ecen 622 tamu active rc filters - texas a&m …s-sanchez/622 lecture 2 active... · active rc...
TRANSCRIPT
ACTIVE RC FILTERS
By Edgar Sánchez-Sinencio 1
1. Basic Building Blocks
• First-Order Filters
• Second-Order Filters, using multiple VCVS
• Second-Order Filter, using one VCVS ( Op Amp)
• State-Variable Biquad
2. Non-Ideal Active – RC Filters
• Using VCVS ( Op Amp) vs. VCCS ( transconductance Amp)
• Second-Order Non-idealities
• Fully Differential Versions
• Fully Balanced, Fully Symmetric Balance Circuits
3. Introduction to Matlab and Simulink for filter Design and
filter approximation techniques
ECEN 622 TAMU
622 (ESS) A1
ACTIVE - RC FILTERS
The basic building block is illustrated below
1
2
1
2
i
o
Z
Z1
A
11
Z
Z
sV
sVsH
cases particularconsider Next we
Z
ZsH
then, A that assume usLet
1
2
rentiator Diffe CRssH
Integrator CsR
1sH
2
21
Vi
Z1 Z2
A Vo
Vi
R1
C2
Vo
Vi
R2C1
Vo
By Edgar Sánchez-Sinencio 2
3
Z1
or
ZF = Z2
Can be:
EXAMPLE: Let
FF
FF
11
11
CsR1
RZ,
CsR1
RZ
Assuming ideal op amp A . Then using (1)
p
zn
FF
111F
1
011
s1
s1K
)CsR1(
CsR1RR
V
VH
(4)
1H
1
F
R
R
z p
zp 1
F
R
R
F
1
C
C
p z
zp
R
sC
1
R
C
R C
sC
sRC1
C
R sRC1
R
By Edgar Sánchez-Sinencio
4
Particular cases are easily derived from (3) and (4)
Integrator: F1 R,0C
1FFF1
F1
RsC
1
CsR
1
R
RH
Differentiator ; 0C,R F1
1F111
F1 CsRCsR
R
RH
Low-Pass: 0C1
FF
1
F
1CsR1
R
R
H
High-Pass: 1R
FF
11
1
F1
CsR1
CsR
R
RH
By Edgar Sánchez-Sinencio
5
One pole and one zero
What are the key differences between Eqs. (4) and (5)?
(5) CsR1
CsR1
C
C
CsR1
sC
sC
CsR1
V
V
11
FF
F
1
11
1
F
FF
i
o
Exercise 1. Obtain the transfer function of the following circuit.
A Vo
R1 RF1 RF2
CF
Vi
C1
Vi
R1 RF CFC1
By Edgar Sánchez-Sinencio
6
Second-Order Filters Based on a Two-Integrator Loop.
• We can design a second-order filter by cascading two inverters. i.e.
)6(
1RCRCsRCRCs
R
R
R
R
RsC1RsC1
R
R
R
R
v
V
2F2F1F1F2F2F1F1F2
2
2F
1
1F
2F2F1F1F
2
2F
1
1F
i
o
What are the locations of the poles?
2F2F1F1F
2F2F1F1F2
2F2F1F1F2F2F1F1Fp
RCRC2
RCRC4RCRCRCRCs
2,1
ViR1
RF1RF2
CF2CF1
R2Vo
By Edgar Sánchez-Sinencio
7
To have complex poles it requires that
?0RCRC2RCRC 2F2F1F1F2
2F2F2
1F1F
Which it is impossible to satisfy. Therefore, cascading two first-order filter yield a
second-order filter with only real poles.
The general form of the second order two-integrator loop has the following topology.
(7a)
Z
Z
Z
Z1
Z
Z
Z
Z
H
2
2F
1
1F
2
2F
3
1F
-1
Vi
Vo
Z3 ZF1
Z2
ZF2
Z1
By Edgar Sánchez-Sinencio
8
Note the similarity of Eq. (7a) with (2). Also observe that A “-1” needs to be inserted
before or after the second inverter to yield a negative feedback loop.
