Piecewise Function Homework: Solve Fourier Coefficient b1

AI Thread Summary
The discussion revolves around finding the Fourier coefficient b1 for a given periodic signal, with a focus on correctly defining the piecewise function. The user expresses uncertainty about their function definition and the impact of shifting the time variable. They initially define the function as f(t) = 1 for 0<t<1 and f(t) = t for 1<t<2, but later realize that the intervals between 2 and 4 should also be considered, even if they integrate to zero. A suggestion is made to redefine the function using the original time intervals from 1 to 4 to ensure accuracy in the Fourier series calculation. Ultimately, the user aims to clarify their function to correctly compute the Fourier coefficient b1.
DrOnline
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Homework Statement


Given the following periodic signal:
2ztbmu8.jpg


What is the Fourier coefficient b1.

I'm not asking for help with the Fourier series. I feel my integration is solid.

What I need help with is creating the correct function!

The Attempt at a Solution



I first declare t=5 as now to be t=0, and T=6 becomes t=2, etc. Am I allowed to do this? It also seems a bit iffy because the period starts at t=4...

f(t) =
1, 0<t<1
t, 1<t<2

This gives me, as far as I can tell, the correct values for 0<t<2, compared to the graph.

Unsure about the "t" function there...

I am doubting whether this is correct, me and a friend have done the series 4-5 times now using this function, and we get the wrong answer, so we think the function we wrote is wrong.

Can somebody help me make sense of this?
 
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f(t) =
1 for t in the interval of [1+4(t-1), 2+4(t-1)]
etc.
 
DrOnline said:

Homework Statement


Given the following periodic signal:
2ztbmu8.jpg


What is the Fourier coefficient b1.

I'm not asking for help with the Fourier series. I feel my integration is solid.

What I need help with is creating the correct function!

The Attempt at a Solution



I first declare t=5 as now to be t=0, and T=6 becomes t=2, etc. Am I allowed to do this? It also seems a bit iffy because the period starts at t=4...

f(t) =
1, 0<t<1
t, 1<t<2

Don't forget ##f(t)=0,~2<t<4##.

This gives me, as far as I can tell, the correct values for 0<t<2, compared to the graph.

Unsure about the "t" function there...

I am doubting whether this is correct, me and a friend have done the series 4-5 times now using this function, and we get the wrong answer, so we think the function we wrote is wrong.

Can somebody help me make sense of this?

I don't understand the ##1.5## on your picture. But if you calculate the FS for the above function on ##(0,4)## extended periodically, it will converge to the required graph. But the function you are using and the one given are not the same since they disagree for the same value of ##t##. One is a translated version of the other and the FS may not look the same.
 
The 1.5 is the average value of the interval from "my" 1 to 2, I wrote that to demonstrate how to calculate the average of the function value to a friend.

You tell me to not forget the interval between 2 and 4, but they do not matter, because they will integrate to zero, this is how I see it. I know it is an inaccurate way when I describe the piecewise function, but I only listed the ones relevant for the FS. I do account for them in the period variables when I do the calculations.

My task is only to find the Fourier coefficient b1, so I don't know if that is affected by me messing with the times.

Tomorrow I think I will re-do this and using the times actually on the paper, use the first period, 1 to 4.

f(t)=
0, 0<t<1
1, 1<t<2
-1+t, 2<t<3
0, 3<t<4

I think I understand what you mean about messing with the times, perhaps it make the actual FS not match the actual graph as asked for in the task.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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