FE Electrical Communications Practice Problems

Answer: D) 15 V

The Handbook's Amplitude Modulation on p. 376 writes the signal as \(x_{AM}(t) = A_c[1 + a\,m_n(t)]\cos(2\pi f_c t)\) with the normalized message \(m_n(t)\) bounded by one in magnitude. The bracket is the envelope, and its largest value occurs when \(\cos(2\pi f_m t) = 1\):

$$x_{peak} = A_c(1 + a) = 10(1 + 0.5) = 15\text{ V}$$

The envelope swings between \(A_c(1 - a) = 5\) V and \(A_c(1 + a) = 15\) V about the unmodulated carrier amplitude of 10 V, so the modulation index can be read from an oscilloscope trace as \(a = (E_{max} - E_{min})/(E_{max} + E_{min}) = 10/20 = 0.5\). With \(a \le 1\) the envelope never crosses zero and follows the message faithfully, which is what allows the simple envelope detector the Handbook mentions; for \(a > 1\) the envelope would fold over and the detector output would be distorted. The power efficiency at \(a = 0.5\), from the Handbook's expression \(\eta = a^2\langle m_n^2\rangle/(1 + a^2\langle m_n^2\rangle)\) with \(\langle m_n^2\rangle = 1/2\) for a sinusoid, is only 11%; most of the transmitted power is in the carrier.

5 V is the minimum of the envelope, \(A_c(1 - a)\).

10 V is the unmodulated carrier amplitude.

12.5 V applies half the index, \(A_c(1 + a/2)\).

Answer: D) \(f_0 = 100\) Hz, \(\omega_0 = 200\pi\) rad/s

The Handbook's Fourier Series on p. 54 is written for a periodic function of period \(T = 2\pi/\omega_0\), and p. 374 restates the fundamental as \(f_0 = 1/T_0 = \omega_0/2\pi\):

$$f_0 = \dfrac{1}{T} = \dfrac{1}{0.01} = 100\text{ Hz}, \qquad \omega_0 = 2\pi f_0 = 200\pi\text{ rad/s} = 628\text{ rad/s}$$

One cycle every 10 ms is 100 cycles per second, and each cycle advances the phase by \(2\pi\) radians. The Fourier series of the signal contains only the harmonics \(n f_0 = 100, 200, 300, \ldots\) Hz, with angular frequencies \(n\omega_0\), whatever the waveform; the waveform decides the coefficients \(a_n\) and \(b_n\), not the frequencies. A common slip is to carry \(2\pi\) into the frequency in hertz or to leave it out of the angular frequency, which the units catch: hertz are cycles per second and radians per second are \(2\pi\) times as many. The 100 Hz fundamental is also the spacing of the spectral lines, and its reciprocal, the period, is the interval over which the Handbook's coefficient integrals are taken.

\(f_0 = 100\) Hz, \(\omega_0 = 100\pi\) rad/s has the frequency right and drops the factor 2 in \(\omega_0 = 2\pi f_0\).

\(f_0 = 50\) Hz, \(\omega_0 = 100\pi\) rad/s halves both, as if the period were 20 ms.

\(f_0 = 200\) Hz, \(\omega_0 = 400\pi\) rad/s doubles both.

Answer: B) \(0\)

The inner product is the sum of the chip-by-chip products:

$$c_1 \cdot c_2 = (+1)(+1) + (-1)(+1) + (+1)(-1) + (-1)(-1) = 1 - 1 - 1 + 1 = 0$$

The two codes are orthogonal, and that is the whole mechanism of code-division multiplexing: every user transmits at the same time in the same band, and a receiver that correlates the sum with one user's code recovers that user's data while the other users' contributions, being orthogonal, correlate to zero. The codes here are two rows of the order-4 Walsh-Hadamard matrix, and the other two rows, \([+1, +1, +1, +1]\) and \([+1, -1, -1, +1]\), are orthogonal to both, so four synchronous users can share the channel. Each code correlates with itself to its length, 4, which is the processing gain the receiver applies. The orthogonality holds only with chip alignment; codes that arrive misaligned are in general no longer orthogonal, which is why practical systems either synchronize the users or accept residual interference from codes that are merely nearly orthogonal.

\(2\) counts only the two chips where the codes agree.

\(-2\) counts only the two where they differ.

\(4\) is the code length, the correlation of a code with itself.

Answer: C) \(1.5 \times 10^{-4}\)

The bit error rate is the fraction of transmitted bits that arrive in error:

$$BER = \dfrac{150}{1{,}000{,}000} = 1.5 \times 10^{-4}$$

One and a half bits in every ten thousand are wrong. The figure is an estimate of the error probability of the channel, good to about 8% here since 150 events have a statistical spread of \(\sqrt{150} \approx 12\); a specification such as \(10^{-9}\) needs billions of bits to confirm, which is why link tests run for hours. The rate depends on the received signal-to-noise ratio and the modulation, falling steeply as the ratio rises, and the Handbook's Error Coding on p. 378 describes the two responses to it: detection with parity or a cyclic redundancy check, followed by retransmission under the ARQ protocol of p. 379, or forward correction with a block, Hamming or Reed-Solomon code that repairs errors at the receiver. A rate of \(10^{-4}\) is acceptable for digitized voice and far too high for data, where \(10^{-9}\) or better is expected after correction.

\(1.5 \times 10^{-3}\), \(1.5 \times 10^{-5}\) and \(1.5 \times 10^{-6}\) slip one or more decades, miscounting the zeros of a million.

Answer: A) 100 Hz and 200 Hz

The two components are at \(200\pi/2\pi = 100\) Hz and \(600\pi/2\pi = 300\) Hz. The Handbook's Ideal-Impulse Sampling on p. 377 gives the sampled spectrum as \(X_\delta(f) = f_s\sum_k M(f - kf_s)\), copies of the signal spectrum at every multiple of 500 Hz, and exact recovery by a low-pass filter requires \(f_s > 2W\), which would be 600 Hz here. The 100 Hz tone satisfies the condition and its copies at \(\pm 100 + 500k\) all lie outside the 250 Hz filter except the original. The 300 Hz tone does not: its copy from the \(k = -1\) term falls at

$$|300 - 500| = 200\text{ Hz}$$

inside the filter passband, while the original 300 Hz lies outside it and is removed. The reconstruction therefore contains 100 Hz and 200 Hz, the second being an alias, a component at a frequency the input never had. The rule for a tone above half the sampling rate is that it folds about \(f_s/2\), appearing at \(f_s - f\) for \(f\) between \(f_s/2\) and \(f_s\); sampling at 600 Hz or more, or low-pass filtering the input to 250 Hz before sampling, would prevent it, which is the purpose of the anti-aliasing filter in every analog-to-digital converter.

100 Hz and 300 Hz ignores aliasing; the 300 Hz tone cannot be represented at 500 samples per second.

100 Hz and 250 Hz folds the tone to the filter edge instead of to \(f_s - f\).

100 Hz and 400 Hz folds the wrong tone, \(500 - 100\).