Test: Transfer Function- 3 - Electrical Engineering (EE) MCQ

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).

• The transfer function represents the relationship between input and output in a system.
• It is only applicable to linear systems, meaning the output is directly proportional to the input.
• Time-invariant systems are those whose behavior does not change over time, making them suitable for transfer functions.
• Non-linear and time-variant systems cannot be effectively described using transfer functions due to their complexity.

Given transfer function is

Converting above transfer function in time-constant form, we get:

Hence, dc gain is

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

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)

Given, c(t) = te^-t

Now, r(t) = u(t) (unit step input)
∴ R(s) = 1/s
So, 

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, 

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.

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)

Impulse response,
c(t) = u(t)

Using final value theorem