Question 1 |

A random variable X takes values -1 and +1 with probabilities 0.2 and 0.8, respectively. It is transmitted across a channel which adds noise N, so that the random variable at the channel output is Y=X+N. The noise N is independent of X, and is uniformly distributed over the interval [-2,2]. The receiver makes a decision

\hat{X}=\left\{\begin{matrix} -1 & if & Y\leq \theta \\ +1& if & Y> \theta \end{matrix}\right.

where the threshold \theta \in [-1,1] is chosen so as to minimize the probability of error Pr[\hat{X}\neq X]. The minimum probability of error, rounded off to 1 decimal place, is ______.

\hat{X}=\left\{\begin{matrix} -1 & if & Y\leq \theta \\ +1& if & Y> \theta \end{matrix}\right.

where the threshold \theta \in [-1,1] is chosen so as to minimize the probability of error Pr[\hat{X}\neq X]. The minimum probability of error, rounded off to 1 decimal place, is ______.

0.1 | |

0.2 | |

0.3 | |

0.8 |

Question 1 Explanation:

MAP criteria should be used to minimise the probability of error.

\begin{array}{c} f_{Y}(y \mid+1) P(+1) \sum_{-1}^{+1} f_{Y}(y \mid-1) F(-1) \\ P(+1)=0.80 \text { and } P(-1)=0.20 \end{array}

P_{\theta(m \text { in })}= Shaded area =2 \times \frac{1}{20}=0.10

\begin{array}{c} f_{Y}(y \mid+1) P(+1) \sum_{-1}^{+1} f_{Y}(y \mid-1) F(-1) \\ P(+1)=0.80 \text { and } P(-1)=0.20 \end{array}

P_{\theta(m \text { in })}= Shaded area =2 \times \frac{1}{20}=0.10

Question 2 |

A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal f(t)=A(1+m(t))cos 2\pi f_c t, where f_c=600kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is _______ Mbps.

9.25 | |

11.81 | |

25.36 | |

14.25 |

Question 2 Explanation:

\begin{aligned} \text { Nyquist rate } &=2 \times 615 \mathrm{kHz} \\ &=1230 \mathrm{kHz}=1.23 \mathrm{MHz} \\ f_{s} &=1.2 \times 1.23=1.476 \mathrm{MHz} \\ \text { Bits/sample }, n &=\log _{2}(256)=8 \\ \text { So, } \quad A_{D} &=n f_{s}=8 \times 1.476 \\ &=11.808 \simeq 11.81 \mathrm{Mbps} \end{aligned}

Question 3 |

A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density N_0/2. Binary signaling, with 0\rightarrow p(t) and 1\rightarrow q(t), is used for the transmission, along with an optimal receiver that minimizes the bit-error probability.

Let \varphi _1(t),\varphi _2(t) form an orthonormal signal set.

If we choose p(t)=\varphi _1(t)\; and \; q(t)=-\varphi _2(t), we would obtain a certain bit-error probability P_b.

If we keep p(t)=\varphi _1(t), but take q(t)=\sqrt{E}\varphi _2(t), for what value of E would we obtain the same bit-error probability P_b?

Let \varphi _1(t),\varphi _2(t) form an orthonormal signal set.

If we choose p(t)=\varphi _1(t)\; and \; q(t)=-\varphi _2(t), we would obtain a certain bit-error probability P_b.

If we keep p(t)=\varphi _1(t), but take q(t)=\sqrt{E}\varphi _2(t), for what value of E would we obtain the same bit-error probability P_b?

0 | |

1 | |

2 | |

3 |

Question 3 Explanation:

\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\phi_{1}(t)

d_{\min }=2

\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\sqrt{E}\phi_{2}(t)

d_{\min }=\sqrt{(\sqrt{E})^{2}+1}=\sqrt{E+1}

To obtain same bit-error probability, \sigma_{\text {min }} should be same.

\begin{aligned} \text { So, } \quad \sqrt{E+1} &=2 \\ E &=3 \end{aligned}

d_{\min }=2

\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\sqrt{E}\phi_{2}(t)

d_{\min }=\sqrt{(\sqrt{E})^{2}+1}=\sqrt{E+1}

To obtain same bit-error probability, \sigma_{\text {min }} should be same.

\begin{aligned} \text { So, } \quad \sqrt{E+1} &=2 \\ E &=3 \end{aligned}

Question 4 |

A random variable X takes values -0.5 and 0.5 with probabilities 1/4 and 3/4, respectively.
The noisy observation of X is Y = X + Z, where Z has uniform probability density over
the interval (-1, 1). X and Z are independent. If the MAP rule based detector outputs \hat{X} as

\hat{X}=\left\{\begin{matrix} -0.5 \; \; \; Y \lt \alpha \\ 0.5 \; \; \; Y \geq \alpha \end{matrix}\right.

then the value of \alpha (accurate to two decimal places) is _______.

