FE Electrical Communications Practice Problems

Answer: B) 5 kHz

The Handbook's Single-Sideband Modulation on p. 380 keeps either the lower or the upper sideband of the DSB spectrum, and states that if \(M(f) = 0\) for \(|f| > W\) the bandwidth of \(x_{LSB}(t)\) or \(x_{USB}(t)\) is \(W\):

$$B = W = 5\text{ kHz}$$

The two sidebands of a DSB or AM signal are mirror images carrying the same information, so one of them can be filtered off, or cancelled by a phasing modulator, with no loss; the result occupies half the spectrum and, with the carrier also absent, needs a fraction of the transmitter power for the same received message. That is why SSB is the standard for HF voice and for the channel stacking of frequency-division multiplexing, where 4 kHz voice channels are packed side by side at 4 kHz spacing. The receiver must reinsert a carrier of the right frequency, a small error shifting every message component by the same amount, which is tolerable for speech and not for music or data; the Handbook mentions synchronous demodulation or carrier reinsertion followed by envelope detection.

10 kHz is the double-sideband bandwidth, \(2W\).

2.5 kHz halves the single sideband as well.

20 kHz is four times the message bandwidth.

Answer: D) \(\dfrac{1}{a + j2\pi f}\)

The pair is in the Handbook's Fourier Transform Pairs table on p. 57, \(e^{-at}u(t) \leftrightarrow 1/(a + j2\pi f)\), and it follows from the definition on p. 377:

$$X(f) = \int_0^{\infty} e^{-at}e^{-j2\pi f t}\,dt = \int_0^{\infty} e^{-(a + j2\pi f)t}\,dt = \left[\dfrac{-e^{-(a + j2\pi f)t}}{a + j2\pi f}\right]_0^{\infty} = \dfrac{1}{a + j2\pi f}$$

the upper limit vanishing because \(a > 0\) makes \(e^{-at}\) decay. The magnitude \(1/\sqrt{a^2 + (2\pi f)^2}\) is the response of a first-order low-pass filter with corner frequency \(a/2\pi\) Hz, which is exactly what this signal is: the impulse response of an \(RC\) section with \(a = 1/RC\), and the Handbook's statement on p. 378 that impulse response and frequency response form a transform pair is the link. The phase is \(-\arctan(2\pi f/a)\), from 0 at dc toward \(-90^\circ\). The same integral with \(s\) in place of \(j2\pi f\) is the Laplace pair \(1/(s + a)\) on p. 58, the Fourier transform being the Laplace transform on the imaginary axis when the signal is absolutely integrable.

\(\dfrac{1}{a - j2\pi f}\) is the transform of \(e^{at}u(-t)\), the time-reversed signal.

\(\dfrac{2a}{a^2 + (2\pi f)^2}\) is the pair for the two-sided \(e^{-a|t|}\), the sum of the two one-sided transforms.

\(\dfrac{j2\pi f}{a + j2\pi f}\) is the transform of the derivative of \(x(t)\), minus its impulse at the origin.

Answer: C) 8 mJ

By Parseval's theorem, total energy is the area under the energy spectral density \(|X(f)|^2\):

$$E = \int_{-\infty}^{\infty} |X(f)|^2\, df$$

The plot is a triangle with peak height \(4\) (mJ/kHz) and base from \(-2\) kHz to \(+2\) kHz, so the base width is \(4\) kHz.

$$E = \tfrac{1}{2}\,(\text{base})(\text{height}) = \tfrac{1}{2}\,(4\ \text{kHz})(4\ \text{mJ/kHz}) = 8\ \text{mJ}$$

Answer: A) \(r = [+1, -1, -1, -3]\); User 3 data = +1

Step 1: Compute each user's transmitted signal:

$$d_1 \cdot c_1 = (-1)[+1, +1, +1, +1] = [-1, -1, -1, -1]$$

$$d_2 \cdot c_2 = (+1)[+1, -1, +1, -1] = [+1, -1, +1, -1]$$

$$d_3 \cdot c_3 = (+1)[+1, +1, -1, -1] = [+1, +1, -1, -1]$$

Step 2: Composite signal:

$$r = [-1+1+1, -1-1+1, -1+1-1, -1-1-1] = [+1, -1, -1, -3]$$

Step 3: Correlate with User 3's code:

$$r \cdot c_3 = (+1)(+1) + (-1)(+1) + (-1)(-1) + (-3)(-1)$$

$$= 1 - 1 + 1 + 3 = 4$$

Step 4: Normalize: \(\hat{d}_3 = \frac{4}{4} = +1\)

The receiver correctly extracts \(d_3 = +1\).

Answer: D) 144 Mbps

The Handbook's Shannon Channel Capacity Formula on p. 384:

$$C = BW\log_2\left(1 + \dfrac{S}{N}\right) = 36 \times 10^6 \times \log_2 16 = 36 \times 10^6 \times 4 = 144\text{ Mbps}$$

A ratio of 15 is 11.8 dB, a typical satellite link budget, and it supports 4 bits per second per hertz; a real transponder carrying quadrature PSK with rate-3/4 coding delivers about 1.5 bits per second per hertz, 54 Mbps, well inside the bound, the gap being the margin for rain fade, imperfect filtering and finite coding. The 36 MHz transponder width is the standard of the C and Ku bands, and the capacity scales with it directly, so a 72 MHz transponder at the same ratio would double the figure. The logarithm must be base 2; with the tables of a calculator, \(\log_2 16 = \log_{10}16/\log_{10}2 = 1.204/0.301 = 4\). The \(1 +\) inside the logarithm matters when the ratio is small, and at a ratio of 15 leaving it out changes the answer by 2%, not enough to move between the options unless they are close, which is why one of them is.

141 Mbps drops the \(1 +\), using \(\log_2 15 = 3.91\).

43.3 Mbps uses the base-10 logarithm, \(36 \times \log_{10}16\).

36 Mbps is the bandwidth itself, the logarithm taken as 1.