Lecture 4

E12 Linear Physical Systems Analysis

Author

Prof. Emad Masroor

Published

January 29, 2026

First-order Circuits

Linearity of solutions: superposition

  • For a linear differential equation, the sum of two solutions is also a solution.

  • For a linear differential equation, a constant times a solution is also a solution.

Illustrating linearity

  • An RC circuit in which the capacitor is initially at 2 Volts.

  • An RC circuit in which the capacitor is initially at 3 Volts.

\[RC \dot{v} + v = 0, \qquad v(0) = 2 \text{ or } 3 \]

\(v(0) = 2\)

Solution: \(v(t) = 2 e^{-t}\)

\(v(0) = 3\)

Solution: \(v(t) = 3 e^{-t}\)

Can we use solutions to these to predict what happens if \(v(0) = 5\)?

Superposition of two solutions

Code 
import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt

RC = 1
V0 = 3

# Set up time domain
t_f = 2
t_vals = np.linspace(0,t_f,100)

def V_sum(t):
  return 5*np.exp(-1/(RC)*t)

def ode_rhs(t,X):
  return (-1/RC)*X[0]

sol1_ode = solve_ivp(ode_rhs,    [0,t_f],[2],t_eval=t_vals)
sol2_ode = solve_ivp(ode_rhs,    [0,t_f],[3],t_eval=t_vals)
sol3_ode = solve_ivp(ode_rhs,    [0,t_f],[5],t_eval=t_vals)

plt.plot(t_vals, sol1_ode.y[0],'.',label="x_1(t): Solution to x'+cx=0, x(0) = 2")
plt.plot(t_vals, sol2_ode.y[0],'.',label="x_2(t): Solution to x'+cx=0, x(0) = 3")
plt.plot(t_vals, sol1_ode.y[0]+sol2_ode.y[0],'-',label="x_1(t) + x_2(t)")
plt.plot(t_vals, sol3_ode.y[0],'.',label="x_3(t): Solution to x'+cx=0, x(0) = 5",markersize=3)

plt.legend()
plt.grid()
plt.show()

Steady State

\[\dot{x} + a x = f(t) \]

If we call \(f(t)\) the ‘input’ and \(x(t)\) the ‘response’, then:

  • \(x(t)\) eventually reaches a steady state, i.e., \(x(t)\) settles into some long-term behavior that depends on \(f(t)\).
  • The steady-state behavior of \(x(t)\) can be
    • zero,
    • or a constant,
    • or some other function of time, depending on \(f(t)\).

A ‘time scale’ for first-order systems

Consider the first-order system \(RC \dot{v} + v = f(t)\)

If \(f(t) = 0\) and \(v(0) = 5\), the solution is

Code 
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation

fig, ax = plt.subplots()

N = 100
t_end = 10
t = np.linspace(0, t_end, N)

def ft(t):
    return 5*np.exp(-t)
v = ft(t)

fig.set_figwidth(3.0)
fig.set_figheight(3.0)

scat = ax.scatter(t[0], v[0], c="b", s=5, label='v(t)')
ax.set(xlim=[0, t_end], ylim=[-1, 6])
ax.set_xlabel('Time [seconds]',fontsize=9)
ax.set_ylabel('V [volts]', fontsize=9)
plt.grid()
ax.legend()


def update(frame):
    # for each frame, update the data stored on each artist.
    x = t[:frame]
    y = v[:frame]
    # update the scatter plot:
    data = np.stack([x, y]).T
    scat.set_offsets(data)
    ax.set_title(f'time = {t[frame]:.2f} s, v = {v[frame]:.2e}',fontsize=9) 
    return scat



ani = animation.FuncAnimation(fig=fig, func=update, frames=N, interval=30)
writer1 = animation.PillowWriter(fps=15) # Set frames per second (fps)
ani.save("first-order.gif",writer = writer1)

When should we say the system has decayed ‘completely’?

Never truly reaches zero

We need a ‘time scale’ to talk about

  • ‘how quickly this system decays’
  • ‘when the system has decayed completely’

The time constant \(\tau\)

  • We found that the first-order system \(RC \dot{v} + v = 0\) with initial condition \(v(0) = v_0\) has solution \[v(t) = v_0 e^{-\frac{t}{RC}}\]

  • Let’s now define a new parameter \(\tau = RC\) called the time constant such that \[v(t) = v_0 e^{-\frac{t}{\tau}}\]

  • What are the units of \(\tau\)?

