Lecture 2: Discrete-Time Systems and
z-Transform
• Discrete-time signals v.s. continuous-time signals
• Discrete-time systems v.s. continuous-time systems
• z-Transform
• Inverse z-Transform
• z-Transform for solving linear difference equations
Continuous-Time Signals
• Signals that change continuously in time:
𝑒 𝑡 ,
−∞ < 𝑡 < ∞, or 𝑡 ≥ 0
𝑒(𝑡)
𝑡
– Defined at all times t (no gap)
– Take arbitrary values
– Example: temperature, position, velocity,…
Discrete-Time Signals
• Signals (sequences, series) defined only at
discrete times:
𝑒 𝑘 ,
𝑘 = ⋯ , −1,0,1, ⋯ , or 𝑘 = 0,1,2, ⋯
𝑒(𝑘)
…
0 1
2
3
• Defined only at integer times k
• Take arbitrary values
4
𝑘
Origin of Discrete-Time Signals
• May come naturally
– Population of a species in different generations
– Annual growth percentage of GDP
– Results of a numerical algorithm in different rounds of iteration
• From sampling continuous-time signals at regular time intervals
𝑒 𝑡
𝑒(𝑘𝑇)
0 T 2T 3T 4T
𝑒 𝑡 ,
−∞ < 𝑡 < ∞
⇒
𝑒 𝑘𝑇 ,
Or simply 𝑒 𝑘 ,
𝑘 = ⋯ , −1,0,1, ⋯
𝑘 = ⋯ , −1,0,1, ⋯
(if 𝑇 is known in the context)
Signals in Digital Control Systems
continuous-time
signal
𝑇
𝑟(𝑡)
+_
𝑒(𝑡)
discrete-time system
A/D
discrete-time
signal
Computer
continuous-time system
D/A
digital signal
Data
Hold
Plant
𝑦(𝑡)
discrete-time continuous-time
signal
signal
Sensor
• The process of representing an arbitrary continuous value 𝑦 𝑘𝑇 with binary bits
of finite word length is called quantization (A/D)
• Quantized signal is discrete in both time and value, called a digital signal,
• Quantization error can be small with enough bits
Continuous-Time LTI Systems
• System with continuous-time input and output signals
𝑒(𝑡)
𝑦(𝑡)
• Continuous-time linear time-invariant (LTI) system:
• Linear differential equation
Transfer Functions of C-T LTI Systems
• Taking Laplace transform (assuming zero initial conditions):
transfer function:
𝛽𝑚 𝑠 𝑚 + ⋯ + 𝛽1 𝑠 + 𝛽0
𝐻 𝑠 =
𝛼𝑛 𝑠 𝑛 + ⋯ + 𝛼1 𝑠 + 𝛼0
𝐸(𝑠)
𝐻(𝑠)
𝑌(𝑠)
• For LTI system, H(s) is rational (fraction of polynomial functions)
• Impulse response is ℎ 𝑡 = ℒ −1 [𝐻 𝑠 ]
Discrete-Time Systems
• Systems with discrete-time input and output signals
𝑒(𝑘)
𝑦(𝑘)
• Discrete-time linear time-invariant (LTI) system:
𝑦 𝑘 + 𝑎𝑛−1 𝑦 𝑘 − 1 + ⋯ + 𝑎0 𝑦 𝑘 − 𝑛 =
𝑏𝑚 𝑒 𝑘 + 𝑏𝑚−1 𝑒 𝑘 − 1 + ⋯ + 𝑏0 𝑒 𝑘 − 𝑚
• Linear difference equation
• Current output depends on a finite history of input/output
Example of Discrete-Time LTI Systems
• Population dynamics (Fibonacci series)
• Opinion dynamics (on social networks)
D-T Approximation of C-T LTI Systems
𝑒(𝑡)
𝑠+1
𝑠+2
𝑦(𝑡)
Sample every T seconds
𝑒(𝑘𝑇)
?
𝑦(𝑘𝑇)
z-Transform
Given a (single-sided) discrete-time signal
𝑓 𝑘 , 𝑘 = 0,1,2, ⋯
its (single-sided) z-transform is defined by
𝐹 𝑧 =𝒵 𝑓 𝑘
= 𝑓 0 + 𝑓 1 𝑧 −1 + 𝑓 2 𝑧 −2 + ⋯
• 𝐹 𝑧 is only defined on a region of the 𝑧-plane (Domain of Convergence)
• Double-sided 𝑧-transform for double-sided signals
D-T Unit Impulse and Step Functions
Unit impulse function:
𝛿 𝑘 =
•
1,
0,
𝑖𝑓 𝑘 = 0
𝑖𝑓 𝑘 ≠ 0
𝛿(𝑘)
⋯ −1 0 1 2 3 4
𝑘
Delayed unit impulse function 𝛿(𝑘 − 𝑙)
Unit step function
1 𝑘 = 1,
1 𝑘
𝑘 = 0,1,2, …
0
1 2
3
4
𝑘
Other Common 𝑧-transform Pairs
𝑓(𝑘)
𝐹(𝑧)
𝑓(𝑘)
𝐹(𝑧)
• 𝑓(𝑘) are single-sided signals, e.g., 𝑓 𝑘 = 𝑎𝑘 ⋅ 1(𝑘), to be strict
Properties of z-Transform
Linearity: (® and ¯ are scalars)
𝒵 𝛼𝑓1 𝑘 + 𝛽𝑓2 𝑘
Convolution:
= 𝛼𝐹1 𝑧 + 𝛽𝐹2 (𝑧)
= 𝐹1 𝑧 𝐹2 𝑧 , where
𝒵 𝑓1 𝑘 ∗ 𝑓2 𝑘
𝑓1 𝑘 ∗ 𝑓2 𝑘 =
𝑓1 𝑙 𝑓2 𝑘 − 𝑙
𝑙=0,…,𝑘
Another property: 𝒵 𝑘𝑓 𝑘
= −𝑧𝑑𝐹(𝑧)
𝑑𝑧
A list of the properties of z-transform can be found in Table 2-2, pp. 37.
