Signals and Metrics MCQs: 12 Solved Data Communication Questions

Frequency and bandwidth use hertz, bit rate and throughput use bits per second, baud counts signal elements, and delay uses time. A remembered formula fails quickly when the unit or denominator is wrong. These 12 previous-year questions test period, wavelength, bit rate, baud rate, spectral bandwidth, throughput, bit error rate, attenuation, propagation delay and bits in flight. Attempt each item carefully, write the answer with its unit, and only then read the explanation.

Signals and metrics formula map

Bandwidth can mean a frequency span measured in hertz or, in link discussions, a capacity expressed in bits per second. Keep those contexts separate. For a wider subject sequence, use the structured GATE CS coverage.

Period, frequency and wavelength MCQs

Question 1: frequency from a period

Exam attribution: UGC NET 2014

The period of a signal is 10 ms. What is its frequency in Hertz ?

A. 10
B. 100
C. 1000
D. 10000

Correct answer: B. 100

Convert first: 10 ms = 10 × 10^-3 s = 0.01 s. Therefore, f = 1/T = 1/0.01 = 100 Hz. The 1000 Hz choice would result from mistakenly using a period of 1 ms, not 10 ms. See the full question page.

Question 2: the wavelength relationship

Exam attribution: DSSSB 2024

Which of the following is correct formula for calculating wavelength (λ) of a signal in the data communication?

A. λ = propagation speed + frequency
B. λ = propagation speed × period
C. λ = propagation speed + period
D. λ = propagation speed × frequency

Correct answer: B. λ = propagation speed × period

Since λ = v/f and T = 1/f, substitution gives λ = vT. For v = 2 × 10^8 m/s and f = 1 MHz, T = 1 μs, so λ = 2 × 10^8 × 10^-6 = 200 m. Addition and v × f do not produce metres, so they are dimensionally invalid. Check the full question page.

Bit rate and baud rate in a mixed-metric MCQ set

One signal element can carry more than one data bit. That is why bit rate and baud rate can differ.

These three questions isolate bit-rate and baud-rate conversion before the set returns to bandwidth, throughput, error rate and delay. Ethernet signal coding, regeneration, cable limits and access-protocol utilisation belong to the separate Ethernet-focused practice set.

Question 3: data bits per signal element

Exam attribution: UGC NET 2014

An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element ?

A. 0.336 bits/baud
B. 3 bits/baud
C. 120,00,000 bits/baud
D. None of the above

Correct answer: B. 3 bits/baud

Divide the bit rate by the signal elements sent per second: 6000 bits/s ÷ 2000 signal elements/s = 3 bits/signal element. Carrying 3 bits per signal element requires 2^3 = 8 distinct signal levels in an M-ary scheme. Review the full question page.

Question 4: baud rate from bit rate

Exam attribution: UGC NET 2014

The bit rate of a signal is 3000 bps. If each signal unit carries 6 bits, the baud rate of the signal is _______.

A. 500 baud/sec
B. 1000 baud/sec
C. 3000 baud/sec
D. 18000 baud/sec.

Correct answer: A. 500 baud/sec

Rearrange the relationship: baud rate = bit rate/bits per signal unit = 3000/6 = 500 baud. The 18000 choice comes from reversing the required operation. The source says baud/sec, but baud already means signal elements per second. Open the full question page.

Question 5: calculate both baud rate and bit rate

Exam attribution: UGC NET 2013

An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are _______ and _______.

A. 4000 bauds \ sec & 1000 bps
B. 2000 bauds \ sec & 1000 bps
C. 1000 bauds \ sec & 500 bps
D. 1000 bauds \ sec & 4000 bps

Correct answer: D. 1000 bauds \ sec & 4000 bps

The stated 1000 signal units/s is directly 1000 baud. Next, 4 bits/signal unit × 1000 signal units/s = 4000 bits/s, because the signal-unit terms cancel. The ordered pair is therefore 1000 baud, 4000 bps. See the full question page.

Bandwidth and throughput MCQs

Spectral bandwidth is measured in hertz. Observed throughput is measured in bits per second.

Question 6: bandwidth from lower and upper frequencies

Exam attribution: Indian Space Research Organization 2008

What is the bandwidth of the signal that ranges from 40 kHz to 4 MHz

A. 36 MHz
B. 360 kHz
C. 3.96 MHz
D. 396 kHz

Correct answer: C. 3.96 MHz

Put both endpoints in the same unit: 40 kHz = 0.04 MHz. Then B = f_high - f_low = 4 MHz - 0.04 MHz = 3.96 MHz. Subtracting 40 directly from 4 mixes units and has no meaning. See the full question page.

