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4.1
Filter Order:
The order of a filter is the order of its transfer function. If the filter order is
increased (i.e. a first order low-pass to a second-order low-pass), a sharper
transition between the stopband and passband of the filter is possible.
Example: For the two low-pass filters, determine circuit parameters such that (fc) is
100 Hz, and GDC is 3 2 1.586 . Plot transfer function magnitudes and observe
the transition near fc.
C
R
R
vi(t)
C
R
vi(t)
C
Rf
R1
Transfer functions are:
+
vo(t)
-
(K-1)Rf
Rf
+
vo(t)
-
4.2
First-order filter:
Rf
1
R1
H ( j )
p
1
1
RC
GDC = 1+Rf / R1
fc = 1 / (2RC)
Second order filter:
K
p
p2
1 (3 K )
1
1 2
RC RC
For GDC = K = 3 2 1.586
fc = 1 / (2RC) (Not valid for any other value of K.
The value of K was contrived so the cutoff would
come out this way). For a general K value fc =
H ( j )
/ (2RC) , where
2
7 6 K K 2 7 6 K K 2 4
2
Why are there 4 distinct possible values that can be
obtained for the cut-off frequency?
Matlab Plots:
w = [0:1024]*2*pi;
p = j*w;
h1 = (3-sqrt(2)) ./ (1 + (p / (2*pi*100)));
h2 = (3-sqrt(2)) ./ (1 + (sqrt(2)*p / (2*pi*100)) + ( p / (2*pi*100)) .^2);
.
4.3
figure(1)
plot(w/(2*pi),abs(h1),'b-', w/(2*pi), abs(h2), 'b:')
title('first order (-), second order (---)')
xlabel('Hz')
ylabel('gain')
first order (-), second order (---)
1.4
1.2
gain
1
0.8
0.6
0.4
0.2
0
100
200
300
400
500
Hz
600
700
800
900
1000
4.4
Useful Circuits:
Describe the transfer functions of these circuits:
Non-Inverting Op Amp: (gain = 1 + Rf / R1) Inverting Op Amp: (gain = - Rf / R1)
R1
+
vo
vi
R1
-
Rf
vi
+
vo
-
Voltage Follower: (gain = 1)
vi
Rf
+
vo
-
4.5
Cascading Filter Stages with no Loading Effects:
Given two active filter circuits with transfer functions H 1( p) and H 2 ( p) , a circuit
composed of these circuits connected in series has a transfer function equal to the
product of the individual circuit transfer functions H T ( p) H 1 ( p) H 2 ( p) , provided
that the connection of these circuits does not significantly alter the output resistance
of the first circuit or the input resistance of the second circuit (i.e. no loading
effects).
V1o = V2i
V2o = VTo
VTi = V1i
H 1 ( p)
H 2 ( p)
VTo V1o V2 o
HT ( p)
H 1 ( p) H 2 ( p)
VTi V1i V2i
Example: Sketch the transfer function magnitude of a first-order low-pass filter
(with GDC = 2 and fc = 1 kHz) in series with a second order band-reject filter ( with
f0 = 500 Hz, B = 200 Hz, and GDC = G = 1 ).
4.6
Other Useful Circuits:
Describe the transfer functions of these circuits.
The summing amp:
The differential amp:
R1
vi1
Rf
R2
vi2
+
vo
-
R2
R1
vi1
R1
RN
vi2
viN
Rf
Rf
Rf
vo vi1
vi 2
viN
R2
RN
R1
vo
R2
vi 2 vi1
R1
+
vo
R2
-
4.7
Adding Outputs of Filter Stages:
Given two filter circuits with transfer functions H 1( p) and H 2 ( p) , a circuit
composed of these circuits connected in parallel with a common input and their
outputs summed together (through a summing amp) has a transfer function equal to
the scaled sum of the individual transfer functions H ( p) H ( p) H ( p) ,
provided that the connection of these circuits does not significantly alter the output
resistance or input resistance of these circuits (i.e. no loading effects).
