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Preamble

 

AC analysis is a special case of dynamic analysis where we are interested in the time varying steady state waveforms (sinusoids); we ignore transient components. The important point is that when a linear circuit is driven by a sinusoidal AC source, its steady state response (i.e. currents and voltages) is also a sinusoid (of the same frequency, but of different magnitudes and phases). The method of phasors you have seen in first year is a short cut method for solving AC problems, although you could use differential equations or s-domain methods. Before we look at AC analysis proper let's do some bare hands calculations to help our understanding.

Consider the circuit of Figure 33. Here, $v_g(t) = 5 \cos(2t) u(t)$ V, v(0^-)=0 V, R=1 $\Omega$, and C=1 F. We are interested in the capacitor voltage v(t).

  

Figure 33: RC circuit with AC source.

The s-domain circuit is shown in Figure 34. Now

$$\cos(2t) = \frac{1}{2} (e^{j2t} + e^{-j2t}) ,$$

and so

$$V_g(s) = {\cal L}[5 \cos(2t)](s) = \frac{5}{2}( \frac{1}{s-j2} + \frac{1}{s+j2} ) .$$

  

Figure 34: RC circuit in s-domain.

By KVL,

$$V_g(s) = I(s)(1+\frac{1}{s}) ,$$

and solving for current gives

$$I(s) = \frac{s V_g(s)}{s+1} .$$

By Ohm's law, since v(0^-)=0 V,

$$V(s) = I(s) \frac{1}{s} = \frac{ V_g(s)}{s+1} .$$ (39)

Substituting in the formula for V_g(s) we get

$$\begin{array}{rl} V(s) & = \frac{5}{2}( \frac{1}{(s-j2)(s+1)}... ...2j)(s-2j)} + \frac{5}{2(1-2j)(s+2j)} -\frac{1}{s+1} \end{array}$$ (40)

after some algebra and partial fractions. With yet more algebra we get

$$v(t) = \cos(2t) +2\sin(2t) -e^{-t} \ {\rm V}, \ t \geq 0.$$ (41)

Here,

$\cos(2t) +2\sin(2t)$ V is the AC steady state response to the AC source $v_g(t)=5 \cos(2t)$ V.  

-e^-t V is the transient component (which decays away and we will ignore).

Exercise. Check all these calculations!

The term $\cos(2t) +2\sin(2t)$ is indeed a sinusoid of frequency 2 rad/sec, since by trigonometry

$$\begin{array}{rl} \cos(2t) +2\sin(2t) & = \sqrt{5}( \frac{1}{... ...\ \\ & = \sqrt{5} \cos(2t -63.4^\circ) \ {\rm V}. \end{array}$$

(Note that 5 = 1²+2², and $\cos \theta = 1/\sqrt{5}$, $\sin \theta =2/\sqrt{5}$ implies $\theta = 63.4^\circ$.) The capacitor voltage has magnitude $\sqrt{5}$ V, and phase $-63.4^\circ$.    We write

$$v(t) = \sqrt{5} \cos(2t -63.4^\circ) \ {\rm V}$$ (42)

for the AC steady state response.

This type of calculation can get complicated, and it is much simpler to use complex currents and voltages. Suppose we have

$$v_g(t) = 5 e^{j2t} \ {\rm V}$$

and note that the real voltage is given by

$$5 \cos(2t) \ {\rm V} = Re ( 5 e^{j2t}) \ {\rm V} .$$

Then

$$V_g(s) = \frac{5}{s-2j} ,$$

and from (39) we have

$$\begin{array}{rl} V(s) & = \frac{5}{(s-2j)(s+1)} \\ \\ & = \frac{5}{1+2j} ( \frac{1}{s-2j} - \frac{1}{s+1} ) \end{array}$$ (43)

Therefore

$$\begin{array}{rl} v(t) & = \frac{5}{1+2j} e^{j2t} - \frac{5}{... ...j(2t-63.4^\circ)} - \frac{5}{1+2j} e^{-t} \ {\rm V} \end{array}$$ (44)

So the complex AC steady state response is

$$v(t) = \sqrt{5}e^{j(2t-63.4^\circ)} \ {\rm V}$$ (45)

with real part (42).

In equation (45) the term e^j2t is the complex sinusoid, while the complex number

$$1-j2 \ {\rm V} = \sqrt{5}e^{-j63.4^\circ} \ {\rm V}$$ (46)

is the phasor.  

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