Construction ...

Construction of State Equations from SFG

Trending Questions

Q. A second order LTI system is described by the following state equations

\frac{d}{d t} x_{1} (t) - x_{2} (t) = 0

\frac{d}{d t} x_{2} (t) + 2 x_{1} (t) + 3 x_{2} (t) = r (t)

where

x_{1} (t)

and

x_{2} (t)

are the two state variables and

r (t)

denotes the input. The output

c (t) = x_{1} (t)

. The system is

Q. The block diagram of a system with one input

u

and two output

y_{1}

and

y_{2}

is given below.
A state space model ofthe above system in terms ofthe state vector

x

and the output vector

y = {[\begin{matrix} y_{1} & y_{2} \end{matrix}]}^{T}

$˙ x = [1] x + [1] u; y [\begin{matrix} 1 & 2 \end{matrix}] x$
$˙ x = [- 2] x + [1] u; y [\begin{matrix} 12 \end{matrix}] x$
$˙ x = [\begin{matrix} - 2 & 0 0 & - 2 \end{matrix}] x + [\begin{matrix} 11 \end{matrix}] u; y [\begin{matrix} 1 & 2 \end{matrix}] x$
$x = [\begin{matrix} 2 & 0 0 & 2 \end{matrix}] x + [\begin{matrix} 12 \end{matrix}] u; y [\begin{matrix} 1 & 2 \end{matrix}] x$

Q. The state equation for the current

I_{1}

shown in the network shown below in terms of the voltage

V_{x}

and the independent source V, is given by

Q. A signal flow graph of a system is given below

The set of equations that correspond to this signal flow graph is

$\frac{d}{d t} ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ = ⎡ ⎢ ⎣ \begin{matrix} β & - γ & 0 γ & α & 0 - α & - β & 0 \end{matrix} ⎤ ⎥ ⎦ ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 0 & 1 0 & 0 \end{matrix} ⎤ ⎥ ⎦ (\begin{matrix} u_{1} u_{2} \end{matrix})$
$\frac{d}{d t} ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ = ⎡ ⎢ ⎣ \begin{matrix} 0 & α & γ 0 & - α & - γ 0 & β & - β \end{matrix} ⎤ ⎥ ⎦ ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 0 & 1 1 & 0 \end{matrix} ⎤ ⎥ ⎦ (\begin{matrix} u_{1} u_{2} \end{matrix})$
$\frac{d}{d t} ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ = ⎡ ⎢ ⎣ \begin{matrix} - α & β & 0 - β & - γ & 0 α & γ & 0 \end{matrix} ⎤ ⎥ ⎦ ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 0 & 1 0 & 0 \end{matrix} ⎤ ⎥ ⎦ (\begin{matrix} u_{1} u_{2} \end{matrix})$
$\frac{d}{d t} ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ = ⎡ ⎢ ⎣ \begin{matrix} - β & 0 & β γ & 0 & α - γ & 0 & - α \end{matrix} ⎤ ⎥ ⎦ ⎛ ⎜ ⎝ \begin{matrix} x_{1} x_{2} x_{3} \end{matrix} ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 0 & 0 1 & 0 \end{matrix} ⎤ ⎥ ⎦ (\begin{matrix} u_{1} u_{2} \end{matrix})$

Q. The state diagram of a system is shown below is described by the state -variable equations:

˙ x = A X + B u; y = C X + D u

The state -variable equations of the system in the figure above are

$˙ X = [\begin{matrix} - 1 & 0 1 & - 1 \end{matrix}] X + [\begin{matrix} - 1 1 \end{matrix}] u y = [\begin{matrix} 1 & - 1 \end{matrix}] X + u$
$˙ X = [\begin{matrix} - 1 & 0 - 1 & - 1 \end{matrix}] X + [\begin{matrix} - 1 1 \end{matrix}] u y = [\begin{matrix} - 1 & - 1 \end{matrix}] X + u$
$˙ X = [\begin{matrix} - 1 & 0 1 & - 1 \end{matrix}] X + [\begin{matrix} - 1 1 \end{matrix}] u y = [\begin{matrix} - 1 & - 1 \end{matrix}] X - u$
$X = [\begin{matrix} - 1 & - 1 0 & - 1 \end{matrix}] X + [\begin{matrix} - 1 1 \end{matrix}] u y = [\begin{matrix} 1 & - 1 \end{matrix}] X - u$

