Question 1

If the channel bandwidth available to a frequency-modulated (FM) system is doubled while keeping other conditions comparable to amplitude modulation (AM), by what factor does the improvement ratio γFM/γAM (output SNR ratio of FM to AM) typically inc…

Accepted Answer

Introduction / Context:One advantage of wideband FM is improved output signal-to-noise ratio (SNR) compared with AM, for a given received carrier power. This improvement grows with FM system deviation and effectively with the occupied bandwidth. Understanding how γFM/γAM scales with bandwidth guides design trade-offs between spectral efficiency and noise performance.Given Data / Assumptions:γ denotes the ratio of output SNR to input SNR.We compare FM and AM under comparable received power and noise conditions.Bandwidth is increased by a factor of 2.Concept / Approach:For wideband FM, the output SNR improvement over AM is approximately proportional to the square of the peak frequency deviation (or modulation index for single-tone), which in turn scales with the occupied bandwidth (per Carson’s rule intuition). If bandwidth is doubled, the improvement factor scales roughly as the square, yielding approximately four times greater γFM/γAM, ignoring pre-emphasis and receiver details.Step-by-Step Solution:Assume wideband FM regime where SNR_o(FM) ∝ (deviation)^2 for a given noise floor.If bandwidth B → 2B, effective deviation (and thus improvement) scales → improvement factor ≈ (2)^2 = 4.Therefore, γFM/γAM increases by a factor of approximately 4.Verification / Alternative check:Classic FM noise theory shows that, absent threshold effects, FM demodulator output noise decreases as deviation increases; empirical lab setups confirm about quadratic scaling before practical limits (receiver IF bandwidth, pre-emphasis) intervene.Why Other Options Are Wrong:Factors 2 or 3: underestimate the quadratic dependence in the wideband regime.Factor 6: not supported by the square-law trend.No change: contradicts FM noise advantage principles.Common Pitfalls:Applying this scaling in the narrowband FM regime or near the FM threshold where different behaviors occur; ignoring pre-emphasis/de-emphasis which alters high-frequency noise shaping but not the basic wideband scaling trend.Final Answer:Factor of 4

Question 2

In a binary symmetric channel (BSC), if the conditional probability of bit error is p = 0.5 due to extreme noise, what are the channel capacity and the binary entropy function H(p), respectively?

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Introduction / Context:Channel capacity quantifies the maximum reliable information rate over a noisy channel. For a binary symmetric channel (BSC), capacity depends on the crossover probability p. Understanding the edge case p = 0.5 (complete randomness) clarifies the limits of digital communication under severe noise.Given Data / Assumptions:BSC with error probability p = 0.5.Capacity formula for BSC: C = 1 − H(p) bits per channel use (for binary input), where H(p) is the binary entropy function.Concept / Approach:The binary entropy function H(p) = − p log2 p − (1 − p) log2 (1 − p). At p = 0.5, H(0.5) = 1 (maximum uncertainty). Therefore, C = 1 − H(0.5) = 0: no information can be transmitted reliably because the output is statistically independent of the input (complete randomization).Step-by-Step Solution:Compute H(0.5): H(0.5) = −0.5 * log2 0.5 − 0.5 * log2 0.5 = 1.Compute capacity: C = 1 − H(0.5) = 1 − 1 = 0 bits/use.Thus, the pair (Capacity, H(p)) = (0, 1).Verification / Alternative check:Intuition: with p = 0.5, the channel effectively flips the bit randomly; the output conveys no information about the input, hence zero capacity.Why Other Options Are Wrong:Any nonzero capacity contradicts the total randomization at p = 0.5.Entropy values other than 1 at p = 0.5 are incorrect; binary uncertainty is maximal at 0.5.Common Pitfalls:Applying C = log2 M − H without recognizing that M = 2 for binary channels; forgetting that H(p) peaks at p = 0.5.Final Answer:0 and 1

Question 3

A carrier is simultaneously amplitude-modulated by two sinusoidal tones with modulation indices m1 = 0.4 and m2 = 0.3. What is the resultant (total) modulation index for the AM signal?

