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Let

$$\mathscr{F}\Big\{x_1(t)+x_2(t)\Big\}=X_1(f)+X_2(f)$$

I think that in general

$$\big|X_1(f)+X_2(f)\big|^2\leq\big|X_1(f)\big|^2+\big|X_2(f)\big|^2$$

but I was wondering if

$$X_1(f)X_2(f)=0,\qquad\forall f$$

it is true that

$$\big|X_1(f)+X_2(f)\big|=\big|X_1(f)\big|+\big|X_2(f)\big|$$

so one can find the Fourier transform magnitude of the sum of the signals if the individual magnitudes are known.

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  • $\begingroup$ @MattL. for the inequation its square magnitudes and for the equation its just magnitudes. $\endgroup$
    – giannisl9
    Nov 24, 2023 at 10:48
  • $\begingroup$ @MattL. may you please elaborate? $\endgroup$
    – giannisl9
    Nov 24, 2023 at 10:50
  • $\begingroup$ @MattL. thank you, I am looking forward to it! $\endgroup$
    – giannisl9
    Nov 24, 2023 at 11:00
  • $\begingroup$ You are wrong on the vector triangle inequality. In last equation square notation is still there. $\endgroup$
    – abhilash
    Nov 24, 2023 at 12:30

1 Answer 1

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In general we have

$$\big|X_1(f)+X_2(f)\big|\neq \big|X_1(f)\big|+\big|X_2(f)\big|\tag{1}$$

However, the condition $X_1(f)X_2(f)=0$ $\forall f$ implies that for any $f$, either $X_1(f)$ or $X_2(f)$ or both must be zero. Note that this is the case even though $X_1(f)$ and $X_2(f)$ are generally complex-valued.

So for the condition $X_1(f)X_2(f)=0$, the inequality $(1)$ becomes an equality because either one of the two functions or both vanish.

Let's use $F_1$ to denote the frequency region where $X_1(f)=0$, and similarly for $F_2$ and $X_2(f)$. Then we have

$$\big|X_1(f)+X_2(f)\big|=\begin{cases}\big|X_1(f)\big|,&f\in F_2\\\big|X_2(f)\big|,&f\in F_1\end{cases}$$

Hence, for $X_1(f)X_2(f)=0$, $f\in F_1\cup F_2$, the inequality $(1)$ trivially becomes an equality because at least one of the summands on the right-hand side of $(1)$ is zero.

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  • $\begingroup$ What if they are orthogonal signals? $\endgroup$
    – abhilash
    Nov 24, 2023 at 12:19
  • $\begingroup$ Orthogonality does not imply that the product is zero. Just that the sum or integral over the product is zero. $\endgroup$
    – Hilmar
    Nov 24, 2023 at 12:45

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