Online Test 2 - Digital Logic

ExplanationIn six binary digit form number is

 for n independent Boolean variables, each taking one particular Boolean value, there are 2n different possible combinations. A Boolean function has to assign one Boolean value to each one of these combinations.
This brings the number of different possible Boolean functions of n variables to a total of  $2^{2^n}$

 Z = A$\bar B$C=  $\dfrac{(AC)(\bar B)}{\overline {AC} + B}$Let F
$\therefore$ 5 NAND gates required.

2x + 8 = 5y.x = $\dfrac{5y - 8}{2}$………….. (1)From the above option, if y = 8, x = 16.So the equivalent relationship(28)₁₆ = (50)₈

Propagation delay in 4-bit ripple counter = 4 T_d = 40 ns. Propagation delay in synchronous counter = T_d = 10 ns.

Shift register is used as serial and parallel converter and counter is used as frequency division. Decoder is used as addressing memory chips.

Explanation Y = $\bar A$$\bar B$I₀ + $\bar A$BI₁ + A$\bar B$I₂ + ABI₃= $\bar A$$\bar B$×1 + $\bar A$BC + A$\bar B$×0 + ABC= $\bar A$$\bar B$ + $\bar A$BC + ABC= $\bar A$$\bar B$ + BC

A₁ = A, A₀ = B, I₀ = I₂ = A, I₁ = I₃ = BY = $\bar A_1$$\bar A_0$I₀ + $\bar A_1$A₀ I₁ + A_{1^{$\bar A_0$}}I₂ + A₁ A₀ I₃Y = ^{$\bar A$$\bar B$}A + ^{$\bar A$}BB + A^{$\bar B$}A + ABBY = ^{$\bar A$}B + A^{$\bar B$} + AB = ^{$\bar A$}B + A = A + B

J-K flip flop excitation table is given as,

So, (d) option is correct.

Given truth table:

Table 2 (J and K column constructed using table 1)

J =$\bar Y$

K = X

 According to the definition of flip-flop, it consists of a latch used to store 1-bit of information. Race around condition occurs in J-K flip-flop, if the output continuously toggles. D-type flip-flop's output is equal to input.

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Explanation Let P = 1011 and Q = 1001. Given Y_n = P_n ^$\oplus$Q_n^$\oplus$R_nZ_n = R_n Q_n ^{$\oplus$ $\bar p_n$} R_n ^$\oplus$ Q_{n^{$\bar p_n$}}

                      Where, Y = P + Q + RZ = RQ + $\bar PR$ +  $Q\bar P$   From the circuit given, we have R_n + 1 = Z_n, 1$\le$n$\le$3Z₄ = Y_s (MSB). Hence, the output is 00010, which shows a 4-bit subtractor P – Q. 

0 = $\bar X_1\bar X_2$, 2 = X₁$\bar X_2$,  5 = X₂^{$\bar X_1$},  E = X₁X₂ G = 2 + 5 + E = X₁ + X₂ a = 1b = 0 + 2 = $\bar X_2$c = 0 + 5 = $\bar X_1$d = 1e = 0 + 2 + E = X₁ + X₂ f = 0 + 5 + E = X₁ X₂

$W = R + \bar PQ + \bar RS$

$X = PQ\bar R\bar S+\bar P\bar Q\bar R\bar S+ \bar P\bar Q\bar R\bar S$

$Y = RS + \overline{PR + P\bar Q + \bar P \bar Q} = RS + \bar PR.\overline{PR}.\overline{P\bar Q}\overline{\bar P\bar Q}=RS + (\bar P + bar R)(\bar P + Q)(P + Q) = RS + (\bar P + \bar RQ)(P +Q) = RS + \bar pQ + \bar RPQ + Q\bar R$

$Z=R+S+\overline{PQ+\bar P.\bar Q.\bar R + P\bar Q.\bar S}=R+S+\overline{PQ}.\overline{\bar p\bar Q\bar S} = R + S + (\bar P + Q)(P+Q+R).(\bar P + Q + S)= R + S+ (\bar P + \bar Q(Q +S)).(P+ Q+ R)= R +S +\bar P Q + \bar P R+\bar QPS + \bar QRS(P + Q + R) . (\bar P + Q + S)= R + S + (\bar P + \bar Q (Q + S)) . (P + Q + R)= R + S + \bar P Q + \bar PR + \bar PPS + \bar PRS $

From above, we can conclude that $W = Z, X =\bar Z$

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Complementing the output (Q – P) results in a <span style="font-family:" arial;="" font-size:="" small;="">correct  option.

$ P = 111 0111 01 $
$\bar P = 000 10010 $
$\bar P+ 1 = 000 100 11 \qquad Q = 11100110$
$\bar Q = 000 11 001 \qquad \bar Q + 1 = 000 11 010 \qquad Q – P = 00011010$

$P – Q = 00010011$

$g(x, y, z) = F0 + F1 = i\bar z + i\bar z $

$I_1 = F_2. F_3 = x\bar y.xy = 0g(x,y,z) =\bar z + z = 1$

$=\bar ABC + AB\bar C= B (A\bar C + \bar AC)=B (A + C)(\bar A + \bar C)$

$J_A = \bar ABC + AB\bar C = B (\bar C + A\bar C) $

$K_A = A + \bar A\bar C$

$D = A\bar B\bar C + BC $

$J_A = \bar ABC + AB\bar C = B (\bar AC + A\bar C)$

$K_A = A + \bar A\bar C $

$D = A\bar B\bar C + BC$