01․ The open loop transfer function for unity feedback system is given by
Find the steady state error for a ramp input of magnitude 10?

Applying final value theorem, Steady state error = Lts→0 S E(s) Steady state error for different types and different inputs as follows:

02․ The system function N(s) = V(s)/I(s) = (s+3)/(4s+5). The system is initially at rest. If the excitation i(t) is a unit step, which of the following is the final value?

Steady state value can be obtained by applying final value theorem to the function.
Lim

_{s→0}s C(s) = Lim_{s→0}s R(s)H(s) Where, C(s) = Steady state value R(s) = Input H(s) = Network response Final value = Lim_{s→0}s * (s+3)/(4s+5)*1/s = 3/503․ Find the Laplace transform for sinωt?

04․ Given a unity feedback system with G(s) = k/s(s+4), the value of k for damping ratio of 0.5 is

Characteristic equation is s² + 4s +k = 0
Compared to second ordered standard characteristic equation,
Natural frequency ω

_{n}= √k 2ξω_{n}= 4 Given ξ = 0.5 ω_{n}= 4 = √k Therefore, k = 1605․ The type number of the control system with G(s) = k(s+2)/(s(s²+2s+3) is

Every transfer function representing a control system has certain type and order. The steady state analysis depends on the type of the control system. The type of the system is obtained from open loop transfer function. The number of open loop poles occurring at origin determines the type of the control system
Therefore, for given system type number is 1.

06․ For type 1 system with parabolic input, the steady state error is

Applying final value theorem, Steady state error = Lt

_{s→0}S E(s) Steady state error for different types and different inputs as follows:07․ what is the Laplace transform of a function δ(t-2)?

L(δ(t)) = 1
Given function is δ(t-2), it represents the time delay by two units.
It can be written as,
L(δ(t-2)) = e

^{-2s}08․ The impulse response of a linear system is e

^{-t}, (t > 0). The corresponding transfer function isTransfer function = L(impulse response)
The transfer function in terms of Laplace transform of impulse response is known as weighting function.
L(e

^{-t}) = 1/(s+1)09․ A lead compensator used for a closed loop controller has the following transfer function For such a lead compensator

Zero at transfer function = - a, Pole at transfer function = - b. For a lead compensator, the zero is nearer to the origin as compared to the pole.

10․ The transfer functions of two compensators are given below. Which one is correct?
which one is correct?

For a lead compensator, the zero is nearer to the origin and for lag compensator the pole is nearer to the origin. Here in C

_{1}numerator (zero) s = - 1, denominator (pole) = - 10. So, s = - 1 is nearer to the origin. In C_{2}numerator (zero) s = - 10, denominator (pole) = - 1. So, s = - 10 is further from the origin. So, C_{1}is lead compensator and C_{2}is lag compensator.<<<56789>>>