# The algorithm of phasor analysis

1. Transform EMS, current and voltage functions into phasors.
2. Use any method or circuit analysis to derive the complex values of unknown phasors.
3. Transform the derived phasors into functions in the time domain.
4. Plot the derived functions in the time domain.
5. Calculate complex power, apparent power, active power and reactive power.

# The Case of phasor analysis of RC-circuit

## Transform EMS, current and voltage functions into phasors

Phasor of input excitation: $$\underline{E}=E_me^{j\Psi_E}$$ Input impedance: $$\underline{Z}_{in}=R-\frac{j}{\omega C}$$

## Derive the complex values of unknown phasors

Phasor of input current: $$\underline{I}_{in}=\frac{\underline{E}}{\underline{Z}_{in}}=\frac{E_me^{j\Psi_E}}{\left| \underline{Z}_{in}\right|e^{j\Psi_Z}}=\frac{E_m}{\left|\underline{Z}_{in}\right|}e^{j\left(\Psi_E-\Psi_Z \right )}$$ Magnitude of impedance: $$\left| \underline{Z}_{in}\right|=\sqrt{R^2+\left(\frac{1}{\omega C} \right )^2}$$ Impedance phase: $$\Psi_Z=arctan\left(-\frac{1}{\omega CR} \right )$$

## Transform the derived phasors into functions in the time domain

\begin{aligned} & \underline{I}_{in}=10^{-4}\cdot\left(5+j5 \right )& \rightarrow & i_{in}=\left|\underline{I}_{in}\right|sin\left(\omega t+\Psi_i \right )\\ & \underline{U}_R=5+j5 & \rightarrow & u_R=\left|\underline{U}_R\right|sin\left(\omega t+\Psi_{u_R} \right )\\ & \underline{U}_C=5-j5 & \rightarrow & u_C=\left|\underline{U}_C\right|sin\left(\omega t+\Psi_{u_C} \right ) \end{aligned} Calculate the amplitudes: $$I_{in.m}=\left|\underline{I}_{in}\right|=\sqrt{\left(5\cdot 10^{-4} \right )^2+\left(5\cdot 10^{-4} \right )^2}=7.07\cdot 10^{-4}\left[A \right ]$$ $$U_{R.m}=\left|\underline{U}_R\right|=\sqrt{\left(5 \right )^2+\left(5 \right )^2}=7.07\left[V \right ]$$ $$U_{C.m}=\left|\underline{U}_C\right|=\sqrt{\left(5 \right )^2+\left(5 \right )^2}=7.07\left[V \right ]$$ Calculate the phases: $$\varphi_{i_n}=arctan\left(\frac{0.0005}{0.0005} \right )=0.785\left[rad \right ]$$ $$\varphi_{U_R}=arctan\left(1 \right )=0.785\left[rad \right ]$$ $$\varphi_{U_C}=arctan\left(-1 \right )=-0.785\left[rad \right ]$$

## Calculate complex power, apparent power, active power and reactive power

Calculate the RMS values: $$E=\frac{E_m}{\sqrt{2}}=7.07\left[V \right ]\qquad I_{in}=\frac{\left|\underline{I}_{in}\right|}{\sqrt{2}}=5\cdot 10^{-4}\left[A \right ]$$ Calculate the active power: $$P=\left(I_{in} \right )^2R=2.5\cdot 10^{-3}\left[W \right ]$$ Calculate the reactive power: $$Q=-Q_C=-\left(I_{in} \right )^2\cdot\left(\frac{1}{\omega C} \right )=-2.5\cdot 10^{-3}\left[VAr \right ]$$ Calculate complex power: $$S=E\cdot I_{in}e^{-j\Psi_{in}}=7.07\cdot 5\cdot 10^{-4}e^{-j0.785}=\left(2.5-j2.5 \right )\cdot 10^{-3}$$ $$P=2.5\cdot 10^{-3}\left[W \right ]\qquad Q=-2.5\cdot 10^{-3}\left[VAr \right ]$$ Calculate the apparent power: $$S=\sqrt{P^2+Q^2}=3.5\cdot 10^{-3}\left[VA \right ]$$

