# Find DC Voltage of a Ramp, Step Wave

1. Sep 21, 2007

### djpardi

1. The problem statement, all variables and given/known data
-Find The DC voltage of the signal below created by the function generator
-Find RMS Voltage
-if the V waveform is applied across 1000 Ohm load what is the power dissipated in the load?

The wave form has axis symmetry which means only cosine terms

2. Relevant equations
If the signal was a unit step function only all I would need is the average value to determine the DC component. How do I get that value of a Unit Ramp Function?

Would it be an RMS value?

If so does Vrms=Vp/sqrt(2)?

Since there are several numbers of different wave forms would I get the DC component by using Fourier Series?

3. The attempt at a solution
Fourier Series:
The wave form has axis symmetry which means only cosine terms

The area on the top side is smaller than the area on the bottom side so the DC component will be negative.

If it was a pulse, the average voltage would be, Vav=(duty cycle)(peak value)+(1-duty cycle)(Vb)

If it was a sin it would be Vrms=Vp/sqrt(2)

Im not sure what the average voltage of a ramp wave is!

Im gonna guess, a0 = 1/T0* int(f(t)dt, t,-T0/2, +T0/2)

But I cant figure out how to apply it!

Last edited: Sep 22, 2007
2. Sep 22, 2007

### antonantal

If you are given a periodic signal, no matter what the shape of the signal is, the DC component is always the average value of the signal over a period.
If $$f(x)$$ is the function representing the signal waveform, there is a theorem which states that the average value of $$f(x)$$ over a period of $$T$$ is $$\frac{1}{T}\int_{T}f(x)dx$$

You can always use the Fourier series to find any component of a periodic signal. But if you look at the formula for calculating the first coefficient ( $$a_0$$ ) in the Fourier series you will see that it's exactly the integral above.

The integral can be calculated very simple if you know the relation between the integral of a function and the area between the graph of that function and the x-axis.

3. Sep 22, 2007

### DefaultName

What if you don't know the value of the period, but you know f(x) and you find dx? What if T is some random value, do you just leave your answer in terms of T?

4. Sep 22, 2007

### antonantal

Yes, if you don't know T, you can calculate the average value in terms of T. But in general, in problems you are given the period and in real life you can measure it.

5. Sep 22, 2007

### djpardi

ok great thanx for the quick response!

6. Sep 22, 2007

### djpardi

Ok, So I was in the right direction but im still a little confused. Through Fourier Series I get RMS to be the following

SQRT(1/To int([f(t)]^2dt

and I plug the Fourier Series Function f(t) into the above equation to get the DC voltage(Integral we initial discussed):

1/To int([A0]^2dt =Ao^2

Does that mean that in this case the DC voltage is 1 since Ao is 1? To me that doesn't make sense.

7. Sep 22, 2007

### antonantal

The Root Mean Square has nothing to do with the Fourier series.

The DC voltage is the mean value of the signal over a period not the root mean square.

f(t) is the signal. There is no such thing as Fourier Series Function.