Evaluate Digital vs Analog Transmission - Grade 11 Physics

1. Which conclusion best evaluates the likely fidelity of the two transmissions?
   Answer: The analog signal may have higher fidelity because the digital sampling rate is below the required minimum of $12.0\ \text{kHz}$. Whenever you see a question mixing digital and analog transmission, the key concept to apply is the Nyquist-Shannon Sampling Theorem: to accurately reconstruct a signal, the sampling frequency must be at least twice the highest frequency present in the signal. This minimum is called the Nyquist rate. Here, the music signal reaches $6.0\ \text{kHz}$, so the required minimum sampling rate is: $f_{Nyquist} = 2 \times 6.0\ \text{kHz} = 12.0\ \text{kHz}$ The digital transmitter samples at only $10.0\ \text{kHz}$, which falls below this threshold. Sampling too slowly causes aliasing — higher frequencies get misrepresented as false lower frequencies — meaning the reconstructed audio will be distorted. The analog transmitter, passing all frequencies up to $7.0\ \text{kHz}$, comfortably covers the full $6.0\ \text{kHz}$ signal range without this problem. So C is correct: the analog signal may have higher fidelity because the digital rate is insufficient. A is wrong because simply exceeding the signal's highest frequency isn't enough — the sampling rate must exceed twice that frequency. $10.0\ \text{kHz} > 6.0\ \text{kHz}$ looks convincing but misses the Nyquist requirement entirely. B is wrong for a similar reason: comparing raw frequency limits without applying the factor-of-two rule leads to a false equivalence. D is the most dangerous distractor — it inverts the Nyquist rule. The required rate is twice the highest frequency, not half of it. Study tip: Memorize the Nyquist rate as $f_s \geq 2f_{max}$. On exam questions, watch for sampling rates that "seem" high enough — always double-check against twice the signal's maximum frequency.
2. Which comparison of the maximum transmission-related measurement uncertainty is correct?
   Answer: The analog uncertainty is about $2.0\ ^\circ\text{C}$, while the digital uncertainty is at most $0.05\ ^\circ\text{C}$. When comparing analog and digital transmission uncertainty, your job is to trace how noise or rounding affects the final temperature reading, not just the signal itself. In the analog system, the voltage scale is $0.10\ \text{V}$ per $1\ ^\circ\text{C}$, meaning $1\ \text{V}$ corresponds to $10\ ^\circ\text{C}$. Channel noise can shift the received voltage by up to $0.20\ \text{V}$. To convert that voltage error into a temperature error: $\frac{0.20\ \text{V}}{0.10\ \text{V}/^\circ\text{C}} = 2.0\ ^\circ\text{C}$. That's a significant uncertainty introduced purely by transmission noise. In the digital system, every bit arrives correctly — so there is zero transmission noise. The only uncertainty comes from rounding each measurement to the nearest $0.10\ ^\circ\text{C}$. Rounding to the nearest unit introduces a maximum error of half that unit: $\frac{0.10}{2} = 0.05\ ^\circ\text{C}$. This makes answer B correct: analog uncertainty ≈ $2.0\ ^\circ\text{C}$, digital uncertainty ≤ $0.05\ ^\circ\text{C}$. A is wrong because it dramatically underestimates the analog uncertainty — it confuses the voltage noise value ($0.20\ \text{V}$) with a temperature error without doing the conversion. C incorrectly assigns $2.0\ ^\circ\text{C}$ to the digital system, ignoring that perfect bit transmission means no transmission-related error beyond rounding. D flips the two systems entirely, assigning the small uncertainty to analog and the large one to digital — the opposite of reality. A useful tip: whenever a signal is transmitted as voltage, always convert the noise into the measured quantity's units before comparing uncertainties. Don't stop at the voltage level.
