Solving Audio Stereo Problem: Expression of (M^2) + (D^2)

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Homework Help Overview

The problem involves expressing the sum of squares of two components, M and D, derived from a column vector related to stereo audio. The original poster seeks to express (M^2) + (D^2) in terms of two variables, L and R, based on their definitions as projections onto specific vectors.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of the expression for (M^2) + (D^2) and whether it aligns with the definitions of M and D. There is also exploration of the geometric interpretation of these components and their relationship to vector projections.

Discussion Status

The discussion is ongoing, with participants questioning the assumptions about the relationships between M, D, and the vectors involved. Some guidance has been offered regarding the algebraic manipulation of the expressions, but multiple interpretations and clarifications are still being explored.

Contextual Notes

There are constraints regarding the approval of attachments, which may impact the sharing of relevant information. Participants are also grappling with the definitions and implications of vector projections in the context of the problem.

Xkaliber
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Homework Statement



The problem is given in the attachment below. The relevant information is provided under the "Stereo Audio" heading. Additionally, I have a couple of pieces of extra information not listed in the Stereo Audio section:

M is the component of (column vector) [R, L] on (column vector) [1, 1]
(This notation means that R and L are two entries in a column vector. The "R" being the upper entry and the "L" being the lower entry. The same applies for the 1 and 1)

D is the component of (column vector) [R, L] on (column vector) [-1, 1]


For the first problem, it asks me to express (M^2) + (D^2) in terms of L and R. Am I correct in writing (M^2) + (D^2) = [(L+R)^2]/4 +[(L-R)^2]/4?
 

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Is there a reason that my attachment has not been approved? How long does it usually take?
 
It takes a while for any link to another site to be approved. (Some brave soul with really really good virus protection has to check it out.)

I don't see that there is really any "Linear Algebra" involved here.

Yes, if M= (L+ R)/2 and D= (L-R)/2 then M2+ D2= (L+R)2/4+ (L-R)2/4. It would be a good idea, I think, to multiply those squares out and combine the fractions.
 
Yeah, that's what it seemed to me as well.

Additionally, I can't seem to be able to picture exactly what this scenario would look like graphed. First, we are given M= (L+ R)/2, which appears to be half the length of the vector [R, L]. However, at the same time I am told M is the component of vector [R, L] on [1, 1] (basically, a projection of [R, L] onto the [1, 1] vector) However, it seems these cannot be true at the same time, at least in my mind. So is it possible to graph these based on the given information?
 
Last edited:
Xkaliber said:
Yeah, that's what it seemed to me as well.

Additionally, I can't seem to be able to picture exactly what this scenario would look like graphed. First, we are given M= (L+ R)/2, which appears to be half the length of the vector [R, L]. However, at the same time I am told M is the component of vector [R, L] on [1, 1] (basically, a projection of [R, L] onto the [1, 1] vector) However, it seems these cannot be true at the same time, at least in my mind. So is it possible to graph these based on the given information?
No, (L+ R)/2 is certainly not "half the length of the vector [R,L]"! That would be [itex](1/2)\sqrt{L^2+ R^2}[/itex].
Drawing a right triangle diagram, you should be able to see that the projection of vector u on vector v has length [itex]|u|sin\theta[/itex] where [itex]\theta[/itex] is the angle between u and v. [itex]Since u.v= |u||v|sin\theta[/itex], we can write that as u.v/|v|. In the particular case that u= [L, R] and v= [1, 1] that is [itex](L+ V)/\sqrt{2}[/itex].
 
So is M= (L+ R)/2 a sum of two vectors or the sum of their magnitudes?
 

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