Complex Representation of AC Signals

Sinusoidal variables:

To represent a sinusoidal voltage $v(t)=V_m cos(\omega t+\phi)$ or current $i(t)=I_m sin(\omega t+\psi)$, three values are needed:

• Amplitude, the peak value $V_m$ or $I_m$,
• Frequency $\omega=2\pi f=2\pi/T$ ($T=1/f$ is the period),
• Initial phase $\phi$.

Sinusoidal functions are closely related to complex exponentials due to Euler's formula:

$$\left\{ \begin{array}{l} e^{ j\theta}=\cos\theta+j \sin\the... ...\sin\theta=(e^{j\theta}-e^{-j\theta})/2j \end{array} \right.$$

As exponential functions can be much more conveniently manipulated than sinusoidal functions, a sinusoidal function is often considered as the real or imaginary part of the corresponding complex variable:

$$A\;\cos(\omega t+\phi)=Re[A\;e^{j\omega t+\phi}],\;\;\;\; A\;\sin(\omega t+\phi)=Im[A\;e^{j\omega t+\phi}]$$

Complex Arithmetic

A complex number can be represented in two different formats in either the Euclidean and polar coordinate system:

• Euclidean representation:

  $z=x+jy$ where $x$ and $y$ are the real (horizontal) and imaginary (vertical) part of complex variable $z$, respectively.

• Polar representation:

  $z = \vert z\vert\;e^{j\angle z}$ where $\vert z\vert$ and $\angle z$ are the magnitude and phase angle, respectively.

The two representations can be converted from one to the other:

• From $(x,y)$ to $(\vert z\vert,\angle z)$:
  
  $$\left\{ \begin{array}{ll} \vert z\vert=\sqrt{x^2+y^2} & \mbox... ...gle z=tan^{-1} (y/x) & \mbox{phase angle} \end{array} \right.$$
  
  

• From $(\vert z\vert,\angle z)$ to $(x,y)$:
  
  $$z=\vert z\vert\;e^{j\angle z}=\vert z\vert(\cos\angle z+j\;\sin\angle z)=x+jy$$
  
  
  due to Euler identity, i.e.,
  
  $$\left\{ \begin{array}{ll} x=\vert z\vert\;\cos\angle z & \mbo... ...ert\;\sin\angle z & \mbox{imaginary part} \end{array} \right.$$
  
  

The arithmetic operations of two complex numbers $z=x+jy=\vert z\vert\;e^{j\angle z}$ and $w=u+jv=\vert w\vert\;e^{j\psi}$ are listed below:

• Add/Subtract:
  
  $$z+w=(x+u)+j(y+v),\;\;\;\;z-w=(x-u)+j(y-v)$$
  
  

• Multiply:
  
  $$z\;w=(x+jy)(u+jv)=(xu-yv)+j(xv+yu)=\vert z\vert\;\vert w\vert\;e^{j(\angle z+\angle w)}$$
  
  

• Divide:
  
  $$\frac{z}{w}=\frac{x+jy}{u+jv}=\frac{(x+jy)(u-jv)}{(u+jv)(u-jv... ... =\frac{\vert z\vert}{\vert w\vert}\;e^{j(\angle z-\angle w)}$$
  
  

• Rotation:

  A complex number (vector) $z = \vert z\vert\;e^{j\angle z}$ multiplied by $e^{j\alpha}$ will become $z\;e^{j\angle z}=\vert z\vert\;e^{j(\angle z+\alpha)}$, i.e., rotated by an angle of $\alpha$. In particular, As $e^{j\pi/2}=j$ and $e^{-j\pi/2}=-j$, they can be considered as $90^\circ$ rotation factors. Any complex number multiplied by $j$ or $-j$ will be rotated counter clockwise or clockwise by 90 degrees.

• Complex Conjugate:

  The complex conjugate of $z=x+jy=\vert z\vert e^{j\angle z}$ is $z^*=x-jy=\vert z\vert e^{-j\angle z}$. In general, $z^*$ can be obtained by negating every $j$ in the expression of $z$ (replacing $j$ by $-j$). The magnitude of a complex number $z=x+jy$ can be found by:
  

  $$\sqrt{zz^*}=\sqrt{(x+jy)(x-jy)}=\sqrt{x^2+y^2}=\vert z\vert$$
  
  

• Reciprocal:
  
  $$\vert z^{-1}\vert=\frac{1}{\vert z\vert}=\frac{1}{\sqrt{x^2+y... ...\;\; \angle(z^{-1})=\angle(\frac{1}{z})=0-\angle{z}=-\angle z$$
  
  

Addition/Subtraction of Sinusoids:

Consider the addition/subtraction of two sinusoids of the same frequency $\omega$:

$$\left\{ \begin{array}{l} x_1(t)=A_1\cos(\omega t+\varphi_1)\\ x_2(t)=A_2\cos(\omega t+\varphi_2) \end{array} \right.$$

As the frequency of the sum is obviously the same as $\omega$, we can write the sum as:

$$x(t)=x_1(t)+x_2(t)=A\cos(\omega t+\varphi)$$

and we need to find $A$ and $\varphi$ in terms of $A_1$, $A_2$, $\varphi_1$ and $\varphi_2$.

$$x(t) = x_1(t)+x_2(t)=A_1\cos(\omega t+\varphi_1)+A_2\cos(\omega t+\varphi_2)$$

$$= A_1\cos\omega t\cos\varphi_1-A_1\sin\omega t\sin\varphi_1 +A_2\cos\omega t\cos\varphi_2-A_2\sin\omega t\sin\varphi_2$$

$$= (A_1\cos\varphi_1+A_2\cos\varphi_2)\cos\omega t -(A_1\sin\varphi_1+A_2\sin\varphi_2)\sin\omega t$$

Here we have used the trigonometric identity $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$.

