Understanding Complex Integrals: A Step-by-Step Example with Truncated Waves

[Kazza_765]

I'm working through an example regarding the spectral content of a truncated wave, and came across this in the textbook.

$$\frac{1}{\sqrt{2\pi}}\int_{-T}^T cos(\omega_0-\omega)t + isin(\omega_0-\omega)t dt$$

Because the sine term is an odd function and the cosine is even, the integral reduces to

$$\frac{2}{\sqrt{2\pi}}\int_{0}^T cos(\omega_0-\omega)t dt$$

I have no idea what they mean by the cosine being even and the sine odd. If anyone can explain this step to me that would be great.

 

[assyrian_77]

Ok, I'll try to explain in simple terms. You know that an integral represents the area under the curve/graph, right? If the graph or parts of it are in the negative y-plane (under the x-axis), the area is negative. If a graph is symmetrical (looking the same on both sides of the y-axis), the function is called even. The cosine is like that (draw it and see for yourself). The sine, on the other hand, is odd, which means that one part of it is in the negative y-plane and the other in the positive. The two sections have the same area, but the one that is in the negative y-plane has a negative area and therefore, the areas cancel and the integral of is zero. Hopefully this helped some.

 

[shmoe]

If f(x) is an even function then f(x)=f(-x) for all x.
If f(x) is an odd function then f(x)=-f(-x) for all x.

For your question, on the interval [-T,0] do a change of variables to change t to -t and do some fiddling.

Also, think about the graphs of any odd function (sin(x) for example) on an interval symmetric about the origin, you should be able to 'see' the cancellation (similar for an even function)

 

[kahless2005]

Ordinary Diff. Eq.?

This looks a lot like a problem in my Diff. Eq. book. From that, an numeric constant r is found to be $a+bi$. Then the equation is translated to ${e^a(cos(b*t) + sin(b*t))}$.
from that, the equation can be shortned to ${e^a*cos(b*t)}$.


Im not sure if this applies here, but I hope this helps.

 

[Kazza_765]

Thankyou for the responses. I've got it now.