The Autocorrelation function is given by exp . Then the PSD is given by
X is a binary memory less source with p (x = 0) = 0.3. If the source is transmitted over a BSC with crossover probability = 0.1, then the capacity of the channel is
Capacity of the channel is
C = 1 H(0.1)=1
= 1 0.469 = 0.531
Two binary random variables X and Y are distributed according to the joint distributions p(x = y = 0) = p(x = 0, y = 1) = p(x = y = 1) = 1/3^{ }. Then the entropy
H(x) is
The marginal probabilities are given by
In a Binary digital communication system, the threshold voltage of comparator is 'x' volts and voltage at input of comparator is uniformly distributed in the range of x0.5 to x+0.25 when '0' is transmitted and it is in the range x0.25 to x+0.5 when bit '1' is transmitted, the average probability of error in the system is_____________
Probability density function of f (z/0) is
When '0' is transmitted, it will be detected as '1' with probability
⇒
Average probability of error = 0.
If probability of error of PSK signal is 'x', then what will be probability of error of FSK and ASK signal. (Assume x =
As Qfunction is not linear operator, we cannot find the probability errors of FSK and ASK with probability error of PSK
If input to matched filter is s(t) = f(t)+ 2f (t)δ(t  0.5) + f(t  2.5)*8δ(t+ 0.5)
where f(t) is shown below, then determine the impulse response of matched filter h(t) ; 2 5 ≤ t < 3
(Note: δ is Delta function and * is convolution operator)
2f (t)6(t  0.5) = 2f(0.5) = 2x1 = 2 f(t2.5)*6(t+ 0.5) = f(t(2.5 0.5))
=f(t2)
s(t)=f(t)+f(t2)+2
= 2f +6 ; 2 5 t 5 3
Match the following:
Modulation Type of signaling format to be
technique used for modulation
ASK (a) ONOFF
PSK (b) Non return to zero
FSK (c) Return to zero
DPSK
QPSK
A.S.K signal will be present when ^{1}1' is transmitted and absent when '0' is transmitted hence ON 
OFF
In P.S.K for phase reversal a non return to zero format will be used
In FSK
f_{i} (t) = f_{c} (t)+ kfA_{v} for '1'
f_{c} (t) kfA_{v} for '0'
hence non return to zero
In D.P.S.K the output of differential enconder will be in ON  OFF format which is than modified using Amplitude level shifting to get NRZ format
In Q.P.S.K
For '10'
Ac cos 27tfct  Ac si n 2/rfct . Hence NRZ
Consider a BSC with crossover probability of error P. Suppose that R is the number of bits in a source code word that represents one of 2^{R} possible levels at the output of quantizer. If R = 5, P = 0.01, then the probability of having 5 or less bits in error in a code word is_____________
P(5 or less errors in 5 bits) =
A black and white television picture may be viewed as consisting of approximately 3x10^{5} elements, each of which may occur one of 10 distinct brightness levels with equal probability. Transmission rate is 30 picture frames/ second and SNR is 30 dB. Then the minimum bandwidth required to support the transmission of the resulting video signal is MHz
The information in each element is log210 bits. The information in each picture is [31og210]x10^{5} bits. The transmitted information rate is
[9 log210] x10^{6}bits / second.
The channel must have this capacity. From the information capacity theorem,
A random variable X has
Q. The value of constant K is
f_{x}(x) = 2kx ; 0 ≤ x ≤ 10 ⇒
A random variable X has
Q. The value of P(4 < x 5 ≤ 8) is
A telegraphic source has two symbols dot and dash. The dot duration is 0.2sec. The dash is twice as long as the dot and half as probable.
Q. The entropy of source is
P(Dot) + P(Dash) =1
P(Dot)+ = 1
P(Dot) = 2/3P(Dash) = 1/3
= 0.92 bits / symbol
A telegraphic source has two symbols dot and dash. The dot duration is 0.2sec. The dash is twice as long as the dot and half as probable.
Q. The information rate is
Number of symbols / sec =
= 3.75
Information rate = 3.75 x0.92= 3.45bits / sec
Consider a cascaded block of an LTI system
The impulse response of S_{1.} is h_{1.} t = δ t 2 The impulse response of 5_{2} is h_{2} t = t.e^{r}u t
The input x(t) is wide sense stationary random process with R_{xx} 0 = 3 and variance = 1
Q. The mean of the random process w(t) is
Consider a cascaded block of an LTI system
The impulse response of S_{1.} is h_{1.} t = δ t 2 The impulse response of 5_{2} is h_{2} t = t.e^{r}u t
The input x(t) is wide sense stationary random process with R_{xx} 0 = 3 and variance = 1
Q. If w(t) is input to system 5_{2}, then mean of process Y(t) will be
CDF of a random variable x is given below.
Q. E[x] is
Find PDF from CDF, PDF =
CDF of a random variable x is given below.
Q. Variance is
Variance
Consider a discrete memory less source X with two symbols x1 and x2 and P (x_{i}) = 0.9, P (x_{2}) = 0.1. These symbols are encoded as below.
Q. Find the source efficiency η
average code length L per symbol is = (0.9)1 + (0.1)1 = 1bit
0.9 log_{2} 0.9  0.1log_{2} 0.1 = 0.469bits / symbol
code efficiency = 0.469 = 46.9%
Consider a discrete memory less source X with two symbols x1 and x2 and P (x_{i}) = 0.9, P (x_{2}) = 0.1. These symbols are encoded as below.
Q. Find the redundancy γ
redundancy r = 1 η =1 0.469 = 0.531 = 53.1%
The input to a linear delta modulator having a stepsize D = 0.628 is a sine wave with frequency f_{m} and peak amplitude E_{m}. If the sampling frequency f_{s} = 40kHz, the combination of the sinewave frequency and peak amp, where slope overload will take place is
E_{m} f_{in} E_{m} f_{in}
(a) 0.3V 8kHz (b) 1.5V 4kHz
(c) 1.5V 2kHz (d) 3.0V 1kHz
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