Compare Digital and Analog Signals - Grade 11 Physics

1. Which conclusion follows from the channel limitations?
   Answer: The analog voice fits within the channel's frequency limit, but the uncompressed digital bit rate of $64~kb/s$ exceeds the channel's reliable limit of $10~kb/s$. When a question compares analog and digital transmission over the same channel, you need to apply two separate limits: a frequency limit for analog signals, and a bit-rate limit for digital signals. These are different constraints, so you must check each independently. For the analog voice, the signal contains frequencies up to $4~\text{kHz}$, and the channel passes frequencies up to $5~\text{kHz}$. Since $4~\text{kHz} < 5~\text{kHz}$, the analog voice fits comfortably within the channel. For the digital voice, calculate the bit rate: $8{,}000~\text{samples/s} \times 8~\text{bits/sample} = 64{,}000~\text{b/s} = 64~\text{kb/s}$. The channel's maximum reliable bit rate is approximately twice its bandwidth: $2 \times 5~\text{kHz} = 10~\text{kb/s}$. Since $64~\text{kb/s} \gg 10~\text{kb/s}$, the digital signal far exceeds the channel's capacity. This makes C the correct conclusion. A is wrong because it confuses two different quantities — the sampling rate ($8~\text{kHz}$) is not the same as a bit rate, and comparing it to the $10~\text{kHz}$ figure is meaningless here. The real issue is bit rate, not sampling rate in Hz. B has the logic completely backwards. The analog voice at $4~\text{kHz}$ is within the $5~\text{kHz}$ frequency limit, not above it. B misreads which signal is the problem. D is a false blanket statement — digital transmission does not always require less bandwidth than analog; in fact, uncompressed digital signals often require significantly more. Study tip: Always track units carefully — frequency (Hz) applies to analog limits, while bit rate (b/s) applies to digital limits. Mixing them up is the most common trap on questions like this.
2. Which comparison is most accurate?
   Answer: The parity check may miss the two-bit error, while the analog signal contains distortion during the noise pulse. When comparing digital and analog error handling, focus on what each system can do versus what it actually does by design. Even parity works by ensuring the total number of $1$s in an 8-bit group (7 data + 1 parity) is even. If exactly two bits are flipped by noise, the count of $1$s changes by $+2$ or $-2$ — both of which leave the total parity even. The receiver's parity check sees "even number of 1s" and concludes the data arrived correctly, missing the error entirely. Meanwhile, an analog signal is a continuous waveform — a noise pulse physically distorts the waveform during its duration, embedding real distortion that a receiver cannot automatically separate from the original signal without additional processing. This makes B the most accurate comparison: the parity check can fail to catch two-bit errors, and the analog signal genuinely contains distortion. A is wrong on both counts: parity does not necessarily detect even-numbered bit flips, and analog systems have no built-in mechanism to "automatically remove" noise — the distortion persists in the signal. C is wrong because even parity only detects certain errors (odd numbers of flipped bits); it cannot correct any errors on its own. Error correction requires additional redundancy, like Hamming codes. D overstates both cases. The digital message is not necessarily unchanged — it is silently corrupted. And the analog message is not necessarily completely unreadable — it may still be intelligible despite distortion. Your study tip: always distinguish between error detection and error correction — parity bits only detect odd-numbered bit errors; they fix nothing.
