# Complex Vectors expression

## Homework Statement

Consider the unit vector, $$\hat{v}(t)$$, expressed in instantaneous form:
$$\hat{v}(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y}$$ (#0)

The Vector will rotate counterclockwise in the x-y plane with angular velocity $$\omega$$.

Since both components are sinusoidally time varying, and since there is a 90degrees phase shift between the components, we can express this vector as:
$$\hat{v} = (1 + j0)\hat{x} + (0 + j)\hat{y}$$ (#1)

2. Relevant Question
How was the equation above (equation #1) defined?

3. Thought Process
By Euler's Identity, phasors can be written as,
$$Ae^{j\phi} = {Acos(\phi), Asin(\phi)}$$ (real, and imaginary parts respectively)

Can we relate the identity above somehow to change equation (#0) into something like $$\hat{v}(t) = cos(\omega t)\hat{x} + jsin(\omega t)\hat{y}$$

And if we take the function of "t" out from equation (#0), why wouldn't equation (#1) become:
$$\hat{v} = cos(\omega)\hat{x} + jsin(\omega )\hat{y}$$

Thanks,

JL

Last edited:

cepheid
Staff Emeritus
Gold Member
I think you might be confusing a couple of things here. We have a real vector in the (real) 2D cartesian coordinate plane (which is referred to as $\mathbb{R}^2$ because it is the set of all ordered pairs generated by the intersection of two real number lines).

This real vector, in turn has components that are oscillatory functions of time. Because these components are oscillatory functions of time, they can each be described (abstractly) using a phasor, which is a vector rotating in the complex plane, (which is known as $\mathbb{C}$, not $\mathbb{R}^2$).

So the phasor is a complex number that describes how the x component of v(t) or the y component of v(t) oscillates in time. It is an entirely separate vector from v(t) itself.

How is the red equation derived? Well, cos(ωt) can be described using a complex quantity as follows:

cos(ωt) = Re[ej(ωt+φ)]​

where, in this case, φ = 0. When we express the phasor (either in polar or rectangular form), we typically take away the time-dependence. But we don't just "get rid of" t. We set t = 0. In other words, the phasor tells you the amplitude of the oscillation and the INITIAL phase (φ). So, in that form, the phasor would be:

ej(φ) = ej(0)

which is, in rectangular form:

ej(0) = 1*cos(0) + j*sin(0) = 1 + 0j = 1​

Similarly:

sin(ωt) = cos(ωt - π/2) = Re[ej(ωt+φ)]​

where, in this case, φ = -π/2

So, setting t = 0, we have the phasor given by:

e-j(π/2) = -j​

We know this by inspection, but if you like, we can explicitly convert it to rectangular form like we did for the first case:

e-j(π/2) = 1*cos(π/2) - j*sin(π/2) = 0 - j​

So, setting t = 0, and replacing the vector components with their corresponding phasors (in rectangular form), we obtain:

v(0) = (1 + 0j)x + (0 - j)y

where it is UNDERSTOOD that these vector components are now being expressed as phasors, and if you wanted to get the actual vector components vx(0) = 1, vy(0) = 0, you would have to take the REAL parts of those phasors.

Last edited:
Cepheid,

That's a great explanation, thanks a lot.

Jeffrey