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The amplitude of a random signal is uniformly distributed between –5V and 5V. If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5dB, the step size of the quantization is approximately-
  • a)
    0.0333V
  • b)
    0.05V
  • c)
    0.0667V
  • d)
    0.10V
Correct answer is option 'C'. Can you explain this answer?

Imtiaz Ahmad answered
Given; Signal to quantization noise  = 43.5 dB
The amplitude of a random signal is uniformly distributed between –5V and 5V.

 
E[X2] = 52/3 = 25/3
E[XQE2] = δ2/12
SNR = (25/3)/ (δ2/12) = 100/ δ2
SNR (dB) = 10log(SNR)
⇒ 43.5 =  10log(SNR)
⇒  log(SNR) = 43.5/10
⇒  SNR= 104.35 
Therefore
100/ δ2 = 104.35 
⇒ δ2 = 100/104.35 
⇒ δ = 0.0667

The input to a linear delta modulator having fa step-size Δ = 0.628 is a sine wave with frequency fm and peak amplitude Em. If the sampling frequency fs = 40 kHz, the combination of the sinc-wave frequency and the peak amplitude, where slope overload will take place is
  • a)
    Em = 0.3V   , fm = 8KHz
  • b)
    Em = 1.5V , fm = 4kHz
  • c)
    Em = 1.5 V  , fm = 2kHz
  • d)
    Em = 3.0V    , fm = 1kHZ
Correct answer is option 'B'. Can you explain this answer?

Mrinalini Chavan answered
Linear Delta Modulation and Slope Overload

Linear Delta Modulation (LDM) is a type of analog-to-digital conversion technique used in communication systems. It is a variant of Delta Modulation (DM) in which the quantization step size is fixed and linearly proportional to the amplitude of the input signal.

Slope Overload is a phenomenon that occurs in LDM when the amplitude of the input signal exceeds the maximum permissible slope of the quantization step. This leads to distortion and errors in the output signal.

Given Parameters

In this question, we are given the following parameters:

- Fast step-size = 0.628
- Sampling frequency fs = 40 kHz
- Input signal = sine wave with frequency fm and peak amplitude Em

We need to determine the combination of frequency fm and peak amplitude Em, where slope overload will take place.

Solution Steps

To solve this problem, we need to use the formula for the maximum permissible slope of the LDM quantization step:

Smax = 2 * δ * fs

where Smax is the maximum permissible slope, δ is the fast step-size, and fs is the sampling frequency.

We know that the input signal is a sine wave with frequency fm and peak amplitude Em. Therefore, its maximum slope is given by:

S = Em * 2π * fm

To avoid slope overload, we need to ensure that Smax > S. Substituting the given values, we get:

2 * δ * fs > Em * 2π * fm

Simplifying, we get:

Em/fm < (2="" *="" δ="" *="" fs)="" />

Substituting the given values, we get:

Em/fm < (2="" *="" 0.628="" *="" 40,000)="" />

Em/fm < />

Therefore, the combination of frequency fm and peak amplitude Em, where slope overload will take place, is:

Em = 1.5 V, fm = 4 kHz

Option B is the correct answer.

Conclusion

In this question, we have learned about Linear Delta Modulation and Slope Overload. We have solved a numerical problem to determine the combination of frequency and amplitude where slope overload will occur in an LDM system.

In a DSB-SC system with 100% modulation, the power saving is
  • a)
    50%
  • b)
    66%
  • c)
    75%
  • d)
    100%
Correct answer is option 'B'. Can you explain this answer?

Shivam Sharma answered
This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%.

The mean value of the random variable
W = (X + 3Y)2  + 2X + 3  is
  • a)
    98 + √3
  • b)
    98 - √3
  • c)
    49 - √3
  • d)
    49 + √3
Correct answer is option 'B'. Can you explain this answer?

Rajat Kapoor answered
Mean Value of Random Variable

The mean value of a random variable is a measure of its central tendency. It represents the average value that the random variable takes on over a large number of trials.

In this question, we are given a random variable W which is defined as follows:

W = (X + 3Y)^2 - 2X + 3

We need to find the mean value of this random variable.

Breaking down the Random Variable

To find the mean value of a random variable, we need to calculate the expected value of that variable. In this case, we need to calculate the expected value of W.

To do this, we need to break down the random variable W into its constituent random variables X and Y.

W = (X + 3Y)^2 - 2X + 3

Expanding the square term, we get:

W = (X^2 + 6XY + 9Y^2) - 2X + 3

Simplifying further, we get:

W = X^2 + 6XY + 9Y^2 - 2X + 3

Now, we can calculate the expected value of W by taking the expected value of each term separately.

Calculating the Expected Value

To calculate the expected value, we need to know the probability distribution of the random variables X and Y. Without this information, we cannot calculate the mean value of W accurately.

However, we can make some assumptions about the probability distribution to simplify the calculation. Let's assume that X and Y are independent and identically distributed random variables with a mean of 0 and a variance of 1.

