# The response of an initially relaxed linear constant parameter network to a unit impulse applied at t=0 is 4e-2t u(t). The response of this network to a unit step function will be

1.  2[1-e-2t]u(t)

2.  4[e-t-e-2t]u(t)

3.  Sin 2t

4.  (1-4e-4t)u(t)

4

2[1-e-2t]u(t)

Explanation :
No Explanation available for this question

# The closed-loop transfer function of a control system is given by For a unit step input the output is

1.  -3e-2t+4e-t-1

2.  -3e-2t-4e-t+1

3.  Zero

4.  Infinity

4

-3e-2t+4e-t-1

Explanation :
No Explanation available for this question

# The Laplace transform of a continuous-time signal x(t) is; If the Fourier transform of this signal exists, then x(t) is

1.  e2t u(t)-2e-t u(t)

2.  -e2tu(t)+2e-t u(t)

3.  -e2t u(t)+2e-t u(t)

4.  -e2t u(t)-2e-t u(t)

4

e2t u(t)-2e-t u(t)

Explanation :
No Explanation available for this question

# Laplace transform of the signal x(t)=tu(t)*cos2πt u(t) will be

1.

2.

3.

4.

4

Explanation :
No Explanation available for this question

# Laplace transform of the signal will be

1.

2.

3.

4.

4

Explanation :
No Explanation available for this question

# The time signal x(t) corresponding to will be

1.  (2e-2t+e-t)u(t)

2.  (2e-t-e-2t)u(t)

3.  (2e-2t-e-t)u(t)

4.  (2e-t+e-2t)u(t)

4

(2e-t-e-2t)u(t)

Explanation :
No Explanation available for this question

# For the transform pair below, the time signal y(t) will be Y(s) = (s+1)X(s)

1.  [cos 2t–2sin 2t]u(t)

2.

3.  [cos 2t+2sin 2t]u(t)

4.

4

[cos 2t–2sin 2t]u(t)

Explanation :
No Explanation available for this question

# The bilateral Laplace transform of x(t) = e-t u(t+2) will be

1.

2.

3.

4.

4

Explanation :
No Explanation available for this question

# The bilateral Laplace transform of x(t)=e½u(t)+e-t u(t)+et u(-t) will be

1.

2.

3.

4.

4

Explanation :
No Explanation available for this question

# The response of LTI system forwill be

1.  [1-e-10t]u(t)

2.  e-10tu(t)

3.  u(t)

4.  [1-2e-10t]u(t)

4