Integral Approximation: Tau <<T

AI Thread Summary
When tau is much less than T, the discussion focuses on the validity of two integral relations involving a function v(t) and exponential terms. It is emphasized that proving one relation requires demonstrating the periodicity of the integrand with respect to T/(1+a). A substitution for v(t) leads to a simplified integral form, which is suggested to be sufficient for establishing the equality. The conclusion indicates that if the final integral form is correct, then the equality holds true.
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Hello,

If tau<<T which of the following relations are true:

\int_{\tau/(1+a)}^{(T+\tau)/(1+a)}v(t)e^{-j2\pi f_0 at}e^{-j2\pi\frac{k}{T}t[1+a]}\,dt=\int_{0}^{T/(1+a)}v(t)e^{-j2\pi f_0 at}e^{-j2\pi\frac{k}{T}t[1+a]}\,dt

or

\int_{\tau/(1+a)}^{(T+\tau)/(1+a)}v(t)e^{-j2\pi f_0 at}e^{-j2\pi\frac{k}{T}t[1+a]}\,dt\simeq\int_{0}^{T/(1+a)}v(t)e^{-j2\pi f_0 at}e^{-j2\pi\frac{k}{T}t[1+a]}\,dt

Thanks
 
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Your assumtion is not enough to show one of the relations is true. It seems you need to show that the integrand is periodic with T/(1+\alpha).
 
jashua said:
Your assumtion is not enough to show one of the relations is true. It seems you need to show that the integrand is periodic with T/(1+\alpha).

OK, the integral after substituting for v(t) will look like:

\int_{\tau/(1+a)}^{(T+\tau)/(1+a)}e^{j2\pi\frac{m-k}{T}t[1+a]}\,dt

Is that enough to tell now?

Thanks
 
If that is the final version of the integral then it is enough to say that the equality holds.
 
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