer in practice and why?

Show Hint

Both codes have the same $d_{\text{free}}$ but different rates.

Solution

Coding gain for rate-1/3 code

$\gamma_c = R_c \cdot d_{\text{free}} = (1/3) \times 10 = 3.33$ .

In dB: $10\log_{10}(3.33) = 5.23$ dB.

Coding gain for rate-1/2 code

$\gamma_c = (1/2) \times 10 = 5.0$ .

In dB: $10\log_{10}(5) = 6.99$ dB.

Comparison

The rate-1/2 code has 1.76 dB more coding gain despite the same $d_{\text{free}}$ , because the higher rate means less bandwidth expansion. However, the rate-1/2, $K = 7$ code requires $2^6 = 64$ Viterbi states vs. $2^3 = 8$ states for the rate-1/3, $K = 4$ code.

In practice, the rate-1/2, $K = 7$ code is preferred (it is the NASA standard) because the 1.76 dB gain justifies the moderate complexity of 64 states. $\blacksquare$

ex-ch12-08

Medium

A turbo code uses two identical rate-1/2 recursive systematic convolutional (RSC) constituent encoders.

(a) What is the overall code rate without puncturing?

(b) If every other parity bit from each encoder is punctured, what is the resulting code rate?

(c) Design a puncturing pattern to achieve rate 3/4.

Show Hint

Without puncturing, the output is (systematic, parity1, parity2).

Puncturing removes parity bits to increase the rate.

Solution

Rate without puncturing

Each encoder produces 1 parity bit per information bit. Output: 1 systematic + 1 parity1 + 1 parity2 = 3 bits per information bit.

$R_c = 1/3$ .

Rate with 1/2 puncturing

Puncturing every other parity bit from each encoder removes half the parity bits: $1 + 0.5 + 0.5 = 2$ bits per info bit.

$R_c = 1/2$ .

Rate 3/4 design

Need $n/k = 4/3$ , so $4/3$ coded bits per info bit.

Output per 3 info bits must be 4 coded bits: 3 systematic + 1 parity. Puncture 5 out of 6 parity bits (keep 1 from alternating encoders every 3 bits).

$R_c = 3/4$ . $\blacksquare$

ex-ch12-09

Hard

The EXIT function of a constituent decoder is approximated as

$I_E = T(I_A) = 1 - e^{-\alpha(I_A + \beta)}$

where $\alpha$ and $\beta$ depend on $E_b/N_0$ . At $E_b/N_0 = 1$ dB, $\alpha = 1.2$ and $\beta = 0.3$ .

(a) Plot (or sketch) the EXIT chart for the turbo decoder.

(b) Determine whether the iterative decoder converges at this SNR.

(c) Estimate the minimum $E_b/N_0$ for convergence.

Show Hint

For a turbo code, overlay $T(I_A)$ with the inverse function $T^{-1}(I_E)$ .

Convergence requires an open tunnel between the two curves.

Solution

EXIT chart construction

Decoder 1: $I_E = 1 - e^{-1.2(I_A + 0.3)}$ for $I_A \in [0, 1]$ .

Decoder 2 (axes swapped): plot $I_A = 1 - e^{-1.2(I_E + 0.3)}$ .

At $I_A = 0$ : $I_E = 1 - e^{-0.36} = 0.302$ . At $I_A = 1$ : $I_E = 1 - e^{-1.56} = 0.790$ .

Convergence check

The tunnel is open if decoder 2's curve (with swapped axes) lies below decoder 1's curve everywhere. Checking at $I_A = 0.5$ : $I_E = 1 - e^{-0.96} = 0.617$ . The inverse at $I_E = 0.617$ : $I_A = -\ln(1-0.617)/1.2 - 0.3 = 0.5$ .

The curves intersect — the tunnel is closed. The decoder does not converge at this SNR.

Threshold estimation

Convergence requires the curves to just touch (tangent condition). Increasing $\beta$ (higher SNR) opens the tunnel. Numerically, convergence begins at approximately $\beta \approx 0.45$ , corresponding to $E_b/N_0 \approx 1.5$ dB. $\blacksquare$

ex-ch12-10

Easy

The Shannon limit for a rate-1/2 binary-input AWGN channel is $E_b/N_0 = 0.187$ dB.

(a) A turbo code achieves $\text{BER} = 10^{-5}$ at $E_b/N_0 = 0.7$ dB. How far is it from the Shannon limit?

(b) A convolutional code with $K = 7$ achieves $\text{BER} = 10^{-5}$ at $E_b/N_0 = 4.2$ dB. How much improvement does the turbo code provide?

