Canonical Systems
=================

In classical control theory two systems are the canonical systems: the
first and second order systems (higher order systems can be
constructed by cascading these canonical systems).


First Order Systems
-------------------

A first order system is characterized with a differential equation
having only first order derivatives. Although simple these first order
systems are of great importance. E.g. a mass-damper system and a mass
heating system are examples of first order systems. And often higher
order systems can be approximated with a first order system.

The canonical first order system is characterized with transfer function:

.. math::
   H(s) = K \frac{1}{1+\tau s}

where $\tau$ is the **time constant** of the system and $K$ is the
**DC gain** of the system.

The impulse response can be derived using the Laplace transform pair

.. math::
   e^{-at}u(t) \LTright \frac{1}{a+s}

We can rewrite $H(s)$ as

.. math::
   H(s) = K \frac{1}{1+\tau s} = \frac{K}{\tau}\frac{1}{1/\tau + s}

and thus the **impulse response of a first order system** is:

.. math::
   h(t) = \frac{K}{\tau} e^{-t/\tau}u(t)


To calculate the step response of the first order system we use the
fact that the Laplace transform of the unit step equals $1/s$. And
thus the Laplace transform of the step response is

.. math::
   \frac{K}{s(1+\tau s)}

This Laplace transform is not in our table of common transform
pairs. But if we calculate the partial fractions of this we are
lookking for factors $A$ and $B$ such that

.. math::
   \frac{K}{s(1+\tau s)} = K\left(\frac{A}{s} + \frac{B}{1+\tau s}\right)

meaning that

.. math::
   A(1+\tau s) + Bs = 1

leading to $A=1$ and $B=-\tau$. So the Laplace transform of the step
response equals:

.. math::
   \frac{K}{s(1+\tau s)} = K\left( \frac{1}{s} - \frac{\tau}{1+\tau s}\right) =
   K\left( \frac{1}{s} - \frac{1}{1/\tau+ s}\right)

Now we have two terms that are in the table of Laplace transform pairs
and the **step response of a first order system** is:

.. math::
   K\left(u(t) - e^{-t/\tau}u(t)\right) = K(1-e^{-t/\tau})u(t)



.. exec_python:: canonical_first  control
   :linenumbers:
   :code: shutter
   :Code_label: Show code for figure
   :results: hide

   import numpy as np
   import matplotlib.pyplot as plt
   from scipy import signal
   import control

   plt.clf()

   tau = 1
   H = control.tf([1], [tau, 1])
   t, ystep = control.step_response(H)
   plt.plot(t, ystep, label=r'$\tau=1$');
   
   tau=2
   H = control.tf([1], [tau, 1])
   t, ystep = control.step_response(H)
   plt.plot(t, ystep, label=r'$\tau=2$');
   plt.legend();
   plt.savefig('source/figures/firstorder_stepresponse.png')


.. figure:: /figures/firstorder_stepresponse.png
   :align: center

   **Step Response of First Order System.**

   
Second Order Systems
--------------------

A canonical second order system is characterized with transfer function

.. math::
   H(s) = \frac{\w_0^2}{s^2 + 2 \beta \w_0 s + \w_0^2}

where $\w_0$ is the undamped (natural) frequency and $\beta$ is the
damping parameter.
   
This system has two poles that determine the behaviour of the
system. Here study its impulse response and step response.

The poles can be calculated with the quadratic formula:

.. math::
   p_{1,2} &= \frac{-2\beta\w_0 \pm \sqrt{4\beta^2\w_0^2 - 4 \w_0^2}}{2}\\
   &= -\beta\w_0 \pm \w_0\sqrt{\beta^2-1}\\
   &= \w_0\left( -\beta \pm \sqrt{\beta^2-1} \right)

We can distinguish four different situations depending on the value
of $\beta$.

$\beta=0$ Undamped
  For an undamped system we get:

  .. math::
     H(s) = \frac{\w_0^2}{s^2+\w_0^2}

  such that the Laplace transform of the step response equals

  .. math::
     \frac{H(s)}{s} = \frac{\w_0^2}{s(s^2+\w_0^2)}

  We want to write this like

  .. math::
     \frac{A}{s} + \frac{Bs}{s^2+\w_0^2}

  leading to

  .. math::
     As^2 + A\w_0^2 + Bs^2 = \w_0^2

  such that

  .. math::
     A=1, \quad B=-1

  The Laplace transform of the step response then becomes:

  .. math::
     \frac{H(s)}{s} = \frac{1}{s} - \frac{s}{s^2+\w_0^2}

  From the table of Laplace transform pairs we find the **step response
  of an undamped second order system**:

  .. math::
     \left( 1 - \cos(\w_0 t) \right) u(t)

  Evidently this is where the term *undamped* comes from: a step at the
  input leads to a sinusoidal response with infinite duration.


