# Homework Help: Piecewise function - create function from graph

1. Nov 7, 2012

### DrOnline

1. The problem statement, all variables and given/known data

Hope I named the problem accurately.

I'm trying to calculate the average value, and the RMS value of this voltage.

3. The attempt at a solution

I've been given the graph, and I know how to calculate average and RMS. That's not the problem.

What I am wondering, is if my suggested functions are correct.

I have actually solved this before, but that was a year ago, and when I revisit it now, I am not 100% sure.

-t?
4-4t?
1-t?

Because I was told back then, I believe, that the function for the third domain, which I write is "-t", should be "4-4t" or something similar.

Can somebody explain to me why -t does not work, and why 4-4t works? I can see that if I continue that line from 4, past 3, and towards 0, it hits u=4. And since I want it to hit the t-line at t=4...

I kinda get it, but.. Still seems weird to me! I could nod in agreement if I saw 1-t as well.

Can somebody help me make sense of this?

2. Nov 7, 2012

### Ray Vickson

Do not "guess"---reason it out! Your u(t) is OK for t ≤ 3. What should it be for t between 3 and 4? Well, it should be linear, dropping from u(3) = 1 to u(4) = 0. So, what would be the equation of a line through (3,1) and (4,0)?

RGV

3. Nov 7, 2012

### DrOnline

I was banking heavily on guesswork, to be honest. ;)

Right. Well I did write -t. From the starting position of the third domain, (1,0), it goes to (0,1), from the perspective of that specific domain! Add one t, drop one u..

But I'm going with 4-4t, because that makes the function valid outside of the domain too.

Honestly, I just don't know for sure. It's not like I am desperate for somebody to solve my work here, I have enough done to hand this in, but I need to know how to think, for the future. I was wondering if somebody can explain the reasoning to me.

Does the function for each domain have to pass through the exact values of the domain, even if applied outside of the domain? If so, 4-4t.

I'm going with 4-4t, Mr. Vickson, is that good?

4. Nov 7, 2012

### Ray Vickson

You are are asking the wrong person: you should ask *yourself*. If you use u(t) = 4 - 4t, do you get u = 1 at t = 3? Do you get u = 0 at t = 4?

RGV

5. Nov 8, 2012

### DrOnline

Thanks. Well I reached the conclusion the right function was 4-t.

f(3) = 4-3 = 1
f(4) = 4-4 = 0

I don't rightly know why I was having so much problems with this, seems straight forward now.

6. Nov 8, 2012

### SammyS

Staff Emeritus
If the function is extended "backwards" to -4 ≤ t ≤ 0, then the third piece of that is f(t) = -t for -1 ≤ t ≤ 0 .

7. Nov 8, 2012

### DrOnline

Right. I understand. But what is your point? Sounds a bit dismissive, but what are you getting at?

8. Nov 8, 2012

### SammyS

Staff Emeritus
I didn't mean to be dismissive.

What meant by my previous post was you weren't that far off base when you initially used f(t) = -t for the third "piece" of you function.

9. Nov 9, 2012

### DrOnline

Oh, I meant was that perhaps what *I* write sounded dismissive, "But what is your point?". Hehe, misunderstanding.

Yes, I wasn't far off base, what I did was move the third part of the function to instead of go from t = 3 to 4, to t = 0 to 1, but I didn't realize that would cause problems with the definite integral.

Or indeed, just by testing f(t) = -t, input 3 or 4, I see now that it doesn't match the actual curve! I just lacked training.

I get it now. I just needed help to clear my mind or get some input on how to break this down, wouldn't want to mess up on something basic like this on the exams. My average and RMS calculations went fine.