To convert $3\sin(\omega t) - 2\cos(\omega t)$ into phasor form, we combine the sinusoidal terms into a single expression of the form:

$A \sin(\omega t + \phi)$

Steps:

1. Express $\sin(\omega t)$ and $\cos(\omega t)$ in terms of $\sin(\omega t + \phi)$:

   • Recall the trigonometric identity: $\sin(\omega t + \phi) = \sin(\omega t)\cos(\phi) + \cos(\omega t)\sin(\phi)$

   We need to rewrite $3\sin(\omega t) - 2\cos(\omega t)$ in this form.

2. Find the magnitude $A$:

   • Let $A = \sqrt{a^2 + b^2}$, where $a = 3$ and $b = -2$. $A = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$

3. Find the phase angle $\phi$:

   • $\phi = \arctan\left(\frac{b}{a}\right) = \arctan\left(\frac{-2}{3}\right)$.
   • $\phi \approx -33.69^\circ$ (in degrees) or $\phi \approx -0.588$ radians.

4. Rewrite the expression:

   • Substitute $A$ and $\phi$ into $A\sin(\omega t + \phi)$: $\sqrt{13}\sin\left(\omega t - 33.69^\circ\right)$

Phasor Representation:

In phasor form, this can also be expressed as a complex number:

$\mathbf{P} = A e^{j\phi} = \sqrt{13} e^{-j33.69^\circ}$

or equivalently:

$\mathbf{P} = \sqrt{13} \left(\cos(-33.69^\circ) + j\sin(-33.69^\circ)\right)$

Final Result:

$\boxed{3\sin(\omega t) - 2\cos(\omega t) = \sqrt{13}\sin\left(\omega t - 33.69^\circ\right)}$

Would you like a detailed derivation or clarification on any part? 😊

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Relative Questions:

1. What is the significance of phasor form in signal analysis?
2. How can we convert phasor form back into its time-domain representation?
3. What are other methods to find the magnitude and angle of a sinusoidal expression?
4. How does the sign of the angle affect the phase shift direction?
5. Can this conversion process be extended to include damped sinusoids?

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Tip:

When converting sinusoidal functions to phasor form, always ensure the coefficients are properly handled for signs and quadrants when calculating the angle $\phi$.