Lecture 23

E12 Linear Physical Systems Analysis

Author

Prof. Emad Masroor

Published

April 14, 2026

Plan for week 14

  • Comprehensive Final Assigment in class on Thursday to synthesize everything we have learned

  • No HW will be assigned on week 14

  • Tuesday:

    • Instructor analyzes an electrical system on board

  • Thursday:

    • You will analyze a mechanical system in class and turn in your work, including some/all of:
  • State variables
  • Differential Equation
  • Laplace Transform
  • Block Diagram
  • Transfer Function
  • Free Response
  • Step Response
  • Impulse Response
  • Frequency Response
  • Bode Plots

Input-Output Equations

In Lecture 12 we saw that the governing differential equations for linear physical systems can be written in the following form:

\[\dot{\boldsymbol{x}} = \boldsymbol{A} \boldsymbol{x} + \boldsymbol{B} \boldsymbol{u}\]

  • \(\boldsymbol{x}\) is the state vector (\(n \times 1\))
  • \(\boldsymbol{A}\) is the system matrix (\(n \times n\))
  • \(\boldsymbol{B}\) is the input control matrix (\(n \times m\))
  • \(\boldsymbol{u}\) is the input vector (\(m \times 1\))

\[{\boldsymbol{y}} = \boldsymbol{C} \boldsymbol{x} + \boldsymbol{D} \boldsymbol{u}\]

  • \(\boldsymbol{y}\) is the output vector (\(p \times 1\))
  • \(\boldsymbol{C}\) is the state output matrix (\(p \times n\))
  • \(\boldsymbol{D}\) is the control output matrix (\(p \times m\))

Linear Input-Output Equations & State Variables

\[\dot{\boldsymbol{x}} = \boldsymbol{A} \boldsymbol{x} + \boldsymbol{B} \boldsymbol{u} \tag{1}\]

\[{\boldsymbol{y}} = \boldsymbol{C} \boldsymbol{x} + \boldsymbol{D} \boldsymbol{u} \tag{2}\]

Equation 1 is a differential equation for the state variables.

  • The (rate of change of) state variables depends on (1) a linear combination of the state variables themselves, and (2) a linear combination of the input variables.

Equation 2 is an algebraic equation for the output variables.

  • The output variables depend on (1) a linear combination of the state variables and (2) a linear combination of the input variables.
  • If the output variables we want are the state variables, then \(\boldsymbol{C}\) is the identity matrix \(\boldsymbol{C} = \boldsymbol{1}\) and \(\boldsymbol{D} = \boldsymbol{0}\).

Side note: Nonlinear equations

\[\dot{\boldsymbol{x}} = \boldsymbol{A} \boldsymbol{x} + \boldsymbol{B} \boldsymbol{u}\]

  • If \(\dot{\boldsymbol{x}} = ...\) is a nonlinear function of \(\boldsymbol{x}\) or of \(\boldsymbol{u}\),
    • the above equation cannot be written.
    • the differential equation can still be numerically solved using ode45 or solve_ivp.
    • Cannot use the Laplace Transform

\[{\boldsymbol{y}} = \boldsymbol{C} \boldsymbol{x} + \boldsymbol{D} \boldsymbol{u}\]

  • If \({\boldsymbol{y}} = ...\) is a nonlinear function of \(\boldsymbol{x}\) or of \(\boldsymbol{u}\),
    • the above equation cannot be written.
    • we can still use the nonlinear equation to obtain required outputs.

Multiple-Input, Multiple-Output systems

There are multiple options for how to write down the equations of a system. Let’s look at two.

