Digital Signal Processing(DSP)

Dr.shiref abo elnour

 (Lectures 2, 3)
Discrete Time Linear Time Invariant System

What will be covered today

➢ What is a Discrete Time System?
➢ Properties of Discrete Time System
➢ What is a LTI System?
➢ Properties of LTI System?
➢ What is a Convolution?
 Discrete-Time System

➢ Discrete-Time System is a transformation or operator that

maps input sequence x[n] into an output sequence y[n].
• y[n]=T{x[n]};
x[n], Y[n]: discrete-time signal

Discrete-Time System
EX. THE IDEAL DELAY SYSTEM
y[n]= x[n -n𝑑 ,], -∞<n<∞
•If n𝑑 is a positive integer: the delay of the system, Shift the
input sequence to the right by n𝑑 , samples to form the
output .
*If n𝑑 is a negative integer: the system will shift the input
to the left by |n𝑑 | samples, corresponding to a time
advance.
PROPERTIES OF DISCRETE TIME SYSTEM
LINEAR SYSTEMS
➢If
x1 [n] T{.} 𝑦1 [𝑛]

x2 [n] T{.} 𝑦2 [𝑛]

➢And only If:

x1 [n] + x2 [n] T{.} 𝑦1 [𝑛] + 𝑦2 [𝑛] Additivity property

ax[𝑛] T{.} 𝑎𝑦[𝑛] Homogeneity or scaling property

➢ Principle of superposition:
x3 [n]=ax1 [n]+bx2 [n] T{.} y3 [n]=ay1 [n]+by2 [n]
 EXAMPLE OF LINEAR SYSTEM
𝑛
 Ex. Accumulator system 𝑦[𝑛] = ෍ 𝑥[𝑘]
for arbitrary 𝑥1 [n] 𝑎𝑛𝑑 𝑥2 [n] 𝑘=−∞

𝑛 𝑛

𝑦1 [𝑛] = ෍ 𝑥1 [𝑘] 𝑦2 [𝑛] = ෍ 𝑥2 [𝑘]

𝑘=−∞ 𝑘=−∞

 𝑤ℎ𝑒𝑛 𝑥3 [n]=𝑎𝑥1 [n]+b𝑥2 [n]

𝑛 𝑛

𝑦3 [𝑛] = ෍ 𝑥3 𝑘 = ෍ (𝑎𝑥1 𝑘 + 𝑏𝑥2 𝑘 )

𝑘=−∞ 𝑘=−∞
𝑛 𝑛

= 𝑎 ෍ 𝑥1 𝑘 + 𝑏 ෍ 𝑥2 𝑘 = 𝑎𝑦1 𝑛 + 𝑏𝑦2 [𝑛]

𝑘=−∞ 𝑘=−∞
EXAMPLE NONLINEAR SYSTEMS

Method: find one counterexample

For y[n] = (𝑥 [n]) 2
Counterexample 12 + 12 ≠ (1 + 1) 2
For y[n] = 𝑙𝑜𝑔10 (|𝑥 𝑛 |)
The system is non linear
 PROPERTIES OF DISCRETE -TIME SYSTEM
TIME- INVARIANT SYSTEMS
Shift – Invariant Systems
𝑥1 [𝑛] T{·} 𝑦1 [𝑛]

𝐱 𝟐 𝐧 = 𝐱 𝟏 𝐧 − 𝐧𝟎 T{·} 𝒚𝟐 [𝒏]=𝒚𝟏 [𝒏 − 𝐧𝟎 ]
EXAMPLE OF TIME − INVARIANT SYSTEM

Ex. the accumulator as a time − Invariant system

𝑦[𝑛] = ෍ 𝑥[𝑘]
𝑘=−∞

𝑥1 𝑛 = 𝑥 𝑛 − 𝑛0
𝒏 𝒏

𝒚𝟏 𝒏 = ෍ 𝒙𝟏 𝒌 = ෍ 𝒙 𝒌𝟏 = 𝒚[𝒏 − 𝒏𝟎 ]
𝒌=−∞ 𝒌=−∞
 1

2
𝑦2 = 𝑦1 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑡𝑖𝑚𝑒 𝑖𝑛𝑣𝑎𝑟𝑖𝑎𝑛𝑡

𝑏. 𝑦2 (n) = 2x (3n) =2x2 (3n - n0 )

𝑦1 𝑛 − 𝑛0 = 2𝑥1 (3 𝑛 − 𝑛0 = 2𝑥1 (3𝑛 − 3𝑛0 )

Since 𝑦1 ≠ 𝑦2 the system is time variant.

