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An electrical waveform is an electrical quantity (such as voltage, current or power) that can vary over time. Waveforms can deliver power. For example, in the United States, the voltage across the terminals of a power outlet is a sinusoidal waveform with frequency 60 Hz, which is equivalent to a period of 1/60 s. Waveforms can also convey information. In lab, you use a waveform called pulse width modulation (PWM) to spin a motor at a desired speed. Waveforms that transmit information and are also called signals.

Figure 1

We can analyze circuits involving time-varying waveforms because Ohm's law, KVL, KCL and the power formulas are all valid at every instant of time. For example, if a voltage waveform $V(t)$ is applied across a resistor $R$, then the resistor's current waveform $I(t)$ and power waveform $P(t)$ satisfy the following equations.

\begin{align}

I(t) &= \frac{V(t)}{R} & & \text{at each time }t \\

P(t)&=V(t)I(t)=(I(t))^2 R=\frac{(V(t))^2}{R} & & \text{at each time }t

\end{align}

We often wish to summarize the overall behavior of a waveform by a single number. For a periodic power waveform of period $T$, we calculate the average power $P_{\text{avg}}$ as follows. First integrate $P(t)$ from $0$ to $T$ to obtain the "area under the curve" in one period. Then divide the result by $T$ to give the average (or mean) value during the period.

\begin{align}

P_{\text{avg}}&=\frac{1}{T} \int_0^T P(t) dt \label{WVF-API}

\end{align}

To summarize a periodic voltage or current waveform of period $T$, we do not use the average. Instead, we use the root mean square (RMS) voltage $V_{\text{rms}}$ or current $I_{\text{rms}}$. RMS is literally the square root of the mean of the square of the waveform.

\begin{align}

V_{\text{rms}}&=\sqrt{\frac{1}{T} \int_0^T (V(t))^2 dt} \label{WVF-VRM} \\

I_{\text{rms}}&=\sqrt{\frac{1}{T} \int_0^T (I(t))^2 dt} \label{WVF-IRM}

\end{align}

The reason for this definition is that it gives the average power across a resistor $R$ similar formulas as for the nonvarying case.

\begin{align}

P_{\text{avg}}=V_{\text{rms}}I_{\text{rms}}=\frac{(V_{\text{rms}})^2}{R}=(I_{\text{rms}})^2 R \label{WVF-PFS}

\end{align}

We derive $P_{\text{avg}}=(V_{\text{rms}})^2/R$ below. The other derivations are similar.

\begin{align}

P_{\text{avg}}&=\frac{1}{T} \int_0^T P(t) dt & & \text{by the definition of } P_{\text{avg}} \text{ in } \eqref{WVF-API} \\

&= \frac{1}{T} \int_0^T \frac{(V(t))^2}{R} dt & & \text{since } P(t)=\frac{(V(t))^2}{R} \text{ at each time }t \\

&= \frac{1}{R} \frac{1}{T} \int_0^T (V(t))^2 dt \\

&= \frac{(V_{\text{rms}})^2}{R} & & \text{by the definition of } V_{\text{rms}} \text{ in } \eqref{WVF-VRM}

\end{align}

Note that, in general, $V_{\text{avg}}$ and $I_{\text{avg}}$ have no relationship with $P_{\text{avg}}$.

PWM is a technique of delivering a specific amount of average power to a motor (or other device) so that it operates at a desired rate. We show that the average power delivered by PWM to a resistor is directly proportional to the PWM's duty cycle.

Figure 2

We begin evaluating $V_{\text{rms}}$ by plotting $(V(t))^2$ and visually integrating it over one period.