Q. For the given circuit, which one of the following is the correct state equation?

$\frac{d}{d t} [\frac{v}{i}] = [\begin{matrix} - 4 & 4 - 2 & - 4 \end{matrix}] [\frac{v}{i}] + [\begin{matrix} 0 & 4 4 & 0 \end{matrix}] [\begin{matrix} i_{1} i_{2} \end{matrix}]$
$\frac{d}{d t} [\frac{v}{i}] = [\begin{matrix} - 4 & - 4 - 2 & - 4 \end{matrix}] [\frac{v}{i}] + [\begin{matrix} 4 & 0 0 & 4 \end{matrix}] [\begin{matrix} i_{1} i_{2} \end{matrix}]$
$\frac{d}{d t} [\frac{v}{i}] = [\begin{matrix} - 4 & - 4 - 2 & 4 \end{matrix}] [\frac{v}{i}] + [\begin{matrix} 4 & 4 4 & 0 \end{matrix}] [\begin{matrix} i_{1} i_{2} \end{matrix}]$
$\frac{d}{d t} [\frac{v}{i}] = [\begin{matrix} 4 & - 4 - 2 & - 4 \end{matrix}] [\frac{v}{i}] + [\begin{matrix} 0 & 4 4 & 4 \end{matrix}] [\begin{matrix} i_{1} i_{2} \end{matrix}]$

Q. The following equation defines a separately excited dc motor in the form of a differential equation

\frac{d^{2} ω}{d t^{2}} + \frac{B}{J} \frac{d ω}{d t} + \frac{K^{2}}{L J} ω = \frac{K}{L J} V_{a}

The above equation may be organized in the statespace form as follows

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{d^{2} ω}{d t^{2}} \frac{d ω}{d t} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ = P ⎡ ⎣ \begin{matrix} \frac{d ω}{d t} ω \end{matrix} ⎤ ⎦ + Q V_{a}

Where the P matrix is given by

$⎡ ⎢ ⎣ \begin{matrix} - \frac{B}{J} & - \frac{K^{2}}{L J} 1 & 0 \end{matrix} ⎤ ⎥ ⎦$
$⎡ ⎢ ⎣ \begin{matrix} - \frac{K^{2}}{L J} & - \frac{B}{J} 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$⎡ ⎢ ⎣ \begin{matrix} 0 & 1 - \frac{K^{2}}{L J} & - \frac{B}{J} \end{matrix} ⎤ ⎥ ⎦$
$⎡ ⎢ ⎣ \begin{matrix} 1 & 0 - \frac{B}{J} & - \frac{K^{2}}{L J} \end{matrix} ⎤ ⎥ ⎦$

Q. The signal flow graph of a system is shown below:

The state variable representation of the system can be

$˙ x = [\begin{matrix} 1 & 1 - 1 & 0 \end{matrix}] x + [\begin{matrix} 02 \end{matrix}] u$

$y = [\begin{matrix} 0 & 0.5 \end{matrix}] x$
$˙ x = [\begin{matrix} - 1 & 1 - 1 & 0 \end{matrix}] x + [\begin{matrix} 02 \end{matrix}] u$

$y = [\begin{matrix} 0 & 0.5 \end{matrix}] x$
$˙ x = [\begin{matrix} 1 & 1 - 1 & 0 \end{matrix}] x + [\begin{matrix} 02 \end{matrix}] u$

$y = [\begin{matrix} 0.5 & 0.5 \end{matrix}] x$
$˙ x = [\begin{matrix} - 1 & 1 - 1 & 0 \end{matrix}] x + [\begin{matrix} 02 \end{matrix}] u$

$y = [\begin{matrix} 0.5 & 0.5 \end{matrix}] x$