Data-Communication

Question 1
If the period of a signal is 100 ms. Then its frequency in Hertz is___
A
10
B
100
C
1000
D
10000
Question 1 Explanation: 
Given data,
Frequency = 1/Period.
Frequency = 1/100 x 10-3 Hz
= 10 Hz
Question 2
If the channel is band limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of channel?
A
16.15kbps
B
23.24 kbps
C
40.12 kbps
D
24.74 kbps
Question 2 Explanation: 
→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=B log​ 2​ (1+S/N).
where S and N are the signal and noise power, respectively, at the output of the channel.
→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel. =6 * log​ 2​ (1+16)
=6 * 4.08746284125034
=24.74 kbps
Question 3
At 100% modulation, the power in each sideband is__ of that of arrier.
A
50%
B
40%
C
60%
D
25%
Question 3 Explanation: 
→ Power in sidebands may be calculated as modulation index.
→ Each sideband is 25%. Total modulation index=1.
Question 4
The capacity relationship is given by
A
C=Wlog​ 2​ (1+S/N)
B
C=2Wlog​ 2​ (1+S/N)
C
C=Wlog​ 2​ (1-S/N)
D
C=Wlog​ 10​ (1+S/N)
Question 4 Explanation: 
→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=W log​ 2​ (1+S/N).
where S and N are the signal and noise power, respectively, at the output of the channel.
→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.
Question 5
Twelve 1 ​ Ω ​ resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is
A
5/6 ​ Ω
B
1/6 ​ Ω
C
6/5 ​ Ω
D
3/2 ​ Ω
Question 5 Explanation: 
Question 6
The final value theorem is used to find the
A
Steady state value of the system output
B
Initial value of the system output
C
Transient behaviour of the system output
D
None of these
Question 6 Explanation: 
The final value theorem is used to find the steady state value of the system output
Question 7
For the discrete signal x[n]=a​ n​ u[n], a>0 the z-transform is
A
(z+a)/z
B
(z-a)/z
C
z/(z-a)
D
z/(z+a)
Question 7 Explanation: 
Let x[n] be causal signal given by x[n] = a​ n​ u[n]
→ The Z-Transform of x[n] is given by

The ROC is defined by | az −1 | < 1 or | z| > | a| .
Region of Convergence(RoC)
Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)
1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)
2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)
3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).
Question 8
​ For a periodic signal v(t)=30 sin 100t+10cos 300t +6 sin (500t + ​ π ​ /4), the fundamental frequency in rad/s
A
100
B
300
C
500
D
None of the above
Question 8 Explanation: 
Given, the signal
V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)
So, we have
ω 1 = 100 rads
ω 2 = 300 rads
ω 3 = 500 rads
∴ The respective time periods are
T 1 = 2π/ω1 = 2π/100 sec
T 2 = 2π/ω2 = 2π/300 sec
T 3 = 2π/ω3 = 2π/500 sec
So, the fundamental time period of the signal is
LCM (T 1 , T 2 , T 3 ) =LCM (2π,2π,2π)/HCF (100,300,500)
as T 0 =2π/100
∴ The fundamental frequency, ω0 =2π/T 0 = 100 rad/s
Question 9
​ If the number of bits per sample in a PCM system is increased from a n to n+1, the improvement in signal to quantization noise ratio will be
A
3dB
B
6db
C
2n dB
D
n dB
Question 9 Explanation: 
(S/N q)dB
= (1.76 + 6 n) dB
(SQNR) 1 = 1 .76 + 6 n
(SQNR) 2 = 1 .76 + 6 n(n + 1 ) = 1 .76 + 6 n + 6
(SQNR) 2 − (SQNR) 1 = 1 .76 + 6 n + 6 − 1 .76 − 6 n = 6 dB
So for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.
Question 10
A carrier A​ c​ Cos(w​ C​ )t is frequency modulated by a signal E​ m​ Cos(w​ m​ )t. The modulation index is m​ f​ . the expression for the resulting FM signal is
A
A​ c,​ Cos[w​ c​ t+ m​ f​ Sin(w​ m​ )t]
B
A​ c​ Cos[w​ c​ t+ m​ f​ Cos(w​ m​ )t]
C
A​ c​ Cos[w​ c​ t+ ​ π ​ m​ f​ Sin w​ m​ t]
D
A​ c​ Cos[w​ c​ t+ 2 ​ π ​ m​ f​ E​ m​ Cos(w​ m​ )t/w​ m​ ]
Question 10 Explanation: 
There are 10 questions to complete.

Access quiz wise question and answers by becoming as a solutions adda PRO SUBSCRIBER with Ad-Free content

Register Now

If you have registered and made your payment please contact solutionsadda.in@gmail.com to get access