Lecture 7

E12 Linear Physical Systems Analysis

Prof. Emad Masroor

2026-02-10

First-order systems’ response to pulse input

What’s a pulse?
A pulse can be constructed by superimposing scaled and shifted Heaviside Step Functions
Recall what those are: \[ u_s(t) = \begin{cases} 1 & t > 0 \\ 0 & t < 0 \end{cases} \]
A pulse of width \(D\) and height \(A\) can be constructed using two (re-scaled and shifted) step functions

First-order systems’ response to pulse input

We wish to find \(x(t)\) when the input \(f(t)\) is the sum of two step functions. \[ \dot{x} + ax = \overbrace{A u_s(t) - A u_s(t-D)}^{f(t)} \]

Approach: Use linearity!
Separately find solutions of \(\dot{x} + ax = A u_s(t)\) and \(\dot{x} + ax = - A u_s(t-D)\), keeping \(x(0) = 0\).
Add up the two solutions. This is the solution to \(\dot{x} + ax = f(t)\)

Step Response of First Order System

What is the output of a first-order system if the input is the unit step function \(u_s(t)\) ? \[\dot{x} + ax = u_s(t) = \begin{cases} 1 & t > 0 \\ 0 & t < 0 \end{cases}, \quad x(0) = 0 \tag{1}\]
Physically, this is equivalent to the following circuit:

In-class task: What’s \(x(t)\) from Equation 1 ?

Step Response of First Order System

The correct way to write the step response of a first-order system \(\dot{x} + ax = u_s(t)\) is \[ x(t) = \begin{cases} \displaystyle \frac{\left( 1- e^{-at} \right)}{a} & t > 0 \\ 0 & t < 0 \end{cases} \]
This is equivalent to \[ x(t) = \displaystyle u_s(t) \frac{\left( 1- e^{-at} \right)}{a} \]

Step Response of Shifted First Order System

What is the output of a first-order system if the input is the shifted unit step function \(u_s(t-t_1)\) ? \[\dot{x} + ax = u_s(t-t_1) = \begin{cases} 1 & t > t_1 \\ 0 & t < t_1 \end{cases}, \quad x(0) = 0 \tag{2}\]
Physically, this is equivalent to the following circuit:

In-class task: If \(x(t)\) for an un-shifted unit step input is \(\displaystyle \begin{cases} \displaystyle \frac{\left( 1- e^{-at} \right)}{a} & t > 0 \\ 0 & t < 0 \end{cases}\), what’s \(x(t)\) from Equation 2 ?

Shifted Step Response of First Order System

The correct way to write the shifted unit step response of a first-order system \(\dot{x} + ax = u_s(t-t_1)\) is \[ x(t) = \begin{cases} \displaystyle \frac{\left( 1- e^{-a(t-t_1)} \right)}{a} & t > t_1 \\ 0 & t < t_1 \end{cases} \]
This is equivalent to \[ x(t) = \displaystyle u_s(t-t_1) \frac{\left( 1- e^{-a(t-t_1)} \right)}{a} \]

Summarizing the step response and shifted step response

Step Response: \[ x(t) = \displaystyle u_s(t) \frac{\left( 1- e^{-at} \right)}{a} \]

Shifted Step Response: \[ x(t) = \displaystyle u_s(t-t_1) \frac{\left( 1- e^{-a(t-t_1)} \right)}{a} \]

Putting these together to get pulse response

The output when the input is a step function, a.k.a. the step response, is a building block for pulse response.
Approach: ‘piece together the answer’ (valid but tricky)
Use Laplace Transform and the time-shifting property

The time-shifting property and Laplace Transforms

\[ \begin{aligned} \mathcal{L}\left[ f(t-D) \cdot u_s(t-D) \right] &= e^{-sD} F(s) \\ \mathcal{L}^{-1}\left[ e^{-sD} F(s) \right] &= f(t-D) \cdot u_s(t-D) \end{aligned} \]

The Laplace Transform of a function shifted forward in time by \(D\) seconds equals to the Laplace Transform of that function (unshifted), multiplied by \(e^{-sD}\)
In general, the functions should be zero for \(t < 0\) before you shift them!

The pulse response

\[\dot{x} + 2 x = 3 u_s(t) - 3 u_s(t-2.5)\] Solve for \(x(t)\) when \(f(t)\) is as shown to the left, i.e., a pulse of height \(3\) and duration \(2.5\)

In-class task: Find solution with one step function as input

First, we approach \(\dot{x} + 2 x = 3 u_s(t)\)
Applying Laplace Transforms, we get \[ sX(s) + 2 X(s) = \frac{3}{s} \implies X(s) = \frac{3}{s(s+2)} \]
Which can be broken up into partial fractions \[ \frac{3}{2} \frac{1}{s} - \frac{3}{2} \frac{1}{s+2} \]
And then re-converted into the sum of two terms in the time domain:

\(\displaystyle \frac{3}{2} u_s(t)\), a unit step function multiplied by a constant, and
\(\displaystyle - \frac{3}{2} e^{-2t}\), when \(t > 0\)

So the result is \[ x(t) = \begin{cases} \displaystyle \frac{3}{2} \left( 1 - e^{-2t} \right) & t > 0 \\ 0 & t < 0 \end{cases} \]

