Exercises

All PSK and rectangular QAM signals lie in the span of $\{\cos(2\pi f_c t),\, \sin(2\pi f_c t)\}$, so $N = 2$.

Exception: BPSK uses only $\cos(2\pi f_c t)$, so $N = 1$.

(a) BPSK: $N = 1$. (b) QPSK: $N = 2$. (c) 8-PSK: $N = 2$. (d) 16-QAM: $N = 2$.

Each of the $M$ tones is a distinct basis function:

(e) 4-FSK: $N = 4$. (f) 8-FSK: $N = 8$.

In general, orthogonal $M$-FSK has $N = M$. $\blacksquare$

$\|s_1\| = A\sqrt{T}$, so $\varphi_1(t) = 1/\sqrt{T}$.

$\langle s_2, \varphi_1 \rangle = A \int_0^T \cos(2\pi t/T) / \sqrt{T}\, dt = 0$.

So $u_2(t) = s_2(t) - 0 = A\cos(2\pi t/T)$.

$\|u_2\|^2 = A^2 \int_0^T \cos^2(2\pi t/T)\,dt = A^2 T/2$.

$\varphi_2(t) = \sqrt{2/T}\, \cos(2\pi t/T)$.

$\mathbf{s}_1 = [A\sqrt{T},\; 0]^T$

$\mathbf{s}_2 = [0,\; A\sqrt{T/2}]^T$

$\mathbf{s}_3 = \mathbf{s}_1 + \mathbf{s}_2 = [A\sqrt{T},\; A\sqrt{T/2}]^T$

Signal space dimension is $N = 2$ (not 3, since $s_3 = s_1 + s_2$). $\blacksquare$

Choose $s_1$ if $\|r - s_1\|^2 < \|r - s_2\|^2$, which simplifies to: choose $s_1$ if $r > 0$.

The boundary is $r = 0$ (the perpendicular bisector of the segment from $-\sqrt{E_b}$ to $+\sqrt{E_b}$).

$P_e = Q\!\left(\sqrt{2E_b/N_0}\right) = Q(\sqrt{20}) = Q(4.47) \approx 3.9 \times 10^{-6}$.

The MAP boundary shifts to

$r_0 = \frac{N_0}{4\sqrt{E_b}}\ln\frac{P(s_1)}{P(s_2)} = \frac{N_0}{4\sqrt{E_b}}\ln\frac{0.7}{0.3}$

$= \frac{N_0}{4\sqrt{E_b}} \times 0.847$

The boundary moves toward $s_2$ (less likely signal), slightly reducing overall error probability. $\blacksquare$

$\mathcal{R}_1 = (-\infty, -2d]$, $\mathcal{R}_2 = (-2d, 0]$, $\mathcal{R}_3 = (0, 2d]$, $\mathcal{R}_4 = (2d, \infty)$.

Outer points ($\pm 3d$): error if noise exceeds $d$ in the inward direction. $P_e(\text{outer}) = Q(d/\sqrt{N_0/2}) = Q(\sqrt{2d^2/N_0})$.

Inner points ($\pm d$): error if noise exceeds $d$ in either direction. $P_e(\text{inner}) = 2Q(\sqrt{2d^2/N_0})$.

Average: $P_s = \frac{1}{4}[Q + 2Q + 2Q + Q] = \frac{3}{2}Q(\sqrt{2d^2/N_0})$.

$E_s = 5d^2$, $E_b = E_s/\log_2 4 = 5d^2/2$. So $d^2 = 2E_b/5$.

$P_s = \frac{3}{2}Q\!\left(\sqrt{\frac{4E_b}{5N_0}}\right)$. $\blacksquare$

$R_b = R_s \log_2 M = 1 \times 10^6 \times 2 = 2$ Mbps.

$E_b = E_s / \log_2 M = 2/2 = 1$ mJ.

$P = E_s \cdot R_s = 2 \times 10^{-3} \times 10^6 = 2$ kW. $\blacksquare$

$E_s = 2(16-1)d^2/3 = 10d^2$, so $d = \sqrt{E_s/10}$. $d_{\min} = 2d = 2\sqrt{E_s/10} = \sqrt{2E_s/5}$.

