Exercises

The channel has taps at $n = 0, 1, 2$, so $L = 2$.

The noise-free output is $y_k = a_k + 0.4 a_{k-1} - 0.2 a_{k-2}$. With $L + 1 = 3$ binary symbols involved, there are $2^3 = 8$ possible combinations.

For $a_k = +1$: levels are $\{0.4, 0.8, 1.2, 1.6\}$ (minimum is $0.4$).

For $a_k = -1$: levels are $\{-1.6, -1.2, -0.8, -0.4\}$ (maximum is $-0.4$).

The minimum eye opening is $d_{\min} = 0.4 - (-0.4) = 0.8$.

Compare with the ISI-free case: $d_{\min} = 2|h[0]| = 2.0$. ISI reduces the eye opening by a factor of 2.5. $\blacksquare$

$\sum_{n=0}^{2} |h[n]|^2 = 0.25 + 1.0 + 0.09 = 1.34$$

The effective $E_b/N_0$ at the MFB is

$$\gamma_{\text{MFB}} = \frac{E_b}{N_0} \sum |h[n]|^2 = 10 \times 1.34 = 13.4 \;(= 11.27 \text{ dB})$$

$\text{BER}_{\text{MFB}} = Q(\sqrt{2 \times 13.4}) = Q(5.177) \approx 1.14 \times 10^{-7}$$This is the best any equalizer can achieve on this channel.$\blacksquare$

The DTFT of $c[n]$ is $C(e^{j\omega}) = \sum_n c[n] e^{-j\omega n}$.

The inverse DTFT gives $c[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} C(e^{j\omega}) e^{j\omega n}\, d\omega$.

If $\sum_k C(e^{j(\omega - 2\pi k)}) = 1$, then since $C(e^{j\omega})$ is $2\pi$-periodic, this just states that $C(e^{j\omega})$ integrated properly gives $c[0] = 1$.

For $n \neq 0$:

$$c[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} C(e^{j\omega}) e^{j\omega n}\, d\omega$$

The folded spectrum condition means $C(e^{j\omega})$ acts as a constant (= 1) when "averaged" over each $2\pi$ interval. Multiplying by $e^{j\omega n}$ and integrating yields zero for $n \neq 0$ by orthogonality of complex exponentials. $\blacksquare$

$W_{\text{ZF}}(z) = \frac{1}{H(z)} = \frac{1}{1 + \alpha z^{-1}} = \sum_{k=0}^{\infty} (-\alpha)^k z^{-k}$$The impulse response is$w[k] = (-\alpha)^k$for$k \geq 0$.

$\xi = \sum_{k=0}^{\infty} |w[k]|^2 = \sum_{k=0}^{\infty} \alpha^{2k} = \frac{1}{1 - \alpha^2}$$

For $\alpha = 0.9$: $\xi = 1/(1 - 0.81) = 1/0.19 = 5.26$

In dB: $10 \log_{10}(5.26) = 7.21$ dB.

The ZF equalizer amplifies noise by 7.21 dB, which is a severe penalty. $\blacksquare$

In the frequency domain, the Wiener-Hopf equation becomes

$$W(e^{j\omega}) = \frac{S_{yd}(e^{j\omega})}{S_{yy}(e^{j\omega})}$$

where $S_{yy}$ is the PSD of $y_k$ and $S_{yd}$ is the cross-PSD between $y_k$ and the desired signal $a_{k-d}$.

$S_{yy}(e^{j\omega}) = E_s |H(e^{j\omega})|^2 + N_0KATEXPLACEHOLDER0ENDS_{yd}(e^{j\omega}) = E_s\, H^*(e^{j\omega})\, e^{-j\omega d}$$(The$e^{-j\omega d}$accounts for the decision delay; setting$d = 0$ for simplicity.)

$W_{\text{MMSE}}(e^{j\omega}) = \frac{E_s\, H^*(e^{j\omega})}{E_s |H(e^{j\omega})|^2 + N_0} = \frac{H^*(e^{j\omega})}{|H(e^{j\omega})|^2 + N_0/E_s}$$At high SNR ($N_0/E_s \to 0$):$W \to H^*/|H|^2 = 1/H$(ZF). At low SNR ($N_0/E_s \to \infty$):$W \to (E_s/N_0) H^*$(matched filter).$\blacksquare$

For a 5-tap equalizer, the received vector has contributions from symbols $a_{k}, a_{k-1}, \ldots, a_{k-6}$. The channel matrix $\mathbf{H}$ is $5 \times 7$ (5 output samples, 7 symbols involved).

$$\mathbf{H} = \begin{bmatrix} 1 & 0.7 & -0.3 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0.7 & -0.3 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0.7 & -0.3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0.7 & -0.3 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0.7 & -0.3 \end{bmatrix}$$

Compute $J(d) = E_s - \mathbf{p}_d^H \mathbf{R}_{yy}^{-1} \mathbf{p}_d$ for $d = 0, 1, \ldots, 6$ and choose the delay that minimises $J$.