3. Which statement identifies the most defensible advantage of the digital system?
   Answer: Parity can reveal many altered blocks, allowing retransmission instead of accepting an undetected distorted value. When a question compares digital and analog systems, focus on what error-detection actually does — not what it magically prevents. Parity checking is a detection mechanism, not a prevention or correction mechanism. Keeping that distinction sharp will guide you straight to the right answer. A parity bit is an extra bit added to a data block so that the total number of 1s is always even (or always odd, depending on the scheme). When the ground station receives the block, it recounts the 1s. If the count doesn't match the expected parity, the station knows something changed during transmission and requests a retransmission. This is exactly what answer B describes — parity reveals corrupted blocks so they can be resent rather than silently accepted as valid data. In an analog system, noise distorts the voltage continuously and there's no built-in mechanism to even recognize that distortion occurred. That's the core advantage. Answer A is wrong because parity detects errors, it does not correct them — and it doesn't catch every error either (an even number of flipped bits can slip through undetected). The word "guarantees" and "corrected" make this a double overstatement. Answer C is wrong for a similar reason: parity does nothing to stop noise from flipping bits in the first place. Noise is a physical phenomenon; a parity bit has no power over it. Answer D is completely unrelated — parity bits add overhead to a block, which if anything slightly reduces efficiency; they have no effect on propagation speed. Your strategy: watch for answer choices that upgrade a detection tool into a prevention or correction tool — that's a classic distractor pattern in digital communications questions.
4. Which comparison of the final copies is correct?
   Answer: The analog copy gains about $1.0\ \text{mV}$ of noise, while the digital copy retains its original quantization error. When you see a question comparing analog and digital copying, focus on two separate physics principles: how independent noise accumulates statistically, and what happens when a digital signal is regenerated at each stage. For the analog copy, each stage adds independent noise of $\sigma = 0.20\ \text{mV}$. Because the noise sources are independent, they add in quadrature — you sum the variances, not the amplitudes. After 25 stages, the total RMS noise is $\sigma_{total} = \sigma\sqrt{n} = 0.20\sqrt{25} = 0.20 \times 5 = 1.0\ \text{mV}$. That's the key calculation confirming answer D's first claim. For the digital copy, regeneration means each stage reads the bits, identifies them perfectly, and outputs a clean new signal. No new noise accumulates — the only imperfection is the quantization error introduced during the original analog-to-digital conversion, which stays fixed throughout all 25 stages. So the digital copy's error never grows, exactly as D states. Answer A is wrong because it claims the analog copy accumulates $5.0\ \text{mV}$, which would only be correct if you added amplitudes directly ($0.20 \times 25$) — a violation of how independent noise combines. A also incorrectly claims digital copies accumulate voltage-level error. Answer B is wrong because independent noise does not cancel exactly — cancellation would require the noises to be perfectly correlated and opposite in sign. Answer C is wrong on both counts: repeated copying doesn't average away analog noise (it grows), and digital regeneration doesn't add quantization levels. Study tip: Always use $\sigma_{total} = \sigma\sqrt{n}$ for independent noise accumulation — noise adds in quadrature, not linearly.
5. Which statement correctly evaluates the proposal?
   Answer: The proposal exactly uses the channel capacity and gains error detection, but leaves no rate available for other data. When a question asks you to evaluate a digital transmission proposal, your first move should always be to calculate the actual total bit rate after overhead is added — then compare it to the channel capacity. Here, the system generates $800\ \text{kbit/s}$ of data. Engineers add one check bit per four data bits, meaning for every 4 bits of data, 1 extra bit is transmitted — a ratio of $\frac{1}{4}$. The overhead is: $800 \times \frac{1}{4} = 200\ \text{kbit/s}$ Total transmission rate: $800 + 200 = 1000\ \text{kbit/s} = 1.0\ \text{Mbit/s}$ This exactly matches the channel's maximum capacity. The system gains error detection, but since the channel is fully saturated, no bandwidth remains for anything else. That makes D the correct answer. A is wrong because adding one check bit per four data bits increases the rate by only 25%, not doubles it (which would be 100%). One check bit per four is a 1:4 ratio, not 1:1. B misreads the problem — "one check bit for every four data bits" means one per four bits, not per four thousand bits. That misreading gives a negligibly small overhead of just $0.2\ \text{kbit/s}$, which is far too low. C invents a false rule. Error checking does not universally add half the original rate. The actual overhead depends entirely on the specific scheme used (here, 1-in-4 = 25%). Study tip: On overhead/rate problems, always translate the ratio into a fraction and multiply by the original rate. Never assume a fixed percentage — the problem always tells you the scheme.