Figure 3

\begin{align}

V_{\text{rms}}&=\sqrt{\frac{1}{T} \int_0^T (V(t))^2 dt} & & \text{by the definition of } V_{\text{rms}} \text{ in } \eqref{WVF-VRM}\\

&= \sqrt{\frac{1}{T} A^2 T_{\text{on}}} & &\text{from Fig. 3} \\

&= A\sqrt{\frac{T_{\text{on}}}{T}} \label{WVF-PV2}

\end{align}

We now obtain average power $P_{\text{avg}}$ delivered to a resistor $R$ under the PWM voltage waveform $V(t)$ shown in Fig. 2.

\begin{align}

P_{\text{avg}}&= \frac{(V_{\text{rms}})^2}{R} = \left( A\sqrt{\frac{T_{\text{on}}}{T}} \right)^2 \frac{1}{R} = \frac{A^2}{R} \frac{T_{\text{on}}}{T} \label{WFV-PPW}

\end{align}

Therefore, the RMS of a PWM waveform is proportional to the square root of its duty cycle, while the average power it delivers to a resistor is directly proportional to its duty cycle.

Sinusoidal waveforms are common in electrical engineering. For example, an AC generator produces a sinusoidal voltage as a result of its rotational motion.

Figure 4

First we apply integration to show that $V_{\text{rms}}=A/\sqrt{2}$ for a sinusoidal voltage with amplitude $A$.

\begin{align}

V_{\text{rms}}&=\sqrt{\frac{1}{T} \int_0^T A^2 \sin^2 \left( \frac{2 \pi t}{T} \right) dt} & &\text{since } V(t)=A \sin \left( \frac{2 \pi t}{T} \right) \\

&= \sqrt{\frac{1}{T} \int_0^T A^2 \frac{1}{2} \left( 1-\cos\left( 2 \frac{2 \pi t}{T} \right) \right) dt} & &\text{since } \sin^2\theta=\frac{1}{2}(1-\cos 2\theta) \\

&= \frac{A}{\sqrt{2}} \sqrt{\frac{1}{T} \int_0^T \left( 1-\cos\left( \frac{4 \pi t}{T} \right) \right) dt} \\

&= \frac{A}{\sqrt{2}} \sqrt{\frac{1}{T} \left[ t- \frac{T}{4 \pi}\sin\left( \frac{4 \pi t}{T} \right) \right]_0^T} & &\text{since } \int \cos\theta d\theta=\sin\theta+C \\

&= \frac{A}{\sqrt{2}} \sqrt{\frac{1}{T} T} & &\text{since } \sin(4\pi)=\sin(0)=0 \\

&= \frac{A}{\sqrt{2}} \label{WVF-AS2}

\end{align}

Now that we know $V_{\text{rms}}$, we can obtain $P_{\text{avg}}$ absorbed by a resistor $R$ under the applied voltage $V(t)$ shown in Fig. 4.

\begin{align}

P_{\text{avg}}&= \frac{(V_{\text{rms}})^2}{R} = \left( \frac{A}{\sqrt{2}} \right)^2 \frac{1}{R} = \frac{A^2}{2R} \label{WFV-PSN}

\end{align}

Another way to determine $P_{\text{avg}}$ directly is to derive and plot $P(t)$ and then read off the average value $P_{\text{avg}}$ graphically.

\begin{align}

P(t)&=\frac{(V(t))^2}{R} \\

&=\frac{A^2}{R} \sin^2 \left( \frac{2 \pi t}{T} \right) & &\text{since } V(t)=A \sin \left( \frac{2 \pi t}{T} \right) \\

&=\frac{A^2}{R} \frac{1}{2} \left( 1-\cos\left( 2 \frac{2 \pi t}{T} \right) \right) & &\text{since } \sin^2\theta=\frac{1}{2}(1-\cos 2\theta) \\

&=\frac{A^2}{2R} - \frac{A^2}{2R} \cos\left( \frac{4 \pi t}{T} \right) \label{WVF-EQP}

\end{align}

Figure 5

The voltage across the terminals of a power outlet in the United States is referred to as 120 V. But this value is the RMS voltage of the sinusoidal waveform, so in fact the amplitude $A=120\sqrt{2}$, which is approximately 170 V.

Figure 6

Notice that $V_{\text{avg}}=0$ for this waveform and is therefore unrelated to $P_{\text{avg}}=A^2/(2R)$.

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