 # Phasor and sinusoid Calculus Example Problem #2

### Using Phasors, determine i(t) for the following integro-differential equation:

$$10\int_{}^{} i\;dt + \frac{di}{dt} + 6i = 5\cos(5t + 22^{\circ})$$ By inspection we can see that the angular frequency is: $$\omega = 5\frac{rad}{s}$$ ...and from the intro to sinusoid and phasor calculus page we know that the derivative of i(t) is equivalent to the following in phasor form: $$\frac{di}{dt} \iff j\omega \mathbb{I}$$ ...and that the integral of i(t) is equivalent to the following in phasor form: $$\int_{}^{} i \; dt = \frac{\mathbb{I}}{j\omega}$$ ...where: $$\mathbb{I} = phasor \; form \; of \; the \; sinusoid \; i(t)$$ This allows us to rewrite the equation as: $$\frac{10\mathbb{I}}{j5} + j5\mathbb{I} + 6\mathbb{I} = 5\cos(5t+22^{\circ}) \qquad(Expression \; 1)$$

Note that the right side of the equation is equivalent to the following: $$5\cos(5t+22^{\circ}) = 5\angle22^{\circ}$$ ...which allows us to rewrite expression #1 as:

$$\frac{10\mathbb{I}}{j5} + j5\mathbb{I} + 6\mathbb{I} = 5\angle22^{\circ}$$

# Goal: To isolate to phasor term on the left side of the equation and convert it to i(t) in sinusoid form

$$\mathbb{I} \Big( \frac{2}{j} + j5 + 6 \Big) = 5\angle22^{\circ} \qquad (Expression \; 2)$$

Recall from our rules for complex numbers in the review of complex numbers page that: $$2 \Big( \frac{1}{j} \Big) = -2j$$ ..which means that we can rewrite expression #2 as:

$$\mathbb{I}(6 - j2 + j5) = 5\angle22^{\circ}$$ $$\mathbb{I}(6 + 3j) = 5\angle22^{\circ} \qquad (Expression \; 3)$$

For the 6+3j term: $$r = \sqrt{6^2+3^2}$$ $$r = 3\sqrt{5}$$ $$\phi_{ref} = tan^{-1} \Big( \frac{3}{6} \Big) \;\;\;,lies \; 1^{st} \; quadrant$$ $$\phi_{ref} = 26.57^{\circ} = \phi$$ Therefore: $$6+3j = 3\sqrt{5} \angle 26.57^{\circ}$$ ...which allows us to rewrite expression #3 as:

$$\mathbb{I} = \frac{ 5\angle22^{\circ}}{3\sqrt{5} \angle 26.57^{\circ}} \qquad (Expression \; 4)$$

Recall the property of division for complex numbers from our review of complex numbers page $$\frac{z_1}{z_2} = \frac{r_1}{r_2} \angle (\phi_1 - \phi_2)$$ ...and use it on expression #4:

$$\mathbb{I} = \Big( \frac{5}{3\sqrt{5}} \Big) \angle (22^{\circ} - 26.57^{\circ})$$ $$\mathbb{I} = \Big( \frac{\sqrt{5}}{3} \Big) \angle (-4.57^{\circ})$$ ...which is equivalent to the following when expressed in exponential form: $$\mathbb{I} = \Big( \frac{\sqrt{5}}{3} \Big)e^{j(-4.57^{\circ})} \qquad(Expression \; 5)$$ We now need to multiply expression #5 by e^jwt and take the real part in order to obtain the sinusoid i(t). Recall that omega equals 5. $$i(t) = R_e \Big\{ \Big( \frac{\sqrt{5}}{3} \Big)e^{j(-4.57^{\circ})} e^{j5t} \Big\}$$ $$i(t) = R_e \Big\{ \Big( \frac{\sqrt{5}}{3} \Big) e^{j(5t-4.57^{\circ})} \Big\} \qquad(Expression \; 6)$$

Recall Euhler's Identity: $$e^{\pm j\phi} = \cos\phi \pm j\sin\phi$$ ...and apply it to expression #6:

$$i(t) = R_e \Big\{ \frac{\sqrt{5}}{3}\cos(5t-4.57^{\circ}) + j\frac{\sqrt{5}}{3} \sin(5t-4.57^{\circ}) \Big\}$$ ...which we exaluate to obtain the:

$$i(t) = 0.745\cos(5t-4.57^{\circ})$$