Part 10 - Steady-state sinusoidal analysis

In this part we’ll properly define and understand what alternating-current (AC) is.

Alternating-current

As we have seen, AC is defined using sinusoidal functions,

V(t) = V_m \sin(\omega t),

where,

\begin{align*} V_m & \quad \text{Amplitude } [\mathrm{V}] \newline \omega & \quad \text{Angular frequency } [\mathrm{rad/s}] \newline T & \quad \text{Period } [\mathrm{s}] \newline f & \quad \text{Frequency } [\mathrm{Hz}] \end{align*}

We can write,

T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}.

If we want to write this even more generally we also add a constant,

V(t) = V_m \sin(\omega t + \theta).

We say that this sine function leads the previous by $\theta$ . The other way around works as well, the previous function lags this function by $\theta$ .

Root Mean Square

The Root Mean Square, or RMS, gives us a sensible comparison of our AC current/voltage to the equivalent DC current/voltage.

Definition 1 (Root mean square)

This is useful for the power consumption - we define the RMS as,

V_{RMS} = \sqrt{\dfrac{1}{T} \int_0^T V^2(t) \, dt}.

Which means, in our sinusoidal case,

\begin{align*} V_{RMS} & = \dfrac{V_m}{\sqrt{2}}, \newline I_{RMS} & = \dfrac{I_m}{\sqrt{2}}. \end{align*}

Which means,

P_{Avg} = \dfrac{(V_{RMS})^2}{R} = (I_{RMS})^2 R.

Radians and degrees

If you’re like me and always forget how to convert radians to degrees or the other way around - here’s a quick little reminder,

\text{rad} = \text{deg} \cdot \dfrac{\pi}{180}.

Along with,

\begin{align*} \sin(\omega t) & = \cos(\omega t - 90^\circ), \newline \cos(\omega t) & = \sin(\omega t + 90^\circ). \end{align*}

This is really important since, if we want to compare the phases of two wave functions - we need to:

Both must be written in either sine or cosine.
Both must be written with positive amplitude.
Each must have the same constant frequency.

Phasors

Now we’re going to talk about this beautiful idea that happens to perfectly fit our electrical needs!

Definition 2 (Phasor)

A phasor is a complex number representing a sinusoidal function as a vector.

The magnitude of the phasor (meaning, the magnitude of the vector) equals the amplitude of the sinusoidal function:

|\mathbf{V}| = V_m.

The angle of the phasor equals the phase of the sinusoidal function:

\angle \mathbf{V} = \theta.

If you’re not familiar with complex numbers and their different representations, here is a quick overview:

Figure 1: The unit circle in the complex plane.

Note: In maths we usually denote the imaginary unit with $i$ , in electrical engineering, we rather use $j$ to avoid confusion with the current.

Rectangular form,

\begin{align*} z & = x + jy, \newline z & = |z|(\cos(\varphi) + j \sin(\varphi)). \end{align*}

Polar form,

\begin{align*} z & = |z| \angle \theta^\circ, \newline z & = |z| e^{j \theta^\circ}. \end{align*}

Polar to rectangular form,

\begin{align*} z & = r \angle \theta^\circ, \newline x & = r \cos(\theta^\circ), \newline y & = r \sin(\theta^\circ), \newline z & = x + j y. \end{align*}

Rectangular to polar form,

\begin{align*} z & = x + jy, \newline r & = \sqrt{x^2 + y^2}, \newline \theta & = \tan^{-1}\left(\dfrac{y}{x}\right) = \arctan\left(\dfrac{y}{x}\right), \newline z & = r \angle \theta^\circ, \newline z & = r e^{j \theta^\circ}, \newline z & = r(\cos(\theta) + j \sin(\theta)). \end{align*}

Why do we learn both forms and how to convert them? Because, if we want to perform multiplication or division - it’s much easier done in Polar form.

If we want to perform addition or subtraction, it’s easier done in rectangular form.

Multiplication,

a = (b \angle c^\circ) \cdot (d \angle e^\circ) = bd \angle (c^\circ + e^\circ).

Division,

a = \dfrac{b \angle c^\circ}{d \angle e^\circ} = \dfrac{b}{d} \angle (c^\circ - e^\circ).

Phasor relations

Now that we’ve defined phasors, let’s look at the relationship between different concepts we have encountered.

Resistance

As we know from Ohm’s law: $V = RI$ . If our $V$ and $I$ both are phasors - it means that they are in parallel:

Figure 2: Voltage and current phasors of a resistor.

Inductor

As we know, $V = L \dfrac{dI}{dt}$ . Now, if both our $V$ and $I$ are phasors, it means that taking the derivative of $I$ will make it either sine or cosine.

This means in the end that $V$ and $I$ will be $90^\circ$ apart!

Figure 3: Voltage and current phasors of an inductor.

As you may or may not know, but multiplying something with the imaginary unit rotates it by $90^\circ$ . This means that,

V = j\omega LI.

Capacitor

We know that, $I = C \dfrac{dV}{dt}$ . The same logic applies here, since taking the derivative of a sinusoidal function will shift it by $90^\circ$ this means that,

I = j\omega CV.

Figure 4: Voltage and current phasors of a capacitor.

Summary

This means that we now have two different ways to calculate these equations - in the time domain or in the frequency domain:

Time domain,

\begin{align*} V & = RI, \newline V & = L \dfrac{dI}{dt}, \newline V & = \dfrac{1}{C} \int I \, dt. \end{align*}

Frequency domain,

\begin{align*} V & = RI, \newline V & = j\omega LI, \newline V & = \dfrac{1}{j\omega C} I. \end{align*}

Impedance

The definition of impedance can be quite dense and technical - but in this series we’ll settle with the definition that,

Definition 3 (Impedance)

Z = \dfrac{V}{I}

This means that,

\begin{align*} Z_R & = R, \newline Z_L & = j\omega L, \newline Z_C & = \dfrac{1}{j\omega C}. \end{align*}

Impedance is therefore a complex number, but in the unit Ohm!

We call the imaginary component of impedance reactance.

A purely resistive impedance is equal to the resistance. A purely reactive impedance has no real component.

Impedances in series and parallel follow the same rules as resistors! So we can combine and simplify our circuits even more now!

But the most important thing here is that, we can, just from impedance, understand what dominates in a circuit:

\begin{align*} Z_{eq} & = 1.5 + \dfrac{(j)(1 - 2j)}{j + 1 - 2j} \newline & = 1.5 + \dfrac{2 + j}{1 - j} \newline & = 1.5 + \dfrac{(2 + j)(1 + j)}{(1 - j)(1 + j)} \newline & = 1.5 + \dfrac{1 + j3}{2} \newline & = 2.5 \angle 36.87^\circ \ \mathrm{k\Omega}. \end{align*}

From this, we can see that, since we have a positive impedance angle, means that it’s mostly inductive than anything else. This tells us a lot about the circuit.

Since inductors resist change in the current, it will ultimately mean that the current is late, or lags behind the voltage!

Which is exactly what we’ll see if we calculate $I(t)$ :

\begin{align*} I & = \dfrac{V}{Z_{eq}} \newline & = \dfrac{40 \angle -90^\circ}{2.5 \angle 36.87^\circ} \newline & = 16 \angle -126.9^\circ \ \mathrm{mA}. \end{align*}

Which means,

I(t) = 16 \cos(3000t - 126.9^\circ) \ \mathrm{mA}.

Just as we predicted, the current lags behind by a lot!

As we can see, our new tools and knowledge about how we can use complex numbers are quite powerful!