basics properties of signal and system by uet

Chapter 2
Spring 2024 EE220 Signals & Systems
Dr. Muhammad Rehan Chaudhry
Biomedical Engineering Department
UET Lahore Narowal Campus
May 1, 2024
Dr. Muhammad Rehan Chaudhry Chapter 2 Spring 2024 EE220 Signals & Systems 1 / 60

Linear Time-Invariant Systems
1 Systems that are both linear and time invariant are called Linear
Time-Invariant (LTI) systems.
2 With systems that are linear and time invariant, and using the
impulse function in CT and DT, we can produce an important and
useful mechanism for characterizing those system.
3 In this lecture we shall develop in detail the representation of both
continuous-time and discrete-time signals as a linear combination of
delayed impulses and then utilizing this property to represent the
output of lineart time-invariant systems.
4 The resulting representation is referred to as convolution.

Introduction (from Oppenheim)
A linear system: the response to a linear combination of inputs is the
same linear combination of the individual responses.
Time invariance: If the input is shifted in time by some amount, then
the output is simply shifted by the same amount.
For a linear system, if the system inputs can be decomposed as a
linear combination of some basic inputs and the system response is
known for each of the basic inputs, then the response can be
constructed as the same linear combination of the responses to each
of the basic inputs.
Signals can be decomposed as a linear combination of basic signals in
a variety of ways (e.g., Taylor series expansion that expresses a
function in polynomial form.) However, in the context of signals and
systems, it is important to choose the basic signals in the expansion
so that in some sense the response is easy to compute.
For systems that are both linear and time-invariant, there are two
particularly useful choices for these basic signals: delayed impulses
and complex exponentials.

Introduction (from Oppenheim)
In this lecture we develop in detail the representation of both
continuous-time and discrete-time signals as a linear combination of
delayed impulses and the consequences for representing linear,
time-invariant systems. The resulting representation is referred to as
convolution.

Introduction
Using the convolution we can express the response of an LTI system
to an arbitrary input in terms of the system’s response to the unit
impulse.
An LTI system is completely characterized by its response to a single
signal, namely, its response to the unit impulse.
In discrete time, we have the convolution sum. In continuous time,
we have the convolution integral.

Strategy for Exploiting Linearity and Time Invariance
Decompose the input signal to a linear combination of basic signals.
Chose basic signals so that the response is easy to compute
(analytical convenience).
1 Delayed impulses → convolution
2 Complex exponentials → Fourier analysis

A DT Signal as a Superposition of Weighted Delayed
Impulses
We can express a DT signal as a linear combination of weighted
delayed impulses.
If we have a liner system, and a signal expressed as above as a linear
combination of basic signals, the response would be the same linear
combinaton fo the responses for individual basic signals.

A DT Signal as a Superposition of Weighted Delayed Impulses
n
−4 −3 −2 −1 0 1 2 3 4
x[n]
x[−2]
x[−1]
x[0]
x[1]
x[2]
x[n] = x[−2]δ[n + 2]
+ x[−1]δ[n + 1]
+ x[0]δ[n]
+ x[1]δ[n − 1]
+ x[2]δ[n − 2]
x[n] =
∞
X
k=−∞
x[k]δ[n − k].
A linear combination of weighted
delayed impulses.

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
x[−2]
x[−1]
x[0]
x[1]
x[2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−2]δ[n + 2]
x[−2]
x[n] = x[−2]δ[n + 2]
+ x[−1]δ[n + 1]
+ x[0]δ[n]
+ x[1]δ[n − 1]
+ x[2]δ[n − 2]
x[n] =
∞
X
k=−∞
x[k]δ[n − k].
delayed impulses.

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
x[−2]
x[−1]
x[0]
x[1]
x[2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−2]δ[n + 2]
x[−2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−1]δ[n + 1]
x[−1]
x[n] = x[−2]δ[n + 2]
+ x[−1]δ[n + 1]
+ x[0]δ[n]
+ x[1]δ[n − 1]
+ x[2]δ[n − 2]
x[n] =
∞
X
k=−∞
x[k]δ[n − k].
delayed impulses.

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
x[−2]
x[−1]
x[0]
x[1]
x[2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−2]δ[n + 2]
x[−2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−1]δ[n + 1]
x[−1]
n
−4 −3 −2 −1 0 1 2 3 4
x[0]δ[n]
x[0]
x[n] = x[−2]δ[n + 2]
+ x[−1]δ[n + 1]
+ x[0]δ[n]
+ x[1]δ[n − 1]
+ x[2]δ[n − 2]
x[n] =
∞
X
k=−∞
x[k]δ[n − k].
delayed impulses.

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
x[−2]
x[−1]
x[0]
x[1]
x[2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−2]δ[n + 2]
x[−2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−1]δ[n + 1]
x[−1]
n
−4 −3 −2 −1 0 1 2 3 4
x[0]δ[n]
x[0]
n
−4 −3 −2 −1 0 1 2 3 4
x[1]δ[n − 1]
x[1]
x[n] = x[−2]δ[n + 2]
+ x[−1]δ[n + 1]
+ x[0]δ[n]
+ x[1]δ[n − 1]
+ x[2]δ[n − 2]
x[n] =
∞
X
k=−∞
x[k]δ[n − k].
delayed impulses.