$E_b = E_s/\log_2 16 = E_s/4$, so $E_s = 4E_b$. $d_{\min} = \sqrt{8E_b/5} = \sqrt{1.6\, E_b}$.

QPSK: $d_{\min} = \sqrt{2E_s} = \sqrt{2 \times 2E_b} = 2\sqrt{E_b}$.

16-QAM: $d_{\min} = \sqrt{1.6\, E_b} \approx 1.265\sqrt{E_b}$.

QPSK has $2/1.265 = 1.58$ times larger $d_{\min}$ at the same $E_b$, or equivalently $20\log_{10}(1.58) = 4.0$ dB advantage. This is the power penalty for doubling spectral efficiency from 2 to 4 bits/s/Hz. $\blacksquare$

One valid Gray mapping:

Adjacent pairs and their Hamming distance: 000-001: 1, 001-011: 1, 011-010: 1, 010-110: 1, 110-111: 1, 111-101: 1, 101-100: 1, 100-000: 1. All adjacent pairs differ by exactly one bit. Valid.

$\text{BER} \approx \text{SER}/\log_2 8 = \text{SER}/3$. Therefore $\text{SER} \approx 3 \times 10^{-5}$. $\blacksquare$

Square $M$-QAM $= \sqrt{M}$-PAM (I) $\times$ $\sqrt{M}$-PAM (Q). The I and Q noise components are independent.

$P(\text{correct}) = P(\text{I correct}) \times P(\text{Q correct}) = (1 - P_{\sqrt{M}})^2$.

$P_s = 1 - (1 - P_{\sqrt{M}})^2$.

For $\sqrt{M}$-PAM with minimum distance $2d$ and $\sqrt{M}$ levels:

$P_{\sqrt{M}} = \frac{2(\sqrt{M}-1)}{\sqrt{M}} Q\!\left(\sqrt{\frac{2d^2}{N_0}}\right)$

With $E_s = 2(M-1)d^2/3$: $d^2 = 3E_s/(2(M-1))$.

$P_{\sqrt{M}} = \frac{2(\sqrt{M}-1)}{\sqrt{M}} Q\!\left(\sqrt{\frac{3E_s}{(M-1)N_0}}\right)$

For $M = 4$: $\sqrt{M} = 2$, $P_2 = Q(\sqrt{2E_s/N_0})$.

$P_s = 1 - (1 - Q(\sqrt{2E_s/N_0}))^2 = 2Q(\sqrt{2E_s/N_0}) - Q^2(\sqrt{2E_s/N_0}) \approx 2Q(\sqrt{2E_s/N_0})$ at high SNR.

With $E_s = 2E_b$: $P_s \approx 2Q(\sqrt{4E_b/N_0})$, and $P_b \approx Q(\sqrt{2E_b/N_0})$ (BPSK-equivalent). $\blacksquare$

$\Delta f = R_s = 100$ kHz. $W \approx 4 \times 100 = 400$ kHz.

$R_b = R_s \log_2 4 = 200$ kbps. $\eta = R_b/W = 200/400 = 0.5$ bits/s/Hz.

QPSK: $W \approx R_s = 100$ kHz (Nyquist). $\eta_{\text{QPSK}} = 200/100 = 2$ bits/s/Hz.

QPSK is 4 times more bandwidth-efficient, but 4-FSK can operate at lower $E_b/N_0$ (better power efficiency). $\blacksquare$

$\phi(0) = 0$ $\phi(T_b) = 0 + (+1)\pi/2 = \pi/2$ $\phi(2T_b) = \pi/2 + (-1)\pi/2 = 0$ $\phi(3T_b) = 0 + (+1)\pi/2 = \pi/2$ $\phi(4T_b) = \pi/2 + (+1)\pi/2 = \pi$ $\phi(5T_b) = \pi - \pi/2 = \pi/2$

Starting from $\phi(0) = 0$ (a multiple of $\pi/2$), each phase increment is $\pm\pi/2$. The sum of any number of $\pm\pi/2$ terms is always a multiple of $\pi/2$.

In general, the phase at bit boundary $k$ is $\phi(kT_b) = \frac{\pi}{2}\sum_{i=0}^{k-1} b_i \pmod{2\pi}$.

