Modern Control Systems (MCS)
Lecture-23-24
Time Response Discrete Time Control Systems
Steady State Errors
Dr. Imtiaz Hussain
Assistant Professor
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
1
Lecture Outline
• Introduction
• Time Response of DT System
– Examples
• Final Value Theorem
– Examples
• Steady State Errors
2
Introduction
• The time response of a discrete-time linear system is the
solution of the difference equation governing the system.
• For the linear time-invariant (LTI) case, the response due
to the initial conditions and the response due to the
input can be obtained separately and then added to
obtain the overall response of the system.
• The response due to the input, or the forced response, is
the convolution summation of its input and its response
to a unit impulse.
3
Example-1
• Given the discrete-time system
𝑦 𝑘 + 1 − 0.5𝑦 𝑘 = 𝑢 𝑘
• Find the impulse response of the system.
Solution
• Taking z-transform
𝑧𝑌 𝑧 − 0.5𝑌 𝑧 = 𝑈 𝑧
𝑌(𝑧)
1
=
𝑈(𝑧) 𝑧 − 0.5
4
Example-1
• Since U(z)=1
1
𝑌(𝑧) =
𝑧 − 0.5
• Taking Inverse z-Transform
𝑦 𝑘 = (0.5)𝑘−1 ,
𝑘≥0
5
Example-2
• Given the discrete time system
𝑦 𝑘+1 −𝑦 𝑘 =𝑢 𝑘+1
• find the system transfer function and its response to a
sampled unit step.
Solution
• The transfer function corresponding to the difference
equation is
𝑧𝑌 𝑧 − 𝑌 𝑧 = 𝑧𝑈 𝑧
𝑌(𝑧)
𝑧
=
𝑈(𝑧) 𝑧 − 𝑧
6
Example-2
𝑧
𝑌(𝑧) =
𝑈(𝑧)
𝑧−1
• Since U z =
𝑧
𝑧−1
𝑧
𝑧
𝑌(𝑧) =
×
𝑧−1 𝑧−1
• Taking Inverse z-Transform (time advance Property)
𝑧
𝑌(𝑧) = 𝑧
(𝑧 − 1)2
𝑦 𝑘 = 𝑘 + 1,
𝑘≥0
7
Home Work
• Find the impulse, step and ramp response functions for
the systems governed by the following difference
equations.
1. 𝑦 𝑘 + 1 − 0.5𝑦 𝑘 = 𝑢 𝑘
2. 𝑦 𝑘 + 2 − .01𝑦 𝑘 + 1 + 0.8𝑦 𝑘 = 𝑢(𝑘)
8
Final Value Theorem
• The final value theorem allows us to calculate the limit of a
sequence as k tends to infinity, if one exists, from the ztransform of the sequence.
• If one is only interested in the final value of the sequence, this
constitutes a significant short cut.
• The main pitfall of the theorem is that there are important
cases where the limit does not exist.
• The two main case are
1. An unbounded sequence
2. An oscillatory sequence
9
Final Value Theorem
• If a sequence approaches a constant limit as k tends to
infinity, then the limit is given by
𝑓 ∞ = lim 𝑓 𝑘
𝑘→∞
𝑧−1
𝑓 ∞ = lim
𝐹 𝑧
𝑧→1 𝑧
𝑓 ∞ = lim(𝑧 − 1)𝐹 𝑧
𝑧→1
10
Example-3
• Verify the final value theorem using the z-transform of a
decaying exponential sequence and its limit as k tends to
infinity.
Solution
• The z-transform of an exponential sequence is
𝑧
𝐹 𝑧 =
𝑧 − 𝑒 −𝑎𝑇
• Applying final value theorem
𝑧−1
𝑧−1
𝑧
𝑓 ∞ = lim
𝐹 𝑧 = lim
𝑧→1 𝑧
𝑧→1 𝑧 𝑧 − 𝑒 −𝑎𝑇
𝑓 ∞ =0
11
Example-4
• Obtain the final value for the sequence whose ztransform is
Solution
𝑧 2 (𝑧 − 𝑎)
𝐹 𝑧 =
(𝑧 − 1)(𝑧 − 𝑏)(𝑧 − 𝑐)
• Applying final value theorem
𝑧−1
𝑧 2 (𝑧 − 𝑎)
𝑓 ∞ = lim
𝑧→1 𝑧 (𝑧 − 1)(𝑧 − 𝑏)(𝑧 − 𝑐)
1−𝑎
𝑓 ∞ =
(1 − 𝑏)(1 − 𝑐)
12
Home work
• Find the final value of following z-transform functions if it
exists.
1. 𝐹(𝑧) =
𝑧
𝑧 2 −1.2𝑧+0.2
2. 𝐹(𝑧) =
𝑧
𝑧 2 −0.3𝑧+2
13
Steady State Error
• Consider the unity feedback block diagram shown in
following figure.
• The error ratio can be calculated as
𝐸(𝑧)
1
=
𝑅(𝑧) 1 + 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
• Applying the final value theorem yields the steady-state
error.
