Data-Communication
Question 1 |
If the period of a signal is 100 ms. Then its frequency in Hertz is___
10 | |
100 | |
1000 | |
10000 |
Question 1 Explanation:
Given data,
Frequency = 1/Period.
Frequency = 1/100 x 10-3 Hz
= 10 Hz
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?
16.15kbps | |
23.24 kbps | |
40.12 kbps | |
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
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.
50% | |
40% | |
60% | |
25% |
Question 3 Explanation:
→ Power in sidebands may be calculated as modulation index.
→ Each sideband is 25%. Total modulation index=1.
→ Each sideband is 25%. Total modulation index=1.
Question 4 |
The capacity relationship is given by
C=Wlog 2 (1+S/N) | |
C=2Wlog 2 (1+S/N) | |
C=Wlog 2 (1-S/N) | |
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.
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
5/6 Ω | |
1/6 Ω | |
6/5 Ω | |
3/2 Ω |
Question 5 Explanation:
Question 6 |
The final value theorem is used to find the
Steady state value of the system output | |
Initial value of the system output | |
Transient behaviour of the system output | |
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
(z+a)/z | |
(z-a)/z | |
z/(z-a) | |
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 dened 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).
→ 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 dened 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
100 | |
300 | |
500 | |
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
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
3dB | |
6db | |
2n dB | |
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.
= (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 c, Cos[w c t+ m f Sin(w m )t] | |
A c Cos[w c t+ m f Cos(w m )t] | |
A c Cos[w c t+ π m f Sin w m t] | |
A c Cos[w c t+ 2 π m f E m Cos(w m )t/w m ] |
Question 10 Explanation: