Hello.
We all know that DeMorgan's Law is as follows:
(A∪B)' = A'∩B'
and
(A∩B)' = A'∪B'
where ' refers to the complement of a set and A and B are both sets.
We also know that this can be extended to more than two terms.
My question is whether or not the following is true:
(A∩B∪C)' = A'∪B'∩C'...
Hello. I am an EE student with two semester left. I can optionally take a course on Data Structures from the CS department this next semester but am unsure if the material is old school or is still relevant today.
The course covers the following topics with programming assignments in C...
Hello all,
Stanford is offering FREE online Machine Learning and Artificial Intelligence courses running from October 10 to December 18.
Would anyone like to register and start an online PF study group with me?
Here are the links:
http://www.ml-class.org
http://www.ai-class.com
Homework Statement
Using the definition of the z-transform, show that if X(z) is the z-transform of x(n) = x_{R}(n) +jx_{I}(n), then:
Z\{x^{*}(n)\}=X^{*}(z^{*})Homework Equations
z-tranform definition:
Z\{x(n)\}=X(z)=\sum x(n)z^{-n}
The Attempt at a Solution
x(n) = x_{R}(n) + jx_{I}(n)...
So taking the z-transform and evaluating at z = e^(j*Omega) is the definition of the DTFT, right? Isn't the DTFT more prone to instability than the z-transform?
Hmm.. I will think about this more. Maybe someone else is better suited to answer this question as my DSP class begins covering the...
Homework Statement
Prove that the system is either T.I. or is not T.I.
Homework Equations
y(n) = x(n)*h(n)
x(n) is the input signal
y(n) is the output signal
h(n) is the system
The Attempt at a Solution
Inputing x(n-n0) into the system I get out:
as the output x(n-n0)*h(n)...
So in a discrete domain the functions are only defined at integer values of the independent variables.
For the continuous domain the functions are defined at real number values (or complex values in the case of the z-domain).
For an example of a transformation that maps from a discrete domain...
Okay, I'll try that.
But is there a way to simplify:
A'B' + CA + CB
to A'B' + C?
Or can you just get stuck when simplifying Boolean expressions?
Because when I work out the K-map on:
A'B' + CA + CB
I end up with:
A'B' + C.
Okay, so I am using this concept in the following problem but am getting stuck.
I am trying to simplify the following expression (A' is the complement of A) :
Y= A'B'C' + A'B'C + A'BC + AB'C + ABC
Y= A'B'(C' + C) + AB'C + A'BC + ABC
Y= A'B' + A'BC + AB'C + ABC
Y = A'B' + AB'C + BC(A + A')...