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Rate 1/3 convolutional encoder: encode 10011

Short answer

Using the generators (0,1,1), (1,0,1), (1,1,0) to compute c1, c2, c3 per input bit and flushing with two zeros, the encoded sequence is 011 101 110 011 110 011 110.

The 1/3 rate convolutional encoder with constraint length equal to 3 has following three generator sequences each of length 3. (g₀⁽¹⁾, g₁⁽¹⁾, g₂⁽¹⁾) = (0, 1, 1) (g₀⁽²⁾, g₁⁽²⁾, g₂⁽²⁾) = (1, 0, 1) (g₀⁽³⁾, g₁⁽³⁾, g₂⁽³⁾) = (1, 1, 0) Determine the encoded sequence for the following input message (m₀, m₁, m₂, m₃, m₄) = (1 0 0 1 1)

Rate 1/31/3: each input bit gives 3 output bits. Constraint length K=3K=3, so the register holds mi,mi−1,mi−2m_i,m_{i-1},m_{i-2}. Each output is the modulo-2 convolution of the input with its generator:

ci(j)=∑l=02gl(j)mi−l(mod2)c_i^{(j)}=\sum_{l=0}^{2}g_l^{(j)}m_{i-l}\pmod 2

With the given generators:

ci(1)=0⋅mi+mi−1+mi−2=mi−1⊕mi−2ci(2)=mi⊕mi−2ci(3)=mi⊕mi−1\begin{aligned} c_i^{(1)}&=0\cdot m_i+m_{i-1}+m_{i-2}=m_{i-1}\oplus m_{i-2}\\ c_i^{(2)}&=m_i\oplus m_{i-2}\\ c_i^{(3)}&=m_i\oplus m_{i-1} \end{aligned} 
 m -->[ m_i ]-->[ m_i-1 ]-->[ m_i-2 ]
        |           |           |
        |           +----(+)----+------> c1 = mi-1+mi-2
        +----------(+)----------+------> c2 = mi+mi-2
        +----(+)----+------------------> c3 = mi+mi-1
     commutator: c1 c2 c3 per input bit

Encoding

Message (m0…m4)=10011(m_0\dots m_4)=10011, then two 0s (K−1K-1) to flush the register. Register starts at 00.

iimim_imi−1m_{i-1}mi−2m_{i-2}c(1)c^{(1)}c(2)c^{(2)}c(3)c^{(3)}
0100011
1010101
2001110
3100011
4110110
5011011
6001110

Check by polynomial (transform-domain) method

m(D)=1+D3+D4m(D)=1+D^3+D^4; g(1)(D)=D+D2g^{(1)}(D)=D+D^2, g(2)(D)=1+D2g^{(2)}(D)=1+D^2, g(3)(D)=1+Dg^{(3)}(D)=1+D.

c(1)(D)=(1+D3+D4)(D+D2)=D+D2+D4+D6 ⇒0110101c(2)(D)=(1+D3+D4)(1+D2)=1+D2+D3+D4+D5+D6 ⇒1011111c(3)(D)=(1+D3+D4)(1+D)=1+D+D3+D5 ⇒1101010\begin{aligned} c^{(1)}(D)&=(1+D^3+D^4)(D+D^2)=D+D^2+D^4+D^6\ \Rightarrow 0110101\\ c^{(2)}(D)&=(1+D^3+D^4)(1+D^2)=1+D^2+D^3+D^4+D^5+D^6\ \Rightarrow 1011111\\ c^{(3)}(D)&=(1+D^3+D^4)(1+D)=1+D+D^3+D^5\ \Rightarrow 1101010 \end{aligned}

Interleaving the three sequences bit by bit gives the same result.

Encoded sequence: 011 101 110 011 110 011 110 (21 bits; the first 15 bits, 011 101 110 011 110, correspond to the message itself and the last 6 come from flushing.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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