Introduction to Electric Circuits

136 Resonance

Dividing the numerator and the denominator by joJCR this becomes

Z = (L/CR)/[1 + (LC/jwCR){(1/LC) - w2}] (6.24)

Now at resonance, L/CR = Zd, the dynamic impedance; 1/wCR (-o~L/R)

= Q; and LC = 1/oJ0 2. Substituting these into Equation (6.24) we obtain

Z = Zd/[1 4- (Q/j(.o02)(~Oo 2- 0)2)1

Rearranging to put the 'j' in the numerator we have

Z = Zd//[1 + (jQ/jwo (0) 2- ~o02)] (6.25)

For frequencies close to resonance, we could say that ~o = O)o + &o where &o

is a small frequency deviation. In that case

(o2_ ~o0 2)_ (O)o + &o) 2- O)o 2

= Wo^2 + 2Wo6W + 6w 2- w o^2

= 2w06w + 60-12

= 2~o06o~ (since 6o02 --, 0)

Substituting this in Equation (6.25) we have

z- Zd/[1 + Zd/[1 + j2Q(&o/o 0)]

The magnitude of this is

Z- Z,~/V'{1 + [2Q(6oJ/w0] 2} (6.26)

The bandwidth of a parallel resonant circuit is defined to be the frequency

range between the two frequencies for which Z//Zd -- 1/V'2, and from Equation

(6.26) Z/Zd- 1/V'{1 + [2Q(3o)/o~o)] 2, so to obtain the bandwidth we put

1/V'{1 + [2Q(3o)/o)0)] 2= l/X/2. Thus, inverting both sides and taking the

square root, we get

1 + [2Q(6w/wo)] 2= 2

[2Q(&o/o)0)] 2= 1

2Q(aw/Wo) = 1

6W/Wo = 1/2Q

aw = Wo/2Q

The bandwidth is the band of frequencies from -6o~ to + 6oJ (i.e. 26oJ). Thus

B - 26w = 2w0/2Q = wo/Q

Finally

B- a~o/Q (6.27)

Example 6.10

Determine (1) the bandwidth of the circuit in Example 6.9, (2) the effect on the
bandwidth of this circuit of adding a resistance of 50 11 in series with the coil.