In-class task: Find solution with the other step function as input

Next, we approach \(\dot{x} + 2 x = -3 u_s(t-2.5)\)
Applying Laplace Transforms, we get \[ sX(s) + 2 X(s) = -\frac{3}{s}e^{-2.5s} \implies X(s) = \frac{1}{s+2} \frac{-3}{s} e^{-2.5s} \]
Which can be broken up into partial fractions \[ -\frac{3}{2} \frac{1}{s} e^{-2.5s} + \frac{3}{2} \frac{1}{s+2} e^{-2.5s} \]
Each of these can be brought back into the time domain by shifting some function to the right by \(2.5\) seconds. \[ \underbrace{-\frac{3}{2} \frac{1}{s}}_{\text{Shift this}} e^{-2.5s} + \underbrace{\frac{3}{2} \frac{1}{s+2}}_{\text{Shift this}} e^{-2.5s} \]
We read the highlighted terms off the table and find \[ \begin{aligned} \mathcal{L}^{-1} \left[ -\frac{3}{2} \frac{1}{s} \right] &= -\frac{3}{2} u_s(t) \\ \mathcal{L}^{-1} \left[ \frac{3}{2} \frac{1}{s+2} \right] &= \frac{3}{2} e^{-2t} \quad \text{when } t > 0 \end{aligned} \]
Then we shift the result to the right, term by term: \[ x(t) = {-\frac{3}{2} u_s(t-2.5)} + {\frac{3}{2} e^{-2(t-2.5)} u_s(t-2.5)} \]
We now have the response of the system to the other step function.

So the result is \[ x(t) = -\frac{3}{2} u_s \left(t-5/2 \right) + \frac{3}{2} u_s \left( t-5/2 \right ) e^{-2(t-5/2)} \]

Putting it together

A note about the length of a pulse

Depending on the duration of a pulse, the output of the system may look very different.

What matters is how the duration compares with \(\tau\)

Response of a First-order system to \(\delta(t)\) input

\(\delta(t)\) is also known as the ‘unit impulse’ function, or the Dirac Delta function.
\(\delta(t)\) is the derivative of the unit step function, i.e.

The unit impulse response of a system equals the time-derivative of the (unit) step response

Consider the system \(\dot{x} + a x = f(t)\) where the input is \(f(t)\).

If the input is a step, \(f(t) = u_s(t)\) the output is called the step response: \[x(t) = \begin{cases} \displaystyle \frac{1 - e^{-at}}{a} & t > 0 \\ 0 & t < 0 \end{cases}\]
To obtain \(x(t)\) when the input is an impulse, \(f(t) = \delta(t)\),
differentiate the unit step response w.r.t time: \[x(t) = \begin{cases} \displaystyle e^{-a t} & t > 0 \\ 0 & t < 0 \end{cases}\] which is known as the impulse response

In-class task: What is the impulse response of the system below in terms of \(m\) and \(b\)?

Shifted Impulse Response

The impulse need not be located at \(t=0\).
We can easily move the impulse response to a later time.

Response of the system to a unit impulse applied at \(t = 0\):

\[ x(t) = \begin{cases} \displaystyle \frac{1}{m}e^{-t b / m} & t > 0 \\ 0 & t < 0 \end{cases} \]

Response of the system to a unit impulse applied at \(t = t_1\):

\[ x(t) = \begin{cases} \displaystyle \frac{1}{m}e^{-(t- t_1) b / m} & t > t_1 \\ 0 & t < t_1 \end{cases} \]

Using a shifted impulse function to represent impact

Consider the model \[m \dot{x} + b x = f(t), \quad v(0) = v_0 \neq 0\]
Let the force experienced by the object be a whack at a time \(t_1 > 0\).
Then, we can use the input \(f(t) = A \delta(t-t_1)\) where \(A\) is the area under the force-time curve.

But what is an impulse?

Recall Newton’s 2nd Law \(m\dot{v} = f\)
What if the force is applied over a period of time?
Integrate! from \(t_1\) to \(t_2\), we get \[m v_2 - m v_1 = \int_{t_1}^{t_2} f(t) dt\]
Impulse: The change in momentum of an object, i.e., \(\int f dt\)
A sudden change at \(t = t_1\) can be modeled by any of the following forces. (Thinner the ‘better’!)

Same quantity of impulse, applied in different ways; response slightly different
Idealized impulse is \(\delta(t)\) (often called just ‘impulse’ or ‘unit impulse’)

The Shifted Ramp function and its uses

A shifted ramp function with slope \(c\) and delay \(D\) can be constructed like: \[f(t) = \begin{cases} u_s(t - D) c(t - D) & t > D \\ 0 & t < D \end{cases}\]

Using shifted ramp functions

In-class task: Write down a formula for the following function using superposition of 4 (shifted and scaled) ramp functions

\[f(t) = \begin{cases} u_s(t - D) c(t - D) & t > D \\ 0 & t < D \end{cases}\]

The Transfer Function of a system and the Impulse Response of that system

Consider the system \[m \dot{v} + b v = f_a(t)\] where \(v(t)\) is the output and \(f_a(t)\) the input.

Find the transfer function

\[ \begin{aligned} m s V(s) + b V(s) = F_a(s)& \\ V(s) = \frac{1}{ms + b} F_a(s)& \\ \implies \frac{V(s)}{F_a(s)} = \underbrace{\boxed{\frac{1}{m s + b}}}_{\text{Transfer Function}}& \end{aligned} \]

Find \(v(t)\) when \(f_a(t) = \delta (t)\)

\[ \begin{aligned} m s V(s) + b V(s) &= \mathcal{L} \left[ \delta (t) \right] \\ m s V + b V &= 1 \\ \implies V(s) = \underbrace{\boxed{\frac{1}{ms+b}}}_{\text{Impulse Response}} \end{aligned} \]

Tip

The Transfer Function of a system is also the Laplace Transform of that system’s (unit) impulse response.