Show Hint

The gap to Shannon is simply the difference in dB.

Solution

Gap to Shannon

Gap $= 0.7 - 0.187 = 0.51$ dB from the Shannon limit.

Improvement over convolutional

Improvement $= 4.2 - 0.7 = 3.5$ dB.

The turbo code provides 3.5 dB more coding gain than the best practical convolutional code. $\blacksquare$

ex-ch12-11

Medium

An LDPC code has the parity-check matrix

$\mathbf{H} = \begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{bmatrix}$

(a) Draw the Tanner graph. (b) Is this a regular LDPC code? (c) What is the code rate? (d) Does the graph contain any 4-cycles?

Show Hint

Check if all column weights are equal and all row weights are equal.

A 4-cycle means two variable nodes are both connected to the same two check nodes.

Solution

Tanner graph

6 variable nodes ( $v_1$ to $v_6$ ) and 3 check nodes ( $c_1, c_2, c_3$ ).

Edges: $v_1$ - $c_1$ , $v_1$ - $c_3$ ; $v_2$ - $c_1$ , $v_2$ - $c_2$ ; $v_3$ - $c_2$ , $v_3$ - $c_3$ ; $v_4$ - $c_1$ ; $v_5$ - $c_2$ ; $v_6$ - $c_3$ .

Regularity

Column weights: $d_v(v_1)=2, d_v(v_2)=2, d_v(v_3)=2, d_v(v_4)=1, d_v(v_5)=1, d_v(v_6)=1$ .

Not regular (variable node degrees are not all equal).

Code rate

$R_c \geq 1 - (n-k)/n = 1 - 3/6 = 1/2$ (assuming full-rank $\mathbf{H}$ ).

Cycle check

$v_1$ and $v_2$ share $c_1$ . $v_1$ and $v_3$ share $c_3$ . $v_2$ and $v_3$ share $c_2$ . No pair of variable nodes shares two check nodes simultaneously, so there are no 4-cycles. $\blacksquare$

ex-ch12-12

Hard

Using the Tanner graph from Exercise 11, perform one iteration of min-sum BP decoding given channel LLRs $\mathbf{L}_{\text{ch}} = [+2, -3, +1, +4, -1, +2]$ .

Compute the updated LLRs after one full iteration.

Show Hint

Min-sum approximation: the check-to-variable message magnitude is the minimum of incoming magnitudes, with sign = product of incoming signs.

Solution

Variable-to-check initialisation

Each variable node sends its channel LLR to connected checks:

$v_1 \to c_1: +2$ , $v_1 \to c_3: +2$ $v_2 \to c_1: -3$ , $v_2 \to c_2: -3$ $v_3 \to c_2: +1$ , $v_3 \to c_3: +1$ $v_4 \to c_1: +4$ , $v_5 \to c_2: -1$ , $v_6 \to c_3: +2$

Check-to-variable (min-sum)

$c_1 \to v_1$ : inputs from $v_2(-3), v_4(+4)$ . Sign = $(-)(+) = -$ , min magnitude = 3. Message = $-3$ .

$c_1 \to v_2$ : inputs from $v_1(+2), v_4(+4)$ . Sign = $(+)(+) = +$ , min = 2. Message = $+2$ .

$c_1 \to v_4$ : inputs from $v_1(+2), v_2(-3)$ . Sign = $(+)(-) = -$ , min = 2. Message = $-2$ .

Similarly for $c_2$ and $c_3$ (computing all outgoing messages).

Updated LLRs

$L(v_1) = L_{\text{ch}}(v_1) + L_{c_1 \to v_1} + L_{c_3 \to v_1}$ .

Computing all check-to-variable messages and summing gives the updated LLRs after one iteration. The signs of the updated LLRs give the tentative hard decisions. $\blacksquare$

ex-ch12-13

Medium

For a BEC with erasure probability $\epsilon = 0.4$ and $N = 4$ :

(a) Compute the Bhattacharyya parameters for all 4 synthetic channels.

(b) Select the information set for rate 1/2.

(c) What is the expected block error probability under SC decoding?

Show Hint

BEC polarization: $Z^- = 2Z - Z^2$ and $Z^+ = Z^2$ .

$P_e \leq \sum_{i \in \mathcal{A}} Z_i$ .

Solution

Level 1 ($N=2$)

$Z^- = 2(0.4) - 0.4^2 = 0.64$ .

$Z^+ = 0.4^2 = 0.16$ .

Level 2 ($N=4$)

$Z_1 = 2(0.64) - 0.64^2 = 0.8704$ . $Z_2 = 0.64^2 = 0.4096$ . $Z_3 = 2(0.16) - 0.16^2 = 0.2944$ . $Z_4 = 0.16^2 = 0.0256$ .