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
x[−2]
x[−1]
x[0]
x[1]
x[2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−2]δ[n + 2]
x[−2]
n
−4 −3 −2 −1 0 1 2 3 4
x[−1]δ[n + 1]
x[−1]
n
−4 −3 −2 −1 0 1 2 3 4
x[0]δ[n]
x[0]
n
−4 −3 −2 −1 0 1 2 3 4
x[1]δ[n − 1]
x[1]
n
−4 −3 −2 −1 0 1 2 3 4
x[2]δ[n − 2]
x[2]
x[n] = x[−2]δ[n + 2]
+ x[−1]δ[n + 1]
+ x[0]δ[n]
+ x[1]δ[n − 1]
+ x[2]δ[n − 2]
x[n] =
∞
X
k=−∞
x[k]δ[n − k].
delayed impulses.

Convolution Sum
x[n] =
∞
X
k=−∞
x[k]δ[n − k], input.

Convolution Sum
x[n] =
∞
X
k=−∞
Linear system

Convolution Sum
x[n] =
∞
X
k=−∞
Linear system
y[n] =
∞
X
k=−∞
x[k]hk[n] where hk[n] is the output due to the delayed impulse.
δ[n − k] → hk[n].

Convolution Sum
x[n] =
∞
X
k=−∞
Linear system
y[n] =
∞
X
k=−∞
δ[n − k] → hk[n].
If time invariant

Convolution Sum
x[n] =
∞
X
k=−∞
Linear system
y[n] =
∞
X
k=−∞
δ[n − k] → hk[n].
If time invariant
hk[n] = h0[n − k] where h0 is the response of the system for an impulse at 0.
h0[n] = h[n] define.

Convolution Sum
x[n] =
∞
X
k=−∞
Linear system
y[n] =
∞
X
k=−∞
δ[n − k] → hk[n].
If time invariant
hk[n] = h0[n − k] where h0 is the response of the system for an impulse at 0.
h0[n] = h[n] define.
If LTI
y[n] =
∞
X
k=−∞
x[k]h[n − k] convolution sum.

Convolution Sum: Summary
The convolution of the sequence x[n] and h[n] is given by
y[n] =
∞
X
k=−∞
x[k]h[n − k], (1)
which we represent symbolically as
y[n] = x[n] ∗ h[n] (2)

Example
Computer y[n] = x[n] ∗ h[n] for x[n] and h[n] as shown in Figure 1.
n
−4 −3 −2 −1 0 1 2 3 4
x[n]
1 1
2
n
−4 −3 −2 −1 0 1 2 3 4
h[n]
1
2
Figure: Computing convolution

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
1 1
2
n
−4 −3 −2 −1 0 1 2 3 4
h[n]
1
2
k
−4 −3 −2 −1 0 1 2 3 4
x[k]
1 1
2
k
−4 −3 −2 −1 0 1 2 3 4
h[−k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[−2 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[−1 − k]
2
1

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
1 1
2
n
−4 −3 −2 −1 0 1 2 3 4
h[n]
1
2
k
−4 −3 −2 −1 0 1 2 3 4
x[k]
1 1
2
k
−4 −3 −2 −1 0 1 2 3 4
h[−k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[−2 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[−1 − k]
2
1
x[k]
1 1
2
x[k]
1 1
2

n
−4 −3 −2 −1 0 1 2 3 4
x[n]
1 1
2
n
−4 −3 −2 −1 0 1 2 3 4
h[n]
1
2
k
−4 −3 −2 −1 0 1 2 3 4
x[k]
1 1
2
k
−4 −3 −2 −1 0 1 2 3 4
h[−k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[−2 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[−1 − k]
2
1
x[k]
1 1
2
x[k]
1 1
2
y[−2] = 0
y[−1] = 1 × 1 = 1

k
−4 −3 −2 −1 0 1 2 3 4
h[−k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[1 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[2 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[3 − k]
2
1

k
−4 −3 −2 −1 0 1 2 3 4
h[−k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[1 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[2 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[3 − k]
2
1
x[k]
1 1
2
x[k]
1 1
2
x[k]
1 1
2
x[k]
1 1
2

k
−4 −3 −2 −1 0 1 2 3 4
h[−k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[1 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[2 − k]
2
1
k
−4 −3 −2 −1 0 1 2 3 4
h[3 − k]
2
1
x[k]
1 1
2
x[k]
1 1
2
x[k]
1 1
2
x[k]
1 1
2
y[0] = 2 × 1 + 1 × 1 = 3
y[1] = 2 × 1 + 1 × 2 = 4
y[2] = 2 × 2 = 4
y[3] = 0

n
−4 −3 −2 −1 0 1 2 3 4
y[n]
1
3
4 4
y[n] =



















0 n ≤ −2
1 n = −1
3 n = 0
4 n = 1
4 n = 2
0 n ≥ 3.

