Enhanced architecture and implementation of spectrum shaping codes

Remove the remainder operation

Next, we propose an improved MASP algorithm with no remainder and square (MASP-NRS). According to the MASP formula Eq. (7), we compute a 360-degree remainder, obtain the related sine-cosine value, and perform an addition. A sine-cosine accumulation requires n clocks. The parallel execution for the accumulation is complex, and serial operation is employed instead. The time complexity of the remainder operation is O(n), while a sine-cosine accumulation requires n remainders. It leads to the time complexity of accumulation O(n²). To reduce this time complexity, we propose an algorithm to remove the remainder operation that includes the following methods.

Improvement 1: Reduce the number of codewords involved in accumulation. The ${\sum }_{i=\left(l-1\right)n+1}^{nl}{w}_{i}cos\left(2\pi f_{s}i\right)$ and ${\sum }_{i=\left(l-1\right)n+1}^{nl}{w}_{i}sin\left(2\pi f_{s}i\right)$ of the current l-th codeword need to be added to the ${\sum }_{i=1}^{nl}{w}_i^{\ast }cos\left(2\pi f_{s}i\right)$ and ${\sum }_{i=\left(l-1\right)n+1}^{nl}{w}_i^{\ast }sin\left(2\pi f_{s}i\right)$ of the previous (l − 1) codewords. We have (13)${\displaystyle \begin{array}{c}R{e}_s^l=\sum _{i=1}^{n\left(l-1\right)}{w}_i^{\ast }cos\left(2\pi f_{s}i\right)+\sum _{i=\left(l-1\right)n+1}^{nl}{w}_{i}cos\left(2\pi f_{s}i\right),\hfill \\ I{m}_s^l=\sum _{i=1}^{n\left(l-1\right)}{w}_i^{\ast }sin\left(2\pi f_{s}i\right)+\sum _{i=\left(l-1\right)n+1}^{nl}{w}_{i}sin\left(2\pi f_{s}i\right).\hfill \end{array}}$ The values of $R{e}_s^l$ and $I{m}_s^l$ increase as the number of codewords increases. After accumulating 64 codewords, we reset $R{e}_s^l$ and $I{m}_s^l$ to 0 to limit these values.

Improvement 2: Eliminate the remainder of the operation and use ROM storage instead. Let the shaping code utilize two dual servo frequencies, f₁ and f₂. A codeword w has n bits that is multiplied by four groups of sine-cosine cos(2πf₁i), sin(2πf₁i), cos(2πf₂i), sin(2πf₂i), and 0 ≤ i ≤ n − 1. Each sine-cosine group contains n data points. A total of 64×4×n sine-cosine values are stored.

Improvement 3: Adopt small bit-width. The initial sine-cosine values need to be transformed from decimals to integers to calculate on the FPGA. Multiply the initial sine-cosine values by 15 and round to the nearest integer number, which is approximately equivalent to moving the values left by four bits. As a result, including the sign bit, the bit width of four groups of sine-cosine values is 5, with a maximum value of 15.

Improvement 4: The parallel operation. Each item in codeword w has a value of either -1 or 1. The select operations can multiply w by sine and cosine. We can acquire n sine-cosine values from ROM at the same time and perform parallel selection operations to complete the sine-cosine accumulation in a single clock cycle. Thus, we eliminate the remainder operation, reducing the accumulation time complexity from O(n²) to O(1).

Remove the square operation

Equation (7) involves a square operation, which has a calculating cost of O(n²) and is challenging to compute. We provide a segmented line search algorithm with dynamic error. The search algorithm seeks segmented points, which are combined to produce a segmented curve. Applying the curve, we obtain an approximate estimation. The operation of this curve only involves deterministic shifts and additions/subtractions with a complexity of O(1). The complexity of the proposed search algorithm is two orders of magnitude lower than that of the square operation. The key features of the algorithm are the usage of dynamical error and the balanced coefficient of mean square error. The search algorithm is described in Algorithm 1 .

 
_____________________________________________________________________________ 
 
Algorithm 1 Segmented line search algorithm with dynamical error 
Require: f(x): the square function; x: the independent variable; 
Ensure: a set of segmented points; 
  1:  x ∈ [x0, x1, ⋅⋅⋅, xn−1], k=1; 
  2:  xb = x0, xv = x2, sp0 = x0; 
  3:  for j = 2; j <= n − 1; j + +  do 
 4:       while xb < xi < xv do 
 5:            compute  ˆ f (xi) = f(xv)−f(xb) 
    xv−xb    (xi − xb) + f(xb); 
  6:       end while 
 7:       compute error =  v−1 
   ∑ 
 i=b+1       [ ˆ f (xi)−f(xi)]2 
            ____________________________________ 
(v−b−1)e| ˆ f (xi)−f(xi)|α 
       β        ; 
  8:       if error ≥|f(xv) − f(xb)|μ then 
 9:            spk = xv−1; 
10:            xb = xv; 
11:            xv = xv+2; 
12:            k = k + 1; 
13:       else 
14:            xb = xb; 
15:            xv = xj; 
16:       end if 
17:  end for 
18:  return sp = [sp0, sp1, sp2, ...]; 
_____________________________________________________________________________    