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Introduction / Context:In multi-tone amplitude modulation, each tone contributes its own sidebands. The overall modulation depth must consider the combined effect to assess power distribution, distortion risk (overmodulation), and detector linearity. Knowing how to compute the net modulation index avoids clipping and excessive envelope distortion.Given Data / Assumptions:Two independent sinusoidal modulators with indices m1 = 0.4 and m2 = 0.3.Standard linear AM with independent tones.No phase correlation that would alter envelope peak beyond RMS combination.Concept / Approach:For independent sinusoidal modulations, the effective modulation index m_total is the root-sum-square of the individual indices: m_total = sqrt(m1^2 + m2^2). This reflects the combined RMS modulation depth. It is commonly used to ensure the total modulation does not exceed unity (to prevent envelope overmodulation and distortion in envelope detectors).Step-by-Step Solution:Compute squares: m1^2 = 0.4^2 = 0.16; m2^2 = 0.3^2 = 0.09.Sum: 0.16 + 0.09 = 0.25.Take square root: sqrt(0.25) = 0.5 → total modulation index = 0.5.Verification / Alternative check:Envelope simulation or phasor addition of sidebands yields identical conclusions for uncorrelated tones. Practical AM broadcast standards often keep combined indices below 1 to avoid audible distortion.Why Other Options Are Wrong:0.7: simple arithmetic sum; overestimates true combined index for independent tones.0.35 or 0.1: incorrect combinations that do not follow RMS summing.0.58: not the correct RMS for given values.Common Pitfalls:Summing indices linearly instead of by RMS; ignoring the overmodulation threshold when multiple program components are present.Final Answer:0.5

Question 4

In radar and pulsed communication systems, the acronym PRF stands for which fundamental timing parameter?

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Introduction / Context:PRF (Pulse Repetition Frequency) is a key timing parameter in radar and pulsed systems defining how often pulses are transmitted. It influences maximum unambiguous range, Doppler ambiguity, and average transmitted power. Recognizing PRF helps in interpreting radar performance metrics and link timing constraints in pulsed communication links.Given Data / Assumptions:Pulsed radar transmits short pulses at a regular repetition rate.The question focuses on the meaning of the acronym PRF.Concept / Approach:Pulse Repetition Frequency is the number of pulses transmitted per second (Hz). Its inverse, Pulse Repetition Interval (PRI), is the time between pulse starts. Higher PRF reduces maximum unambiguous range but improves Doppler measurement tolerance; lower PRF increases unambiguous range but may lower average power or affect Doppler resolution strategies.Step-by-Step Solution:Define PRF: count of transmit pulses per second.Relate to PRI: PRI = 1 / PRF; these determine timing/range ambiguities.Select the correct expansion: Pulse Repetition Frequency.Verification / Alternative check:Standard radar texts and specifications consistently define PRF = pulses per second; radar planners choose PRF considering range/Doppler ambiguities.Why Other Options Are Wrong:Power/Return/Response factors are not standard expansions for PRF.Peak Radar Frequency is unrelated; carrier frequency is separate from PRF.Common Pitfalls:Confusing PRF with carrier frequency or bandwidth; PRF is a timing parameter independent of RF carrier frequency in pulse radars.Final Answer:Pulse Repetition Frequency

Question 5

In a facsimile (fax) reproduction system, the time to scan one line is 2 ms. If the blanking interval occupies 4% of this time, what is the effective visible trace time?

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Introduction / Context:Facsimile (fax) transmission systems scan each line of an image within a given time slot. Part of this period is used for blanking or retrace, during which no visible trace is recorded. The visible trace time determines how long actual image information is being transmitted and displayed.Given Data / Assumptions:Total line scan time = 2 ms.Blanking interval = 4% of total time.Concept / Approach:Visible time = Total scan time − Blanking time. Blanking time = (percentage blanking) * (total time). For 4% blanking, blanking time = 0.04 * 2 ms = 0.08 ms. Hence, visible trace = 2 ms − 0.08 ms = 1.92 ms.Step-by-Step Solution:Blanking time = 0.04 * 2 ms = 0.08 ms.Visible trace time = 2 ms − 0.08 ms = 1.92 ms.Thus, effective visible trace time = 1.92 ms.Verification / Alternative check:Cross-check: If visible = 96% of 2 ms, 0.96 * 2 ms = 1.92 ms. Both approaches confirm the same answer.Why Other Options Are Wrong:2.08 ms: would imply more than the total line period, impossible.50 ms: unrelated large value, not feasible.0.08 ms: corresponds to the blanking time, not the visible time.19.2 ms: incorrect scaling by factor of 10.Common Pitfalls:Mixing blanking duration with visible trace duration; misinterpreting percentage application.Final Answer:1.92 ms

Question 6

The maximum rate at which information can be transmitted reliably through a communication channel primarily depends on which parameter?