Let us consider the following filter where
2F2F
2F2F
1F1F221133
RsC1
RZ,
sC
1Z,RZ,RZ,RZ
Thus Eq. (7a) yields:
2o
o2
21o
22F11F2F2F
2
22F31F
2F2F
22F
11F
2F2F
22F
31F
sQ
sRCRC
1
RC
ss
RCRC
1
H
RsC1
RR
RsC
11
RsC1
RR
RsC
1
H
By Edgar Sánchez-Sinencio
9
By injecting in different current summing nodes a general biquad filter can be obtained.
22F11F2F2F
2
3113211F222F31F
3
o
2F2F
22F
11F
2F2F
12F
2F3
2F2F
2
2F2
2F2F
RR
31F3
o
RCRC
1
RC
ss
KRsCVKRsCVRCRC
1V
V
RsC1
RR
RsC
11
RsC1
VR
RK
RsC1
R
RK
RsC1RsC
1V
V
22F
R1CF1 r
r
RF2
CF2
Vo
R2
R3VLP=V3
VBP=V2
-VBP=V1
r/K2
RF2/K3
Vo2Vo1
By Edgar Sánchez-Sinencio
10
Exercise 2. Obtain the expressions of Vo1 and Vo2.
More general biquad expressions and topologies can be obtained by adding a summer.
o301o20in10oT VKVKVKV
Exercise 3. Draw an active-RC topology of the block diagram show above.
Exercise 4 a) For only obtain Vo and Vo1 when instead of the resistor RF2/K3 a
capacitor K4 CF2 is used. b) For only obtain Vo1 when the resistor R3 is replaced
by a capacitor KHPCF1.
0V1
0V3
S
Vin
Vo1
Vo
-K10
-K20
-K30
By Edgar Sánchez-Sinencio
11
By using also the positive input of the op amp other useful filters can be obtained.
V2
V1
Z1 Z2
ZR
ZC
1
2
1
22
CR
R1
1
2
o
Z
Z1
A
11
Z
Z1V
ZZ
ZV
Z
Z
V
Example. Phase shifter Z2=R2=R1, Z1=R1, ZR=R 21C VV withA and sC
1Z
sRC1
sRC1
sRC1
sRC2sRC12
sRC1
sRC1
V
V
1
o
By Edgar Sánchez-Sinencio
12
Sallen and Key Bandpass Filter
K
Vi
Vo
C1
R2
C2R1
R3
K is a non-inverting amplifier
Using Nodal Analysis
(3) V
(2) 0
2
122 V-
(1) 11
2o
21
1322
31211
KV
RsCVsC
R
V
R
VVsC
RRCCsV io
21231
31
132
3
1
3
22
2
11
11
CCRRR
RR
CR
s
R
R
R
RK
CRs
CR
sK
sV
sVsH
i
o
By Edgar Sánchez-Sinencio
13
A particular case is for R1=R2 =R3=R, C1=C2=C
Then
Q
24K and
2RC
Q and givena for or
K4
2Q;
RC
2
o
o
2
2o
Exercise 5. Prove the transfer function is a BP filter of the following circuit
212122121
212
11
i
o
CCRR1K
1s
CR
1
CRR
RR
1K
1s
CR
s
1K
K
sV
sVsH
-K
Vi
R1 C2C1
Vo
R2
By Edgar Sánchez-Sinencio
A4In the past before IC fabrication, active filters implementation preferred one op amp
structure. One very popular type is the Sallen and Key unity gain implementations.
QRC
CQC
CC
RRR
sQ
s
sH
o
oo
oLP
21
4 2
2
1
21
22
2
One also popular topology is the Rauch Filter
22113211
2
1322
CRCR
1
R
1
R
1
R
1
C
ss
CRCR
1
sH
LP Rauch Filer
LP Sallen - Key
ViVo
R2 R1
C1
C2
1
ViVo
C2
C1
R3
R1
R2
By Edgar Sánchez-Sinencio 14
15
Another technique for analysis and design based on state-variable uses building blocks.
CIRCUIT REPRESENTATION
V1
V2
R2
R1C1
Vo1
S 1/sV1
V2
22CR
1
11CR
1
Vo1
Note: V1 and V2 can take
any value including Vo.
