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Transcript
FILTER CIRCUITS
The operation of many electronic instruments requires the amplification of a selected
range of frequencies while suppressing others. The behaviour of an amplifier for a
given frequency can be controlled by adding reactive elements - capacitors and
inductors. Such circuits are usually called filters because they “filter” out frequencies
except those lying in a selected range. The response of any electronic circuit depends
to some degree on the frequency of operation. In an amplifier, for example, the gain
will be constant over some range but fall off outside this range of frequencies. The
frequency behaviour is called its frequency response characteristic and is usually
displayed graphically. The ratio of output voltage to input voltage as it varies over a
given frequency range is know as the circuit’s transfer function.
add graphs
Low-pass
High-pass
Band-pass
Notch
FILTERS
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FILTERS
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Simple RC SERIES CIRCUIT
complex circuit
Complex representation
 Potential differences
 Currents
 Impedances
Z=R+jX
XC = -1 /  C
XL =  L
simple circuit
1
1
Z1
vs
Z1
vs
2
3
vS  VSo e jt
5
Z5
4
Z3
Z2
Applied emf
vs
Z4
load
Z3
1
Z2 and Z3 in parallel
Z4 
Z1 and Z4 in series
Z5  Z1  Z4
1
1

Z 2 Z3

Z 2 Z3
Z 2  Z3
Currents and potential differences
v
i5  S  i1  i4
Z5
v1  i1 Z1
v4  i4 Z4  v2  v3
v
i2  2
Z2
i3 
v3
Z3
Checks
i1  i4  i2  i3
Power
p1  v1 i1
FILTERS
vS  v1  v4
p2  v2 i2
p3  v3 i3
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5
LOW PASS FILTER
Control data
1
Z1 = R1 = 1000 
R1
C2 = 1 F
vs
Z2 = - j / ( C2)
Z3 = R3 = 1000 
3
2
R3
C2
VSo = 10 V
fmin = 10 Hz fmax = 104 Hz
This circuit is called a low-pass filter because it passes low frequency signals with
less attenuation than high frequency signals. A passive RC circuit such as this one
does not have a sharp cut-off above a certain frequency, rather, the attenuation
changes gradually. This frequency is known as the cut-off frequency (break-point
frequency or corner frequency) fo.
For very low frequencies, the capacitive reactance is so large compared to the
resistance values that it is essentially an open circuit. Thus the circuit behaves as if the
capacitor was removed. The circuit then acts as a simple voltage divider and the
voltage gain Av is
Av 
v3
R3

vs R1  R3
or
 R3 
Av  20 log10 
 dB
 R1  R3 
Since the capacitor has very little effect upon the circuit at low frequencies, the load
voltage v3 and supply voltage vS are in phase ( ~ 0).
For very high frequencies, the capacitive reactance is much smaller than R3, thus
shorting it out. Therefore, R3 has no effect on the circuit and the circuit again acts as a
simple voltage divider and the voltage gain Av is
v
Z2
 j / C
1  jR1C
Av  3 


vs
Z1  Z 2
R1  j /  C 1  ( R1C ) 2
Since f is large ( = 2f) we can assume that R1C >> 1, then
Av 
FILTERS
1
R1C
and
 1 
Av (dB)  20log10 
  20log10  f   constant
 R1C 
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Therefore, at high frequencies, the voltage gain falls off in inverse proportion to the
frequency. This is shown as a straight line segment with a slope of –20 when the
voltage gain Av (dB) is plotted against log10(f). In many applications, it is necessary to
use multiple filter stages to give a greater fall off in gain with frequency.
Since the capacitor dominates the circuit at high frequencies, the phase difference
between the load voltage v3 and the supply voltage vS is  ~ –90° (-/2 rad).
The cut-off frequency fo is defined when the gain is
Av 
Z3
1
1
 max gain  
2
2 Z1  Z3
 1

