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Electronics and Communication Engineering - Exam Questions Papers

Exercise : Exam Questions Papers - Exam Paper 7
16.
(11)2 + (11)3 = (11)?
4
5
6
none
Answer: Option
Explanation:

(10)2 + (11)3 = (11)7

⇒ 2 + 1 + 3 + 1 = x + l ⇒ x = 6.

17.
For a given transfer function: Damping ratio is:
0.707
0.695
1.414
0.798
Answer: Option
Explanation:

1 + G(s)H(s) = s2 + 2s + 2

ωn = 2 , 2εeωn = 2, εe = = 0.707.

18.
The energy of constant-amplitude complex valued exponential sequence is __________ (where A is the constant amplitude)
0
1
∞
A2
Answer: Option
Explanation:

Let the constant amplitude complex valued exponential sequence.

x(n) = Ae-jωn

Energy = |x(n)|2 = |Ae-jωn|2

A2 |(e-jωn)|2 = A2(∞) = ∞

Hence energy in infinite.

19.
Find the electric charge required on the earth and moon to balance their gravitational attraction if the charge on the earth is 10 times that on the moon.
1.73 TC
17.3 GC
17.3 TC
1.73 GC
Answer: Option
Explanation:

Using law of gravitations, the force between the two point masses is F =

M1 = mass of moon = 6.7 x 1022 Kg

M2 = mass of earth = 6 x 1024 Kg

r = distance between masses = 380 mm

G = universal gravitational constant = 6.7 x 10-11 Nm2/Kg2

Let Q1 and Q2 be the charge on moon and earth respectively. Now the gravitational force must be balanced by the force of repulsion

Q1Q2 = 4pε0M1M2G

but Q2 = 10Q1

= 10 x 4pω0M1M2G

∴ Q2 = 173 x 1012

Q1 = 17.3 x 1012 C

Q1 = 17.3 TC.

20.
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. The minimum bandwidth required to transmit this signal will be
77 K bits/sec
92 K bits/sec
84.8 K bits/sec
82.8 K bits/sec
Answer: Option
Explanation:

Nyquist rate = 2 x 4 kHz = 8 kHz

Sampling rate = 8 kHz + 1.2 kHz = 9.2 kHz

Number of bits required = log2512 = 9

Thus bandwidth = 9 x 9.2 = 82.8 k bits/sec.

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