first order low pass filter ecse-2010ssawyer/circuitsspring2020_all/... · 2016. 4. 25. · 5 1...
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1
ECSE-2010
Lecture 24 continued
� First order low pass filter
� First order high pass filter
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inv
R
CCv
+
−
C C inAs 0 (DC): Z ; v vω → → ∞ =
C CAs : Z 0; v 0ω → ∞ → =
AC
inv
R
CCv
+
−
Let's look at this in the s-Domain
C
in No Initial Stored Energy
V (s)Find H(s)
V (s)=
AC
2
C c
c
V (s) 1 RC 1H(s)
V(s) s 1 RC s 1 s
ωττ ω
= = = =+ + +
FIRST ORDER FILTERS
c
c
H(j ) H(j ) /j
ωω ω θω ω
= =+
c
2 2c
H(j ) Gainωω
ω ω= =
+
1
c
tan Phase Shiftωθω
−= − =
c
2 2c
H(j ) Gainωω
ω ω= =
+
c 10
1At 10 , Gain ; 20log (Gain) 20 dB
10ω ω= ≈ ≈ −
c 10
1At 100 , Gain , 20log (Gain) 40 dB
100ω ω= ≈ ≈ −
c 10
1At 1000 , Gain , 20log (Gain) 60 dB
1000ω ω= ≈ ≈ −
Gain decreases at a Slope 20 dB decade= −
1st ORDER LOW PASS FILTER
1st ORDER LOW PASS FILTER
c
2 2c
H(j ) Gainωω
ω ω= =
+
0
cω ω.1 1 10 100.01
10log scale
1020log H
Gain in dB
•3−
20−
•
•
•40−
Asymptotes
20 dB/decade−
Bode Plot of Gain
••
c
2 2c
H(j ) Gainωω
ω ω= =
+
0
cω ω.1 1 10 100.01
10log scale
1020log H
Gain in dB
•
20−
•
•
•40−
Asymptotes
20 dB/decade−
Bode Plot of Gain
••
Passband Stopband
1st ORDER LOW PASS FILTER
3
1
c
tan Phase Shiftωθω
−= − =
1st ORDER LOW PASS FILTER
cFor "low frequencies" ( ): 0ω ω θ<< →0
cFor "high frequencies" ( ): 90ω ω θ>> → −0
cFor : 45ω ω θ= = −
C
Most of the change in occurs for .1 10ωθω
≤ ≤
C
Corner Frequencies at .1, 10ωω
=
00
cω ω.1 1 10 100.01
10log scale
θ
•
045−•
•
•
090−
Asymptotes
••
1st ORDER LOW PASS FILTER
1
c
tan Phase Shiftωθω
−= − =
•
Bode Plot of θ
1st ORDER LOW PASS FILTER
R
Coutv
+
−inv
+
−
out c
in c
V (s) 1 RCH(s)
V (s) s 1 RC s
ωω
= = =+ +
Low Pass Filter=>
Low Frequencies "Pass"; High Frequencies "Stopped"
BETTER 1st ORDER LOW PASS
invR
C
1R FR
outv
F
1
cF
1 c
R 1 RCH(s) (1 )
R s 1 RC
R (1 )
R s
ωω
= ++
= ++
CCV+
CCV−
4
1st ORDER HIGH PASS FILTER
out
in c
V (s) s sH(s)
V (s) s 1 RC s ω= = =
+ +High Pass Filter=>
inv
+
−R
C
outv
+
−
High Frequencies "Pass"; Low Frequencies "Stopped"
10As 0, Gain 0, 20log (Gain)ω → → → −∞
2 2c
H(j ) Gainωω
ω ω= =
+
cc 2 2
c
.01At .01 , Gain .01
(.01)
ωω ωω
= = ≈+
1020log (Gain) 40 dB= −
1st ORDER HIGH PASS FILTER
2 2c
H(j ) Gainωω
ω ω= =
+
c 10At .01 , Gain .01; 20log (Gain) 40 dBω ω= ≈ ≈ −
c 10At .1 , Gain .1, 20log (Gain) 20 dBω ω= ≈ ≈ −
Gain increases at a Slope 20 dB decade= +
1st ORDER HIGH PASS FILTER
c 10
1At , Gain , 20log (Gain) 3 dB
2ω ω= = = −
1st ORDER HIGH PASS FILTER
2 2c
H(j ) Gainωω
ω ω= =
+
0
cω ω.1 1 10 100.01
10log scale
1020log H
Gain in dB
3−
20−
•
•
•40−
Asymptotes20 dB/decade+
Bode Plot of Gain
••
5
1st ORDER HIGH PASS FILTER
2 2c
H(j ) Gainωω
ω ω= =
+
0
cω ω.1 1 10 100.01
10log scale
1020log H
Gain in dB
20−
•
•
•40−
Asymptotes20 dB/decade+
Bode Plot of Gain
••
Stopband Passband
0 1
c
90 tan Phase Shiftωθω
−= − =
1st ORDER HIGH PASS FILTER
0cFor "low frequencies" ( ): 90ω ω θ<< →
0cFor "high frequencies" ( ): 0ω ω θ>> →0
cFor : 45ω ω θ= = +
C
Most of the change in occurs for .1 10ωθω
≤ ≤
C
Corner Frequencies at .1, 10ωω
=
00
cω ω.1 1 10 100.01
10log scale
θ
•
045+•
•
•
090+
Asymptotes
••
1st ORDER HIGH PASS FILTER
0 1
c
90 tan Phase Shiftωθω
−= − =
•
Bode Plot of Phase Shift
BETTER 1st ORDER HIGH PASS
invC
R
1RFR
outvCCV+
CCV−
F
1
F
1 c
R sH(s) (1 )
R s 1 RC
R s (1 )
R s ω
= ++
= ++
6
1. Find poles, zeros
2. In each region between sequential poles/zeros, can use H(jω) or H(s)
a) Draw line in first region based on three possibilities based on frequency range
b) Repeat for every region
3. Add corrections at poles/zeros
a) n*pole n(-3db) correction at that pole
b) n*zero n(+3db) correction at that zero
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We have three possibilities