S
V1
V2
V3
V4
V5
K4
Vo2
-K1
-K2
K3
K5
V1
V2
V3
V4
V5
R1
R2
R3
R4
RF
Vo2
R5
By Edgar Sánchez-Sinencio
16
Let us apply to a two-integrator loop plus Mason’s Rule.
For Second-topology
S 1/s 1/s1oVinV 3oV
-Ko1-1
-KQ
Ko2
2oV
-K1
BP KKsKs
sVK
s
KK
s
K1
Vs
K
V
HPKKsKs
VsK
s
KK
s
K1
VKV
2o1oQ2
in1
22o1oQ
in1
o2
2o1oQ2
in2
1
22o1oQ
in11o
By Edgar Sánchez-Sinencio
17
Next we show that we can go from an Active-RC representation into a
block diagram or vice versa.
KHN Biquad Filter
V1
R4
R1
R5 R6 R7
R2
C1 C2
VHP VBP VLP
S 1/s 1/sVi
3
5
R
R
-R5/R4
VLPVHP VBP
16CR
1
27CR
1
Q43
5
21
1 KR||R
R1
RR
R
By Edgar Sánchez-Sinencio
18
2716
45
16
Q2
2
3
5
22716
45
16
Q
3
5
i
HP
CRCR
RRs
CR
Ks
sR
R
sCRCR
RR
s
1
CR
K1
R
R
V
V
By Edgar Sánchez-Sinencio
ACTIVE RC FILTERS
By Edgar Sánchez-Sinencio 19
1. Basic Building Blocks
• First-Order Filters
• Second-Order Filters, using multiple VCVS
• Second-Order Filter, using one VCVS ( Op Amp)
• State-Variable Biquad
2. Non-Ideal Active – RC Filters
• Using VCVS ( Op Amp) vs. VCCS ( transconductance Amp)
• Second-Order Non-idealities
• Fully Differential Versions
• Fully Balanced, Fully Symmetric Balance Circuits
3. Introduction to Matlab and Simulink for filter Design and
filter approximation techniques
ECEN 622 TAMU
ELEN 622 (ESS)
Op Amp Non-Idealities
A
iV1Z
2Z
A
V0
oV
21
1
1
2
1
2
1
2
where
11
11
ZZ
Z
A
Z
Z
A
Z
Z
A
Z
Z
sH
Integrator
Case 1 sAs
GB;
sCZ;RZ
2
211
1
11
1
1
1
GBRC;
GBssRCGB
RCs
GB
sRCsH
o
GBtanjH;
RCjGB
RCjH
90
2
1 1
2
Non-Ideal Active-RC Integrators
By Edgar Sánchez-Sinencio 20
o.
GB;
jAtan
GBtan
75
10
1i.e.
1 o11
M
o
o
oo
o
GB
j
GB
jH
1
1
11
2
2
error 5
10 i.e.
1
11
2
2
2
2
0
%~
.GB
GB
GB
M
o
o
M
By Edgar Sánchez-Sinencio 21
It follows that the ideal -6 dB/octave roll-off expected from an ideal integrator
changes to -12 dB/octave at the frequency of the parasitic pole given by
CRs tp
1
which may be approximated by,
CRs tt
1 for
dB
A
o
0
-6 dB/octave
1/AoC
R
1/CR
-12 dB/octave
t
22By Edgar Sánchez-Sinencio
In general
jXR
jT
1
then we define the integrator Q-factor by
jAQ
R
XQ
tI
I
GBQ t
L
23By Edgar Sánchez-Sinencio
Making an analogy of QL of an inductor
L
oL
R
LQ
RLL
Lossy Part
For an integrator one can obtain
jAGB
GBGB
RC
RCQI
112
Miller Integrator
C
Rvi
voS
GB
By Edgar Sánchez-Sinencio 24
How can we compensate this degradation of performance?
a)
CGBRC
1
jAQI
C
R
RC
vi vo
s
GBsA
If we make
Ideally we obtain
RCsV
sV
i
o 1
This integrator yields a positiveC
R
R1
R2
A1
A2
vivo
By Edgar Sánchez-Sinencio 25
ACTIVE – RC INTEGRATOR: Pole Shift and Predistortion
622 (ESS)
26
(1b)
1sRCsA
1sRC
1
V
V
(1a) sRC
1
sRC
11
sA
11
sRC
1
V
V
i
o
i
o
)s(A
+
-Vo
Vi
R C
A(s)
; where Ao is the DC gain and 3dB the dominant pole in
open loop.