# The Case of phasor analysis of RL-circuit

## Transform EMS, current and voltage functions into phasors

Input phasor excitation: $$\underline{E}=E^me^{j\Psi_E}$$ Input impedance: $$\underline{Z}_{in}=R+j\omega L$$

## Derive the complex values of unknown phasors

Input phasor current: $$\underline{I}_{in}=\frac{\underline{E}}{\underline{Z}_{in}}=\frac{E_me^{j\Psi_E}}{\left| \underline{Z}_{in}\right|e^{j\Psi_Z}}=\frac{E_m}{\left|\underline{Z}_{in}\right|}e^{j\left(\Psi_E-\Psi_Z \right )}$$ Magnitude of impedance: $$\left|\underline{Z}_{in} \right|=\sqrt{R^2+\left(\omega L \right )^2}$$ Impedance phase: $$\Psi_Z=arctan\left(\frac{\omega L}{R} \right )$$

## Transform the derived phasors into functions in the time domain

\begin{aligned} & \underline{I}_{in}=0.2-j0.4 & \rightarrow & i_{in}=\left|\underline{I}_{in}\right|sin\left(\omega t+\Psi_i \right )\\ & \underline{U}_R=2-j4 & \rightarrow & u_R=\left|\underline{U}_R\right|sin\left(\omega t+\Psi_{u_R} \right )\\ & \underline{U}_L=8+j4 & \rightarrow & u_L=\left|\underline{U}_L\right|sin\left(\omega t+\Psi_{u_C} \right ) \end{aligned}

## Calculate complex power, apparent power, active power and reactive power

Calculate the RMS values: $$I_{in}=\frac{\left|\underline{I}_{in} \right|}{\sqrt{2}}=0.316\left[A \right ]\qquad E=\frac{E_m}{\sqrt{2}}=7.07\left[V \right ]$$ Calculate the active values: $$P=\left(I_{in} \right )^2\cdot R \approx 1\left[W \right ]$$ Calculate the reactive values: $$Q_L=\left(I_{in} \right )^2\cdot \omega L \approx 2\left[VAr \right ]$$ Calculate the complex power: $$S=E\cdot I_{in}e^{-j\Psi_{i_{in}}}=1+j2$$ Calculate the apparent power: $$S=\sqrt{P^2+Q^2}=2.236\left[VA \right ]$$

# Frequency response of a network

Analysis of a circuit with varying frequency of a sinusoidal sources is called the frequency response of a circuit. A circuit performed the frequency selection is called filters because of their ability to filter out certain input signals on the basis of frequency. Transfer function H(j$$\omega$$) is the magnitude and phase of the output voltage to the magnitude and phase of the input voltage of a circuit.

