# Test: Transfer Function- 2

## 15 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: Transfer Function- 2

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Attempt Test: Transfer Function- 2 | 15 questions in 45 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Topicwise Question Bank for Electrical Engineering for Electrical Engineering (EE) Exam | Download free PDF with solutions
QUESTION: 1

### Match List - I (Roots in s-plane) with List - II (Corresponding impulse response) and select the correct answer using the codes given below the lists: List - I    List - II    Codes: Solution:

For A: Two conjugate poles on imaginary axis i.e. it’s impulse response = sin at (marginally stable).
For B: Two poles at origin i.e. impulse response = t u(t) (unstable system).
For C: Two complex conjugate poles with negative real parts. So, its impulse response = er-at sin bt (stable system).
For D: Two imaginary complex double roots. So, impulse response = t sin bt. (unstable system).

QUESTION: 2

Solution:
QUESTION: 3

### The transfer function is applicable to

Solution:

Transfer function is applicable to only linear and time-variant systems.

QUESTION: 4

The DC gain of the system represented by the following transfer function is Solution:

Given transfer function is Converting above transfer function in time-constant form, we get: Hence, dc gain is QUESTION: 5

The unit impulse response of a control system is given by c(t) = -te-t + e-t. Its transfer function is

Solution:

Given, c(t) = -te-t + e-t (impulse response)
We know that,   QUESTION: 6

The impulse response of an initially relaxed linear system is e-3t u(t). To produce a response of te-3t u(t), the input must be equal to

Solution:

We know that, impulse response For given system, Now, c(t) = t e-3t So, R{s) = C(s) x (s + 3) or, r(t) = Required input = e-3t u(t)

QUESTION: 7

A linear time invariant system, initially at rest when subjected to a unit step input gave response c(t) = te-t(t ≥ 0). The transfer function of the system is

Solution:

Given, c(t) = te-t Now, r(t) = u(t) (unit step input)
∴ R(s) = 1/s
So, QUESTION: 8

A linear time invariant system having input r(t) and output y(t) is represented by the differential equation The transfer function of the given system is represented as

Solution:

Taking Laplace transform of the given differential equation on both sides, we get
2s2 Y(s) + sY(s) + 5 Y(s) = R(s) + 2e-s R(s)
or, QUESTION: 9

The open loop transfer function of a system is A closed loop pole will be located at s = -12
when the value of K is

Solution:  QUESTION: 10

The singularities of a function are the points in the s-plane at which the function or its derivates

Solution:
QUESTION: 11

Assertion (A): The final value theorem cannot be applied to a function given by Reason (R): The function s F(s) has two poles on the imaginary axis of s-plane.

Solution:

The final value theorem is valid only if sF(s) does not has any poles on the jω axis and in the right half of the s-plane. Hence, both A and R are true and R is a correct explanation of A.

QUESTION: 12

When two time constant elements are cascaded non-interactively then, the overall transfer function of such an arrangement

Solution:
QUESTION: 13

The unit impulse response of a unity feedback control svstem whose Open Iood transfer function Solution:

Given, O.L.T.F.  For unit impulse response, R(s) = 1 or, c(t) = 2(e-t - te-t) + te-t
= 2e-t - te-t = (2- t) e-t
Thus, impulse response, c(t) = (2 - t) e-t u(t)

QUESTION: 14

The impulse response of a linear time invariant system is a unit step function. The transfer function of this system would be

Solution:

Impulse response,
c(t) = u(t)  QUESTION: 15

Consider the function where F(s) is the Laplace transform of f(t) is equal to

Solution:

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