Chapter 1: Digital Filters
(4 weeks)
- The z-Transform
- Digital filters structures, Transfer function, stability and implementation
- Digital filter design
- Minimum-Phase systems
- Multirate filtering
1. The z-Transform
1.1. Definitions
Similar to the Laplace transform which deals with continuous Linear Time-Invariant Systems
(LTIS),
the z- transform provides a way to analyze discrete-time signals and systems in the frequency
domain. The z-transform of a discrete-time signal 𝑥(𝑘) is defined as:
𝑋(𝑧) = ∑∞
𝑘=−∞ 𝑥(𝑘)𝑧
−𝑘
(1.1)
where 𝑧 is a complex-valued variable from the z-plan. It is expressed as:
𝑧 = 𝑟𝑒 𝑗𝜔 (1.2)
The values of z for which the z-transform converges is called the region of convergence (ROC).
The z-transform defined in equation (1.1) is called bilateral (two-sided) z-transform in contrast
to the unilateral (one sided) z-transform given by:
𝑋(𝑧) = ∑∞
𝑘=0 𝑥(𝑘)𝑧
−𝑘
(1.3)
Examples
𝒙(𝒌) = 𝜹(𝒌)
0
𝑋(𝑧) = ∑ 1𝑧 −𝑘 = 1. 𝑧 0 = 1
𝑘=0
ROC=the entire z-plane
𝒙(𝒌) = {𝟏
⏟ , 𝟐, 𝟎, 𝟑}
↑
1
3
𝑋(𝑧) = ∑ 𝑎𝑘 𝑧 −𝑘 = 1. 𝑧 0 + 2. 𝑧 −1 + 0. 𝑧 −2 + 3. 𝑧 −3
𝑘=0
𝑧 3 + 2𝑧 2 + 3
= 1 + 2𝑧 −1 + 3. 𝑧 −3 =
𝑧3
ROC=the entire z-plane except z=0
𝒙(𝒌) = 𝒂𝒌 𝒖(𝒌)
∞ ∞
𝑋(𝑧) = ∑ 𝑎𝑘 𝑧 −𝑘 = ∑ (𝑎𝑧 −1 )𝑘
𝑘=0 𝑘=0
1−𝑟 𝑁+1 𝑁→∞ 1
Rappelez : ∑𝑁 𝑘
𝑘=0 𝑟 = → = 1−𝑟 si |𝑟| < 1
1−𝑟
Donc :
1 𝑧
𝑋(𝑧) = 1−𝑎𝑧 −1 = 𝑧−𝑎 , |𝑎𝑧 −1 | < 1 → |𝑧| > |𝑎|
𝒙(𝒌) = −𝒂𝒌 𝒖(−𝒌 − 𝟏)
−1 ∞ ∞ ∞
𝑋(𝑧) = − ∑ 𝑎𝑘 𝑧 −𝑘 = − ∑ (𝑎𝑧 −1 )−𝑙 = − ∑ 𝑎−𝑙 𝑧 𝑙 = 1 − ∑ (𝑎−1 𝑧)𝑙
𝑘=−∞ 𝑙=1 𝑙=1 𝑙=0
1 𝑧
=1− −1
=
1−𝑎 𝑧 𝑧−𝑎
𝑦𝑖𝑒𝑙𝑑𝑠
and |𝑎−1 𝑧| < 1 → |𝑧| < |𝑎|
1.2. Z- transform properties
• Linearity
𝑍{𝑥1 (𝑘)} = 𝑋1 (𝑧), 𝑍{𝑥2 (𝑘)} = 𝑋2 (𝑧)
𝑍{𝑎𝑥1 (𝑘) + 𝑏𝑥2 (𝑘} = 𝑎𝑋1 (𝑧) + 𝑏𝑋2 (𝑧), 𝑅𝑂𝐶 = 𝑅𝑥1 ∩ 𝑅𝑥2 (1.4)
Example
𝒙(𝒌) = (𝟐)𝒌 𝒖(𝒌) + (−𝟑)𝒌 𝒖(𝒌)
𝑥1 (𝑘) = 2𝑘 𝑢(𝑘) , 𝑥2 (𝑘) = (−2)𝑘 𝑢(𝑘)
2
𝑧
𝑋1 (𝑧) = 𝑅𝑥1 = |𝑧| > 2
𝑧−2
𝑧
𝑋2 (𝑧) = , 𝑅 = |𝑧| > 3
𝑧 + 3 𝑥2
The intersection of 𝑅𝑥1 and 𝑅𝑥2 is |𝑧| > 3. Thus, the z-transform of the sequence 𝑥(𝑘) is:
𝑧 𝑧
𝑋(𝑧) = + , |𝑧| > 3
𝑧−2 𝑧+3
• Time shifting
𝑍{𝑥(𝑘 − 𝑛)} = 𝑧 −𝑛 𝑋(𝑧) (1.5)
The ROC of 𝑧 −𝑘0 𝑋(𝑧) is the same as that of 𝑋(𝑧) except 𝑧 = 0 for 𝑛 > 0 or 𝑧 = ∞ for
𝑛 < 0.
