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I know that PSD is the FT of the auto-correlation function. Does this mean that if we multiply a signal with a real-valued coefficient (to alter its amplitude), the PSD of this new signal would be the original PSD of the signal, multiplied by the coefficient squared?

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he PSD of this new signal would be the original PSD of the signal, multiplied by the coefficient squared?

Yes.

It's easy enough to show. The autocorrelation of a discrete signal $x[n]$ is given as

$$r_{xx}[n] = \sum_k x[k]\cdot x[k+n]$$

If we scale $y[n] = a \cdot x[n]$ the autocorrelation of the scaled signal becomes

$$r_{yy}[n] = \sum_k y[k]\cdot y[k+n] = \sum_k a\cdot x[k]\cdot a\cdot x[k+n] = a^2 \cdot r_{xx}[n]$$

If the autocorrelation scales with $a^2$ so does the PSD since the Fourier Transform is a linear, i.e. $\mathbb{F}(a \cdot x[n]) = a \cdot \mathbb{F}(x[n])$

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