Discrete Signals and Systems Lecture

Chapter One
Discrete-Time Signals and Systems
Lecture #3
Rediet Million
AAiT, School Of Electrical and Computer Engineering
rediet.million@aait.edu.et
March, 2018
(Rediet Million) DSP-Lecture #3 March, 2018 1 / 27

1.2.Discrete-Time Signals and Systems
1.2.2. Discrete-Time Systems
A discrete-time system is deﬁned mathematically as a transformation
(processor or algorithm) that transforms an input sequence x(n) into
an output sequence y(n) according to some input/output rule.
y(n) = T{x(n)}
Discrete-time system examples:
An ideal delay: y(n) = x(n − nd ) , nd is the delay of the system.
Compressor (down-sampler): y(n) = x(Mn)
A rate expander(up-sampler): y(n) = x( n
M )
Modulator:y(n) = x(n)cos(nωo)

Discrete-Time Systems...
An accumulator system is deﬁed as
y[n] =
n−1
k=−∞
x[k] + x(n) i.e y(n) = y(n − 1) + x(n)
A moving average system is deﬁed by
y[n] = 1
M1+M2+1
M2
k=M1
x(n − k)
- This system computes the nth sample of the output sequence as the
average of 1
M1+M2+1 samples of the input sequence around the nth sample.

Moving average smoothes out rapid variations e.g
x(n)= s(n)signal +d(n)noise
using 5 point moving average
y[n] = 1
5
4
k=0
x(n − k)

Classiﬁcation of DT Systems
1.Linear system
A system is linear if it satisﬁes superposition principle.
Additive property:
T{x1(n) + x2(n)} = T{x1(n)} + T{x2(n)} = y1(n) + y2(n)
Homogeneity or scaling property:
T{ax(n)} = aT{x(n)} = ay(n) where a is arbitrary constant.
These two properties are combined into the principle of
superposition:
T{ax1(n) + bx2(n)} = aT{x1(n)} + bT{x2(n)} = ay1(n) + by2(n)

Classiﬁcation of DT Systems...
Example-1
- Accumulator:
y[n] =
n
k=−∞
x[k]
x(n) = a · x1(n) + b · x2(n)
y[n] =
n
k=−∞
[a · x1(k) + b · x2(k)]
=
n
k=−∞
[a · x1(k)] +
n
k=−∞
[b · x2(k)]
= a ·
n
k=−∞
[x1(k)] + b ·
n
k=−∞
[x2(k)]
= a · y1(n) + b · y2(n)
Thus,the system is linear.
Example-2
- Oﬀset accumulator:
y[n] = C +
n
k=−∞
x[k]
y[n] = C +
n
k=−∞
[a · x1(k) + b · x2(k)]
= C+a·
n
k=−∞
[x1(k)]+b·
n
k=−∞
[x2(k)]
= a · y1(n) + b · y2(n)
Thus,the system is non-linear
unless C=0.

2.Time(shift)-invariant system
Time shift or delay of the input sequence causes a corresponding shift
in the output sequence.
if x(n) −→ y(n) implies that
x(n − no) −→ y(n − no)
Example :
Take the system y(n) = n · x(n)
If x(n) = x1(n − no) then
y(n) = n · x1(n − no)
=
y(n − no) = (n − no) · x1(n − no)
⇒ Hence, the system is not shift invariant system.

Discrete-time systems...
3.Causal system
A system is causal if the output value at n = no depends only on the
input sequence values for n ≤ no.
In other words, the output of the system at any time depends only on
present and past values of the input x(n)
i.e[x(n), x(n − 1), x(n − 2), ..x(n − no)]
but does not depends on future inputs
i.e[x(n + 1), x(n + 2), ..x(n + no)] → non-causal system.
Example :
-Forward diﬀerence system,y(n) = x(n − 1) − x(n), is causal system.
-Backward diﬀerence system,y(n) = x(n + 1) − x(n), is non-causal
system.

4.Stable system
A system is said to be stable in the Bounded Input-Bounded Output
sense if, for any input that is bounded,|x(n)| ≤ A < ∞ , the output
will be bounded,|y(n)| ≤ B < ∞.
If for some bounded input x(n) the output is unbounded (infinite)
,then the system is classified as unbounded.
Example : Consider the accumulator system y[n] =
n
k=−∞
x[k]
When x(n) = u(n) ,which is clearly bounded by A = 1. For this
input,the output of the accumulator is
y[n] =
n
k=−∞
u[k] =
(n + 1) , n ≥ 0
0 , n < 0
There is no finite choice for B such that ,(n − 1) ≤ B < ∞ for all n.
⇒ Thus,the system is unstable.

