Control Systems Test 6

State transition matrix, ϕ(t) = e^At

From the properties of state transition matrix,

\(\phi \left( 0 \right) = {e^{A\left( 0 \right)}} = I\)

1. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}&0\\ 1&{{e^{ - 2t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 1&0 \end{array}} \right] \ne I} \end{array}\)

2. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^t}}&t\\ 1&{2{e^{ - t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 1&2 \end{array}} \right] \ne I} \end{array}\)

3. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^t} + {e^{ - t}}}&0\\ 0&{2{e^{ - t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&2 \end{array}} \right] \ne I} \end{array}\)

4. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ 0&{{e^{ - 2t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = I} \end{array}\)

Hence option (d) can be state transition matrix.

Given \(\left| {sI - A} \right| = {s^2} + 5s + 3\)

where 's' is the Eigen Value of the given matrix.

The characteristic Equation is obtained by solving for the determinent of sI - A matrix, i.e 

\(sI - A = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} a&1\\ { - 3}&b \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {s - a}&{ - 1}\\ 3&{s - b} \end{array}} \right]\)

|sI - A| = (s - a) (s - b) + 3

= s² - (a + b) s + ab + 3

Comparing this with the given characteristic Equation s² + 5s + 3 = 0, we get;

ab + 3 = 3

⇒ ab = 0

Also, -(a + b) = 5

⇒ a + b = -5

⇒ a + b + ab = -5 + 0 = -5

ẋ = Ax + Bu

y = Cx + Du

Given that, D ≠ 0

y = Cx + Du

The given system is

Ẋ = AX + Bu

Where,

\(A = \left[ {\begin{array}{*{20}{c}}{ - 1}&2\\0&2\end{array}} \right]\:and\:B = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\)

We determine stability using characteristic equation (i.e. poles or eigenvalues of the system).

|sI - A| = 0

\(\left| {\left[ {\begin{array}{*{20}{c}}s&0\\0&s\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 1}&2\\0&2\end{array}} \right]} \right| = 0\)

\(\left| {\begin{array}{*{20}{c}}{s + 1}&{ - 2}\\0&{s - 2}\end{array}} \right| = 0\)

(s + 1)(s - 2) = 0

s = -1 and s = 2

i.e. the system has one pole in right half of the s-plane.

Hence, the system is unstable.

Now, the controllability matrix is given by:

C_M = [B AB]

Where

\(\left[ {AB} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&2\\0&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\2\end{array}} \right]\)

So,

\({C_M} = \left[ {\begin{array}{*{20}{c}}0&2\\1&2\end{array}} \right]\)

|C_M| = -2 ≠ 0

Hence, the system is Controllable.

For the given system,

We have

\(A = \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&4 \end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right],\;C = \left[ {4,\;0} \right]\)

The transfer function of the system is given by:

\(G\left( s \right) = \frac{{Y\left( s \right)}}{{U\left( s \right)}} = C{\left( {sI - A} \right)^{ - 1}}B\)

So, we obtain

\(\left[ {sI - A} \right] = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {s - 2}&0\\ 0&{s - 4} \end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {s - 2} \right)\left( {s - 4} \right)}}\left[ {\begin{array}{*{20}{c}} {\left( {s - 4} \right)}&0\\ 0&{\left( {s - 2} \right)} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\left( {s - 2} \right)}}}&0\\ 0&{\frac{1}{{\left( {s - 4} \right)}}} \end{array}} \right]\)

Therefore, the transfer function of the system is:

\(\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \left[ {4\;0} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\left( {s - 2} \right)}}}&0\\ 0&{\frac{1}{{\left( {s - 4} \right)}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right] \)

\(= \left[ {4\;0} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\left( {s - 2} \right)}}}\\ {\frac{1}{{\left( {s - 4} \right)}}} \end{array}} \right]\)

\(\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{4}{{\left( {s - 2} \right)}}\)

Here, input is unit impulse, i.e. U(s) = 1.