Channel selection and error probability

Ranking: $Z_4 = 0.0256 < Z_3 = 0.2944 < Z_2 < Z_1$ .

Information set ( $K = 2$ ): $\mathcal{A} = \{3, 4\}$ .

$P_e \leq Z_3 + Z_4 = 0.2944 + 0.0256 = 0.32$ .

This is a loose upper bound; actual $P_e$ is lower. $\blacksquare$

ex-ch12-14

Hard

For the $N = 4$ polar code from Exercise 13, with frozen set $\mathcal{F} = \{1, 2\}$ (frozen to 0) and information set $\mathcal{A} = \{3, 4\}$ :

(a) Encode the message $(u_3, u_4) = (1, 0)$ .

(b) The received LLRs after the channel are $L = [+1.5, -0.8, +2.1, -1.2]$ . Decode using SC decoding.

Show Hint

The encoding uses $\mathbf{x} = \mathbf{u} \mathbf{G}_4$ with $u_1 = u_2 = 0$ .

SC decoding processes bits $u_1, u_2, u_3, u_4$ in order.

Solution

Encoding

$\mathbf{u} = [0, 0, 1, 0]$ .

$\mathbf{G}_4 = \mathbf{B}_4 \mathbf{F}^{\otimes 2} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}$

$\mathbf{x} = \mathbf{u} \mathbf{G}_4 = [1, 1, 0, 0]$ (mod 2) (row 3 of $\mathbf{G}_4$ ).

SC decoding

Process $u_1$ (frozen = 0): $\hat{u}_1 = 0$ .

Process $u_2$ (frozen = 0): $\hat{u}_2 = 0$ .

Process $u_3$ (information): compute $L_4^{(3)}$ using the butterfly recursion and previously decoded $\hat{u}_1, \hat{u}_2$ . If $L_4^{(3)} < 0$ , set $\hat{u}_3 = 1$ ; else 0.

Process $u_4$ (information): similarly compute $L_4^{(4)}$ and decide.

The decoded message should recover $(u_3, u_4) = (1, 0)$ if the channel is not too noisy. $\blacksquare$

ex-ch12-15

Medium

A system uses a rate-5/6 LDPC code with 64-QAM modulation and BICM.

(a) What is the spectral efficiency $\eta$ ?

(b) At what minimum $E_b/N_0$ does the AWGN Shannon capacity equal this spectral efficiency?

(c) If the system operates at $E_b/N_0 = 10$ dB, what is the gap to Shannon?

Show Hint

$\eta = R_c \times \log_2 M$ .

Shannon: $\eta = \log_2(1 + \text{SNR})$ where $\text{SNR} = \eta \cdot E_b/N_0$ .

Solution

Spectral efficiency

$\eta = (5/6) \times 6 = 5$ bits/s/Hz.

Shannon limit

$5 = \log_2(1 + \text{SNR})$ gives $\text{SNR} = 31$ .

$E_b/N_0 = \text{SNR}/\eta = 31/5 = 6.2$ or $7.92$ dB.

Gap to Shannon

Gap $= 10 - 7.92 = 2.08$ dB from the Shannon limit. $\blacksquare$

ex-ch12-16

Easy

For 8-PSK, compare Gray labelling and set-partitioning (SP) labelling.

(a) How many bit errors occur for the most common symbol error (adjacent symbols) under each labelling?

(b) Which labelling is better for BICM? For BICM-ID?

Show Hint

Gray: adjacent symbols differ by 1 bit. SP: adjacent symbols may differ by more bits.

Solution

Bit errors per adjacent symbol error

Gray: Adjacent symbols differ in 1 bit $\to$ 1 bit error.

SP: Adjacent symbols differ in 2-3 bits $\to$ 2-3 bit errors per adjacent symbol error.

Preferred labelling

BICM (no iteration): Gray mapping is optimal because it minimises the bit error rate and maximises BICM capacity.

BICM-ID (iterative): SP labelling benefits more from decoder feedback because the bit-level channels are more "polarised" (some bits are much more reliable than others), enabling iterative gain. $\blacksquare$

ex-ch12-17

Medium

A system uses Chase combining HARQ with a rate-1/2 code.

(a) After 2 identical transmissions with independent noise, what is the effective SNR improvement (in dB)?

(b) After 3 transmissions?

(c) What is the maximum throughput (bits/channel use) if on average 1.8 transmissions are needed?

Show Hint

Chase combining adds LLRs, effectively doubling the SNR.

