# Fourier transform magnitude of the sum of two signals

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.

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

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.

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