Explaining the Coefficient of a 45-Degree Rotation Vector

[jeff1evesque]

Homework Statement

A complex vector is written as,
$$\hat{v}(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} = \hat{x} + j\hat{y}$$,

where $$\omega$$ is the angular velocity, and the vector rotates counterclockwise in the x-y plane.If a unit vector is rotated in the x-y plane but is phase shifted by 45degrees, then:

$$\hat{v}(t) = (\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}})\hat{x} + \frac{1}{\sqrt{2}} - j\frac{1}{\sqrt{2}})\hat{x} \Rightarrow (cos(\omega t + 45^{\circ})\hat{x} + (cos(\omega t - 45^{\circ})\hat{y}$$

Can someone explain to me why there are terms $$\frac{1}{\sqrt{2}}$$ in the equation above. I always thought a 45 degree triangle had sides of $$\sqrt{2}, \sqrt{2}, 2$$, but not sure how the coefficient $$\frac{1}{\sqrt{2}}$$ is obtained.thanks,JL

 

[tiny-tim]

Hi JL!

(you've got your x's and y'x mixed up … and you can get LaTeX to write big brackets "to fit" by typing \left( and \right) )

Because cos45º = sin45º = 1/√2 (a 45 degree triangle also has sides of 1, 1/√2, 1/√2)

 

[Mene]

The coefficient \frac{1}{\sqrt{2}} in the equation above represents the amplitude of the vector after it has been rotated by 45 degrees. This can be understood by considering the unit circle, where the magnitude of the vector at any given point on the circle is equal to 1. When we rotate the vector by 45 degrees, the magnitude of the vector becomes \frac{1}{\sqrt{2}} at the point where it intersects the x-axis. This is because the x and y components of the vector are now equal in magnitude (due to the 45 degree rotation), and using the Pythagorean theorem, we can calculate the magnitude of the vector to be \sqrt{\frac{1}{2}^2 + \frac{1}{2}^2} = \frac{1}{\sqrt{2}}. Therefore, the coefficient \frac{1}{\sqrt{2}} is necessary to maintain the magnitude of the vector at 1 after the 45 degree rotation. This is a fundamental property of complex numbers and is essential for understanding the behavior of rotating vectors in the x-y plane.