\hat{X}=\left\{\begin{matrix} -0.5 \; \; \; Y \lt \alpha \\ 0.5 \; \; \; Y \geq \alpha \end{matrix}\right.

then the value of \alpha (accurate to two decimal places) is _______.

-0.5 | |

-1 | |

1 | |

0.5 |

Question 4 Explanation:

\begin{array}{c} P\left(x_{0}\right)=\frac{1}{4} ; P\left(x_{1}\right)=\frac{3}{4} \\ \text { MAP criteria, } f_{Y}\left(y \mid x_{0}\right) P\left(x_{0}\right) \overset{x_{2}}{\underset{x_{1}}{\gtrless}}\left(y \mid x_{1}\right) P\left(x_{1}\right) \end{array}

\text{So}\quad \alpha=-0.5

Question 5 |

A binary source generates symbols X \in \{-1,1\} which are transmitted over a noisy channel. The probability of transmitting X= 1 is 0.5. Input to the threshold detector is R= X + N.
The probability density function f_{N}(n) of the noise N is shown below.

If the detection threshold is zero, then the probability of error (correct to two decimal places) is __________.

If the detection threshold is zero, then the probability of error (correct to two decimal places) is __________.

0.12 | |

0.2 | |

0.8 | |

1 |

Question 5 Explanation:

Let s_{0} and s_{1} be the transmitted symbols representing the transmitted value {-1 , 1} respectively and let r_{0} and r_{1} be the received symbols.

Probability of error,

\begin{aligned} P_{e} &=P\left(s_{1}\right) P\left(r_{0} \mid s_{1}\right)+P\left(s_{0}\right) P\left(r_{1} \mid s_{0}\right) \\ P\left(r_{0} \mid s_{1}\right) &=P\left(r_{1} \mid s_{0}\right)=\frac{1}{2} \times 1 \times \frac{1}{4}=\frac{1}{8}\\ \text{Given that,}\\ P\left(s_{0}\right)&=P\left(s_{1}\right)=\frac{1}{2}\\ \text{So}\quad P_{e}&=\frac{1}{2}\left(\frac{1}{8}+\frac{1}{8}\right)=\frac{1}{8}=0.125 \end{aligned}

Probability of error,

\begin{aligned} P_{e} &=P\left(s_{1}\right) P\left(r_{0} \mid s_{1}\right)+P\left(s_{0}\right) P\left(r_{1} \mid s_{0}\right) \\ P\left(r_{0} \mid s_{1}\right) &=P\left(r_{1} \mid s_{0}\right)=\frac{1}{2} \times 1 \times \frac{1}{4}=\frac{1}{8}\\ \text{Given that,}\\ P\left(s_{0}\right)&=P\left(s_{1}\right)=\frac{1}{2}\\ \text{So}\quad P_{e}&=\frac{1}{2}\left(\frac{1}{8}+\frac{1}{8}\right)=\frac{1}{8}=0.125 \end{aligned}

Question 6 |

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 40dB. The minimum number of bits per sample needed to achieve the desired SQNR is _______

5 | |

6 | |

7 | |

8 |

Question 6 Explanation:

For sinusoidal input is applied to a PCM system,

\mathrm{SQNR}=6 n+1.8

n= number of bits per sample

Given that, the required \mathrm{SQNR}=40 \mathrm{dB}

\begin{aligned} \mathrm{so}, 6 n+1.8 &\geq 40 \\ n &\geq \frac{38.2}{6}\\ n_{\min }&=7 \\ \because \; &n \text{ is always an integer value} \end{aligned}

\mathrm{SQNR}=6 n+1.8

n= number of bits per sample

Given that, the required \mathrm{SQNR}=40 \mathrm{dB}

\begin{aligned} \mathrm{so}, 6 n+1.8 &\geq 40 \\ n &\geq \frac{38.2}{6}\\ n_{\min }&=7 \\ \because \; &n \text{ is always an integer value} \end{aligned}

Question 7 |

In binary frequency shift keying (FSK), the given signal waveform are

u_{0}(t)=5cos(20000\pi t);o\leq t\leq T , and

u_{1}(t)=5cos(22000\pi t);o\leq t\leq T,

Where T is the bit-duration interval and t is in seconds. Both u_{0}(t) and u_{1}(t) are zero outside the interval 0\leq t \leq T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u_{0}(t) and u_{1}(t) uncorrelated is

u_{0}(t)=5cos(20000\pi t);o\leq t\leq T , and

u_{1}(t)=5cos(22000\pi t);o\leq t\leq T,

Where T is the bit-duration interval and t is in seconds. Both u_{0}(t) and u_{1}(t) are zero outside the interval 0\leq t \leq T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u_{0}(t) and u_{1}(t) uncorrelated is

