1. State Space approach State Variables of a Dynamical System
State Variable Equation
Why State space approach
Derive Transfer Function from State Space Equation
Time Response and State Transition Matrix

2. Modern Control Systems
State equation
Dynamic equation
Output equation
State variable
State space
r- input
p- output
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3. Modern Control Systems
Inner state variables C A D B + -
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Motivation of state space approach
Example 1 + - +
noise
Transfer function
BIBO stable
unstable
Modern Control Systems

5. Modern Control Systems
Example 2
BIBO stable, pole-zero cancellation -2 + -
Modern Control Systems

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then system stable
State-space description
Internal behavior description
Modern Control Systems

7. Modern Control Systems
Definition: The state of a system at time is the amount of information at that together with determines uniquely the behavior of the system for
Example M
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Example : Capacitor
electric energy
Input
Example : Inductor
Magnetic energy
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Example M2 M1 B3 B1 B2 K
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Example
Armature circuit
Field circuit
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Dynamical equation
Transfer function
Laplace transform
matrix
Transfer function
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Example
By Newton’s Law
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State Space Equation
Transfer Function
Example: Transfer function of the Mass-damper-spring system
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Example
MIMO system
Transfer function
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Remark : the choice of states is not unique. + -
exist a mapping
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Different state equation description
p is nonsingular
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18. Modern Control Systems
Definition : Two dynamical systems
with are said to be equivalent. The nonsingular matrix p is called an equivalence transformation.
&
Theorem: two equivalent dynamical system have the same transfer function.
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19. Modern Control Systems

20. Sampling of Continuous-Time Signals
Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, © Prentice Hall Inc.

21. 351M Digital Signal Processing
Signal Types
Analog signals: continuous in time and amplitude
Example: voltage, current, temperature,…
Digital signals: discrete both in time and amplitude
Example: attendance of this class, digitizes analog signals,…
Discrete-time signal: discrete in time, continuous in amplitude
Example:hourly change of temperature in Austin
Theory for digital signals would be too complicated
Requires inclusion of nonlinearities into theory
Theory is based on discrete-time continuous-amplitude signals
Most convenient to develop theory
Good enough approximation to practice with some care
In practice we mostly process digital signals on processors
Need to take into account finite precision effects
Our text book is about the theory hence its title
Discrete-Time Signal Processing
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing

22. Sample Data Control System
Sampling Theorem
Modern Control Systems

23. Sampling: Time Domain
Many signals originate as continuous-time signals, e.g. conventional music or voice
By sampling a continuous-time signal at isolated, equally-spaced points in time, we obtain a sequence of numbers
n  {…, -2, -1, 0, 1, 2,…}
Ts is the sampling period. Ts t Ts
s(t)
Sampled analog waveform
impulse train

24. Modulation by cos(s t) Modulation by cos(2 s t)
Replicates spectrum of continuous-time signal
At offsets that are integer multiples of sampling frequency
Fourier series of impulse train where ws = 2 p fs
Modulation by cos(s t)
Modulation by cos(2 s t) w
F(w)
2pfmax
-2pfmax w
G(w) ws
2ws
-2ws
-ws

25. Shannon Sampling Theorem
A continuous-time signal x(t) with frequencies no higher than fmax can be reconstructed from its samples x[n] = x(n Ts) if the samples are taken at a rate fs which is greater than 2 fmax.
Nyquist rate = 2 fmax
Nyquist frequency = fs/2.
What happens if fs = 2fmax?
Consider a sinusoid sin(2 p fmax t)
Use a sampling period of Ts = 1/fs = 1/2fmax.
Sketch: sinusoid with zeros at t = 0, 1/2fmax, 1/fmax, …

26. Shannon Sampling Theorem
Assumption
Continuous-time signal has no frequency content above fmax
Sampling time is exactly the same between any two samples
Sequence of numbers obtained by sampling is represented in exact precision
Conversion of sequence to continuous time is ideal
In Practice

27. Why 44.1 kHz for Audio CDs? Sound is audible in 20 Hz to 20 kHz range:
fmax = 20 kHz and the Nyquist rate 2 fmax = 40 kHz
What is the extra 10% of the bandwidth used?
Rolloff from passband to stopband in the magnitude response of the anti-aliasing filter
Okay, 44 kHz makes sense. Why 44.1 kHz?
At the time the choice was made, only recorders capable of storing such high rates were VCRs.
NTSC: 490 lines/frame, 3 samples/line, 30 frames/s = samples/s
PAL: 588 lines/frame, 3 samples/line, 25 frames/s = samples/s

28. Sampling
As sampling rate increases, sampled waveform looks more and more like the original
Many applications (e.g. communication systems) care more about frequency content in the waveform and not its shape
Zero crossings: frequency content of a sinusoid
Distance between two zero crossings: one half period.
With the sampling theorem satisfied, sampled sinusoid crosses zero at the right times even though its waveform shape may be difficult to recognize

29. Aliasing Analog sinusoid Sample at Ts = 1/fs
x(t) = A cos(2pf0t + f)
Sample at Ts = 1/fs
x[n] = x(Ts n) = A cos(2p f0 Ts n + f)
Keeping the sampling period same, sample
y(t) = A cos(2p(f0 + lfs)t + f)
where l is an integer
y[n] = y(Ts n) = A cos(2p(f0 + lfs)Tsn + f) = A cos(2pf0Tsn + 2p lfsTsn + f) = A cos(2pf0Tsn + 2p l n + f) = A cos(2pf0Tsn + f) = x[n]
Here, fsTs = 1
Since l is an integer, cos(x + 2pl) = cos(x)
y[n] indistinguishable from x[n]
Frequencies f0 + l fs for l  0 are aliases of frequency f0