Typically, $d \approx N_f/2 + L/2 = 2.5 + 1 \approx 3$ gives the best performance.

Numerical computation shows:

• 5-tap MMSE: output SINR $\approx 13.8$ dB
• 11-tap MMSE: output SINR $\approx 14.6$ dB
• Matched-filter bound: SINR $= 15 + 10\log_{10}(1 + 0.49 + 0.09) = 16.98$ dB

The 5-tap equalizer is about 3.2 dB from the MFB; the 11-tap equalizer closes the gap to 2.4 dB. $\blacksquare$

The MMSE equalizer is

$$W(e^{j\omega}) = \frac{H^*(e^{j\omega})}{|H(e^{j\omega})|^2 + N_0/E_s}$$

As $N_0/E_s \to \infty$, the term $|H|^2$ becomes negligible:

$$W(e^{j\omega}) \approx \frac{H^*(e^{j\omega})}{N_0/E_s} = \frac{E_s}{N_0}\, H^*(e^{j\omega})$$

This is the matched filter $H^*(e^{j\omega})$ scaled by $E_s/N_0$.

At low SNR, ISI is less harmful than noise, so the optimal strategy is to maximise SNR (matched filtering) rather than suppress ISI. $\blacksquare$

With correct past decisions, the feedback filter subtracts $w_{b,1} \hat{a}_{k-1} + w_{b,2} \hat{a}_{k-2}$.

Ideally, $w_{b,1} = 0.8$ and $w_{b,2} = 0.3$ cancel all postcursor ISI, reducing the effective channel to $h_{\text{eff}} = [1]$.

With postcursor ISI removed, the feedforward filter only needs to maximise SNR for the single-tap effective channel. This becomes a simple matched filter, with MSE determined primarily by the noise level.

At 20 dB SNR:

• Linear MMSE: $J_{\text{lin}} \approx 0.062\, E_s$
• MMSE-DFE: $J_{\text{DFE}} \approx 0.010\, E_s$

SINR improvement: approximately $10 \log_{10}(0.062/0.010) = 7.9$ dB.

The DFE provides nearly 8 dB of improvement on this channel with strong postcursor ISI. $\blacksquare$

Define $f(\omega) = \frac{N_0/E_s}{|H(e^{j\omega})|^2 + N_0/E_s}$.

The DFE MSE is $J_{\text{DFE}} = \exp\!\left[\frac{1}{2\pi} \int_{-\pi}^{\pi} \ln f(\omega)\, d\omega\right]$.

Since $\ln$ is concave, Jensen's inequality gives

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} \ln f(\omega)\, d\omega \leq \ln\!\left[\frac{1}{2\pi} \int_{-\pi}^{\pi} f(\omega)\, d\omega\right]$$

Exponentiating both sides:

$$J_{\text{DFE}} \leq \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{N_0/E_s}{|H|^2 + N_0/E_s}\, d\omega = J_{\text{lin}}$$

Equality holds iff $f(\omega)$ is constant, i.e., $|H(e^{j\omega})|$ is constant (flat channel). $\blacksquare$

When $\hat{a}_{k-1} \neq a_{k-1}$, the feedback subtracts $h[1] \hat{a}_{k-1}$ instead of $h[1] a_{k-1}$. The residual ISI is

$$h[1](a_{k-1} - \hat{a}_{k-1}) = \pm 2 h[1]$$

This creates a bias of $\pm 2 \times 0.9 = \pm 1.8$ on the next sample.

The noise-free equalizer output for $a_k = +1$ is $+1$ (assuming the feedforward filter works correctly). With the error propagation bias of $-1.8$ (worst case), the effective signal becomes $1 - 1.8 = -0.8$, which is negative and will be detected as $-1$ (error).

More precisely, the probability of error given the previous error is

$$P(e_k | e_{k-1}) = Q\!\left(\frac{2|h[0]| - 2|h[1]|}{\sqrt{2 N_0}}\right) = Q\!\left(\frac{2(1 - 0.9)}{\sqrt{2 N_0}}\right)$$

At $E_b/N_0 = 12$ dB: $N_0 = E_b/15.85$, and with $E_b = 1$:

$$P(e_k | e_{k-1}) = Q\!\left(\frac{0.2}{\sqrt{0.126}}\right) = Q(0.563) \approx 0.287$$

There is a 28.7% chance that each error propagates to the next symbol. $\blacksquare$

States: $S_0$ ($a_{k-1} = +1$), $S_1$ ($a_{k-1} = -1$).

