GATE EE 2002 | GATE EE Year Wise Previous Years Questions

GATE EE 2002

Subjective

-0

A single input single output system with $$y$$ as output and $$u$$ as input, is described by $$${{{d^2}y} \over {d{t^2}}} + 2{{dy} \over {dt}} + 10y = 5{{d\,u} \over {dt}} - 3\,u$$$
For the above system find an input $$u(t),$$ with zero initial condition, that produces the same output as with no input and with the initial conditions.
$${{d\,y\left( {{0^ - }} \right)} \over {dt}} = - 4,\,\,\,y\left( {{0^ - }} \right) = 1$$

GATE EE 2002

MCQ (Single Correct Answer)

-0.3

What is the $$rms$$ value of the voltage waveform shown in Fig? GATE EE 2002 Signals and Systems - Continuous Time Periodic Signal Fourier Series Question 11 English

GATE EE 2002 Signals and Systems - Continuous Time Periodic Signal Fourier Series Question 11 English

$$200/\pi \,V$$

$$100/\pi \,V$$

$$200$$ $$V$$

$$100$$ $$V$$

GATE EE 2002

MCQ (Single Correct Answer)

-0.3

Fourier Series for the waveform, $$f(t)$$ shown in Fig. is GATE EE 2002 Signals and Systems - Continuous Time Periodic Signal Fourier Series Question 12 English

GATE EE 2002 Signals and Systems - Continuous Time Periodic Signal Fourier Series Question 12 English

$${8 \over {{\pi ^2}}}\left[ {\sin \left( {\pi t} \right) + {1 \over 9}\sin \left( {3\,\pi t} \right) + {1 \over {25}}\sin \left( {5\,\pi t} \right) + ........} \right]$$

$${8 \over {{\pi ^2}}}\left[ {\sin \left( {\pi t} \right) - {1 \over 9}\cos \left( {3\,\pi t} \right) + {1 \over {25}}\sin \left( {5\,\pi t} \right) + ........} \right]$$

$${8 \over {{\pi ^2}}}\left[ {\cos \left( {\pi t} \right) + {1 \over 9}\cos \left( {3\,\pi t} \right) + {1 \over {25}}\cos \left( {5\,\pi t} \right) + ........} \right]$$

$${8 \over {{\pi ^2}}}\left[ {\cos \left( {\pi t} \right) - {1 \over 9}\sin \left( {3\,\pi t} \right) + {1 \over {25}}\sin \left( {5\,\pi t} \right) + ........} \right]$$

GATE EE 2002

MCQ (Single Correct Answer)

-0.3

$$s(t)$$ is step response and $$h(t)$$ is impulse response of a system. Its response $$y(t)$$ for any input $$u(t)$$ is given by

$${d \over {d\,t}}\int\limits_0^t s \left( {t - \tau } \right)\,u\left( \tau \right)\,d\,\tau $$

$$\int\limits_0^t s \left( {t - \tau } \right)\,u\left( \tau \right)\,d\,\tau $$

$$\int\limits_0^t {\int\limits_0^\tau s \left( {t - {\tau _1}} \right)\,u\left( {{\tau _1}} \right)\,d{\tau _1}} \,d\tau $$

$${d \over {d\,t}}\int\limits_0^t h \left( {t - \tau } \right)\,u\left( \tau \right)\,d\,\tau $$

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