  .. exec_python:: canonical_second_undamped  control
     :linenumbers:
     :code: shutter
     :Code_label: Show code for figure
     :results: hide

     import numpy as np
     import matplotlib.pyplot as plt
     from scipy import signal
     import control
     
     plt.clf()
     
     w0 = np.pi
     b = 0
     t = np.linspace(0, 20, 1000)
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, label=r'$\omega_0=\pi$');
     
     w0 = np.pi/2
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, label=r'$\omega_0=\pi/2$');
     plt.legend();
     
     plt.savefig('source/figures/secondorder_undamped.png')


  .. figure:: /figures/secondorder_undamped.png
     :align: center

     **Step Response of Second Order Undamped System.**


 
    
  
$0<\beta<1$ Underdamped
  For $\beta$ in between zero and one the system is underdamped. It
  will oscillate but the amplitude will decrease over time.

  .. exec_python:: canonical_second_undamped  control
     :linenumbers:
     :code: shutter
     :Code_label: Show code for figure
     :results: hide

     import numpy as np
     import matplotlib.pyplot as plt
     from scipy import signal
     import control

     plt.clf()
     
     w0 = np.pi
     t = np.linspace(0, 10, 1000)
	       
     for b in [0.20, 0.50]:
	H = control.tf([w0**2], [1, 2*b*w0, w0**2])
	_, ystep = control.step_response(H, t)
	plt.plot(t, ystep, label=r'$\beta={}$'.format(b));
   
     plt.legend();
   
     plt.savefig('source/figures/secondorder_underdamped.png')


  .. figure:: /figures/secondorder_underdamped.png
     :align: center

     **Step Response of Second Order Underdamped System.**





$\beta = 1$ Critically Damped
  In case $\beta=1$ there are two identical real valued poles:

  .. math::
     p_{1,2} = -\w_0

  and we can write

  .. math::
     H(s) = \frac{\w_0^2}{(s+\w_0)(s+\w_0)}
     = \frac{\w_0}{s+\w_0}\frac{\w_0}{s+\w_0}
     = \frac{1}{1+s/\w_0} \frac{1}{1+s/\w_0}

  showing that a critically damped second order system is the cascade
  of two identical first order system each with time constant
  $1/\w_0$.

  .. exec_python:: canonical_second_criticallydamped  control
     :linenumbers:
     :code: shutter
     :Code_label: Show code for figure
     :results: hide

     import numpy as np
     import matplotlib.pyplot as plt
     from scipy import signal
     import control

     plt.clf()
     w0 = np.pi
     b = 1
     t = np.linspace(0, 5, 1000)
     
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, label=r'$\omega=\pi$');

     w0 = np.pi/2
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, label=r'$\omega=\pi/2$');
     plt.legend();
   
     plt.savefig('source/figures/criticallydamped2ndstep.png')


  .. figure:: /figures/criticallydamped2ndstep.png
     :align: center

     **Step Response of Second Order Critically Damped System.**

     
$\beta>1$ Overdamped
  For $\beta>1$ we have two poles both of which are real valued:

  .. math::
     p_{1,2} = \w_0\left(-\beta \pm \sqrt{\beta^2-1} \right)


  .. exec_python:: canonical_second_criticallydamped  control
     :linenumbers:
     :code: shutter
     :Code_label: Show code for figure
     :results: hide

     import numpy as np
     import matplotlib.pyplot as plt
     from scipy import signal
     import control

     plt.clf()

     w0 = np.pi

     b = 1
     t = np.linspace(0, 10, 1000)
     
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, '--', label=r'$\beta={}$ (critically damped)'.format(b));

     b = 2
     t = np.linspace(0, 10, 1000)
     
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, label=r'$\beta={}$'.format(b));

     b = 4
     H = control.tf([w0**2], [1, 2*b*w0, w0**2])
     _, ystep = control.step_response(H, t)
     plt.plot(t, ystep, label=r'$\beta={}$'.format(b));
     plt.legend();
     
   
     plt.savefig('source/figures/overdamped2ndstep.png')


  .. figure:: /figures/overdamped2ndstep.png
     :align: center

     **Step Response of Second Order Overdamped System.** The
     critically damped response is shown for comparison.