3. Which comparison of the systems is correct?
   Answer: System X has finer amplitude resolution and a lower bit rate, and both sampling rates can avoid aliasing. When comparing digital systems, you need to evaluate three things: amplitude resolution, bit rate, and aliasing protection. Amplitude resolution depends on the number of bits per sample. More bits means more quantization levels ($2^n$), so finer resolution. System X uses 10 bits → $2^{10} = 1024$ levels; System Y uses 8 bits → $2^8 = 256$ levels. System X has finer amplitude resolution. Aliasing is avoided when the sampling rate is at least twice the highest signal frequency (Nyquist criterion): $f_s \geq 2f_{max} = 2 \times 9~\text{kHz} = 18~\text{kHz}$. System X samples at 20 kHz ✓ and System Y at 30 kHz ✓ — both exceed 18 kHz, so both avoid aliasing. Bit rate = sampling rate × bits per sample. System X: $20{,}000 \times 10 = 200{,}000~\text{bps}$. System Y: $30{,}000 \times 8 = 240{,}000~\text{bps}$. System X has the lower bit rate. So the correct answer is A: System X has finer amplitude resolution, a lower bit rate, and both systems satisfy the Nyquist criterion. Choice B is wrong because it correctly identifies System X's resolution advantage but incorrectly claims only System Y avoids aliasing — System X's 20 kHz also clears the 18 kHz threshold. Choice C is wrong because System Y actually has coarser resolution (fewer bits). Choice D incorrectly attributes finer resolution to System Y and wrongly excludes System Y from avoiding aliasing. Study tip: Always calculate the Nyquist threshold ($2f_{max}$) explicitly before judging aliasing — don't assume a lower sampling rate automatically fails.
4. Which choice gives the lowest available sampling rate that can avoid aliasing and the resulting digital bit rate?
   Answer: $8.0~kHz$ and $64~kb/s$, because the sampling rate must be at least twice the highest frequency. Whenever you see a question about digital sampling, the key concept to apply is the Nyquist-Shannon Sampling Theorem: to accurately reconstruct a signal without aliasing, you must sample at a rate at least twice the highest frequency present. Aliasing is the distortion that occurs when a signal is under-sampled, causing different frequencies to become indistinguishable. Here, the highest signal frequency is $3.6~\text{kHz}$, so the minimum required sampling rate is $2 \times 3.6 = 7.2~\text{kHz}$. From the available options $(4.0, 6.0, 8.0~\text{kHz})$, the lowest rate that meets or exceeds $7.2~\text{kHz}$ is $8.0~\text{kHz}$. The resulting bit rate is then: $\text{Bit Rate} = 8.0~\text{kHz} \times 8~\text{bits} = 64~\text{kb/s}$ This makes C correct. Choice A is wrong on two levels: $4.0~\text{kHz}$ is below the Nyquist minimum, so aliasing would occur, and the reasoning that "one sample per cycle is sufficient" is a classic misconception — you need two samples per cycle minimum. Choice B selects $6.0~\text{kHz}$, which only needs to exceed the signal frequency — another common misconception. $6.0~\text{kHz}$ is greater than $3.6~\text{kHz}$ but still below the required $7.2~\text{kHz}$ Nyquist rate, so aliasing would still result. Choice D uses the correct sampling rate of $8.0~\text{kHz}$ but incorrectly calculates the bit rate as $28.8~\text{kb/s}$, perhaps confusing it with $3.6 \times 8$. Remember: Nyquist = $2 \times f_{\text{max}}$, and bit rate = sampling rate × bits per sample. Memorize both formulas together.
5. An analog value of $2.0~V$ and a digital $0$ are sent through their respective systems. What is received after four sections?
   Answer: The analog value is $3.2~V$, while the digital value is regenerated as $0~V$ after every section. Whenever you see a question comparing analog and digital signal transmission, focus on what each system does between sections — analog systems accumulate errors, while digital systems reset them. For the analog path, each of the four cable sections adds $+0.30~V$ to whatever voltage it receives. Starting at $2.0~V$: $2.0 + 4(0.30) = 2.0 + 1.2 = 3.2~V$ The offset compounds because each relay amplifies and forwards the already-degraded signal, including its accumulated error. For the digital path, the repeater receives the signal before forwarding it and makes a decision: is the input below or above $2.5~V$? A digital $0$ is transmitted as $0~V$. After the first cable section, it arrives at $0 + 0.30 = 0.30~V$ — still below $2.5~V$, so the repeater regenerates it as exactly $0~V$. This resets the offset entirely. The same thing happens after every section, so the output is always a clean $0~V$. This confirms answer A is correct. Answer B is wrong because the analog signal definitely does not stay at $2.0~V$ — the offset accumulates with each section. Answer C wrongly applies the same behavior to both signals, ignoring that the digital repeater actively regenerates the signal and eliminates offset. Answer D is wrong because $0.30~V$ never exceeds the $2.5~V$ threshold, so the digital signal is never decoded as a $1$. On signal questions, always ask: does the system regenerate or just relay? That distinction is the entire point of digital repeaters.