## Magnitude plot

The magnitude of the system function of the network plotted against frequency. $$H\left(j\omega \right )=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}=\frac{a_n\left(j\omega \right )^n+a_{n-1}\left(j\omega \right )^{n-1}+\cdots +a_2\left(j\omega \right )^2+a_1j\omega+a_0}{b_m\left(j\omega \right )^m+a_{m-1}\left(j\omega \right )^{m-1}+\cdots +b_2\left(j\omega \right )^2+b_1j\omega+b_0}$$ $$A\left(j\omega \right )=Re\left(A\left(j\omega \right ) \right )+Im\left(A\left(j\omega \right ) \right )$$ $$B\left(j\omega \right )=Re\left(B\left(j\omega \right ) \right )+Im\left(B\left(j\omega \right ) \right )$$ For example, if $$n=m=3$$: $$H\left(j\omega \right )=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}=\frac{a_3\left(j\omega \right )^3+a_2\left(j\omega \right )^2+a_1j\omega+a_0}{b_3\left(j\omega \right )^3+b_2\left(j\omega \right )^2+b_1j\omega+b_0}$$ $$j^2=-1\qquad j^3=-j$$ $$H\left(j\omega \right )=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}=\frac{-a_3j\omega^3-a_2\omega^2+a_1j\omega+a_0}{-b_3j\omega^3-b_2\omega^2+b_1j\omega+b_0}$$ \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=-a_2\omega^2+a_0 & Im\left(A\left(j\omega \right ) \right )=-a_3\omega^3+a_1\omega\\ & Re\left(B\left(j\omega \right ) \right )=-b_2\omega^2+b_0 & Im\left(B\left(j\omega \right ) \right )=-b_3\omega^3+b_1\omega\ \end{aligned} $$H\left(j\omega \right )=Re\left(H(j\omega) \right )+Im\left(H\left(j\omega \right ) \right )$$ $$Re\left(H\left(j\omega \right ) \right )=\frac{Re\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )+Im\left(A\left(j\omega \right ) \right )\cdot Im\left(B\left(j\omega \right ) \right )}{\left(Im\left(B\left(j\omega \right ) \right ) \right )^2+\left(Re\left(B\left(j\omega \right ) \right ) \right )^2}$$ $$Im\left(H\left(j\omega \right ) \right )=\frac{Im\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )-Im\left(B\left(j\omega \right ) \right )\cdot Re\left(A\left(j\omega \right ) \right )}{\left(Im\left(B\left(j\omega \right ) \right ) \right )^2+\left(Re\left(B\left(j\omega \right ) \right ) \right )^2}$$ $$H\left(\omega \right )=\left|H\left(j\omega \right )\right|=\sqrt{\left(Re\left(H\left(j\omega \right ) \right ) \right )^2+\left(Im\left(H\left(j\omega \right ) \right ) \right )^2}$$ $$H\left(\omega \right )=\sqrt{\frac{\left(Re\left(A\left(j\omega \right ) \right ) \right )^2+\left(Im\left(A\left(j\omega \right ) \right ) \right )^2}{\left(Re\left(B\left(j\omega \right ) \right ) \right )^2+\left(Im\left(B\left(j\omega \right ) \right ) \right )^2}}$$

## Phase angle plot

The phase angle of the system function of the network plotted against frequency $$\varphi\left(\omega\right)=arctan\left(\frac{Im\left(H\left(j\omega \right ) \right )}{Re\left(H\left(j\omega \right ) \right )} \right )$$ $$\varphi\left(\omega\right)=arctan\left(\frac{Im\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )-Im\left(B\left(j\omega \right ) \right )\cdot Re\left(A\left(j\omega \right ) \right )}{Im\left(A\left(j\omega \right ) \right )\cdot Im\left(B\left(j\omega \right ) \right )+Re\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )}\right )$$

## Group delay

Group delay $$\tau\left(\omega\right)\left[s\right]$$ is the time delay of the amplitude envelopes of the various sinusoidal components of a signal through a device under test, and is a function of frequency for each component. $$\tau\left(\omega \right )=-\frac{d\varphi\left(\omega\right)}{d\omega}=-\frac{d\left(arctan\left(\frac{Im\left(H\left(j\omega \right ) \right )}{Re\left(H\left(j\omega \right ) \right )} \right ) \right )}{d\omega}$$ $$\tau\left(\omega \right )=-\frac{\frac{\left(\frac{d\left(Im\left(H\left(j\omega \right ) \right ) \right )}{d\omega} \right )\cdot Re\left(H\left(j\omega \right ) \right )-Im\left(H\left(j\omega \right ) \right )\cdot\left(\frac{d\left(Re\left(H\left(j\omega \right ) \right ) \right )}{d\omega}\right)}{\left(Re\left(H\left(j\omega \right ) \right ) \right )^2}}{1+\left(\frac{Im\left(H\left(j\omega \right ) \right )}{Re\left(H\left(j\omega \right ) \right )} \right )^2}$$

