GATE ECE 2001 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2001

MCQ (Single Correct Answer)

-0.6

The Inverting Op-Amp shown in the figure, has an Open-loop gain of 100. The closed loop gain $${{{V_o}} \over {{V_s}}}$$ is GATE ECE 2001 Analog Circuits - Operational Amplifier Question 54 English

GATE ECE 2001 Analog Circuits - Operational Amplifier Question 54 English

-8

-9

-10

-11

GATE ECE 2001

MCQ (Single Correct Answer)

-0.6

In the figure assume the Op-Amps to be ideal. The output V₀ of the circuit is GATE ECE 2001 Analog Circuits - Operational Amplifier Question 53 English

GATE ECE 2001 Analog Circuits - Operational Amplifier Question 53 English

10 $$\cos $$ (100 t)

10 $$\int\limits_0^t {\cos \,\,\left( {100\tau } \right)} \,\,d\tau $$ v

$${10^{ - 4}}\int\limits_0^t {\cos \,\,\left( {100\tau } \right)} \,\,d\tau $$

$${10^{ - 4}}{d \over {dt}}\cos \left( {100t} \right)$$

GATE ECE 2001

MCQ (Single Correct Answer)

-0.6

An Amplifier using an Op-Amp with a Slew-Rate SR = V/$$\mu $$sec, has a gain of 40 dB. If this amplifier has to faith fully amplify sinsoidal signals from dc to 20 KHz. Without introducing any Slew Rate induced distortion, then the input signal level must not exceed.

795 mV

395 mV

79.5 mV

39.5 mV

GATE ECE 2001

MCQ (Single Correct Answer)

-0.3

The PDF of a Gaussian random variable X is given by $${p_x}(x) = \,{1 \over {3\sqrt {2\pi } }}\,\exp \,[ - \,{(x - 4)^2}/18]$$.

The probability of the event {X = 4} is

1/2

$$1/\left( {3\sqrt {2\,\pi } } \right)$$

1/4

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