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Introduction / Context:Shannon’s communication theory defines limits for information transfer. While many factors affect link quality, the fundamental determinant of capacity is the channel bandwidth and the signal-to-noise ratio (SNR). This question highlights the role of bandwidth.Given Data / Assumptions:Channel capacity formula: C = B * log2(1 + S/N).B = channel bandwidth.S/N = signal-to-noise ratio.Concept / Approach:For fixed SNR, increasing bandwidth linearly increases the number of bits per second that can be carried. Carrier frequency does not directly dictate capacity; losses and power affect SNR but not the structural limit. Bandwidth is the key parameter determining potential data rate.Step-by-Step Solution:Recall Shannon formula: C = B * log2(1 + S/N).Observe proportionality: C ∝ B for fixed S/N.Therefore, the maximum reliable information rate depends on channel bandwidth.Verification / Alternative check:Modern high-data-rate systems (optical, Wi-Fi, 5G) exploit very wide bandwidth allocations, confirming this principle.Why Other Options Are Wrong:Carrier frequency: only indirectly affects propagation, not capacity formula.Transmission loss: affects received SNR but not the fundamental dependency.Transmitted power: influences SNR but not as directly as bandwidth for maximum rate.Antenna gain: improves link margin, not fundamental capacity scaling.Common Pitfalls:Confusing carrier frequency allocation with actual usable bandwidth.Final Answer:Available channel bandwidth

Question 7

A telephone exchange has 9000 subscribers. If the number of calls originated during the busy hour is 10,000 in one hour, what is the average calling rate per subscriber?

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Introduction / Context:Traffic engineering in telephony requires understanding the offered traffic load, measured in Erlangs or as an average call rate. Busy-hour call attempts per subscriber are key for designing capacity of switches and trunks.Given Data / Assumptions:Total subscribers = 9000.Total calls in peak hour = 10,000.Concept / Approach:Calling rate per subscriber = total calls / number of subscribers. This provides the average number of originated calls per subscriber during the busy hour.Step-by-Step Solution:Total calls = 10,000.Subscribers = 9000.Rate = 10,000 / 9000 = 1.111… calls per subscriber.Rounded or expressed as ≈ 1.11 calls per subscriber; however in many engineering exam contexts, they approximate it as 10/9 ≈ 1.11.Verification / Alternative check:Divide precisely: 10,000 ÷ 9000 = 1.111. Thus 10/9 is the exact expression.Why Other Options Are Wrong:0.9 or 0.81 or 0.1: numerical errors, not the correct ratio.1 call: underestimates slightly; true ratio is greater than 1.Common Pitfalls:Misinterpreting the calling rate as calls per second instead of per subscriber per hour; forgetting to normalize by subscriber count.Final Answer:10/9 calls per subscriber

Question 8

In the figure , the output y(t) will be

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Question 9

When it is desired to match a transmission line with its load across a range of frequencies, which technique is generally most effective?

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Introduction / Context:Transmission line matching is critical to minimize reflections and standing waves. Single-stub tuning works well at a single frequency, but broadband matching requires techniques that remain effective across multiple frequencies.Given Data / Assumptions:Objective: match line and load across a range of frequencies.Candidate methods: balun, stub(s), amplifier, etc.Concept / Approach:Single stubs provide point frequency matching; double-stub tuners provide flexibility to correct mismatch at two points, offering better broadband coverage. Baluns are for balanced-to-unbalanced conversion, not broadband impedance matching. Broadband amplifiers are not used as a substitute for matching networks.Step-by-Step Solution:Recognize broadband requirement.Single stub → narrowband; double stub → wider range compensation.Therefore, the best method is to use a double stub tuner.Verification / Alternative check:Microwave engineering references confirm: double and triple stub tuners are used for broadband matching in practice.Why Other Options Are Wrong:Balun: not for broadband impedance matching, used for balanced-unbalanced conversion.Single stub: effective but narrowband.Broadband amplifier: not a true matching technique.Quarter-wave transformer: broadband limited, works best at a center frequency.Common Pitfalls:Confusing balun's role; assuming broadband amplifier can replace matching network.Final Answer:Using a double stub tuner

Question 10

If an FM signal with modulation index mf is passed through a frequency tripler circuit, what will be the new modulation index of the output signal?

