# AC Circuit Analysis - Sources with Different Frequencies

In AC circuit analysis, if the circuit has sources operating at different frequencies, Superposition theorem can be used to solve the circuit. Please note that AC circuits are linear and that is why Superposition theorem is valid to solve them.

## Problem

Determine $i_x(t)$ where $i_s(t)=4 \cos (4t) ~ A$ and $v_s(t)=2 \sin (2t) ~ V$ .

## Solution with AC Circuit Analysis

Since sources are operating at different frequencies, i.e. $4 \frac{rad}{s}$ and $2 \frac{rad}{s}$ , we have to use the Superposition theorem. That is to say that we need to determine contribution of each source on $i_x(t)$ . Then, the final answer is to obtained by adding the individual responses in the time domain. Please note that, since the impedances depend on frequency, we need to have a different frequency-domain circuit for each frequency.

### Contribution of the current source

To find the contribution of the current source, we need to turn off other source(s). So, we need to turn off the voltage source. This is very similar to DC circuits that we discussed before:

Voltage sources become a short circuit when turned off.

#### Frequency domain

We first convert the circuit to the frequency domain:
$i_s(t)=4 \cos (4t) ~ A \rightarrow I_s=4\angle 0^{\circ} (\omega = 4 \frac{rad}{s})$
$L=1\text{H} \rightarrow Z_L=j \omega L=j4$
$C=1\text{F} \rightarrow Z_C=\frac{1}{j \omega C}=\frac{1}{j4}=-j0.25$
$R=1 \Omega \rightarrow Z_R=R=1$
$i_{x_1}(t) \rightarrow I_{x_1}$

Using current divider:
$I_{x_1}=\frac{Z_C}{Z_C+R}I_s$
$=\frac{-j0.25}{-j0.25+1} \times 4$
$=\frac{-j0.25}{1-j0.25} \times 4$
$=\frac{-j0.25 \times (1+j0.25)}{(1-j0.25)\times (1+j0.25)} \times 4$
$=\frac{-j0.25 + 0.0625)}{1^2+0.25^2} \times 4$
$=\frac{-j0.25 + 0.0625)}{1.0625} \times 4$
$=0.235-j0.941$
$=0.97\angle -76^{\circ}$

#### Time domain

Conversion to time-domain:
$i_{x_1}(t)=0.97 \cos (4t-76^{\circ}) ~ \text{A}$
Please note that the sinusoidal function for $i_s(t)$ is cosine and consequently, cosine must be used in converting $I_{x_1}$ to the time domain.

### Contribution of the voltage source

To find the contribution of the voltage source, the current source needs to be turned off. As mentioned before:

To turn off a current source it should be replaced by an open circuit

#### Frequency domain

$v_s(t)=2 \sin (2t) ~ V \rightarrow V_s=2\angle 0^{\circ} (\omega = 2 \frac{rad}{s})$
$L=1\text{H} \rightarrow Z_L=j \omega L=j2$
$C=1\text{F} \rightarrow Z_C=\frac{1}{j \omega C}=\frac{1}{j2}=-j0.5$
$R=1 \Omega \rightarrow Z_R=R=1$
$i_{x_2}(t) \rightarrow I_{x_2}$

As the inductor branch is open, this is a very simple circuit with three elements in series: $R$ , $Z_C$ and $V_s$ . Therefore,
$I_{x_2}=\frac{V_s}{R+Z_C}$
$=\frac{2}{1-j0.5}$
$=\frac{2\times (1+j0.5)}{(1-j0.5)\times (1+j0.5)}$
$=\frac{2+j}{1^2+0.5^2}$
$=\frac{2+j}{1.25}$
$=1.6+j0.8$
$=1.789\angle 26.6^{\circ}$

#### Time domain

$i_{x_2}(t)=1.789 \sin (2t+26.6^{\circ}) ~ \text{A}$
(why $\sin$ ?)

Now that we have determined both $i_{x_1}(t)$ and $i_{x_2}(t)$ in time domain, we can go ahead and add them up to find $i_x(t)$ . Please note that we could not add $I_{x_1}$ and $I_{x_2}$ because they are not phasors with the same frequency.
$i_{x}(t)=i_{x_1}(t)+i_{x_2}(t)=0.97 \cos (4t-76^{\circ}) + 1.789 \sin (2t+26.6^{\circ}) ~ \text{A}$

# Mesh (Current) Analysis Problem

Solve the circuit by mesh analysis and find the current $I_x$ and the voltage across $R_2$ .

# Solution

## Mesh Analysis

There are four meshes in the circuit. So, we need to assign four mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.

# Stability using Routh Stability Criterion

Determine the stability of the system whose characteristics equation is:

$a(s) = 2s^5 + 3s^4 + 2s^3 + s^2 + 2s + 2$ .

# Solution

All coefficients are positive and non-zero; therefore, the necessary condition for stability is satisfied. Let's write the Routh array:

$\begin{array}{l | c c c} s^5 & 2 & 2 & 2 \\ s^4 & 3 & 1 & 2 \\ s^3 & & & \\ s^2 & & & \\ s^1 & & & \\ s^0 & & &\end{array}$

# Problem

Find $I_x$ and $I_y$ :

# Solution

Three resistors are in series and their equivalent, $6\Omega$ , is parallel with the voltage source. So, according to the Ohm's law: $I_y=-\frac{6V}{6 \Omega}=-1 A$ . The negative sign comes from the direction $I_y$ .
Applying KCL at the bottom node:
$-(-2A)+I_x+I_y=0 \rightarrow I_x=-1 A$ .
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# Find Equivalent Impedance - AC Steady State Analysis

Determine the driving-point impedance of the network at a frequency of $2$ kHz:

# Solution

Lets first find impedance of elements one by one:

## Resistor $R$

The resistor impedance is purely real and independent of frequency.

$Z_R=R=20 \Omega$

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## Problem

Find $I_x$ and $I_y$ :

# Superposition method - Circuit with two sources

Find $I_x$ using superposition rule:

# Solution

## Superposition

The superposition theorem states that the response (voltage or current) in any branch of a linear circuit which has more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are turned off (made zero).

# Find Thevenin's and Norton's Equivalent Circuits

Find Thevenin's and Norton's Equivalent Circuits:

Suppose that $R_1=5\Omega$ , $R_2=3\Omega$ and $I_S=2 A$ .

# Solution

The circuit has both independent and dependent sources. In these cases, we need to find open circuit voltage and short circuit current to determine Norton's (and also Thevenin's) equivalent circuits.

# Solve Using Current Division Rule

Find current of resistors, use the current division rule.

Suppose that $R_1=2 \Omega$ , $R_2=4 \Omega$ , $R_3=1 \Omega$ , $I_S=5 A$ and $V_S=4 V$

Solution:
$R_2$ and $R_3$ are parallel. The current of $I_S$ is passing through them and it is actually divided between them. The branch with lower resistance has higher current because electrons can pass through that easier than the other branch. Using the current division rule, we get
$V_S=10V$ , $I_S=4 A$ , $R_1=2 \Omega$ , $R_2=6 \Omega$ , $R_3=1 \Omega$ , $R_4=2 \Omega$ .