Test: Control Systems - 2 - Electrical Engineering (EE) MCQ

A linear time invariant (LTI) system provides the same output for the same input irrespective of when input is given. LTI systems are also used to predict the system's long term behavior.

Hence the correct answer is an option (b).

The abrupt change in input of a step function takes place at t=0, and the step's size is1 unit.

Let the size of the step function be A, where A=1.
The Laplace transform of a function is given by:
L[f(t)] = F(s) = ∫₀^∞f(t) e^-st dt
F(s) = ∫0^∞Af(t) e^-st dt
F(s) = -A/s (e^-∞-e^-0)
F(s) = A/s
We know, A =1
F(s) = 1/s
Hence the correct answer is an option (d).

Static systems do not have any feedback system. Hence, the output depends only on the present input.

Hence, the correct answer is an option (d).

The given function can be written as:

=  A/s + B/ (s+1) + C/(s+5)
1 = A(s+1)(s+5) +Bs(s+5) +Cs(s+1)
To calculate the value of A, put s= 0, we get:
1 = A(1)(5)
A = 1/5 = 0.2
Now, to calculate the value of B, put s=-1, we get:
1 = B (-1)(4)
B = -1/4 = -0.25
Similarly, put s=-5, we get:
1 = C (-5) (-4)
C = 1/20 = 0.05
Substituting the value of A, B, and C in F(s), we get:
F(s) = A/s + B/ (s+1) + C/(s+5)
F(s) = 0.2/s - 0.25/ (s+1) + 0.05/(s+5)
We know, Laplace transform of 1/(s + a) = e^-at and 1/s = 1.
f(t) = -0.2-0.25e^-t + 0.05e^-5t
Hence the correct answer is option (b).

The given equation is: s² + 2s + 2 = 0

It is a second-order differential equation. The Laplace transform of a standard form of a second-order differential equation is:

Comparingthe value we get:

Thus, the system is underdamped.
Hence, the correct answer is an option (d).

The standard transfer function can be written as:

The given equation is: 4/ (s² + 4s + 4)
Comparing the values, we get:
K= 1

The settling time for a 2 percent band is calculated as:

Settling time = 2 seconds

The settling time for a 5 percent band is calculated as:

Settling time = 1.5 seconds
Hence, the correct answer is an option (c).

 The step error can be calculated as:
ess = sR(s)/ (1 + G(s))
R(s) = 1/s (in case of unity feedback)
G(s) = 1/ (s+2)(s+3)
ess = (s x 1/s )/ (1 + (1/ (s+2)(s+3)))
ess = 1/(1 + kp)
Where, kp is the step error coefficient
Kp can be calculated as:

Hence, the correct answer is an option (d).

The RL circuit comprises of the resistor and inductor connected in series.
The equation can be written as:

1 = RI(s) + sLI(s)
1 = I(s) [R + sL]
I(s) = 1 / (R +sL)
Taking the inverse Laplace, we get:

The equation clearly depicts that the impulse response is a decaying exponential function.
Hence, the correct answer is option (d).

The given characteristic equation is: s³ + 3s² + 4s + 1 = 0.

To find the number of roots, we need to create a Routh table, as shown below:

There are no roots and no significant changes in the RHS plane, as shown in the above table. Hence, all three roots lie in the LHS plane.

Hence, the correct answer is an option (b).

The given characteristic equation is: s⁴ + 5s³ + 3s² + 6s + 5 = 0. We first need to find the roots by creating the Routh's array table.

Routh's array table is shown below:

In the first column of the above table, we have two sign changes. It means that two roots are in the RHS plane. Hence, the system is unstable.

Hence the correct answer is an option (a).

For the above equation, we need to find the roots by creating the Routh's array table. The given equation is: s³ + 3s² + 4s + A = 0

The table is given below:

There is no change in sign in the first column of the Routh table. It means that all roots lies in the left half of the s-plane.

Putting A and (12 - A)/3 > 0, we get:

A > 0 (or 0 < A)

(12 - A)/3 > 0

12 - A > 0

12 > A (or A < 12)

From the above equations, we get two values of A, i.e., A > 0 and A < 12. It means that A lies between 0 and 12, as shown below:

0 < A < 12

Hence, the correct answer is an option (a).

The number of asymptotes in a given system is equal to the number of branches approaching infinity. So, the formula to calculate the number of asymptotes is P -Z. Here, P and Z represent the poles and zeroes.

We know that poles and zeroes are calculated by equating the denominator and numerator to zero. So, for the given open-loop transfer function, we get:
P = 3
Z = 3
So, the number of zeroes at infinity = 3 - 3 = 0
Hence, the correct answer is an option (b).

We will first calculate the number of branches approaching infinity and then the asymptotes. With the help of asymptotes, we will calculate the value of centroid.

The given transfer function of the system is G(s) = K / [(s + 1) (s + 4 + 4j) (s + 4 - 4j)].

The number of asymptotes is equal to the number of branches approaching infinity. There are no zeroes but three poles.

So, P - Z = 3

Let's calculate the value of the poles by equating the denominator equal to zero. We get:

Poles located at: -1,

The angle of asymptotes is calculated by:

Θ = (2q + 1) 180 / (P - Z)

Here, q = 0, 1, 2…

The number of asymptotes is equal to the number of branches approaching infinity.

So, we will calculate the asymptotes at value 0, 1, and 2.

For q = 0,

Θ = 180⁄3 = 60degrees

For q = 1,

Θ = (2+1) 180⁄3 = 180degrees

For q = 2,

Θ = (4+1) 180⁄3 = 300degrees

Centroid is defined as a common point where all the asymptotes intersect on the real axis.

σ = / (P - Z)

σ = (- 1 - 4 - 4 - 0) / 3

σ = (-9)⁄3

σ = -3

Hence, the correct answer is an option (d).

The gain margin indicates the additional gain provided to the system without affecting its stability.

The total phase shift of a second-order system is approximately equal to 180 degrees, which leads to the infinite frequency. Thus, the gain margin is also infinite.

Hence, the correct answer is an option (b).

The given open-loop feedback control system is: G(s)H(s) = 1 / [(s + 1)³]

The phase angle for the cross-over frequency can be calculated by:

-3 tan^-1 ω = -180°

We get the value of ω = √3

Now, the gain margin is the magnitude of the transfer function, as shown below:

= 1 / |G(s)H(s)| = |(j√3 + 13|
= 8

Hence, the correct answer is an option (c).