GATE ECE 1991 | Continuous Time Linear Invariant System Question 50 | Signals and Systems

The voltage across an impedance in a network is V(s) = Z(s) I(s), where V(s), Z(s) and $${\rm I}$$(s) are the Laplace Transforms of the corresponding time functions V(t), z(t) and i(t).

The voltage v(t) is

$$v\left( t \right) = z\left( t \right)\,.\,i\left( t \right)$$

$$v\left( t \right) = \int\limits_0^t {i\left( \tau \right)\,z\left( {t - \tau } \right)d\tau } $$

$$v\left( t \right) = \int\limits_0^t {i\left( \tau \right)z\left( {t + \tau } \right)d\tau } $$

$$v\left( t \right) = z\left( t \right) + i\left( t \right)$$

GATE ECE 1990

MCQ (Single Correct Answer)

-0.6

The impulse response and the excitation function of a linear time invariant casual system are shown in Fig. a and b respectively. The output of the system at t = 2 sec. is equal to GATE ECE 1990 Signals and Systems - Continuous Time Linear Invariant System Question 31 English

GATE ECE 1990 Signals and Systems - Continuous Time Linear Invariant System Question 31 English

1/2

3/2

GATE ECE 1990

MCQ (Single Correct Answer)

-0.6

The response of an initially relaxed linear constant parameter network to a unit impulse applied at $$t = 0$$ is $$4{e^{ - 2t}}u\left( t \right).$$ The response of this network to a unit step function will be:

$$2\left[ {1 - {e^{ - 2t}}} \right]u\left( t \right)$$

$$4\left[ {{e^{ - t}} - {e^{ - 2t}}} \right]u\left( t \right)$$

$$\sin 2t$$

$$\left( {1 - 4{e^{ - 4t}}} \right)u\left( t \right)$$

GATE ECE 1988

MCQ (Single Correct Answer)

-0.6

The transfer function of a zero-order hold is

$${{1 - \exp ( - Ts)} \over s}$$

$$1/s$$

$$1/\left[ {1 - \exp \left( { - Ts} \right)} \right]$$

GATE ECE Subjects

Signals and Systems

Network Theory

Control Systems

Digital Circuits

General Aptitude

Electronic Devices and VLSI

Analog Circuits

Engineering Mathematics

Microprocessors

Communications

Electromagnetics