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IB Maths AI 1.10 Notes

This page contains our IB Maths AI notes for 1.10. By reading each one of these notes, you will fully cover the content for IB Maths AI 'Advanced complex numbers'.

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Sinusoidal functions using phasors

This section develops two linked ideas.

  1. First, sinusoidal functions with the same frequency can be combined into a single sinusoidal function with a new amplitude and phase shift. 
  2. Second, complex numbers can be interpreted geometrically on an Argand diagram, where addition behaves like vector addition and multiplication produces a stretch and a rotation.

Suppose two sinusoidal functions have the same frequency but different phase shifts:

f1(t)=acos(ωt+α) and f2(t)=bcos(ωt+β)f_1(t)=a\cos(\omega t+\alpha)\text{ and }f_2(t)=b\cos(\omega t+\beta)

Their sum can always be written in the form:

f1(t)+f2(t)=Rcos(ωt+ϕ)f_1(t)+f_2(t)=R\cos(\omega t+\phi)

for some amplitude RR and phase shift ϕ\phi.

The key idea is that each sinusoidal term can be treated like a vector, or equivalently a complex number, with modulus equal to its amplitude and argument equal to its phase angle. Adding the functions is then the same as adding these vectors.

To combine expressions of the form acos(ωt+α)+bcos(ωt+β)a\cos(\omega t+\alpha)+b\cos(\omega t+\beta), treat acisαa\operatorname{cis}\alpha and bcisβb\operatorname{cis}\beta as complex numbers, where cisθ=cosθ+isinθ\operatorname{cis}\theta=\cos\theta+i\sin\theta. Then, add them:

acisα+bcisβ=Rcisϕa\operatorname{cis}\alpha+b\operatorname{cis}\beta=R\operatorname{cis}\phi

Then the required sinusoidal form is Rcos(ωt+ϕ)R\cos(\omega t+\phi).

Two AC voltage sources are connected in a circuit, where V1=10cos(40t)V_1=10\cos(40t) and V2=20cos(40t+π18)V_2=20\cos\left(40t+\frac{\pi}{18}\right). Find the total voltage in the form V=Acos(40t+B)V=A\cos(40t+B).

Represent the two voltages by the complex numbers z1=10cis0z_1=10\operatorname{cis}0 and z2=20cisπ18z_2=20\operatorname{cis}\frac{\pi}{18}.

Add them: z=z1+z2=10+20(cosπ18+isinπ18)z=z_1+z_2=10+20\left(\cos\frac{\pi}{18}+i\sin\frac{\pi}{18}\right).

So z=(10+20cosπ18)+i(20sinπ18)z=\left(10+20\cos\frac{\pi}{18}\right)+i\left(20\sin\frac{\pi}{18}\right).

Using cosπ180.9848\cos\frac{\pi}{18}\approx0.9848 and sinπ180.1736\sin\frac{\pi}{18}\approx0.1736, we get z29.696+i3.472z\approx29.696+i3.472.

Now find the modulus: A=z=29.6962+3.472229.898A=|z|=\sqrt{29.696^2+3.472^2}\approx29.898.

Now find the argument: B=arg(z)=tan1(3.47229.696)0.116B=\arg(z)=\tan^{-1}\left(\frac{3.472}{29.696}\right)\approx0.116 radians.

Hence V29.9cos(40t+0.116)V\approx29.9\cos(40t+0.116). This phase shift is approximately 6.676.67^\circ.

This method only works directly when the sinusoidal functions have the same frequency. If the frequencies are different, the sum cannot in general be written as a single sinusoidal function of the same type.

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