So, we have the output as:

\(Y\left( s \right) = \frac{4}{{\left( {s - 2} \right)}}\)

Taking inverse Laplace transform, we get output in the time domain as:

y(t) = 4e^2t

Given the state space representation,

\(\dot x = \left[ {\begin{array}{*{20}{c}} { - 3}&1\\ 0&{ - 2} \end{array}} \right]x\)

The steady-state value is given as:

\({x_{ss}} = \mathop {\lim }\limits_{t \to \infty } x\left( t \right)\)

Initial condition:

\(x\left( 0 \right) = \left[ {\begin{array}{*{20}{c}} {10}\\ { - 10} \end{array}} \right]\)

So, we obtain the matrix as:

\(A = \left[ {\begin{array}{*{20}{c}} { - 3}&1\\ 0&{ - 2} \end{array}} \right]\)

And ẋ = Ax

Taking Laplace transform on both sides, we get:

sX(s) – x(0) = AX(s)

X(s) (sI - A) = x(0)

X(s) = (sI - A)^-1x(0)

By the final value theorem, steady-state value is given by:

\({x_{ss}} = \mathop {\lim }\limits_{t \to \infty } x\left( t \right) = \mathop {\lim }\limits_{s \to 0} sX\left( s \right)\)

\({x_{ss}} = \mathop {\lim }\limits_{s \to 0} s\left[ {{{\left( {sI - A} \right)}^{ - 1}}x\left( 0 \right)} \right]\)    ---(1)

So, we obtain:

\(\left( {sI - A} \right) = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} { - 3}&1\\ 0&{ - 2} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {s + 3}&{ - 1}\\ 0&{s + 2} \end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 2}&1\\ 0&{s + 3} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{1}{{s + 3}}}&{\frac{1}{{\left( {s + 2} \right)\left( {s + 3} \right)}}}\\ 0&{\frac{1}{{s + 2}}} \end{array}} \right]\)

Substituting it in equation (1), we get the steady-state value as:

\({x_{ss}} = \mathop {\lim }\limits_{s \to 0} s\left[ {\begin{array}{*{20}{c}} {\frac{1}{{s + 3}}}&{\frac{1}{{\left( {s + 2} \right)\left( {s + 3} \right)}}}\\ 0&{\frac{1}{{s + 2}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {10}\\ { - 10} \end{array}} \right]\)

\({x_{ss}} = \left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]\)

The given system is described as:

ẋ(t) = Ax(t) + Bu(t)

and y(t) = Cx(t)

Where,

\(A = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 2}\end{array}} \right];B = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right];C = \left[ {1\;1} \right]\)

Now, the state model is defined as:

ẋ(t) = Ax(t) + Bu(t)       ---(1)

y(t) = Cx(t) + Du(t)       ---(2)

Taking Laplace transform of equation (1) with initial condition zero, we get:

sX(s) = AX(s) + BU(s)

X(s) [sI - A] = BU(s)

This can be written as:

X(s) = (sI - A)^-1B U(s)       ---(3)

Taking the Laplace transform of equation (2), we get:

Y(s) = CX(s) + DU(s)

Substituting value of X(s) from equation (3), we get:

Y(s) = C(sI - A)^-1 BU(s) + DU(s)

\(\frac{{Y\left( s \right)}}{{U\left( s \right)}} = C\left[ {{{\left( {sI - A} \right)}^{ - 1}}} \right]B + D\)

Here, D = 0.

Hence, the output transfer function will be:

\(T\left( s \right) = \frac{{Y\left( s \right)}}{{U\left( s \right)}} = C\left[ {{{\left( {sI - A} \right)}^{ - 1}}} \right]B\)       ---(4)

For the given system, we have:

\(\left( {sI - A} \right) = \left[ {\begin{array}{*{20}{c}}s&0\\0&s\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 2}\end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}}{s + 1}&0\\0&{s + 2}\end{array}} \right]\)

So,

\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\left[ {\begin{array}{*{20}{c}}{s + 2}&0\\0&{s + 1}\end{array}} \right]\)

Substituting it in equation (4), we get:

\(\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \left[ {1\;1} \right]\left[ {\begin{array}{*{20}{c}}{\frac{1}{{s + 1}}}&0\\0&{\frac{1}{{s + 2}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}}{\frac{1}{{s + 1}}}&{\frac{1}{{s + 2}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\)

\(\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{s + 1}}\)