Example
Consider and input x[n] and a unit impulse response h[n] given by
x[n] = αn
u[n]
h[n] = u[n],
(3)
which 0 < α < 1. Find y[n] and sketch.

−10 −5 5 10
0.5
1
n
x[n] = αnu[n]
−10 −5 5 10
0.5
1
n
u[n]
n
0.5
1
k
h[n − k]
Figure: The signals x[n] and h[n] for the
example.
x[k]h[n − k] =
(
αk, 0 ≤ k ≤ n,
0, otherwise.

−10 −5 5 10
0.5
1
n
x[n] = αnu[n]
−10 −5 5 10
0.5
1
n
u[n]
n
0.5
1
k
h[n − k]
Figure: The signals x[n] and h[n] for the
example.
x[k]h[n − k] =
(
αk, 0 ≤ k ≤ n,
0, otherwise.
Thus for n ≥ 0,
y[n] =
n
X
k=0
αk
y[n] =
1 − αn+1
1 − α
for n ≥ 0.
y[n] =
1 − αn+1
1 − α
u[n].

Continuous-Time Systems: The Convolution Integral
1 Similar to what we did in DT, in this section we obtain a complete
characterization of a continuous-time LTI system in terms of its unit
impulse response.
2 In discrete time, the key to developing the convolution sum was the
sifting property of the DT unit impulse—i.e., mathematical
representation of a signal as a superposition of scaled and shifted unit
impulse functions.
3 We begin by considering the staircase approximation x̂(t) of a CT
signal x(t).

−2∆−∆ 0 ∆ 2∆
0.5
1
t
x(t)
Figure: Staircase approximation of a CT signal.

−2∆−∆ 0 ∆ 2∆
0.5
1
t
x(t)
−2∆−∆ 0 ∆ 2∆
x(−2∆)
t
x(−2∆)δ∆(t + 2∆)∆

−2∆−∆ 0 ∆ 2∆
0.5
1
t
x(t)
−2∆−∆ 0 ∆ 2∆
x(−2∆)
t
x(−2∆)δ∆(t + 2∆)∆
−2∆−∆ 0 ∆ 2∆
x(−∆)
t
x(−∆)δ∆(t + ∆)∆

−2∆−∆ 0 ∆ 2∆
0.5
1
t
x(t)
−2∆−∆ 0 ∆ 2∆
x(−2∆)
t
x(−2∆)δ∆(t + 2∆)∆
−2∆−∆ 0 ∆ 2∆
x(−∆)
t
x(−∆)δ∆(t + ∆)∆
−2∆−∆ 0 ∆ 2∆
x(0)
t
x(0)δ∆(t)∆

−2∆−∆ 0 ∆ 2∆
0.5
1
t
x(t)
−2∆−∆ 0 ∆ 2∆
x(−2∆)
t
x(−2∆)δ∆(t + 2∆)∆
−2∆−∆ 0 ∆ 2∆
x(−∆)
t
x(−∆)δ∆(t + ∆)∆
−2∆−∆ 0 ∆ 2∆
x(0)
t
x(0)δ∆(t)∆
−∆−∆ 0 ∆ 2∆
x(∆)
t
x(∆)δ∆(t − ∆)∆

The approximation that we saw can be expressed as a linear combination
of delayed impulses. Define
δ∆(t) =
(
1
∆ , 0 ≤ t < ∆
0, otherwise.
Since ∆δ∆(t) has unit amplitude, we have
x̂(t) =
∞
X
k=−∞
x(k∆)δ∆(t − k∆)∆.
Here, for any value of t, only one term in the summation on the right hand
side is nonzero.
x(t) = lim
∆→0
∞
X
k=−∞
x(k∆)δ∆(t − k∆)∆.
As ∆ → 0, the summation approaches an integral. Consequently,
x(t) =
Z ∞
−∞
x(τ)δ(t − τ)dτ
This is known as the sifting property of the continuous time impulse.

Example:
Use the sifting property to express u(t) in terms of δ(t).

Example:
Use the sifting property to express u(t) in terms of δ(t).
u(t) =
Z ∞
−∞
u(τ)δ(t − τ)dτ =
Z ∞
0
δ(t − τ)dτ

The Continuous-Time Unit Impulse Response and the
Convolution Integral Representation of LTI Systems
Let’s define ĥk∆(t) as the response of an LTI system to the input
δ∆(t − k∆).
ŷ(t) =
∞
X
k=−∞
x(k∆)ĥk∆(t)∆.
Since the pulse δ∆(t − k∆) corresponds to a shifted unit impulse as
∆ → 0, the response ĥk∆(t) to this input pulse becomes the response to
an impulse in the limit. If we let h1_τ(t) denote the response at time t to
a unit impulse δ(t − τ) located at time τ, then
y(t) = lim
∆→0
∞
X
k=−∞
x(k∆)hk∆(t)∆.
As a ∆ → 0, the summation on the right-hand side becomes an integral.
y(t) =
Z ∞
−∞
x(τ)hτ (t)dτ