Proof :
∞
𝑍{𝑥(𝑘 − 𝑛)} = ∑ 𝑥(𝑘 − 𝑛)𝑧 −𝑘
𝑘=−∞
Let 𝑚 = 𝑘 − 𝑛, then: 𝑍{𝑥(𝑚)} = ∑∞
𝑚=−∞ 𝑥(𝑚)𝑧
−𝑚−𝑛
= 𝑧 −𝑛 ∑∞
𝑚=−∞ 𝑥(𝑚)𝑧
−𝑚
= 𝑧 −𝑛 𝑋(𝑧)
Example
𝒙(𝒌) = 𝒖(𝒌) − 𝒖(𝒌 − 𝟓)
𝑧 𝑧 𝑧 − 𝑧 −4 𝑧5 − 1
𝑋(𝑧) = 𝑍𝑇{𝑢(𝑘)} − 𝑍𝑇{𝑢(𝑘 − 5)} = − 𝑧 −5 = = 5
𝑧−1 𝑧−1 𝑧−1 𝑧 − 𝑧4
ROC: |z| > 1
• Time reversal
Let 𝑍{𝑥(𝑘)} = 𝑋(𝑧) and its ROC: 𝑟1 < |𝑧| < 𝑟2
Then:
1 1
𝑍{𝑥(−𝑘)} = 𝑋(𝑧 −1 ) , ROC: < |𝑧| < 𝑟 (1.6)
𝑟2 1
Proof:
∞
𝑍{𝑥(−𝑘)} = ∑ 𝑥(−𝑘)𝑧 −𝑘
𝑘=−∞
∞
Let 𝑚 = −𝑘, then: 𝑍{𝑥(𝑚)} = ∑∞ 𝑚 −1 −𝑚
𝑚=−∞ 𝑥(𝑚)𝑧 = ∑𝑘=−∞ 𝑥(𝑚)(𝑧 ) = 𝑋(𝑧 −1 )
1 1
ROC: 𝑟1 < |𝑧 −1 | < 𝑟2 or equivalently: 𝑟 < |𝑧| < 𝑟
2 1
Example
3
𝒙(𝒌) = 𝒖(−𝒌)
1 1
𝑍{𝑢(𝑘)} = 1−𝑧 −1 ⇒ 𝑍{𝑢(−𝑘)} = 1−𝑧 ROC: |𝑧| < 1
• Differentiation in the z-transform
𝑑𝑋(𝑧)
𝑇𝑍{𝑘𝑥(𝑘)} = −𝑧 , 𝑅𝑂𝐶 = 𝑅𝑥 (1.7)
𝑑𝑧
Proof:
𝑑𝑋(𝑧)
𝑍{𝑥(𝑘)} = 𝑋(𝑧) = ∑∞
𝑘=−∞ 𝑥(𝑘)𝑧
−𝑘
→ = ∑∞
𝑘=−∞ −𝑘𝑥(𝑘)𝑧
−𝑘−1
=
𝑑𝑧
−𝑧 −1 ∑∞
𝑘=−∞ 𝑘𝑥(𝑘)𝑧
−𝑘
= − 𝑧 −1 𝑍{𝑘𝑥(𝑘)}
𝑑𝑋(𝑧)
Thus: 𝑇𝑍{𝑘𝑥(𝑘)} = −𝑧 𝑑𝑧
Example
𝒙(𝒌) = 𝒌𝒂𝒌 𝒖(𝒌)
𝑧
Let 𝑦(𝑘) = 𝑎𝑘 𝑢(𝑘), then 𝑌(𝑧) = 𝑧−𝑎 |𝑧| > 𝑎
𝑑𝑌(𝑧) 𝑧−𝑎−𝑧 𝑎𝑧
𝑇𝑍{𝑘𝑦(𝑘)} = −𝑧 = −𝑧 2
=
𝑑𝑧 (𝑧 − 𝑎) (𝑧 − 𝑎)2
• Multiplication by an Exponential Sequence
𝑧
𝑇𝑍{𝑎𝑘 𝑥(𝑘)} = 𝑋( ) , 𝑅𝑂𝐶 = |𝑎|𝑅𝑥
𝑎
• Initial Value theorem (for causal signals)
𝑥(0) = lim 𝑋(𝑧)
𝑧→∞
• Final value theorem
lim 𝑥(𝑘) = lim(𝑧 − 1)𝑋(𝑧)
𝑘→∞ 𝑧→1
• Convolution of two sequences
𝑍{𝑥1 (𝑘)} = 𝑋1 (𝑧), 𝑍{𝑥2 (𝑘)} = 𝑋2 (𝑧)
𝑍{𝑥1 (𝑘) ∗ 𝑥2 (𝑘} = 𝑋1 (𝑧)𝑋2(𝑧) (1.8)
The ROC is at least the intersection of 𝑅𝑥1 and 𝑅𝑥2
𝒙(𝒌) = (𝟐)𝒌 𝒖(𝒌) ∗ (−𝟑)𝒌 𝒖(𝒌)
𝑥1 (𝑘) = 2𝑘 𝑢(𝑘) , 𝑥2 (𝑘) = (−2)𝑘 𝑢(𝑘)
𝑧
𝑋1 (𝑧) = 𝑧−2 𝑅𝑥1 = |𝑧| > 2
𝑧
𝑋2 (𝑧) = , 𝑅 = |𝑧| > 3
𝑧 + 3 𝑥2
Thus:
4
𝑧 𝑧 𝑧2 2 𝑧 3 𝑧
𝑋(𝑧) = ∙ = = +
𝑧 − 1/2 𝑧 + 1/3 (𝑧 − 1/2)(𝑧 + 1/3) 15 𝑧 − 1/2 10 𝑧 + 1/3
