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Past Questions — 2015
A rectangular waveguide has width a. Then the cutoff frequency (with corresponding wavelength λ) is
A)
a= λ/2
B)
a= λ/C
C)
a= λ/4
D)
a=2λ
For a rectangular waveguide, the cutoff condition for the dominant mode (TE10) occurs when the guide width equals half the cutoff wavelength. Hence, a = λ/2. The other options do not match the standard cutoff relation for rectangular waveguides.
Satellite communication uses
A)
3 to 6 GHz
B)
3 to 10 MHz
C)
10 to 50 GHz
D)
100 to 200 GHz
Satellite communication commonly uses microwave frequencies, especially the C-band around 3.7 to 4.2 GHz for downlink and 5.925 to 6.425 GHz for uplink. Hence the general range given as 3 to 6 GHz is correct. The other options are not suitable: 3 to 10 MHz is far too low for satellite links, while 10 to 50 GHz and 100 to 200 GHz are much higher bands and not the standard basic range asked in such exam questions.
The rate at which information can be carried through a communication channel depends on
A)
carrier frequency
B)
bandwidth
C)
transmission loss
D)
transmitted power
The information-carrying capacity of a communication channel is primarily determined by its bandwidth. According to Shannon’s channel capacity concept, a wider bandwidth allows more information to be transmitted per unit time. Carrier frequency does not directly decide capacity, while transmission loss and transmitted power affect signal quality and range, not the basic channel capacity in the same direct way as bandwidth.
If a line is terminated in characteristic impedance Z0; the input impedance measured at the input will be
A)
Z0
B)
Z0 /2
C)
zero
D)
infinite
For a transmission line terminated in its characteristic impedance Z0, there is no reflection at the load. Hence the input impedance seen at the sending end equals the characteristic impedance itself, i.e., Z_in = Z0. Therefore, the correct option is (a). The other options are incorrect because Z0/2, zero, and infinite input impedance occur only under special mismatched or open/short-circuited conditions, not when the line is properly matched.
The characteristic impedance of a cable is about
A)
2 Ω
B)
5 Ω
C)
50 Ω
D)
300 Ω
The characteristic impedance of a cable is typically around 50 Ω for standard transmission/coaxial cables. This is a widely used nominal value in electrical and communication engineering. 2 Ω and 5 Ω are far too low, while 300 Ω is more typical of certain twin-lead transmission lines, not the usual cable characteristic impedance asked here.
The velocity of sound waves in air
A)
is constant at all temperatures
B)
varies directly as temperature
C)
varies inversely as absolute temperature
D)
varies directly as square root of absolute temperature
The speed of sound in air is given by v = \u221a(\u03b3RT/M), so it is proportional to the square root of the absolute temperature (in Kelvin). Hence, as temperature increases, sound velocity increases as \u221aT. Option (a) is wrong because velocity changes with temperature, (b) is wrong because it is not directly proportional to T, and (c) is wrong because it does not vary inversely with temperature.
In the 8085 microprocessor the RST6 instruction transfers the program execution to the following locations
A)
30H
B)
24H
C)
48H
D)
60H
In the 8085 microprocessor, the RST instructions are restart instructions that transfer control to fixed memory locations. The vector address is calculated as 8 times the RST number. For RST 6, the address is 6 × 8 = 48H. Hence, execution jumps to 48H. The other options correspond to different RST vectors: RST 3 = 18H, RST 4 = 20H, RST 5 = 28H, and RST 7 = 38H.
The number of pins in microprocessor 8086 is
A)
30
B)
40
C)
50
D)
60
The Intel 8086 microprocessor is a 16-bit microprocessor packaged in a 40-pin DIP (Dual In-line Package). Therefore, the number of pins is 40. The other options are incorrect because 30, 50, and 60 are not the standard pin count of the 8086.
The number of interrupt lines in 8085 microprocessor is
A)
3
B)
4
C)
5
D)
7
The 8085 microprocessor has 5 interrupt lines: TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR. Hence the correct answer is 5. Options 3, 4, and 7 are incorrect because they do not match the standard interrupt structure of the 8085.
For AM receivers the standard IF frequency is
A)
106 KHz
B)
455 KHz
C)
1.07 MHz
D)
10.7 MHz
In standard AM broadcast receivers, the intermediate frequency (IF) is conventionally 455 kHz. This value is widely used because it provides a good compromise between selectivity, image-frequency rejection, and ease of filter design. Option (d) 10.7 MHz is the standard IF for FM receivers, not AM. Option (c) 1.07 MHz is too high for standard AM IF, and (a) 106 kHz is not the usual standard.