The Continuous-Time Unit Impulse Response and the
Convolution Integral Representation of LTI Systems
Let’s define ĥk∆(t) as the response of an LTI system to the input
δ∆(t − k∆).
ŷ(t) =
∞
X
k=−∞
x(k∆)ĥk∆(t)∆.
Since the pulse δ∆(t − k∆) corresponds to a shifted unit impulse as
∆ → 0, the response ĥk∆(t) to this input pulse becomes the response to
an impulse in the limit. If we let h1_τ(t) denote the response at time t to
a unit impulse δ(t − τ) located at time τ, then
y(t) = lim
∆→0
∞
X
k=−∞
x(k∆)hk∆(t)∆.
As a ∆ → 0, the summation on the right-hand side becomes an integral.
y(t) =
Z ∞
−∞
x(τ)hτ (t)dτ
x(t) =
R ∞
−∞ x(τ)δ(t − τ)dτ

k∆ (k + 1)∆
x(k∆)hk∆(t)∆
τ
x(τ)hτ (t)
Figure: Graphical illustration

In addition to being linear, the system is time-invariant, the response of
the LTI system to the unit impulse δ(t − τ)
hτ (t) = h0(t − τ).
Defining unit impulse response h(t) as
h(t) = h0(t),
we have
y(t) =
Z ∞
−∞
x(τ)h(t − τ)dτ
which is referred to as the convolution integral or the superposition
integral. This corresponds to the representation of a continuous-time LTI
system in terms of its response to a unit impulse.
y(t) = x(t) ∗ h(t).
As in discrete time, a continuous-time LTI system is completely
characterized by its impulse response—i.e., by its response to a.single
elementary signal, the unit impulse δ(t).

Example: Let x(t) be the input to an LTI system with unit impulse
response h(t), where
x(t) = e−at
u(t), a > 0
and
h(t) = u(t).

0
1
τ
h(τ)
Figure: Calculation of convolution integral for the example

0
1
τ
h(τ)
0
1
τ
x(τ)

0
1
τ
h(τ)
0
1
τ
x(τ)
t 0
1
t < 0
τ
h(t − τ)

0
1
τ
h(τ)
0
1
τ
x(τ)
t 0
1
t < 0
τ
h(t − τ)
0 t
t > 0
τ
h(t − τ)

For t < 0, the product x(τ) and h(t − τ) is zero, consequently y(t) is zero.
For t > 0,
x(τ)h(t − τ) =
(
e−aτ , 0 < τ < t,
0, otherwise.
y(t) =
Z t
0
e−aτ
dτ = −
1
a
e−aτ
t
0
=
1
a
(1 − e−at
)
Thus for all t,
y(t) =
1
a
(1 − e−at
)u(t)

0
1
a
t
y(t) = 1
a
(1 − e−at)u(t)
Figure: Response

Example: Consider the convolution of the following two signals:
x(t) =
(
1, 0 < t < T,
0, otherwise.
and
h(t) =
(
t, 0 < t < 2T,
0, otherwise.

0 T
1
τ
x(τ)
Figure: x(τ) and h(t − τ)

0 T
1
τ
x(τ)
t − 2T t 0
2T
t < 0
τ
h(t − τ)

0 T
1
τ
x(τ)
t − 2T t 0
2T
t < 0
τ
h(t − τ)
t − 2T t
0
2T
0 < t < T
τ
h(t − τ)

0 T
1
τ
x(τ)
t − 2T t 0
2T
t < 0
τ
h(t − τ)
t − 2T t
0
2T
0 < t < T
τ
h(t − τ)
t − 2T t
T < t < 2T
τ
h(t − τ)

0 T
1
τ
x(τ)
t − 2T t 0
2T
t < 0
τ
h(t − τ)
t − 2T t
0
2T
0 < t < T
τ
h(t − τ)
t − 2T t
T < t < 2T
τ
h(t − τ)
t − 2T t
2T
2T < t < 3T
τ
h(t − τ)

0 T
1
τ
x(τ)
t − 2T t 0
2T
t < 0
τ
h(t − τ)
t − 2T t
0
2T
0 < t < T
τ
h(t − τ)
t − 2T t
T < t < 2T
τ
h(t − τ)
t − 2T t
2T
2T < t < 3T
τ
h(t − τ)
t − 2T t
0
2T
t > 3T
τ
h(t − τ)

t − 2T t 0
t
2T
0 < t < T
y(t) = 1
2
t2
τ
x(τ)h(t − τ)
t − 2T t
0 T
t − T
t
2T
T < t < 2T
y(t) = Tt − 1
2
T2
τ
x(τ)h(t − τ)
t − 2T
t − T
2T
2T < t < 3T
y(t) = − 1
2
t2 + Tt + 3
2
T2
τ
x(τ)h(t − τ)
Figure: Product x(τ)h(t − τ) for values of t for which this product is not
identically zero.

0 T 2T 3T
t
y(t)
Figure: y(t) = x(t) ∗ h(t)
y(t) =















0, t < 0,
1
2 t2, 0 < t < T,
Tt − 1
2 T2, T < t < 2T,
−1
2 t2 + Tt + 3
2 T2, 2T < t < 3T,
0, 3T < t.

