Specifications
Table Of Contents
- Introduction
- LTI Models
- Operations on LTI Models
- Model Analysis Tools
- Arrays of LTI Models
- Customization
- Setting Toolbox Preferences
- Setting Tool Preferences
- Customizing Response Plot Properties
- Design Case Studies
- Reliable Computations
- GUI Reference
- SISO Design Tool Reference
- Menu Bar
- File
- Import
- Export
- Toolbox Preferences
- Print to Figure
- Close
- Edit
- Undo and Redo
- Root Locus and Bode Diagrams
- SISO Tool Preferences
- View
- Root Locus and Bode Diagrams
- System Data
- Closed Loop Poles
- Design History
- Tools
- Loop Responses
- Continuous/Discrete Conversions
- Draw a Simulink Diagram
- Compensator
- Format
- Edit
- Store
- Retrieve
- Clear
- Window
- Help
- Tool Bar
- Current Compensator
- Feedback Structure
- Root Locus Right-Click Menus
- Bode Diagram Right-Click Menus
- Status Panel
- Menu Bar
- LTI Viewer Reference
- Right-Click Menus for Response Plots
- Function Reference
- Functions by Category
- acker
- allmargin
- append
- augstate
- balreal
- bode
- bodemag
- c2d
- canon
- care
- chgunits
- connect
- covar
- ctrb
- ctrbf
- d2c
- d2d
- damp
- dare
- dcgain
- delay2z
- dlqr
- dlyap
- drss
- dsort
- dss
- dssdata
- esort
- estim
- evalfr
- feedback
- filt
- frd
- frdata
- freqresp
- gensig
- get
- gram
- hasdelay
- impulse
- initial
- interp
- inv
- isct, isdt
- isempty
- isproper
- issiso
- kalman
- kalmd
- lft
- lqgreg
- lqr
- lqrd
- lqry
- lsim
- ltimodels
- ltiprops
- ltiview
- lyap
- margin
- minreal
- modred
- ndims
- ngrid
- nichols
- norm
- nyquist
- obsv
- obsvf
- ord2
- pade
- parallel
- place
- pole
- pzmap
- reg
- reshape
- rlocus
- rss
- series
- set
- sgrid
- sigma
- sisotool
- size
- sminreal
- ss
- ss2ss
- ssbal
- ssdata
- stack
- step
- tf
- tfdata
- totaldelay
- zero
- zgrid
- zpk
- zpkdata
- Index

11 Reliable Computations
11-4
Conditioning and Numerical Stability
Two of the key concepts in numerical analysis are the conditioning of problems
and the stability of algorithms.
Conditioning
Consider the linear system given by
A =
0.7800 0.5630
0.9130 0.6590
b =
0.2170
0.2540
Thetruesolutionisx = [1, –1]' and youcan calculate it approximately using
MATLAB.
x = A\b
x =
1.0000
–1.0000
format long, x
x =
0.99999999991008
–0.99999999987542
Of course, in real problems you almost never have the luxury of knowing the
true solution. This problem is very ill-conditioned. To see this, add a small
perturbation to
A
E =
0.0010 0.0010
–0.0020 –0.0010
and solve the perturbed system
xe = (A+E)\b
xe =
–5.0000
7.3085
Ax b=
AE+
()
xb=