# The Case of frequency analysis of RC-circuit

## RC-circuit

$$Z_{in}\left(j\omega \right )=R-\frac{j}{\omega C}=\frac{j\omega CR+1}{j\omega C}$$

$$I_{in}\left(j\omega \right )=\frac{E\left(j\omega \right )}{Z_{in}\left(j\omega \right )}=E\left(j\omega \right )\frac{j\omega C}{j\omega CR+1}$$

$$U_C\left(j\omega \right )=I_{in}\left(j\omega \right )\frac{1}{j\omega C}=E\left(j\omega \right )\frac{1}{j\omega CR+1}$$

Calculate transfer function: $$H_{u_c}\left(j\omega \right )=\frac{U_C\left(j\omega \right )}{E\left(j\omega \right )}=\frac{1}{j\omega CR+1}=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}$$

## Magnitude response

Let's express the components of transfer function: \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=1 & Im\left(A\left(j\omega \right ) \right )=0&\\ & Re\left(B\left(j\omega \right ) \right )=1 & Im\left(B\left(j\omega \right ) \right )=\omega CR&\\ & Re\left(H_{u_c}\left(j\omega \right ) \right )=\frac{1}{\left(\omega CR \right )^2+1} & Im\left(H_{u_c}\left(j\omega \right ) \right )=\frac{-\omega CR}{\left(\omega CR\right)^2+1}& \end{aligned} Magnitude response can be plotted by the following function: $$H{u_c}\left(\omega \right )=\sqrt{\frac{\left(Re\left(A\left(j\omega \right ) \right ) \right )^2+\left(Im\left(A\left(j\omega \right ) \right ) \right )^2}{\left(Re\left(B\left(j\omega \right ) \right ) \right )^2+\left(Im\left(B\left(j\omega \right ) \right ) \right )^2}}=\sqrt{\frac{1}{1+\left(\omega CR \right )^2}}$$ The transfer function magnitude is decreased by the factor 0.707 from its maximum value is called cut-off frequency $$f_c$$. A Low-Pass filter passes signals at frequencies lower than the cutoff frequency from the input to the output.

## Phase response

Let's express the components of transfer function: \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=1 & Im\left(A\left(j\omega \right ) \right )=0&\\ & Re\left(B\left(j\omega \right ) \right )=1 & Im\left(B\left(j\omega \right ) \right )=\omega CR&\\ \end{aligned} $$\varphi\left(\omega \right )=arctan\left(\frac{Im\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )-Im\left(B\left(j\omega \right ) \right )\cdot Re\left(A\left(j\omega \right ) \right )}{Im\left(A\left(j\omega \right ) \right )\cdot Im\left(B\left(j\omega \right ) \right )+Re\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )} \right )$$ Phase response can be plotted by the expression: $$\varphi_{u_c}\left(\omega \right )=arctan\left(-\omega CR \right )$$

## Group delay

$$\varphi\left(\omega \right )=arctan\left(-\omega CR \right )$$ Group delay can be plotted by the following function: \begin{aligned} \tau\left(\omega \right )&=-\frac{d\varphi\left(\omega \right )}{d\omega}&=&-\frac{d\left(arctan\left(-\omega CR \right ) \right )}{d\omega}\\ & =-\frac{-CR}{1+\left(-\omega CR \right )^2}&=&\frac{CR}{1+\omega^2\left(CR \right )^2} \end{aligned}

## RC-circuit

$$U_R\left(j\omega \right )=I_{in}\left(j\omega \right )\cdot R=E\left(j\omega \right )\frac{j\omega CR}{j\omega CR+1}$$ Calculate transfer function: $$H_{u_R}\left(j\omega \right )=\frac{U_R\left(j\omega \right )}{E\left(j\omega \right )}=\frac{j\omega CR}{j\omega CR+1}=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}$$