Note: Can ALSO analyze using H(s) instead of
H jω( ) ∝ ωn+n 20⋅
db
decslope
H jω( ) ∝constant K, constant 20 log⋅ K⋅
H jω( ) ∝1
ωn n− 20⋅db
decslope
Hjω( )
1. Crossing a n*pole is a change of phase of –n*90 deg (absolute change)
a) Changing over approximately two decades
b) Specifically, 0.1 and 10 times ωc
2. Crossing a n*zero is a change of phase +n*90 deg (absolute change)
a) Changing over approximately two decades
b) Specifically, 0.1 and 10 times ωc
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1. Crossing a n*pole is a change of phase of –n*90 deg (absolute change)
a) Changing over approximately two decades
b) Specifically, 0.1 and 10 times ωc
2. Crossing a n*zero is a change of phase +n*90 deg (absolute change)
a) Changing over approximately two decades
b) Specifically, 0.1 and 10 times ωc
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SLA have slopes of +/- n*45 deg per decade slope (+ is a zero, - is a pole)
7
ECSE-2010
Lecture 19.2
� Cascaded filters/Parallel filters
� First order Bandpass filter
� First order Bandstop (Notch filter)
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cL
cH cL
sH(s) K
s s
ωω ω
= + +
cL cH
H H L L
Let's Design Such that
R C R C
ω ω⇒
�
�
cL
2 2 2 2cH cL
H(j ) Kωωω
ω ω ω ω
= + +
High Pass Low Pass
>
>
dBK
ωcHω cLω
10log scale
Gain in dB20 dB/decade+
BANDPASS FILTER
cH.1ω cL10ω
dBK 20−
20 dB/decade−
Passband
Stopband Stopband
cL cHω ω�
cL cHBandwidth ω ω= −
>
8
stFor Low Frequencies Looks Like a 1 Order Low Pass⇒
cLAL H
B cL cH
R sH(s) H (s) H (s) 1
R s s
ωω ω
= + = + + + +
stFor High Frequencies Looks Like a 1 Order High Pass⇒
cH cL
L L H H
Let's Design Such that
R C R C
ω ω⇒
�
�
>
>
dBK
ωcLω cHω
10log scale
Gain in dB
BANDGAP OR NOTCH FILTER
cH cLω ω�
dB20
dec−
Stopband
Passband Passband
dB20
dec+
cL cHω ω
cH cLBandwidth ω ω= −
>
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a) ( ) ( )1000
1000
+=
ssH
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a) ( ) ( )1000
1000
+=
ssH
9
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-40
-30
-20
-10
0
Mag
nitu
de (
dB)
101
102
103
104
105
-90
-45
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/s)
>> sys=tf([1000],[1 1000]);>> h=bodeplot(sys);grid>>
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( ) ( )2
2
1000
1000
+=
ssHb)
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( ) ( )2
2
1000
1000
+=
ssHb)
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-80
-60
-40
-20
0
Mag
nitu
de (
dB)
101
102
103
104
105
-180
-135
-90
-45
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/s)
>> sys=tf([1000000],[1 2000 1000000]);>> h=bodeplot(sys);grid
10
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( ) ( )( )1000010010
2
++=
ss
ssHc)
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( ) ( )( )1000010010
2
++=
ss
ssHc)
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( ) ( )( )1000010010
2
++=
ss
ssHc)
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( ) ( )( )1000010010
2
++=
ss
ssHc)
11
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-60
-40
-20
0
20
40
Mag
nitu
de (
dB)
100
101
102
103
104
105
106
0
45
90
135
180
Pha
se (
deg)
Bode Diagram
Frequency (rad/s)
>> sys=tf([10 0 0],[1 10100 1000000]);>> h=bodeplot(sys);grid>> setoptions(h,'MagLowerLimMode','manual','MagLowerLim',-60)
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2) Bode plot-multiple stages
R1
100
R21k
L10.1
1
2
L2
0.001
12
U1
OPAMP
+
-
OUT
R310k
R4
90k
0
00
VoutVin
H1(s) H2(s) H3(s)
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12
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-30
-20
-10
0
10
20
Mag
nitu
de (
dB)
101
102
103
104
105
106
107
108
-90
-45
0
45
90
Pha
se (
deg)
Bode Diagram
Frequency (rad/s)
>> sys=tf([10000000 0],[1 1001000 1000000000]);>> h=bodeplot(sys);grid