(b) 𝒚(𝒏) = 𝒏𝒙(𝒏)
𝑦 𝑛 − 𝑛0 = 𝑛𝑥 𝑛 − 𝑛0
𝑦1 = 𝑛 − 𝑛0 𝑥 𝑛 − 𝑛0
y(𝑛 − 𝑛0 ) ≠ 𝑦1
PROPERTIES OF DISCRETE -TIME SYSTEM
CAUSALITY
 A system is causal if, every choice 𝑛0 ,the output sequence value at index
𝑛 = 𝑛0 depends only on the input sequence value for 𝑛 ≤ 𝑛0
in past and present value.

 i.e if 𝑥1 [𝑛] = 𝑥2 𝑛 , 𝑓𝑜𝑟 𝑛 ≤ 𝑛0

Then 𝑦1 [𝑛] = 𝑦2 [𝑛] , 𝑓𝑜𝑟 𝑛 ≤ 𝑛0

EX. The forward and backward difference
systems
 Forward difference system is not causal

 y[n]=x[n+1]-x[n]

 Backward difference system is causal

 y[n]=x[n]-[n-1]
EX. The forward and backward difference
systems
 PROPERTIES OF DISCRETE-TIME
SYSTEMSSTABILITY
 Bounded-Input Bounded-Output (BIBO) Stability:
every bounded input sequence produces a bounded output sequence.

If |x[n]| <𝑩𝒙 <∞, for all n

Then |x[n]| <𝑩𝒚 <∞, for all n
Ex. Testing for Stability or Instability
𝑛

 Accumulator system 𝑦[𝑛] = ෍ = 𝑥[𝑘]

𝑘=−∞

0 𝑛<0
x[n]= u[n]= ൜ : 𝑏𝑜𝑢𝑛𝑑𝑒𝑑
1 𝑛≥0
𝑛
0
𝑦 𝑛 = ෍ 𝑥 𝑘 =൜ : 𝑛𝑜𝑡 𝑏𝑜𝑢𝑛𝑑𝑒𝑑
𝑛+1
𝑘=−∞
Digital Signal Processing(DSP)
Dr.shiref abo elnour
➢ Linear Time-Invariant (LIT) Systems
2
x[n]𝛿 𝑛 = 𝑥 0 = 2
1
x[n]𝛿 𝑛 + 2 = 𝑥 −2 = 2 -1 2
-2 0 1
x[n]𝛿 𝑛 + 1 = 𝑥 −1 = −1
x[n]𝛿 𝑛 − 1 = 𝑥 1 = 1
x[n]𝛿 𝑛 − 2 = 𝑥 2 = 2

x[n]= x[0]𝛿 0 +x[-2]𝛿 𝑛 + 2 + x[-1]𝛿 𝑛 + 1 +x[1]𝛿 𝑛 − 1 + x[2]𝛿 𝑛 − 2

In general form
∞

𝑥 𝑛 = ෍ 𝑥 𝑘 𝛿[𝑛 − 𝑘]
𝑘=−∞

x[n] y[n]
y[n]= H[x(n)] H{x[n]}
∞

=𝐻[ ෍ 𝑥 𝑘 𝛿 𝑛−𝑘 ]
𝑘=−∞
∞ ∞ ∞

=[ ෍ 𝐻𝑥 𝑘 𝛿 𝑛−𝑘 = ෍ 𝑥 𝑘 𝐻 𝛿 𝑛 − 𝑘 = ෍ 𝑥 𝑘 ℎ[𝑛 − 𝑘]
𝑘=−∞ 𝑘=−∞ 𝑘=−∞
Linear Time-Invariant (LIT) Systems
Shift-invariant system

𝛿𝑛 h[n]

𝛿[𝑛 − 𝑛0 ] h[n-𝑛0 ]
 Representation of general sequence as a linear
combination of delayed impulse
LTI systems: Convolution
𝛿𝑛 h[n]

𝛿[𝑛 − 𝑛0 ] h[n-k]

➢ Principle of superposition
∞ ∞

𝑦 𝑛 = 𝐻 { ෍ 𝑥 𝑘 𝛿[𝑛 − 𝑘]} = ෍ 𝑥 𝑘 𝐻{𝛿 𝑛 − 𝑘 }

𝑘=−∞ 𝑘=−∞

= ෍ 𝑥 𝑘 ℎ 𝑛 − 𝑘 = 𝑥 𝑛 ∗ ℎ[𝑛]
𝑘=−∞
 EX: Find y[n] of the following signal x[n] , h[-n]
h[-n]

K= -1 0.5
1
0.5
1
0.5 0.5

-1 1 -1 2
× -1 0 =
0 1
2
- 0.5 - 0.5
-1

x[-1]𝛿[𝑛 + 1] h[𝑛 + 1] x[-1] h[𝑛 + 1]

