Upload
others
View
7
Download
0
Embed Size (px)
Citation preview
ECE 3042 Experiment 6 Active Filter Design
1. Design a 3rd order Butterworth low-pass filters having a dc gain of unityand a cutoff frequency, fc, of 10.28 kHz.
fc 10.28kHz:= K 1:= The transfer function is given on page 72 j 1−:=
TB f( ) K1
jffc⋅ 1+
⋅1
jffc⋅⎛⎜
⎝⎞⎟⎠
2j
ffc⋅+ 1+
⋅:= This is the product of a 1st order LPF & 2nd orderLPF.
1 103× 1 104× 1 105× 1 106×1 10 4−×
1 10 3−×
0.01
0.1
1
10
TB f( )
f
2 103× 6 103× 1 104×
0.6
0.8
1
1.2
1.4
TB f( )
f
To implement this with a circuit cascade the circuit on page 82 with that on page 83.For each the dc gain is unity (K) which means that RF is zero (a wire) and the resistorfrom the inverting input to ground is infinity (nothing there).
Start by picking two of the capacitors to be 0.01μF. For the Butterworth filter ωc & ωo arethe same.
So for the circuit on page 82 C 0.01μF:= R11
2 π⋅ fc⋅ C⋅:= R1 1.548 103
× Ω=This is a 1st order LPF
For the circuit shown on page 83, the 2nd order LPF, pick C1 0.01μF:= C2 0.1C1:=then Eq. 6.84 applies with Q=1 and ωo=ωc
Q 1:=ωo 2 π⋅ fc⋅:=
R11
2 Q⋅ ωo⋅ C2⋅1 1 4 Q2
⋅C2
C1−+
⎛⎜⎝
⎞⎟⎠
⋅:= R21
2 Q⋅ ωo⋅ C2⋅1 1 4 Q2
⋅C2
C1−−
⎛⎜⎝
⎞⎟⎠
⋅:=
R1 1.374 104× Ω= R2 1.745 103
× Ω= C1 1 10 8−× F= C2 1 10 9−
× F=
Now that all of the component values have been determined it's time to simulate it withSPICE
0.001
0.01
0.1
1
10
103 104 105
Butterworth Filter
Gai
n (L
inea
r)
Frequency (Hz)
The next step is to head to the lab and build the circuits. Use 741s, LF351s, LF347s.The resistors for your use are 5%. The caps are 20 %. Try to find 3 caps reasonablyclose to the design values.
φ f( )180π
arg TB f( )( )⋅:=τp f( )
1−2 π⋅
φ f( )f
⋅:= τg f( )1−
2 π⋅ fφ f( )d
d⎛⎜⎝
⎞⎟⎠
⋅:=
1 103× 1 104× 1 105×0
1 10 3−×
2 10 3−×τp f( )
τg f( )
f
2. Design a 3rd order unity dc gain Chebyshev LPF with a -3 frequency of 10.28 kHz and 0.5db ripple in the pass band.
db 0.5:= f3 10.28kHz:= K 1:= n 3:= t3 x( ) 4 x3⋅ 3 x⋅−:= j 1−:=
Eq. 6.47 must be solved to obtain the relationship between ωc & ω3which requires that Eq. 6.46 be solved which is a cubic polynomial g(x)ε 10
db
10 1−:=
g x( ) 4 x3⋅ 3 x⋅−1ε
−:= Since t3(0)=0 To solve this form the vector v
This is done with the Insert Matrix selection from the toolbar.The roots are now obtained with the polyroots(.) function.With MathCad the index on arrays in a matrix begin with 0.
v
1−
ε
3−
0
4
⎛⎜⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎟⎠
:=Since this is a cubic polynomial it will have at least one real rootand its the one required.