6 1 𝑘 6 1 𝑘
⟹ 𝑥(𝑘) = ( ) 𝑢(𝑘) + (− ) 𝑢(𝑘)
10 2 15 3
Frome these properties we define some common Z transforms as presents the following
image:
1.3. The inverse of z-transform
1.3.1. Partial-fraction expansion
This approach is useful if 𝑋(𝑧) is rational function:
𝐵(𝑧) 𝑏0 + 𝑏1 𝑧 −1 … + 𝑏𝑚 𝑧 −𝑚
𝑋(𝑧) = =
𝐴(𝑧) 1 + 𝑎1 𝑧 −1 … + 𝑎𝑛 𝑧 −𝑛
Case 1: 𝑋(𝑧) is a proper rational function (𝑚 < 𝑛 and 𝑎𝑛 ≠ 0)
5
To simplify the decomposition, it is preferable to eliminate negative powers of 𝑧 by
multiplying the numerator and the denominator by 𝑧 𝑛 which leads to:
𝑏0 𝑧 𝑛 + 𝑏1 𝑧 𝑛−1 … + 𝑏𝑚 𝑧 𝑛−𝑚
𝑋(𝑧) =
𝑧 𝑛 + 𝑎1 𝑧 𝑛−1 … + 𝑎𝑛
By dividing 𝑋(𝑧) by z , we get a new function which is proper also. The partial-fraction
𝑋(𝑧)
expansion is equivalent to express the function as a sum of simple fractions.
𝑧
𝑋(𝑧)
According to the roots of the denominator of the function , two cases could be considered:
𝑧
- All roots are simple:
𝑋(𝑧) 𝐴1 𝐴2 𝐴𝑛
= + +⋯ (1.9)
𝑧 𝑧−𝑝1 𝑧−𝑝2 𝑧−𝑝𝑛
where 𝑝1 , 𝑝1 , ⋯ 𝑝𝑛 are 𝑛-simple poles of . The coefficients 𝐴1 , 𝐴1 , ⋯ 𝐴𝑛 could be evaluated as:
𝑋(𝑧)
𝐴𝑘 = (𝑧 − 𝑝𝑘 ) | 𝑘 = 1,2, ⋯ 𝑛 (1.10)
𝑧 𝑧=𝑝𝑘
Example
𝑧 1
|𝑧|
𝑋(𝑧) = <
2𝑧 2 − 3𝑧 + 1 2
𝑋(𝑧) 1 1 𝑎1 𝑎2
= 2 = = +
𝑧 2𝑧 − 3𝑧 + 1 2(𝑧 − 1) (𝑧 − 1) 𝑧 − 1 𝑧 − 1
2 2
1 1
𝑎1 = | = 1, 𝑎2 = | = −1
1 2(𝑧 − 1) 𝑧=1/2
2 (𝑧 − 2)
𝑧=1
𝑧 𝑧
𝑋(𝑧) = −
𝑧 − 1 𝑧 − 1/2
1 1 𝑘
Puisque |𝑧| < 2 donc |𝑧| < 1 ⟹ 𝑥(𝑘) = −𝑢(−𝑘 − 1) + (2) 𝑢(−𝑘 − 1)
- Multiple- order roots
𝑋(𝑧)
If has multiple-order root 𝑝𝑖 with multiplicity 𝑟, then:
𝑧
𝑋(𝑧) 1𝐴 𝐴 𝐴
= 𝑧−𝑝 + (𝑧−𝑝2 )2 + ⋯ (𝑧−𝑝𝑟 )𝑟 (1.11)
𝑧 𝑖 𝑖 𝑖
where:
1 𝑑𝑘 𝑋(𝑧)
𝐴𝑟−𝑘 = 𝑘! 𝑑𝑧 𝑘 [(𝑧 − 𝑝𝑖 )𝑟 ]| (1.12)
