 # Phasors (Sinusoid Example Problems)

### Example 1:

Given the following sinusoid, $$45cos(5 \pi t + 36^{\circ})$$ ...calculate the amplitude, phase, angular frequency, period and frequency.

From our previous discussion of sinusoids we can easily see that Amplitude = 45 and the phase is 36 degrees. Also note that: $$angular \; frequency = \omega = 5\pi$$ also: $$\omega = 2\pi f = \frac{2\pi}{T}$$ ...where T is the period of the sinusoid. Therefore by rearranging the above equation: $$T = \frac{2\pi}{\omega}$$ $$T = \frac{2\pi}{5\pi}$$

$$T = 400 ms = period$$

We also know that: $$f = frequency = \frac{1}{T}$$ therefore: $$f = \frac{1}{400ms}$$

$$f = 2.5Hz = frequency$$

### Example 2:

Calculate the phase angle between: $$v_1 = -10cos(\omega t + 50^{\circ})$$ $$v_2 = 12sin(\omega t - 10^{\circ})$$ ...using both the method of trig identities and the graphical approach. Determine which sinusoid is leading.

# Method #1 (trig identities)

We start by expressing both sinusoids in sine form with positive amplitudes (v2 is already in this form): $$v_1 = -10cos(\omega t + 50^{\circ})$$ Using our trig identities: $$v_1 = 10sin(\omega t + 50^{\circ} - 90^{\circ})$$ $$v_1 = 10sin(\omega t - 40^{\circ})$$ and: $$v_2 = 12sin(\omega t - 10^{\circ})$$ From this we can see that the phase difference is 30 degrees

$$\phi = 30^{\circ}$$

The negative signs in front of the phase angles indicate a rightward shift graphically. Since v1 is shifted farther to the right than v2, its starting point occurs later in time (relative to v2). This means that:

$$v2 \; "leads" \; v1 \; by \; 30 \; degrees.$$

# Method #2 (Graphical Approach)

Start by plotting v1 and v2 on our "improvised" coordinate system: From inspection we can see that:

$$\phi = 30^{\circ}$$

We also again notice that v2 "leads" v1 by 30 degrees.

Now that we have a decent understanding of sinusoids, let's also briefly review complex numbers.

Continue on to Review of complex numbers....