u polyroots v( ):= u
0.584− 0.522i−
0.584− 0.522i+
1.167
⎛⎜⎜⎝
⎞⎟⎟⎠
= x u2 0, := x 1.167=
fcf3x
:= fc 8.805 103×
1s
= The transfer function is Eq. 6.56 whichrequires h, a2, a1, & b1
h tanh1n
asinh1ε
⎛⎜⎝
⎞⎟⎠
⋅⎛⎜⎝
⎞⎟⎠
:= ε 0.349= h 0.531= θ113
π
2⋅:=
a2h
1 h2−
:= a11
1 h2−
sin θ1( )( )2−:= b1
12
11
h tan θ1( )⋅( )2+⋅:=
a2 0.626= a1 1.069= b1 1.706= The transfer function then becomes
TC f( ) K1
j f⋅a2 fc⋅
1+
⋅1
j f⋅a1 fc⋅
⎛⎜⎝
⎞⎟⎠
2 1b1
j f⋅a1 fc⋅
⋅+ 1+⎡⎢⎣
⎤⎥⎦
⋅:= y f( ) 20 log TC f( )( )⋅:=
For a check of the solution the magnitude of the transfer function in db will be plotted vs f
1 103× 1 104× 1 105×4−
3−
2−
1−
0
y f( )
f100 1 103× 1 104×1−
0.5−
0
0.5
1
y f( )
f Note that it is3 dB down at10.28 kHz
Equiripple box 3 db box
Now the circuit must be designed. This requires a 1st order lpf cascaded with a 3rd order lpf.This can be done by cascading the circuit shown in Fig. 6.9.a (page 83) with Fig. 6.11 (page83).
For the 1st order filter, since the dc gain is unity, K=1, pick RF=0 (a wire) and R1 infinity.
Pick C1 0.01μF:=
R11
2 π⋅ a2⋅ fc⋅ C1⋅:= R1 2.885 103
× Ω=
For the 2nd order filter, K=1, pick RF=0 & R3 open ckt.
fo fc a1⋅:= ωo 2 π⋅ fo⋅:= Q b1:=
Pick C1 0.1μF:= C2 0.01 C1⋅:=
R11
2 Q⋅ ωo⋅ C2⋅1 1
4 Q2⋅ C2⋅
C1−+
⎛⎜⎜⎝
⎞⎟⎟⎠
⋅:= R1 9.614 103× Ω= C1 1 10 7−
× F=
R21
2 Q⋅ ωo⋅ C2⋅1 1
4 Q2⋅ C2⋅
C1−−
⎛⎜⎜⎝
⎞⎟⎟⎠
⋅:= R2 297.455 Ω= C2 1 10 9−× F=
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1000 104 105
Chebyshev
Gai
n
Frequency
-70
-60
-50
-40
-30
-20
-10
0
10
1000 104 105
Chebyshev
Gai
n
Frequency
φ f( )180π
arg TC f( )( )⋅:=τp f( )
1−
2 π⋅
φ f( )f
⋅:= τg f( )1−
2 π⋅ fφ f( )d
d⎛⎜⎝
⎞⎟⎠
⋅:=
1 103× 1 104× 1 105×0
1 10 3−×
2 10 3−×
3 10 3−×
τp f( )
τg f( )
f
Page 86, Fig 6-14
Second Order Sallen Key BPF
fc 10.28 kHz⋅:= K 1:= Q 5:= j 1−:=fo fc:=
T f( ) K
1Q
j⋅ffo
⋅
jffo
⋅⎛⎜⎝
⎞⎟⎠
2 1Q
j⋅ffo
⋅+ 1+
⋅:= φ f( )180π
arg T f( )( )⋅:=
1 103× 1 104× 1 105×0.01
0.1
1
100−
50−
0
50
100
T f( ) φ f( )
f
R4 3 kΩ⋅:= R5 3 kΩ⋅:= C1 10 nF⋅:= C2 10 nF⋅:= ωo 2 π⋅ fo⋅:=
R1 1.8 kΩ⋅:= R2 1.8 kΩ⋅:= R3 1.8 kΩ⋅:= Ko 1R5
R4+:=
Given
KR2
R1 R2+
Ko R3⋅ C2⋅
R1 R2⋅
R1 R2+
⎛⎜⎝
⎞⎟⎠
C1 C2+( )⋅ R3 C2⋅ 1Ko R1⋅
R1 R2+( )−⎡⎢⎣
⎤⎥⎦
⋅+
⋅=
ωo1
R1 R2⋅
R1 R2+R3⋅ C1⋅ C2⋅
=
Q
R1 R2⋅
R1 R2+R3⋅ C1⋅ C2⋅
R1 R2⋅
R1 R2+
⎛⎜⎝
⎞⎟⎠
C1 C2+( ) R3 C2⋅ 1Ko R1⋅
R1 R2+−
⎛⎜⎝
⎞⎟⎠
⋅+
=
Find R1 R2, R3, ( )1.548 104
×
1.178 103×
2.189 103×
⎛⎜⎜⎜⎜⎝
⎞⎟⎟⎟⎟⎠
Ω=
0.01
0.1
1
103 104 105
Sallen Key BPFG
ain
Frequency
Second Order Infitite Gain Multiple Feedback BPF, page 87, Exp 6
fo 10.28 kHz⋅:= Q 5:= K 1:= ωo 2 π⋅ fo⋅:= j 1−:=
