1
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
Consider a linear system whose state space Representation is $$\mathop x\limits^ \bullet \left( t \right) = AX\left( t \right).$$
If the initial state vector of the system is $$x\left( 0 \right) = \left[ {\matrix{ 1 \cr { - 2} \cr } } \right],$$
then the system response is $$x\left( t \right) = \left[ {\matrix{ {{e^{ - 2t}}} \cr { - 2{e^{ - 2t}}} \cr } } \right].$$
If the initial state vector of the system changes to $$x\left( 0 \right) = \left[ {\matrix{ 1 \cr { - 1} \cr } } \right],$$
then the system response becomes $$x\left( t \right) = \left[ {\matrix{ {{e^{ - t}}} \cr { - {e^{ - t}}} \cr } } \right].$$

The eigen value and eigen vector pairs $$\left( {{\lambda _{i,}}{V_i}} \right)$$ for the system are

A
$$\left[ { - 1,\left[ {\matrix{ 1 \cr { - 1} \cr } } \right]} \right]and\left[ { - 2,\left[ {\matrix{ 1 \cr { - 2} \cr } } \right]} \right]$$
B
$$\left[ { - 2,\left[ {\matrix{ 1 \cr { - 1} \cr } } \right]} \right]and\left[ { - 1,\left[ {\matrix{ 1 \cr { - 2} \cr } } \right]} \right]$$
C
$$\left[ { - 1,\left[ {\matrix{ 1 \cr { - 1} \cr } } \right]} \right]and\left[ {2,\left[ {\matrix{ 1 \cr { - 2} \cr } } \right]} \right]$$
D
$$\left[ {1,\left[ {\matrix{ 1 \cr { - 1} \cr } } \right]} \right]and\left[ { - 2,\left[ {\matrix{ 1 \cr { - 2} \cr } } \right]} \right]$$
2
GATE ECE 2007
MCQ (Single Correct Answer)
+1
-0.3
X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is:
A
100111
B
001000
C
000111
D
101001
3
GATE ECE 2007
MCQ (Single Correct Answer)
+1
-0.3
The Boolean function Y=AB+CD is to be realized using only 2-input NAND gates. The minimum number of gates required is
A
2
B
3
C
4
D
5
4
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
The Boolean expression Y= $$\overline A \,\overline B \,\overline C \,D + \overline A BC\overline D + A\overline {B\,} \overline C \,D + AB\overline C \,\overline D $$
A
Y = $$\overline A \,\overline B \,\overline C \,D + \overline A B\overline C + A\overline C D$$
B
Y = $$\overline A \,\overline B \,\overline C \,D + BC\overline D + A\overline B \overline C \,D$$
C
Y=$$ \overline A \,BC\,\overline D + \overline B \,\overline C D + A\overline B \overline C \,D$$
D
Y= $$\overline A \,BC\,\overline D + \overline B \,\overline C D + AB\overline C \,\overline D $$
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