22-Elec-A3 Signals and Communications · December 2013

Question 6 of 6: Transmitter Power Budget for AM and DSB

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

Question 6: Transmitter Power Budget for AM and DSB (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.

Find. The message power \(S_x\); the largest achievable power per sideband under DSB modulation subject to both rating limits; and sketches of the modulated carrier for AM at \(\mu = 0.5\) and \(\mu = 1\) and for DSB, with envelopes shown.

Approach. Compute the mean-square value of the tone; express both the average transmitted power and the peak envelope power of a DSB signal in terms of the carrier amplitude \(A_c\); see which of the two ratings binds first; then halve the resulting transmitted power between the two sidebands.

1. Message power (part a). For a sinusoidal message \(x(t) = A_m\cos\omega_m t\), the power is the mean square: $$S_x = \overline{x^2(t)} = \frac{A_m^2}{2} = \frac{1^2}{2}$$ $$\boxed{\,S_x = 0.5\,}$$
2. Express DSB transmitted power in terms of \(A_c\). The DSB (suppressed-carrier) signal is \(x_c(t) = A_c\,x(t)\cos\omega_c t\). Averaging, and using \(\overline{\cos^2\omega_c t} = \tfrac12\) with \(\omega_c \gg \omega_m\): $$S_T = \overline{A_c^2 x^2(t)\cos^2\omega_c t} = \frac{A_c^2 S_x}{2} = \frac{A_c^2}{4}$$
3. Express the peak envelope power in terms of \(A_c\). The envelope of the DSB signal is \(A_c|x(t)|\), whose maximum is \(A_{max} = A_c A_m = A_c\). The peak-envelope-power rating therefore reads $$A_{max}^2 = A_c^2 \le 8000\ \text{W} \;\Longrightarrow\; A_c^2 \le 8000$$
4. Determine which rating binds. Substituting the peak-power ceiling into the average-power expression: $$S_T\big|_{\text{PEP limit}} = \frac{A_c^2}{4} \le \frac{8000}{4} = 2000\ \text{W}$$ Since 2000 W is below the 3000 W average-power rating, the peak envelope power is the binding constraint and the transmitter cannot be driven to its full average rating on this message: $$\boxed{\,S_{T,\max} = \min(3000,\ 2000) = 2\ \text{kW}\,}$$
5. Split between the sidebands (part b). A DSB signal contains no carrier component; all transmitted power resides in the two sidebands, and by symmetry they share it equally: $$P_{sb} = \frac{S_{T,\max}}{2} = \frac{2000}{2}$$ $$\boxed{\,P_{sb} = 1\ \text{kW per sideband}\,}$$
6. Construct the part (c) envelopes. For AM the transmitted signal is \(x_c(t) = A_c[1+\mu x(t)]\cos\omega_c t\), so a \(\pm 1\) square-wave message makes the envelope alternate between the two levels \(A_c(1+\mu)\) and \(A_c(1-\mu)\):
   • i. \(\mu = 0.5\): envelope alternates between \(1.5A_c\) and \(0.5A_c\). It never reaches zero, so the envelope still traces the message and an envelope detector recovers it correctly (no overmodulation).
   • ii. \(\mu = 1\): envelope alternates between \(2A_c\) and \(0\). This is the critical modulation case: the carrier is fully switched off during the negative half of the message, the limit of distortionless envelope detection.
   • iii. DSB: \(x_c(t) = A_c x(t)\cos\omega_c t\), so the envelope \(A_c|x(t)| = A_c\) is constant. The message information is carried entirely in a \(180^\circ\) phase reversal of the carrier at each square-wave transition, which an envelope detector cannot see.

The three sketches make the essential trade-off visible. AM spends power on a carrier term that conveys no information but leaves the envelope proportional to the message, allowing a diode envelope detector; DSB spends nothing on the carrier and is therefore more power-efficient, but the constant envelope with phase reversals forces the receiver to use coherent detection with a locally regenerated carrier.