22-Elec-A3 Signals and Communications · May 2016

Question 2 of 6: Quadrature Modulation, Coherent Demodulation and Envelope Detection

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Quadrature Modulation, Coherent Demodulation and Envelope Detection (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A quadrature-multiplexed signal carrying an AM-with-carrier channel on the in-phase axis and a suppressed-carrier channel on the quadrature axis: in-phase envelope $A(1+m_1(t))$ with $A\gg 1$ and $|m_1|\le 1$; quadrature term $m_2(t)$ with $|m_2|\le 1$; both messages band-limited to $W\ll f_c$ and statistically independent.

Find. (a) a receiver that recovers both messages exactly, (b) the ideal envelope-detector output and its audible character, (c) the signal-to-distortion ratio for $m_1$ and (d) whether any $m_2$ component survives envelope detection.

Approach. Part (a) is a coherent quadrature (I/Q) receiver — multiply by $2\cos$ and $2\sin$ and low-pass filter. Parts (b)–(d) follow from writing $x(t)$ in envelope-and-phase form and expanding the square root for $A\gg 1$.

1. Recognise the structure of $x(t)$. The signal is the sum of two carriers in phase quadrature, so it may be written as $$x(t)=a(t)\cos(2\pi f_c t)+b(t)\sin(2\pi f_c t),\qquad a(t)=A(1+m_1(t)),\quad b(t)=m_2(t).$$ Because $a$ and $b$ occupy the same band around $f_c$, only a phase-sensitive (coherent) receiver can separate them.
2. Coherent detection of the in-phase channel. Multiplying by $2\cos(2\pi f_ct)$ and using $2\cos^2\theta=1+\cos 2\theta$, $2\sin\theta\cos\theta=\sin 2\theta$, $$2x(t)\cos(2\pi f_c t)=a(t)+\underbrace{a(t)\cos(4\pi f_ct)+b(t)\sin(4\pi f_ct)}_{\text{centred at }2f_c}.$$ Since $W\ll f_c$, a low-pass filter of cut-off $W$ removes everything at $2f_c$ and leaves $a(t)=A(1+m_1(t))$. Blocking the DC term $A$ and scaling by $1/A$ gives $$\boxed{\;\hat m_1(t)=m_1(t)\ \text{exactly}\;}$$
3. Coherent detection of the quadrature channel. Multiplying instead by $2\sin(2\pi f_c t)$ and low-pass filtering leaves the quadrature amplitude directly, $$\boxed{\;\hat m_2(t)=b(t)=m_2(t)\ \text{exactly}\;}$$ with no DC term to remove. The block diagram above shows the complete receiver: a power split, two multipliers driven by phase-locked quadrature carriers, two low-pass filters, and a DC block plus $1/A$ scaling on the $m_1$ arm. The carrier itself is available for locking because the in-phase channel transmits a large residual carrier of amplitude $A$ — a phase-locked loop tracking that line supplies both references, which is the practical reason the designer left the carrier in.
4. Form the envelope. An ideal envelope detector outputs the magnitude of the complex envelope, which is unaffected by the carrier phase: $$e(t)=\sqrt{a^{2}(t)+b^{2}(t)}=\sqrt{A^{2}\left(1+m_1(t)\right)^{2}+m_2^{2}(t)}.$$ Note that this is a genuinely non-linear operation — it mixes the two channels, which is precisely what the coherent receiver of parts (a) avoids.
5. Expand for $A\gg 1$. Factoring out the dominant in-phase term and using $\sqrt{1+\varepsilon}\approx 1+\varepsilon/2$ for small $\varepsilon$, $$e(t)=A\left(1+m_1\right)\sqrt{1+\frac{m_2^{2}}{A^{2}(1+m_1)^{2}}} \approx A\left(1+m_1(t)\right)+\frac{m_2^{2}(t)}{2A\left(1+m_1(t)\right)}.$$ With $|m_1|\le 1$ the leading term never changes sign (no over-modulation), and since $A\gg 1$ the second term is $O(1/A)$. Keeping only first order, $$\boxed{\;e(t)\approx A+A\,m_1(t)+\frac{m_2^{2}(t)}{2A}\;}$$ so the detector delivers a DC pedestal $A$, a clean replica of $m_1$ scaled by $A$, and a small additive term proportional to $m_2^{2}$.
6. Interpret the sound. After the usual DC block, a listener hears $m_1(t)$ essentially undistorted, plus a faint background contributed by $m_2^{2}(t)$. Squaring an audio waveform is not a linear distortion: it doubles and sums the frequencies present in $m_2$ (a tone at $f$ becomes DC plus a tone at $2f$), so the second channel is not intelligible — it is heard as a quiet, harsh, rectified-sounding rasp or rumble modulating in step with the loudness of the second programme, not as crosstalk speech. That is the answer the examiner is after: channel 1 clearly, channel 2 as low-level unintelligible distortion.
7. Signal-to-distortion ratio for $m_1$ (part c). Comparing the wanted term $A\,m_1$ with the distortion term $m_2^{2}/2A$ at their worst-case amplitudes ($|m_1|_{\max}=|m_2|_{\max}=1$), the amplitude ratio is $$\frac{|A\,m_1|}{\left|m_2^{2}/2A\right|}=\frac{A}{1/(2A)}=2A^{2},$$ so as a power (mean-square) ratio $$\boxed{\;\mathrm{SDR}_{m_1}\approx\left(2A^{2}\right)^{2}=4A^{4}\;}$$ For a representative $A=10$ this is $4\times 10^{4}$, i.e. 46.0 dB — large, as it must be, because $A\gg 1$ is exactly the condition that makes envelope detection viable here. The ratio improves as the fourth power of $A$: each doubling of the carrier constant buys 12 dB.
8. Is there an $m_2$ component (part d)? Inspecting the expansion, $m_2$ appears only as $m_2^{2}$; there is no term linear in $m_2$ at any order, because the envelope depends on the quadrature component through $b^{2}$ alone. Formally, the correlation of the detector output with $m_2$ is zero for a zero-mean message. Hence $$\boxed{\;\text{no linear }m_2\ \text{component: } \mathrm{SDR}_{m_2}=0\ (-\infty\ \text{dB})\;}$$ Everything the envelope detector produces from the second channel is distortion. Recovering $m_2$ requires the coherent quadrature branch of part (a) — which is the engineering point of the whole question.