Phasors

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Introduction to phasors

Many real world signals can be described by a time depending sinusoidal signals such as: x(t)=Acos(ωot+ϕ) In this equation is A represents the amplitude, from which the dimension is equal to the dimension of the signal x(t), ω0 [rad/sec] the radian (or angular) frequency and ϕ [rad] the phase.

Visualisation of a phasor.
Visualisation of a phasor.

By using the Euler equations and replacing θ by ωot+ϕ we obtain the following time depending complex exponential function z(t): z(t)=Aej(ωot+ϕ)=Acos(ωot+ϕ)+jAsin(ωot+ϕ)

Such time depending complex exponential, which is depicted as a time depending complex vector in Fig. 1, is called a phasor. The projection of the phasor on the real axis behaves like a cosine signal, while the projection on the imaginary axis behaves like a sine function. In other words we can generalize a time depending sinusoidal signal by a phasor, since it describes both a sine and cosine function at the same time. Alternatively we write the cosine and sine function in phasor notation as follows using the Euler equations: Acos(ωot+ϕ)=e{Aej(ωot+ϕ)}=Aej(ωot+ϕ)+Aej(ωot+ϕ)2Asin(ωot+ϕ)=m{Aej(ωot+ϕ)}=Aej(ωot+ϕ)Aej(ωot+ϕ)2j



Phasor addition rule

We can write a phasor as the following product: Aej(ωot+ϕ)=Aejϕejω0t

The second part ejω0t is the phasor component, containing the radian frequency ω0. The first part Aejϕ is a complex number, in which A represents the amplitude and ϕ the phase of the phasor. By using this fact it becomes obvious that the addition of two phasors, A1ej(ω0t+ϕ1) and A2ej(ω0t+ϕ2), with the same radial frequency ω0 result in one new phasor Aej(ω0t+ϕ) having the same frequency ω0. This can be shown as follows: A1ej(ω0t+ϕ1)+A2ej(ω0t+ϕ2)=A1ejϕ1ejω0+A2ejϕ2ejω0=(A1ejϕ1+A2ejϕ2)ejω0

The amplitude A and phase ϕ of the resulting phasor Aej(ω0t+ϕ) can be calculated by using the complex addition rule for the complex numbers A1ejϕ1 and A2ejϕ2 of the individual phasors as follows: Aejϕ=A1ejϕ1+A2ejϕ2

This phasor addition example is depicted in Fig. 2.

The addition of phasors.
Visualisation of the addition operation on phasors with equal frequencies.

We can use this property when a signal x(t) consists of the sum of two sinusoidal signals, both with the same frequency ω0, which goes as follows:

x(t)=A1cos(ω0t+ϕ1)+A2cos(ω0t+ϕ2)=(A12ejϕ1+A22ejϕ2)ejω0t+(A12ejϕ1+A22ejϕ2)ejω0t

By defining Aejϕ=A1ejϕ1+A2ejϕ2 we obtain: x(t)=(A2ejϕ)ejω0t+(A2ejϕ)ejω0t=Acos(ω0t+ϕ)

Thus the amplitude A and phase ϕ of the resulting sinusoidal signal x(t) are found by complex addition of the amplitude and phase of the individual sinusoidal components. \
This example can be extended to the addition of N sinusoidal signals, all with the same radian frequency ω0, resulting in one sinusoidal signal with the same radian frequency ω0 as follows:

x(t)=k=1NAkcos(ω0t+ϕk)=Acos(ω0t+ϕ)

in which the amplitude A and phase ϕ can be calculated by using complex addition rules as follows: Aejϕ=k=1NAkejϕk

In words this results in the so called phasor addition rule:

The sum of original sinusoidal signals, all with the same radian frequency ω0, results in one sinusoidal signal with the same radian frequency ω0. Amplitude and phase of this resulting signal can be found by adding the complex representation of amplitude and phase of the individual original sinusoidal signals.

Example


The function x(t) consists of the sum of the following 3 sinusoidal signals: x(t)=5cos(ω0t+32π)+4cos(ω0t+23π)+4cos(ω0t+13π) Express x(t) in the form x(t)=Acos(ω0t+ϕ) by finding the numerical values of A and ϕ.
The different cosine functions can be written as the real part of phasors. x(t)=e{5ej(ω0t+32π)}+e{4ej(ω0t+23π)}+4e{ej(ω0t+13π)} The phasor component (ejω0t) can be separated: x(t)=e{ejω0t(5ej3π2+4ej2π3+4ejπ3)} The complex numbers, containing the amplitude and phase of the 3 individual sinusoidal signals, can be added together by using complex addition rules. Since we have to add complex numbers it is easiest to rewrite these complex numbers in Cartesian notation: x(t)=e{ejω0t(j5+4(12+j123)+4(12+j123))=e{ejω0t((435)j)}=e{ejω0t((435)ejπ2)}=e{(435)ejω0t+π2} Now this answer can be written as the final answer: x(t)=(435)cos(ω0t+π2)