T f( ) K−
1Q
j⋅ffo
⋅
jffo
⋅⎛⎜⎝
⎞⎟⎠
2 1Q
j⋅ffo
⋅+ 1+
:=
1 103× 1 104× 1 105×0.01
0.1
1
T f( )
f
1 103× 1 104× 1 105×
100−
0
100
180 arg T f( )( )⋅
π
fC1 0.01 μF⋅:= C2 C1:= R1 10 kΩ⋅:= R2 240 Ω⋅:= R3 20 kΩ⋅:=
Given
R3 C1⋅
R1 C1 C2+( )⋅K=
1
R1 R2⋅ R3⋅ C1⋅ C2⋅
R1 R2+
ωo=
R3 C1⋅ C2⋅ R1 R2+( )⋅
R1 R2⋅( )C1 C2+
Q=
Find R1 R2, R3, ( )7.741 103
×
157.98
1.548 104×
⎛⎜⎜⎜⎝
⎞⎟⎟⎟⎠
Ω=
0.01
0.1
1
103 104 105
Infinite Gain Multiple Feedback BPF
Gai
n
Frequency
General Bi Quadratic Filter
T s( )
sωo
⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
sωo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
sωo
⎛⎜⎝
⎞⎟⎠
2 1Q
sωo⋅+ 1+
⎡⎢⎢⎢⎢⎢⎣
⎤⎥⎥⎥⎥⎥⎦
=
fo 10.28kHz:= N 2000:= i 0 N 1−..:= fstart 1kHz:= j 1−:=
fstop 100kHz:=
f i fstartfstopfstart
⎛⎜⎝
⎞⎟⎠
i
N 1−
⋅:=
Allpass
Q 3:= α 1:= β 0:= γ 1:=
T f( )
ffo
j⋅⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
j f⋅fo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
jffo
⎛⎜⎝
⎞⎟⎠
2 1Q
j f⋅fo
⋅+ 1+
:=
1 103× 1 104× 1 105×0.1
1
10
T fi( )
fi1 103× 1 104× 1 105×
100−
0
100
180arg T fi( )( )
π⋅
fi
Notch α 1:= β 0.5:= γ 1:= Q 3:= fo 10.28kHz:=
T f( )
ffo
j⋅⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
j f⋅fo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
jffo
⎛⎜⎝
⎞⎟⎠
2 1Q
j f⋅fo
⋅+ 1+
:=
1 103× 1 104× 1 105×
50−
0
50
180arg T fi( )( )
π⋅
f i1 103× 1 104× 1 105×
0.1
1
10
T fi( )
fi
Bandpass α 0:= β 0.5:= γ 0:= fo 10.28kHz:= Q 3:=
T f( )
ffo
j⋅⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
j f⋅fo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
jffo
⎛⎜⎝
⎞⎟⎠
2 1Q
j f⋅fo
⋅+ 1+
:=
1 103× 1 104× 1 105×0.1
1
10
T fi( )
fi
1 103× 1 104× 1 105×
50−
0
50
180arg T fi( )( )
π⋅
f i
Highpass α 0.5:= β 0:= γ 0:= fo 10.28kHz:= Q 3:=
T f( )
ffo
j⋅⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
j f⋅fo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
jffo
⎛⎜⎝
⎞⎟⎠
2 1Q
j f⋅fo
⋅+ 1+
:=
1 103× 1 104× 1 105×0.1
1
10
T fi( )
fi1 103× 1 104× 1 105×0
50
100
150
180arg T fi( )( )
π⋅
f i
Lowpass α 0.5:= β 0.5:= γ 1:= fo 10.28kHz:= Q 3:=
T f( )
ffo
j⋅⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
j f⋅fo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
jffo
⎛⎜⎝
⎞⎟⎠
2 1Q
j f⋅fo
⋅+ 1+
:=
1 103× 1 104× 1 105×0.1
1
10
T fi( )
fi1 103× 1 104× 1 105×
150−
100−
50−
0
180arg T fi( )( )
π⋅
fi
Third Order Elliptic Filter with 0.5 db ripple unity dc gain with stop band to cutoff ratio of 1.5
with a min attenuation in the stop band of 21.9 dB from page 79 in the lab manual
T s( )1
s0.76695ωc
1+
s1.6751 ωc⋅
⎛⎜⎝
⎞⎟⎠