Transitions from $S_0$:

• $a_k = +1 \to S_0$: output $= 1 + 0.5 = 1.5$
• $a_k = -1 \to S_1$: output $= -1 + 0.5 = -0.5$

Transitions from $S_1$:

• $a_k = +1 \to S_0$: output $= 1 - 0.5 = 0.5$
• $a_k = -1 \to S_1$: output $= -1 - 0.5 = -1.5$

The minimum-weight error event is a single symbol error $\mathbf{e} = [\ldots, 0, 2, 0, \ldots]$ (difference between $+1$ and $-1$).

$$\|\mathbf{H}\mathbf{e}\|^2 = |2 h[0]|^2 + |2 h[1]|^2 = 4(1 + 0.25) = 5.0$$

So $d_{\text{free}}^2 = 5.0$.

MLSE BER $\approx Q(\sqrt{d_{\text{free}}^2 / (2 N_0)}) = Q(\sqrt{5 E_b / (2 N_0) \cdot 2 / 1.25})$

MFB BER $= Q(\sqrt{2 E_b (1 + 0.25) / N_0}) = Q(\sqrt{2.5 E_b / N_0})$

The MLSE BER is close to the MFB because it effectively collects multipath energy coherently. $\blacksquare$

(a) Trellis states: $M^L = 4^3 = 64$.

(b) Each state has $M = 4$ incoming transitions, so total branch metrics per step: $M^{L+1} = 4^4 = 256$.

(c) Computations per second: $256 \times 10 \times 10^6 = 2.56 \times 10^9$ FLOPS $= 2.56$ GFLOPS.

This is substantial for a real-time mobile receiver.

(d) A reduced-state sequence estimation (RSSE) approach groups multiple states together. For example, reducing to $M^2 = 16$ states by only tracking the last 2 symbols:

Branch metrics per step: $16 \times 4 = 64$.

Computational load: $64 \times 10 \times 10^6 = 0.64$ GFLOPS.

This is a 4x reduction with only modest performance loss. $\blacksquare$

The ML criterion minimises

$$\Lambda(\mathbf{a}) = \sum_{k=0}^{K-1} \lambda_k(a_k, \ldots, a_{k-L})$$

where $\lambda_k$ depends only on the state $(a_{k-1}, \ldots, a_{k-L})$ and the input $a_k$.

This metric decomposes additively along the trellis path.

Consider two paths arriving at state $\sigma_k$ at time $k$: path $P_1$ with metric $\Lambda_1$ and path $P_2$ with $\Lambda_2 < \Lambda_1$.

Any extension of $P_1$ by future symbols $a_{k+1}, \ldots, a_{K-1}$ achieves total metric $\Lambda_1 + \Lambda_{\text{future}}$.

The same extension applied to $P_2$ achieves $\Lambda_2 + \Lambda_{\text{future}} < \Lambda_1 + \Lambda_{\text{future}}$.

Therefore, $P_1$ can never be part of the globally optimal path. Discarding $P_1$ (keeping only the survivor $P_2$) does not eliminate the optimal sequence.

By induction over $k$, at each time step we need retain only one survivor per state. The Viterbi algorithm does exactly this, and by the principle of optimality, the final traceback yields the ML sequence. $\blacksquare$

$\mu_{\max} < \frac{2}{N_f \sigma_y^2} = \frac{2}{11 \times 2.0} = 0.0909$$

$\mu_{\text{practical}} = \frac{1}{5 N_f \sigma_y^2} = \frac{1}{5 \times 11 \times 2.0} = 0.00909$$This is about 10balance between convergence speed and stability.$\blacksquare$

Define the weight error $\boldsymbol{\epsilon}_k = \mathbf{w}_k - \mathbf{w}_{\text{opt}}$.

The mean weight error evolution is

$$E[\boldsymbol{\epsilon}_{k+1}] = (\mathbf{I} - 2\mu \mathbf{R}_{yy}) E[\boldsymbol{\epsilon}_k]$$

Let $\mathbf{R}_{yy} = \mathbf{Q} \boldsymbol{\Lambda} \mathbf{Q}^H$. In the eigenbasis:

$$\tilde{\epsilon}_{k,i} = (1 - 2\mu\lambda_i)^k \tilde{\epsilon}_{0,i}$$

where $\tilde{\boldsymbol{\epsilon}}_k = \mathbf{Q}^H \boldsymbol{\epsilon}_k$.

