Find the Fourier Series of the Function

[CptJackWest]

Homework Statement

For the function f(t)={t+2, -2<t<0
{2, 0<t<2
{f(t+4) all t

Homework Equations

a(n)=2/T*Integration[f(t)cos(n*w*t)dt]

b(n)=2/T*Integration[f(t)sin(n*w*t)dt]

xcos(ax)= 1/a^2(cos(ax)+ax*sin(ax))

The Attempt at a Solution

T=4
w=2*pi/T
w=pi/2


a(n)=2/T*Integration[f(t)cos(n*w*t)dt]
a(n)=2/4*Integration[(t+2)cos((n*pi*t)/2)dt + 2/4*Integration[2cos((n*pi*t)/2)dt]]


That is as far as i got.
I know the next step is to integrate each term with respect to t, but not sure what to do with the (t+2) out the front. We were told to use the formula sheet and replace ( this the third formula above). Any help would be great
Cheers, Jack

 

[tiny-tim]

Hi Jack!

(have a pi: π and an omega: ω and an integral: ∫ )

xcos(ax)= 1/a^2(cos(ax)+ax*sin(ax))
…
… not sure what to do with the (t+2) out the front. We were told to use the formula sheet and replace ( this the third formula above).

Now find the equivalent formula for xsin(ax)

 

[CptJackWest]

Thanks for the help,
equivalent formula for xsin(ax)= 1/a^2(sin(ax)+ax*cos(ax))
But what do i do with the t+2 out the front of the cos((n*pi*t)/2).
That is the bit i need help with!

 

[tiny-tim]

Write x instead of t and use the formula!

 

[CptJackWest]

But won't it integrate differently because it is in the formula x is the same inside the cos as well as in front. But in my equation I have a t and a t+2.


xcos(ax)= 1/a^2(cos(ax)+ax*sin(ax))
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a(n)=2/4*Integration[(t+2)cos((n*pi*t)/2)dt + 2/4*Integration[2cos((n*pi*t)/2)dt]]
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