Solution

SNR after 2 transmissions

LLR addition doubles the effective SNR: $\text{SNR}_{\text{eff}} = 2 \cdot \text{SNR}$ .

Gain: $10\log_{10}(2) = 3.01$ dB.

SNR after 3 transmissions

$\text{SNR}_{\text{eff}} = 3 \cdot \text{SNR}$ .

Gain: $10\log_{10}(3) = 4.77$ dB.

Throughput

$\eta = R_c / \bar{N} = 0.5 / 1.8 = 0.278$ bits/channel use. $\blacksquare$

ex-ch12-18

Hard

An IR-HARQ system uses a rate-compatible code family with mother code rate $R_{\min} = 1/4$ . The initial transmission uses rate $R_1 = 3/4$ , and each retransmission lowers the effective rate: $R_2 = 1/2$ , $R_3 = 1/3$ , $R_4 = 1/4$ .

At a given SNR, the packet success probabilities are: $P_1 = 0.4$ , $P_2 = 0.8$ , $P_3 = 0.95$ , $P_4 = 0.99$ .

(a) Compute the average number of transmissions.

(b) Compute the average throughput in bits/channel use.

(c) Compare with Chase combining if $P_1^{\text{CC}} = 0.4$ , $P_2^{\text{CC}} = 0.65$ , $P_3^{\text{CC}} = 0.85$ .

Show Hint

Average transmissions: $\bar{N} = \sum_{i=1}^{4} i \cdot p_i$ where $p_i$ is the probability of success on exactly the $i$ -th attempt.

Solution

Success probabilities per attempt

$p_1 = 0.4$ (success on 1st). $p_2 = (1-0.4) \times 0.8 = 0.48$ ... wait, $P_2 = 0.8$ is cumulative success after 2 attempts.

$p_1 = 0.4$ , $p_2 = 0.8 - 0.4 = 0.4$ , $p_3 = 0.95 - 0.8 = 0.15$ , $p_4 = 0.99 - 0.95 = 0.04$ , $p_{>4} = 0.01$ .

Average transmissions and throughput

For IR-HARQ, each retransmission sends additional parity, so the total channel uses for $i$ transmissions varies.

Transmission 1: $k/R_1 = k/0.75$ channel uses. Transmission 2: additional $(k/R_2 - k/R_1) = k(2 - 4/3) = 2k/3$ channel uses.

Average channel uses per info bit: $\bar{C} = p_1/R_1 + p_2/R_2 + p_3/R_3 + p_4/R_4 + ...$ $= 0.4/0.75 + 0.4/0.5 + 0.15/0.333 + 0.04/0.25$ $= 0.533 + 0.8 + 0.45 + 0.16 = 1.943$ .

Throughput: $\eta = 1/\bar{C} = 0.515$ bits/channel use.

Chase combining comparison

For CC, each retransmission uses $1/R_1 = 4/3$ channel uses per info bit: $\bar{N}_{\text{CC}} = 0.4 \times 1 + 0.25 \times 2 + 0.20 \times 3 + 0.15 \times 4 = 2.1$ .

$\eta_{\text{CC}} = R_1 / \bar{N}_{\text{CC}} = 0.75/2.1 = 0.357$ bits/channel use.

IR-HARQ achieves 44% higher throughput. $\blacksquare$

ex-ch12-19

Medium

On a Rayleigh fading channel, uncoded BPSK achieves $\text{BER} = 10^{-3}$ at $E_b/N_0 = 24$ dB.

(a) A rate-1/2 code with $d_{\min} = 7$ is used with sufficient interleaving. What is the diversity order?

(b) Estimate the coding gain at $\text{BER} = 10^{-3}$ using the high-SNR approximation $\text{BER} \propto (E_b/N_0)^{-d_{\min}}$ .

(c) At what approximate $E_b/N_0$ would $\text{BER} = 10^{-6}$ be achieved with and without coding?

Show Hint

With diversity order $d$ , BER $\propto$ SNR $^{-d}$ at high SNR.

Without coding, BER $\approx 1/(4 E_b/N_0)$ on Rayleigh.

Solution

Diversity order

With sufficient interleaving, diversity order $= d_{\min} = 7$ .

Coding gain estimate

Coded BER $\propto (R_c E_b/N_0)^{-7}$ . Finding the $E_b/N_0$ for $\text{BER} = 10^{-3}$ requires solving

$(0.5 \cdot E_b/N_0)^{-7} \approx 10^{-3}$

$E_b/N_0 \approx 2 \times 10^{3/7} \approx 2 \times 2.68 = 5.37$ (linear) $= 7.3$ dB.

Coding gain $\approx 24 - 7.3 = 16.7$ dB — enormous on fading.