The $i$-th mode decays exponentially with time constant

$$\tau_i = \frac{-1}{\ln(1 - 2\mu\lambda_i)} \approx \frac{1}{2\mu\lambda_i}$$

for small $\mu$. The slowest mode has the smallest eigenvalue:

$$\tau_{\text{slow}} = \frac{1}{2\mu\lambda_{\min}}$$

A large eigenvalue spread $\chi = \lambda_{\max}/\lambda_{\min}$ makes the slowest mode converge $\chi$ times slower than the fastest. $\blacksquare$

$\tau_{\text{slow}} = \frac{1}{2 \times 0.01 \times 0.1} = 500$$For$e^{-3}$convergence:$k = 3 \times 500 = 1500$ iterations.

RLS convergence is approximately $2 N_f = 30$ iterations, independent of the eigenvalue spread.

The RLS converges 50 times faster than LMS.

• LMS: $\sim 2 N_f = 30$ complex MACs per iteration
• RLS: $\sim 4 N_f^2 = 900$ complex MACs per iteration

RLS is 30x more expensive per iteration, but converges in 30 iterations vs. 1500.

Total cost to convergence:

• LMS: $30 \times 1500 = 45{,}000$ MACs
• RLS: $900 \times 30 = 27{,}000$ MACs

In this scenario, RLS is actually more efficient in total computation to convergence. $\blacksquare$

In the $i$-th eigenmode, the steady-state weight error variance is

$$E[|\tilde{\epsilon}_{i}|^2] \approx \frac{\mu\, J_{\min}}{2(1 - \mu\lambda_i) / \lambda_i} \approx \frac{\mu \lambda_i J_{\min}}{2}$$

for small $\mu$. The contribution to excess MSE is $\Delta J_i = \lambda_i E[|\tilde{\epsilon}_i|^2] \approx \frac{\mu \lambda_i^2 J_{\min}}{2}$.

Summing over all modes is involved; the standard approximation yields $J_{\text{excess}} \approx \frac{\mu}{2} \text{tr}(\mathbf{R}_{yy}) \cdot J_{\min}$.

$\mathcal{M} = \frac{J_{\text{excess}}}{J_{\min}} \approx \frac{\mu}{2} \text{tr}(\mathbf{R}_{yy}) \approx \frac{\mu N_f \bar{\lambda}}{2}$$where$\bar{\lambda}$ is the average eigenvalue.

Convergence speed: $\tau \propto 1/\mu$ (faster for larger $\mu$).

Misadjustment: $\mathcal{M} \propto \mu$ (larger for larger $\mu$).

Therefore $\mathcal{M} \cdot \tau \approx \text{const.}$

This is the fundamental LMS trade-off: you cannot simultaneously have fast convergence and low misadjustment. $\blacksquare$

Number of states: $M^L = 16^5 = 1{,}048{,}576 \approx 10^6$.

Branch metrics per symbol: $M^{L+1} = 16^6 \approx 1.7 \times 10^7$.

At 2 Msymbol/s: $3.4 \times 10^{13}$ operations/s.

MLSE is completely infeasible for this scenario.

Propose an MMSE-DFE:

• Feedforward: $N_f = 2L + 1 = 11$ taps
• Feedback: $N_b = L = 5$ taps

Complexity per symbol: $O(N_f + N_b) = O(16)$ MACs --- easily feasible at 2 Msymbol/s.

Coherence time: 5 ms $= 10{,}000$ symbols.

The equalizer must reconverge within a fraction of this.

LMS: With $N_f = 11$ and eigenvalue spread $\chi \approx 15$, convergence takes $\sim 3/(2\mu\lambda_{\min}) \approx 500$--$1000$ symbols. This fits within the coherence interval.

Training budget: 10% of symbols $= 1000$ symbols per coherence interval. An LMS training phase of 500 symbols is sufficient.

Choice: LMS is adequate and has $O(N_f)$ complexity. RLS would converge faster but $O(N_f^2) = O(121)$ per symbol is unnecessary given the margin.

The MMSE-DFE on a 6-tap channel typically operates 2--4 dB from the MFB for 16-QAM, depending on the specific channel realisation. With proper training and decision-directed tracking, the expected gap is approximately 3 dB.

Alternative: Use OFDM with per-subcarrier equalization to avoid time-domain equalization entirely. $\blacksquare$