# Transition diagram28 f combined excitation map 28 g

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E.Transition diagram28F.Combined excitation map28G.Individual excitation maps28H.Output map29I.Logic diagram29J.Structural modeling29K.Dataflow modeling30L.Behavioral modeling30M.Result Waveforms32Quine-Mccluskey using......................................................................................................................33A.Question 133B.Question 234
Boolean function1)Consider boolean function f (a, b, c, d) =∏M(0,4,7,11,14)d(6,8,9,13)Implement f in the following forms:(a) minimized form of SOP (sum of product).(b) minimized form of POS (product of sum).(c) minimized form of ALL-NAND.(d) minimized form of ALL-NOR.(e)using2 × 1 multiplexors only (Shannon decomposition).(f) using4 x 1 multiplexors(g) using the Reed-Muller (R-M) approach to obtain the minimized EXOP (XOR of productterms) expansion.(h) multi-level logic (use weak division with a divisor of your choice)(i) multi-level logic (use strong division with a divisor of your choice)Answer:(a)SOP: f = m(1,2,3,5,10,12,15)+d(6,8,9,13) = ac’+c’d+abd+a’b’d+b’cd’abcd0001111000001X0111XX111010101X01k-map for f (a, b, c, d)(b)POS: f = ∏M(0,4,7,11,14)d(6,8,9,13) = (a+c+d)(a+b’+c’)(b’+c’+d)(a’+b+d’)(c)ALL-NAND: shown on the submitted paper(d)ALL-NOR: shown on the submitted paper(e)2 × 1 multiplexors:1.f(a,b,c,d) = d’f(a,b,c,0)+df(a,b,c,1)= d’(ac’+b’c)+d(c’+ab+a’b’)2.abc000111100dd11110dd’ab 010c+dc’+d’1c'dc'+d
(f)4 x 1 multiplexorsf(a,b,c,d) = c’d’f(a,b,0,0)+c’df(a,b,0,1)+cd’f(a,b,1,0)+ cdf(a,b,1,1)=c’d’a+c’d1+cd’b’+cd(a’b’+ab)(g) Reed-Muller (R-M) approach:f = m(1,2,3,5,10,12,15)+d(6,8,9,13)= m(0,4,7,11,14)d(6,8,9,13)f0=f4=f7=f11=f14=0, f1=f2=f3=f5=f10=f12=f15=1 and f6=0, f8=f9=f13=1.shown on the submitted paper(h)weak division:f=ac’+c’d+abd+a’b’d+b’cd’g=a+df/a={c’,bd}f/d={c’,a’b’}quotient: f/g=f/a∩f/d={c’,bd}∩{c’,a’b’}={c’}remainder: f=g•(f/g)+r→ r = f-g(f/g)= ac’+c’d+abd+a’b’d+b’cd’-ac’+dc’= abd+a’b’d+b’cd’Final result: f=g•(f/g)+r=(a+d)•c’+abd+a’b’d+b’cd’
(i)strong division: Function and divisor: f=ac’+c’d+abd+a’b’d+b’cd’, g=a+dCalculate fDCand f~:fDC=g(a+d)=g’a+g’d+ga’d’ ; f~=f+fDCabcd00011110001101111111111011abcd00011110001x1x011x1x1x1x111xx1xx101x1abcd0001111000xx110111111111101xx1k-map for f(a,b,c,d)k-map for f(a,b,c,d) when g=0k-map for f(a,b,c,d) when g=1quotient: c’remainder:abd+a’b’d+b’cd’Final result: f~=f+fDC=f=c’(a+d)+abd+a’b’d+b’cd’Asynchronous circuitB.Problem Assumption1.Since the train can run from different direction, thus totally two defined section and a crossing need to be monitored by four sensors. Two sensors keep watch on north direction which implied by input X1, other two sensors monitor south direction which implied by input X2. The first sensor is responsible for the defined section, and the second sensor pay attention on the crossing.2. When the input’s voltage level is high, it means the train is entering the deifned section or crossing. Thus the gate should off and red light should flash, meanwhile the output’s voltage levelwill be high which implemented by output Z1. 3. There are eight stable states have been designed as following, they are A, B, C, D, E, F, G, H. Moreover, two feedback variables Y1fand Y2f should be included,in that case, two excitation variablesY1eand Y2e