Accepted Answer

Introduction / Context:Frequency multipliers are commonly used in FM transmitter chains to shift carrier frequency while maintaining modulation. Understanding how frequency deviation and modulation index scale through multipliers is fundamental for FM system design.Given Data / Assumptions:Input modulation index = mf.Frequency multiplier factor = 3 (tripler).Concept / Approach:Modulation index mf = frequency deviation / modulating frequency. Through a frequency multiplier of factor N, both carrier frequency and frequency deviation are multiplied by N, while modulating frequency remains the same. Hence, the modulation index is also multiplied by N.Step-by-Step Solution:Given multiplier factor N = 3.New deviation = N * old deviation.New mf = N * old mf = 3mf.Verification / Alternative check:Transmission chain design examples: a baseband FM oscillator with mf, followed by doublers/triplers, scales mf accordingly.Why Other Options Are Wrong:mf: ignores multiplication effect.mf/3 or mf/9: contradict actual scaling (inverse relation would only occur if modulating frequency also changed, which it does not).sqrt(3) * mf: not the relation for frequency multipliers.Common Pitfalls:Forgetting that both carrier and deviation scale equally, preserving deviation ratio multiplication.Final Answer:3mf

Question 11

In acoustics and sound system design, can direct sound waves and reflected sound waves interfere with each other and affect the listener’s experience?

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Introduction / Context:Sound propagation in rooms or auditoriums involves both direct sound from the source and reflections from walls, floor, ceiling, and objects. The superposition of these sound paths can produce constructive or destructive interference, affecting clarity and perceived loudness.Given Data / Assumptions:Sound waves can overlap in time and space.Basic wave principle: superposition.Concept / Approach:When direct and reflected sound waves arrive at the listener with different path lengths, they may be in phase (constructive interference) or out of phase (destructive interference). This leads to reinforcement or cancellation at certain frequencies, producing comb filtering effects in acoustics.Step-by-Step Solution:Identify sources: direct wave from loudspeaker; reflected wave from wall.Apply principle of superposition: total sound = sum of both waveforms.Depending on phase difference, result can be louder or quieter.Verification / Alternative check:Measured impulse responses in rooms clearly show reflections combining with direct sound, producing interference patterns in frequency response.Why Other Options Are Wrong:False: violates basic wave physics.Only outdoor/indoor or below certain frequencies: interference can occur in any environment with multiple sound paths.Common Pitfalls:Assuming interference is only an optical phenomenon; sound waves obey the same wave superposition principles.Final Answer:True

Question 12

In the standing wave pattern formed on a transmission line due to impedance mismatch, how are the positions of voltage nodes and current antinodes, as well as current nodes and voltage antinodes, related?

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Introduction / Context:Standing waves form on a transmission line when the line is not properly matched with its load, causing reflections. Understanding the relationship between voltage and current standing wave patterns is essential in RF engineering and impedance matching applications.Given Data / Assumptions:Transmission line with forward and reflected waves.Lossless line assumption.Concept / Approach:Voltage and current distributions on a standing wave pattern are spatially shifted by 90 degrees. Where voltage reaches minimum (node), current reaches maximum (antinodes). Where voltage reaches maximum (antinodes), current is at minimum (node). Hence, voltage nodes always coincide with current antinodes and vice versa.Step-by-Step Solution:Reflections superimpose forward and backward waves.At some points, destructive interference of voltages → voltage nodes.At those same points, currents reinforce → current antinodes.Thus, node of one corresponds to antinode of the other.Verification / Alternative check:Microwave textbooks confirm that voltage and current standing wave patterns are spatially offset by λ/4, supporting this relationship.Why Other Options Are Wrong:Option A: Nodes do not coincide; one is maximum when the other is minimum.Option B: Antinodes also do not coincide; they are shifted.Option D: Both statements are incorrect together.Option E: Not applicable as option C is correct.Common Pitfalls:Confusing optical interference patterns with RF standing waves; assuming voltage and current behave identically along the line.Final Answer:Voltage nodes coincide with current antinodes, and current nodes coincide with voltage antinodes

Question 13

For reducing or attenuating high-frequency components in an electrical circuit, which type of component should be used?