21+
s1.0720 ωc⋅
⎛⎜⎝
⎞⎟⎠
2 12.3672
s1.0720 ωc⋅⋅+ 1+
⋅=
this requires that a first order filter be cascaded with a bi quad
for a notch at fp this means
fc 10.28kHz:= γ 1:= β12
:= fo 1.072 fc⋅:=
α
γ1.0721.6751
⎛⎜⎝
⎞⎟⎠
2+
20.705=:= Q 2.3672:=
T2 f( )
ffo
j⋅⎛⎜⎝
⎞⎟⎠
22 α⋅ γ−( )⋅
1Q
j f⋅fo
⎛⎜⎝
⎞⎟⎠
⋅ 2 β⋅ γ−( )⋅+ γ+
jffo
⎛⎜⎝
⎞⎟⎠
2 1Q
j f⋅fo
⋅+ 1+
:=
for the first order filterf1 0.7669 fc⋅:=
T1 f( )1
1 jff1⋅+
:=T f( ) T1 f( ) T2 f( )⋅:=
1 103× 1 104× 1 105×
0.2
0.4
0.6
0.8
T fi( )
fi
φ f( )180π
arg T f( )( )⋅:=
τg f( )1
2 π⋅−
fφ f( )d
d⎛⎜⎝
⎞⎟⎠
⋅:=
τp f( )φ f( )2 π⋅ f⋅
−:=
M f( ) 20 log T f( )( )⋅:=
1 103× 1 104× 1 105×30−
20−
10−
0
200−
100−
0
100
200
M fi( ) φ f(
fi
φ f( )180π
arg T f( )( )⋅:=τp f( )
φ f( )2 π⋅ f⋅
−:=
τg f( )1
2 π⋅−
fφ f( )d
d⎛⎜⎝
⎞⎟⎠
⋅:=
1 103× 1 104× 1 105×3−
2−
1−
0
1
2
4− 10 3−×
2− 10 3−×
0
2 10 3−×
4 10 3−×
6 10 3−×
M fi( )
fi
Third Order Bessel Filter f3 10.28kHz:=fc
f31.7557
:=
a2 2.3222:= a1 2.5415:= b1 0.69105:=
T f( )1
jf
a2 fc⋅⋅ 1+
1
jf
a1 fc⋅⋅⎛⎜
⎝⎞⎟⎠
2 1b1
j⋅f
a1 fc⋅⋅+ 1+
⋅:=
M f( ) 20 log T f( )( )⋅:= φ f( )180π
arg T f( )( )⋅:=
φ f( )180π
arg T f( )( )⋅:=τp f( )
φ f( )2 π⋅ f⋅
−:=
τg f( )1
2 π⋅−
fφ f( )d
d⎛⎜⎝
⎞⎟⎠
⋅:=
1 103× 1 104× 1 105×3−
2−
1−
0
1
2
2− 10−×
1− 10−×
0
1 10 3−×
2 10 3−×
M fi( )
fi
n 3=f3fc
1.7557= a1 2.5415:= b1 0.69105:=
a2 2.3222:=
T s( )1
sa2 ωc⋅
1+
1
sa1 ωc⋅
⎛⎜⎝
⎞⎟⎠
2 1b1
sa1 ωc⋅⋅+ 1+
⋅= 1st Order LPF2nd Order LPF
f3 10.28kHz:= fcf3
1.7557:= ωc 2 π⋅ fc⋅:=
C3 1nF:= R31
a2 ωc⋅ C3⋅1.171 104
× Ω=:=
Use Sallen Key
C1 0.01μF:= C2 0.1 C1⋅:=
R11
2 b1⋅ a1⋅ ωc⋅ C2⋅1 1 4 b1
2⋅
C2C1⋅−+
⎛⎜⎜⎝
⎞⎟⎟⎠
⋅ 1.47 104× Ω=:=
R21
2 b1⋅ a1⋅ ωc⋅ C2⋅1 1 4 b1
2⋅
C2C1⋅−−
⎛⎜⎜⎝
⎞⎟⎟⎠
⋅ 778.221 Ω=:=
5
5
4
4
3
3
2
2
1
1
D D
C C
B B
A A
0 0
00
Title
Size Document Number Rev
Date: Sheet of
Active Filters 1
Third Order Elliptic Filter
A
1 1Tuesday, March 11, 2008
Title
Size Document Number Rev
Date: Sheet of
Active Filters 1
Third Order Elliptic Filter
A
1 1Tuesday, March 11, 2008
Title
Size Document Number Rev
Date: Sheet of
Active Filters 1
Third Order Elliptic Filter
A
1 1Tuesday, March 11, 2008
Third Order Elliptic Filter
V
R2
14.44k
R2
14.44k
C10.295nC10.295n
R3
14.44k
R3
14.44k
C2
1n
C2
1n
R4
68.4k
R4
68.4k
U1
OPAMP
U1
OPAMP
+
-
OUTC3
0.705n
C3
0.705n
R568.4kR568.4k
R8
20.2k
R8
20.2k
U2
OPAMP
U2
OPAMP
+
-
OUT
V11Vac0Vdc
V11Vac0Vdc
U3
OPAMP
U3
OPAMP
+
-
OUT
R7
14.44k
R7
14.44k
R1
14.44k
R1
14.44kC41nC41n
Date/Time run: 03/23/08 10:30:41** Profile: "SCHEMATIC1-sweep" [ G:\brewdoc\ece3042hwk\WA2_Sprg08_Filter\Elliptic\elliptic-pspicefiles\s...