Discrete-Time Systems| Classiﬁcation of DT Systems
(#2) Class exercises & Assignment
1) For each of the following systems,determine whether the system is
linear system and shift invariant system .
a.y(n) = x(n)sin(2πn
7 + π
6 )
b.y(n) = log [x(n)]
c.y(n) = 2x(n) + 3
d.y(n) = [x(n)]2
e.y(n) = ex(n)
2) For each of the following systems,determine whether the system is
stable and causal system .
a.y(n) = [x(n + 1)]2
b.y(n) = ex(n)
x(n−1)
c.y(n) = cos[x(n + 2)]

1.3.LTI System and Discrete-Time Fourier Transform
1.3.1 Linear Time-Invariant (LTI) Systems
This is a particularly important class of systems that combines the
properties of linearity and time-invariance.
Important due to convenient representations and signiﬁcant
applications.
A linear time-invariant system is completely characterized by its
impulse response h(n).
Given a system :
If x(n) = δ(n) then y(n) ≡ h(n) → impulse response

Linear Time-Invariant (LTI) Systems...
For example take a simple system:
Using signal decomposition we know that we can express any sequence
by unit sample (impulse) δ(n) as :
x(n) =
∞
k=−∞
x(k)δ(n − k)

When we apply the above sequence x(n) as an input to LTI system
- Due to linearity :
y(n) = T[x(n)] = T[
∞
k=−∞
x(k)δ(n − k)]
y(n) = T[x(n)] =
∞
k=−∞
x(k)T[δ(n − k)]
- Due to Time(shift) invariance :
T[δ(n)] = h(n) ⇒ T[δ(n − k)] = h(n − k)
-Therefore,
y(n) = T[x(n)] =
∞
k=−∞
x(k)h(n − k) ≡ x(n) ∗ h(n)
=
∞
k=−∞
h(k)x(n − k).... show the proof !
This expression is called convolution sum.

Steps involved to perform the convolution sum
y(n) = x(n) ∗ h(n) =
∞
k=−∞
x(k)h(n − k)
1 Consider x(k) and h(k) as a function of k.
2 Obtain the sequence h(n − k)
Reﬂect h(k) about the origin to get h(−k).
Shift the reﬂected sequence by n(to the right if n > 0 and to the
left if n < 0).
3 Multiply the two sequence x(k) & h(n − k) and sum the product for
all values of K.

Examples
1. Analytically convolution of two signals:
x(n) = anu(n) =
an , n ≥ 0
0 , n < 0
and h(n) = u(n)
Using direct evaluation of the convolution sum we ﬁnd
y(n) = x(n) ∗ h(n) =
∞
k=−∞
x(k)h(n − k) =
∞
k=−∞
aku(k)u(n − k)
From the deﬁnition of unit step u(k) = 0 for k < 0 and u(n − k) = 0 for
k > n.Thus,
y(n) =
n
k=0
ak =
1 − an+1
1 − a
= (
1 − an+1
1 − a
)u(n)
Some useful formulas:
∞
n=0
an =
1
1 − a
for |a| < 1
N−1
n=0
an =
1 − aN
1 − a
N−1
n=0
n =
1
2
N(N − 1)
∞
n=0
nan =
a
(1 − a)2
for |a| < 1

Examples
2.Graphical convolution of two signals:

Examples

Properties of LTI-systems
LTI systems are completely characterized by the impulse response
function,h(n). Therefore, we can determine causality and stability
from h(n):
Causality: h(n) = 0 for n < 0.
Stability:
∞
n=−∞
|h(n)| < ∞.
For the convolution operator,y(n) = x(n) ∗ h(n),there are several
properties of interest :
Commutative: x(n) ∗ h(n) = h(n) ∗ x(n)
Associative: [x(n) ∗ h1(n)] ∗ h2(n) = x(n) ∗ [h1(n) ∗ h2(n)]
Distributive: x(n) ∗ [h1(n) + h2(n)] = x(n) ∗ h1(n) + x(n) ∗ h2(n)
Discrete-time convolution will be discussed detail in chapter three.

