$$\dot{x} = \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix}x + \begin{bmatrix} 2 \\ 0 \end{bmatrix}u$$
$$y = \begin{bmatrix} 1.5 & 0.625 \end{bmatrix}x$$
Which of the following expressions correctly represents the transfer function $\frac{Y(s)}{U(s)}$ of the system with zero initial conditions?
Transfer Function from State-Space Equations
The transfer function $G(s)$ of a linear time-invariant (LTI) system described by state-space equations $\dot{x} = Ax + Bu$ and $y = Cx + Du$ is given by the formula:
$G(s) = C(sI - A)^{-1}B + D$
Where:
- $A = \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix}$
- $B = \begin{bmatrix} 2 \\ 0 \end{bmatrix}$
- $C = \begin{bmatrix} 1.5 & 0.625 \end{bmatrix}$
- $D = 0$ (implicitly, as not provided)
- $I$ is the identity matrix.
Calculating $sI - A$
First, compute the matrix $sI - A$:
$sI - A = s\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} s+4 & 1.5 \\ -4 & s \end{bmatrix} $
Calculating the Inverse $(sI - A)^{-1}$
Next, find the determinant of $sI - A$:
$\det(sI - A) = (s+4)(s) - (1.5)(-4) = s^2 + 4s + 6$
The inverse is:
$(sI - A)^{-1} = \frac{1}{\det(sI - A)} \text{adj}(sI - A) = \frac{1}{s^2+4s+6} \begin{bmatrix} s & -1.5 \\ 4 & s+4 \end{bmatrix} $
Calculating the Transfer Function $G(s)$
Now, compute $C(sI - A)^{-1}B$:
$C(sI - A)^{-1}B = \begin{bmatrix} 1.5 & 0.625 \end{bmatrix} \left( \frac{1}{s^2+4s+6} \begin{bmatrix} s & -1.5 \\ 4 & s+4 \end{bmatrix} \right) \begin{bmatrix} 2 \\ 0 \end{bmatrix} $
$= \frac{1}{s^2+4s+6} \begin{bmatrix} 1.5 & 0.625 \end{bmatrix} \begin{bmatrix} s & -1.5 \\ 4 & s+4 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} $
Multiply the matrices:
$= \frac{1}{s^2+4s+6} \begin{bmatrix} (1.5)s + (0.625)(4) & (1.5)(-1.5) + (0.625)(s+4) \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} $
$= \frac{1}{s^2+4s+6} \begin{bmatrix} 1.5s + 2.5 & -2.25 + 0.625s + 2.5 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} $
$= \frac{1}{s^2+4s+6} \begin{bmatrix} 1.5s + 2.5 & 0.625s + 0.25 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} $
Perform the final multiplication:
$= \frac{1}{s^2+4s+6} [ (1.5s + 2.5)(2) + (0.625s + 0.25)(0) ] $
$= \frac{1}{s^2+4s+6} [ 3s + 5 ] $
$G(s) = \frac{3s+5}{s^2+4s+6} $
Since $D=0$, this is the final transfer function.
Important Questions from State Space Representation
A discrete system is represented by the difference equation
\(\left[ {\begin{array}{*{20}{c}} {{X_1}\left( {k + 1} \right)}\\ {{X_2}\left( {k + 1} \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&{a - 1}\\ {a + 1}&a \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{X_1}\left( k \right)}\\ {{X_2}\left( k \right)} \end{array}} \right]\)
It has initial conditions X1(0) = 1; X2(0) = 0. The pole locations of the system for a = 1, areFor the network given below:-
The state equation is:-Arrange the following in the sequence of steps for writing STATE EQUATION:-
A. Choose a normal tree for a network
B. Take either voltages or charges across the capacitor, which are tree branches and eithercurrents or fluxes through inductors, which are co-tree chords
C. Write independent KVL, KCL and branch voltage current relations (VCR).
D. Rearrange the vector matrix differential equation for state variable representation.
E. Eliminate all non-state variables of the network
Choose the correct answer from the options given below:- Consider the state-space representation of a system
$ \dot{x} = Ax + Bu $
where $x$ is the state vector, $u$ is the input, $A$ is the system matrix and $B$ is the input matrix. Choose the matrix $A$ from the following options such that the system has a pole at the origin. - The state-space model of a system is given as
$$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix} u(t) $$
where u(t) is a unit step input occurring at t = 0 and $x(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
The time response of the system is
