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Why does the function call "function [hdr, record] = edfReadOddur( 'tinnaprime.edf',varargin )" give an error?
This is the error message: Error: File: edfReadOddur.m Line: 1 Column: 40 Unexpected MATLAB expression.
8 years 前 | 1 个回答 | 0
1
个回答提问
edf to Matlab conversion?
I need to convert edf to Matlab. I have tried to use edfRead. The function call is function [hdr, record] = edfread('aesa1.ed...
8 years 前 | 1 个回答 | 0
1
个回答已回答
Distinguishing matrices in for loops
Combining the suggestions of David and KSSV I believe that I have found a solution to my problem. I have been trying to write a ...
Distinguishing matrices in for loops
Combining the suggestions of David and KSSV I believe that I have found a solution to my problem. I have been trying to write a ...
8 years 前 | 0
| 已接受
提问
Distinguishing matrices in for loops
I want to distinguish matrices in a for loop. The following does not work: T=[0,1,-0.1,0.8,0;0,0,0,1,0;-0.2,-1,0,-0.2,0;0...
8 years 前 | 1 个回答 | 0
1
个回答提问
Logistic Function Transform of vector values
I need to transform the elements of a vector by a logistic function into a vector with elements with values between 0 and 1. I ...
8 years 前 | 1 个回答 | 0
1
个回答已回答
Test based on inequality of two vectors does not succeed.
function stationarystates(S0,T) %This function is a simple model of a Markov chain % S0 is the initial state % T is the t...
Test based on inequality of two vectors does not succeed.
function stationarystates(S0,T) %This function is a simple model of a Markov chain % S0 is the initial state % T is the t...
9 years 前 | 0
提问
Test based on inequality of two vectors does not succeed.
stationarystates([1,0],[0.8,0.2;0.6,0.4]) function stationarystates( S0,T ) %This function is a simple model of a Mar...
9 years 前 | 2 个回答 | 0
2
个回答提问
Correct and incorrect answer from linsolve
The following gives a correct answer: A=[-2/5,1/5;2/5,-1/5;1,1] B=[0;0;1] linsolve(A,B) ans =0.3333 0.6667 The...
9 years 前 | 1 个回答 | 0
1
个回答已回答
I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.
I found the answer to my question: 3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1 3*x/5-5/5*x+y/5=0 2*x/5+4/5*y-5/5*y=0 x+y=1 -2/5...
I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.
I found the answer to my question: 3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1 3*x/5-5/5*x+y/5=0 2*x/5+4/5*y-5/5*y=0 x+y=1 -2/5...
9 years 前 | 0
| 已接受
提问
I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.
3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1
9 years 前 | 1 个回答 | 0
