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where does that exponential term gone, is this because it is a constant term or it has to do something with stablity?

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  • $\begingroup$ Lol really? Imagine it's a vector with length 1, no matter the angle it will always be 1. 😂 $\endgroup$
    – gurghet
    Commented Nov 4, 2017 at 19:45
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    $\begingroup$ @gurghet be nice. $\endgroup$ Commented Nov 4, 2017 at 21:43

2 Answers 2

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The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as $z = r e^{j \phi}$ where $|z|=r$ is its magnitude and $\arg z = \phi$ is the argument. Using this note that

$$ |e^{-j\Omega \lambda}| = 1 $$

which is why it "disappeared".

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Simply because, when $C$ is a real number, $|e^{jC}|=1$. Here, $C=\Omega\lambda$, apparently real numbers.

Another interpretation is: if you change the phase of a signal (like a mere time-shift), it turns into a constant modulus in the Fourier domain, and provides some invariance, used for instance recently in data classification with Invariant Scattering Convolution Networks.

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