Category: Microwave Engineering Objective Questions

250+ TOP MCQs on Transistor Multipliers and Answers

Microwave Engineering Multiple Choice Questions on “Transistor Multipliers”.

1. Transistor multipliers are more efficient compared to diode multipliers from all operational aspects.
A. True
B. False
Answer: A
Clarification: Transistor multipliers offer better bandwidth and the possibility of conversion efficiency greater than 100% (conversion gain). FET multipliers also require less input and DC power than diode multipliers.

2. A major characteristic property required by frequency multipliers for frequency multiplication to happen is:
A. High gain
B. High conversion efficiency
C. Non linearity
D. None of the mentioned
Answer: C
Clarification: Non linearity property of devices like transistors and diodes is exploited in frequency multipliers. In transistor amplifiers, FET are used since several nonlinearities exist in FET can be used for harmonic generation.

3. If the input power for a frequency doubler is 10.7 mW and the output measured after the frequency doubling process is 21 mW, then the conversion gain for the frequency doubler is:
A. 4.5 dB
B. 8.4 dB
C. 9.8 dB
D. 2.9 dB
Answer: D
Clarification: The conversion gain for a frequency doubler is given by the expression, P2/Pavail. Here P2 is the power measured at the output of the frequency doubler and Pavail is the power input. Substituting the given values in the equation, the conversion gain is 2.9 dB.

4. An ideal _______ produces an output consisting of the sum and difference frequencies of the two input signals.
A. Mixer
B. Amplifier
C. Product modulator
D. Phase modulator
Answer: A
Clarification: A mixer is a three port device that uses a mixer or a time varying element to achieve frequency conversion. An ideal mixer produces an output consisting of the sum and difference frequencies of the two input signals.

5. A mixer consists of a non-linear device that produces various harmonics of the input frequency.
A. True
B. False
Answer: A
Clarification: A non linear device can generate a wide variety of harmonics and other products of the input frequencies. A filter is used to extract only the desired frequency components.

6. A mixer can be used for both up conversion and down conversion at the transmitter and receiver respectively.
A. True
B. False
Answer: A
Clarification: At the transmitter, mixer is used to convert the baseband signal to a broadband signal with use of a high frequency local oscillator. At the receiver, a mixer is used to convert the received broadband signal to baseband signal using a local oscillator. This is called down conversion.

7. A mixer having high conversion loss is said to have very high:
A. Gain
B. Loss
C. Bandwidth
D. None of the mentioned
Answer: B
Clarification: Conversion loss for a mixer is defined as the ratio of available RF input power to the available IF output power. A higher value of conversion loss implies that large amount of power is lost in down converting the frequency from RF to IF range. This makes them less efficient.

8. The IS-54 digital cellular telephone system uses a receive frequency band of 869-894 MHz, with a first IF frequency range of 87 MHz, one possible range of local oscillator frequency is:
A. 956 to 981 MHz
B. 750 to 784 MHz
C. 869 to 894 MHz
D. None of the mentioned
Answer: A
Clarification: The two possible local oscillator frequency range is given by fLO = fRF ± fIF.. fLO is the local oscillator frequency, fRF is the received frequency and fIF is the intermediate frequency range. Substituting the given values in the above equation, one possible frequency range is 956 to 981 MHz.

9. The curve of FET transconductance v/s gate-to-source voltage is a straight line through origin.
A. True
B. False
Answer: B
Clarification: The gate to source voltage of an FET is slowly increased from negative voltage towards zero. As the voltage applied becomes more positive, the transconductance increases up to a certain level and then remains a constant. This property of a FET is used in single-ended FET mixer.

10. RF input matching and RF-LO isolation can be improved through the use of:
A. Balanced mixer
B. Single-ender diode mixer
C. Single ended FET mixer
D. Image reject mixer
Answer: A
Clarification: RF input matching and RF-LO isolation can be improved through the use of balanced mixer. It consists of two single ended mixers combined with a hybrid junction.


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250+ TOP MCQs on Parallel Plate Waveguide and Answers

Microwave Engineering Multiple Choice Questions on “Parallel Plate Waveguide”.

1. The modes of wave propagation that a parallel plate waveguide can support are:
A. TEM, TE, TM modes
B. TM, TE modes
C. TEM, TM modes
D. TEM, TE modes
Answer: A
Clarification: Parallel plate waveguide is the simplest type of waveguide that can support TE and TM modes. It can also support a TEM mode since it is formed from two flat conducting plates.

