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Dr. J. Tani, Prof. Dr. E. Frazzoli
151-0591-00 Control Systems I (HS 2018) Exercise Set 6
Topic: Transfer functions 1: Definitions and properties
Discussion: 02. 11. 2017
[email protected], 5th November 2018 Learning objectives: The student can
- Understand the significance of the control block representation
- Can calculate the transfer function for a graphical representation with control blocks
- Understand what g(s) represents
- Can calculate Y(s) based on g(s) and U(s)
- Can derive a realisation in state space form given a transfer function
- Can calculate the poles and zeroes of a transfer function
- Can classify a transfer function as causal or non-causal
Exercise 1 (Control Block Calculation)
There are various ways of combining system blocks in control systems that you should be familiar with. The most important of which are standard feed forward and feedback models. After this exercise you should be familiar with the following three ways of combining systems Σ and be able to use this concept to derive the system equations of more complex combined systems.
Case A - Feed forward serial connection of two systems:
Figure 1: Feed forward serial.
The associated input-output behavior is described by
Σ = Σ 2 · Σ 1 (1)
Case B - Feed forward parallel connection of two systems:
Figure 2: Feed forward parallel.
The associated input-output behavior is described by
Σ = Σ 1 + Σ 2 (2)
Case C - Feedback parallel connection of two systems: Feedback can be considered the cornerstone of this course. It is usually illustrated in the way shown below. Please especially take not of the sign of the connection from output y to error term e. This sign, in opposite order can again be found in the system equation 3 below.
Figure 3: Feedback.
The associated input-output behavior is described by
Σ = Σ 1 · (1 ∓ Σ 1 · Σ 2 )−^1 (3)
Diagram Q1:
Diagram Q2:
Diagram Q3:
a) Derive the system equations Σ from reference input r to output y for diagram Q1.
b) Homework: Derive the system equations Σ from input u to output y for diagram Q2.
c) Homework: Derive the system equations Σ from input u to output y for diagram Q3.
Exercise 3 (Realisation of Transfer Functions)
In the last exercise, we derived the transfer function based on the state space description of a given system. In the following exercise, we want to follow the inverse operation. In theory, there exists an infinite number of possible state space descriptions that fulfil the equation
g ( s ) = C ( sI − A )−^1 B + D (9)
Of course, an infinite number of solutions contains several solutions that are impractical to work with. In general, we are interested in so-called minimal realisations. These are state space descriptions that yield the given transfer function and are controllable and observable. The most common minimal realisation is the controllable canonical form. Given a general transfer function in the form of
g ( s ) = bn − 1 · sn −^1 + bn − 2 · sn −^2 + ... + b 0 sn^ + an − 1 · sn −^1 + ... + a 0
the controllable canonical form is the state space description with the following matrices:
A =
− a 0 − a 1 − a 2 ... − an − 2 − an − 1
^ (11)
B =
C =
[ b 0 b 1 ... bn − 1
] (13)
D =
[ d
] (14)
a) We are given the transfer function
g ( s ) =
s^3 + 3 · s^2 + s + 5 s ( s + 3)( s + 5)
Bring the transfer function to the general form described above.
b) Derive the controllable canonical form for the given system.
c) As previously stated, the minimal realisation of a system is controllable and observable. Does this mean that we can only realise systems with a transfer function derived from a controllable and observable state space description? Hint: Was controllability or observability a necessary condition to derive a transfer function or a minimal realisation?
Exercise 4 (Homework: Poles and Zeroes of a Transfer function)
In this exercise, we will calculate the poles and zeroes of a transfer function. Mathematically, a zero is described as a frequency, for which the transfer function g ( s ) = 0. In other words, if a transfer function can be described as a fraction of two polynomials
g ( s ) = Z ( s ) P ( s )
the zeroes of this transfer function can be found where
z = [ s : Z ( s ) = 0]
Poles can be found where
p = [ s : P ( s ) = 0]
Calculate the poles and zeroes for the following transfer functions
a)
g ( s ) = s^2 + 5 s + 3 s ( s + 3)( s + 5)
b)
g ( s ) =
( s + 5)^3 ( s + 2)( s − 1)
c)
g ( s ) = ( s + 3) ( s^2 + 6 s + 9)( s − 2)
d)
g ( s ) =
s^2 + 2 σs + ( σ^2 + ω^2 )