Example: Find y(t), the convolution of the following two signals:
x(t) = e2t
u(−t),
and
x(t) = u(t − 3).

0
1
τ
x(τ) = e2τ u(−τ)
Figure: Convolution considered in the example.

0
1
τ
x(τ) = e2τ u(−τ)
t − 3 0
1
τ
h(t − τ)

0
1
τ
x(τ) = e2τ u(−τ)
t − 3 0
1
τ
h(t − τ)
3
0
1
2
1
t
y(t)

When t − 3 ≤ 0, the product of x(τ) and h(t − τ) is nonzero for
−∞ < τ < t − 3, and the convolving integral becomes
y(t) =
Z t−3
−∞
e2τ
dτ =
1
2
e2(t−3)
.
F0r t − 3 ≥ 0, the product of x(τ)h(t − τ) is nonzero for −∞ < τ < 0,
and the convolving integral becomes
y(t) =
Z 0
−∞
e2τ
dτ =
1
2
.

Recapitulation
1 In discrete time the representation takes the form of the convolution
sum, while its continuous-time counterpart is the convolution integral:
y[n] =
∞
X
k=−∞
x[k]h[n − k] = x[n] ∗ h[n]
y(t) =
Z ∞
−∞
x(τ)h(t − τ)dτ = x(t) ∗ h(t)
2 Characteristics of an LTI system are completely determined by its
impulse response (h(t) in CT, h[n] in DT.).

The Commutative Property
DT
x[n] ∗ h[n] = h[n] ∗ x[n] =
∞
X
k=−∞
h[k]x[n − k].
CT
x(t) ∗ h(t) = h(t) ∗ x(t) =
Z ∞
−∞
h(τ)x(t − τ)dτ.
Verify the commutative property for DT.

DT
x[n] ∗ h[n] = h[n] ∗ x[n] =
∞
X
k=−∞
h[k]x[n − k].
CT
x(t) ∗ h(t) = h(t) ∗ x(t) =
Z ∞
−∞
x[n] ∗ h[n] =
∞
X
k=−∞
x[k]h[n − k].
Substituting r = n − k, or
equivalently k = n − r

DT
x[n] ∗ h[n] = h[n] ∗ x[n] =
∞
X
k=−∞
h[k]x[n − k].
CT
x(t) ∗ h(t) = h(t) ∗ x(t) =
Z ∞
−∞
x[n] ∗ h[n] =
∞
X
k=−∞
x[k]h[n − k].
Substituting r = n − k, or
equivalently k = n − r
x[n]∗h[n] =
∞
X
r=−∞
x[n−r]h[r] = h[n]∗x[n].

The Distributive Property
Convolution distributes over addition.
DT
x[n] ∗ (h1[n] + h2[n]) = x[n] ∗ h1[n] + x[n] ∗ h2[n].
CT
x(t) ∗ (h1(t) + h2(t)) = x(t) ∗ h1(t) + x(t) ∗ h2(t).
x(t)
h1(t)
h2(t)
+
y1(t)
y2(t)
y(t)
x(t) h1(t) + h2 y(t)
Figure: Distributive property.

The Associative Property
DT
x[n] ∗ (h1[n] ∗ h2[n]) = (x[n] ∗ h1[n]) ∗ h2[n].
CT
x(t) ∗ (h1(t) ∗ h2(t)) = (x(t) ∗ h1(t)) ∗ h2(t).
As a consequence,
y[n] = x[n] ∗ h1[n] ∗ h2[n]
and
y(t) = x(t) ∗ h1(t) ∗ h2(t).
are unambiguous.
Using the commutative property together with the associative property we
can see that the order in which they are cascaded does not matter as far
as the overall system impulse response is concerned.

x[n] h1[n] h2[n] y[n]
Figure: Associative property.

x[n] h1[n] h2[n] y[n]
x[n] h[n] = h1[n] ∗ h2[n] y[n]

x[n] h1[n] h2[n] y[n]
x[n] h[n] = h1[n] ∗ h2[n] y[n]
x[n] h[n] = h2[n] ∗ h1[n] y[n]

x[n] h1[n] h2[n] y[n]
x[n] h[n] = h1[n] ∗ h2[n] y[n]
x[n] h[n] = h2[n] ∗ h1[n] y[n]
x[n] h2[n] h1[n] y[n]

LTI Systems with and without Memory
1 A system is memoryless if its output at any time depends only on the
value of the input at that same time.
2 The only way that this can be true for a discrete-time LTI system is if
h[n] = 0 for n ̸= 0.
3 In this case the impulse response has the form
h[n] = Kδ[n],
where K = h[0] is a constant.
4 The convolution sum reduces to the relation
y[n] = Kx[n].
5 If a discrete-time LTI system has an impulse response h[n] that is not
identically zero for n ̸= 0, then the system has memory.
6 For CT:
h(t) = Kδ(t).
y(t) = Kx(t).

lnvertibility of LTI Systems
An LTI system is invertible only if an inverse system exists that, when
connected in series with the original system, produces an output equal to
the input to the first system.
x(t) h(t) h1(t) w(t) = x(t)
y(t)
x(t) Identity system δ(t) x(t)
Figure: Inverse of a CT LTI system.

