Fourier coefficients of cos(4πt)

[Jncik]

Homework Statement

suppose that we have the continuous time signal

x(t) = cos(4πt) with fundamental period of T=1/2

Homework Equations

a_{k} = \frac{1}{T} \int_{T}{}x(t)e^{-j\omega_{0}kt}dt

where \omega_{0} is obviously \frac{2\pi}{1/2} = 4\pi

well the problem is that this integration becomes

<br /> a_{k} = \frac{1}{T} \int_{T}{}x(t)e^{-j\omega_{0}kt}dt =<br /> 2\int_{0}^{\frac{1}{2}}cos(4 \pi t)e^{-j4 \pi kt}dt=<br /> ...=0<br />

I find as a result 0, the calculation is complicated but using wolfram alpha's calculator I have the same result

here is the picture

[PLAIN]http://img52.imageshack.us/img52/6729/36554157.gif

as you can see

we have

1 - e^{-2\pi j k} = 1 - ( cos(2\pi k) - jsin(2\pi k)) = 1 - 1 = 0

hence it must be 0
now my book says the following

The nonzero FS coefficients of x(t) are a_{1} = a_{-1} = 1/2

but I have 0 as a result, and if I plug 1 or -1 I will get 0 again, what's wrong here?

thanks in advance

 

[tiny-tim]

Hi Jncik!

πj²1² + π = πj²(-1)² + π = 0

 

[Jncik]

thanks tiny-tim, I didn't notice that :)

 

[LCKurtz]

And the numerator is 0 also, giving an indeterminate form. You can best see the problem if you calculate the sine/cosine coefficients. You will see that the integrals for a_n and b_n have a different form when n = 1, requiring a separate and easier integration. If you work out the complicated formulas for other n and plug n = 1 in those answers, you will get the same of indeterminate answer.

 

[Jncik]

And the numerator is 0 also, giving an indeterminate form. You can best see the problem if you calculate the sine/cosine coefficients. You will see that the integrals for a_n and b_n have a different form when n = 1, requiring a separate and easier integration. If you work out the complicated formulas for other n and plug n = 1 in those answers, you will get the same of indeterminate answer.

thanks what I did for n = 1 first I tried to find out the result using the integration but I used these two formulas

2\int_{0}^{\frac{1}{2}}cos(4 \pi t)e^{-4 \pi t} dt = 2\int_{0}^{\frac{1}{2}}cos(4 \pi t)cos(4 \pi t) dt - 2j\int_{0}^{\frac{1}{2}}cos(4 \pi t)sin(4 \pi t) dt = ...

because as you said it wouldn't work using just the integration without expanding the form of e^{-j 4 \pi t}

sadly our professor doesn't want the trigonometric expansion but prefers the series with the e^...

after solving this I thought it would be better if I just used Euler's formula about the cos(4πt)

it is known that

cos(4\pi t) = \frac{e^{j 4 \pi t} + e^{-j 4 \pi t}}{2}

hence the coefficients are a_{1} = a_{-1} = \frac{1}{2}

but I am not sure if this was what the exercise wanted cause this solution takes less than 5 seconds to finish

 

[tiny-tim]

… but I am not sure if this was what the exercise wanted cause this solution takes less than 5 seconds to finish

ruined your day, did it? ​

 

[Jncik]

ruined your day, did it? ​

haha yes :p

 

[LCKurtz]

after solving this I thought it would be better if I just used Euler's formula about the cos(4πt)(

it is known that

cos(4\pi t) = \frac{e^{j 4 \pi t} + e^{-j 4 \pi t}}{2}

hence the coefficients are a_{1} = a_{-1} = \frac{1}{2}

but I am not sure if this was what the exercise wanted cause this solution takes less than 5 seconds to finish

On a quiz I would expect that is exactly what the professor wanted. The point being for the student to recognize that trig functions like that are their own (finite) Fourier Series.