CIS 540: Spring 2014 Midterm
March 13, 2014
1. Reactions of this process is shown below:
Await dependencies:
Output y has no dependency
Output z depends on input x
The component is non-deterministic, since in some states two transitions are enabled due to
CIS 540 Spring 2014: Homework 2 Solution
1. For the transition system T, the state variables are x, y of type nat and z of type bool. The initial state
is x = 0, y = 0, z = 0. The transitions of the system are: (a, b, 0) (a+1, b, 1) and (a, b, 1) (a, b+1,
CIS 540: Principles of Embedded Computation
Homework 1 Solution, Spring 2013
Problem 1
Solution: From condition (1) we know that the inc and dec should not be high simultaneously in a
given round. So the possible outputs are: only inc=1, only dec=1, or no
CS 1567
Many views of the world
Coordinate systems and sensor data
Mapping to a common reference
Robot coordinates (x,y,theta)
Defined by user in software
Origin set dynamically
X,Y and theta are also user defined
X
Origin set to start
point of path
theta
Basics of Computer Vision for
Robotics
Digital Representation of
Images
Analog Image
Digitized Image
Quantize the image into 2D array
Each element is a pixel
One pixel array is greyscale
i.e. 8 bits per pixed
Each pixel can be 0 -> 255
0 is black
255 is w
Nestl USA Diversity Leadership Symposium Virtual Info Sessions
Nestl invites you to join us for at least one of the following LIVE virtual information sessions! Hear
firsthand from Nestl recruiters and Nestl employees about:
Nestl and the Diversity Leader
Basics of PID Control
how to make the robot start and stop where
you want it to
PID is an acronym for
P - proportional
I - Integral
D- Differential
Some gures from PID without a PHD , Tim Wescott
http:/www.embedded.com/2000/0010/0010feat3.htm
Cl
CIS 540 Spring 2014: Homework 8 Solutions
May 2, 2014
Problem 7.15
Initially we have
0
0
RA =
0
0
0
0
0
0
0
0
0
0
0
0
.
0
0
The DBM that represents the invariant of state A is
0 0 0 0
5 0
inv
RA =
3 0 .
0
Then RA can be computed as
inv
RA = Canonica
CIS 540 Spring 2014: Homework 7
April 25, 2014
1. Representing the system in the form S = AS , we get
s1
s2
0
2
=
1
3
s1
s2
The eigen values of the matrix are 1 and 2. The corresponding eigen vectors are [1 1]T and
T
[1 2] .
P =
1
2
1
1
2
1
and P 1 =
1
1
CIS 540 Spring 2014: Homework 5 Solutions
April 2, 2014
1. The two formulas are equivalent. To prove this we must show implication in both the directions.
Let us rst assume that a trace
in the current state and
2
(1
satises
2 ).
Now
(1
2 )
means that
1