The University of Toleo s7ms_il7.fm - EECS: igital Logic esign r. nthony. Johnson Stuent Name_ igital Logic esign Miterm # Problems Points. 3. 4 3. 6 4. Total 5 Was the eam fair? yes no /6/7
The University of Toleo s7ms_il7.fm - EECS: igital Logic esign r. nthony. Johnson Stuent Name_ Problem 3 points For full creit, mark your answers yes, no, or not applicable for all offere choices!. The bit strings shown below are vali representations of negative numbers in the four-bit two s complement representation of negative numbers? yes no not applicable _. Shown below is the truth table of a boolean function F(,). Given net to the truth table is a list of boolean function names. For the names in the list, inicate which names are, an which names are not, the name of the function F(,). F(,) yes no not applicable XNOR, N, NOR, NN,.3 SOM (sum of minterms) representation of oolean functions lens itself to irect implementation using the following types of two-level logic circuits: yes no not applicable NN-NN, NOR-NOR, OR-N, N-OR. /6/7
The University of Toleo s7ms_il7.fm - 3 EECS: igital Logic esign r. nthony. Johnson Stuent Name_ Problem 4 points Positional representations of the function (i), i=,,3, in various raies are shown in Table. i function (i) (i) -8 4 8 (i) s ecimal representation -8 97 Table eight-bit base-two representation of (i) -(i) 3-4E 6-78 Problem statement Using the values of function (i), i =,,3, emonstrate an ability to:. convert by han the liste values of (i), i =,,3, to ecimal representation;. convert by han the liste values of (i), i =,,3, to eight-bit two s complement representation; 3. perform by han the subtraction of numbers in two s complement representation. Problem Solution For full creit, eplicit emonstration of unerstaning the following solution steps is epecte. 3. Epress (i), for i =,,3, in ecimal representation, an epress both, (i) an -(i) in the eightbit base-two representation which uses the two s complement notation for negative numbers. Show your computation on the opposite page, an enter the results into Table. Hint#:irect conversion from octal an heaecimal to binary representation is easier, an shoul be applie. Stuents are avise to avoi an inirect, e.g. octal ecimal binary conversion. No partial creit will be given for a correct conversion from an erroneous ecimal representation.. Using the eight-bit base-two representation an the two s complement notation for negative numbers, show the calculation of the ifference ()-(3) when only an aer circuit is available. Hint#:This part of the problem is consiere inepenent of the results of calculation performe in part.. No partial creit will be earne for a correct calculation proceure applie to erroneous binary representations of () an -(3) - whether taken from Table, or calculate inepenently. = () + = -(3) = - Result of calculation uner. to be grae: /6/7
The University of Toleo s7ms_il7.fm - 4 EECS: igital Logic esign r. nthony. Johnson Stuent Name_ Problem 3 6 points Given is an incompletely specifie logical/switching function F (,,C,) in the ecimal lists of sumsof minterms an on t cares representation (3-) F (,,C,) = Σm(,, 4, 6, 8,, 3) + (3, 7,, 5) (3-) Problem statement On the eample of the given function F, emonstrate an ability to:. erive the Truth table an Karnaugh map representations of F,. use the Karnaugh map metho to erive a minimal number of literals SOP epression of F, 3. esign the two-level N-OR an NN-NN implementations of function F, as specifie in sections 3.3 an 3.4 below. Hint # For full creit: all equations, all answers to questions, all circuit moels an other graphical representations are epecte to be entere into the space esignate for them; all shown numerical results must be precee by the symbolic an numeric epressions whose evaluation prouces these numerical results. Problem Solution For full creit, eplicit emonstration of unerstaning the following solution steps is epecte. 3. Prepare the truth table an the Karnaugh map representation of the function F an show your result in the space reserve for Figures 3-(a).an 3-(b). C F (a) C (b) F = + + (c) Figure 3- Representation forms of the function F. (a)truth table. (b)karnaugh map. (c)minimum number of literals SOP representation of F. /6/7
The University of Toleo s7ms_il7.fm - 5 EECS: igital Logic esign r. nthony. Johnson Stuent Name_ 3. pply the Karnaugh map minimization metho to erive the minimum number of literals sum-of-proucts (SOP) representation of the function F. Enter the erive algebraic representation of F in the space reserve for Figure 3-(c). 3.3 In the space reserve for Figure 3-(a), prepare a logic circuit moel of the two-level N-OR form of implementation of the erive, Figure 3-(c) epression of function F. 3.4 In the space reserve for Figure 3-(b), prepare a logic circuit iagram of the two-level NN-NN form of implementation of the erive, Figure 3-(c) epression of function F F F (a) (b) Figure 3- Two-level implementations of the algebraic epression of Figure 3-(c) of function F. (a) N- OR implementation. (b)nn-nn implementation of F. /6/7
The University of Toleo s7ms_il7.fm - 6 EECS: igital Logic esign r. nthony. Johnson Stuent Name_ Problem 4 points Given is the epression (4-) of a logical function F. F (X,Y,Z) = X Y Z+X Y Z+X Y Z (4-) Problem Statement emonstrate an ability to:. apply the algebraic manipulation metho to erive the minimum number of literals SOP/ POS (sum-of-proucts, or prouct-of-sums) representation of a logic/switching function F. Hint # For full creit, give answers to all questions, prepare all require circuit iagrams, write all equations for which the space has been reserve, an show all symbolic an numerical epressions whose evaluation prouces shown numerical results. Problem Solution For full creit, eplicit emonstration of unerstaning the following solution steps is epecte. 4. Using the algebraic manipulation metho, erive from the epression (4-) the minimum number of literals SOP/POS representation of the logical function (4-). Show your manipulation below, or on the opposite page, an enter the result in the space reserve for equation (4-). F = X Y Z + X Y Z + X Y Z = = X Y (Z+Z) + X+Y+Z = = X Y + X+Y+Z= = X Y + X+Y+Z = X+Y+Z T4(a) Simplifie epression of F to be grae: F = X+Y+Z (4-) /6/7