MES Jr. Grade • 2015 • Paper-III
If word length is 16 bits the decimal equivalent of largest number which can be stored is
A)
32767
B)
32766
C)
16383
D)
16384
For a 16-bit signed word, 1 bit is used for the sign and 15 bits for magnitude. The largest positive number is therefore 2^15 - 1 = 32767. Hence option (a) is correct. The other options are smaller values and do not represent the maximum storable number for a 16-bit signed word.
MES Jr. Grade • 2015 • Paper-III
What is the need for altitude control for a satellite? Name two altitude control components in satellite communication.
MES Jr. Grade • 2015 • Paper-III
A matched load is one in which
A)
reflection is zero
B)
reflection is unity
C)
partial reflection
D)
none of these
A matched load means the load impedance equals the characteristic impedance of the transmission line. In this condition, no signal is reflected back from the load, so the reflection coefficient is zero. Hence, option (a) is correct. Option (b) is wrong because unity reflection occurs in an open or short circuit, not a matched load. Option (c) is wrong because partial reflection happens only when the load is mismatched.
MES Jr. Grade • 2015 • Paper-III
Which signaling scheme is most affected by noise?
A)
ASK
B)
FSK
C)
PSK
D)
QAM
Amplitude Shift Keying (ASK) is most affected by noise because its information is carried in the signal amplitude. Noise commonly disturbs amplitude, so ASK is more vulnerable than FSK and PSK. QAM also uses amplitude, but in basic digital modulation comparisons, ASK is considered the most noise-sensitive scheme.
MES Jr. Grade • 2015 • Paper-III
Define antenna beam width and bandwidth. Differentiate clearly between the two.
MES Jr. Grade • 2015 • Paper-III
A duplexer is used to
A)
Couple two antennas to a transmitter
B)
Isolate the antenna from the local oscillator
C)
Prevent interference between two antennas connected to receiver
D)
Use an antenna for reception or transmission without interference
A duplexer is a microwave device that allows a single antenna to be used for both transmitting and receiving, while isolating the transmitter and receiver so they do not interfere with each other. Option (a) is incorrect because a duplexer does not couple two antennas to a transmitter. Option (b) is incorrect because isolation from the local oscillator is not the main function of a duplexer. Option (c) is incorrect because it is not specifically about two antennas connected to a receiver. Hence, (d) is correct.
MES Jr. Grade • 2015 • Paper-III
In a Klystron amplifier the input cavity is called
A)
buncher
B)
catcher
C)
pierce gun
D)
collector
In a klystron amplifier, the input cavity is the **buncher cavity**. It modulates the velocity of the electron beam, causing electrons to bunch together. The output cavity is called the **catcher cavity**, which extracts energy from the bunched beam. A **pierce gun** is the electron gun that produces the beam, and the **collector** collects the spent electrons.
MES Jr. Grade • 2015 • Paper-III
Which of the following is analog in nature?
A)
PCM
B)
PWM
C)
delta modulation
D)
differential PCM
PWM (Pulse Width Modulation) is analog in nature because the information is represented by varying the width of pulses continuously according to the message signal. In contrast, PCM, delta modulation, and differential PCM are all digital pulse modulation techniques, since they encode the signal into discrete levels or binary form.
MES Jr. Grade • 2015 • Paper-III
Impedance inversion may be obtained with
A)
A short circuited stub
B)
A quarter wave line
C)
An open circuited stub
D)
A half wave line
Impedance inversion is a property of a quarter-wave transmission line. For a line of length \(\lambda/4\), the input impedance becomes \(Z_{in}=\frac{Z_0^2}{Z_L}\), so a high load impedance appears as low at the input and vice versa. This is called impedance inversion. A short-circuited or open-circuited stub can provide specific reactances, and a half-wave line repeats the load impedance, not invert it.
MES Jr. Grade • 2015 • Paper-III
A sinusoidal voltage of 2 kV peak value is amplitude modulated to give 20% modulation. The peak value of each side band term is
A)
800V
B)
400V
C)
200V
D)
100V
For single-tone AM, the peak value of each sideband component is (mVc)/2, where m is the modulation index and Vc is the carrier peak value. Here, Vc = 2 kV = 2000 V and m = 20% = 0.2. So each sideband peak = (0.2 × 2000)/2 = 200 V. Hence, the correct option is (c).
MES Jr. Grade • 2015 • Paper-III
Most electromagnetic transmission in rectangular waveguide uses
A)
TE10 mode
B)
TE01 mode
C)
TE20 mode
D)
TE11 mode
In a rectangular waveguide, the dominant mode is TE10. It has the lowest cutoff frequency, so it is the mode most commonly used for electromagnetic transmission. TE01 and TE20 have higher cutoff frequencies, and TE11 is not the dominant mode in a rectangular waveguide.
MES Jr. Grade • 2015 • Paper-III
Which principle of operation of cavity wave meters is used in microwave networks?
A)
phase shift
B)
resonance
C)
polarisation shift
D)
gyration
Cavity wave meters work on the principle of resonance. A cavity resonator is coupled to the microwave network, and when its resonant frequency matches the signal frequency, a strong response is observed. This makes it useful for measuring microwave frequency. The other options are not the operating principle of cavity wave meters: phase shift is related to transmission effects, polarisation shift is unrelated, and gyration refers to ferrite devices like gyrators/circulators.
