Any one can solve this:
Simplify the boolean expression
Z=A+A'B + A'B'C+ A'B'C'D
What will be the final answer of this question.
Z = A + A'B + A'B'C + A'B'C'D <br>
Z = A + A'(B + B'C + B'C'D) (distributivity)
Z = A + A'(B + B'(C + C'D)) (distributivity)
Z = A + A'(B + B'(C(D+1) + C'D)) (null law)
Z = A + A'(B + B'(CD + C + C'D)) (distributivity)
Z = A + A'(B + B'(C + D(C + C'))) (distributivity)
Z = A + A'(B + B'(C + D)) (inverse law)
Z = A + A'(B + C + D) (same 4 steps applied above)
Z = A + B + C + D (same as above)
So, whole expression actually can be solved with this rule:
A + A'B = A + B (absorption law)
Related
I am solving a fourth order equation in matlab using the solve function.My script looks like this:
syms m M I L Bp Bc g x
m = 0.127
M = 1.206
I = 0.001
L = 0.178
Bc = 5.4
Bp = 0.002
g = 9.8
eqn = ((m + M)*(I + m*L^2) - m^2*L^2)*x^4 + ((m + M)*Bp + (I + m*L^2)*Bc)*x^3 + ((m + M)*m*g*L + Bc*Bp)*x^2 + m*g*L*Bc*x == 0
S = solve(eqn, x)
In the answer, I should get 4 roots, but instead I get such strange expressions:
S =
0
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 1)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 2)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 3)
The first root, which is 0, is displayed clearly. Is it possible to make the other three roots appear as numbers as well? I looked for something about this in the documentation for the solve function, but did not find it.
I have to restrict it with the folowings:
P(-1) = f(-1), P(0)=f(0), P(1)=f(1), P'(1)=f'(1)
Let the polynomial be
ax³ + bx² + cx + d
By the given equations,
- a + b - c + d = f(-1)
d = f(0)
a + b + c + d = f(1)
3a +2b + c = f'(1)
You should be able to solve.
Using DeMorgan's theorem show that:
a. (A + B)'(A' +B)' = 0
b. A + A'B + A'B' = 1
Boolean expression F = x'y + xyz':
Derive an algebraic expression for the complement F'
Show that F·F' = 0
Show that F + F' = 1
Please Help me
Assuming you know how DeMorgan's law works and you understand the basics of AND, OR, NOT operations:
1.a) (A + B)'(A' + B)' = A'B'(A')'B' = A'B'AB' = A'AB'B' = A'AB' = 0 B' = 0.
I used two facts here that hold for any boolean variable A:
AA' = 0 (when A = 0, A' = 1 and when A = 1, A' = 0 so (AA') has to be 0)
0A = 0 (0 AND anything has to be 0)
1.b) A + A'B + A'B' = A + A'(B + B') = A + A' = 1.
I used the following two facts that hold for any boolean variables A, B and C:
AB + AC = A(B + C) - just like you would do with numeric variables and multiplication and addition. Only here we work with boolean variables and AND (multiplication) and OR (addition) operations.
A + A' = 0 (when A = 0, A' = 0 and when A = 1, A' = 0 so (A + A') has to be 1)
2.a) Let's first derive the complement of F:
F' = (x'y + xyz')' = (x'y)'(xyz')' = (x + y')((xy)' + z) = (x + y')(x' + y' + z) = xx' + xy' + xz + x'y' + y'y' + y'z = 0 + xy' + xz + x'y' + y' + y'z = xy' + xz + y'(x + 1) + y'z = xy' + xz + y' + y'z = xy' + xz + y'(z + 1) = xy' + y' + xz = y'(x + 1) = xz + y'.
There is only one additional fact that I used here, that for any boolean variables A and B following holds:
A + AB = A(B + 1) = A - logically, variable A completely determines the output of such an expression and part AB cannot change the output of entire expression (you can check this one with truth tables if it's not clear, but boolean algebra should be enough to understand). And of course, for any boolean variable A + 1 = A.
2.b) FF' = (x'y + xyz')(xz + y') = x'yxz + x'yy' + xyz'xz + xyz'y'.
x'yxz = (xx')yz = 0xz = 0
xyy'= x0 = 0
xyz'xz = xxy(zz') = xy0 = 0
xyz'y' = xz'(yy') = xz'0 = 0
Therefore, FF' = 0.
2.c) F + F' = x'y + xyz' + xz + y'
This one is not so obvious. Let's start with two middle components and see what we can work out:
xyz' + xz = x(yz' + z) = x(yz' + z(y + y')) = x(yz' + yz + y'z) = x(y(z + z') + y'z) = x(y + y'z) = xy + xy'z.
I used the fact that we can write any boolean variable A in the following way:
A = A(B + B') = AB + AB' as (B + B') evaluates to 1 for any boolean variable B, so initial expression is not changed by AND-ing it together with such an expression.
Plugging this back in F + F' expression yields:
x'y + xy + xy'z + y' = y(x + x') + y'(xz + 1) = y + y' = 1.
I not understand why the first boolean expression on the question can be simplified into the last. Please help me.
XY + Z(X ⊕ Y)
= XY + Z(X¬Y + ¬XY) // Expand XOR operation
= XY + X¬YZ + ¬XYZ // Distribute AND over OR
= XYZ + XY¬Z + X¬YZ + ¬XYZ // Expand XY
= (XYZ + XY¬Z) + (XYZ + X¬YZ) + (XYZ + ¬XYZ) // Copy XYZ and add parens
= XY + XZ + YZ // Remove trivial X+¬X = 1's
Reassembly is the reverse of disassembly.
Hi I'v solved this halfway, please help me for the rest.
so far I've got..
x'yz + xy'z + xyz' + xyz
z(x'y + xy') + xy(z'+z)
z(x'y + xy') + xy
i don't understand how to solve the z(x'y + xy') part of this expression.. please somebody help..
x'y + xy' is XOR. So you can simplify to z(x+y) + xy because z(x'y + xy') + xy is z when x != y and xy when x == y.
The (x+y) after z is necessary to disallow influence of z when x == y == 0.