Sum of two sinusoids

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Let $A_1 , A_2 , w_1, w_2 $ are positives numbers If for all real t , $s_1(t)=A_1\cos(w_1t+\phi_1)$ , $s_2(t)=A_2\cos(w_2t+\phi_2)$ and $s=s_1 +s_2$

We know if $w_1 = w_2$, we can write $s$ as $s(t)=A\cos(wt+\phi), \forall t\in \mathbb R$.

If $w_1\neq w_2$, I'm trying to prove that, we can't write $s$ as $s(t)=A\cos(wt+\phi), \forall t\in \mathbb R$.

My attempt was deleted (any comment)

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A Rigorous Mathematical Proof

If $\forall t > 0, A\cos(wt + \phi) = A_1\cos(w_1t + \phi_1) + A_2\cos(w_2t + \phi_2)$, then for $\forall x > 0$, \begin{equation} A \int_0^x \cos^2(wt + \phi)\, dt = A_1\int_0^x \cos(w_1t + \phi_1)\cos(wt + \phi)\, dt + A_2\int_0^x \cos(w_2t + \phi_2)\cos(wt + \phi)\, dt. \end{equation}

But, \begin{align*} A \int_0^x \cos^2(wt + \phi)\, dt &= A \int_0^x \frac{1+\cos(2wt + 2\phi)}{2}\, dt \\ &= \frac{A}{2} x + g_1(x), \end{align*} with a bounded function $g_1$ on $\mathbb{R}^*_+$.

If $w \neq w_1$ and $w \neq w_2$, then \begin{equation*} A_1\int_0^x \cos(w_1t + \phi_1)\cos(wt + \phi)\, dt + A_2\int_0^x \cos(w_2t + \phi_2)\cos(wt + \phi)\, dt = g_2(x), \end{equation*} with a bounded function $g_2$ on $\mathbb{R}^*_+$. We deduce that $\forall x > 0, \frac{A}{2} x + g_1(x) = g_2(x)$. This is possible only if $A = 0$, leading to a contradiction. Therefore, $w = w_1$ or $w = w_2$.

Now, if, for example, $w = w_1$, we have $$\forall t > 0, A\cos(wt + \phi) - A_1\cos(wt + \phi_1) = A_2\cos(w_2t + \phi_2)$$ But the fundamental period of $A\cos(wt + \phi) - A_1\cos(wt + \phi_1)$ is known as $\frac{2\pi}{w}$, and the fundamental period of $A_2\cos(w_2t + \phi_2)$ is $\frac{2\pi}{w_2}$. Therefore, we conclude that $w = w_2$.

The proof generalizes for the sum of n sinusoidal signals.

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To prove that you can't write $s(t) = A\cos(wt + \phi)$ when $w_1 \neq w_2$, let's start with the assumption that $s(t) = A\cos(wt + \phi)$ for all $t \in \mathbb{R}$.

First, we express $s(t)$ as a sum of two cosine functions:

$$ A\cos(wt + \phi) = A_1\cos(w_1t + \phi_1) + A_2\cos(w_2t + \phi_2) $$

Now, consider the high-frequency terms. Take the derivative of both sides with respect to $t$:

$$ -Aw\sin(wt + \phi) = -A_1w_1\sin(w_1t + \phi_1) - A_2w_2\sin(w_2t + \phi_2) $$

The left side represents a high-frequency oscillation with angular frequency $w$, while the right side is the sum of two high-frequency oscillations with angular frequencies $w_1$ and $w_2$.

If $w_1 \neq w_2$, there is no way to express the left side as a combination of the right-side terms, as their angular frequencies are different. This implies that you can't write $s(t)$ in the form $A\cos(wt + \phi)$ for all $t \in \mathbb{R}$ when $w_1 \neq w_2$.

Edit:

$$ \textbf{Claim:} \text{ If } w_1 \neq w_2, \text{ you cannot express } s(t) \text{ in the form } A\cos(wt + \phi) \text{ for all } t \in \mathbb{R}. $$

Proof:

  1. Assume that $s(t) = A\cos(wt + \phi)$ for all $t \in \mathbb{R}$.

  2. Take the derivative of both sides with respect to $t$: $$ s'(t) = -Aw\sin(wt + \phi) $$

  3. Now, express $s(t)$ as a sum of two cosine functions: $$ s(t) = A_1\cos(w_1t + \phi_1) + A_2\cos(w_2t + \phi_2) $$

  4. Take the derivative of this expression: $$ s'(t) = -A_1w_1\sin(w_1t + \phi_1) - A_2w_2\sin(w_2t + \phi_2) $$

  5. The left side, $s'(t)$, represents a high-frequency oscillation with angular frequency $w$, while the right side is the sum of two high-frequency oscillations with angular frequencies $w_1$ and $w_2$.

  6. If $w_1 \neq w_2$, the angular frequencies on the right side are different.

  7. To express $s'(t)$ as a combination of the right-side terms, it would be necessary to find $A_1$, $A_2$, $\phi_1$, $\phi_2$ such that the right side matches the left side for all $t$ in $\mathbb{R$.

  8. However, since $w_1 \neq w_2$, the high-frequency oscillations on the right side will have different angular velocities, and there will be no combination of $A_1$, $A_2$, $\phi_1$, $\phi_2$ that can make the right side match the left side for all $t$.

  9. This implies that you cannot write $s(t)$ in the form $A\cos(wt + \phi)$ for all $t \in \mathbb{R}$ when $w_1 \neq w_2$.