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Introduction / Context:In electronic circuits, filtering and attenuation of high frequencies is often required to suppress noise or to achieve desired frequency response. Different components respond differently to frequency, making some more suitable for attenuating high-frequency signals.Given Data / Assumptions:We want to attenuate high frequencies.We can choose among capacitors, inductors, and resistors.Concept / Approach:Inductors present reactance XL = 2πfL, which increases with frequency. Thus, as frequency increases, an inductor increasingly resists current flow, attenuating high-frequency signals. Conversely, capacitive reactance decreases with frequency, making capacitors pass high frequencies rather than attenuate them.Step-by-Step Solution:XL = 2πfL.As frequency f increases, XL increases linearly.Therefore, inductors block or attenuate high-frequency signals effectively.Verification / Alternative check:Low-pass filters are often constructed with series inductors and shunt capacitors to block high frequencies and pass low frequencies.Why Other Options Are Wrong:Shunt capacitance: passes high frequencies, not attenuates them.Series capacitance: blocks low frequencies, passes high, opposite effect.Resistance: attenuates all frequencies equally, not selective.Parallel resonance: attenuates only at a narrow band, not all high frequencies.Common Pitfalls:Assuming resistance is sufficient for selective attenuation; forgetting inductors inherently resist higher frequencies.Final Answer:Inductance

Question 14

Which of the following methods is considered an indirect way of generating Frequency Modulation (FM)?

Accepted Answer

Introduction / Context:Frequency modulation (FM) can be generated either by direct methods, which vary the oscillator frequency directly with the message signal, or by indirect methods, which build FM using phase modulation or other frequency-altering techniques. Understanding these distinctions is fundamental in communication engineering.Given Data / Assumptions:Four modulator types provided: reactance, Armstrong, varactor, direct FM.Task: identify indirect method.Concept / Approach:Direct FM methods: reactance modulator, varactor diode modulator. These directly vary oscillator capacitance or reactance. Indirect FM: Armstrong modulator, which produces FM through phase modulation of a crystal oscillator, ensuring high frequency stability.Step-by-Step Solution:Reactance bipolar transistor → direct method.Varactor diode → direct method.Reactance FM modulator → direct method.Armstrong modulator → indirect method (via phase modulation of carrier).Verification / Alternative check:Classic Armstrong system described in communication textbooks as the first practical method of generating FM using indirect modulation.Why Other Options Are Wrong:Reactance bipolar transistor modulator: directly varies oscillator frequency.Varactor diode modulator: direct variation of capacitance → direct FM.Reactance FM modulator: another direct variation method.Direct frequency synthesizer: also a direct FM method.Common Pitfalls:Confusing phase modulation used in Armstrong method with direct frequency control; assuming all are direct FM methods.Final Answer:Armstrong modulator

Question 15

Assertion (A): Free space does not interfere with normal radiation and propagation of radio waves Reason (R): Free space has no magnetic or gravitational fields.

Accepted Answer

Since free space does not have magnetic and other fields, waves can propagate without any interference.

Question 16

If the frequency of a wave is 20 Hz, what will be its time period in seconds?

Accepted Answer

Introduction / Context:The relationship between frequency and time period is a fundamental concept in physics and communication systems. Frequency defines how many cycles occur per second, while the time period specifies the duration of one cycle.Given Data / Assumptions:Frequency f = 20 Hz.Formula: T = 1 / f.Concept / Approach:Frequency and time period are reciprocals. A wave with higher frequency has a smaller time period.Step-by-Step Solution:T = 1 / fT = 1 / 20T = 0.05 secondsVerification / Alternative check:20 cycles per second ⇒ each cycle = 1/20 = 0.05 s.Why Other Options Are Wrong:20 secs: corresponds to 0.05 Hz, not 20 Hz.2 secs: corresponds to 0.5 Hz.0.2 secs: corresponds to 5 Hz.0.005 secs: corresponds to 200 Hz.Common Pitfalls:Confusing reciprocal relation and multiplying instead.Final Answer:0.05 secs

Question 17

In electronics, a push–pull amplifier is primarily classified as which type of amplifier?