Question 7: upper frequency from a bandwidth

Exam attribution: Indian Space Research Organization 2007; BEL 2007

If the bandwidth of a signal is 5 kHz and the lowest frequency is 52 kHz, what is the highest frequency?

A. 5 kHz
B. 10 kHz
C. 47 kHz
D. 57 kHz

Correct answer: D. 57 kHz

Rearrange B = f_high - f_low as f_high = B + f_low. Substitution gives 5 kHz + 52 kHz = 57 kHz. The 47 kHz option comes from subtracting in the wrong direction. Check the full question page.

Question 8: observed throughput from frames per minute

Exam attribution: UGC NET 2015

A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network ?

A. 2 Mbps
B. 60 Mbps
C. 120 Mbps
D. 10 Mbps

Correct answer: A. 2 Mbps

The network delivers 15,000 × 8,000 = 120,000,000 bits/minute. Dividing by 60 gives 2,000,000 bits/s = 2 Mbps. This observed throughput is below the 10 Mbps link bandwidth, and 2/10 × 100 = 20% utilisation provides a quick check. Review the full question page.

Bit error rate, attenuation and propagation delay MCQs

Convert hours to seconds and kilometres to metres before using these formulas. Otherwise, even the right relationship produces the wrong scale.

Question 9: bit error rate over ten hours

Exam attribution: Indian Space Research Organization 2011

Data is transmitted continuously at 2.048 Mbps rate for 10 hours and received 512 bits errors. What is the bit error rate?

A. 6.9 e-9
B. 6.9 e-6
C. 69 e-9
D. 4 e-9

Correct answer: A. 6.9 e-9

The total transmitted bits are 2.048 × 10^6 × 10 × 3600 = 73,728,000,000. Thus BER = 512/73,728,000,000 = 6.944... × 10^-9, which rounds to 6.9 × 10^-9. BER is a dimensionless ratio, not a rate in bps. See the full question page.

Question 10: half-power attenuation in decibels

Exam attribution: UGC NET 2026

Assume a signal travels through a transmission medium and its power is reduced to one-half, which means P₂ = 1/2 P₁. What is the power loss in attenuation?

A. -6 db
B. -3 db
C. -1.5 db
D. -1 db

Correct answer: B. -3 db

For a power ratio, 10 log10(P2/P1) = 10 log10(0.5) = 10(-0.3010) = -3.01 dB. Output power relative to input is about -3 dB, while the magnitude of the loss is 3 dB. Using 20 log10 would be wrong because the question gives a power ratio. Check the full question page.

Question 11: propagation time over a cable

Exam attribution: UGC NET 2014

What is the propagation time if the distance between the two points is 48,000 Km ? Assume the propagation speed to be 2.4 × 10⁸ metre/second in cable.

A. 0.5 ms
B. 20 ms
C. 50 ms
D. 200 ms

Correct answer: D. 200 ms

First, 48,000 km = 48,000,000 m. Then t_p = 48,000,000/(2.4 × 10^8) = 0.2 s = 200 ms. This is distance-based propagation delay, which differs from transmission delay based on message size and bit rate. Open the full question page.

Bits in flight and the mixed-metric exam trap

Question 12: number of bits present in a cable

Exam attribution: DSSSB 2018

Suppose a 100 km long cable has a speed of 1.536 Mbps. The signal speed in cable is 2/3 speed of the light. What will be the number of bits in the cable?

A. 500
B. 768
C. 307
D. 400

Correct answer: B. 768

Take the speed of light as 3 × 10^8 m/s, so propagation speed is (2/3)(3 × 10^8) = 2 × 10^8 m/s. The one-way delay is 100,000/(2 × 10^8) = 0.0005 s. Bits in flight are therefore 1.536 × 10^6 × 0.0005 = 768 bits, the one-way bandwidth-delay product. See the full question page.

A higher bit rate or longer delay puts more bits in flight.

For every mixed-metric problem:

1. Identify the requested metric.

2. Normalise all units.

3. Choose the correct denominator.

4. Run a dimensional check.

Continue with address calculation practice or switch to protocol behaviour practice to apply the same disciplined method elsewhere in Computer Networks.

Signals and metrics: the short version and next step

Write the metric and unit, convert every value, substitute once, then check whether the answer's unit matches the question. Classify each wrong attempt as definition, relationship, unit conversion or denominator, and retry only that group after one day. This keeps revision focused: conversion errors need unit practice, while relationship errors require the formula map. Record the corrected unit beside each answer clearly. If you need a sequenced route through the subject, use GATE Guidance by Sanchit Sir. When you are ready for timed mixed practice, use the GATE Test Series.