𝑧 𝑧=𝑝𝑖
Example
1 𝑧3
𝑋(𝑧) = = , |𝑧| > 1
(1 + 𝑧 −1 )(1 − 𝑧 −1 )2 (𝑧 + 1)(𝑧 − 1)2
6
First:
𝑋(𝑧) 𝑧2
==
𝑧 (𝑧 + 1)(𝑧 − 1)2
We have: 𝑝1 = −1 is a simple pole, 𝑝2 = 1 is a double pole. Then:
𝑋(𝑧) 𝐴1 𝐴2 𝐴3
= + +
𝑧 𝑧 + 1 𝑧 − 1 (𝑧 − 1)2
𝑋(𝑧) 1
𝐴1 = (𝑧 + 1) | =
𝑧 𝑧=−1 4
𝑋(𝑧) 1
𝐴3 = (𝑧 − 1)2 | =
𝑧 𝑧=1 2
1 𝑑 𝑋(𝑧) 3
𝐴2 = [(𝑧 − 𝑝𝑖 )2 ]| =
2! 𝑑𝑧 𝑧 𝑧=1 4
Thus:
1 3 1
𝑧 𝑧 𝑧
𝑋(𝑧) = 4 + 4 + 2 2
𝑧 + 1 𝑧 − 1 (𝑧 − 1)
1 3 1
Since |𝑧| > 1 then: 𝑥(𝑘) = 4 (−1)𝑘 𝑢(𝑘) + 4 𝑢(𝑘) + 2 𝑘𝑢(𝑘)
Case 2: 𝑋(𝑧) is an improper rational function (𝑚 ≥ 𝑛)
An improper function can always be expressed as the sum of a polynomial and a proper
function.
𝐵(𝑧) 𝐵1 (𝑧)
𝑋(𝑧) = 𝐴(𝑧) = 𝑐0 + 𝑐1 𝑧 −1 + ⋯ + 𝑐𝑚−𝑛 𝑧 𝑛−𝑚 (1.13)
𝐴(𝑧)
Example
11 1
1 + 3𝑧 −1 + 6 𝑧 −2 + 3 𝑧 −3
𝑋(𝑧) =
5 1
1 + 6 𝑧 −1 + 6 𝑧 −2
1 −1 1
𝑧
𝑋(𝑧) = 1 + 2𝑧 −1 + 6 = 1 + 2𝑧 −1 + 6𝑧
5 1 5 1
1 + 6 𝑧 −1 + 6 𝑧 −2 𝑧2 + 6 𝑧 + 6
1
𝑋(𝑧) = 1 + 2𝑧 −1 + 6𝑧 = 1 + 2𝑧 −1 +
𝑧
+
−𝑧
1 1 1 1
(𝑧 + 3) (𝑧 + 2) 𝑧+3 𝑧+2
1.3.2. Power series expansion
7
The z-transform defined by 𝑋(𝑧) = ∑∞
𝑘=−∞ 𝑥(𝑘)𝑧
−𝑘
is a power series where the sequence
values 𝑥(𝑘) are the coefficient of 𝑧 −𝑘 :
𝑋(𝑧) = ⋯ + 𝑥(−1)𝑧 + 𝑥(0) + 𝑥(1)𝑧 −1 + ⋯ (1.14)
Example
1 1
𝑋(𝑧) = =
1 − 1.5𝑧 −1 + 0.5𝑧 −2 1
(1 − 𝑧 −1 ) (1 − 𝑧 −1 )
2
• |𝑧| > 1 : the sequence 𝑥(𝑘) must be causal, then we perform a power series expansion in
negative power of 𝑧.
3 7 15 −3 3 7 15
𝑋(𝑧) = 1 + 2 𝑧 −1 + 4 𝑧 −2 + 𝑧 + ⋯ 𝑥(𝑘) = {1, 2 , 4 , , …}
8 8
• |𝑧| < 1/2 : the sequence 𝑥(𝑘) must be anticausal, and we perform a power series expansion
in positive power of 𝑧.
𝑋(𝑧) = 2𝑧 2 + 6𝑧 3 + 14𝑧 4 + ⋯ , 𝑥(𝑘) = {… , 14,6,2,0,0}