Translation in Time
• Shift right (time delay): (n is a positive integer)
𝒵 𝑓 𝑘 − 𝑛 ⋅ 1(𝑘 − 𝑛) = 𝑧 −𝑛 𝐹 𝑧
• Shift left:
𝒵 𝑓 𝑘 + 𝑛 ⋅ 1(𝑘) = 𝑧 𝑛 𝐹 𝑧 −
𝑛−1
𝑓 𝑘 𝑧 −𝑘
𝑘=0
• Example: 𝑓 𝑘 = 𝑎𝑘 ⋅ 1(𝑘), 𝑛 = 1
Initial and Final Value Theorems
Initial Value Theorem: 𝑓 0 = lim 𝐹(𝑧)
𝑧→∞
Final Value Theorem: 𝑓 ∞ = lim 𝑧 − 1 𝐹(𝑧)
𝑧→1
• FVT only holds if 𝑓 ∞ exists
Example: 𝑓 𝑘 = 𝑎𝑘 ⋅ 1(𝑘)
Inverse z-Transform
Inverse z-transform of 𝐹 𝑧 is the signal 𝑓 𝑘 , 𝑘 = 0,1, …
whose 𝑧-transform is exactly 𝐹 𝑧
Can be found in different ways:
• Power series method (long division)
• Inversion-formula method
• Partial fraction expansion
Power Series (Long Division) Method
For rational function 𝐹 𝑧 , e.g., 𝐹 𝑧 =
1
𝑧 2 −3𝑧+2
Inversion Formula Method
The inverse 𝑧-transform of 𝐹 𝑧 is given by
𝑓 𝑘 =
𝑝 Residue
of 𝐹 𝑧 𝑧 𝑘−1 at poles 𝑝 of 𝐹 𝑧 𝑧 𝑘−1
 residue at a simple pole 𝑝:
𝑅𝑒𝑠𝑖𝑑𝑢𝑒
𝑧=𝑝
= 𝑧 − 𝑝 𝐹 𝑧 𝑧 𝑘−1
𝑧=𝑝
 residue at a pole 𝑝 of multiplicity m:
𝑅𝑒𝑠𝑖𝑑𝑢𝑒
Example:
𝐹 𝑧 =
1
𝑧 2 − 3𝑧 + 2
1
𝐹 𝑧 =
𝑧(𝑧 − 1)
𝑧=𝑝
=
1
𝑑 𝑚−1
[
𝑚−1 ! 𝑑𝑧 𝑚−1
𝑧−𝑝
𝑚
𝐹 𝑧 𝑧 𝑘−1 ]
𝑧=𝑝
Partial Fraction Expansion Method
𝑧
• If 𝐹 𝑧 = 𝑧−𝑝
, then 𝑓 𝑘 = 𝒵 −1 𝐹 𝑧 = 𝑝𝑘 , 𝑘 = 0,1,2, …
• If 𝐹 𝑧 =
𝑝𝑧
𝑧−𝑝 2
, then 𝑓 𝑘 = 𝒵 −1 𝐹 𝑧 = 𝑘 ⋅ 𝑝𝑘 , 𝑘 = 0,1,2, …
• ….
