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I'm trying to apply Fourier analysis to a specific problem I have. I have essentially an integral like the following

$$ \int_{\Omega} f(t) g(t) dt $$

And I'm trying to assume that $g$ is a narrow band signal (namely the Fourier transform is defined on a compact set). I want to prove that if $g$ is really narrow band then

$$ \int_{\Omega} f(t) g(t) dt \approx \left( \int_{\Omega} f(t) dt \right) \left( \int_{\Omega} g(t) dt \right) $$

To prove this, I made this assumption

$$ g(t) = g_a(t) = \frac{1}{a} \int_{-\infty}^{+\infty} \hat{g}(\omega)rect \left( \frac{\omega}{a} \right) e^{-j2\pi \omega t} d \omega $$

Where $a$ is a positive real parameter. When $a \to 0$ this integral turn out to be

$$ \frac{1}{a} \int_{-\infty}^{+\infty} \hat{g}(\omega)rect \left( \frac{\omega}{a} \right) e^{-j2\pi \omega t} d \omega \to \hat{g}(0) $$

which means

$$ g(t) = g_a(t) \to \hat{g}(0) = \int_{-\infty}^{+\infty} g(t) dt $$

and therefore

$$ \int_{\Omega} f(t) g(t) dt \approx \int_{\Omega} f(t) \hat{g}(0) dt = \hat{g}(0) \int_{\Omega} f(t) dt = \int_{-\infty}^{+\infty} g(t) dt \int_{\Omega} f(t) dt $$

which is not exactly what I want , but it's close enough. The final question is:

is the model

$$ g_a(t) = \frac{1}{a} \int_{-\infty}^{+\infty} \hat{g}(\omega)rect \left( \frac{\omega}{a} \right) e^{-j2\pi \omega t} d \omega $$

the correct model for a narrow band signal? I'm not entirely sure because of the factor $1/a$ I've introduced. But the idea why it makes sense is the following.

If the signal has narrow band, that means the variation in time is really slow, that means the narrower is the band the more the signal tend to be constant, and in theory if it would be constant then I'd be able to factorize the function $g(t)$ from the integral.

Clarification : probably in this discussion is more correct to say "low frequency".

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  • $\begingroup$ Note that in general "narrowband" does not imply slow time variation. That's only the case where you have a signal that is both narrowband and baseband (its spectrum exists around $f=0$), which would make it highly correlated with DC. A narrowband passband signal does not have to change slowly over time. $\endgroup$ – MBaz Feb 14 '18 at 17:26
  • $\begingroup$ Sure, I'll clarify what I meant. I meant base band anyway, or low frequency. $\endgroup$ – user8469759 Feb 14 '18 at 17:28
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Note that you can also use Parseval's theorem. Assuming that $f(t)$ and $g(t)$ are both real-valued we have:

$$\int_{-\infty}^{\infty}f(t)g(t)dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)G^*(\omega)d\omega\tag{1}$$

If $g(t)$ is a narrow-band low-pass signal, i.e., if we can model $G(\omega)$ as $G(\omega)\approx G(0)\text{rect}(\omega/\Delta\omega)$, and if we can assume that $F(\omega)$ is approximately constant (equal to $F(0)$) in the narrow interval $\Delta\omega$, then we can write

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)G^*(\omega)d\omega\approx\frac{1}{2\pi}F(0)G^*(0)\Delta\omega\tag{2}$$

where $\Delta\omega$ is twice the bandwidth of $g(t)$ (because the bandwidth is defined as the support at positive frequencies). With

$$F(0)=\int_{-\infty}^{\infty}f(t)dt$$

and

$$G^*(0)=G(0)=\int_{-\infty}^{\infty}g(t)dt$$

we obtain from $(1)$ and $(2)$

$$\int_{-\infty}^{\infty}f(t)g(t)dt\approx\frac{\Delta\omega}{2\pi}\int_{-\infty}^{\infty}f(t)dt\int_{-\infty}^{\infty}g(t)dt\tag{3}$$

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  • $\begingroup$ How did you get (2)? $\endgroup$ – user8469759 Feb 14 '18 at 17:42
  • $\begingroup$ Also, any chance it can be derived something similar where the integration region is a symmetric set of the form $[-t_0,t_0]$? $\endgroup$ – user8469759 Feb 14 '18 at 17:44
  • $\begingroup$ @user8469759: Eq (2) just follows from the fact that if the bandwidth is very small, then both functions can be approximated (in that narrow interval) by their value at $\omega=0$. Integrating from $-t_0$ to $t_0$ won't work because if $g(t)$ is a narrow band signal, it will have a large support in the time domain. $\endgroup$ – Matt L. Feb 14 '18 at 17:51
  • $\begingroup$ Still not sure I'm following, are you applying any calculus theorem? That's more of my question. I get the intuition, But I can think many cases where that approx would fail. $\endgroup$ – user8469759 Feb 14 '18 at 18:07
  • $\begingroup$ @user8469759: You just model $G(\omega)$ as $G(0)\cdot\text{rect}(\omega/\Delta\omega)$, and you approximate $F(\omega)$ as constant (=$F(0)$) in that narrow interval, nothing fancy. $\endgroup$ – Matt L. Feb 14 '18 at 18:43

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