We generally use expression Eq. (14) to calculate the mean square error. (14)${\displaystyle \begin{array}{c}error=\frac{\sum _{i=1}^n{\left[\hat{f}\left(x_i\right)-f\left(x_i\right)\right]}^2}{n}\hfill \end{array}}$

where $\hat{f}\left(x_i\right)$ represents the predicted value and f(x_i) the actual value.

The large, varied item significantly influences the error expression Eq. (14), but the little various item has a minor impact. So, in Algorithm 1 , we propose a balanced-coefficient mean square error expression Eq. (15) to accurately describe the importance of each item. (15)${\displaystyle \begin{array}{c}erro{r}^{\ast }=\frac{\sum _{i=1}^n{\left[\hat{f}\left(x_i\right)-f\left(x_i\right)\right]}^2}{n\left[e^{\frac{{\left|\hat{f}\left(x_i\right)-f\left(x_i\right)\right|}^{\alpha }}{\beta }}\right]}\hfill \end{array}}$ where α and β are called fast and slow decay factors, respectively.

Multiple segmented line Eq. (16) can be generated when segmented points are provided. The product of x and k₀, k₁, …,  can be replaced by shift operation on x. The complexity of calculating x2 is O(n2), whereas applying Eq. (16) and combining with the shift operation to compute the square of x decreases the complexity to O(1). As a result, we can rewrite Eq. (7) as Eq. (17). (16)${\displaystyle \begin{array}{c}\hat{f}\left(x\right)=\left\{\begin{array}{c}{k}_{0}x+b_0,s{p}_0\le x\le s{p}_1\hfill \\ k_{1}x+b_1,s{p}_1<x\le s{p}_2\hfill \\ k_{2}x+b_2,s{p}_2<x\le s{p}_3\hfill \\ ⋮\hfill \end{array}\right.\hfill \end{array}}$ (17)${\displaystyle \begin{array}{c}\sum _{s=1}^t\left\{\hat{f}\left[R{e}_s^{l-1}+\sum _{i=\left(l-1\right)n+1}^{nl}{w}_{i}cos\left(2\pi f_{s}i\right)\right]+\hat{f}\left[I{m}_s^{l-1}+\sum _{i=\left(l-1\right)n+1}^{nl}{w}_{i}sin\left(2\pi f_{s}i\right)\right]\right\}\hfill \end{array}}$

The values of the cosine and sine functions range from −1 to 1 in Eq. (17). It can obtain large values of ${\left[R{e}_s^{l-1}+{\sum }_{i=\left(l-1\right)n+1}^{nl}{w}_{i}cos\left(2\pi f_{s}i\right)\right]}^2$ and ${\left[I{m}_s^{l-1}+{\sum }_{i=\left(l-1\right)n+1}^{nl}{w}_{i}sin\left(2\pi f_{s}i\right)\right]}^2$, when the length n of a codeword is large and the binary bits w_i are all positive 1. Note that we use the MASP algorithm only for comparison. Thus, we can simultaneously reduce the sum of the two trigonometric functions without affecting the comparison. Then, we can modify Eq. (17) as (18)${\displaystyle \begin{array}{c}\sum _{s=1}^t\left\{\hat{f}\left[\frac{R{e}_s^{l-1}+\sum _{i=\left(l-1\right)n+1}^{nl}{w}_{i}cos\left(2\pi f_{s}i\right)}{num}\right]+\hat{f}\left[\frac{I{m}_s^{l-1}+\sum _{i=\left(l-1\right)n+1}^{nl}{w}_{i}sin\left(2\pi f_{s}i\right)}{num}\right]\right\}.\hfill \end{array}}$ Where num expresses an integer.

The implementation of removing reminder operation

Let the shaping code utilize two kinds of dual servo frequencies, f₁ = 1/90 and f₂ = 1/60. Each sine-cosine group contains 80 data points. We store 64 × 4× 80 groups of sine-cosine values, requiring a total of 64 × 4 × 80 × 5 = 12.5 KB.

Next, using a pipelined operation, we implement the sine-cosine accumulation in Eq. (18). Figure 2 illustrates the pipeline structure.