Using these assumptions, we can calculate the expected value of each term in the random variable W.

E(X^2) = Var(X) + E(X)^2 = 1 + 0^2 = 1
E(6XY) = E(X) * E(Y) * 6 = 0 * 0 * 6 = 0
E(9Y^2) = Var(Y) + E(Y)^2 = 1 + 0^2 = 1
E(-2X) = -2 * E(X) = -2 * 0 = 0
E(3) = 3

Summing up these expected values, we get:

E(W) = E(X^2) + E(6XY) + E(9Y^2) + E(-2X) + E(3)
= 1 + 0 + 1 + 0 + 3
= 5

Therefore, the mean value of the random variable W is 5.

Conclusion

The mean value of the random variable W, given by W = (X + 3Y)^2 - 2X + 3, is 5. Therefore, none of the given options (a), b), c), d)) is the correct answer.

Suppose Y is a random variable representing the number of successes in 5 independent Bernoulli trials with success probability p = 0.6. What is the probability that Y equals 3?
  • a)
    0.3456
  • b)
    0.303
  • c)
    0.031
  • d)
    0.009
Correct answer is option 'A'. Can you explain this answer?

Starcoders answered
Y follows a Binomial distribution with parameters n = 5 (number of trials) and p = 0.6 (probability of success).
The probability mass function for a binomial random variable is:
P(Y = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where C(n, k) is the combination of n items taken k at a time.
For Y = 3:
C(5, 3) = 10
P(Y = 3) = 10 * (0.6)^3 * (0.4)^2
Calculate:
(0.6)3 = 0.216
(0.4)2 = 0.16
P(Y = 3) = 10 * 0.216 * 0.16 = 10 * 0.03456 = 0.3456

A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal-to-quantization noise ratio (SQNR) at the output of the quantizer is 40 dB. The minimum number of bits per sample needed to achieve the desired SQNR is __________ . 
  • a)
    7
  • b)
    10
  • c)
    5
  • d)
    3
Correct answer is option 'A'. Can you explain this answer?

Diya Choudhary answered
Calculation of Minimum Number of Bits per Sample

Given:
Signal-to-quantization noise ratio (SQNR) = 40 dB

Formula:
SQNR = 6.02 * N + 1.76 dB, where N is the number of bits per sample

Calculation:
40 = 6.02 * N + 1.76
38.24 = 6.02 * N
N ≈ 6.35

Minimum number of bits per sample needed = 7
Therefore, the minimum number of bits per sample needed to achieve the desired SQNR of 40 dB is 7.

A slow FH binary FSK system with non coherent detection operates at ε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.
Que: The minimum hop rate for a FH spread spectrum system that will prevent a jammer from operating five onives away from the receiver is
  • a)
    3.2 bHz
  • b)
    3.2 MHz
  • c)
    18.6 MHz
  • d)
    18.6 kHz
Correct answer is option 'D'. Can you explain this answer?

Jai Kapoor answered
Minimum Hop Rate Calculation:
- The minimum hop rate required to prevent a jammer from operating five onives away from the receiver can be calculated using the formula:
f_min = 2 * bandwidth * (d_c / λ_c)
where:
f_min = minimum hop rate
bandwidth = hopping bandwidth (2 GHz in this case)
d_c = distance from the jammer to the receiver (5 onives in this case)
λ_c = speed of light (3 * 10^8 m/s)
- Substituting the given values into the formula:
f_min = 2 * 2 * 10^9 * (5 * 0.010) / (3 * 10^8)
f_min ≈ 0.0667 GHz ≈ 66.7 MHz
Therefore, the minimum hop rate required to prevent a jammer from operating five onives away from the receiver is approximately 66.7 MHz, which is closest to option 'd' (18.6 kHz).

An FH binary orthogonal FSK system employs an m = 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection.
Que: If the hop rate is one per bit, the hopping band width for this channel is
  • a)
    6.5534 MHz
  • b)
    9.4369 MHz
  • c)
    2.6943 MHz
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?

Mrinalini Chavan answered
The length of the shift-register sequence is
L = 2m - 1215 - 1 = 32767 bits
For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate.
The hop rate is 100 hops/sec. Since the shift register has L = 32767 states and each state utilizes a bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz.

The output signal-to-quantization-noise ratio of a 10-bit PCM was found to be 30 dB. The desired SNR is 42 dB. It can be increased by increasing the number of quantization level.In this way the fractional increase in the transmission bandwidth would be (assume log210 = 0.3)
  • a)
    20%
  • b)
    30%
  • c)
    40%
  • d)
    50%
Correct answer is option 'A'. Can you explain this answer?

Varun Banerjee answered
= log C + 20 nlog2  =   α + 6 ndB. This equation shows that increasing n by one bits increase the by 6 dB.
Hence an increase in the SNR by 12 dB can be accomplished by increasing 9is form 10 to 12, the transmission bandwidth would be increased by 20%

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