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RLC Band-pass Filters
V0  j  
ZR
Vi  j 
ZC  Z L  Z R
HV  j  
R
1
 jL  R
jC
jRC

2
1   j  LC  jRC
A


 

1  j 1  j 
1 
1 

1
1
1
 LC;

 RC
12
HV  j  
A

1   
 1 
2

1   
 2 
2
e



j  arctan   arctan 
 1 
 2
2

 

1
2
Band-pass Filters and Resonant Circuits
HV  j  

  
1  Q 2   0 
 0  
1
LC
0 L
R
  
1  jQ  0 
 0  
1
0 
Q
1

0
B
2
e
 
arctan Q   0
 0 
Resonant frequency

0
2  1
Quality factor



RLC Resonant
when   0
L 
1
C
jL  
1
jC
Z L   ZC
I  j  

V ( j )
Z R  Z L  ZC
At resonance:
• current is max.
• Zeq =R
• current and voltage
are in phase.
• the higher Q, the
narrower the resonant
peak.
V
1 

R  j  L 


C


V
V
I 

2
R
1 

2
R   L 

C 

Applications:
tuning circuit
Concept Check: ac Circuit Resonance
The current in an RLC series circuit leads the ac voltage source.
To bring the circuit to resonance, should the capacitance be
increased or decreased?
Answer: ICE, it is capacitive, reduce the impedance of the
capacitor, i.e. increase the capacitance. Alternatively, inductance
can be increased to offset the impedance of the capacitor.
Ex: Notch filter, p303
Decibel Measure
•
•
Amplifier gain and filter loss are often specified in decibels (dB), a logarithmic
measure of ratios. Most generally dB are specified for power ratios,
– power gain in dB= 10 log10 (Pout/Pin)
– More generally in this course, we are interested in voltage gain;
since P ~ V2, voltage gain in dB= 20 log10 |Vout/Vin|
Advantages of decibel measures:
– Ease of handling large quantities that can vary over many orders of
magnitude.
• However, keep this compression firmly in mind! For example, the
Richter scale for earthquake intensity is logarithmic -- a 7 on the
Richter scale actually has an amplitude 10 times more powerful than a
6, corresponding to a factor of about 31-32 times more energy.
– In cascaded amplifier/filter systems, the overall gain is the product
of each stage's gain. Since log(A.B) = log(A) + log(B), if one
wishes to consider the overall gain of several stages, one simple
adds the gain of each in dB measure.
– As we will see below, the frequency dependence of an amplifier or
filter is most often summarized on a Bode plot. For a Bode plot,
the log of the |gain| is plotted against the log of the frequency. Thus
in dB measure the vertical axis becomes a linear axis.
Bode Plots
Consider a RC low-pass filter:
Let us plot filter gain vs. the dimensionless frequency ' = RC/0
'
Vout/Vin
0.01
1.000 =
0 dB
0.1
0.995 =
0 dB
1
0.707 =
- 3 dB
10
0.100 =
- 20 dB
100
0.010 =
- 40 dB
• At low enough frequencies the gain flattens off at unity,
0 dB.
• At large enough frequencies, the gain is falling off at
the rate of - 20 dB per decade of frequency (a factor of
10 increase in frequency). This fall-off is also often
referred to as - 6 dB per octave (a factor of 2 increase in
frequency.) Convince yourself of the equivalence!
'
Vout/Vin
Bode Plots: cont.
0.01
1.000 =
0 dB
0.1
0.995 =
0 dB
1
0.707 =
- 3 dB
10
0.100 =
- 20 dB
100
0.010 =
- 40 dB
• cutoff frequency, where ' = 1, i.e. 0=1/RC, the
gain is - 3 dB. The - 3 dB point is considered to be the
breakpoint.
•The Bode plot for the RC low-pass filter is often sketched
by drawing a horizontal line up to the breakpoint followed
by a line falling off at - 20 dB per decade as shown by the
blue line in the graph below. The actual gain curve is
shown in red for comparison; the largest error in the
approximation is at the breakpoint. This type of
approximate plot is known as an asymptotic Bode plot.
Bode Plot: High-pass filter
3db point
20 dB/Decade