Step 1: In Fig. 2, we use r_f_kcos_w_i and r_f_ksin_w_i to denote the product of w_i with cos(2πf_ki) and sin(2πf_ki), k = 1, 2, and 0 ≤ i ≤ 79. According to the value of w_i, we use selectors to determine the 80 values of r_f₁cos_w_i, 0 ≤ i ≤ 79.

Step 2: Accumulate r_f₁cos_w_i. If 80 data points are added two by two, the four-stage accumulation will need 40, 20, 10, and 5 addition operations, respectively. Thus, five operands are remaining after the four-stage addition. However, adding these five operands two by two is inconvenient. We construct a segmented accumulation equation because the r_f₁cos_w_i has a small five-bit width Eq. (19). Applying the equation, the first stage accumulation of 80 data requires only 32 addition operations. (19)${\displaystyle \begin{array}{c}\left\{\begin{array}{c}r\text{\_}1\text{\_}ai\mathrm{=}r\text{\_}{f}_{1}cos\text{\_}{w}_{i\ast 2}+r\text{\_}{f}_{1}cos\text{\_}{w}_{i\ast 2+1},0\le i\le 15,\hfill \\ r\text{\_}1\text{\_}ai\mathrm{=}r\text{\_}{f}_{1}cos\text{\_}{w}_{i\ast 2}+r\text{\_}{f}_{1}cos\text{\_}{w}_{i\ast 2+1}+\hfill \\ r\text{\_}{f}_{1}cos\text{\_}{w}_{i+48},16\le i\le 31.\hfill \end{array}\right.\hfill \end{array}}$

Step 3: In Fig. 2, the variables r_1_a0, …, r_1_a31, r_2_a0, …, r_2_a15, r_3_a0, …, r_3_a7, r_4_a0, …, r_4_a3, r_5_a0, …, r_5_a1, r_6_a, comprise a six-step sine-cosine cumulative pipeline. A two-by-two addition is then performed, i.e., (20)${\displaystyle \begin{array}{c}\left\{\begin{array}{c}r\text{\_}2\text{\_}a0=r\text{\_}1\text{\_}a0+r\text{\_}1\text{\_}a1,\hfill \\ r\text{\_}3\text{\_}a0=r\text{\_}2\text{\_}a0+r\text{\_}2\text{\_}a1,\hfill \\ r\text{\_}4\text{\_}a0=r\text{\_}3\text{\_}a0+r\text{\_}3\text{\_}a1,\hfill \\ r\text{\_}5\text{\_}a0=r\text{\_}4\text{\_}a0+r\text{\_}4\text{\_}a1.\hfill \end{array}\right.\hfill \end{array}}$ The cumulative result r_6_a = r_5_a0 + r_5_a1. Each r_6_a of the current codeword needs to be added to the accumulated values of the previous codewords (denoted by Re₁, Im₁, Re₂ and Im₂) to yield r_7_Re₁, r_7_Im₁, r_7_Re₂ and r_7_Im₂.

Step 4: Following accumulation, an operation instead of the square is performed, which is introduced in the next section. The accumulation of an encoded codeword is completed after 14 cycles. At the 5th clock, calculate the next encoded codeword. In Fig. 2, the buffer indicates a cache of one clock.

The implementation of removing square operation

For Eq. (18), a symbol contains 80 bits. Due to the k-constrained algorithm, there will not be five consecutive 1’s and five consecutive 0’s, and a symbol contains no more than 80*80% 1’s. In addition, the sine-cosine values are represented by integers in the range of 0 to 15. In extreme cases, 80*80% 1’s are required to multiply with these sine-cosine values. The sine-cosine values involved in the multiplication are considered as the mean value, 7.5, and then the multiplication result is 80*80% *7.5. The result of the current symbol needs to be added to that of the previous symbol, so the accumulated result can be 80*80% *7.5*2 = 960. To simplify calculating the square of large number, the num in Eq. (18) is set to 16. Thus, 80*80% *7.5*2/16 = 960/16 = 60. The division by 16 yields the same result as a 4-bit right shift. In order to prevent some accumulated results divided by 16 from exceeding 60, we add an overflow control. If any results are greater than 60, the results are set to 60.