Example
Consider the following relationship of a pure time shift:
y(t) = x(t − t0)
Is the corresponding system memoryless? What is the inverse system of the system?

Example
y(t) = x(t − t0)
Delay if t0 > 0, and advance if t0 < 0. E.g., if t0 > 0 then the output at time t equals the input
at the earlier time t − t0. If t0 = 0, the system is the identity system and that is memoryless.
For any other value of t0, the system has memory.

Example
y(t) = x(t − t0)
Delay if t0 > 0, and advance if t0 < 0. E.g., if t0 > 0 then the output at time t equals the input
at the earlier time t − t0. If t0 = 0, the system is the identity system and that is memoryless.
For any other value of t0, the system has memory.
Setting the input equal to δ(t), the impulse response can be obtained:
h(t) = δ(t − t0).
Therefore,
x(t − t0) = x(t) ∗ δ(t − t0).
That is, the convolution of a signal with a shifted impulse simply shifts the signal. To recover the
input from the output, i.e., to invert the system, all that is required is to shift the output back.
h1(t) = δ(t + t0).
Then
h(t) ∗ h1(t) = δ(t − t0) ∗ δ(t + t0) = δ(t).

Example
Determine y[n], and find the inverse system of the following LTI system with impulse response
h[n] = u[n].

Example
h[n] = u[n].
y[n] = x[n] ∗ h[n] =
∞
X
k=−∞
x[k]h[n − k].

Example
h[n] = u[n].
y[n] = x[n] ∗ h[n] =
∞
X
k=−∞
x[k]h[n − k].
y[n] =
∞
X
k=−∞
x[k]u[n − k].
As u[n − k] is 0 for n − k < 0 and 1 for n − k ≥ 0,
y[n] =
n
X
k=−∞
x[k].

Example
h[n] = u[n].
y[n] = x[n] ∗ h[n] =
∞
X
k=−∞
x[k]h[n − k].
y[n] =
∞
X
k=−∞
x[k]u[n − k].
y[n] =
n
X
k=−∞
x[k].
This is the summer or the accumulator. As we saw before, the system is invertible, and its
inverse is
x[n] = y[n] − y[n − 1].

Example
h[n] = u[n].
y[n] = x[n] ∗ h[n] =
∞
X
k=−∞
x[k]h[n − k].
y[n] =
∞
X
k=−∞
x[k]u[n − k].
y[n] =
n
X
k=−∞
x[k].
This is the summer or the accumulator. As we saw before, the system is invertible, and its
inverse is
x[n] = y[n] − y[n − 1].
Choosing the input y[n] = δ[n],
h1[n] = δ[n] − δ[n − 1].

Example ctd.
Verify that h1[n] = δ[n] − δ[n − 1] indeed is the inverse of h[n] = u[n].

Example ctd.
Verify that h1[n] = δ[n] − δ[n − 1] indeed is the inverse of h[n] = u[n].
h[n] ∗ h1[n] = u[n] ∗ {δ[n] − δ[n − 1]}
= u[n] ∗ δ[n] − u[n] ∗ δ[n − 1]
= u[n] − u[n − 1]
= δ[n]

Causality for LTI Systems
1 The output of a causal system depends only on the present and past
values of the input to the system.
2 For a DT LTI system, y[n] must not depend on x[k] for k > n.
3 For this to be true, all of the coeﬀicients h[n − k] that multiply values
of x[k] for k > n must be zero.
4 This then requires that the impulse response of a causal discrete-time
LTI system satisfy the condition
h[n] = 0 for n < 0.
5 The impulse response of a causal LTI system must be zero before the
impulse occurs, which is consistent with the intuitive concept of
causality.
6 More generally, causality for a linear system is equivalent to the
condition of initial rest; i.e., if the input to a causal system is 0 up to
some point in time, then the output must also be 0 up to that time.
7 The equivalence of causality and the condition of initial rest applies
only to linear systems.

Causality for LTI Systems
1 A continuous-time LTI system is causal if
h(t) = 0 for t < 0.
2 Causality of an LTI system is equivalent to its impulse response being
a causal signal.

For a causal DT LTI system, the condition h[n] = 0 for n < 0 implies
that the convolution sum becomes
y[n] =
n
X
k=−∞
x[k]h[n − k].
and as
y[n] = h[n] ∗ x[n] =
∞
X
k=−∞
h[k]x[n − k].
y[n] =
∞
X
k=0
h[k]x[n − k].