Then (1b) becomes
(2)
RCsAs
RC
A
RCRC
1Ass
RC
A
V
V
dB3dB3o
2
dB3o
dB3dB3dB3o
2
dB3o
i
o
Let GB=Ao3dB
G. Daryanani, “Principles of Active Network Synthesis and Design,” John Wiley and Sons, 1976.
dB3
dB3o
s
A)s(ALet
By Edgar Sánchez-Sinencio
27
The roots of the denominator are
(3a)
RCGB
41
2
GB
2
GBP
2/1
2dB3
2,1
Using the approximation 1–X)1/2≅ 1 −X/2 for X<<1, then
)b3(
RCGB
21
2
GB
2
GBP
2dB3
2,1
Thus the roots yield
GBRCA
1GBP
RCA
1P
o2
o1
By Edgar Sánchez-Sinencio
28
The Bode Plot Looks Like
By Edgar Sánchez-Sinencio
29
PREDISTORTION; FREQUENCY COMPENSATION
In order to relax the bandwidth op amp requirement one can use a RC or CC on
the Miller Integrator. That is
B. Wu and Y. Chiu, “A 40nm CMOS Derivative-Free IF Active-RC BPF with Programmable Bandwidth and Center Frequency Achieving Over 30dBm IIP3”,
IEEE JSSC, Vol. 50, No. 8, pp 1772-1784, August 2015.
GB
1R Cor
GB
1C R Use CC
Equivalent
Vi
RC
R
C
VoA(s)
Vi
R RCC
Vo
Vi
R
C
Vo
Vi R
C
Vo
CC
CC
By Edgar Sánchez-Sinencio
30
Vi
Vi
Z1
Z1 Z2
ZF
A
Zo ZL
ZL
Vo
Vo
Gm
Using VCVS vs. VCIS in Active-RC Filters
The motivation is to use OTA (VCIS) instead of more power hungry Op Amp (VCVS)
1
2
i
o
L1
21
1m
2m
m
L1
21
1
2m
1
2
i
o
1
F
i
o
1
F
1
F
i
o
oo
Z
Z
V
V
ZZ
ZZ
Z
1g
Z
1g
g
ZZ
ZZ
Z
1
1
Zg
11
Z
Z
V
V
Z
Z
V
V
then,A If
Z
Z1
A
11
Z
Z
V
V
0ZRFor
By Edgar Sánchez-Sinencio
Using VCVS
𝑉𝑖 − 𝑉𝑥𝑍1
+𝑉𝑜 − 𝑉𝑥𝑍𝐹
= 0
𝑉𝑜 = −𝐴𝑉𝑥
Vx
Signal flow graph:
VxVi Vo−𝐴
𝑉𝑥 = 𝑉𝑖𝑍𝐹
𝑍1 + 𝑍𝐹+ 𝑉𝑜
𝑍1𝑍1 + 𝑍𝐹
β
𝑍𝐹𝑍1 + 𝑍𝐹
𝑍1𝑍1 + 𝑍𝐹
Using Mason’s rule:
𝑉𝑜𝑉𝑖=
−𝐴𝑍𝐹
𝑍1 + 𝑍𝐹
1 + 𝐴𝑍1
𝑍1 + 𝑍𝐹
=−𝑍𝐹/𝑍1
1 +1𝐴
1 +𝑍𝐹𝑍1
Thus as 𝐴 → ∞ the gain becomes −𝑍𝐹/𝑍1
By Edgar Sánchez-Sinencio 31
Using VCCS
𝑉𝑖 − 𝑉𝑥𝑍1
+𝑉𝑜 − 𝑉𝑥𝑍2
= 0
𝑉𝑜 − 𝑉𝑥𝑍2
+𝑉𝑜𝑍𝐿
= −𝐺𝑚𝑉𝑥
Vx
Signal flow graph:
VxVi Vo
1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿
𝑉𝑥 = 𝑉𝑖𝑍2
𝑍1 + 𝑍2+ 𝑉𝑜
𝑍1𝑍1 + 𝑍2
β
𝑍2𝑍1 + 𝑍2
𝑍1𝑍1 + 𝑍2
Using Mason’s rule:
𝑉𝑜𝑉𝑖=
𝑍2𝑍1 + 𝑍2
1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿
1 −1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿
𝑍1𝑍1 + 𝑍2
=−𝑍2/𝑍1
1 +1
𝐺𝑚𝑍2 − 11 +
𝑍2𝑍1
1 +𝑍2𝑍𝐿
What conditions do we need to impose on𝐺𝑚 for proper operation?
