Aufgaben:Exercise 4.10: Turbo Encoder for UMTS and LTE

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Turbo encoder for UMTS and LTE

The mobile communications standards  $\rm UMTS$  and  $\rm LTE$  each use a turbo code that is largely identical to the encoder described in the  "The Basics of Turbo Codes"  chapter.

• The  $1/n$ convolutional code is systematic, meaning that the encoded sequence  $\underline{x}$  includes the information sequence  $\underline{u}$  as a component.

• The parity-check sequences  $\underline{p}_1$  and  $\underline{p}_2$  are based on the same transfer function:

$$G_1(D) = G_2(D) = G(D).$$

• $\underline{p}_1$  and  $\underline{p}_2$  however,  use different input sequences  $\underline{u}$  and  $\underline{u}_{\pi}$,  respectively.  Here,  ${\rm \Pi}$  marks the interleaver,  for UMTS and LTE mostly a  $S$–random interleaver.

Given filter structure

The main difference compared to the description in the theory part results from a different transfer function  $G(D)$  given by the recursive filter structure drawn on the left.

Hints:

• The exercise belongs to the chapter  "Basics of Turbo Codes".

• Knowledge is expected about
  • the  "Algebraic and Polynomial Description of Convolutional Codes",
  • the  "Code Description with State and Trellis Diagram".

• For further guidance on how to do this, see  $\text{Exercise 4.8}$  and  $\text{Exercise 4.9}$.

• The information sequence  $\underline{u}$  is partially specified by its  $D$–transform for easier description in the subtasks.  For example:

$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad U(D) = D+ D^2\hspace{0.05cm},$$

$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad U(D) = D+ D^8\hspace{0.05cm}.$$

Questions

1

What are the characteristics of the considered turbo code  $($memory  $m$,  influence length  $\nu$,  rate $R)$?

$ m \hspace{0.2cm} = \ $

$ \nu \hspace{0.3cm} = \ $

$R \hspace{0.2cm} = \ $

2

What are the  (identical)  transfer functions  $G_1(D) = G_2(D) = G(D)$?

	$G(D) = (1 + D + D^3)/(1 + D^2 + D^3)$.
	$G(D) = (1 + D^2 + D^3)/(1 + D + D^3)$.

3

What is the impulse response  $\underline{g}$?

	$\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \hspace{0.05cm} \text{...}\hspace{0.05cm})$
	$\underline{g} = (1, \, 1, \, 1, \, 1, \, 0, \, 0, \, 1, \, 0, \, 1, \, 1, \, 1, \, 0, \, 0, \, 1, \, 0, \hspace{0.05cm} \text{...}\hspace{0.05cm})$.
	$\underline{g}$  continues to infinity.

4

Are there periodic components within the impulse response  $\underline{g}$ ?

	Yes,  with period  $P = 7$.
	Yes,  with period  $P = 8$.
	No.

5

Let $U(D) = D + D^2$. Which statements are true?

	The initial sequence  $\underline{p}$  contains a periodic component.
	The period  $P$  is unchanged from  $\underline{g}$ .
	The Hamming weight of the input sequence is  $w_{\rm H}(\underline{u}) = 2$.
	The Hamming weight of the output sequence is  $w_{\rm H}(\underline{p}) = 6$.

6

Which statements are true for  $U(D) = D + D^8$ ?

	The initial sequence  $\underline{p}$  contains a periodic component.
	The period  $P$  is unchanged from  $\underline{g}$ .
	The Hamming weight of the input sequence is  $w_{\rm H}(\underline{u}) = 2$.
	The Hamming weight of the output sequence  is $w_{\rm H}(\underline{p}) = 6$.

Solution

Solution

Polynomial division for subtask (3): $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$

(1)  The code parameters are  $k = 1$  and  $n = 3$   ⇒   code rate  $\underline{R = 1/3}$.

• The memory is  $\underline{m = 3}$.

• The influence lengths result in  $\nu = 1, \ \nu_2 = 4$  and  $\nu_3 = 4$   ⇒   Total influence length  $\underline{\nu = 9}$.

(2)  As the comparison of the  "recursive filter"  on the data page with the  "filter structure"  in the theory section for fractional–rational  $G(D)$,  the  proposed solution 1 is correct.

