Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
513 views
in General by (115k points)
closed by
An analog baseband signal, band-limited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities p1 = p4 = 0.125 and p2 = p3. The information rate (bits/sec) of the message source is __________

1 Answer

0 votes
by (152k points)
selected by
 
Best answer

Concept:

The Entropy of a source, which is also the average information content is given by:

\(H = \sum P\left( {{x_i}} \right){\log _2}\left( {\frac{1}{{P\left( {{x_i}} \right)}}} \right)\)

Where P(xi) = Probability of each codeword.

Application:

We have P1 = P4 = 0.125

Now, P1 + P2 + P3 + P4 = 1

Since P2 = P3, the above equation becomes:

\(0.12 + 2{{\rm{P}}_2} + 0.125 = 1\)

2P2 = 0.75

P2 = P3 = 0.375

Average information will be:

H = - P1log2P1 - P2log2P2 - P3log2P3 - P4log2P4

H = - 0.125.log2 0.125 - 0.375.log2 0.375 - 0.375.log0.375 - 0.125.log0.125

H = 0.375 + 0.531 + 0.531 + 0.375

H = 1.812 bits/symbol.

Sampling rate is the Nyquist rate, i.e.

r = 2 × 100 = 200 samples/sec

Thus, the information rate will be: r × H

= 200 × 1.812

= 362.4 bits/sec.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...