$f_{\text{inst}} = f_c + b_k/(4T_b)$:

Bit 1 ($b = +1$): $f_c + 1/(4T_b)$ Bit 2 ($b = -1$): $f_c - 1/(4T_b)$ Bit 3 ($b = +1$): $f_c + 1/(4T_b)$ Bit 4 ($b = +1$): $f_c + 1/(4T_b)$ Bit 5 ($b = -1$): $f_c - 1/(4T_b)$

The frequency toggles between two values separated by $\Delta f = 1/(2T_b)$. $\blacksquare$

$W_{\text{MSK}} \approx 1.18 R_b = 1.18$ MHz.

$W_{\text{GMSK}} \approx 0.86 R_b = 0.86$ MHz.

Saving: $(1.18 - 0.86)/1.18 = 27$%.

GMSK with $BT = 0.3$ introduces controlled ISI. For an MLSE receiver, the degradation relative to MSK is about 0.5 dB. For a simple discriminator detector, it is about 1-2 dB.

There is no irreducible BER floor from the Gaussian filtering itself (the ISI is deterministic and can be equalised), but practical implementations with simple detectors may exhibit a floor at very high SNR. $\blacksquare$

(a) $W = 1.0 \times 20/2 = 10$ MHz. Excess: 0 MHz. (b) $W = 1.25 \times 20/2 = 12.5$ MHz. Excess: 2.5 MHz. (c) $W = 1.5 \times 20/2 = 15$ MHz. Excess: 5.0 MHz. (d) $W = 2.0 \times 20/2 = 20$ MHz. Excess: 10 MHz.

The excess bandwidth $\alpha R_s/2$ is the price paid for the roll-off transition band, which provides faster time-domain tail decay and robustness to timing errors. $\blacksquare$

With $\alpha = 0.5$ and $R_s = 1/T_s$:

• Flat region: $|f| \leq (1-0.5)/(2T_s) = 0.25/T_s$
• Transition: $0.25/T_s < |f| \leq 0.75/T_s$
• Zero: $|f| > 0.75/T_s$

Adjacent copies $P(f)$ and $P(f - 1/T_s)$ overlap in the range $0.25/T_s < f < 0.75/T_s$.

In the flat region: only one copy contributes $P(f) = T_s$. Sum $= T_s$.

In the transition band at frequency $f$, the two contributions are: $P(f) = \frac{T_s}{2}[1 + \cos(\pi T_s(f - 0.25/T_s)/0.5)]$ $P(f - 1/T_s) = \frac{T_s}{2}[1 + \cos(\pi T_s(1/T_s - f - 0.25/T_s)/0.5)]$

Since the cosine argument of the second term is $\pi$ minus the first, $\cos(\theta) + \cos(\pi - \theta) = 0$, giving:

$P(f) + P(f - 1/T_s) = T_s/2 + T_s/2 = T_s$.

The Nyquist criterion is satisfied. $\blacksquare$

$W = (1 + 0.35) \times 5/2 = 3.375$ MHz. Total occupied bandwidth: $2W = 6.75$ MHz (double-sided).

$L = 10$ symbol periods $\times$ 4 samples/symbol $= 40$ samples, plus 1 for the center tap $= 41$ taps.

The RRC pulse decays as $|p(t)| \sim 1/|t|$ for large $|t|$ (slower than the RC pulse, which decays as $1/|t|^3$).

For $L = 10$: the residual ISI power from the truncated tails is approximately $-40$ dB relative to the main lobe.

For $L = 20$: the residual is approximately $-50$ dB.

The ISI degradation at $\text{BER} = 10^{-5}$ is approximately 0.1 dB for $L = 10$ and negligible for $L = 20$. In practice, $L = 10$ to 12 is sufficient for most applications. $\blacksquare$

Ideal eye opening = $d_{\min}$ (normalised to 1.0). Actual = 0.6. The effective $d_{\min}$ is reduced by factor 0.6.

SNR penalty $= -20\log_{10}(0.6) = 4.4$ dB.

This means 4.4 dB more $E_b/N_0$ is needed to achieve the same BER as the ideal case.

Maximum timing error $= 0.7T_s / 2 = 0.35T_s$.