MES Jr. Grade • 2015 • Paper-III
The highest priority of interrupt in 8085 microprocessor system is
A)
RST 7.5
B)
RST 6.5
C)
INTR
D)
TRAP
In the 8085 microprocessor, TRAP has the highest interrupt priority. It is a non-maskable, vectored interrupt, so it cannot be disabled by the interrupt enable/disable instructions. The priority order is: TRAP > RST 7.5 > RST 6.5 > RST 5.5 > INTR. Hence, among the given options, TRAP is correct. INTR has the lowest priority among these interrupts.
MES Jr. Grade • 2015 • Paper-III
The beam of a radar should be
A)
Very narrow
B)
Very wide
C)
Neither too narrow nor too wide
D)
Either (b) or (c)
A radar beam should be very narrow so that it can focus energy in a specific direction, improve angular resolution, and accurately detect the target’s direction. A wide beam would spread energy over a larger area, reducing directionality and precision. Hence, the correct choice is (a) Very narrow.
MES Jr. Grade • 2015 • Paper-III
Show that TEM waves cannot propagate through a hollow waveguide.
MES Jr. Grade • 2015 • Paper-III
What is difference between Microprocessor and Microcontroller?
MES Jr. Grade • 2015 • Paper-III
Why is refractive index of the core more than the cladding in optical fibers?
MES Jr. Grade • 2015 • Paper-III
For accurate clock period in 8085 it is preferable to use
A)
Crystal
B)
Crystal of LC resonant circuit
C)
Crystal of RC circuit
D)
LC or RC circuit
In the 8085 microprocessor, the most accurate and stable clock period is obtained using a crystal oscillator. A crystal provides a very precise frequency with minimal drift due to temperature or supply variations. LC and RC circuits are less stable and less accurate, so they are not preferred for accurate clock generation.
MES Jr. Grade • 2015 • Paper-III
What are TE and TM modes? Why is TE called the dominant mode in rectangular waveguide?
MES Jr. Grade • 2015 • Paper-III
Describe the flag registers in 8085 microprocessor.
MES Jr. Grade • 2015 • Paper-III
In a klystron amplifier the force on all electrons
A)
Causes acceleration of all electrons
B)
Causes retardation of all electrons
C)
Causes acceleration of some electrons and retardation of others
D)
None of the above
In a klystron amplifier, the RF field in the buncher cavity interacts with the electron beam and produces velocity modulation. Depending on the phase of the RF field at the instant an electron passes through, some electrons are accelerated while others are retarded. This creates electron bunching, which is essential for amplification. Hence, the force does not act the same way on all electrons.
MES Jr. Grade • 2015 • Paper-III
Which of the following is taken as reference antenna for directive gain?
A)
½ wave dipole
B)
elementary doublet
C)
isotropic antenna
D)
infinitesimal dipole
Directive gain is defined with respect to a reference antenna that radiates equally in all directions, i.e., an isotropic antenna. Hence, the directive gain compares the radiation intensity of the given antenna to that of an isotropic radiator. A half-wave dipole is commonly used as a reference for gain in some contexts, but for directive gain the standard reference is the isotropic antenna. Therefore, option (c) is correct.
MES Jr. Grade • 2015 • Paper-III
Which of the following parameters is negligible in transmission lines?
A)
R
B)
L
C)
C
D)
G
In transmission line modeling, the shunt conductance G represents leakage current through the insulation between conductors. For overhead transmission lines, this leakage is extremely small, so G is usually neglected. Resistance R, inductance L, and capacitance C are all significant and are included in line analysis.
MES Jr. Grade • 2015 • Paper-III
In a PCM system, if we increase the quantization levels from 2 to 8, how do the relative bandwidth requirements vary?
A)
Get doubled
B)
Get tripled
C)
Remain same
D)
Become eight times
In PCM, the number of bits per sample is n = log2(L), where L is the number of quantization levels. Increasing levels from 2 to 8 changes n from 1 bit to 3 bits. Since PCM bit rate and hence required bandwidth are proportional to the number of bits per sample, the bandwidth becomes 3 times. Therefore, the relative bandwidth requirement gets tripled. Options (a), (c), and (d) are incorrect because the increase is not by a factor of 2, unchanged, or 8 in this case.
MES Jr. Grade • 2015 • Paper-III
A 400 W carrier is modulated to a depth of 75 percent. The total power in modulated wave will be
A)
38.5 W
B)
400 W
C)
512.5 W
D)
615.5 W
For standard AM, total transmitted power is
P_t = P_c \left(1 + \frac{m^2}{2}\right),
where P_c is carrier power and m is modulation index.
Given:
P_c = 400 W, m = 75% = 0.75
So,
P_t = 400 \left(1 + \frac{0.75^2}{2}\right)
= 400 \left(1 + \frac{0.5625}{2}\right)
= 400 (1 + 0.28125)
= 400 \times 1.28125
= 512.5 W.
Hence, the correct answer is 512.5 W. The other options do not match the AM power formula.