1
GATE ECE 2013
MCQ (Single Correct Answer)
+2
-0.6
The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage Vwx1 = 100V is applied across WX to get an open circuit voltage YZ1 V across YZ. Next, an ac voltage VYZ2 =100V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 / VWX1 , VWX2 / VYZ2 are respectively, GATE ECE 2013 Network Theory - Network Theorems Question 13 English
A
125 / 100 and 80 / 100
B
100 / 100 and 80 / 100
C
100 / 100 and 100 / 100
D
80 / 100 and 80 / 100
2
GATE ECE 2013
MCQ (Single Correct Answer)
+2
-0.6
A system is described by the differential equation $$${{{d^2}y} \over {d{t^2}}} + 5{{dy} \over {dt}} + 6y\left( t \right) = x\left( t \right)$$$
Let x(t) be a rectangular pulse given by $$$x\left( t \right) = \left\{ {\matrix{ {1\,\,\,\,\,\,\,\,\,0 \le \,t\, \le 2} \cr {0\,\,\,\,\,otherwise} \cr } } \right.$$$

Assuming that y(0) = 0 $${{dy} \over {dt}} = 0$$ at t = 0, the Laplace transform of y(t) is

A
$${{{e^{ - 2s}}} \over {s\left( {s + 2} \right)\left( {s + 3} \right)}}$$
B
$${{1 - {e^{ - 2s}}} \over {s\left( {s + 2} \right)\left( {s + 3} \right)}}$$
C
$${{{e^{ - 2s}}} \over {\left( {s + 2} \right)\left( {s + 3} \right)}}$$
D
$${{1 - {e^{ - 2s}}} \over {\left( {s + 2} \right)\left( {s + 3} \right)}}$$
3
GATE ECE 2013
MCQ (Single Correct Answer)
+2
-0.6
The DFT of a vector [a b c d] is the vector [α β γ δ ]. Consider the product GATE ECE 2013 Signals and Systems - Discrete Fourier Transform and Fast Fourier Transform Question 10 English The DFT of the vector [ p q r s] is a scaled version of
A
$$\left[ {{\alpha ^2}{\beta ^2}{\Upsilon ^2}{\delta ^2}} \right]$$
B
$$\sqrt {\alpha \,} \,\sqrt \beta \,\sqrt \gamma \,\sqrt \delta $$
C
$$\left[ {\alpha + \beta \,\beta + \delta \, + \gamma \,\gamma \, + \alpha } \right]$$
D
$$\left[ {\delta {\rm{ }}\beta \,\gamma \,\delta } \right]$$
4
GATE ECE 2013
MCQ (Single Correct Answer)
+1
-0.3
The impulse response of a system is h(t) = t u(t). For an input u(t - 1), the output is
A
$${{{t^2}} \over 2}u\left( t \right)$$
B
$${{t\left( {t - 1} \right)} \over 2}u\left( {t - 1} \right)$$
C
$${{{{\left( {t - 1} \right)}^2}} \over 2}u\left( {t - 1} \right)\,$$
D
$${{{t^2} - 1} \over 2}u\left( {t - 1} \right)$$
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