0.25ms | |

0.5ms | |

0.75ms | |

1.0ms |

Question 7 Explanation:

\begin{aligned} u_{0}(t) &=5 \cos 20000 \pi t \\ u_{1}(t) &=5 \cos 22000 \pi t \\ f_{1} &=11000 \mathrm{Hz} \text { and } f_{2}=10000 \mathrm{Hz} \end{aligned}

For FSK waveforms to be uncorrelated,

\begin{aligned} f_{1}-f_{2} &=n \frac{R_{b}}{2} ; n=1,2,3, \ldots \\ R_{b} &=\frac{2\left(f_{1}-f_{2}\right)}{n}=\frac{2000}{n}-\text { bits } / \mathrm{sec} \\ R_{b(\max )} &=2000 \mathrm{bits} / \mathrm{sec}\\ &\quad \quad \because \text{minimum value of n=1 }\\ T_{b(\min )}&=\frac{1}{R_{b(\max )}}=0.5 \mathrm{ms} \end{aligned}

For FSK waveforms to be uncorrelated,

\begin{aligned} f_{1}-f_{2} &=n \frac{R_{b}}{2} ; n=1,2,3, \ldots \\ R_{b} &=\frac{2\left(f_{1}-f_{2}\right)}{n}=\frac{2000}{n}-\text { bits } / \mathrm{sec} \\ R_{b(\max )} &=2000 \mathrm{bits} / \mathrm{sec}\\ &\quad \quad \because \text{minimum value of n=1 }\\ T_{b(\min )}&=\frac{1}{R_{b(\max )}}=0.5 \mathrm{ms} \end{aligned}

Question 8 |

Which one of the following statements about differential pulse code modulation (DPCM) is true?

The sum of message signal sample with its prediction is quantized | |

The message signal sample is directly quantized, and its prediction is not used | |

The difference of message signal sample and a random signal is quantized | |

The difference of message signal sample with its predictions is quantized |

Question 8 Explanation:

In DPCM, difference of message signal sample
with its prediction is quantized.

Question 9 |

In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is inter symbol interference zero?

A | |

B | |

C | |

D |

Question 9 Explanation:

Given,

Symbol rate =2000 symbol/sec

So R_{b}=2 \mathrm{kHz}

Nyquist's criterion for ISI free transmission is.

\sum_{n=-\infty}^{\infty} P\left(f-n R_{b}\right)=\text { Constant }

So, only option (b) satisfies this condition for given R_{b}

Symbol rate =2000 symbol/sec

So R_{b}=2 \mathrm{kHz}

Nyquist's criterion for ISI free transmission is.

\sum_{n=-\infty}^{\infty} P\left(f-n R_{b}\right)=\text { Constant }

So, only option (b) satisfies this condition for given R_{b}

Question 10 |

The bit error probability of a memoryless binary symmetric channel is 10^{-5}. If 10^{5} bits are sent over this channel, then the probability that not more than one bit will be in error is __________

0.73 | |

1.5 | |

0.25 | |

2.5 |

Question 10 Explanation:

bit error probability, p=10^{-5}

Number of bits transmitted,

n=10^{5}

If n-bits are transmitted, probability of getting error in 'r' bits is given by

p_{e}={ }^{n} C_{r} \rho^{r}(1-p)^{n-r}

The probability that not more than one bit will be in error is

\begin{array}{l} ={ }^{n} C_{0} p^{0}(1-p)^{-0}+{ }^{n} C_{\{(p)}^{1}(1-p)^{p-1} \\ =(1-p)^{n}+n p(1-p)^{n-1} \\ =\left(1-10^{-5}\right)^{10^{5}}+10^{5} \times 10^{-5}\left(1-10^{-5}\right)^{10^{5}-1} \\ =0.7357 \end{array}

Number of bits transmitted,

n=10^{5}

If n-bits are transmitted, probability of getting error in 'r' bits is given by

p_{e}={ }^{n} C_{r} \rho^{r}(1-p)^{n-r}

The probability that not more than one bit will be in error is

\begin{array}{l} ={ }^{n} C_{0} p^{0}(1-p)^{-0}+{ }^{n} C_{\{(p)}^{1}(1-p)^{p-1} \\ =(1-p)^{n}+n p(1-p)^{n-1} \\ =\left(1-10^{-5}\right)^{10^{5}}+10^{5} \times 10^{-5}\left(1-10^{-5}\right)^{10^{5}-1} \\ =0.7357 \end{array}

There are 10 questions to complete.