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Frequency Design

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Frequency response of generalised systems. Stability margins ... can be displayed on an argand diagram in the form of a polar plot (Real Imaginary) ... – PowerPoint PPT presentation

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Title: Frequency Design


1
Frequency Design
  • CY2A7
  • (old course code CY2A6 CY2A2 for past
    exams/notes)
  • Introduction to frequency design.
  • Crash course on Matlab code.
  • Modelling basic systems
  • Frequency response of generalised systems.
  • Stability margins
  • Separating Magnitude and Phase responses
  • Approximations by hand
  • Compensation techniques
  • - Proportional, Phase lead, Phase lag/lead.
  • 9 PI, PD PID controllers.
  • 10 Modelling a complete system in Matlab,
    including controller design to compensate for
    undesired time response.

2
Why Do We Study Frequency Response?
Magnitude Time (sec)
Magnitude (dB) Frequency Response
(Radians/sec)
3
Why Do We Study Frequency Response?
Some inputs are step or ramp e.g., climb a
hill Other inputs have a frequency e.g. bumpy
ground The magnitude and phase of the input
will be different on different surfaces Its
important to predict the Frequency Response of
our system to different inputs so that we know
how it will behave
4
Types of Frequency Response
Input Function Sketch Use Impulse
?(t) Transient response
Modeling Step u(t) Transient
response Steady-state error Ramp
tu(t) Steady-state error Sinusoid sin
?t Transient response Modelling
Steady-state error
f(t) t
f(t) t
f(t) t
f(t) t
5
Domains
Time Frequency plots Logarithmic plots
f(t) t
j? s-plane ?
Im F-plane Re
Magnitude (units) Frequency (rads/s)
Phase (degrees) Frequency (rads/s)
6
Radian

7
Magnitude Phase
input e(t) 8 sin(2t) output
c(t) 2.528sin(2t-108.4)
gtgt t00.0016.24 gtgt i 8sin(2t) gtgt o
2.528sin(2t-108.4)
8
Forms for Frequency Response
Sinusoids M sin(?t?) can be represented in a
number of useful formats 1. Polar form Mi
??i Where Mi (A2 B2)1/2 ?i -
tan-1 (B / A) 2. Rectangular form A - j B 3.
From Eulers formula Mi ej ?i
9
Intro to Frequency Response Theory
Consider a linear time invariant system g.
r(t) g(t) c(t) Input sinusoid
r with frequency ? and amplitude A r(t) A
sin(?t) Output sinusoid c with frequency ?
and new amplitude K c(t) K sin(?t?)
Frequency stays the same Amplitude and
phase may change! M K/A Magnitude of
amplitude change ? Phase Difference
10
Intro to Frequency Response Theory
  • Magnitude Phase relationship between a
    sinusoidal input and output is known as the
    FREQUENCY RESPONSE of a system.
  • Testing such systems is normally done by keeping
    A constant and varying ? thus finding K (and
    hence M) ?.

11
Intro to Frequency Response Theory
Consider a linear time invariant system. r(t)
Asin(?t) r() G() c()
c(t) Ksin(?t?) K New Amplitude ?
Phase Difference r(t)
g(t) c(t) Convert
r(t) to R(s) Calculate system output C(s) convert
to c(t). C(s)
to c(t)
12
Input, Plant, Output
Input
So applying a sinusoidal input r(t) A sin
?t Input r(t) e(t) open loop only 8
sin(2t) on this road surface! where A8 and ? 2
rads/s Modelling the suspension system to
give the transfer function G(s) 2 s2
3s2 Need to move from the time domain to the
frequency domain - Laplace operator s j?. We
need to know about G(j?) (frequency response). We
need to calculate the magnitude phase at the
known frequency magnitude phase
G (j?) G (j?)ejt G (j?) ??
The output can be written as c(t) AG(j?)
sin(?t ?) The output can now be determined
as c(t) 2.528sin(2t-108.4) for this
input and system
Modelling
Time response
13
Magnitude Phase
We need to know about G(j?) (frequency
response). G (j?) G (j?)ejt G
(j?) ?? So applying a sinusoidal input c(t)
A sin ?t the output can be written as c(t)
AG(j?) sin(?t ?) Given the transfer
function G(s) 2 and the s2 3s2 input
e(t) 8 sin(2t) the frequency response can
be obtained using the above equations where A8
and ? 2 G(j?) s 2j 2
(2j)2 3(2j) 2 G(j?) s 2j 2 1
- 1 - 3 j -2 6j -1
3j 10 10 Using trig gives
G(j?) 0.316?-108.4 The output can now
be written asc(t) 2.528sin(2t-108.4)
Im 0.1 ?
R 0.3
14
General Analysis
We can also consider the transfer function for
all frequencies, instead of just specific
one G(s) can also be written as G(s)
1 1 (1 T1s) ( 1
T2s) (1 s) (1 0.5s) So the example has
two time constants T1 ls and T20.5s. Hence
after 4 seconds (4 longest time constant for the
1st order component) the system will have reached
steady state. Given the complex transfer
function G(j?) we can obtain the steady state
response for any sinusoidal input given that the
system is stable. G(j?) 0 lt ? lt ?
Frequency response function, can be displayed
on an argand diagram in the form of a polar plot
(RealImaginary)

15
First-order example
Given the system
e(t) c(t) G(s) l ? RC 1
RCs Choose RC such that ? l second
giving G(s) 1 1
s Converting this equation into a polar form
gives G(j?) 1 (1 ?2)-l/2 ? -tan-1
(?) 1 j? Where M (A2
B2)1/2 (12 ?2)-l/2 ? - tan-1 (B / A)
-tan-1 (?/1)
R
C
16
Frequency Implicit
Evaluating at convenient frequencies
gives a set of polar coordinates for the
system ? Magn Phase
0 1 0 0.5 0.894 -26.6 1 0.707
-45 2 0.477 -63.4 3 0.316 -71.6 5
0.196 -78.6 10 0.1 -84.3
Im
R
? 0 ? 10
? 0.5 ? 1
17
Frequency Explicit
Evaluating at convenient frequencies
gives a set of polar coordinates for the
system ? Magn Phase
0 1 0 0.5 0.894 -26.6 1 0.707
-45 2 0.477 -63.4 3 0.316 -71.6 5
0.196 -78.6 10 0.1 -84.3
The magnitude and phase an also be plotted
separately Magn 1.0 .707
? Phase ? -45 -90

18
Frequency Design Resources
References Engineering Systems M.
Hargreaves modelling control
Prentice-Hall ISBN 0582234190 An
Introduction To Control Systems K. Warwick
World Scientific
Control Systems Engineering N. Nise J.
Wiley Sons ISBN 0471366013 Modern
Control Systems R. Dorf R. Bishop Addison-W
esley Modern Control Engineering K.
Ogata Prentice-Hall ISBN 0132613891
Linear Control Systems Analysis J. DAzzo C.
Houpis and Design McGraw-Hill Also be of
interest http//140.78.137.200/ifac50/home.html h
ttp//www.engin.umich.edu/group/ctm/freq/nyq.html
http//uk.youtube.com/watch?vaZNnwQ8HJHU
Oscillations http//uk.youtube.com/watch?vHpovwbP
GEoo Rubens tube
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