(#3) Class exercises & Assignment
For each of the following pairs of sequences,x(n), represents the input to
an LTI system with impulse response h(n).Determine each output y(n)
using discrete time convolution.
a.
b. h(n) =
1 , 0 ≤ n ≤ (N − 1)
0 , otherwise
x(n) = anu(n)
a.
b. h(n) =
αn , 0 ≤ n ≤ 10
0 , otherwise
x(n) =
1 , 0 ≤ n ≤ 5
0 , otherwise

1.3.2 Linear Constant Coefficient Difference Equation
From a computational point of view sometimes convolution sum
representation is not very efficient.
For example,take a system that has a unit sample response
h(n) = αnu(n):
Using convolution sum we can express y(n) as :
y(n) =
∞
k=0
αnx(n − k)
But we can express the output more efficiently in terms of past
values of the output in addition to the current and past values of the
input .
y(n) = αy(n − 1) + x(n)
- This equation is a special case of what is known as a linear constant
coefficient difference equation (LCCDE).

Linear Constant Coefficient Difference Equation...
The general form of LCCDE is :
y(n) =
M
k=0
b(k)x(n − k) −
N
k=1
a(k)y(n − k)
- Where the coefficients a(k) and b(k) are constants that define the
system.
- If a(k) = 0 → recursive difference equation.
- If a(k) = 0 → non-recursive difference equation.
Before solving LCCDE,it is necessary to specify a set of an initial
conditions.
For LTI system that is described by a difference equation, the unit
sample response,h(n), is found by solving the difference equation for
x(n) = δ(n) assuming initial condition is zero.

For a non-recursive system ,a(k) = 0, the difference equation becomes
y(n) =
M
k=0
b(k)x(n − k) ⇒ h(n) =
N
k=0
b(k)δ(n − k)
Thus, h(n) is finite in length and the system is referred to as a
finite impulse response(FIR) system.
However, if a(k) = 0, the unit sample response h(n) is infinite in
length and the system is referred to as an infinite impulse response
(IIR)system.
There are several difference methods that one may use to solve
LCCDEs for a general input x(n).
Homogeneous and particular solution approach.
Z-transform approach→ will discussed in chapter 4.

Homogeneous and Particular Solutions
Given an LCCDE, the general solution is a sum of two parts:
y(n) = yh(n) + yp(n)
- yh(n) is known as the homogeneous solution and yp(n) is the
particular solution.
Homogeneous solutions (zero input response):
The homogeneous solution is the response of the the system to the
initial conditions, assuming that the input x(n) = 0.
The solution is found by solving the diﬀerence equation as :
N
k=0
a(k)y(n − k) = 0
when a0 = 1
y(n) +
N
k=1
a(k)y(n − k) = 0

Homogeneous and Particular Solutions...
The general method of solution to the homogeneous equation is to
assume a homogeneous solution, y(n), as
yh(n) = λn
If the assumed solution is a solution, then it has to satisfy the equation:
N
k=0
akλn−k
= 0
assuming a0 = 1
λn + a1λn−1 + a2λn−2 + .....aNλn−N = 0
λn−N[λN + a1λN−1 + a2λN−2 + .....aN] = 0
The polynomial in braces is called the characteristics polynomial.
Because it is of degree N, it may have N roots (distinct or repeated),
which may be either real or complex.

Homogeneous and Particular Solutions...
Distinct roots: The form of the most general solution will be a linear
combination of those of each characteristics equations as
yh(n) = c1λn
1 + c2λn
2 + c3λn
3 + ...cNλn
N
- ci is determined from the initial conditions.
Repeated roots: If some of the roots of the characteristics equations
are repeated then the form of the solution will be modiﬁed as
yh(n) = c11λn
1 + c12nλn
1 + c13n2
λn
1 + ... + c1mnm−1
λn
1 + c2λn
2 + ...cNλn
N

Homogeneous and Particular Solutions
Particular solutions (zero state response):
It is the solution response to the applied input signal.
To determine the particular solution yp(n) is to assume that the
response will have the same form as the input itself within a multiplier
constant.

(#4 ) Class exercises & Assignment
1) Determine the total response,y(n),of the following causal systems
when the forcing function is x(n). Assume initial conditions are zero.
a.y(n) =
5
6
y(n − 1) −
1
6
y(n − 2) + x(n) , x(n) = 2nu(n)
b.y(n) − 3y(n − 1) − 4y(n − 2) = x(n) + 2x(n − 1) , x(n) = δ(n)
2) Determine the impulse response and the unit step response of the
system described by the following diﬀerence equations for n ≥ 0.
a.y(n) = 0.7y(n − 1) − 0.1y(n − 2) + 2x(n) − x(n − 2)
b.y(n) = 0.6y(n − 1) − 0.08y(n − 2) + x(n)

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