Test: State Variable Analysis- 2 - Electrical Engineering (EE) MCQ

Eigen values of A are given by

or, (s - σ)² + ω² = 0
or,  s = σ ± jω
or,  s = σ + jω and s = σ - jω

Given, 

Transfer function of the given system is

Now, 

∴
= s² + 3s + 2
Now,

Now,

The poles of the system can be obtained from eigen values of the system matrix. Hence, statement-1 is false.
State space techniques can be applied to linear or non-linear, time variant or time invariant systems.
Hence, statement-4 is false.

From given block diagram, the state equations can be written as:
x₁ = -x₁ + u and x₂ = -2x₂ + 2u
in matrix form,

Also, y = x₁ + x₂
In matrix form,

Thus,



Since, |Q_c| ≠ 0 and |Q₀| ≠ 0, therefore given system is both controllable and observable.

Given,






So, x(t) = ϕ(t).x(0)
= state transition equation

From given block diagram, the state equations can be written as:

Also, output equation is
y = 2x₁ + x₂
In matrix form, we have:

and y = [2  1]x(t)
Now,


∴ Transfer function,

Let us select the state variables as V_c and i_L.
Applying KVL in the mesh-2, we have:

or,
.............(1)
Also, by applying KCL at the given node, we get:

or,
..........(2)
From equations (1) and (2), state equations in matrix form can be written as:

The state equations from the given signal flow graph can be written as:

In matrix form,

Also, output is
y(t) = 6x₁ + x_{2
In matrix form}

Here, 




So, IQ_CI = 0 - 0 + 2(0 - 4)
= - 8 ≠ 0
Hence, the system is controllable.
Also, Q₀ = [C^T A^TC^T (A^T)² C^T]
Here,


and


So,

= 1 ≠ 0
Since |Q₀| ≠ 0, therefore given system is observable.

T.F.,


Hence, zeros are at:

So, zeros are at:

or,  -17 -5s + s² + 9s + 20 = 0
or,  s² + 4s + 3 = 0
or,  (s+1) (s+3) = 0
or,  s = -1,- 3