Temperature: 27.0
Date: March 23, 2008 Page 1 Time: 10:33:44
(A) sweep (active)
Frequency
1.0KHz 3.0KHz 10KHz 30KHz 100KHzV(R2:2)
0V
0.2V
0.4V
0.6V
0.8V
1.0V
Goal CutOff Frequency=10.28k
Third Order Elliptic Filter
(10.261K,945.407m)
(5.8374K,943.807m)
(17.219K,5.8725m)
Date/Time run: 03/23/08 10:30:41** Profile: "SCHEMATIC1-sweep" [ G:\brewdoc\ece3042hwk\WA2_Sprg08_Filter\Elliptic\elliptic-pspicefiles\s...
Temperature: 27.0
Date: March 23, 2008 Page 1 Time: 10:39:27
(A) sweep (active)
Frequency
1.0KHz 3.0KHz 10KHz 30KHz 100KHzdb(V(R2:2))
-50
-40
-30
-20
-10
0
10
(10.286K,-519.469m)
(5.7033K,-503.150m)
(17.378K,-46.230)
(26.975K,-21.928)
Third Order Elliptic Filter
Date/Time run: 03/23/08 10:40:03** Profile: "SCHEMATIC1-sweep" [ G:\brewdoc\ece3042hwk\WA2_Sprg08_Filter\Elliptic\elliptic-pspicefiles\s...
Temperature: 27.0
Date: March 23, 2008 Page 1 Time: 10:57:50
(A) sweep (active)
Frequency
1.0KHz 3.0KHz 10KHz 30KHz 100KHz-(P(V(R2:2))/Frequency)*(PI/180)/(2*PI) G(V(R2:2))
0
20u
40u
60u
80u
100u
(1.6090K,25.867u)
(10.290K,73.578u)
(10.290K,33.617u)
Phase and Group Delays for 3rd Order Elliptic Filter
Group Delay
Phase Delay
5
5
4
4
3
3
2
2
1
1
D D
C C
B B
A A
0
00
Title
Size Document Number Rev
Date: Sheet of
Active Filters 1
Third Order Bessel Filter
A
1 1Wednesday, March 12, 2008
Title
Size Document Number Rev
Date: Sheet of
Active Filters 1
Third Order Bessel Filter
A
1 1Wednesday, March 12, 2008
Title
Size Document Number Rev
Date: Sheet of
Active Filters 1
Third Order Bessel Filter
A
1 1Wednesday, March 12, 2008
Third Order Bessel Filter
V
R3
11.71k
R3
11.71k
C1
0.01u
C1
0.01u
U1
OPAMP
U1
OPAMP
+
-
OUT
C21nC21n
U2
OPAMP
U2
OPAMP
+
-
OUTC31nC31n
V11Vac0Vdc
V11Vac0Vdc
R1
14.7k
R1
14.7k
R2
778.221
R2
778.221
Date/Time run: 03/23/08 11:04:29** Profile: "SCHEMATIC1-sweep" [ G:\brewdoc\ece3042hwk\WA2_Sprg08_Filter\Bessel\bessel-pspicefiles\schem...
