# GATE1991-5-a

1.6k views

Analyse the circuit in Fig below and complete the following table
$${\begin{array}{|c|c|c|}\hline \textbf{a}& \textbf{b}& \bf{ Q_n} \\\hline 0&0\\\ 0&1 \\ 1&0 \\ 1&1 \\ \hline \end{array}}$$

edited
0
@Arjun Sir here question is incomplete
0
why?
0
where is c) question?
0
yes.. Actually that was moved to a new question- answer is yet to be moved. Each linked question was made separate so as to be included in exam when created.
1

The output of the circuit given as :

$Q=aQ_{n-1}+ab+bQ_{n-1}$

Hence,
$Q_{n}=Q_{n-1}(a+b)+ab$

$00 \implies Q_{n-1}(0+0) + 0.0 = Q_{n-1}(0) + 0 = 0+0 = 0$

$01 \implies Q_{n-1}(0+1) + 0.1= Q_{n-1} (1)+ 0 = Q_{n-1}+0 = Q_{n-1}$

$10 \implies Q_{n-1}(1+0) + 1.0 = Q_{n-1} (1) + 0 = Q_{n-1}+0= Q_{n-1}$

$11 \implies Q_{n-1}(1+1)+ 1.1 =Q_{n-1}(1) + 1 =Q_{n-1}+1 = 1$

$${\begin{array}{cc|c} \textbf{a}& \textbf{b}& \bf{ Q_n} \\\hline 0&0&0\\ 0&1& Q_{n-1} \\ 1&0& Q_{n-1} \\ 1&1 &1\\ \end{array}}$$

edited
1
How the propagation delay is calculated..??

Here delay should be due to one OR gate and one AND gate.
0

@vaishali Part c is different. See this

https://gateoverflow.in/26442/gate1991_5-c

0

11 =>  Qn-1(1+1) + 1.1 = Qn-1 (1) + 1 = Qn-1+1 = 1 hwo it is equals to 1

0
1+x=1 here x=Q n-1 .
0

## Related questions

1
553 views
When two $4-bit$ numbers $A = a_3a_2a_1a_0$ and $B=b_3b_2b_1b_0$ are multiplied, the bit $c_1$ of the product $C$ is given by ________
Consider the number given by the decimal expression: $16^3*9 + 16^2*7 + 16*5+3$ The number of $1’s$ in the unsigned binary representation of the number is ______