• Idea: for general rational 𝐹 𝑧 =
𝑏(𝑧) 𝑏𝑚 𝑧𝑚 +⋯+𝑏1 𝑧+𝑏0
= 𝑎 𝑧𝑛 +⋯+𝑎 𝑧+𝑎
𝑎(𝑧)
𝑛
1
0
try to find the partial fraction expansion of 𝐹(𝑧)
𝑧
Simple Case
If 𝐹 𝑧 =
𝑏𝑚 𝑧 𝑚 +⋯+𝑏1 𝑧+𝑏0
𝑧−𝑝1 ⋯(𝑧−𝑝𝑛 )
have poles that are all distinct and nonzero:
𝐹(𝑧)
𝑐0
𝑐1
𝑐𝑛
=
+
+ ⋯+
𝑧
𝑧
𝑧 − 𝑝1
𝑧 − 𝑝𝑛
⇒ 𝑓 𝑘 = 𝑐0 𝛿 𝑘 + 𝑐1 𝑝1
where 𝑐0 = 𝐹 𝑧
Example: 𝐹 𝑧 =
𝑧=0
𝑘
+ ⋯ + 𝑐𝑛 𝑝𝑛 𝑘 ,
𝐹 𝑧
𝑐𝑖 = (𝑧 − 𝑝𝑖 )
𝑧
1
𝑧 2 − 3𝑧 + 2
𝑧=𝑝𝑖
𝑘 = 0,1, …
Matlab command: residue
Repeated Poles Case
𝑏𝑚 𝑧 𝑚 + ⋯ + 𝑏1 𝑧 + 𝑏0
If 𝐹 𝑧 /z has repeated poles, e.g., 𝐹 𝑧 =
𝑧 − 𝑝1 3 (𝑧 − 𝑝4 )
𝐹(𝑧)
𝑐0
𝑐1
𝑐2
𝑐3
𝑐4
=
+
+
+
+
𝑧
𝑧
𝑧 − 𝑝1
𝑧 − 𝑝1 2
𝑧 − 𝑝1 3 𝑧 − 𝑝4
where the constants are given by
𝑐0 = 𝐹 𝑧
𝑐1 =
𝑧=0
1 𝑑2
3 𝐹(𝑧)
𝑧−𝑝
1
2! 𝑑𝑧 2
𝑧
𝑑
𝑐2 = 𝑑𝑧
𝑐3 =
𝑐4 = (𝑧 − 𝑝4 )𝐹(𝑧)
𝑧
𝐹(𝑧)
𝑧−𝑝1 3 𝑧
𝐹(𝑧)
𝑧−𝑝1 3 𝑧
𝑧=𝑝1
𝑧=𝑝1
𝑧=𝑝1
𝑧=𝑝4
(𝑝𝑖 ≠ 0)
Example
𝐹 𝑧 =
𝑧
𝑧+1 𝑧−1
2
1
𝐹 𝑧 =
𝑧(𝑧 − 1)
Complex Conjugate Poles Case
If 𝐹 𝑧 has poles that are complex conjugate pairs, use the
following 𝑧-transform pairs:
Solution of Linear Difference Equations
Given a linear difference equation
𝑦 𝑘 + 𝑎𝑛−1 𝑦 𝑘 − 1 + ⋯ + 𝑎0 𝑦 𝑘 − 𝑛 =
𝑏𝑚 𝑒 𝑘 + 𝑏𝑚−1 𝑒 𝑘 − 1 + ⋯ + 𝑏0 𝑦 𝑘 − 𝑚
For given input 𝑒 𝑘 , compute output y 𝑘 , for 𝑘 = 0,1,2, …
– Simplest approach is by iterations (using computer)
– Need to know the initial conditions
Zero-State Response via z-Transform
• Assuming zero initial conditions:
𝑦 −1 = 𝑦 −2 = ⋯ = 𝑦 −𝑛 = 0,
𝑒 −1 = 𝑒 −2 = ⋯ = 𝑒 −𝑚 = 0
• Take the z-transform of the difference equation to obtain
𝑏𝑚 + 𝑏𝑚−1 𝑧 −1 + ⋯ + 𝑏0 𝑧 −𝑚
𝑌 𝑧 =
𝐸(𝑧)
1 + 𝑎𝑛−1 𝑧 −1 + ⋯ + 𝑎0 𝑧 −𝑛
• Find 𝑦 𝑘 (zero-state response) by computing the inverse 𝑧-transform of 𝑌 𝑧
Example (Example 2.9, pp. 39):
𝑦 𝑘 = 𝑒 𝑘 − 𝑒 𝑘 − 1 − 𝑦(𝑘 − 1)
with input 𝑒 𝑘 =
1,
0,
if 𝑘 is even
if 𝑘 is odd
Zero-Input Response via z-Transform
Output of the previous example: 𝑦 𝑘 = 𝑒 𝑘 − 𝑒 𝑘 − 1 − 𝑦(𝑘 − 1)
under zero input 𝑒 𝑘 = 0, 𝑘 = 0,1,2, … and non−zero initial condition
𝑦 −1 = 1,
𝑒 −1 = −1
Idea: instead of dealing with double-sided signals y(k) and e(k), define
𝑦 𝑘 =𝑦 𝑘−1 ,
𝑒 𝑘 = 𝑒(𝑘 − 1)
Full Response
• Superposition Law: For an LTI system, full response is the sum
of zero-state and zero-input responses
• Example: Find the output of the system 𝑦 𝑘 = 𝑒 𝑘 − 𝑒 𝑘 − 1 − 𝑦(𝑘 − 1)
1,
if 𝑘 is even
under nonzero input 𝑒 𝑘 = 0,
if 𝑘 is odd
and nonzero initial condition: 𝑦 −1 = 1, 𝑒 −1 = −1