• Analysis of the above circuit
indicates that,
XC = 1/2pfC @ 0 W
• Thus, Vo = Vi at high
frequencies.
• At f = 0 Hz, XC = , Vo = 0V.
• Between the two extremes, the
ratio, AV = Vo / Vi will vary.
As frequency increases, the
capacitive reactance decreases and more of the input
voltage appears across the output
terminals.
The output and input voltages are
related by the voltage – divider rule:
Vo = RVi / ( R – jXC)
the magnitude of Vo = RVi / ÖR2 + XC2
• For the special case where XC =
R,
Vo =RVi / RÖ2 = (1/Ö2) Vi
AV = Vo / Vi = (1/Ö2) = 0.707
• The frequency at which this
occurs is determined from,
XC = 1/2pf1C = R
where, f1 = 1/ 2pRC
• Gain equation is written as,
AV = Vo / Vi
= R / (R – jXC) = 1/ ( 1 – j(1/wCR)
= 1 / [ 1 – j(f1 / f)]
• In the magnitude and phase
form,
AV = Vo / Vi
= [1 /Ö 1 + (f1/f)2 ] Ð tan-1 (f1 / f)
• In the logarithmic form, the gain in dB is
AV = Vo / Vi = [1 /Ö 1 + (f1/f)2 ]
= 20 log 10 [1 /Ö 1 + (f1/f)2 ]
= - 20 log 10 Ö [ 1 + (f1/f)2]
= - 10 log10 [1 + (f1/f)2]
• For frequencies where f
<< f1 or (f1/ f)2 the equation can be approximated by
AV (dB) = - 10 log10 [ (f1 / f)2]
= - 20 log10 [ (f1 / f)] at f
<< f1
• At f = f1 ;
f1 / f = 1 and
– 20 log101 = 0 dB
• At f = ½ f1;
f1 / f = 2
– 20 log102 = - 6 dB
• At f = ¼ f1;
f1 / f = 4
– 20 log102 = - 12 dB
• At f = 1/10 f1;
f1 / f = 10
– 20 log1010 = - 20dB
• The above points can be plotted
which forms the Bode – plot.
• Note that, these results in a
straight line when plotted in a logarithmic scale.
Although the above calculation
shows at f = f1, gain is 3dB, we know that f1 is
that frequency at which the gain
falls by 3dB. Taking this point, the plot differs
from the straight line and
gradually approaches to 0dB by f = 10f1.
Observations
from the above calculations:
• When there is an octave change
in frequency from f1 / 2 to f1, there exists
corresponding change in gain by
6dB.
• When there is an decade change
in frequency from f1 / 10 to f1, there exists
corresponding change in gain by 20 dB.
0 comments:
Post a Comment