6. Which conclusion follows from the stated channel limits?
   Answer: The analog method is feasible, but the uncompressed digital method requires a data rate greater than the channel limit. When a question mixes analog and digital communication systems, your job is to evaluate each method independently against its own channel constraint — don't let one system's numbers bleed into the other's analysis. For the analog method: the audio signal reaches up to $4.0\ \text{kHz}$, and the analog channel passes up to $5.0\ \text{kHz}$. Since $5.0\ \text{kHz} > 4.0\ \text{kHz}$, the channel can carry the full signal. The analog method is feasible. For the digital method: calculate the required bit rate first. With a sampling rate of $8.0\ \text{kHz}$ and $8$ bits per sample, the data rate is $8{,}000 \times 8 = 64{,}000\ \text{bit/s} = 64\ \text{kbit/s}$. The digital channel only supports $48\ \text{kbit/s}$. Since $64\ \text{kbit/s} > 48\ \text{kbit/s}$, and compression is forbidden, the digital method exceeds the channel capacity. This confirms C is correct. A is wrong because it confuses the sampling rate ($8.0\ \text{kHz}$) with the analog channel's frequency limit ($5.0\ \text{kHz}$) — these are unrelated quantities. Sampling rate doesn't determine analog feasibility. B is wrong because satisfying the Nyquist sampling condition (sampling at twice the audio's maximum frequency) only ensures accurate reconstruction of the signal — it says nothing about whether the digital channel has enough bandwidth to carry the resulting bit stream. D is wrong because it invents a requirement that doesn't exist; analog channels are rated by frequency range, not sampling rate. Study tip: Always compute the actual bit rate (samples/sec × bits/sample) and compare it directly to the channel's capacity in kbit/s — don't assume a valid sampling rate guarantees the channel can handle the data.
7. Which conclusion best compares the two transmission methods under these conditions?
   Answer: The digital method can recover every bit exactly, while the analog output still contains the added voltage noise. When comparing digital and analog transmission, the key question to ask is: does the receiver have a way to strip out noise, or does it pass noise along to the output? Here's the critical insight. The digital receiver applies a decision threshold at $2.5\ \text{V}$. Any received voltage above $2.5\ \text{V}$ is decoded as a 1, and anything below is decoded as a 0. Now check the worst-case scenario: a transmitted 0 ($0\ \text{V}$) rises by $1.8\ \text{V}$ to $1.8\ \text{V}$, which is still below $2.5\ \text{V}$ — correctly decoded as 0. A transmitted 1 ($5.0\ \text{V}$) drops by $1.8\ \text{V}$ to $3.2\ \text{V}$, still above $2.5\ \text{V}$ — correctly decoded as 1. The threshold acts like a filter: as long as noise never pushes a signal past the threshold, the digital receiver recovers every bit perfectly. The analog receiver, however, reproduces whatever voltage it receives — noise included — so its output always carries the error. This confirms B as correct. A is wrong because the analog receiver doesn't "correct" the signal just because noise is smaller than the transmitted voltage — it still adds that noise to the output. C is wrong because the noise margin is sufficient: $2.5\ \text{V}$ separates each level from the threshold, and the maximum noise of $1.8\ \text{V}$ never crosses it. D is wrong because the analog receiver has no threshold mechanism at all — it cannot eliminate noise. A useful rule of thumb: digital systems are noise-resistant because they classify signals rather than copy them. Whenever a question asks you to compare digital and analog under noisy conditions, always check whether the noise exceeds the decision threshold before drawing any conclusions.