In single tone AM modulation, the transmission efficiency for m=1 is
A)
33.3%
B)
50%
C)
100%
D)
16.65%
For single-tone AM, transmission efficiency is the ratio of sideband power to total transmitted power: η = m^2 / (2 + m^2). For m = 1, η = 1 / (2 + 1) = 1/3 = 33.3%. Hence the correct option is (a). Option (b) 50% is not correct for standard AM, (c) 100% is impossible in conventional AM, and (d) 16.65% is too low.
Give the principle of super-hydrodyne receiver.
The positive peak of an AM wave is 16V and the minimum value is 4V. Assuming single tone modulation, the modulation index is
A)
0.6
B)
0.25
C)
4
D)
0.36
For an AM wave, the modulation index is m = (Vmax - Vmin) / (Vmax + Vmin). Here Vmax = 16 V and Vmin = 4 V. So, m = (16 - 4) / (16 + 4) = 12 / 20 = 0.6. Hence, the correct option is (a).
Narrowband FM signal can be considered to be equal to
A)
PM
B)
AM
C)
SSB
D)
DSB
For narrowband FM, the modulation index is very small, so the FM signal can be approximated by a carrier plus two sidebands. This makes it equivalent to a double sideband (DSB) signal with a carrier component. It is not exactly AM, PM, or SSB. Hence, the correct answer is DSB.
In the FM wave described by equation v = 15 sin (4 × 10^8 t + 3 sin 1100 t), the maximum frequency deviation is
A)
175 Hz
B)
525 Hz
C)
58.33 Hz
D)
3 Hz
For an FM wave of the form v = A sin(ωc t + β sin ωm t), the coefficient of sin ωm t is the modulation index β. Here, β = 3 and ωm = 1100 rad/s. The maximum frequency deviation is Δf = β fm, where fm = ωm / (2π) = 1100 / (2π) ≈ 175 Hz. So Δf = 3 × 175 = 525 Hz. Hence, the correct answer is 525 Hz.
INTELSAT stands for
A)
International Telecommunications Satellite
B)
India Telecommunications Satellite
C)
Inter Telecommunications Satellite
D)
None of these
INTELSAT is the abbreviation for International Telecommunications Satellite. It was the global satellite organization used for international communications. The other options are incorrect because INTELSAT does not refer to India Telecommunications Satellite or Inter Telecommunications Satellite.
A waveguide section in a microwave circuit acts as
A)
LP filter
B)
Bandpass filter
C)
HP filter
D)
Bandstop filter
A waveguide has a cutoff frequency: signals below cutoff cannot propagate, while those above cutoff can. This is the behavior of a high-pass filter. Therefore, a waveguide section in a microwave circuit acts as an HP filter. LP filter, bandpass filter, and bandstop filter do not match the basic cutoff nature of a waveguide.
Explain the phenomenon of differential negative resistance in Gunn diode.
What do you mean by modulation? Why do we need to modulate a signal?
What is a Magic Tee? Obtain its Scattering matrix.
An audio signal of FM radio station is sampled at a rate of 15% above the Nyquist rate. Signal is sampled in 512 levels and bandwidth of the signal is 4 KHz. What will be the minimum bandwidth required to transmit this signal?
To separate channels in FDM receivers we have to use
A)
integration
B)
AND gate
C)
band pass filter
D)
differentiation
In Frequency Division Multiplexing (FDM), each channel is assigned a different frequency band. At the receiver, the channels are separated by selecting the desired frequency band using a band pass filter. Integration, AND gate, and differentiation are not used for separating FDM channels.
What is the numerical aperture in an optical fiber when critical angle is 30°?
A)
0.5
B)
0.707
C)
0.866
D)
0.2
For an optical fiber, numerical aperture (NA) is related to the critical angle c by NA = cos(c) when the outside medium is air. Here, c = 30°, so NA = cos 30° = √3/2 ≈ 0.866. Therefore, the correct option is (c). The other options do not match this standard relation.
The decimal equivalent of binary 1101.11 is
A)
9.75
B)
13.75
C)
11.50
D)
11.75
Convert binary 1101.11 to decimal:
- Integer part 1101 = 1×8 + 1×4 + 0×2 + 1×1 = 13
- Fractional part .11 = 1×1/2 + 1×1/4 = 0.5 + 0.25 = 0.75
So, total = 13 + 0.75 = 13.75.
Hence, the correct option is (b).
In 8085 stack pointer is
A)
4 bit register
B)
8 bit register
C)
16 bit register
D)
32 bit register
In the 8085 microprocessor, the stack pointer (SP) holds the address of the top of the stack in memory. Since memory addresses in 8085 are 16-bit, the stack pointer must also be a 16-bit register. Therefore, the correct answer is 16 bit register. The other options are incorrect because 4-bit and 8-bit registers cannot address the full memory space, and 32-bit is not used in 8085.