(3)  Correct are the  proposed solutions 2 and 3:

The upper graph illustrates the polynomial division  $(1 + D + D^3) \ / \ (1 + D^2 + D^3)$.  For explanation:

• Cancelled is the representation with the remainder  $D^8 + D^9 = D^7 \cdot (D + D^2)$.  Thus also holds:

$$(D^8 + D^9) \hspace{0.05cm} /\hspace{0.05cm} (1+ D^2+ D^3 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} D^7 \cdot (D+ D^2+ D^3 + D^6) + {\rm remainder_2}$$

• After summarizing:

$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{ ... }\hspace{0.05cm} \hspace{0.05cm}. $$

• The $D$–inverse transform gives the proposed solution 2:

$$\underline{g}= (\hspace{0.05cm}1\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 1\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{ ... }\hspace{0.05cm})\hspace{0.05cm}. $$

• The impulse response continues to infinity   ⇒   Solution proposal 3 is also correct.

State transition diagram and impulse response

(4)  The impulse response can be expressed as follows:

$$\underline{g}= \Big (\hspace{0.03cm}1\hspace{0.03cm}, \big [ \hspace{0.03cm} 1\hspace{0.03cm}, \hspace{0.03cm} 1\hspace{0.03cm}, \hspace{0.03cm} 1\hspace{0.03cm}, \hspace{0.03cm} 0\hspace{0.03cm}, \hspace{0.03cm} 0\hspace{0.03cm}, \hspace{0.03cm} 1\hspace{0.03cm}, \hspace{0.03cm} 0\hspace{0.03cm} \big ]_{\rm per} \Big ) \hspace{0.15cm}\Rightarrow \hspace{0.15cm} \underline{P = 7} \hspace{0.05cm}. $$

• In the state transition diagram  $($right$)$,  the impulse response  $\underline{g}$  is highlighted in yellow. 

• The impulse response results as the parity-check sequence   $\underline{p}$   for the information sequence   $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{ ... })$.

• The transitions in the diagram are labeled  "$u_i\hspace{0.05cm}|\hspace{0.05cm}\underline{x}_i$",  which is equivalent to  "$u_i\hspace{0.05cm}|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".

• The parity-check sequence   $\underline{p} \ (=$ impulse response  $\underline{g})$   thus results from the respective second coder output symbol.

• The impulse response  $\underline{g}$  is represented by the following states:

$$S_0 → [S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 ] → [S_1 → \ ... \ → S_4] → \ \text{ ... } $$

(5)  The following graph shows the solution using the generator matrix  $\mathbf{G}$.  It holds  $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{ ... } )$.

$\underline{p} = (0, \, 1, \, 1, \, \text{ ... } ) \cdot \mathbf{G}$

$\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }) \cdot \mathbf{G}$

It can be seen that  solutions 1, 2 and 3  are correct:

• The present parity-check sequence  $\underline{p}$  has the same period  $P = 7$  as the impulse response  $\underline{g}$.

• The Hamming weight of the  $($limited$)$  input sequence is actually  $w_{\rm H}(\underline{u}) = 2$.

• Proposition 4 is incorrect.  Rather,  for the semi–infinite output sequence:   $w_{\rm H}(\underline{p}) → \infty$.

• In the transition diagram,  the states

$$S_0 → S_0 → S_1 → S_3 → S_7 → S_6 → S_4 → S_1$$

are passed through first. 

• This is followed  $($infinitely often$)$ by the periodic portion

$$S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 → S_1.$$

(6)  The last graph shows the solution for 

$$U(D) = D + D^8$$

$$ \Rightarrow \hspace{0.3cm}\underline{u} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }).$$

• Correct are the  proposed solutions 3 and 4:

• The input sequence  $\underline{u}$  contains two  "ones"  and the output sequence  $\underline{p}$  six  "ones".

• From position 10  the output sequence is  $\underline{p} \equiv\underline{0}$  
  ⇒   proposals 1 and 2 therefore do not apply.

Further notes:

1. For turbo codes,  especially those input sequences  $\underline{u}$  whose  $D$–transforms are representable as  $U(D) = f(D) \cdot [1 + D^{P}]$  are extremely unfavorable.
2. They cause the error floor as seen on the  "Performance of Turbo Codes" page in the theory section.
3. $P$  indicates here the period of the impulse response  $\underline{g}$.  In our example  $f(D) = D$  and  $P = 7$.