$$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix} u(t) $$
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
System State-Space Equations
The given state-space model is:
$ \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) $This expands to the following differential equations:
- $ \dot{x}_1 = x_1 + x_2 + u(t) $
- $ \dot{x}_2 = x_2 + u(t) $
Input Unit Step and Initial Conditions
The input is a unit step function:
$ u(t) = 1 \quad \text{for } t \ge 0 $The initial conditions are:
$ x(0) = \begin{bmatrix} x_1(0) \\ x_2(0) \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $Solving State Variable x2(t)
First, we solve the differential equation for $ x_2(t) $:
$ \dot{x}_2 = x_2 + 1 $Rearranging gives:
$ \dot{x}_2 - x_2 = 1 $This is a first-order linear differential equation. The integrating factor is $ e^{\int -1 dt} = e^{-t} $. Multiplying the equation by $ e^{-t} $ yields:
$ e^{-t}\dot{x}_2 - e^{-t}x_2 = e^{-t} $The left side is the derivative of $ e^{-t}x_2 $:
$ \frac{d}{dt}(e^{-t}x_2) = e^{-t} $Integrating both sides:
$ e^{-t}x_2(t) = \int e^{-t} dt = -e^{-t} + C_2 $Solving for $ x_2(t) $:
$ x_2(t) = -1 + C_2 e^{t} $Using the initial condition $ x_2(0) = 0 $, we find $ C_2 $:
$ 0 = -1 + C_2 e^{0} \implies C_2 = 1 $Therefore, the solution for $ x_2(t) $ is:
$ x_2(t) = e^t - 1 $Solving State Variable x1(t)
Next, we solve the differential equation for $ x_1(t) $:
$ \dot{x}_1 = x_1 + x_2 + 1 $Substitute the expression for $ x_2(t) = e^t - 1 $ and $ u(t) = 1 $, we get:
$ \dot{x}_1 = x_1 + (e^t - 1) + 1 $ $ \dot{x}_1 = x_1 + e^t $Rearranging gives:
$ \dot{x}_1 - x_1 = e^t $The integrating factor is $ e^{\int -1 dt} = e^{-t} $. Multiplying by $ e^{-t} $:
$ e^{-t}\dot{x}_1 - e^{-t}x_1 = e^{-t} e^t = 1 $The left side is the derivative of $ e^{-t}x_1 $:
$ \frac{d}{dt}(e^{-t}x_1) = 1 $Integrating both sides:
$ e^{-t}x_1(t) = \int 1 dt = t + C_1 $Solving for $ x_1(t) $:
$ x_1(t) = (t + C_1) e^t = t e^t + C_1 e^t $Using the initial condition $ x_1(0) = 1 $, we find $ C_1 $:
$ 1 = (0) e^0 + C_1 e^0 \implies C_1 = 1 $Therefore, the solution for $ x_1(t) $ is:
$ x_1(t) = t e^t + e^t = (1+t)e^t $System Time Response
Combining the solutions for $ x_1(t) $ and $ x_2(t) $, the time response of the system is:
$ x(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = \begin{bmatrix} (1+t)e^t \\ e^t - 1 \end{bmatrix} $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 and output equations for a control system are:
$$\dot{x} = \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix}x + \begin{bmatrix} 2 \\ 0 \end{bmatrix}u$$
$$y = \begin{bmatrix} 1.5 & 0.625 \end{bmatrix}x$$
Which of the following expressions correctly represents the transfer function $\frac{Y(s)}{U(s)}$ of the system with zero initial conditions?