2. The fringe effect can be neglected in a parallel plate waveguide is because of:
A. The dielectric material used
B. Width of the plates is greater than the distance between the plates
C. Material of the parallel plate waveguide used
D. None of the mentioned
Answer: B
Clarification: The strip width W of the parallel plate waveguide is assumed to be much greater than the separation d, hence the fringe effect or the fringing fields can be neglected.

3. The characteristic impedance of a parallel plate waveguide is given by:
A. η*D/W
B. η*W/D
C. D/ η*W
D. η*√(D/W)
Answer: A
Clarification: Characteristic impedance of a waveguide is the ratio of voltage and current. Defining voltage and current in the integral form of electric field and magnetic field respectively and solving the characteristic impedance is η*D/W. here η is the intrinsic impedance of the medium in the waveguide, D is the distance between the plates and W is the width of the rectangular plate.

4. If the width of a parallel plate waveguide is 30 mm and the distance between the parallel pates is 5 mm, with an intrinsic impedance of 377Ω, then the characteristic impedance of the wave is:
A. 50 Ω
B. 62.833 Ω
C. 100 Ω
D. None of the mentioned
Answer: B
Clarification: The expression for intrinsic impedance of a parallel plate waveguide is η*D/W. substituting the given values of intrinsic impedance and distance between plates and width of the plates, intrinsic impedance is 62.833Ω.

5. In TM mode, if the direction of wave propagation is in ‘z’ direction, then:
A. HZ=0
B. EZ=0
C. EY=0
D. HY=0
Answer: A
Clarification: In TM mode (transverse magnetiC., when the wave propagation is along Z direction, magnetic field is absent in Z direction since the fields are transverse in nature. Hence HZ=0.

6. The wave impedance of a TM mode in a parallel plate waveguide is a:
A. Function of frequency
B. Independent of frequency
C. Proportional to square of frequency
D. Inversely proportional to square of frequency
Answer: A
Clarification: The wave impedance of a parallel plate waveguide in TM mode is β/k which is a function of frequency. The wave impedance is real for f>fC and purely imaginary for fC.

7. In a parallel plate waveguide, for a propagating mode, the value of β is:
A. Real
B. Complex
C. Imaginary
D. Cannot be defined
Answer: A
Clarification: The phase velocity and guide wavelength for a parallel plate waveguide are defined only for propagating modes. Propagating modes are those modes for which β are always positive. Hence β is always real for a parallel plate waveguide.

8. For TM2 mode, if the distance between two parallel plates of a waveguide are 40 mm, then the cut off wavelength for TM2 mode is:
A. 20 mm
B. 80 mm
C. 40 mm
D. 60 mm
Answer: C
Clarification: The cutoff wavelength of a TMn mode in a parallel plate waveguide is 2d/n, where d is the distance between the plates and n signifies the mode of operation. For the given condition, substituting the given values, cut off wavelength is 40 mm.

9. For a parallel waveguide, the dominant mode for TM propagation is:
A. TM0 mode
B. TM1 mode
C. TM2 mode
D. Dominant mode does not exist
Answer: B
Clarification: The mode of propagation for which the cutoff wavelength for wave propagation is maximum is called dominant mode. In TM mode of propagation, TM0 mode is similar to TEM mode of propagation. Hence, TM1 mode is the dominant mode.

10. Phase velocity of the plane waves in the two direction in a parallel plate waveguide is:
A. 1/√(μ∈)secant θ
B. 1/√(μ∈)cosecant θ
C. 1/√(μ∈)tangent θ
D. 1/√(μ∈)cosine θ
Answer: A
Clarification: The phase velocity of each plane wave along its direction of propagation (θ direction) is 1/√(μ∈), Which is the speed of light in the material filling the guide. But, the phase velocity of the plane waves in the z direction is 1/√(μ∈)secant θ.

11. For a parallel plate waveguide, which of the following is true?
A. No real power flow occurs in the ‘z’ direction
B. Power flow occurs in ‘z’ direction
C. No power flow occurs in any direction
D. Wave propagation in z direction is not possible in any mode
Answer: A
Clarification: The superposition of the two plane waves in Z direction is such that complete cancellation occurs at y=0 and y=d, to satisfy the boundary conditions that EZ=0 at these planes. As f decrease to fc, β approaches 0 so that θ approaches 90⁰. The two plane waves are then bouncing up and down, with no motion in +z direction, and no power flow occurs in the z direction.