6. The input voltage slowly changes from $2.62~V$ to $2.74~V$. In the absence of transmission noise, how do the recorded outputs compare?
   Answer: The digital output remains at $2.5~V$, while the analog output changes continuously by $0.12~V$. When comparing digital and analog recording systems, the key question to ask is: how does each system respond to a small change in input voltage? Digital systems can only output discrete, pre-set levels, while analog systems track the input continuously and exactly. Here, the allowed digital levels are spaced every $0.5~\text{V}$. When the input is $2.62~\text{V}$, the nearest allowed level is $2.5~\text{V}$ (since $2.62$ is closer to $2.5$ than to $3.0$). When the input rises to $2.74~\text{V}$, the nearest allowed level is still $2.5~\text{V}$ (the midpoint between levels is $2.75~\text{V}$, and $2.74 < 2.75$). So the digital output stays locked at $2.5~\text{V}$ throughout. The analog recorder, by contrast, follows the input exactly, changing smoothly by $0.12~\text{V}$. This confirms answer B. Choice A has the outputs reversed — it incorrectly gives the digital recorder the ability to track small changes, which is precisely what quantization prevents. Choice C claims the digital output jumps from $2.5~\text{V}$ to $3.0~\text{V}$, but this would only happen if the input crossed the $2.75~\text{V}$ threshold — it stops just short at $2.74~\text{V}$. Choice D incorrectly states the digital output is at $3.0~\text{V}$ to begin with, which would require the input to already be above $2.75~\text{V}$. A useful strategy: always find the midpoint between adjacent digital levels (here, $\frac{2.5 + 3.0}{2} = 2.75~\text{V}$) to determine which level gets assigned — then check whether the input actually crosses that threshold.
7. Assuming the interference ends before retransmission, which outcome is most likely?
   Answer: The analog link plays a distorted portion immediately, while the digital link can recover it accurately with added delay. When comparing analog and digital transmission, the key distinction is how each system handles corrupted data. Analog signals are continuous — whatever the receiver gets, it plays directly. Digital signals are divided into discrete packets, each with an error-checking code that flags corruption. During interference, an analog receiver has no way to detect or correct damage. It simply outputs whatever waveform arrives, including the distorted portion. The listener hears a crackle, buzz, or garbled sound in real time — but playback is immediate. A digital receiver, by contrast, runs each incoming packet through an error check (like a checksum or CRC). A corrupted packet fails this check, triggering a retransmission request. Since the interference ends before retransmission occurs, the re-sent packet arrives cleanly, and the message is recovered accurately — just slightly delayed. This is exactly what D describes, making it the correct answer. A gets it backwards: it's the digital link that introduces delay while recovering accuracy, and the analog link that plays immediately with distortion — not the other way around. B is wrong because digital systems are specifically designed to detect corruption through error-checking; receivers can absolutely distinguish a bad packet from a good one. C is tempting but incorrect — the interference absolutely damages the analog signal permanently. Short interference duration helps the digital link recover via retransmission, but it doesn't protect the analog link at all. A useful mental shortcut: analog = "what you get is what you hear," digital = "check it, fix it, delay it." This trade-off between immediacy and accuracy appears frequently in communications technology questions.
