Some examplesGoalPrerequisitesHomeworkFurther readingsExercise #6.3.1Solution #1Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Solution #2Slide 31Slide 32Slide 33Slide 34Slide 35Solution #3Slide 37Slide 38Slide 39Slide 40Exercise #6.3.2Solution #1 (truth table)Solution #1 (Karnaugh maps)Exercise #6.3.3Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Exercise #6.3.4Slide 53Slide 54Slide 55Slide 56Exercise #6.3.5PowerPoint PresentationSlide 59Slide 60Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Slide 67Slide 68Slide 69Slide 70Slide 71RemarkRemark (cont’ed)Slide 74Slide 75Slide 76Slide 77Some examplesSome examplesSome examplesSome examplesPaolo PRINETTOPolitecnico di Torino (Italy)University of Illinois at Chicago, IL (USA)[email protected] [email protected] 6.32 6.3 Goal-This lecture guides the students through the solution of some simple examples of manual synthesis of combinational networks.3 6.3 Prerequisites -Lectures 6.1 and 6.24 6.3 Homework -Students are recommended to try to solve the exercise by themselves, before looking at the proposed solutions.5 6.3 Further readings -No particular suggestion6 6.3 Exercise #6.3.1Design a combinational circuit with 1 output U and an input X (3 downto 0) that, when it receives on X an unsigned hexadecimal digit X, provides, on U, a logical value 1 iff:X < 4 or X > 8.x16U7 6.3 Solution #1-We are going to present a complete first solution, i.e., till the netlist of the target circuit.x16U8 6.3 Solution #100 01 11 1000011110U x16UX(1) X(0)X(3) X(2)9 6.3 Solution #100 01 11 1000011110U x16UX(1) X(0)X(3) X(2)HINT:HINT:label each cell with its label each cell with its corresponding decimal valuecorresponding decimal value10 6.3 Solution #1U x16UX(1) X(0)X(3) X(2)00 01 11 1000 0 4 12 801 1 5 13 911 3 7 15 1110 2 6 14 1011 6.3 Solution #100 01 11 1000 101 111 110 1U x16UX(1) X(0)X(3) X(2)X < 4 or X > 8. 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 1012 6.3 Solution #100 01 11 1000 1 101 1 1 111 1 1 110 1 1 1U x16UX(1) X(0)X(3) X(2)X < 4 or X > 8. 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 1013 6.3 Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U x16UX(1) X(0)X(3) X(2)X < 4 or X > 8. 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 1014 6.3 Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U =x16UX(1) X(0)X(3) X(2)15 6.3 X(3)’X(2)’Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U =x16UX(1) X(0)X(3) X(2)16 6.3 X(3)’X(2)’Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’x16UX(1) X(0)X(3) X(2)17 6.3 Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’x16UX(1) X(0)X(3) X(2)18 6.3 X(3) X(2) Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ x16UX(1) X(0)X(3) X(2)19 6.3 X(3) X(2) Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2)x16UX(1) X(0)X(3) X(2)20 6.3 Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2)x16UX(1) X(0)X(3) X(2)21 6.3 X(3) X(0)Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2) x16UX(1) X(0)X(3) X(2)22 6.3 X(3) X(0)Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2) + X(3) X(0)x16UX(1) X(0)X(3) X(2)23 6.3 Solution #100 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2) + X(3) X(0) x16UX(1) X(0)X(3) X(2)24 6.3 Solution #1X(3) X(1)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2) + X(3) X(0) x16UX(1) X(0)X(3) X(2)25 6.3 Solution #1X(3) X(1)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U U = X(3)’X(2)’ + X(3) X(2) + X(3) X(0) + X(3) X(1)x16UX(1) X(0)X(3) X(2)26 6.3 Solution #1U = X(3)’X(2)’ + X(3) X(2) + X(3) X(0) + X(3) X(1)x16U27 6.3 Solution #1U = X(3)’X(2)’ + X(3) X(2) + X(3) X(0) + X(3) X(1)X(3)X(0)X(3)X(1)X(3)X(2)UX(3)X(2)x16UAABBCCDD28 6.3 Solution #1x16UFurther optimization:-by observing that:-one gets:29 6.3 Solution #1X(3)X(0)X(3)X(1)X(3)X(2)UX(3)X(2)x16UAABBCCDD30 6.3 Solution #2x16U-We are going to present now an alternative solution, due to a different cover of the K-map of the circuit.31 6.3 Solution #200 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U = X(3)’X(2)’ + X(3) X(2)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U X(3) X(2) X(3)’X(2)’x16UX(1) X(0)X(3) X(2)32 6.3 Solution #200 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U = X(3)’X(2)’ + X(3) X(2)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U x16UX(1) X(0)X(3) X(2)33 6.3 Solution #200 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U = X(3)’X(2)’ + X(3) X(2) + X(2)’ X(0)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U X(2)’X(0)x16UX(1) X(0)X(3) X(2)34 6.3 Solution #200 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U = X(3)’X(2)’ + X(3) X(2) + X(2)’ X(0)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U x16UX(1) X(0)X(3) X(2)35 6.3 Solution #200 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U = X(3)’X(2)’ + X(3) X(2) + X(2)’ X(0) + X(2)’ X(1)00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1U X(2)’X(1)x16UX(1) X(0)X(3) X(2)36 6.3 Solution #3x16UWe shall now present a solution based on the map composition approach.37 6.3 abcd00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1fSolution #3x16Uf =f =X(3) X(2) X(1) X(0)  abcd38 6.3 abcd00 01 11 1000 1 0 1 001 1 0 1 111 1 0 1 110 1 0 1 1fab00 01 11 1000 - 0 - 001 - 0 - 111 - 0 - 110 - 0 - 1Rcd00 01 11