Circuit Theory/Phasor Analysis

Phasor Analysis

The mathematical representations of individual circuit elements can be converted into phasor notation, and then the circuit can be solved using phasors.

Resistance, Impedance and Admittance

In phasor notation, resistance, capacitance, and inductance can all be lumped together into a single term called "impedance". The phasor used for impedance is $\mathbb {Z}$. The inverse of Impedance is called "Admittance" and is denoted with a $\mathbb {Y}$. $\mathbb {V}$ is Voltage and $\mathbb {I}$ is current.

$\mathbb {Z} ={\frac {1}{\mathbb {Y} }}$

And the Ohm's law for phasors becomes:

$\mathbb {V} =\mathbb {Z} \mathbb {I} ={\frac {\mathbb {I} }{\mathbb {Y} }}$

It is important to note at this point that Ohm's Law still holds true even when we switch from the time domain to the phasor domain. This is made all the more amazing by the fact that the new term, impedance, is no longer a property only of resistors, but now encompasses all load elements on a circuit (capacitors and inductors too!).

Impedance is still measured in units of Ohms, and admittance (like Conductance, its DC-counterpart) is still measured in units of Siemens.

Let's take a closer look at this equation:

[Ohm's Law with Phasors]

$\mathbb {V} =\mathbb {Z} \mathbb {I}$

If we break this up into polar notation, we get the following result:

$M_{V}\angle \phi _{V}=(M_{Z}\times M_{I})\angle (\phi _{Z}+\phi _{I})$

This is important, because it shows that not only are the magnitude values of voltage and current related to each other, but also the phase angle of their respective waves are also related. Different circuit elements will have different effects on both the magnitude and the phase angle of the voltage given a certain current. We will explore those relationships below.

Resistors

Resistors do not affect the phase of the voltage or current, only the magnitude. Therefore, the impedance of a resistor with resistance R is:

[Resistor Impedance]

$\mathbb {Z} =R\angle 0$

Through a resistor, the phase difference between current and voltage will not change. This is important to remember when analyzing circuits.

Capacitors

A capacitor with a capacitance of C has a phasor value:

[Capacitor Impedance]

$\mathbb {Z} =C\angle \left(-{\frac {\pi }{2}}\right)$

To write this in terms of degrees, we can say:

$\mathbb {Z} =C\angle (-90^{\circ })$

We can accept this for now as being axiomatic. If we consider the fact that phasors can be graphed on the imaginary plane, we can easily see that the angle of $-\pi /2$ points directly downward, along the negative imaginary axis. We then come to an important conclusion: The impedance of a capacitor is imaginary, in a sense. Since the angle follows directly along the imaginary axis, there is no real part to the phasor at all. Because there is no real part to the impedance, we can see that capacitors have no resistance (because resistance is a real value, as stated above).

Reactance

A capacitor with a capacitance of C in an AC circuit with an angular velocity $\omega$ has a reactance given by

$\mathbb {X} ={\frac {1}{\omega C}}\angle (-90^{\circ })$

Reactance is the impedance specific to an AC circuit with angular velocity $\omega$.

Inductors

Inductors have a phasor value:

[Inductor Impedance]

$\mathbb {Z} =L\angle \left({\frac {\pi }{2}}\right)$

Where L is the inductance of the inductor. We can also write this using degrees:

$\mathbb {Z} =L\angle (90^{\circ })$

Like capacitors, we can see that the phasor for inductor shows that the value of the impedance is located directly on the imaginary axis. However, the phasor value for inductance points in exactly the opposite direction from the capacitance phasor. We notice here also that inductors have no resistance, because the resistance is a real value, and inductors have only an imaginary value.

Reactance

In an AC circuit with a source angular velocity of $\omega$, and inductor with inductance L.

$\mathbb {X} =\omega L\angle (90^{\circ })$

Impedances Connected in Series

If there are several impedances connected in series, the equivalent impedance is simply a sum of the impedance values:

----[ Z1 ]----[ Z2 ]--- ... ---[ Zn ]---   ==> ---[ Zseries ]---

[Impedances in Series]

$\sum _{series}\mathbb {Z} _{n}=\mathbb {Z} _{series}$

Notice how much easier this is than having to differentiate between the formulas for combining capacitors, resistors, and inductors in series. Notice also that resistors, capacitors, and inductors can all be mixed without caring which type of element they are. This is valuable, because we can now combine different elements into a single impedance value, as opposed to different values of inductance, capacitance, and resistance.

Keep in mind however, that phasors need to be converted to rectangular coordinates before they can be added together. If you know the formulas, you can write a small computer program, or even a small application on a programmable calculator to make the conversion for you.

Impedances in Parallel

Impedances connected in parallel can be combined in a slightly more complicated process:

[Impedances in Parallel]

$\mathbb {Z} _{parallel}={\frac {\prod _{N}Z_{n}}{\sum _{N}Z_{n}}}$

Where N is the total number of impedances connected in parallel with each other. Impedances may be multiplied in the polar representation, but they must be converted to rectangular coordinates for the summation. This calculation can be a little bit time consuming, but when you consider the alternative (having to deal with each type of element separately), we can see that this is much easier.

Steps For Solving a Circuit With Phasors

There are a few general steps for solving a circuit with phasors:

1. Convert all elements to phasor notation
2. Combine impedances, if possible
3. Combine Sources, if possible
4. Use Ohm's Law, and Kirchoff's laws to solve the circuit
5. Convert back into time-domain representation

Unfortunately, phasors can only be used with sinusoidal input functions. We cannot employ phasors when examining a DC circuit, nor can we employ phasors when our input function is any non-sinusoidal periodic function. To handle these cases, we will look at more general methods in later chapters

Network Function

The network function is a phasor, $\mathbb {H}$ that is a ratio of the circuit's input to its output. This is important, because if we can solve a circuit down to find the network function, we can find the response to any sinusoidal input, by simply multiplying by the network function. With time-domain analysis, we would have to solve the circuit for every new input, and this would be very time consuming indeed.

Network functions are defined in the following way:

[Network Function]

$\mathbb {H} ={\frac {\mathbb {Y} }{\mathbb {X} }}$

Where $\mathbb {Y}$ is the phasor representation of the circuit's output, and $\mathbb {X}$ is the representation of the circuit's input. In the time domain, to find the output, we would need to convolute the input with the impulse response. With the network function, however, it becomes a simple matter of multiplying the input phasor with the network function, to get the output phasor. Using this method, we have converted an entire circuit to become a simple function that changes magnitude and phase angle.

Gain

Gain is the amount by which the magnitude of the sinusoid is amplified or attenuated by the circuit. Gain can be computed from the Network function as such:

[Gain]

$Gain=\left|\mathbb {H} (\omega )\right|={\frac {\left|\mathbb {Y} (\omega )\right|}{\left|\mathbb {X} (\omega )\right|}}$

Where the bars around the phasors are the "magnitude" of the phasor, and not the "absolute value" as they are in other math texts. Again, gain may be a measure of the magnitude change in either current or voltage. Most frequently, however, it is used to describe voltage.

Phase Shift

The phase shift of a function is the amount of phase change between the input signal and the output signal. This can be calculated from the network function as such:

[Phase Shift]

$\angle \mathbb {H} (\omega )=\angle \mathbb {Y} (\omega )-\angle \mathbb {X} (\omega )$

Where the $\angle$ denotes the phase of the phasor.

Again, the phase change may represent current or voltage.