Recent content by ConeOfIce

Haha, well I am glad I could help! If you get rid of the bus-hold and use a normal buffer it works fine, and that was what the original design had (and all future designs). There is no reason to use the bus-hold, it was an over sight by an engineer saying they work in the exact same way as...

Thanks, The thing is, is I cannot get rid of the RC section even though I'm sure that would fix it. The entire RC portion with the bus-hold are on large separate units that cannot not be tampered with. For all new products, the whole issue has already been fixed, but this is too give a...

To say we are heading into the weeds is an understatement. Part of the issue with this is I don't what the RC stuff is for. This stuff was designed years ago while part of a different company that we took over, so I have no one to ask about as to why these decisions were made. But I agree...

Hi all, At work we were working on a bus-hold issue that we fixed with a bit of an unconventional method. We had a circuit that looked somewhat like: Buf>--R(33)------R2(2.7k)---->BH .....| ......cap(220p) .....| ......GND Where Buf is a buffer, and BH is a bus-hold buffer. This is...

Ok, I do this. And I also took the dagge of both sides of the equation. So I got (A|a>)^(dagger) = (|a>a)^dagger which gets <a|A* = a*<a|. And using your statement form above I then do A*<a| = a|a> . How does this get me any closer to the answer?

Homework Statement Show that if an operator A has an eigenket |a> to eigenvalue a then the adjoint operator A† has an eigenbra <a*| to eigenvalue a*. How is <a*| related to |a>? Homework Equations A|a> = |a>a | >† = < | The Attempt at a Solution I actually have no clue where to...

Homework Statement Consider the following two vectors v1= (cosx , sinx)(transpose) , v2= (-sinx , cosx)(transpose). Compute the projectors P1, P2 onto the vectors v1 and v2. Homework Equations (a1) (a1*,b1*) (A) <---input (b1) ...(B) This is a matrix that projects on column...

Oh, I forgot the inverses don't always exist...it has been a bit since I have done and linear algebra work. Your solution makes perfect sense, thanks!

Actually I think I just figured it out. X^-1 Y =X^-1 I -> Y=X^-1 then sub that in for Y in the other equation X^-1 Z = I XX^-1 Z = X Z=X Thanks for your help!

Homework Statement Suppose one has n×n square matrices X, Y and Z such that XY = 1and Y Z = 1. Show that it follows that X = Z. The Attempt at a Solution Now I know if the equatoins had been XY and ZY I would do this: XY=ZY -> XY-ZY=0 -> Y(X-Z)=0 -> X-Z=0 -> X=Z I was wondering if...

The question is: Suppose one has n×n square matrices X, Y and Z such that XY = I and Y Z = I. Show that it follows that X = Z. Now if this were XY and ZY, I would just say that: XY=ZY -> XY-ZY=0 ->Y(X-Z)=0 -> X-Z=0 -> X=Z. I am wondering that since the Y is on different sides of the Z...

Sorry, the L was supposed to be the symbol for the Laplace function, does it still not make a difference?

Homework Statement Consider the vector space of square integrable complex-valued functions in one dimension V = L^2(R) = {f(x) : interal|f(x)|^2dx < ∞}. Show that <f|g> = integral f(x)*g(x)dx defines a scalar product on this vector space. The Attempt at a Solution I actually have no...