Av (dB)  20log10 
 max gain 
 2

 1 
Av (dB)  20log10 
  20log10  max gain 
 2
Av (dB)  3  20log10  max gain 
fo 
1
2
 R1  R3 


 R1 R3 C2 


if R3 very large then fo 
1
2
 1 


 R1 C2 


This is referred to as the 3 dB point, since it is the frequency at which the voltage
gain has fallen 3 dB. At this point, the power is ½ of the maximum power. For
frequencies less than the cut-off frequency, the voltage gain is essentially constant and
independent of frequency. Above the cut-off frequency, the voltage gain falls with a
slope of –20 when the voltage gain Av (dB) is plotted against log10(f). Note, that at
this point the phase angle is –45° ( rads), midway in its range from 0° to –90°.
Maximum power transferred to load
Z3  ZTH 
Z1 Z 2
Z1  Z 2
But Z2 is frequency dependent, therefore, the power delivered to the load depends on
the frequency of the source.
FILTERS
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MATLAB FILE
%m260ab.m
%9 feb jan 01
%RC low pass filter
%data
vso = 10;
fmin = 1;
fmax= 1e4;
num = 501;
%control data
z1 = 1000;
c2 = 1e-6;
z3 = 1000;
%test
z1t =
c2t =
z3t =
data
1000;
2e-6;
1000;
%calcualtions
df = (fmax-fmin)/(num-1);
f = fmin : df : fmax;
w = f .* (2*pi);
%control calculations
xc = 1./(w.*c2);
z2 = -j*xc;
z4 = (z2 .* z3);
z4 = z4 ./(z2 +z3);
z5 = z1+z4;
i5 = vso ./ z5;
i1 = i5;
i4 = i5;
v3 = i4 .* z4;
v3db = real(20*log10(v3/vso));
phase3 = angle(v3)/pi;
power3 = abs((v3 .^2) /z3);
%test calculations
xct = 1./(w.*c2t);
z2t = -j*xct;
z4t = (z2t .* z3t);
z4t = z4t ./(z2t +z3t);
z5t = z1t+z4t;
i5t = vso ./ z5t;
i1t = i5t;
i4t = i5t;
v3t = i4t .* z4t;
v3tdb = real(20*log10(v3t/vso));
phase3t = angle(v3t)/pi;
power3t = abs((v3t .^2) /z3t);
FILTERS
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%graphing
figure(1);
plot(log10(f),v3db);
hold on
plot(log10(f),v3tdb,'m');
grid on;
title('Low Pass Filter');
xlabel('log(f)');
ylabel('Voltage gain Av (dB)')
figure(2);
plot(log10(f),phase3);
hold on
plot(log10(f),phase3t,'m');
grid on;
title('Low Pass Filter');
xlabel('log(f)');
ylabel('phase (in pi rad)');
figure(3);
plot(log10(f),power3);
hold on
plot(log10(f),power3t,'m');
grid on;
title('Low Pass Filter');
xlabel('log(f)');
ylabel('Pload (W)')
Low Pass Filter
-5
-10
Voltage gain Av (dB)
-15
-20
-25
-30
-35
-40
-45
FILTERS
0
0.5
1
1.5
2
log(f)
2.5
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3.5
4
9
Low
Pass
Filter
Low
Pass
Filter
0
0
-0.05
-0.05
-0.1
-0.1
-0.15
phase (in pi rad)
phase (in pi rad)
-0.15
-0.2
-0.2
-0.25
-0.25 -0.3
-0.3-0.35
-0.35 -0.4
-0.45
-0.4
-0.5
-0.45
-0.5
0
0
0.5
0.5
1
1
1.5
1.5
2
log(f)
2.5
2
log(f)
3
2.5
3.5
3
4
3.5
4
Low Pass Filter
0.025
0.02
Pload (W)
0.015
0.01
0.005
0
0
0.5
1
1.5
2
log(f)
2.5
Control data results: R1 = R3  AV(max) = - 6 dB
fo = 314 Hz
AV = - 9 dB
3
3.5
4
Pmax = 0.025 W
P = 0.0125 W (1/2 Pmax) high frequencies: slope
AV/log(f) graph = (-26 + 17)/(3.5 – 3.0) = -18 (~-20)
FILTERS
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