For a causal DT LTI system, the condition h[n] = 0 for n < 0 implies
that the convolution sum becomes
y[n] =
n
X
k=−∞
x[k]h[n − k].
and as
y[n] = h[n] ∗ x[n] =
∞
X
k=−∞
h[k]x[n − k].
y[n] =
∞
X
k=0
h[k]x[n − k].
For a causal CT system, h(t) = 0 for t < 0, convolution integral is
y(t) =
Z t
−∞
x(τ)h(t − τ)dτ =
Z ∞
0

Stability for LTI Systems
A system is stable if every bounded input produces a bounded output. Consider an input x[n]
that is bounded in magnitude:
|x[n]| < B for all n.
|y[n]| =
∞
X
k=−∞
h[k]x[n − k]
|y[n]| ≤
∞
X
k=−∞
|h[k]||x[n − k]|
|y[n]| ≤ B
∞
X
k=−∞
|h[k]| for all n
∞
X
k=−∞
|h[k]| < ∞.
If the impulse response is absolutely summable, then y[n] is bounded in magnitude, and hence,
the system is stable.

Stability for LTI Systems
In CT a system is stable if the impulse response is absolutely integrable.
Z ∞
−∞
|h(τ)|dτ < ∞.

Examples Determine whether the following systems are stable: 1. Pure time shift in DT. 2.
Pure time shift in CT. 3. Accumulator in DT. 4. CT counterpart of the accumulator.
Pure time shift in DT:
∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1

∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1
Ans:stable.

∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1
Ans:stable.
Pure time shift in CT:
Z ∞
−∞
|h(τ)|dτ =
Z ∞
−∞
|δ(τ − t0)|dτ = 1

∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1
Ans:stable.
Z ∞
−∞
|h(τ)|dτ =
Z ∞
−∞
|δ(τ − t0)|dτ = 1
Ans: stable.
Accumulator in DT: h[n] = u[n]
∞
X
n=−∞
|u[n]| =
∞
X
n=0
|u[n]| = ∞

∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1
Ans:stable.
Z ∞
−∞
|h(τ)|dτ =
Z ∞
−∞
|δ(τ − t0)|dτ = 1
Ans: stable.
∞
X
n=−∞
|u[n]| =
∞
X
n=0
|u[n]| = ∞
Ans: unstable.

∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1
Ans:stable.
Z ∞
−∞
|h(τ)|dτ =
Z ∞
−∞
|δ(τ − t0)|dτ = 1
Ans: stable.
∞
X
n=−∞
|u[n]| =
∞
X
n=0
|u[n]| = ∞
Ans: unstable.
CT counterpart of the accumulator:
y(t) =
Z t
−∞
x(τ)dτ
The impulse response of the integrator can be
found by letting x(t) = δ(t):
h(t) =
Z t
−∞
δ(τ)dτ = u(t).
Z ∞
−∞
|u(τ)|dτ =
Z ∞
0
dτ = ∞

∞
X
n=−∞
|h[n]| =
∞
X
n=−∞
|δ[n − n0]| = 1
Ans:stable.
Z ∞
−∞
|h(τ)|dτ =
Z ∞
−∞
|δ(τ − t0)|dτ = 1
Ans: stable.
∞
X
n=−∞
|u[n]| =
∞
X
n=0
|u[n]| = ∞
Ans: unstable.
CT counterpart of the accumulator:
y(t) =
Z t
−∞
x(τ)dτ
The impulse response of the integrator can be
found by letting x(t) = δ(t):
h(t) =
Z t
−∞
δ(τ)dτ = u(t).
Z ∞
−∞
|u(τ)|dτ =
Z ∞
0
dτ = ∞
Ans: unstable.

The Unit Step Response of an LTI System
There is another signal that is also used in describing the behavior of LTI systems: the unit step
response, s[n] or s(t), corresponding to the output when x[n] = u[n] or x(t) = u(t).
u[n] h[n] s[n]
Figure: Unit step response.
s[n] = u[n] ∗ h[n]
Commutative property:
s[n] = h[n] ∗ u[n]
s[n] can be viewed as the response to the
input h[n] of a discrete-time LTI system with
unit impulse response u[n].
h[n] u[n] s[n]
Figure: Unit step response.
u[n] is the unit impulse response of the
accumulator. Therefore,
s[n] =
∞
X
k=−∞
h[k]
h[n] can be recovered from s[n] using the
relation
h[n] = s[n] − s[n − 1].

That is, the step response of a discrete-time LTI system is the running
sum of its impulse response. Conversely, the impulse response of a
discrete-time LTI system is the first difference of its step response.
Similarly, in CT, the step response of an LTI system with impulse response
h(t) is given by s(t) = u(t) ∗ h(t), which also equals the response of an
integrator [with impulse response u(t)] to the input h(t). That is, the unit
step response of a continuous-time LTI system is the running integral of its
impulse response, or
s(t) =
Z t
−∞
h(τ)dτ
the unit impulse response is the first derivative of the unit step response, or
h(t) =
ds(t)
d(t)
= s′
(t).