Temperature: 27.0
Date: March 23, 2008 Page 1 Time: 11:07:14
(A) sweep (active)
Frequency
1.0KHz 3.0KHz 10KHz 30KHz 100KHzDB(V(U2:OUT))
-60
-50
-40
-30
-20
-10
0
Third Order Bessel Filter
(10.286K,-3.0162)
Date/Time run: 03/23/08 11:04:29** Profile: "SCHEMATIC1-sweep" [ G:\brewdoc\ece3042hwk\WA2_Sprg08_Filter\Bessel\bessel-pspicefiles\schem...
Temperature: 27.0
Date: March 23, 2008 Page 1 Time: 11:15:21
(A) sweep (active)
Frequency
1.0KHz 3.0KHz 10KHz 30KHz 100KHz(P(V(U2:OUT))/Frequency)*(PI/180)*(-1/(2*PI)) G(V(U2:OUT))
0
10u
20u
30u
40u
Third Order Bessel FilterGroup Delay
Phase Delay
(10.289K,25.407u)
(10.289K,26.885u)(1.5565K,27.188u)
Georgia Institute of Technology
School of Electrical and Computer Engineering
ECE 3042 Microelectronic Circuits Laboratory Verification Sheet
NAME:________________________________________ SECTION:___________________________
GT NUMBER:___________________________________ GTID:______________________________
Experiment 6: Op‐Amp Active Filters
Procedure Time Completed Date Completed Verification (Must demonstrate
circuit)
Points Possible
Points Received
1. Third‐Order Butterworth LPF
16
2. Third‐Order Chebyshev LPF
16
3. Second‐Order Band‐Pass
16
4. Second‐Order Notch
16
5. First‐Order All‐Pass 16
6. Second‐Order All‐Pass
20
Enter your critical frequency below:
fcrit
To be permitted to complete the experiment during the open lab hours, you must complete at least three procedures during your scheduled lab period or spend your entire scheduled lab session attempting to do so. A signature below by your lab instructor, Dr. Brewer, or Dr. Robinson permits you to attend the open lab hours to complete the experiment and receive full credit on the report. Without this signature, you may use the open lab to perform the experiment at a 50% penalty.
SIGNATURE:____________________________________ DATE:____________________________________
ECE 3042 Check‐off Requirements for Experiment 6 Make sure you have made all required measurements before requesting a check‐off. For all check‐offs, you must demonstrate the circuit or measurement to a lab instructor. All screen captures must have a time/date stamp. Do not follow the procedures in the lab manual. Only Bode magnitude plots are required—it is not necessary to measure the phase. Do not follow the lab report format for this experiment. Instead, display two Bode plots and their associated tables per page. 1. Third‐Order Butterworth Low‐Pass Filter
Bode gain magnitude plot over frequency range of 100Hz to 100khz. Measurement of dc gain and ‐3dB frequency. Table comparing these measured values to the design specifications. Show percent error.
2. Third‐Order Chebyshev Low‐Pass Filter Bode gain magnitude plot over frequency range of 100Hz to 100khz. Measurement of dc gain, dB ripple, and ‐3dB frequency. Table comparing these measured values to the design specifications. Show percent error.
3. Second‐Order Bandpass Filter Bode gain magnitude plot over frequency range of 100Hz to 100khz. Measurement of center frequency fo, ‐3dB bandwidth BW, and Q. Q is given by fo /BW. Table comparing these measured values to the design specifications. Show percent error.
4. Second‐Order Notch Filter
• Implement filter with the general biquad circuit from the class notes. Save this filter—it can be modified to obtain the second‐order all‐pass transfer function.
Bode gain magnitude plot over frequency range of 100Hz to 100khz. Measurement of center frequency fo, ‐3dB bandwidth BW, and Q. Q is given by fo/BW. Table comparing these measured values to the design specifications. Show percent error.
5. First‐Order All‐Pass Filter Bode gain magnitude plot over frequency range of 100Hz to 100khz. Measurement of dc gain. Scope screen capture showing measured fcrit, the frequency at which the phase shift between input and output is 90 deg. Connect input to ch1 and output to ch2. Display measured Vpp for each channel, the frequency, and the phase.
Table comparing these measured values to the design specifications. Show percent error. 6. Second‐Order All‐Pass Filter
Bode gain magnitude plot over frequency range of 100Hz to 100khz. Measurement of dc gain. Scope screen capture showing measured fcrit, the frequency at which the phase shift between input and output is 180 deg. Connect input to ch1 and output to ch2. Display measured Vpp for each channel, the frequency, and the phase.
Table comparing these measured values to the design specifications. Show percent error.