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Aim

Theory

Pretest

Procedure

Demo

Simulation

Posttest

References

Contributors

Feedback

Convolutional Codes and Viterbi Decoding

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Beginner

Intermediate

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1. Consider the convolutional code with the encoder provided in Figure (a). The memory of this encoder is:

\\

a: 0 Explanation

Explanation

b: 3 Explanation

Explanation

c: 4 Explanation

Explanation

d: 5 Explanation

Explanation

2. Consider the convolutional code with the encoder provided in Figure (a). The output bit

\mathbf{v_t^{(0)}}

will be given by

v_t^{(0)} = u_t + u_{t-1} + u_{t-2}

Explanation

v_t^{(0)} = u_t + u_{t-2} + u_{t-3}

Explanation

v_t^{(0)} = u_{t-1} + u_{t-2} + u_{t-3}

Explanation

v_t^{(0)} = u_t + u_{t-1} + u_{t-3}

Explanation

3. Consider the convolutional code with the encoder provided in Figure (a). The output bit

\mathbf{v_3^{(1)}}

will be given by

v_3^{(1)} = u_3 + u_2 + u_{1}

Explanation

v_3^{(1)} = u_3 + u_{1} + u_{0}

Explanation

v_3^{(1)} = u_{2} + u_{1} + u_{0}

Explanation

v_3^{(1)} = u_3 + u_{2} + u_{1}

Explanation

4. Consider an arbitrary rate

\mathbf{1/3}

feedforeward encoder with memory

4

. Suppose

\mathbf{\sigma_t = (0101)}

and

\mathbf{u_t = 1}

. Then

\mathbf{\sigma_{t+1}}

will be

\sigma_{t+1} = (1010)

Explanation

\sigma_{t+1} = (0010)

Explanation

\sigma_{t+1} = (10101)

Explanation

\sigma_{t+1} = (1101)

Explanation

5. Consider the

\mathbf{t}

-th section of the trellis diagram illustrated in Figure (b). Let

\mathbf{a_1 = 1, a_2 = 4, a_3 = 0, a_4 = 2}

and

\mathbf{[r_1 \hspace{0.1cm}r_2] = [1 \hspace{0.1cm} 1]}

. What will be

\mathbf{b_1, b_2, b_3, b_4}?

\\

b_1=2, b_2=0, b_3=1, b_4=2

Explanation

b_1=1, b_2=0, b_3=2, b_4=1

Explanation

b_1=3, b_2=0, b_3=2, b_4=1

Explanation

b_1=1, b_2=0, b_3=1, b_4=4

Explanation

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