TipThe time constant \(\tau\) is measured in seconds.

Mathematically speaking, the function \(e^x\) applies to pure numbers \(x\). So \(t\) and \(\tau\) must have the same units. Thus, \(\tau\) is measured in seconds or multiples of it.

Every first-order linear system has a time constant

The time constnat \(\tau\)

measures how long a first-order system takes to settle down to a steady-state value

Code 
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation

fig, ax = plt.subplots(ncols=2)

N = 100
t_end = 10
t = np.linspace(0, t_end, N)
tau = 3

def ft(t):
    return 5*np.exp(-t)

def ft2(t):
    return 5*np.exp(-t/tau)

v = ft(t)
v2 = ft2(t)

fig.set_figwidth(6)
fig.set_figheight(4)



scat1 = ax[0].scatter(t[0], v[0], c="b", s=5)
ax[0].set(xlim=[0, t_end], ylim=[-1, 6])
ax[0].set_xlabel('Time [seconds]',fontsize=10)
ax[0].set_ylabel('V [volts]', fontsize=10)

scat2 = ax[1].scatter(t[0], v[0], c="b", s=5)
ax[1].set(xlim=[0, t_end], ylim=[-1, 6])
ax[1].set_xlabel('Time [seconds]',fontsize=10)

ax[0].grid(True)
ax[1].grid(True)

ax[0].set_title(r"$\dot{x} + x = 0$",fontsize=14)
ax[1].set_title(r"$3\dot{x} + x = 0$",fontsize=14)



def update(frame):
    # for each frame, update the data stored on each artist.
    x = t[:frame]
    y = v[:frame]
    # also for the other plot
    y2 = v2[:frame]
    # update the scatter plot:

    data = np.stack([x, y]).T
    scat1.set_offsets(data)
    scat2.set_offsets(np.stack([x,y2]).T)
    # ax[0].set_title(f'time = {t[frame]:.2f} s, v = {v[frame]:.2e}',fontsize=9) 
    # ax[1].set_title(f'time = {t[frame]:.2f} s, v = {v2[frame]:.2e}',fontsize=9) 
    return (scat1,scat2)


ani = animation.FuncAnimation(fig=fig, func=update, frames=N, interval=30)
writer1 = animation.PillowWriter(fps=10) # Set frames per second (fps)
# plt.show()
ani.save("timeconst-comparison.gif",writer = writer1)

The time constant \(\tau\)

provides convenient sign-posts to talk about how far a system is from reaching ‘steady-state’

  • When \(t = 1\tau\), the response is 63.3% of the way toward reaching steady-state
  • When \(t = 2\tau\), the response is 86.5% of the way toward reaching steady-state
  • When \(t = 5\tau\), the response is 99.4% of the way toward reaching steady-state.
Time Value Decimal Percent
\(t=0 \tau\) \(v=v_0 e^{-0}\) \(v=1.0 v_0\) 100%
\(t=1 \tau\) \(v=v_0 e^{-1}\) \(v=0.3678v_0\) 36.7%
\(t=2 \tau\) \(v=v_0 e^{-2}\) \(v=0.1353 v_0\) 13.5%
\(t=3 \tau\) \(v=v_0 e^{-3}\) \(v=0.0497 v_0\) 4.9%
\(t=4 \tau\) \(v=v_0 e^{-4}\) \(v=0.0183 v_0\) 1.8%
\(t=5 \tau\) \(v=v_0 e^{-4}\) \(v=0.0067 v_0\) 0.6%
  • At what value of time is \(v(t)\) equal to half its initial value? Give your answer in multiples of \(\tau\).
  • At what value of time is \(v(t)\) equal to 0.1 percent of its original value?

In-class design example

A motor shaft continues to spin after the motor is switched off, slowly slowing down due to friction. No more external power is provided to the shaft during this motion. At the moment it’s switched off, it has a rotational speed of 1200 RPM.

  • For a shaft, \(J = \frac{1}{2}m r^2\).
  • Mass = 2.7 kg and radius = 3.0 cm.
  • Design requirement
    • Shaft must spin down to a stop in 3 seconds.
  • Radians should be considered unitless

Calculate how much \(c\) should be to meet the design requirement.

What are the units of \(c\) in SI?