Zero-Input Response
For a linear system (time-invariant or not), if we put nothing into it, we
get nothing out of it.
x(t) = 0 for all t, then
y(t) = 0 for all t,
x[n] = 0 for all n, then
y[n] = 0 for all n,
“Proof”: If the system is linear and
x(t) → y(t), then if we scale

Zero-Input Response
For a linear system (time-invariant or not), if we put nothing into it, we
get nothing out of it.
x(t) = 0 for all t, then
y(t) = 0 for all t,
x[n] = 0 for all n, then
y[n] = 0 for all n,
“Proof”: If the system is linear and
x(t) → y(t), then if we scale
ax(t) → ay(t).
Select the sale factor a = 0.
Not all systems are like this, e.g., even if a battery is not connected to
anything, the output is 1.5 V.

Implications for Causality
The system cannot anticipate the input.
I.e., If
x1(t) = x2(t), for t < t0,
then
y1(t) = y2(t), for t < t0,
Same for DT.

Implications for Causality for a Linear System
For linear systems, if
x(t) = 0, for t < t0,
then
y(t) = 0, for t < t0,
Initial rest: The system does not respond until an input is given.
For a linear system to be causal it must have the property of initial rest.
Why?

Implications for Causality for a Linear System
For linear systems, if
x(t) = 0, for t < t0,
then
y(t) = 0, for t < t0,
Initial rest: The system does not respond until an input is given.
For a linear system to be causal it must have the property of initial rest.
Why? For linear systems zero in → zero out.

Causality for Linear Time Invariant Systems
For LTI systems,
Causality ⇔
h(t) = 0, t < 0
h[n] = 0, n < 0
“Proof”: ⇒: Why does causality imply the above?

For LTI systems,
Causality ⇔
h(t) = 0, t < 0
h[n] = 0, n < 0
“Proof”: ⇒: Why does causality imply the above? Ans:
δ(t) = 0, t < 0
δ[n] = 0, n < 0

For LTI systems,
Causality ⇔
h(t) = 0, t < 0
h[n] = 0, n < 0
δ(t) = 0, t < 0
δ[n] = 0, n < 0
⇐:Why does h(t) = 0, t < 0 (h[n] = 0, n < 0, imply the system is causal?

For LTI systems,
Causality ⇔
h(t) = 0, t < 0
h[n] = 0, n < 0
δ(t) = 0, t < 0
δ[n] = 0, n < 0
⇐:Why does h(t) = 0, t < 0 (h[n] = 0, n < 0, imply the system is causal? Ans:
y[n] =
∞
∑
k=−∞
x[k]h[n − k] y(t) =
∫ ∞
−∞
x(τ)h(t − τ)dτ
y[n] =
n
∑
k=−∞
x[k]h[n − k] y(t) =
∫ t
−∞
x(τ)h(t − τ)dτ
Only values of x[n] up until n are used to compute y[n] (causal). Similar in CT case.

Example: Accumulator
y[n] =
n
X
k=−∞
x[k]
The accumulator is an LTI system. Also, we saw that its impulse response
is
h[n] = u[n].
1 Does the accumulator have memory?
2 Is the accumulator causal?
3 Is accumulator stable in the BIBO
sense?
4 If invertible, what is the inverse?

y[n] =
n
X
k=−∞
x[k]
is
h[n] = u[n].
sense?
1 h[n] ̸= kδ[n].: has memory

y[n] =
n
X
k=−∞
x[k]
is
h[n] = u[n].
sense?
2 h[n] = 0, n < 0: is causal

y[n] =
n
X
k=−∞
x[k]
is
h[n] = u[n].
sense?
2 h[n] = 0, n < 0: is causal
3
Pn
k=−∞ |h[n]| = ∞ : not stable

y[n] =
n
X
k=−∞
x[k]
is
h[n] = u[n].
sense?
2 h[n] = 0, n < 0: is causal
3
Pn
Inverse h[n] = u[n],
h−1[n] =?:
Accumulator can be
expressed as a recursive
difference equation as
y[n] =
n−1
X
k=−∞
x[k] + x[n]
= y[n − 1] + x[n].

δ[n] h[n] h−1[n] δ[n]
u[n]
x2[n] y2[n]
Figure
We know that
u[n] − u[n − 1] = δ[n].
So
x2[n] − x2[n − 1] = y2[n].
δ[n] − δ[n − 1] = h1[n].
Inverse of the accumulator:
1 Does it have memory? Yes.
2 Is the system causal? Yes.
3 Is the system stable in the
BIBO sense? Yes.

Example
y[n] − ay[n − 1] = x[n]
under the assumption of initial rest ⇒ LTI. Memory? Causal? Stable?

Example
y[n] − ay[n − 1] = x[n]
x[n] = δ[n]
h[n] = an
u[n] (Why?)

Example
y[n] − ay[n − 1] = x[n]
x[n] = δ[n]
h[n] = an
u[n] (Why?)
1 Memory? Yes.
2 Causal? Yes.
3 Stable? If |a| < 1, yes.

Example
dy(t)
dt
+ 2y(t) = x(t)

Example
dy(t)
dt
+ 2y(t) = x(t)
h(t) = e−2t
u(t) (Why?)

Example
dy(t)
dt
+ 2y(t) = x(t)
h(t) = e−2t
u(t) (Why?)
1 Memory? Yes.
2 Causal? Yes.
3 Stable? Yes.

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