8. Which evaluation best explains why the analog radio may be preferred at the extreme edge of coverage?
   Answer: Analog quality can degrade gradually, so partially understandable information may remain after the digital link fails abruptly. When comparing analog and digital systems, the key concept to keep in mind is how each system handles signal degradation. Ask yourself: what happens to the usefulness of the information as conditions worsen? Analog signals are continuous waveforms. As noise increases or signal strength drops, the output quality degrades gradually — speech becomes scratchy or muffled, but fragments of recognizable sound can still reach the listener. Digital systems, by contrast, work on precise binary pulses (0s and 1s). Error correction can compensate up to a threshold, but once too many pulses are misread, the system loses entire data packets — a sudden, total failure rather than a slow fade. This is sometimes called the "cliff effect." At the extreme edge of coverage, a partially understandable analog signal may be more useful than complete digital silence, making A the correct answer. B is wrong because analog signals absolutely do not become immune to noise at long distances — the opposite is true. Noise increases with distance. C is wrong on both counts: analog does not reconstruct missing speech exactly (that's a feature of digital error correction), and digital error correction works within a noise tolerance, not only in zero-noise conditions. D is wrong because signal propagation speed is determined by the medium (essentially the speed of light for radio waves), not by amplitude. Analog signals carry no speed advantage over digital ones. A useful tip: whenever a question contrasts analog and digital behavior under stress, think gradual degradation vs. cliff-effect failure — that distinction appears frequently in communications topics at this level.
9. Which evaluation of the two systems is most accurate?
   Answer: The analog system may reveal the change better because the digital quantization interval is about $20\ \text{mV}$. When comparing analog and digital transmission systems, the key question is always: what limits each system's ability to detect small changes? For analog systems, the limit is noise; for digital systems, it's quantization — the finite "step size" between representable values. For the digital system, calculate the quantization interval by dividing the full range by the number of steps. An 8-bit system has $2^8 = 256$ steps, so the step size is $\frac{5.0\ \text{V}}{256} \approx 0.0195\ \text{V} \approx 20\ \text{mV}$. This means the digital system cannot distinguish any voltage change smaller than ~20 mV. Since the sensor needs to detect changes as small as 6 mV, the digital system fails this requirement. The analog system, with noise of only 1 mV, can potentially resolve a 6 mV change — making D the correct answer. A is tempting but wrong. Eight bits sounds impressive, but the math shows the resolution is ~20 mV, which is larger than the required 6 mV — the opposite of what A claims. B is wrong because the shared voltage range is irrelevant. Resolution depends on how finely each system subdivides that range, which differs between the two systems. C is a false absolute. Analog signals do experience noise, but that noise is finite and measurable — in this case, only 1 mV, which is well below the 6 mV threshold. Study tip: Always calculate digital step size using $\frac{\text{range}}{2^n}$ before comparing it to analog noise. Never assume "more bits = automatically better" without checking the actual numbers against the required resolution.
10. Why is the digital system expected to preserve the transmitted information more reliably over the full distance?
    Answer: Each repeater removes small pulse distortions by reconstructing the bit, whereas analog amplification allows noise to accumulate. Whenever you see a question comparing analog and digital transmission, focus on what happens to noise at each stage of the signal's journey — that's the core concept being tested here. In a digital system, each repeater doesn't simply boost whatever signal arrives; it reads the incoming pulse, decides whether it represents a 0 or a 1, and then generates a completely fresh, clean pulse. Any small noise that accumulated in that section gets discarded along with the "used" signal. Because the passage tells you noise stays below the decision threshold in every section, no errors are introduced — and crucially, no noise carries forward. Over hundreds of sections, the output is as clean as the original. An analog amplifier, by contrast, has no way to distinguish the intended waveform from noise. It magnifies everything equally, so noise compounds with every amplification stage, gradually degrading the signal quality over long distances. Answer A captures this mechanism precisely and is correct. Answer B is wrong because repeaters do not change pulse frequency — they reconstruct pulses at the same standard rate. Frequency manipulation is not part of how repeaters improve reliability. Answer C is wrong because digital pulses do not travel faster than analog signals through the same cable medium; transmission speed depends on the medium and electromagnetic properties, not the encoding scheme. Answer D is wrong because it contains two false claims: analog amplifiers do not "lose all energy" from the waveform (they actually add energy, just also to the noise), and digital repeaters don't conserve the original signal energy — they discard it and generate new energy for the fresh pulse. A useful habit: when comparing analog vs. digital, always ask "what happens to noise at each stage?" That framing will guide you to the right answer on most questions of this type.

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