K= -1 x[-1]𝛿[𝑛 + 1] h[𝑛 + 1] x[-1] h[𝑛 + 1]
0.5 1 0.5 1 0.5

-1 1 -1 2
× -1 0 2 = 0 1
- 0.5 - 0.5
-1 -1 - 0.5
K= 0 x[0]𝛿[𝑛] h[𝑛]
x[0] h[𝑛]
1 0.5 0.5
0.5 0.5
2
0.25
2 0.25

0 × 0 1 3
= 0 1 3
- 0.5
- 0.25
K= 1 1 1 h[𝑛 − 1] x[1] h[𝑛 − 1]
x[1]𝛿[𝑛 − 1] 0.5
1
0.5 0.5 0.5

3
0 1
× 0 1 2 3 4= 0 1 2 4
- 0.5 - 0.5
K= 2 x[-1]𝛿[𝑛 − 2] h[𝑛 − 1] x[1] h[𝑛 − 1]
1
0.5 0.5
0.25
2 1 4 -1 2 3 5
0 × 0 2 3 5 = 1.5 0 1 4
- 0.25
- 0.5 - 0.5 - 0.5
0.75

𝑦 𝑛 = −0.5 𝛿 𝑛 + 1 − 0.75𝛿 𝑛 + 1.5𝛿 𝑛 − 1 -1 0 1 2 3 4 5

−0.75𝛿 𝑛 − 3 + 0.75𝛿 𝑛 − 4 − 0.25𝛿 𝑛 − 5 - 0.5 - 0.75 - 0.25
- 0.75
Ex: determine the output of the system
due to the following input
2, 𝑛=0
1, 𝑛=0
4, 𝑛=1
 𝑥𝑛 = ℎ 𝑛 ቐ 0.5, 𝑛 = 1
−2 ,𝑛 = 2
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
 𝑥 𝑛 = 2𝛿 𝑛 + 4𝛿 𝑛 − 1 − 2𝛿 𝑛 − 2 ℎ 𝑛 = 𝛿 𝑛 + 0.5𝛿 𝑛 − 1

x[n] X[n]*h[n] y[n]

 𝑥 𝑛 = 2ℎ 𝑛 + 4ℎ 𝑛 − 1 − 2ℎ 𝑛 − 2
LTI
= 2(𝛿 𝑛 + 0.5𝛿 𝑛 − 1 + 4(𝛿 𝑛 − 1 + 0.5𝛿 𝑛 − 2 ) − 2(𝛿 𝑛 − 2 + 0.5𝛿 𝑛 − 3
= 2𝛿 𝑛 + 𝛿 𝑛 − 1 + 4𝛿 𝑛 − 1 + 2𝛿 𝑛 − 2 − 2𝛿 𝑛 − 2 + 𝛿 𝑛 − 3
= 2𝛿 𝑛 + 5𝛿 𝑛 − 1 + 𝛿 𝑛 − 3
2, 𝑛 = 0
5, 𝑛 = 1
𝑦𝑛 =
1, 𝑛 = 3
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
EX: Evaluate y[n]=x[n]*h[n] where x[n],h[n]
as shown below
X[n] h[n]
2
1 1

-2 -1 0 0 1 2
[n] [n]

h[-n]
2
1

-2 -1 0 [n]
 X[k] h[-k]
2 2
K=-2
1 1 1 1

-2 -1 0 -2 -1 0 [k] -2 -1 0
[k] [k]
X[k]
2 2
K= -1 1
1 1 1
1

-1 0 1
-1 0 1 [k] [k]
-2 -1 X[k]
0
[k]
2 2
K=0 1
1 1 1 1

-2 -1 0 1 2
[k] 0 1 2 0 [k]
[k]
 2
1 1

-2 -1 0 [k]

2
1 1

-1 0 1 [k]
𝑌 𝑛 = 𝛿 𝑛 + 2 + 3𝛿 𝑛 + 1
+ 4𝛿 𝑛 + 3𝛿 𝑛 − 1 + 𝛿[𝑛 − 2 2
1 1

0 1 2
[k]
4
3 3
y[n]
1 1

-2 -1 0 1 2
[k]
 2) Analytical method
X[n] h[n]
2
1 1

-2 -1 0 0 1 2
[n] [n]

