ECE 275: Multi-level synthesis

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\[ \newcommand{\bx}{\bar{x}} \newcommand{\by}{\bar{y}} \newcommand{\bz}{\bar{z}} \newcommand{\bA}{\bar{A}} \newcommand{\bB}{\bar{B}} \newcommand{\bC}{\bar{C}} \newcommand{\bD}{\bar{D}} \]

Announcments

  • Homework 3 is due on Sept 27th, Monday before class.

Hazards: Ex1

Assume all gates have a propagation delay of 10ns.
  1. Identify the transition when static 1 hazard will happen,
  2. draw the corresponding timing diagram,
  3. and modify the circuit to hazard free circuit.

Hazards: Ex1 Soln

\[ f = A\bar{B} + BC \]
\(\bar{A}\) \(A \)
\(\bar{B}\) \(B \) \(\bar{B}\)
\(\bar{C}\) 0 0 0 1
\(C\) 0 1 1 1
  • Identify adjacent non-overlapping terms
  • Identify the variable that is transitioning between them.
  • Identify the longer path of the circuit with that variable.
  • For a SOP circuit, the transition is that makes to 0->1 for longer circuit and 1->0 for shorter circuit.
\(\bar{A}\) \(A \)
\(\bar{B}\) \(B \) \(\bar{B}\)
\(\bar{C}\) 0 0 0 1
\(C\) 0 1 1+ 1 1+1

Hazards: Ex2

Propagation delay of NOT gate=3 ns, AND/OR gate=5 ns

\[ F = (A + C)(\bA+\bD)(\bB+\bC+D)\]

Multi-level synthesis

  • SOP and POS are two level circuits
  • Fan-in is the number of inputs to a gate
  • Two-level circuits have higher fan-in, but smaller propagation delays.
  • Fan-in is typically limited by the technology used.

Approaches Multi-level synthesis

  • Factorization
  • Functional decomposition

Factorization Ex1

\[ f = ABC + ABD + \bar{A}\bar{B}C \]

Factorization Ex2

\[f_1 = ABD + CD \] \[f_2 = AB\bar{D} + C \bar{D} \]

Functional decomposition Ex1

\[ f = \bx_1 x_2 x_3 + x_1 \bx_2 x_3 + \bx_1 \bx_2 x_4 + x_1 x_2 x_4 \]

Functional decomposition Ex1

\[ f = \bx_1 x_2 x_3 + x_1 \bx_2 x_3 + \bx_1 \bx_2 x_4 + x_1 x_2 x_4 \]
\(\bar{x}_1\) \(x_1 \)
\(\bar{x}_2\) \(x_2 \) \(\bar{x}_2\)
\(\bar{x}_3\)\(\bar{x}_4\) 0 0 0 0
\(x_4\) 1 0 1 0
\(x_3\) 1 1 1 1
\(\bar{x}_4\) 0 1 0 1

Functional decomposition Ex2

\[ f = \sum m(0, 6, 8, 4) \]
\(\bar{x}_1\) \(x_1 \)
\(\bar{x}_2\) \(x_2 \) \(\bar{x}_2\)
\(\bar{x}_3\)\(\bar{x}_4\) 1 0 0 1
\(x_4\) 0 0 0 0
\(x_3\) 0 0 0 0
\(\bar{x}_4\) 0 1 1 0

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