12. TE mode is characterized by:
A. EZ=0
B. HZ=0
C. Ex=0
D. Ey=0
Answer: A
Clarification: In TE mode of wave propagation, the electric field is in transverse direction and hence electric field component in the direction of wave propagation is 0. Hence, EZ=0.

13. If in a parallel plate waveguide, PL=4 mW/m and Pₒ=10 mW/m, then what is the conduction loss?
A. 0.5
B. 0.4
C. 0.1
D. 0.2
Answer: D
Clarification: Conductor loss of a parallel plate waveguide is given by PL/2Pₒ. Substituting the given values in the above equation, conductor loss is 0.2.

14. If the distance between the two plates of a parallel plate waveguide is 20 mm and is operating in TE₂ mode, then the cut off frequency of TE₂ mode is:
A. 20 MHz
B. 15 GHz
C. 5 GHz
D. None of the mentioned
Answer: B
Clarification: The cutoff frequency for TEn mode is n/2d√(∈μ) for a parallel plate waveguide. Substituting the given values, the cutoff frequency is 15 GHz.

15. The wave impedance for a non-propagating mode in TE mode is:
A. K/β
B. Imaginary
C. Zero
D. Non-existing
Answer: B
Clarification: Wave impedance of a parallel plate waveguide for TEN modes is k/β. This expression is valid and real only for propagating modes. For non propagating modes, impedance becomes imaginary.


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250+ TOP MCQs on Binomial Multi-section Matching Transformers and Answers

Microwave Engineering Multiple Choice Questions on “Binomial Multi-section Matching Transformers”.

1. The passband response of a binomial matching transformer can be called optimum:
A. if the roll off in the response curve is high
B. if the response is flat in the impedance matched region
C. if the matching network is frequency sensitive
D. none of the mentioned
Answer: B
Clarification: The response curve of a binomial matching transformer ( θ v/s │Г (θ) │) must be flat at the frequency for which impedance matching is performed and for those frequencies that lie in the required bandwidth. This is one of the most important characteristic of a good matching circuit.

2. If a quality binomial matching transformer gives a good flat response curve, it is called “maximally flat”.
A. true
B. false
Answer: A
Clarification: A binomial matching section can be termed efficient when it is less frequency sensitive and gives a constant gain over a wide range of frequencies. This constant gain implies a flat curve over a wide range of frequencies. This is termed as “maximally flat”.

3. The response curve of a binomial matching transformer is plotted for each section of the matching network individually and then analyzed for optimum solution.
A. true
B. false
Answer: B
Clarification: The response curve of a binomial multisection transformer is determined for an N-section transformer by setting the first N-1 derivatives of │Г (θ) │ to zero at the center frequency, fₒ.

4. To obtain a flat curve in the response of a binomial multisection transformer, N-1 derivatives of │Г (θ) │are set to zero. This implies:
A. the frequency sensitivity of the matching section is increased linearly
B. the frequency sensitivity of the matching section is increased exponentially
C. roll off in the curve is increased
D. none of the mentioned
Answer: D
Clarification: The derivatives of │Г (θ) │ show the rate of change of reflection co-efficient with distance. If this derivative is not zero, the matching circuit becomes more sensitive and a higher bandwidth cannot be obtained. Hence to make the matching network frequency independent, the derivatives are set to zero.

5. The condition │Г (θ) │=0 for θ=π/2 of a binomial multi section transformer corresponds to the:
A. upper cutoff frequency
B. lower cutoff frequency
C. center frequency
D. none of the mentioned
Answer: C
Clarification: θ=π/2 corresponds to the center frequency at which │Г (θ) │=0. θ=βl. β=2 π/λ and l=λ/4. Substituting for β and λ in the equation for θ, θ=π/2. This is the center frequency at which impedance matching is done at which the reflection coefficient is zero and perfect match is achieved.