 𝑥 𝑛 =𝛿 𝑛 +𝛿 𝑛+1 +𝛿 𝑛+2
 ℎ 𝑛 = 𝛿 𝑛 + 2𝛿 𝑛 − 1 + 𝛿 𝑛 − 2
 𝑥 𝑛 ∗ ℎ 𝑛 = 𝑥 𝑛 ∗ [𝛿 𝑛 + 2𝛿 𝑛 − 1 + 𝛿 𝑛 − 2 ]
= 𝑥 𝑛 + 2x n − 1 + x n − 2
= 𝛿 𝑛 + 𝛿 𝑛 + 1 + 𝛿 𝑛 + 2 + 2𝛿 𝑛 − 1 + 2𝛿 𝑛 + 2𝛿 𝑛 + 1
+𝛿 𝑛−2 +𝛿 𝑛−1 +𝛿 𝑛
= 4𝛿 𝑛 + 3𝛿 𝑛 + 1 +𝛿 𝑛 + 2 + 3𝛿 𝑛 − 1 + 𝛿 𝑛 − 2
 3 𝑛
EX: Given LTI system with ℎ 𝑛 = 𝑛 ( ) 𝑈
4
find the output of the system at n=5,n=10, when the input
signal x[n]=u[n]
Y[n]=x[n]*h[n]
3
(4)𝑘 U(k)
h[k]
𝟑
1 𝟗
𝟒
𝟏𝟔 𝟐𝟕
𝟔𝟒

1 2 3 4 k
3
(4)𝑛−𝑘 U(n-k) h[n-k]
h[-k]
𝟑 𝟑
𝟗 𝟗 1
1 𝟐𝟕 𝟏𝟔 𝟒
𝟐𝟕 𝟏𝟔 𝟒
𝟔𝟒 𝟔𝟒

k -4 -3 -2 -1 k -3 -2 -1 1 2
 3
( )𝑛−𝑘 U(n-k) h[n-k]
4
𝟑
𝟗 1
𝟐𝟕 𝟏𝟔 𝟒
𝟔𝟒

-3 -2 -1 1 2

 n< 0 𝑦 𝑛 =0 𝑛
𝑛+1
1 − 𝑎
3 𝑛−𝑘
𝑛
𝑦 𝑛 = ෍ 𝑎𝑘 =
 n≥ 0 𝑦 𝑛 = ෍( ) 𝑢(𝑛 − 𝑘) 1−𝑎
4 𝑘=0
𝑘=0

𝑛
3 𝑛−𝑘 =1 𝑛 𝑛 𝑛
3 𝑛 3 −𝑘 3 𝑛 3 −𝑘 3 𝑛 4 𝑘
𝑦 𝑛 = ෍( ) 𝑢(𝑛 − 𝑘) = ෍( ) ( ) = ( ) ෍ ( ) = ( ) ෍ ( )
4 4 4 4 4 4 3
𝑘=0 𝑘=0 𝑘=0 𝑘=0
4𝑛+1 45+1
4 𝑛 1−3 3 5 1−3
=( ) n=5, 𝑦 𝑛 = ( ) 4 = 0.92
4 4 1−3
3 1−
3 410+1
3 10 1−3
n=10, 𝑦 𝑛 = ( ) 4 = 0.21
4 1−3
 Given LTI system with x[n]=(0.5)𝑛 𝑢 𝑛 + 1 − 𝑢 𝑛 − 2 𝑎𝑛𝑑
h(n) =[u(n)-u(n-2)] find the output of the system y[n]
y[n]=x[n]*h[n]

u(n+1) u(n)
------∞
-1 0 1 2 3 0 1 2 3 4

u(n-2)
u(n-2)
2 3 4

2 3 4
u(n)-u(n-2)
u(n+1)-u(n-2)
0 1

-1 0 1
x(n) 2
1
0.5

-1 0 1
 y[n]=x[n]*h[n]

y[2]=0.5
x(n) y(2)

-1 0 1 1 2

h(n)
y(2) y[3]=0

0 1
2 3
h(-n) 3
y[0]=2+1=3 y(n) 1.5
0.5
-1 0

1 2 3
y(1)
y[1]=1+0.5=1.5
0 1
If the response of LIT system to unit step (step
1𝑛
response) is S[n]= 𝑈 𝑛
2
find the unit sample response h[n]
𝛿 𝑛 = 𝑈 𝑛 − 𝑢[𝑛 − 1]
ℎ 𝑛 = 𝑆 𝑛 −𝑆 𝑛−1
1𝑛 1𝑛−1
= 𝑈
𝑛 − 𝑛−1 𝑈
2 2
𝑛=0 ℎ[0] = 1 − 0 = 1
n<0 ℎ 0 =0
1𝑛 1𝑛−1
𝑛>0 ℎ 𝑛 == −
2 2
0 𝑛<0
h[n]= ൞ 1 𝑛=0
1𝑛 1𝑛−1
− 𝑛 >0
2 2