GATE ECE 1988 | GATE ECE Year Wise Previous Years Questions

GATE ECE 1988

MCQ (Single Correct Answer)

-0.6

In a digital communication system, transmissions of successive bits through a noisy channel are assumed to be independent events with error probability p. The probability of at most one error in the transmission of an 8 - bit sequence is

$$7\left( {1 - p} \right)/8 + p/8$$

$${\left( {1 - p} \right)^8} + 8p{\left( {1 - p} \right)^7}$$

$${\left( {1 - p} \right)^8} + {\left( {1 - p} \right)^7}$$

$${\left( {1 - p} \right)^8} + p{\left( {1 - p} \right)^7}$$

GATE ECE 1988

MCQ (Single Correct Answer)

-0.6

A critically damped, continuous-time, second order system, when sampled, will have ( in Z domain)

A simple pole

double pole on real axis

Double pole on imaginary axis

a pair of complex conjugate poles

GATE ECE 1988

MCQ (Single Correct Answer)

-0.6

Consider a characteristic equation given by
s⁴ + 3s³ + 5s² + 6s + K + 10 = 0.
The condition for stability is

K > 5

-10 < K

K > -4

-10 < K < -4

GATE ECE 1988

MCQ (Single Correct Answer)

-0.6

Consider a closed-loop system shown in fig. (a) below. The root locus for it is shown in fig. (b). The closed-loop transfer function for the system is GATE ECE 1988 Control Systems - Root Locus Diagram Question 5 English

GATE ECE 1988 Control Systems - Root Locus Diagram Question 5 English

$$\frac K{1+\left(0.5s+1\right)\left(10s+1\right)}$$

$$\frac K{\left(s+2\right)\left(s+0.1\right)}$$

$$\frac K{1+K\left(0.5s+1\right)\left(10s+1\right)}$$

$$\frac K{K+0.2\left(0.5s+1\right)\left(10s+1\right)}$$

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