If we define:

$$\left\{ \begin{array}{l} A_r=A_1 \cos\varphi_1+A_2 \cos\varph... ...2_r+A^2_j}\\ \varphi=\tan^{-1} (A_j/A_r) \end{array} \right.$$

to represent two variables $(A_r, A_j)$ in terms of two new variables $(A, \varphi)$, then the previous equation becomes:

$$x(t)=A_r \cos\omega t-A_j \sin\omega t= A\cos\varphi\cos\omega t-A\sin\varphi\sin\omega t=A \cos(\omega t+\varphi)$$

The process is illustrated in the figure below:

Phasor representation:

Addition and subtraction of sinusoids based on the trigonometric identities is lengthy and tedious. Now we consider the phasor method. Based on Euler's formula, a sinusoidal time function can be considered as the real (or imaginary) part of a rotating vector. Similarly, the sum of two sinusoids can be considered as the real (or imaginary) part of the vector sum of their corresponding vectors. If they have the same frequency $\omega$, then they can be considered as static instead of rotating (as if observed from a reference frame rotating at the same frequency as the vectors), and their relative positions with respect to each other stay fixed.

Represent the two sinusoids to be added as the real parts of their corresponding complex exponentials:

$$x_i(t)=A_i \cos(\omega t +\varphi_i)=Re [ A_i e^{j(\omega t+\... ... =Re[A_i e^{j\varphi_i} e^{j\omega t}] =Re[X_i e^{j\omega t}]$$

where we have defined the phasor of the sinusoid $x_i(t)$ as

$$X_i= A_i e^{j\varphi_i},\;\;\;\;\;(i=1,2)$$

to represent $x_i(t)$ as a vector in the complex plane with magnitude $A_i$ and initial phase angle $\varphi_i$. The frequency $\omega$ is not represented in the phasor, as all phasors rotate around the origin at the same rate.

Now the addition of these two sinusoids can be carried out as the vector sum of their phasors (in terms of the real and imaginary parts of both $X_1$ and $X_2$):

$$X = X_1+X_2=A_1 e^{j\varphi_1}+A_2 e^{j\varphi_2}$$

$$= (A_1 \cos\varphi_1+A_2 \cos\varphi_2)+j(A_1 \sin\varphi_1+A_2\sin\varphi_2)$$

$$= A_r+j A_j=A e^{j\varphi}$$

and the time function $x(t)=x_1(t)+x_2(t)$ can be obtained as the real part of the rotating version of this phasor:

$$x(t)=Re[X e^{j\omega t}]=Re[(X_1+X_2) e^{j\omega t}]$$

The frequency $\omega$ is not involved in the computation until the last step when $e^{j\omega t}$ is included in the result. This is the same as in the previous method, where the amplitude $A$ and phase $\varphi$ of $x(t)=x_1(t)+x_2(t)$ are functions of the amplitudes ($A_1$, $A_2$) and phases ($\varphi_1$, $\varphi_2$) of $x_1(t)$ and $x_2(t)$, irrelevant to the frequency $\omega$.

In summary, here are the three steps of the phasor method for addition/subtraction of sinusoidal time functions of the same frequency:

• Obtain phasor $X_i=A_ie^{j\varphi_i}$ for each given sinusoid $A_i\cos(\omega t+\varphi_i)$ ($i=1,2$);
• Find vector sum of the phasors:
  
  $$X=X_1\pm X_2=A_1e^{j\varphi_1}\pm A_2e^{j\varphi_2}$$
  
  

• Obtain the sum as a time function:
  
  $$x(t)=x_1(t)+x_2(t)=Re[X e^{j\omega t}]=Re[(X_1\pm X_2)e^{j\omega t}]$$
  
  

Specifically a sinusoidal voltage can be represented as:

$$v(t)=V_m\;cos(\omega t+\phi)=Re[{\bf V}(t)]=Re[V_m e^{j\phi}e... ...phi} \sqrt{2}e^{j\omega t}]=Re[\dot{V} \sqrt{2} e^{j\omega t}]$$

where

$${\bf V}(t)=V_m e^{j(\omega t+\phi)}=\sqrt{2}V_{eff} e^{j(\omega t+\phi)}$$

is the complex variable and $V_m$ is the peak magnitude of the voltage and $V_{eff}=V_m/\sqrt{2}$ is the effective value (RMS), and the phasor representing the voltage is defined as:

$$\dot{V}\stackrel{\triangle}{=}V_{eff} e^{j\phi}=V_{eff} \angle \phi$$

in terms of

• the magnitude, the rms (effective) value $V_{eff}=V_{rms}$,
• the phase $\phi$

The frequency $\omega=2\pi f$ is not explicitly represented by the phasor, as all currents and voltages in the circuit considered here have the same frequency, same as that of the energy source or input of the circuit.

Example A 120V 60Hz AC voltage

$$v(t)=120 \sqrt{2}\; \cos(2\pi f t+50^\circ) =120\sqrt{2}\; \cos(6.28\times 60 t+50^\circ) =170\;\cos(377\;t+50^\circ)$$

is expressed as $\dot{V}=V_{eff} \angle \phi= 120 \angle 50^\circ$ with rms value $V_{eff}=120$ and $\phi=50^\circ$, and the implied frequency $f=60Hz$. All sinusoidal signals, currents as well as voltages can be represented by phasors.