Given the following equation I have to find $a$ and $\phi$:

$$a_1 \cos(\omega t + \phi_1) + a_2 \cos(\omega t +\phi_2) = a \cos(\omega t + \phi)$$

Is there an easy way to construct $a$ and $\phi$? I just did not manage find one at all...

This is what I've got so far:

A substitution $b_1 = \phi_1 - \phi$ and $b_2 = \phi_2 - \phi$ and $x = \omega t + \phi$ yields

$$a \cos(x) = a_1 \cos(x+b_1) + a_2 \cos(x+b_2)$$

Considering this as real parts of phasors we get the equation

$$a e^{ix} = a_1 e^{ix+ib_1}+a_2 e^{ix+ib_2}$$

Here we can cancel $e^{ix}$, which gets us to :

$$a = a_1 e^{ib_1}+a_2 e^{ib_2}$$

But I have no clue how to go from there, so is this the right way, or did I go in a completelx wrong direction?


1 Answer 1


$$a_1 \cos (\omega t+ \phi_1)+a_2 \cos (\omega t+ \phi_2)=a \cos (\omega t+ \phi)$$ $$\Rightarrow a_1(\cos \omega t \cos\phi_1 - \sin\omega t \sin\phi_1)+a_2(\cos \omega t \cos\phi_2 - \sin\omega t \sin\phi_2)=a(\cos \omega t \cos\phi - \sin\omega t \sin\phi)$$ $$\Rightarrow (a_1 \cos\phi_1 + a_2 \cos\phi_2)\cos \omega t - (a_1 \sin\phi_1+a_2\sin\phi_2)\sin\omega t=(a \cos\phi)\cos \omega t - (a \sin\phi)\sin\omega t$$

So by comparing, $$a_1 \cos\phi_1 + a_2 \cos\phi_2=a \cos\phi$$ and $$a_1 \sin\phi_1+a_2\sin\phi_2=a \sin\phi$$

Solving this, you have $$a_1=\frac{a\sin(\phi-\phi_1)}{\sin(\phi_2-\phi_1)}$$ and $$a_2=\frac{a\sin(\phi_1-\phi)}{\sin(\phi_1-\phi_2)}=\frac{a\sin(\phi-\phi_1)}{\sin(\phi_2-\phi_1)}$$ and $$\phi = \arctan \left(\frac{a_1 \sin\phi_1+a_2\sin\phi_2}{a_1 \cos\phi_1 + a_2 \cos\phi_2}\right)$$

  • $\begingroup$ Thank you very much! That is a neat trick=) $\endgroup$
    – flawr
    Nov 19, 2015 at 19:36

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