DeGracia_Miguel_19481379_Lab5

# DeGracia_Miguel_19481379_Lab5 - %matrix form as well...

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%% Miguel De Gracia. 19481379. Lab Assignment 5. Section 22 % %% Problem 1a % x=2; y=3; x x > y x < y x >= y x <= y x == y x ~= y %The problem returns a 1 if the statement is true, and a 0 if the statement %is false. % %% Problem 1b % y = [1 3; 5 2]; y x > y x < y x >= y x <= y x == y x ~= y all(y==x) any(y==x) find(y<=x) %In this case since y is now a matrix, it compares x to every element of y %and performs the same task as the previous problem. For all or any it does %so by comparing it according to each column.Find returns the indeces in %which the logical statement is true. % %% Problem 1c % x = [2 4; 5 1]; x x > y x < y x >= y x <= y x == y x ~= y all(y==x) any(y==x) find(y<=x) %Since both x and y are now matrices, it compares the elements and %evaluates the logical command at that element, returning an answer in

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Unformatted text preview: %matrix form as well. % %% Problem 1d % a=1; b=0; c=1; a a | b a & b a | c & b (a | c) & b a & (c | b) ~a & c ~(a | b | c) %Evaluates the logical operators when | means or, & and. Compares the %statement and evaluates if the values are equal. % %% Problem 1e % 5 > 2+4|~2+5==5 %Evaluates the furthermost expression 5==5 which is true so it becomes a %one, then 2+1 is 3, then now the expression looks like 6|~3, which will %return a 1, and at the end 5>1 is true so the answer is now 1. % %% Problem 2 % f1(-8) f1(-3) f1(4) f1(17) f type f1 t %% Problem 3 % type poly_adder t %% Problem 3a % poly_adder([1 2 3],[2 1 3]) p %% Problem 3b % poly_adder([1 2 3 4],[1 0 4]) p %% Problem 3c % poly_adder([1 5],[3 2 1 5]) p %% Problem 4 %...
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## This note was uploaded on 02/18/2010 for the course ENGINEERIN 7 taught by Professor Patzek during the Spring '08 term at Berkeley.

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DeGracia_Miguel_19481379_Lab5 - %matrix form as well...

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