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I'm representing some function using complex Fourier series and I have to solve this integral:

$\int_{0}^{T} e^{-jnwt}dt$, where $w=\frac{2\pi}{T}$

I got this:

$\int_{0}^{T} e^{-jnwt}dt=...=-\frac{1}{jnw}e^{-jnwT}+\frac{1}{jnw}=\frac{1}{jnw}-\frac{1}{jnw}(\cos(2jn\pi)-\sin(2jn\pi))$

My question is, how cosine and sine terms in brackets behave for different n (n goes from $-\infty$ to $n=+\infty$)?

EDIT: I made mistake, there is of coure no imaginary j as argument of sine and cosine. I'm not concentrated, it's too late :)

hari
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1 Answers1

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You made an error simplifying the result of the integral. You should have gotten:

$\displaystyle\int_{0}^{T}e^{jnwt}\,dt$ $= -\dfrac{1}{jnw}e^{-jnwT}+\dfrac{1}{jnw} $ $= \dfrac{1-e^{-jnwT}}{jnw}$ $= \dfrac{e^{-jnwT/2}(e^{jnwT/2}-e^{-jnwT/2})}{jnw}$ $= \dfrac{2e^{-jnwT/2}\sin(nwT/2)}{nw}$.

The magnitude of the Fourier coefficients is $\dfrac{2\sin(nwT/2)}{nw}$. As $|n| \to \infty$, this tends to $0$ like $n^{-1}$. Due to the sin term, there could be some oscillations.

JimmyK4542
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