## Magnitude response

\begin{aligned} & Re\left(A\left(j\omega \right ) \right )=0 & Im\left(A\left(j\omega \right ) \right )=\omega CR&\\ & Re\left(B\left(j\omega \right ) \right )=1 & Im\left(B\left(j\omega \right ) \right )=\omega CR&\\ & Re\left(H_{u_R}\left(j\omega \right ) \right )=\frac{\left(\omega CR \right )^2}{\left(\omega CR \right )^2+1} & Im\left(H_{u_R}\left(j\omega \right ) \right )=\frac{\omega CR}{\left(\omega CR\right)^2+1}& \end{aligned} Magnitude response can be plotted by the following function: $$H_{u_R}\left(\omega \right )=\sqrt{\frac{\left(\omega CR \right )^2}{1+\left(\omega CR \right )^2}}$$ A High-Pass filter passes signals at frequencies higher than the cutoff frequency from the input to the output

## Phase response

\begin{aligned} & Re\left(A\left(j\omega \right ) \right )=0 & Im\left(A\left(j\omega \right ) \right )=\omega CR&\\ & Re\left(B\left(j\omega \right ) \right )=1 & Im\left(B\left(j\omega \right ) \right )=\omega CR&\\ \end{aligned} $$\varphi\left(\omega \right )=arctan\left(\frac{Im\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )-Im\left(B\left(j\omega \right ) \right )\cdot Re\left(A\left(j\omega \right ) \right )}{Im\left(A\left(j\omega \right ) \right )\cdot Im\left(B\left(j\omega \right ) \right )+Re\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )} \right )$$ $$\varphi_{u_R}\left(\omega \right )=arctan\left(\frac{\omega CR}{\left(\omega CR \right )^2} \right )=arctan\left(\frac{1}{\omega CR} \right )$$

## Group delay

$$\varphi_{u_R}\left(\omega \right )=arctan\left(\frac{1}{\omega CR} \right )$$ Group delay can be plotted by the following function: $$\tau_{u_R}\left(\omega \right )=-\frac{d\varphi\left(\omega \right )}{d\omega}=-\frac{d\left(arctan\left(\frac{1}{\omega CR} \right ) \right )}{d\omega}=\frac{CR}{\left(CR\omega \right )^2+1}$$

# The Case of frequency analysis of RL-circuit

## RL-circuit

$$Z_{in}\left(j\omega \right )=R+j\omega L$$ $$I_{in}\left(j\omega \right )=\frac{E\left(j\omega \right )}{Z_{in}\left(j\omega \right )}=E\left(j\omega \right )\frac{1}{j\omega L+R}$$ $$U_L\left(j\omega \right )=I_{in}\left(j\omega \right )\cdot j\omega L=E\left(j\omega \right) \frac{j\omega L}{j\omega L+R}$$ Calculate transfer function: $$H_{U_L}\left(j\omega \right )=\frac{U_L\left(j\omega \right )}{E\left(j\omega \right )}=\frac{j\omega L}{j\omega L+R}=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}$$

## Magnitude response

Let's express the components of transfer function: \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=0 & Im\left(A\left(j\omega \right ) \right )=\omega L&\\ & Re\left(B\left(j\omega \right ) \right )=R & Im\left(B\left(j\omega \right ) \right )=\omega L&\\ & Re\left(H_{u_L}\left(j\omega \right ) \right )=\frac{\left(\omega L \right )^2}{\left(\omega L \right )^2+R} & Im\left(H_{u_L}\left(j\omega \right ) \right )=\frac{\omega LR}{\left(\omega L\right)^2+R^2}& \end{aligned} Magnitude response can be plotted by the following function: $$H_{u_L}\left(\omega \right )=\sqrt{\frac{\left(\omega L \right )^2}{R^2+\left(\omega L \right )^2}}$$