6. The reflection co-efficient magnitude of a binomial multisection transformer is:
A. 2N│A││cos (θ)│N
B. 2N│A│
C. 2N│cos (θ) │N
D. none of the mentioned
Answer: A
Clarification: The reflection co-efficient of a binomial multisection transformer is dependent on the length of the matching section, operating frequency and load impedance and characteristic impedance. A is a constant defined as A=2-N (ZL– Z0)/ (ZL+ Z0).

7. The reflection coefficient ГN in terms of successive impedances Zn and Zn+1 when multisection transformers are used in a binomial matching transformer is given by:
A. 0.5ln (Zn+1/Zn)
B. ln (Zn+1/Zn)
C. 0.5ln (Zn/Zn+1)
D. (Zn/Zn+1)
Answer: A
Clarification: After binomial expansion of the equation for Г(θ), the maximum power is N, where N is the number of the sections in the transformer. After making suitable approximations so that the approximated values are in well agreement with actual values, the expression for reflection coefficient is 0.5ln (Zn+1/Zn).

8. In the plot of normalized frequency v/s reflection co-efficient for a binomial multisection filter, the curve has a dip at:
A. center frequency
B. upper cutoff frequency
C. lower cutoff frequency
D. none of the mentioned
Answer: A
Clarification: Since the impedance matching circuit is used to match the load to the transmission line, there will be perfect match in the circuit resulting in zero or low reflection. Hence, there is a dip at the center frequency.

9. As the number of sections in the binomial multisection transformer increases the plot of normalized frequency v/s reflection co-efficient has a wider open curve.
A. true
B. false
Answer: A
Clarification: When more number of sections are used for matching, the reflection co-efficient is low for neighboring frequencies as well. Hence, the network can be used for a wide range of operating frequencies. Hence, this increases the bandwidth.

10. A three section binomial transformer is used to match a 100Ω transmission line to a 50Ω transmission line. Then the value of the constant ‘A’ for this design is:
A. -0.0433
B. 0.0433
C. 0.01
D. -0.01
Answer: A
Clarification: ‘A’ is given by the expression 2-(n+1)ln (ZL/Z0), Where N is the number of sections in the matching network. Substituting the given values in the equation for ‘A’, the value of A is -0.0433.


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250+ TOP MCQs on Other Couplers and Answers

Microwave Engineering Quiz on “Other couplers”.

1. Moreno crossed-guide coupler is a waveguide directional coupler consists of four waveguides at right angle.
A. True
B. False
Answer: B
Clarification: Moreno crossed-guide coupler consists of two waveguides at right angles, with coupling provided by two apertures in the common broad wall of the guides.

2. Schwinger reversed phase coupler is a waveguide coupler designed so that the path lengths for the two coupling apertures are the same for_________
A. Coupled port
B. Uncoupled port
C. Back port
D. Isolated port
Answer: B
Clarification: Schwinger reversed phase coupler is a waveguide coupler designed so that the path lengths for the two coupling apertures are the same for uncoupled port so that the directivity is essentially independent of frequency.

3. The in phase combining of power at the coupled port is achieved by means of a _______
A. Matching network
B. A small slot
C. Quarter Wave transformer
D. None of the mentioned
Answer: B
Clarification: The λg/4 slot spacing leads to in-phase combining at the coupled port, but this coupling is very frequency sensitive. This is the opposite situation from that of the multi hole waveguide coupler.

4. Ribblet short-slot coupler consists of two waveguides that are separated by a distance “d”.
A. True
B. False
Answer: B
Clarification: This coupler consists two waveguides with a common side wall. Coupling takes place in the region where part of the common wall has been removed.

5. Riblet-short slot coupler allows only:
A. TE10 mode of propagation
B. TE20 mode of propagation
C. Both a and b
D. None of the mentioned
Answer: C
Clarification: In the part where common wall has been removed in this coupler, both the TE10 and TE20 mode are excited, and by proper design can be made to cause cancellation at the isolated port and addition at the coupled port.

6. Symmetric tapered coupled line couplers offer higher bandwidth when compared to other forms of couplers.
A. True
B. False
Answer: A
Clarification: Multisection coupled line coupler can be extended to a continuous taper, yielding couple line couplers with good bandwidth characteristics.

7. The coupling and directivity of couplers with apertures in planar lines can be adjusted as per the requirement of the application.
A. True
B. False
Answer: A
Clarification: In symmetric tapered coupled line coupler both the conductor width and separation between them can be adjusted to provide a synthesized coupling or directivity response. This can be tested with computer optimization of a stepped-section approximation to a continuous taper.

