0
$\begingroup$

If I have a two variables $x_1$, $x_2$ and two equations. In the first equation the first variable $x_1$ is multiplied with the conjugate of same number which is multiplied with $x_2$, and in the second equation, the exact inverse happens. For example, we have a number $1-2j$ so:

$(1-2j)x_1 + (1+2j)x_2 = y_1$ ... (1)

$(1+2j)x_1 + (1-2j)x_2 = y_2$ ... (2)

I'm confused, what's the proprieties of two above equations, I mean can we solve $x_2$ and $x_1$ if $y_2$ and $y_2$ are known? .. If not, can we get the value of $y_2$ based on the equation (1), it means we still have two variables but with one equation, we need to get the second equation which is very similar to equation (1)

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ I’m voting to close this question because this isn't signal processing. $\endgroup$
    – Marcus Müller
    Jun 20, 2020 at 8:24

1 Answer 1

Reset to default
1
$\begingroup$

can we solve x2 and x1 if y2 and y2 are known?

Yes, provided that you mean $y_1$ and $y_2$ are known. If you have two linear equations your can (in most cases) solve for two unknown variables.

If not, can we get the value of y2 based on the equation (1),

No. $y_2$ doesn't show up in equation (1) at all, so you can't get it from there.

which is very similar to equation (1)

Similarity has very little to do with it. You can always solve for two unknown variables unless the two equations are linear dependent. The two that you have are not.

Improve this answer
$\endgroup$
1
  • $\begingroup$ About solving $x_1$ and $x_2$ in that case, if the multipliers are equal and conjugate, is it possible? When adding the two equations, variables either will both kept or both cancelled. So, you mean in that case, we can't solve $x_1$ and $x_2$ from the above two equations, right? $\endgroup$
    – New_student
    Jun 20, 2020 at 5:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.