𝑧−1
𝑒 ∞ = lim
𝐸 𝑧
𝑧→1 𝑧
14
Steady state Error
• As with analog systems, an error constant is associated with
each input (e.g., Position Error constant and Velocity Error
Constant)
• Type number can be defined for any system from which the
nature of the error constant can be inferred.
• The type number of the system is the number of unity poles in
the system z-transfer function.
15
Position Error Constant 𝐾𝑝
• Error of the system is given as
𝑅(𝑧)
𝐸(𝑧) =
1 + 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
• Where
𝑧
𝑅 𝑧 =
𝑧−1
• Therefore, the steady state error due to step input is given as
𝑒 ∞ =
𝑧−1
1
𝑧
lim
𝑧→1 𝑧 1+𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧) 𝑧−1
𝑒 ∞ = lim
𝑧→1 1
1
+ 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
16
Position Error Constant 𝐾𝑝
𝑒 ∞ = lim
𝑧→1 1
1
+ 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
• Position error constant 𝐾𝑝 is given as
𝐾𝑝 = lim 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
𝑧→1
• Steady state error can be calculated as
1
𝑒 ∞ =
1 + 𝐾𝑝
17
Velocity Error Constant 𝐾𝑣
• Error of the system is given as
𝑅(𝑧)
𝐸(𝑧) =
1 + 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
• Where
𝜏𝑧
𝑅 𝑧 =
𝑧−1
2
• Therefore, the steady state error due to step input is given as
𝑧−1
1
𝜏𝑧
𝑒 ∞ = lim
𝑧→1 𝑧 1 + 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧) 𝑧 − 1
2
𝜏
𝑒 ∞ = lim
𝑧→1 𝑧 − 1 [1 + 𝐺𝑍𝐴𝑆 𝑧 𝐺 𝑧 ]
18
Velocity Error Constant 𝐾𝑣
𝜏
𝑒 ∞ = lim
𝑧→1 𝑧 − 1 [1 + 𝐺𝑍𝐴𝑆 𝑧 𝐺 𝑧 ]
• 𝐾𝑣 is given as
1
𝐾𝑣 = lim 𝑧 − 1 𝐺𝑍𝐴𝑆 𝑧 𝐺 𝑧
𝜏 𝑧→1
• Steady state error due to sampled ramp input is given as
1
𝑒 ∞ =
𝐾𝑣
19
Example-5
• Find the steady-state position error for the digital position
control system with unity feedback and with the transfer
functions
𝐺𝑍𝐴𝑆
𝐾(𝑧 + 𝑎)
𝑧 =
(𝑧 − 1)(𝑧 − 𝑏)
𝐾𝑐 (𝑧 − 𝑏)
𝐶 𝑧 =
𝑧−𝑐
,0 < 𝑎, 𝑏, 𝑐 < 1
1. For a sampled unit step input.
2. For a sampled unit ramp input
Solution
•
𝐾𝑝 and 𝐾𝑣 are given as
𝐾𝑝 = lim 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
𝑧→1
1
𝐾𝑣 = lim 𝑧 − 1 𝐺𝑍𝐴𝑆 𝑧 𝐺 𝑧
𝜏 𝑧→1
20
Example-5
𝐾𝑝 = lim 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
𝑧→1
•
1
𝐾𝑣 = lim 𝑧 − 1 𝐺𝑍𝐴𝑆 𝑧 𝐺 𝑧
𝜏 𝑧→1
𝐾𝑝 can be further evaluated as
𝐾(𝑧 + 𝑎) 𝐾𝑐 (𝑧 − 𝑏)
𝐾𝑝 = lim
𝑧→1 (𝑧 − 1)(𝑧 − 𝑏)
𝑧−𝑐
𝐾(1 + 𝑎) 𝐾𝑐 (1 − 𝑏)
𝐾𝑝 =
=∞
(1 − 1)(1 − 𝑏) 1 − 𝑐
•
Corresponding steady state error is
1
𝑒 ∞ =
=0
1 + 𝐾𝑝
21
Example-5
𝐾𝑝 = lim 𝐺𝑍𝐴𝑆 𝑧 𝐺(𝑧)
𝑧→1
•
1
𝐾𝑣 = lim 𝑧 − 1 𝐺𝑍𝐴𝑆 𝑧 𝐺 𝑧
𝜏 𝑧→1
𝐾𝑣 is evaluated as
1
𝐾(𝑧 + 𝑎) 𝐾𝑐 (𝑧 − 𝑏)
𝐾𝑣 = lim 𝑧 − 1
𝜏 𝑧→1
(𝑧 − 1)(𝑧 − 𝑏) 𝑧 − 𝑐
1 𝐾(1 + 𝑎) 𝐾𝑐 (1 − 𝑏) 𝐾𝐾𝑐 (1 + 𝑎)
𝐾𝑣 =
=
𝜏 (1 − 𝑏)
1−𝑐
𝜏(1 − 𝑐)
•
Corresponding steady state error is
1
𝜏(1 − 𝑐)
𝑒 ∞ =
=
𝐾𝑣 𝐾𝐾𝑐 (1 + 𝑎)
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END OF LECTURES-23-24
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