𝑉𝑜 =1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿
𝑉𝑥
By Edgar Sánchez-Sinencio 32
Using VCCS: Acceptable 𝐺𝑚 Range
Vx
Signal flow graph:
VxVi Vo
1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿
𝑍2𝑍1 + 𝑍2
𝑍1𝑍1 + 𝑍2
𝑉𝑜𝑉𝑖=
−𝑍2/𝑍1
1 +1
𝐺𝑚𝑍2 − 11 +
𝑍2𝑍1
1 +𝑍2𝑍𝐿
Note from the signal flow graph that having a negative feedback loop requires 𝐺𝑚𝑍2 > 1
For the gain to approach the ideal gain of −𝑍2/𝑍1, we need
𝐺𝑚𝑍2 − 1 ≫ 1 +𝑍2
𝑍11 +
𝑍2
𝑍𝐿𝐺𝑚𝑍2 ≫ 1 + 1 +
𝑍2
𝑍11 +
𝑍2
𝑍𝐿
Thus, guaranteeing this second condition automatically guarantees the negative feedback condition
By Edgar Sánchez-Sinencio 33
Using VCCS: Practical Considerations
Vx
In practice, 𝑍𝐿 = 𝑅𝑜|| 𝑍𝐿𝑜𝑎𝑑 where 𝑅𝑜 is the OTA’s output resistance and 𝑍𝐿𝑜𝑎𝑑 is the external load impedance
𝐺𝑚𝑍2 ≫ 1 + 1 +𝑍2𝑍1
1 +𝑍2𝑍𝐿
One should note that 𝐺𝑚 and 𝑅𝑜 are not independent since increasing current to increase 𝐺𝑚 will reduce 𝑅𝑜.To a first order, one can consider 𝐴 = 𝐺𝑚𝑅𝑜 to be constant.
Finally, in cascaded filter designs, the load of one stage is the input resistor of the next one. We can thus assume 𝑍𝐿𝑜𝑎𝑑 = 𝑍1 as a realistic condition (𝑍𝐿𝑜𝑎𝑑 = ∞ places a looser constraint on 𝐺𝑚)
Substituting for 𝑍𝐿 as described above yields the following constraint on 𝐺𝑚:
𝐺𝑚 ≫1
𝑍1
2 1 + 𝑍1/𝑍2 + 𝑍2/𝑍1
1 −1𝐴1 + 𝑍2/𝑍1
By Edgar Sánchez-Sinencio 34
Numerical Example
• Ideal VCVS and VCCS components from Cadence were used to simulate the above circuits.
• Two configurations were tested:1. Unity gain inverting amplifier (𝑍2 = 𝑍1 = 𝑅)2. Lossy integrator with corner frequency 1 MHz (𝑍2 = 𝑅||𝐶)
• To have a fair comparison, the value of 𝐴 was fixed to 30 for both the VCVS and VCCS implementations (thus the VCCS had 𝑅𝑜 = 30/𝐺𝑚). This is a typical value for the voltage gain of a single stage amplifier.
• In all tests, 𝑅 = 100𝑘Ω, the output resistance of the VCVS is set to 1 𝑘Ω and a load capacitance is added to the output of the amplifier to give an output pole at 10 MHz.
