Waveforms

Learn It!

Goal

What is the average (DC) value of the periodic waveform, with two periods shown to the left?

Method 1

Find average using areas

Average value=(Area above x axis - Area below x axis)/(Period)

Since we want to avoid doing calculus if we can. The integral form of finding
the average value can instead be done using geometry (for simple signals)

\( A_{beige}= \frac{1}{2}(\frac{2}{3}T)(4) \)

\( A_{pink}= \frac{1}{2}(\frac{1}{3}T)(2) \) )

\( A_{pink}= \frac{1}{2}(\frac{1}{3}T)(2) \) )

Find the area of the pieces of the signal .

\(V_{ave}=\frac{Beige-pink}{T}\)
\(V_{ave}=\frac{\frac{1}{2}(\frac{2}{3}T)(4)-\frac{1}{2}(\frac{1}{3}T)(2)}{T} \)
\(V_{ave}=\frac{\frac{4}{3}T - \frac{1}{3}T}{T}=\frac{3T}{T}=1\)

Apply the formula for averages

★

\(V_{ave}=1\)

The correct answer is \(1V\)

Method 2

Use similarity to known signal

We can also try to avoid doing so much mathematics. This signals looks an awful
lot like a more symmetrical one, the standard sawtooth wave. In fact, just shifting a standard sawtooth wave
up by one volt gives us the signal in the problem.

Average of Shifted signal=Average of base signal+Shift
\( (F(t)+DC)_{ave}=\underbrace{F(t)_{ave}}_{0 for sawtooth wave}+DC\)

Since a normal sawtooth wave has no DC component, all of the DC component comes from
the shift, which is up by \(1V\).

★

\(V_{ave}=1V\)

Average value=shift from baseline

Method 3

Use Calculus

\(V=\frac{6}{T}t-2\)

If you want to, calculus works well. Since we're not given a functional
form for the waveform, we must construct one. This is a linear function, so it is of the form
\(y=mx+b\). The intercept is \(-2\),the slope is rise over run, which is \(6\) over \(T\)

\(V_{ave}=\frac{1}{T}\int_0^T{\frac{6}{T}t-2}\)

Using the integral definition of average value of a function

\(V_{ave}=\frac{1}{T} \left[\frac{6}{T}t-2t \right]_0^T\)

\(V_{ave}= \frac{1}{T}\frac{1}{2}\frac{6T^2}{T}-\frac{1}{T}2T=1\)

\(V_{ave}= \frac{1}{T}\frac{1}{2}\frac{6T^2}{T}-\frac{1}{T}2T=1\)

Integrals don't get any easier than this. Note that the \(T\) must ALWAYS cancel out.
The average value of a waveform does not depend on the period, only on the shape.

★

\(V_{ave}=1V\)

We get the same result a third time.

There are many ways to solve most problem. But not all of them are equally fast or easy! Recognizing
a common structure in a problem will save a lot of time.