In Algorithm 1 , the slow decay and fast decay factors are set to 1/2 and 4, and the µis 0.06, and then we get the segmented line equation

(21)${\displaystyle \begin{array}{c}\hat{f}\left(x\right)=\left\{\begin{array}{c}2x,s{p}_0\le x\le s{p}_1\hfill \\ 6x-8,s{p}_1<x\le s{p}_2\hfill \\ 11x-28,s{p}_2<x\le s{p}_3\hfill \\ 18x-77,s{p}_3<x\le s{p}_4\hfill \\ 26x-165,s{p}_4<x\le s{p}_5\hfill \\ 35x-300,s{p}_5<x\le s{p}_6\hfill \\ 46x-520,s{p}_6<x\le s{p}_7\hfill \\ 58x-832,s{p}_7<x\le s{p}_8\hfill \\ 70x-1216,s{p}_8<x\le s{p}_9\hfill \\ 83x-1710,s{p}_9<x\le s{p}_{10}\hfill \\ 98x-2385,s{p}_{10}<x\le s{p}_{11}\hfill \\ 113x-3180,s{p}_{11}<x\le s{p}_{12}\hfill \end{array}\right.\hfill \end{array}}$

where the values of sp₀, sp₁, sp₂, ..., sp₁₂ are 0, 2, 4, 7, 11, 15, 20, 26, 32, 38, 45, 53, 60. We can explore other appropriate values in conjunction with optimization algorithms such as particle swarms (Chen et al., 2022). By replacing the multiplication in Eq. (21) with a shift operation, Eq. (21) become (22)${\displaystyle \begin{array}{c}\hat{f}\left(x\right)=\left\{\begin{array}{c}x<<1,s{p}_0\le x\le s{p}_1\hfill \\ x<<2+x<<1-8,s{p}_1<x\le s{p}_2\hfill \\ x<<3+x<<1+x-28,s{p}_2<x\le s{p}_3\hfill \\ x<<4+x<<1-77,s{p}_3<x\le s{p}_4\hfill \\ x<<4+x<<3+x<<1-165,s{p}_4<x\le s{p}_5\hfill \\ x<<5+x<<1+x-300,s{p}_5<x\le s{p}_6\hfill \\ x<<5+x<<3+x<<2+x-520,s{p}_6<x\le s{p}_7\hfill \\ x<<5+x<<4+x<<3+x-832,s{p}_7<x\le s{p}_8\hfill \\ x<<6+x<<2+x<<1-1216,s{p}_8<x\le s{p}_9\hfill \\ x<<6+x<<4+x<<1+x-1710,s{p}_9<x\le s{p}_{10}\hfill \\ x<<6+x<<5+x<<1-2385,s{p}_{10}<x\le s{p}_{11}\hfill \\ x<<6+x<<5+x<<4+x-3180,s{p}_{11}<x\le s{p}_{12}\hfill \end{array}\right.\hfill \end{array}}$ where the < < indicates that the variable x is left-shifted.

As shown in Fig. 3, we compare the segmented line function $\hat{f}\left(x\right)$ with the square function f(x). It is seen that the two curves exhibit a high degree of concordance, suggesting a strong resemblance between them. Using Eq. (23), the correlation coefficient rela between the estimated and actual square values equals 1. (23)${\displaystyle \begin{array}{c}rela=\frac{\sum _{i=1}^n\left\{f\left(x_i\right)-E\left[f\left(x_i\right)\right]\right\}\left\{\hat{f}\left(x_i\right)-E\left[\hat{f}\left(x_i\right)\right]\right\}}{\sqrt{\sum _{i=1}^n{\left\{f\left(x_i\right)-E\left[f\left(x_i\right)\right]\right\}}^2\sum _{i=1}^n{\left\{\hat{f}\left(x_i\right)-E\left[\hat{f}\left(x_i\right)\right]\right\}}^2}}\hfill \end{array}}$ where E[f(x_i)] denotes the expected actual value and $E\left[\hat{f}\left(x_i\right)\right]$ denotes the expected estimated value. (24)${\displaystyle \begin{array}{c}td=\frac{rela}{\sqrt{\frac{1-rel{a}^2}{n-2}}}\hfill \end{array}.}$

Then, we define a variable td according to Eq. (24), consult the t-distribution table, and obtain a p-value of 0 that is less than the significance level (p = 0.05). As a result, the correlation coefficient rela is regarded as significant. The $\hat{f}\left(x\right)$ and f(x) are completely correlated. (25)${\displaystyle \begin{array}{c}{R}^2=1-\frac{\sum _{i=1}^n{\left[\hat{f}\left(x_i\right)-f\left(x_i\right)\right]}^2}{\sum _{i=1}^n{\left[\bar{f}\left(x_i\right)-f\left(x_i\right)\right]}^2}.\hfill \end{array}}$

Next, we examine the R² relationship between $\hat{f}\left(x\right)$ and f(x) as stated in Eq. (25). The calculated value of R² is zero, demonstrating that the variance of the difference between $\hat{f}\left(x\right)$ and f(x) is 0% of the variance of f(x). The variance of the difference between $\hat{f}\left(x\right)$ and f(x) is extremely small, indicating that $\hat{f}\left(x\right)$ and f(x) are quite close in value.