Spring 2018
Waveforms
Average of a Sawtooth Wave
Learn It!
Pre-Requisite Knowledge
Goal
Find the average of the waveform
Find the average (DC component) of the periodic sawtooth waveform shown below.
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)
\( A_{beige}= \frac{1}{2}(\frac{2}{3}T)(4) \)
\( A_{pink}= \frac{1}{2}(\frac{1}{3}T)(2) \) )
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
\(V_{ave}=\frac{1}{T} \left[\frac{6}{T}t-2t \right]_0^T\)
\(V_{ave}= \frac{1}{T}\frac{1}{2}\frac{6T^2}{T}-\frac{1}{T}2T=1\)
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.