Beyond this, the sample falls outside the eye opening and ISI-induced errors become certain for small noise margins.

• Residual ISI from non-ideal pulse shaping or channel distortion
• Timing jitter in the clock recovery circuit
• Bandwidth limitation in the channel or filters
• Truncation of the pulse-shaping filter to insufficient length $\blacksquare$

Shannon limit at $\eta = 1$: $(2^1 - 1)/1 = 1 = 0$ dB.

Gap $= 9.6 - 0 = 9.6$ dB.

Shannon limit at $\eta = 4$: $(2^4 - 1)/4 = 15/4 = 3.75 = 5.74$ dB.

Gap $= 13.5 - 5.74 = 7.76$ dB.

Shannon limit at $\eta = 6$: $(2^6 - 1)/6 = 63/6 = 10.5 = 10.21$ dB.

Gap $= 17.8 - 10.21 = 7.59$ dB.

The Shannon gap for uncoded modulations is 7.5-9.6 dB. Modern coding (LDPC, turbo) reduces this to 1-2 dB. $\blacksquare$

$\eta = R_b/W = 100/20 = 5$ bits/s/Hz.

$R_s = 2W/(1+\alpha) = 40/1.15 = 34.78$ Msymbols/s.

$\log_2 M = R_b/R_s = 100/34.78 = 2.88$.

Since $\log_2 M$ must be an integer for standard constellations, we need $\log_2 M = 3$ (8-PSK) or $\log_2 M = 4$ (16-QAM).

With 8-PSK: achievable rate $= 34.78 \times 3 = 104.3$ Mbps. Sufficient. With 16-QAM: achievable rate $= 34.78 \times 4 = 139.1$ Mbps. Also sufficient.

At BER $= 10^{-6}$:

• 8-PSK requires $\approx 14$ dB: 18 dB available. Sufficient (4 dB margin).
• 16-QAM requires $\approx 14.5$ dB: 18 dB available. Sufficient (3.5 dB margin).

Both 8-PSK and 16-QAM are feasible. 16-QAM provides more margin for implementation loss and would be the preferred choice.

At 18 dB $E_b/N_0$, we could even consider 64-QAM (requires $\approx 18.5$ dB at BER $= 10^{-6}$), which is marginal.

With coding, 64-QAM with rate-3/4 LDPC would require only about 14 dB and deliver $R_b = 34.78 \times 6 \times 0.75 = 156.5$ Mbps. $\blacksquare$

Points at radius $\sqrt{E_s}$, angular separation $2\pi/8 = \pi/4$.

$d_{\min} = 2\sqrt{E_s}\sin(\pi/8) = 2\sqrt{E_s} \times 0.3827 = 0.765\sqrt{E_s}$.

Grid points: $(\pm d, \pm d/2)$ and $(\pm 3d, \pm d/2)$ (scaled for symmetry).

Alternatively, a $2 \times 4$ grid with spacing $2d$: $\{(\pm d, \pm d), (\pm d, \pm 3d)\}$.

$E_s = \frac{1}{8}[4(d^2 + d^2) + 4(d^2 + 9d^2)] = \frac{1}{8}(8d^2 + 40d^2) = 6d^2$.

$d_{\min} = 2d = 2\sqrt{E_s/6} = 0.816\sqrt{E_s}$.

Ratio: $0.816/0.765 = 1.067$, a 0.56 dB improvement.

The optimal 8-point constellation arranges points in two concentric rings: an inner ring of 4 points and an outer ring of 4 points, with the outer ring rotated by $\pi/4$ relative to the inner ring. This approximates hexagonal packing, which maximises $d_{\min}/\sqrt{E_s}$ in two dimensions.

The optimal ratio is $d_{\min}/\sqrt{E_s} \approx 0.833$, giving a gain of $20\log_{10}(0.833/0.765) = 0.73$ dB over 8-PSK.

• 8-PSK: baseline
• Rectangular 8-QAM: +0.56 dB over 8-PSK
• Optimal 8-point: +0.73 dB over 8-PSK

For larger $M$, QAM increasingly outperforms PSK because the circular arrangement wastes the interior. The gap grows roughly as $1.5\log_2(M/4)$ dB for $M > 4$. $\blacksquare$