T
Vi
H 1 ( p)
H 2 ( p)
1
1
2
1 ( p)Vi
H
2
n
n1
2 ( p)Vi
H
1 ( p)Vi 2 H
2 ( p)Vi
Vo 1 H
Vo
1 ( p) 2 H
2 ( p)
H T ( p)
1 H
Vi
V o
2
4.8
Matlab Example:
Given 3 second-order band-pass filters with center frequencies 300, 1000, 3000 Hz,
bandwidths 200, 500, 600 Hz, and gains at resonance of 1, 1.5, 5, respectively.
Connect the filters so they have a common input and their outputs are fed into a
summing amp with input resistors 10k, 8k, and 20 k corresponding
respectively to the outputs of the 3 band-pass filters. The feedback resistor is 80 k.
Plot the resulting transfer function magnitude.
% This script will plot the sum of 3 band-pass filters.
% The range of interest will correspond to the center frequencies,
% so plot from about a decade before 300 to a decade after 3000
% (let's say 10 - 10,000 Hz)
f = logspace(1, 4, 256);
% Create Frequency Axis
p = j*f*2*pi;
% Compute Individual Transfer Functions
h1 = (200*2*pi)*p ./ (p.^2 + 200*2*pi*p + (2*pi*300)^2);
h2 = 1.5*(500*2*pi)*p ./ (p.^2 + 500*2*pi*p + (2*pi*1000)^2);
h3 = 5*(600*2*pi)*p ./ (p.^2 + 600*2*pi*p + (2*pi*3000)^2);
%
Sum Transfer Functions Together
ht = -(80/10)*h1 - (80/8)*h2 - (80/20)*h3;
figure(1)
semilogx(f, 20*log10(abs(ht)), '.b')
% Plot the sum
xlabel('Frequency (Hz)')
4.9
ylabel('TF Magnitude in dB')
grid
hold
% Hold current plot so others can be plotted on top
semilogx(f, 20*log10(abs(h1)), '-b');
semilogx(f, 20*log10(abs(h2)), ':b');
semilogx(f, 20*log10(abs(h3)), '-.b');
Title('Magnitude of the sum of 3 bandpass filters')
hold off
% Remove hold on plot
Magnitude of the sum of 3 bandpass filters
30
20
TF Magnitude in dB
10
0
-10
-20
-30
-40
-50
1
10
2
3
10
10
Frequency (Hz)
4
10
4.10
Matlab Example:
Given 2 first order low-pass filters both with cut-off frequencies at 10000 Hz, and
gains at DC of 10 and 5. Connect the filters in series so that there are no significant
loading effects. Plot the magnitudes of the individual filters and the resultant.
% This script will plot the product of 2 lowpass filters.
% The range of interest will go from near DC to about a decade after
% The cutoff frequency 10000
f = logspace(1, 5, 256);
% Create Frequency Axis
p = j*f*2*pi;
wc = 2*pi*10000;
% Compute Individual Transfer Functions
h1 = 10 ./ ((p/wc) + 1);
h2 = 5 ./ ((p/wc) +1);
%
Multiply Transfer Functions Together
hp = h1.*h2;
figure(1)
semilogx(f, 20*log10(abs(hp)), '.b')
% Plot the product
xlabel('Frequency (Hz)')
ylabel('TF Magnitude in dB')
grid
hold
% Hold current plot so others can be plotted on top
4.11
semilogx(f, 20*log10(abs(h1)), '-b');
semilogx(f, 20*log10(abs(h2)), '-.b');
Title('Magnitude of the product of 2 low-pass filters')
hold off
% Remove hold on plot
Magnitude of the product of 2 low-pass filters
35
30
TF Magnitude in dB
25
20
15
10
5
0
-5
-10
1
10
2
10
3
10
Frequency (Hz)
4
10
5
10