8. _________ is a key component in the scalar or vector network analyzer.
A. Reflectometer
B. Radiometer
C. Frequency meter
D. None of the mentioned
Answer: A
Clarification: Reflectometer is a circuit that uses a directional coupler to isolate and sample the incident and reflected powers from the load. This is one of the key components in a scalar or vector network analyzer.

9. Reflectometer can also be used as a frequency meter.
A. True
B. False
Answer: A
Clarification: Reflectometers can be used to measure the scattering parameters of a two port network. It is also used as SWR meter and also as power monitor in system applications.

10. A basic Reflectometer circuit can be used to measure the _____________ magnitude of the unknown load.
A. Reflection coefficient
D. Standing wave ratio
C. Transmission coefficient
D. None of the mentioned
Answer: A
Clarification: A basic Reflectometer circuit can be used to measure the reflection coefficient magnitude of the unknown load. This is one of the major applications of Reflectometer. If the reflection coefficient is measured, the unknown load can be easily computed.


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250+ TOP MCQs on Metal Oxide Semiconductor FET and Answers

Microwave Engineering Multiple Choice Questions on “Metal Oxide Semiconductor FET”.

1. There exists no difference between the construction of GaAs MESFET and silicon MOSFET except for the material used in their construction.
A. True
B. False
Answer: B
Clarification: There exists a difference between the construction of MESFET and MOSFET. There is a thin insulating layer of silicon dioxide between the gate contact and the channel region. Because the gate is insulated, it does not conduct DC bias current.

2. MOSFETs can provide a power of several hundred watts when the devices are packaged in:
A. Series
B. Parallel
C. Diagonal
D. None of the mentioned
Answer: B
Clarification: MOSFETs can be used at frequencies into the UHF range and can provide powers of several hundred watts when devices are packaged in parallel. Laterally diffused MOSFETs have direct grounding of the source and can operate at low microwave frequencies with high power.

3. High electron mobility transistors can be constructed with the use of single semiconductor material like GaAs that have high electron mobility.
A. True
B. False
Answer: B
Clarification: High electron mobility transistor is a hetero junction FET, meaning that it does not use a single semiconductor material, but instead is constructed with several layers of compound semiconductor materials.

4. The curve of IDS v/s VDS of an FET does not vary with the gate to source voltage applied.
A. True
B. False
Answer: B
Clarification: Curve of IDS v/s VDS of an FET varies with the gate to source voltage applied. As the gate to source voltage applied becomes more positive, the drain to source current goes on increasing for an applied constant gate to source voltage.

5. High-power circuits generally use higher values of:
A. Gate to source current
B. Drain to source current
C. Drain current
C. Gate to source voltage
Answer: C
Clarification: In order to achieve high drain current for high power applications, DC bias voltage must be applied to both gate and the drain, without disturbing the RF signal paths.

6. High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply.
A. True
B. False
Answer: A
Clarification: High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply. The RF chokes provide a very low DC resistance for biasing, and very high impedance at RF frequencies to isolate the signal from the bias supply.

7. Since multiple layers of semiconductor materials is used in high electron mobility transistors, this results in:
A. High gain
B. Power loss
C. Temperature sensitivity
D. Thermal stress
Answer: D
Clarification: The multiple layers in the high electron mobility transistor result in the thermal and mechanical stress in the layers. To avoid this, the layers usually have matched crystal lattice.

8. A major disadvantage of high electron mobility transistor is that:
A. They have low gain
B. High manufacturing cost
C. Temperature sensitive
D. High driving voltage is required
Answer: B
Clarification: High electron mobility transistors are devices containing multiple layers of different semiconductor materials. This complicated structure of HEMT requires sophisticated fabrication techniques leading to relatively high cost.

9. HEMT fabricated using GaN and aluminum gallium nitride on a silicon substrate can be used in :
A. High power transmitters
B. High power receivers
C. RADAR
D. Smart antennas
Answer: A
Clarification: GaN HEMT operate with drain voltages in the range of 20-40 V and can deliver power up to 100 W at frequencies in the low microwave range, making these devices popular for high power transmitters.