8. Which comparison is valid under these conditions?
   Answer: Digital bits are decoded correctly, while analog values generally contain errors caused by the added noise. When comparing analog and digital signal transmission, the key question is: how much noise can the system tolerate before information is lost? The answer is fundamentally different for each mode. In digital mode, a transmitted $0~V$ arrives as anywhere from $-1.8~V$ to $+1.8~V$, and a transmitted $5~V$ arrives as anywhere from $3.2~V$ to $6.8~V$. The receiver's threshold is $2.5~V$, and since $1.8~V < 2.5~V$, no noisy $0~V$ signal ever crosses the threshold, and no noisy $5~V$ signal ever falls below it. Every bit is recovered correctly. In analog mode, a transmitted voltage like $2.3~V$ might arrive as $2.3 \pm 1.8~V$ — anywhere from $0.5~V$ to $4.1~V$. The receiver has no way to know what the original value was; the noise directly corrupts the measurement. This confirms that B is correct. A is wrong because it misapplies the logic. The $2.5~V$ threshold criterion protects digital signals, not analog ones. Analog transmission has no equivalent protection — every noise event shifts the recovered value. C gets it exactly backwards. Analog values are not recovered exactly; they are corrupted by the full $\pm 1.8~V$ noise. Digital bits are the ones that survive correctly. D is wrong because it treats both modes as equally vulnerable, ignoring the threshold mechanism that makes digital transmission noise-resilient within the stated margin. The study tip: digital systems trade signal resolution for noise immunity — the threshold acts as a filter that analog systems simply don't have.
9. Which statement most accurately classifies this transmission?
   Answer: It is digital information carried by a continuous electromagnetic wave using two distinguishable signal states. When a question asks you to classify a transmission as analog or digital, the key distinction is what carries the information, not what the physical medium looks like. The physical wave can be perfectly continuous and still encode digital data — these are two separate layers of the same signal. Here, the transmitter uses two distinct carrier frequencies: one representing 0, the other representing 1. There are only two possible states, and each symbol maps cleanly to one of them. That's the definition of digital information — a finite set of distinguishable states encoding discrete values. The fact that the underlying electromagnetic wave oscillates continuously is a property of the carrier, not the information. This is exactly how real-world digital radio protocols (like FSK — Frequency Shift Keying) work, making D the correct classification. A is tempting but conflates the physical wave with the information it carries. Yes, electromagnetic waves vary continuously, but "continuous variation" describes the carrier's physics, not the data's nature. The information itself has only two discrete states. B introduces a false hybrid category. There's no alternation between analog and digital — the signal is consistently encoding one of two binary states at all times. "Discrete portions" don't exist within individual carrier cycles. C is flatly wrong on the physics. Frequency differences absolutely can and do represent binary information — this is the entire basis of FSK modulation, widely used in telecommunications. As a study tip: always separate the information layer (how many distinct values are possible?) from the physical layer (how does the wave behave?). Digital vs. analog is determined by the information structure, not the waveform's continuity.
10. Which statement best compares the five-stage copying processes?
    Answer: The digital bit has about a $99.0\%$ chance of surviving all five stages correctly, while analog noise accumulates with each copy. When comparing digital and analog copying processes, the key concept is error propagation: does each stage introduce new distortion, or does it reset/regenerate the signal cleanly? For digital copying, each stage independently has a $0.002$ probability of flipping a bit incorrectly, meaning the probability of a correct read at one stage is $0.998$. Across five independent stages, the probability the bit survives all five correctly is: $P = (0.998)^5 \approx 0.990 = 99.0\%$ This confirms A as correct. Meanwhile, analog stages don't regenerate the signal — they amplify whatever they receive, including noise. So each stage adds a small amount of random noise on top of the previous noise, causing cumulative degradation. B is wrong on two counts: $(0.998)^5 \neq 0.998$, so "exactly 99.8%" is mathematically incorrect (that would be just one stage). It also falsely claims analog regeneration removes noise — analog devices cannot distinguish signal from noise the way digital devices can. C is wrong because it dramatically understates digital reliability. A $1.0\%$ survival rate would imply near-certain failure, which contradicts the calculation above. It also incorrectly claims amplification eliminates analog noise — amplifiers boost noise along with the signal. D reverses reality entirely. It's analog distortion that accumulates, not digital. Digital devices regenerate clean, standardized voltage levels at each stage, preventing accumulation. A useful study tip: whenever you see independent probability across multiple stages, multiply the per-stage probabilities together — don't add them or use just one stage's value.

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