Parallel processing (DSP implementation)

Consider a function unit (${\displaystyle F_{0}}$) and three tasks (${\displaystyle T_{0}}$, ${\displaystyle T_{1}}$, and ${\displaystyle T_{2}}$). The required time for the function unit ${\displaystyle F_{0}}$ to process those tasks is ${\displaystyle t_{0}}$, ${\displaystyle t_{1}}$, and ${\displaystyle t_{2}}$, respectively. Then, if we operate these three tasks in a sequential order, the required time to complete them is ${\displaystyle t_{0}+t_{1}+t_{2}}$.

However, if we duplicate the function unit to another two copies (${\displaystyle F}$), the aggregate time is reduced to ${\displaystyle max(t_{0},t_{1},t_{2})}$, which is smaller than in a sequential order.

Mechanism:

• Parallel: duplicated function units working in parallel
  • Each task is processed entirely by a different function unit.
• Pipelining: different function units working in parallel
  • Each task is split into a sequence of sub-tasks, which are handled by specialized and different function units.

Objective:

• Pipelining leads to a reduction in the critical path, which can increase the sample speed or reduce power consumption at the same speed, yielding higher performance per watt.
• Parallel processing techniques require multiple outputs, which are computed in parallel in a clock period. Therefore, the effective sample speed is increased by the level of parallelism.

Consider a condition that we are able to apply both parallel processing and pipelining techniques, it is better to choose parallel processing techniques with the following reasons

• Pipelining usually causes I/O bottlenecks
• Parallel processing is also utilized for reduction of power consumption while using slow clocks
• The hybrid method of pipelining and parallel processing further increase the speed of the architecture

Consider a 3-tap FIR filter:[2]

${\displaystyle y(n)=ax(n)+bx(n-1)+cx(n-2)}$

which is shown in the following figure.

Assume the calculation time for multiplication units is T_m and T_a for add units. The sample period is given by

${\displaystyle T_{\text{sample}}\geq T_{m}+2T_{a}}$

By parallelizing it, the resultant architecture is shown as follows. The sample rate now becomes

${\displaystyle T_{\text{sample}}\geq {\frac {T_{\text{clock}}}{N}}={\frac {T_{m}+2T_{a}}{3}}}$

where N represents the number of copies.

Please note that, in a parallel system, ${\displaystyle T_{\text{sample}}\neq T_{\text{clock}}}$ while ${\displaystyle T_{\text{sample}}=T_{\text{clock}}}$ holds in a pipelined system.

Consider the transfer function of a 1st-order IIR filter formulated as

${\displaystyle H(z)={\frac {z^{-1}}{1-az^{-1}}}}$

where |a| ≤ 1 for stability, and such filter has only one pole located at z = a;

The corresponding recursive representation is

${\displaystyle y(n+1)=ay(n)+u(n)}$

Consider the design of a 4-parallel architecture (N = 4). In such parallel system, each delay element means a block delay and the clock period is four times the sample period.

Therefore, by iterating the recursion with n = 4k, we have

${\displaystyle y(n+4)=a^{4}y(n)+a^{3}u(n)+a^{2}u(n+1)+au(n+2)+u(n+3)}$

${\displaystyle \rightarrow y(4k+4)=a^{4}y(4k)+a^{3}u(4k)+a^{2}u(4k+1)+au(4k+2)+u(4k+3)}$

The corresponding architecture is shown as follows.

The resultant parallel design has the following properties.

• The pole of the original filter is at z = a while the pole for the parallel system is at z = a⁴ which is closer to the origin.
• The pole movement improves the robustness of the system to the round-off noise.
• Hardware complexity of this architecture: N×N multiply-add operations.

The square increase in hardware complexity can be reduced by exploiting the concurrency and the incremental computation to avoid repeated computing.

Another advantage for the parallel processing techniques is that it can reduce the power consumption of a system by reducing the supply voltage.

Consider the following power consumption in a normal CMOS circuit.

${\displaystyle P_{\text{seq}}=C_{\text{total}}\cdot V_{0}^{2}\cdot f}$

where the C_total represents the total capacitance of the CMOS circuit.

For a parallel version, the charging capacitance remains the same but the total capacitance increases by N times.

In order to maintain the same sample rate, the clock period of the N-parallel circuit increases to N times the propagation delay of the original circuit.

It makes the charging time prolongs N times. The supply voltage can be reduced to βV₀.

Therefore, the power consumption of the N-parallel system can be formulated as

${\displaystyle P_{\text{para}}=(NC_{\text{total}})\cdot (\beta V_{0}^{2})\cdot {\frac {f}{N}}=\beta ^{2}P_{\text{seq}}}$

where β can be computed by

${\displaystyle N(\beta V_{0}-V_{t})^{2}=\beta (V_{0}-V_{t})^{2}.\,}$