## Phase response

Let's express the components of transfer function: \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=0 & Im\left(A\left(j\omega \right ) \right )=\omega L&\\ & Re\left(B\left(j\omega \right ) \right )=R & Im\left(B\left(j\omega \right ) \right )=\omega L&\\ \end{aligned} $$\varphi\left(\omega \right )=arctan\left(\frac{Im\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )-Im\left(B\left(j\omega \right ) \right )\cdot Re\left(A\left(j\omega \right ) \right )}{Im\left(A\left(j\omega \right ) \right )\cdot Im\left(B\left(j\omega \right ) \right )+Re\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )} \right )$$ Phase response can be plotted by the following function: $$\varphi_{u_L}\left(\omega \right )=arctan\left(\frac{\omega LR}{\left(\omega L \right )^2} \right )=arctan\left(\frac{R}{\omega L} \right )$$

## Group delay

$$\varphi_{u_L}\left(\omega \right )=arctan\left(\frac{R}{\omega L} \right )$$ Group delay can be plotted by the following function: $$\tau_{u_L}\left(\omega \right )=-\frac{d\varphi\left(\omega \right )}{d\omega}=-\frac{d\left(arctan\left(\frac{R}{\omega L} \right ) \right )}{d\omega}=\frac{RL}{\omega^2L^2+R^2}$$

## RL-circuit

$$U_R\left(j\omega \right )=I_{in}\left(j\omega \right )\cdot R=E\left(j\omega \right )\frac{R}{j\omega L+R}$$ Calculate transfer function: $$H_{U_L}\left(j\omega \right )=\frac{U_R\left(j\omega \right )}{E\left(j\omega \right )}=\frac{R}{j\omega L+R}=\frac{A\left(j\omega \right )}{B\left(j\omega \right )}$$

## Magnitude response

Let's express the components of transfer function: \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=R & Im\left(A\left(j\omega \right ) \right )=0&\\ & Re\left(B\left(j\omega \right ) \right )=R & Im\left(B\left(j\omega \right ) \right )=\omega L&\\ & Re\left(H_{u_R}\left(j\omega \right ) \right )=\frac{R^2}{\left(\omega L \right )^2+R^2} & Im\left(H_{u_R}\left(j\omega \right ) \right )=\frac{-\omega LR}{\left(\omega L\right)^2+R^2}& \end{aligned} Magnitude response can be plotted by the following function: $$H_{u_R}\left(\omega \right )=\sqrt{\frac{R^2}{R^2+\left(\omega L \right )^2}}$$

## Phase response

Let's express the components of transfer function: \begin{aligned} & Re\left(A\left(j\omega \right ) \right )=R & Im\left(A\left(j\omega \right ) \right )=0&\\ & Re\left(B\left(j\omega \right ) \right )=R & Im\left(B\left(j\omega \right ) \right )=\omega L&\\ \end{aligned} $$\varphi\left(\omega \right )=arctan\left(\frac{Im\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )-Im\left(B\left(j\omega \right ) \right )\cdot Re\left(A\left(j\omega \right ) \right )}{Im\left(A\left(j\omega \right ) \right )\cdot Im\left(B\left(j\omega \right ) \right )+Re\left(A\left(j\omega \right ) \right )\cdot Re\left(B\left(j\omega \right ) \right )} \right )$$ Phase response can be plotted by the following function: $$\varphi_{u_R}\left(\omega \right )=arctan\left(-\frac{\omega LR}{R^2} \right )=arctan\left(-\frac{\omega L}{R} \right )$$

## Group delay

$$\varphi_{u_R}\left(\omega \right )=arctan\left(-\frac{\omega L}{R} \right )$$ Group delay can be plotted by the following function: $$\tau_{u_R}\left(\omega \right )=-\frac{d\varphi\left(\omega \right )}{d\omega}=-\frac{d\left(arctan\left(-\frac{\omega L}{R} \right ) \right )}{d\omega}=\frac{RL}{\omega^2L^2+R^2}$$