Accepted Answer

Introduction / Context:A push–pull amplifier is widely used in audio systems and RF transmitters. It consists of two active devices operating in opposite half-cycles of the input waveform.Given Data / Assumptions:Classified by its output function.Push–pull design reduces distortion and improves efficiency.Concept / Approach:The purpose of push–pull is to deliver significant power to a load, hence it is a power amplifier. Voltage or current amplifiers deal with small signal conditioning, not direct load driving.Step-by-Step Solution:Two transistors (NPN and PNP or MOSFET pairs) amplify alternate halves of the waveform.When combined, they produce a larger output signal with reduced harmonics.This makes it efficient for delivering power to speakers or loads.Verification / Alternative check:Datasheets and textbooks define push–pull amplifiers under “power amplifier” categories.Why Other Options Are Wrong:Voltage amplifier: small-signal stages, not output stage.Current amplifier: specialized, not the push–pull use.None of the above: incorrect, it clearly belongs to power class.Common Pitfalls:Confusing “signal amplifier” role with final power delivery.Final Answer:power amplifier

Question 18

Radio broadcasts for public transmission are generally transmitted using which modulation technique?

Accepted Answer

Introduction / Context:Radio broadcasting historically used AM and later FM depending on frequency bands. This question asks about the general method used for traditional long-distance radio broadcasting.Given Data / Assumptions:Broadcast band: Medium wave (MW) and shortwave (SW) radio.Standard practice worldwide until the dominance of FM for high fidelity.Concept / Approach:Amplitude modulation (AM) was used widely because it is simple to implement and robust over long distances. Frequency modulation (FM) is used for VHF high-fidelity audio but not considered “general broadcast” historically.Step-by-Step Solution:Radio broadcasts in MW/SW = AM.FM = higher quality, short range (VHF ~88–108 MHz).Hence the default broadcast type = AM.Verification / Alternative check:ITU and broadcasting standards confirm AM is the general technique for global reach.Why Other Options Are Wrong:FM: not used generally worldwide, only in specific bands.Both: no, systems are distinct.PCM: digital telephony, not classical broadcast.Common Pitfalls:Mixing local FM stations with global definition of “radio broadcast.”Final Answer:amplitude modulation

Question 19

Non-coherent detection is generally not possible for which type(s) of digital modulation?

Accepted Answer

Introduction / Context:Detection techniques are classified as coherent (requires carrier phase reference) or non-coherent (does not require carrier phase reference). Different modulation schemes have different requirements.Given Data / Assumptions:ASK = Amplitude Shift Keying.FSK = Frequency Shift Keying.PSK = Phase Shift Keying.Concept / Approach:PSK encodes data in the phase of a carrier wave. Phase can only be correctly interpreted if the receiver has a reference to the carrier's phase, requiring coherent detection.Step-by-Step Solution:ASK can use envelope detectors (non-coherent).FSK can use frequency discriminators (non-coherent).PSK cannot be decoded without phase synchronization.Verification / Alternative check:Digital communication textbooks confirm that BPSK and QPSK require coherent detection, while non-coherent detection works for ASK/FSK.Why Other Options Are Wrong:ASK: envelope detection works.FSK: slope/energy detectors allow non-coherent decoding.Both (a) and (c): wrong because only (a) strictly requires coherent detection.Common Pitfalls:Assuming FSK also strictly requires coherent detection.Final Answer:PSK

Question 20

How does the energy content of atmospheric noise vary with frequency?

Accepted Answer

Introduction / Context:Atmospheric noise arises due to lightning discharges and other natural processes. It plays a role in determining radio link design, particularly in HF and VHF bands.Given Data / Assumptions:Noise sources: natural (lightning, cosmic noise).Observed across radio frequencies.Concept / Approach:Atmospheric noise power is inversely related to frequency. At lower frequencies, noise is dominant; at higher frequencies, thermal and man-made noise dominate.Step-by-Step Solution:At LF/MF (~30 kHz–3 MHz): atmospheric noise dominates.At HF/VHF and above: atmospheric noise decreases rapidly with frequency.Thus, energy content decreases as frequency increases.Verification / Alternative check:ITU noise curves (P.372) confirm steep decline of atmospheric noise beyond a few MHz.Why Other Options Are Wrong:Independence or increase with frequency contradicts experimental evidence.Temperature has minor effect, not a deciding factor here.Common Pitfalls:Confusing atmospheric noise with thermal noise (which is flat).Final Answer:decreases as frequency is increased

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