1
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
The block diagram of a system with one input u and two outputs y1 and y2 is given below GATE ECE 2011 Control Systems - State Space Analysis Question 23 English

A state space model of the above system in terms of the state vector $$\underline x $$ and the output vector $$\underline y = {\left[ {\matrix{ {{y_1}} & {{y_2}} \cr } } \right]^\tau }$$ is

A
$$\mathop {\underline x }\limits^ \bullet = \left[ 2 \right]\underline x + \left[ 1 \right]u;\underline y = \left[ {\matrix{ 1 & 2 \cr } } \right]x$$
B
$$\mathop {\underline x }\limits^ \bullet = \left[ { - 2} \right]\underline x + \left[ 1 \right]u;\underline y = \left[ {\matrix{ 1 \cr 2 \cr } } \right]x$$
C
$$\mathop {\underline x }\limits^ \bullet = \left[ {\matrix{ { - 2} & 0 \cr 0 & { - 2} \cr } } \right]\underline x + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\underline y = \left[ {\matrix{ 1 & 2 \cr } } \right]x$$
D
$$\mathop {\underline x }\limits^ \bullet = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]\underline x + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\underline y = \left[ {\matrix{ 1 \cr 2 \cr } } \right]x$$
2
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
The logic function implemented by the circuit below is (ground implies a logic "0" GATE ECE 2011 Digital Circuits - Combinational Circuits Question 41 English
A
F= AND (P, Q)
B
F= OR (P, Q)
C
F= XNOR (P, Q)
D
F= XOR (P, Q)
3
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
When the output Y in the circuit below is ‘1’, it implies that data has

GATE ECE 2011 Digital Circuits - Sequential Circuits Question 54 English
A
changed from 0 to 1
B
changed from 1 to 0
C
changed in either direction
D
not changed
4
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
Two D flip-flops are connected as a synchronous counter that goes through the following QB QA sequence $$00 \to 11 \to 01 \to 10 \to 00 \to ......$$
A
$${D_A} = {Q_{B,}}\,{D_B} = {Q_A}$$
B
$${D_A} = {\overline Q _A},\,{D_B} = {\overline Q _B}$$
C
$${D_A} = \left( {{Q_A}\,{{\overline Q }_B}\, + {{\overline Q }_A}\,{Q_B}} \right),\,\,{D_B} = {Q_A}$$
D
$${D_A} = \left( {{Q_A}{Q_B} + {{\overline Q }_A}{{\overline Q }_B}} \right),\,\,{D_B} = {\overline Q _B}$$
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