Introduction to Coding Theory - - Unit 6 - Week 5 - Nptel

(1)

X

Courses » Introduction to Coding Theory

Unit 6 - Week 5

[email protected] ▼

Announcements Course Ask a Question Progress FAQ

Register for Register for Certification exam Certification exam

Course outline

How to access the portal Week 1 Week 2 Week 3 Week 4 Week 5

Week 6 Week 7

Decoding of Convolutional Codes-II: BCJR Algorithm Problem solving session-IV Problem solving session-V Performance Bounds for Convolutional Codes Quiz : Assignment-5 An Introduction To Coding Theory - Week 5 Feedback Assignment-5 solution

Due on 2019-03-06, 23:59 IST.

1) 1 point

2) 1 point

3) 1 point

Assignment-5

The due date for submitting this assignment has passed.

As per our records you have not submitted this assignment.

BCJR algorithm can be applied to

Linear block codes Convolutional codes both

none

No, the answer is incorrect.

Score: 0

Accepted Answers:

both

What is the overall constraint length of the

encoder realized in controller canonical form ?

2 4 6 8

No, the answer is incorrect.

Score: 0

4

What is the overall constraint length of the

encoder realized in observer canonical form ?

2 4 6 8

Score: 0

G(D) = [ D

²

]

1 D

D

²

1 1 + D + D

²

G(D) = [ D

²

]

1 D

D

²

1 1 + D + D

²

A project of In association with

Funded by

Introduction to Coding Theory - - Unit 6 - Week 5 https://onlinecourses.nptel.ac.in/noc19_ee26/un...

1 of 4 Friday 10 May 2019 10:29 AM

(2)

Powered by Week 8

4) 1 point

5) 1 point

6) 1 point

7) 1 point

Accepted Answers:

6

Is the encoder given by generator

matrix catastrophic?

Yes No

Score: 0

Yes

Which of the following is the minimal encoder corresponding to the generator matrix?

None of the above No, the answer is incorrect.

Score: 0

Is the systematic equivalent encoder corresponding to the generator

matrix realizable?

Yes No

No, the answer is incorrect.

Score: 0

No

Consider the (2,1,2) non-systematic feedforward encoder

with . Its minimum delay feedforward inverse is given by

G(D) = [ 1 + D

³ ^{1+ + +}^D_1+D+D² ^D³2^D⁵

]

G(D) = [ 1

_{1+D+ + + +}^{1+ + +}_D^D2² D^D³³ D^D⁴⁵ D⁵

]

[ 1

_1+D+D^1+D 2

] [ 1

_1+D+D^1+D²2

] [ 1

_1+D+D^1+D²3

]

[ 1

_1+D+D^1+D²2

]

G(D) = [ 1 ]

0 1

D

²

D 1 + D

²

G(D) = [1 + D

²

1 + D + D

²

]

[ D

²

] 1 + D

²

[ D

²

]

D

²

[ ] 1 1

Introduction to Coding Theory - - Unit 6 - Week 5 https://onlinecourses.nptel.ac.in/noc19_ee26/un...

2 of 4 Friday 10 May 2019 10:29 AM

(3)

8) 1 point

9) 1 point

Does not exist No, the answer is incorrect.

Score: 0

Consider the (3,1,2) non-systematic feed-forward encoder

with . An information sequence is encoded using

this convolutional encoder and transmitted over a binary symmetric channel with crossover probability of . The event error probability can be approximated by

No, the answer is incorrect.

Score: 0

Given below is a trellis diagram of (3,1,2) convolutional encoder. What is the correct

expression for forward recursion, alpha calculation? All the notations are same as used in the lectures.

[ ] 1 1

G(D) = [ 1 + D 1 + D

²

1 + D + D

²

] p = 10

⁻²

1.28 × 10

⁻⁵

2 × 10

⁻⁶

4 × 10

⁻⁷

2.56 × 10

⁻⁸

1.28 × 10

⁻⁵

(01) = (00, 10) ∗ (10) + (11, 01) ∗ (01) α

l

γ

l−1

α

l−1

γ

l−1

α

l−1

(01) = (10, 01) ∗ (10) + (11, 01) ∗ (11) α

l

γ

l−1

α

l−1

γ

l−1

α

l−1

(01) = (01, 10) ∗ (10) + (01, 00) ∗ (00) α

l

γ

l−1

α

l−1

γ

l−1

α

l−1

Introduction to Coding Theory - - Unit 6 - Week 5 https://onlinecourses.nptel.ac.in/noc19_ee26/un...

3 of 4 Friday 10 May 2019 10:29 AM

(4)

10) 1 point none of the above

No, the answer is incorrect.

Score: 0

For the same encoder given in previous question, what is the correct expression for beta calculation? All the notations are same as used in the lectures.

none of the above No, the answer is incorrect.

Score: 0

(01) = (10, 01) ∗ (10) + (11, 01) ∗ (11) α

l

γ

l−1

α

l−1

γ

l−1

α

l−1

(01) = (01, 10) ∗ (10) + (01, 00) ∗ (00) β

l

γ

l

β

l+1

γ

l

β

l+1

(01) = (10, 01) ∗ (01) + (10, 00) ∗ (00) β

l

γ

l

β

l+1

γ

l

β

l+1

(01) = (10, 01) ∗ (10) + (11, 01) ∗ (11) β

l

γ

l

β

l+1

γ

l

β

l+1

(01) = (01, 10) ∗ (10) + (01, 00) ∗ (00) β

l

γ

l

β

l+1

γ

l

β

l+1

Previous Page End

Introduction to Coding Theory - - Unit 6 - Week 5 https://onlinecourses.nptel.ac.in/noc19_ee26/un...

4 of 4 Friday 10 May 2019 10:29 AM