  • Four state variables: \(x_1\), \(\dot{x}_1\), \(x_2\), \(\dot{x}_2\).
  • Two inputs: \(f_1(t)\) and \(f_2(t)\).
  • We are interested in three outputs:
    1. The position of the first mass \(x_1\)
    2. The distance between the two masses, \(x_2-x_1\)
    3. The total momentum of the system, \(m_1 \dot{x}_1 + m_2 \dot{x}_2\)
  • Four state variables: \(x_1\), \(\dot{x}_1\), \(x_2-x_1\), \(\dot{x}_2\).
  • Two inputs: \(f_1(t)\) and \(f_2(t)\).
  • We are interested in three outputs:
    1. The position of the first mass \(x_1\)
    2. The position of the second mass, \(x_2\)
    3. The total potential elastic energy of the system, \(\frac{1}{2}k {x_1}^2 + \frac{1}{2} k x_2^2\)

Checking the validity of equations

\[ \begin{aligned} \dot{\boldsymbol{x}} &= \boldsymbol{A} \boldsymbol{x} + \boldsymbol{B} \boldsymbol{u} \\ {\boldsymbol{y}} &= \boldsymbol{C} \boldsymbol{x} + \boldsymbol{D} \boldsymbol{u} \end{aligned} \]

  • The number of state variables should equal the number of energy-storing elements
  • The output variables should be a linear combination of the input variables and the state variables.
  • There are no restrictions on the number of input variables and output variables.

Note:

  • Multiplying a matrix with a matrix is the same thing as a dot product
  • A ‘(column) vector’ is the same as a ‘matrix with with 1 column’
  • To correctly multiply \(A B\), a.k.a. \(A \cdot B\), the number of columns of \(A\) must equal the number of rows of \(B\)
  • Matrix multiplication is not commutative, \(A \cdot B \neq B \cdot A\)

Constructing the input control matrix

\[ \frac{d}{dt} \begin{bmatrix} x_1 \\ \dot{x}_1 \\ x_2 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -(k_1+k_2)/m_1 & 0 & k_2/m_1 & 0 \\ 0 & 0 & 0 & 1 \\ k_2/m_2 & 0 & -k_2/m_2 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ \dot{x}_1 \\ x_2 \\ \dot{x}_2 \end{bmatrix} + {\color{red}{\begin{bmatrix} 0 \\ f_1/m_1 \\ 0 \\ f_2/m_2 \end{bmatrix}}} \]

  • We can re-write the equations of this system in the form \[\dot{\boldsymbol{x}} = \boldsymbol{A} \boldsymbol{x} + \boldsymbol{B} \boldsymbol{u}\]

  • where \(\boldsymbol{u}\) is a vector of inputs. Here, \(\boldsymbol{u} = \begin{bmatrix} f_1 \\ f_2 \end{bmatrix}\)

  • to find \(\boldsymbol{B}\) we construct a matrix of size \(n \times m\)

    • here, \(n\) is the number of state variables and \(m\) the number of inputs.

\[ \underbrace{{\color{magenta}{\begin{bmatrix} 0 & 0 \\ 1/m_1 & 0 \\ 0 & 0 \\ 0 & 1/m_2 \end{bmatrix}}}}_{\boldsymbol{\displaystyle B}} \begin{bmatrix} f_1 \\ f_2 \end{bmatrix} = {\color{red}{\begin{bmatrix} 0 \\ f_1/m_1 \\ 0 \\ f_2/m_2 \end{bmatrix}}} \]

Constructing the (linear) output equations

We are interested in three specific outputs:

  1. The position of the first mass \(x_1\)
  2. The distance between the two masses, \(x_2-x_1\)
  3. The total momentum of the system, \(m_1 \dot{x}_1 + m_2 \dot{x}_2\)

  • To construct the output equation \[{\boldsymbol{y}} = \boldsymbol{C} \boldsymbol{x} + \boldsymbol{D} \boldsymbol{u}\]
  • we need to express the output variables of interest (\(\boldsymbol{y}\)) as a linear combination of
      1. the state variables (\(\boldsymbol{x}\)) and
      1. the input variables (\(\boldsymbol{u}\)).
  • Let \(y_1 = x_1\)
  • Let \(y_2 = x_2-x_1\)
  • and \(y_3 = m_1 \dot{x}_1 + m_2 \dot{x}_2\)
  • So we have \[\boldsymbol{C} = \begin{bmatrix} +1 & 0 & 0 & 0 \\ -1 & 0 & +1 & 0 \\ 0 & m_1 & 0 & m_2 \end{bmatrix}, \]
  • multiplying \[ \boldsymbol{x} = \begin{bmatrix} x_1 \\ \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_2 \end{bmatrix} \]

Two outputs in a circuit