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.
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.
a(x) here is a type of equation used in signal
Let
y(x) = x^3 + x^2 + x + 1
a(x) is inputed by the user so for example if user inputs
a(x) = D2 + D + 6
y(x)*a(x) = D2(x^3 + x^2 + x + 1) + D(x^3 + x^2 + x + 1) + 6(x^3 + x^2 + x + 1)
here D2(x^3 + x^2 + x + 1) = 6x + 2 and D(x^3 + x^2 + x + 1) = 3x^2 + 2x + 1
So D is differentiation and D2 is double differential
So i want to know how i would do something like this on MATLAB
Use Matlab symbolic computation toolbox https://www.mathworks.com/products/symbolic.html
Calculus
Evaluate exact analytical solutions for definite or indefinite integral,
calculate derivatives of symbolic expressions or functions,
and approximate functions using series expansions.
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)
I have this equation in x and y:
(x + y)^(1/2) - 6*y*(x + y)^5 - (x + y)^6 + (x - 1)/(2*(x + y)^(1/2)) = 0.
Now I call the solver:
R_c = #(y)solve((x + y)^(1/2) - 6*y*(x + y)^5 - (x + y)^6 + (x - 1)/(2*(x + y)^(1/2)), x, 'Real', true);
which gives me the implicit solutions as a function of y. Now try
R_c(.3)
to find the explicit solution at y = 0.3. MATLAB's answer is:
ans =
0.42846617518653978966562924618638
0.15249587894102346284238111155954
0.12068186494007759990714181154349
However, the last entry in this array is NOT a solution. Test:
double(subs(subs((x + y)^(1/2) - 6*y*(x + y)^5 - (x + y)^6 + (x - 1)/(2*(x + y)^(1/2)), x, .12068186494007759990714181154349), y, .3))
yields
-0.0585.
This is not a rounding error. The other 2 solutions work perfectly and solve the equation correctly. I wonder where MATLAB the third value gets from. Can anyone help?
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.