ISRO 2011 EC SET A
1. In an amplitude modulated system if the total power is 600 W and the power in the carrier is 400W, the modulation index is
a. 0.5
b. 0.75
c. 0.9
d. 1
Solution:
PSide band = PTotal
– Pcarrier
PSide
band = 600 - 400 = 200 W
Solving we get µ = 1
Answer: Option d
2. The channel capacity under the Gaussian noise environment for a discrete memoryless channel with a bandwidth of 4 MHz and SNR 31 is
a. 20 Mbps
b. 4 Mbps
c. 8 Kbps
d. 4 Kbps
Solution:
Channel capacity = C=B log2(1+
SNR)
= 4 M log2(1+31)
= 4 M x 5 = 20 Mbps
Answer: Option: a
3. In satellite communication, frequency modulation is used because satellite channel has
a. High Modulation index
b. Small bandwith and negligible noise
c. Large bandwidth and severe noise
d. Maximum Bandwidth and Minimum noise
Answer: Option d
4. For a 3-µm diameter optical fiber with core and cladding indexes of refraction of 1.545 and 1.510 respectively. The cut off wavelength is
a. 2.3 µm
b. 1.29 µm
c. 1.5 µm
d. 3.24 µm
Solution:
= 1.2808 µm
Answer: option b
Solution:
Answer: Option b
A Process contains only sequential statements.
Answer : option b
212 = 4096 bytes = 4Kbytes
Answer: Option d
Aswer: option d
Thread: the primitive that can execute code. It contains an Instruction pointer and sometimes has its own stack.
Process: An environment in which one or several threads run.
Kernel: Run time OS- takes care of thread scheduling.
Answer: Option d
The ouput toggles when J=K=1.
Answer: Option d
Answer: Option d
Answer: Option c
Answer: Option c
Answer: Option b
Answer: Option c
Answer: Option a
Answer: Option b
Answer: Option c
Answer: Option a
Answer: Option a
Answer: Option d
Answer: Option b
Answer: Option a
Answer: Option c
Answer: Option a
Answer: Option b
Answer: Option b
Answer: Option b
Answer: Option d
Answer: Option a
Answer: Option d
Answer: Option a
Answer: Option c
Answer: Option b
Answer: Option b
Answer: Option b
Answer: Option c
Answer: Option b
Answer: Option c
Answer: Option a
Answer: Option b
Answer: Option a
Answer: Option c
Answer: Option c
Answer: Option d
Answer: Option b
3. In satellite communication, frequency modulation is used because satellite channel has
a. High Modulation index
b. Small bandwith and negligible noise
c. Large bandwidth and severe noise
d. Maximum Bandwidth and Minimum noise
Answer: Option d
4. For a 3-µm diameter optical fiber with core and cladding indexes of refraction of 1.545 and 1.510 respectively. The cut off wavelength is
a. 2.3 µm
b. 1.29 µm
c. 1.5 µm
d. 3.24 µm
Solution:
= 1.2808 µm
Answer: option b
5. A 12 bit ADC is
operating with a 1 µs clock period and total conversion time is seen to be 14 µs
always. The ADC must be of the type
a) Flash
b) Counting type
c)
Integrating type
d) Successive Approximation type
Answer: Option b
A Process contains only sequential statements.
Answer : option b
212 = 4096 bytes = 4Kbytes
Answer: option d
A resistance in parallel with an indeal current source.
Answer: Option b
The dominant mode has lowest cutoff frequency and hence highest cutoff wavelength.
Answer: Option d
The reverse recovery time involves a diode switching from FB to RB. Reverse recovery time will be of several microseconds. Therefore at t=0, the diode would not have recovered from forward bias condition. Hence current will be same as 100 mA.A resistance in parallel with an indeal current source.
Answer: Option b
The dominant mode has lowest cutoff frequency and hence highest cutoff wavelength.
Answer: Option d
Answer: Option d
Aswer: option d
Thread: the primitive that can execute code. It contains an Instruction pointer and sometimes has its own stack.
Process: An environment in which one or several threads run.
Kernel: Run time OS- takes care of thread scheduling.
Answer: Option d
Answer: Option c
Total frequency = 3.6k + 3 x (1.2k) = 7.2kHz
Sampling frequency = 2 x 7.2 k = 14.4 Khz
Answer: Option b
An Addition of 6-dB improvement in SNR is obtained for each bit added to the PCM word.
6.02 x n = 6.02 x 2 = 12.04 = increase by 12 dB
Answer: Option b
QPSK is mostly used in digital data.
Answer: option d
Answer: Option a
Clock
|
J
|
K
|
Qn
|
Q’n
|
Qn+1
|
Q’n+1
|
0th clock
|
0
|
1
|
0
|
1
|
1
|
0
|
1st clock
|
1
|
1
|
1
|
0
|
0
|
1
|
2nd clock
|
0
|
1
|
0
|
1
|
1
|
0
|
3rd clock
|
1
|
1
|
1
|
0
|
0
|
1
|
4th clock
|
0
|
1
|
0
|
1
|
1
|
0
|
5th clock
|
1
|
1
|
1
|
0
|
0
|
1
|
The ouput toggles when J=K=1.
Answer: Option d
Answer: Option c
Answer: Option d
An instruction cycle has one or more machine cycles.
Instruction > Machine
Machine cycle time period is shorter than instruction cycle time period .
Answer: option c
Pnew = 2000 W
Answer: Option d
Answer: Option a
Answer: Option c
Answer: Option c
Pre-emption : The ability of the operating sytem to pre-empt or stop a currently scheduled task in favour of a higher priority task.
Answer: Option c
Answer: Option a
1200k + 2(450k) = 1200 + 900 = 2100kHz
Answer: Option d
Answer: Option c
Answer: option d
Answer: Option b
Answer: Option c
Answer: Option c
Anwer: Option b
First 3 choices are wrt I/o mapped I/o
Answer: Option d
Answer: Option a
Answer: Option b
Answer: Option c
Answer: Option a
Answer: Option b
Answer: Option b
Answer: Option c
Answer: Option b
Answer: Option C
Answer: Option b
Answer: Option c
Answer: Option a
Answer: Option d
Answer: Option c
Answer: Option c
Answer: Option a
Answer: Option d
Answer: Option b
Answer: Option a
Answer: Option c
Answer: Option a
Answer: Option b
Answer: Option b
Answer: Option d
Answer: Option a
Answer: Option d
Answer: Option a
Answer: Option c
Answer: Option b
Answer: Option b
Answer: Option b
Answer: Option c
Answer: Option b
Answer: Option c
Answer: Option a
Answer: Option b
Answer: Option a
Answer: Option c
Answer: Option d
Answer: Option b