10. The scattering parameter S11 for GaN HELMT increases with increase in frequency of operation
A. True
B. False
Answer: B
Clarification: For GaN, the S11 parameter of the amplifier decreases with increase in frequency of operation. Experimental results have shown that S11 parameter was 0.96 at 0.5 GHz of frequency and 0.88 at 4 GHz of frequency.


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250+ TOP MCQs on System Aspects of Antennas and Answers

Microwave Engineering Multiple Choice Questions on “System Aspects of Antennas”.

1. If an antenna has a directivity of 16 and radiation efficiency of 0.9, then the gain of the antenna is:
A. 16.2
B. 14.8
C. 12.5
D. 19.3
Answer: A
Clarification: Gain of an antenna is given by the product of radiation efficiency of the antenna and the directivity of the antenna. Product of directivity and efficiency thus gives the gain of the antenna to be 16.2.

2. Gain of an antenna is always greater than the directivity of the antenna.
A. True
B. False
Answer: B
Clarification: Gain of an antenna is always smaller than the directivity of an antenna. Gain is given by the product of directivity and radiation efficiency. Radiation efficiency can never be greater than one. So gain is always less than or equal to directivity.

3. A rectangular horn antenna has an aperture area of 3λ × 2λ. Then the maximum directivity that can be achieved by this rectangular horn antenna is:
A. 24 dB
B. 4 dB
C. 19 dB
D. Insufficient data
Answer: C
Clarification: Given the aperture dimensions of an antenna, the maximum directivity that can be achieved is 4π A/λ2, where A is the aperture area and λ is the operating wavelength. Substituting the given values in the above equation, the maximum directivity achieved is 19 dB.

4. A rectangular horn antenna has an aperture area of 3λ × 2λ. If the aperture efficiency of an antenna is 90%, then the directivity of the antenna is:
A. 19 dB
B. 17.1 dB
C. 13 dB
D. 21.1 dB
Answer: B
Clarification: Given the aperture dimensions of an antenna, the directivity that can be achieved is ap4π A/λ2, where A is the aperture area and λ is the operating wavelength, ap is the aperture efficiency. Substituting the given values in the above equation, the directivity achieved is 17.1 dB.

5. If an antenna has a directivity of 16 and is operating at a wavelength of λ, then the maximum effective aperture efficiency is:
A. 1.27λ2
B. 2.56λ2
C. 0.87λ2
D. None of the mentioned
Answer: A
Clarification: Maximum effective aperture efficiency of an antenna is given by D λ2/4π, D is the directivity of the antenna. Substituting in the equation the given values, the maximum effective aperture is 1.27λ2.

6. A resistor is operated at a temperature of 300 K, with a system bandwidth of 1 MHz then the noise power produced by the resistor is:
A. 3.13×10-23 watts
B. 4.14×10-15 watts
C. 6.14×10-15 watts
D. None of the mentioned
Answer: B
Clarification: For a resistor noise power produced is given by kTB, where T is the system temperature and B is the bandwidth. Substituting in the above expression, the noise power produced is 4.14×10-15 watts.

7. With an increase in operating frequency, the background noise temperature:
A. Increases
B. Decreases
C. Remains constant
D. Remains unaffected
Answer: A
Clarification: The plot of frequency v/s background noise temperature shows that with the increase of the signal frequency, the background noise temperature increases. Also, with the increase of the elevation angle from the horizon, background noise temperature increases.

8. The noise temperature of an antenna is given by the expression:
A. radTb + (1-raD. Tp
B. (1-raD. TP
C. radTb
D. None of the mentioned
Answer: A
Clarification: The noise temperature of an antenna is given by the expression radTb + (1-raD. Tp. here, Tb is the brightness temperature and Tp is the physical temperature of the system. rad is the radiation efficiency. Noise temperature of a system depends on these factors.

9. Low is the G/T ratio of an antenna, higher is its efficiency.
A. True
B. False
Answer: B
Clarification: In the G/T ratio of an antenna, G is the gain of an antenna and T is the antenna noise temperature. Higher the G/T ratio of an antenna better is the performance of the antenna.

10._________ has a constant power spectral density.
A. White noise
B. Gaussian noise
C. Thermal noise
D. Shot noise
Answer: A
Clarification: Thermal noise has a power spectral density for a wide range of frequencies. Its plot of frequency v/s noise power is a straight line parallel to Y axis.


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