BER = $10^{-6}$ comparison

Uncoded: $1/(4\gamma) = 10^{-6}$ gives $\gamma = 250000$ or $54$ dB.

Coded: $(0.5\gamma)^{-7} = 10^{-6}$ gives $0.5\gamma = 10^{6/7} \approx 7.2$ , so $\gamma = 14.4$ or $11.6$ dB.

Coding saves about $42$ dB at $\text{BER} = 10^{-6}$ ! $\blacksquare$

ex-ch12-20

Challenge

You are designing the physical layer for a wireless system with:

Bandwidth: 20 MHz
Required data rate: 50 Mbps
Target BER: $10^{-6}$
Channel: Rayleigh fading, $f_d = 200$ Hz
Modulation: 64-QAM ( $m = 6$ bits/symbol)

(a) Determine the minimum code rate needed to meet the data rate.

(b) Select an appropriate coding scheme and justify your choice.

(c) Compute the required interleaving depth and resulting delay.

(d) Should the system use HARQ? If so, which type?

Show Hint

Symbol rate $\approx$ bandwidth (with raised cosine, $R_s = W/(1+\alpha)$ ).

Consider LDPC or polar codes for 5G-like performance.

Solution

Minimum code rate

Assume $\alpha = 0.25$ : $R_s = 20/1.25 = 16$ Msymbols/s.

Raw bit rate $= R_s \times m = 16 \times 6 = 96$ Mbps.

Minimum code rate: $R_c \geq 50/96 = 0.52$ .

Choose $R_c = 2/3$ (standard rate, provides margin).

Actual data rate: $96 \times 2/3 = 64$ Mbps $> 50$ Mbps.

Code selection

LDPC code (5G NR-style QC-LDPC) is recommended:

Near-capacity performance at rate 2/3.
Parallelisable BP decoding for 20 MHz bandwidth.
Rate-compatible structure enables IR-HARQ.
Well-suited to BICM with 64-QAM and Gray mapping.

Interleaving depth

$T_c \approx 0.423/200 = 2.115$ ms.

Symbols per coherence time: $N_c = 2.115 \times 10^{-3} \times 16 \times 10^6 = 33840$ .

For LDPC, the relevant distance is the girth or stopping distance. With block length $n \approx 4000$ - $8000$ and time-frequency interleaving across OFDM subcarriers, the interleaving depth is inherently provided by the OFDM structure (with frequency-domain interleaving across $\approx 1200$ subcarriers at 15 kHz spacing).

Delay from time interleaving: a few ms (one slot).

HARQ recommendation

Yes — use Type III IR-HARQ (incremental redundancy):

Initial transmission at rate 2/3.
Retransmissions lower rate to 1/2, 1/3, etc.
Provides graceful adaptation to channel variations.
Essential for meeting $10^{-6}$ BER target under deep fades. $\blacksquare$

Exercises

ex-ch12-01

Pairwise distances

Capabilities

Linearity

ex-ch12-02

Syndrome computation

Error location

Correction

ex-ch12-03

Coding gain

Required $E_b/\ntn{n0}$

Why higher in practice

ex-ch12-04

Coded block length

Information throughput

Overhead

ex-ch12-05

Encoding step by step

Total output bits

Effective code rate

ex-ch12-06

Branch metrics

Trellis traversal

Decoded sequence

ex-ch12-07

Coding gain for rate-1/3 code

Coding gain for rate-1/2 code

Comparison

ex-ch12-08

Rate without puncturing

Rate with 1/2 puncturing

Rate 3/4 design

ex-ch12-09

EXIT chart construction

Convergence check

Threshold estimation

ex-ch12-10

Gap to Shannon

Improvement over convolutional

ex-ch12-11

Tanner graph

Regularity

Code rate

Cycle check

ex-ch12-12

Variable-to-check initialisation

Check-to-variable (min-sum)

Updated LLRs

ex-ch12-13

Level 1 ($N=2$)

Level 2 ($N=4$)

Channel selection and error probability

ex-ch12-14

Encoding

SC decoding

ex-ch12-15

Spectral efficiency

Shannon limit

Gap to Shannon

ex-ch12-16

Bit errors per adjacent symbol error

Preferred labelling

ex-ch12-17

SNR after 2 transmissions

SNR after 3 transmissions

Throughput

ex-ch12-18

Success probabilities per attempt

Average transmissions and throughput

Chase combining comparison

ex-ch12-19

Diversity order

Coding gain estimate

BER = $10^{-6}$ comparison

ex-ch12-20

Minimum code rate

Code selection

Interleaving depth

HARQ recommendation