Expert: Alon Mandes - 2/23/2009

Question
QUESTION: hello alon,
need some help with the following question.
Determine the fourier series of the signal f(t) = 1+t/pi for -pi<t<0.
1-t/pi for 0<t<pi.
any help would be greatly appreciated
thanks keith

ANSWER: The Fourier series for f(t) -π<t<π is ao/2 +∑a(n)cos(nt)+b(n)sin(nt)
Where n goes from 1 to ∞ , & :
ao= (1/π)∫f(t)dt.
a(n)= (1/π)∫f(t)cos(nt)dt.
b(n)= (1/π)∫f(t)sin(nt)dt.
The integration is from -π to π .

Ok, let's start :
ao= (1/π)∫1+(t/π) dt + (1/π)∫1-(t/π) dt
= (1/π)[t+(t²/2π)]|-π -> 0| + (1/π)[t-(t²/2π)]|0 -> π|
= (1/π)[0+π-π²/2] + (1/π)[π+π²/2-0] = 2

a(n)= (1/π)∫[1+(t/π)]cos(nt)dt+(1/π)∫[1-(t/π)]cos(nt)dt
   = (2/π)∫[1+(t/π)]cos(nt)dt ("because cos & f(t) are even funcn")
   = (2/π)∫cos(nt)dt + (2/π²)∫tcos(nt)dt
   =    0          + (2/π²)* [ cos(nt)+(nt)sin(nt) ]/n²
("The 1st part is zero because sin(0)=0 & sin(π)=0 , the 2nd part
was performed by 'integration by parts' ") Hence,
a(n)=(2/n²π²)* [ cos(πn)-1 ] Which can be also written as :
{-4/n²π² if n is odd & 0 if n is even}.
As for b(n) I will leave it for you to calculate it. you should get
b(n)=0 for all n. Therefore :
f(t) ~ 1 + ∑ )=(4/n²π²)* [cos(πn)-1]cos(nt) {n=1,3,5,7,9,11,...} . In another form:
f(t) ~ 1 + ∑ [4/(2n-1)²π²]cos(nt) {n=1,2,3,4,5,6,7,8,9,...}

Alon.



---------- FOLLOW-UP ----------

QUESTION: hello again alon,

thanks for your speedy answer, im trying to go through my books and understand how you found the answer for a(n).
would you mind going through it step by step so that i can see how the answer came about.
mant thanks keith


ANSWER: 1. a(n)=(1/π)∫[1+(t/π)]cos(nt)dt {t goes from -π<t<π} =
(1/π)∫[1+(t/π)]cos(nt)dt  {t goes from -π<t<0}   +
(1/π)∫[1-(t/π)]cos(nt)dt  {t goes from 0<t<π}
  = (2/π)∫[1+(t/π)]cos(nt)dt {t goes from 0<t<2π}
("because cos & f(t) are even function..Integral on an even
function over a symmetric region is twice the integral on the
positive region")
  = (2/π)∫cos(nt)dt + (2/π²)∫tcos(nt)dt
  =    0          + (2/π²)* [ cos(nt)+(nt)sin(nt) ]/n²
  =    1st          +  2nd
("The 1st part is zero because sin(0)=0 & sin(π)=0 , the 2nd part
was performed by 'integration by parts' ") Hence,
a(n)=(2/n²π²)* [ cos(πn)-1 ].
The form " [ cos(πn)-1 ] " can be written as :
" -2 if n is odd & 0 if n is even ".
Thus,
(2/n²π²)* [ cos(πn)-1 ]=(-4/n²π²) if n is odd & o if n is even.
Therefore,
f(t) ~ 1 + ∑ [4/(2n-1)²π²]cos(2n-1)t {n=1,2,3,4,5,6,7,8,9,11...}
Explanation :
If h(n) is a series in which n=1,3,5,7,9,11,13,...
Then we can claim that the function h(2n-1) has the same values
as h(n) only this time n=1,2,3,4,5,6,7,8,9,10,11,12,..

Alon.




---------- FOLLOW-UP ----------

QUESTION: hello again Alon,

with reference to the previous question I need to plot the signal f(t) and the sum of the first four terms of the fourier series on a graph.
I know that the signal f(t) is an even saw tooth/triangular wave however I am not to sure how to derive the first four terms of the fourier series from the answer given.
Could you show me how the series would be written when n = 1,3,5,7. and any pointers on plotting the sum of these terms on a graph.
many thanks keith

Answer
In terms of fourier series, our signal f(t) will be :
f(t) ~ 1 + [4/π²]cos(t) + [4/9π²]cos(3t) + [4/25π²]cos(5t) +
      [4/49π²]cos(7t) + [4/81π²]cos(9t) + .........
In order to graph the signal in fourier term which is also called
Tregonometrical Polynom , use Microsoft Excel data base, the only
variable here is Time.

Alon.