(Part 1): Expressing the Derivative of a Sinusoid in Phasor Form:

Given the following sinusoid: $$ v(t) = R_e \{ \mathbb{V}e^{j\omega t} \} = V_m \cos(\omega t + \phi) $$ ...which when expressed in phasor form is equivalent to the following: $$ \mathbb{V} = V_me^{j\phi} = V_m \angle \phi $$ The derivative of the sinusoid v(t) is: $$ \frac{d}{dt}[v(t)] = -V_m \omega \sin (\omega t + \phi) $$ If we use trig identities to convert this to cosine form we get: $$ \frac{d}{dt}[v(t)] = \omega V_m \cos (\omega t + \phi + 90^{\circ}) $$

Now recall that: $$ V_m \cos (\omega t + \phi) = R_e \{ V_me^{j(\omega t + \phi)} \} $$ ...and use it to obtain the following for this example:

$$ \frac{d}{dt}[v(t)] = R_e \{ \omega V_m e^{j(\omega t + \phi + 90^{\circ})} \} $$ $$ \frac{d}{dt}[v(t)] = R_e \{ \omega V_m e^{j\omega t} e^{j\phi} e^{j90^{\circ}} \} \qquad(Expression \; 1)$$

Recall that: $$ \mathbb{V} = V_me^{j\phi} $$ ...and apply it to expression #1:

$$ \frac{d}{dt}[v(t)] = R_e \{ \omega e^{j\omega t} e^{j90^{\circ}} \mathbb{V} \} \qquad(Expression \; 2)$$

Recall Euhler's Identity: $$ e^{j\phi} = \cos\phi + j\sin\phi $$ ...and in expression #2 we can use it in the following manner: $$ e^{j90^{\circ}} = \cos 90^{\circ} + j\sin 90^{\circ} = j $$ ...which we use to rewrite expression #2 as:

$$ \frac{d}{dt}[v(t)] = R_e \{ j\omega \mathbb{V} e^{j\omega t} \} \qquad(Expression \; 3)$$ In order to convert expression #3 to the phasor domain, we remove the time factor e^jwt, giving us:

$$ \frac{dv}{dt} \iff j\omega \mathbb{V} \qquad(Expression \; A)$$

(Part 2): Expressing the Integral of a Sinusoid in Phasor Form:

Using the same given definitions in the beginning of part 1 of this page we have: $$ \int_{}^{} v(t) \; dt = V_m \int_{}^{} \cos(\omega t + \phi) dt $$ $$ \int_{}^{} v(t) \; dt = \frac{V_m}{\omega} \sin(\omega t + \phi) $$ If we use trig identities to convert this to cosine form we get: $$ \int_{}^{} v(t) \; dt = \frac{V_m}{\omega} \cos (\omega t + \phi -90^{\circ}) \qquad(Expression \; 4)$$

Recall that: $$ V_m \cos(\omega t + \phi) = R_e \{ V_me^{j(\omega t + \phi)} \} $$ ...and apply it to expression #4 which gives us:

$$ \int_{}^{} v(t) \; dt = R_e \Big[ \frac{V_m}{\omega} e^{j(\omega t + \phi - 90^{\circ})} \Big] $$ $$ \int_{}^{} v(t) \; dt = R_e \Big[ \frac{V_m}{\omega} e^{j\omega t} e^{j\phi} e^{-90^{\circ}j} \Big] \qquad(Expression \; 5)$$

Recall that: $$ \mathbb{V} = V_me^{j\phi} $$ ...and apply it to expression #5:

$$ \int_{}^{} v(t) \; dt = R_e \Big[ \frac{1}{\omega} \mathbb{V} e^{j\omega t} e^{-90^{\circ}j} \Big] \qquad(Expression \; 6)$$

Recall Euhler's Identity: $$ e^{j\phi} = \cos\phi + j\sin\phi $$ ...and in expression #6 we can use it in the following manner: $$ e^{-90^{\circ}j} = \cos (-90^{\circ}) - j\sin 90^{\circ} = -j $$ However, we remember that the 5th property of complex numbers from our Review of complex numbers page states that: $$ -j = \frac{1}{j} $$ ...which we use to rewrite expression #6 as:

$$ \int_{}^{} v(t) \; dt = R_e \Big[ \frac{\mathbb{V}}{j\omega} e^{j\omega t} \Big] \qquad(Expression \; 7)$$ In order to convert expression #7 to the phasor domain, we remove the time factor e^jwt, giving us:

$$ \int_{}^{} v(t) \; dt \iff \frac{\mathbb{V}}{j\omega} \qquad(Expression \; B) $$

Summary:

For Expression A:

The derivative of v(t) with respect to time can be replaced by the: $$ multiplication \; of \; j\omega \; in \; the \; phasor \; domain. $$

For Expression B:

The integral of v(t) with respect to time can be replaced by the: $$ division \; of \; j\omega \; in \; the \; phasor \; domain. $$

Both expressions A and B are useful in finding steady-state solutions to AC circuits because they do not require you to know the initial values of the variables involved.

Next we will look at an example of using expressions A and B to solve a differential equation.

Continue on to using phasors to solve a differential equation....