TipGoverning equation

\[ J \frac{d \omega}{dt} + c \omega = T \]

\(T\): Torque \(\omega\): Angular velocity \(c\): Amount of (dynamic) friction

Steady-state response for first-order systems

Let’s consider 4 types of input:

  1. \(f(t) = 0\)
  2. \(f(t) = \text{const.}\)
  3. \(f(t) = c t\)
  4. \(f(t) = \cos(\omega t)\)
  5. (lab 1) \(f(t) =\) an impulse

Transient vs. steady-state

\[\dot{x} + ax = f(t)\]

Input Transient Response Steady-State Response
\(f(t) = 0\) \(x_0 e^{-at}\) \(0\)
\(f(t) = k\) \(e^{-at} \left( x_0 -k \right)\) \(k\)
\(f(t) = c t\)
\(f(t) = \cos \omega t\)
\(f(t) =\) impulse (lab 1)

First-order Response to zero input

First-order Response to constant input

First-order Response to ramp input

First-order Response to periodic input

Response of first-order systems

Input Response
\(f(t) = 0\)
\(f(t) = k\)
\(f(t) = c t\)
\(f(t) = \cos \omega t\)

Transient and steady-state

  1. Transient and steady-state components of a solution \(x(t)\) can be mathematically defined
  2. Approximately, we can say that steady-state sets in after \(5\tau\) units of time have passed.

Numerical Solution of differential equations

When \(\dot{x} = f(x,t)\) is too hard to integrate symbolically, we can use a computer.

Analytical Numerical
is written mathematically as \(x(t) =\) some function of time. is stored in a computer as a series of points \((t_k, x_k)\)
Once analytical solution is known, you can calculate \(x\) for any \(t\) you want. interpolation must be used between calculated points
Accurate and as precise as can be Usually accurate and ‘precise enough’
Only wrong if you make an algebra mistake Can be wrong even if there are no algebra mistakes and eqaution is correctly entered into computer program

A program to numerically solve IVPs

numerical_solution.m
function dxdt = rhs1(t,x)
    % return the value of f(x,t)
    dxdt = t - x;
end

% Time span over which to integrate
tspan = linspace(0,6,30);

x0 = 3;

% Call ode45 and assign solution to x.
[t,x] = ode45(@rhs1, tspan, x0);

% Plot
plot(t, x, 'o-')
1
The right hand side of \(\dot{x} = f(x,t)\)
2
Integrate from \(t=0\) to \(t=6\) and calculate at 30 points
3
Initial condition
4
Syntax for ode45. 
numerical_solution.py
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp

def rhs1(t,x):
    # return the value of f(x,t)
    return t - x[0]

# Time span over which to integrate
tspan = np.linspace(0,6,30) 

x0 = [3]

sol = solve_ivp(rhs1,tspan[[0,-1]],x0,t_eval=tspan)

# Plot the solution
plt.plot(sol.t,sol.y[0],'o-'); plt.show()
1
Packages must be imported
2
The right hand side of \(\dot{x} = f(x,t)\)
3
Initial condition has to be given as a list or array
4
Provide start/end points and t_eval=
5
Plot using attributes t and y of object of type OdeResult returned by solve_ivp. 
sol = NDSolve[{x'[t] + x[t] == t, x[0] == 3}, x[t], {t, 0, 6}]
Plot[x[t] /. sol, {t, 0, 6}]

\[ \dot{x} = t - x, \quad x(0) = 3 \]

Solving a differential equation with a computer

Download code from the website at https://emadmasroor.github.io/E12-S26/Resources/numsol/ for MATLAB or Python.

Modify the code to solve:

Initial Value Problem

\[ \begin{align} 5 \dot{x} + x &= 1.5 \\ x(0) &= 0 \end{align} \] from \(t=0\) to \(t=20\).

True Solution

\[ x(t) = 1.5 \left( 1- e^{-t/5}\right) \]

First-order system with periodic input

Use ode45 or solve_ivp to calculate how the voltage across a capacitor in an RC circuit responds to a sinusoidally varying input

  • Use \(R = 50 k\Omega\) and \(C= 1 \mu F\)
  • Capcitor starts out with voltage of \(2.0\)
  • Let the input voltage be \(v_i(t) = 1.0 \cos \omega t\) with \(\omega =1\)

Bonus question: How does the amplitude and phase differ in \(x(t)\) and \(f(t)\) ?