# Phasor Notation Problems (Converting from sinusoids to phasors)

### Ex 1) Convert the following sinusoid to phasor notation:

$$v(t) = -4\sin(30t + 50^{\circ})$$

# Step 1) Express the sinusoid in positive cosine form so that it can be written as the real part of a complex number.

We will use the graphical approach in order to convert the sine function having a negative amplitude into a cosine function having a positive amplitude:

From the graph above we can see that: $$-4\sin(30t + 50^{\circ}) = 4\cos(30t + 50^{\circ} + 90^{\circ})$$ Therefore: $$v(t) = 4\cos(30t + 140^{\circ})$$ And from our previous explanation of phasor notation we know that: $$v(t) = R_e \{ 4e^{j(30t+140^{\circ})} \}$$ $$= R_e \{ 4e^{j30t} e^{j140^{\circ}} \} \qquad (Expression \; 1)$$

# Step 2) Remove the time factor from expression #1. Whatever is left is the phasor.

Now recall Expression #4 from the previous page $$\mathbb{V} = V_me^{j\phi}$$ and apply it to the expression #1 to give us the following: $$\mathbb{V} = 4e^{j140^{\circ}} \qquad (Expression \; 2)$$ Notice that the e^(jwt) term (e^(j30t) in this case) has been removed.

Converting expression #2 to polar form gives us: $$\mathbb{V} = 4 \angle 140^{\circ}$$

### Ex 2) Convert the following sinusoid to phasor notation:

$$v(t) = -8\cos(16t+15^{\circ})$$

# Step 1) Express the sinusoid in positive cosine form so that it can be written as the real part of a complex number.

We will once again use the graphical approach to convert this cosine function having a negative amplitude into a cosine function having a positive amplitude.

From the above graphical illustration we can see that: $$-8\cos(16t+15^{\circ}) = 8\cos(16t-165^{\circ})$$ ...and when we express this as the real part of a complex number we get: $$v(t) = R_e \{ 8e^{j(16t-165^{\circ})} \}$$ or: $$v(t) = R_e\{ 8e^{j16t}e^{j(-165^{\circ})} \} \qquad (Expression \; 3)$$

# Step 2) Remove the time factor from expression #3. Whatever is left is the phasor.

Now recall Expression #4 from the previous page $$\mathbb{V} = V_me^{j\phi}$$ and apply it to the expression #3 to give us the following: $$\mathbb{V} = 8e^{j(-165^{\circ})} \qquad (Expression \; 4)$$ Notice that the e^(jwt) term (e^(j16t) in this case) has been removed.

Converting expression #4 to polar form gives us: $$\mathbb{V} = 8 \angle (-165^{\circ})$$

Now let's look a some example problems that go the other way and convert phasors to sinusoids.

Continue on to Converting phasors to sinusoids....