• With these numbers, the constraint on 𝐺𝑚 is 𝐺𝑚 ≫ 54 𝜇𝑆
• The value of 𝐺𝑚was swept from 30 𝜇𝑆 to 600 𝜇𝑆 and the simulation results are shown in the following slides.
By Edgar Sánchez-Sinencio 35
Simulation Results: Inverting Amplifier
VCVS Response
Increasing 𝐺𝑚
VCVS Response
Increasing 𝐺𝑚
By Edgar Sánchez-Sinencio 36
Simulation Results: Lossy Integrator
VCVS Response
Increasing 𝐺𝑚
VCVS Response
Increasing 𝐺𝑚
By Edgar Sánchez-Sinencio 37
Conclusions
• It is possible to use VCCS (OTA) instead of VCVS (Opamp) in active-RC filters in order to avoid using costly buffer stages.
• Proper performance requirements place a lower limit on the transconductance of the OTA used.
• Using a transconductance of 10-15x the minimum requirement yields a comparable performance to a design employing an Opampimplementation.
By Edgar Sánchez-Sinencio 38
It can be shown that for equal s
GBsA , the Tow-Thomas filter has the following
deviations
GB
k
GBQ;GB
Q
GBQQQ
o
o
o
oooo
oo
oa
2
2
and
414
or
41
1
Vi
QR R
R/k
A1A2
A3R
VBP
CCC1
C2
VL
r
r
-VL
C
rr
VBP -VL
Improved integrator version with positive QI
2
2
1
22
11
GB
kQ
GB
GBkQQ
ooo
o
oa
Improved version by replacing noninverting integrator:
By Edgar Sánchez-Sinencio 39
Single Ended
Fully-Differential Version
How to generate Fully-Differential Filters based on Single-Ended
Version?
CRsV
V
i
o 1
+
-
iVR
oV
C
+
-iV
oV
R C
+
-
iV R
oV
C
+
-
-
+R
C
R
iV
iV
oVoV
C
By Edgar Sánchez-Sinencio 40
Particular Case. Assume no is Available.vi-
Symmetric conditions
Read fully balanced - fully
symmetric circuits from 607.
XQ RRRRR
11111
2303
+
-
-
++
-
C
C
R
R R
RoV
iV
oV
iV
-
++
--
+
R2
R3
R01
RQ
RX
R03
R02
C1 C
2
VBPV
HPV
LP
iV
State -Variable
Filter
+
-
iV
oV
R
R-
iV
C
C
oV
iV
-+
By Edgar Sánchez-Sinencio 41
KHN State Variable Two-Integrator Filter
Use Mason’s Rule:
Next we consider the fully-differential version of the KHN filter.
,1
20202
CRK
,03
303
R
RK
R
RK 3
101
01
1
CRK
2
332
R
RK
030201012
020132
2
03020101
2020132
1KKKsKKs
KKK
s
KKK
s
KK
sKKK
V
V
QQi
LP
03K
32K
QK
01K+
BPViV s
1
02K
LPV
s1
HPV
By Edgar Sánchez-Sinencio 42
KHN Fully-Differential Version
-+-+ +
--+
+
-+-
XR 03R
3R QR
01R
1C
1C3R
2R
2R
QR
03R
XR
02R
02R
2C
2C
iV
iV
LPV
LPV
By Edgar Sánchez-Sinencio 43
How can we take advantage of improved combination of ± QI in fully differential versions?
QRR
R/k
A1
C1
+
-A2
+
-
C2R
RR/k C1
QRR
C2
inV
inV
oV
oV
QRR
R/k
A1
C1
+
-A2
+
-
C2R
RR/k C1
QRR
C2
inV
inV
oV
oV
SAME!
By Edgar Sánchez-Sinencio 44
622 (ESS)
Effects of Non-Ideal Op Amps on the Tow-Thomas Biquad
)3,2,1i(A When i
2121
21221
32o
2121
2121o2
AA
1
A
1
A
11
AQA
1
QA
1
AA
2
A
21
1
AA
1
A
1
A
11
AA
1Q3
A
Q1
A
1Q21
QsssD
are finite, the denominator becomes of the transfer function yields:
R
( )
( )
( )
( )
( )
( )A1=∞
A2=∞A3=∞
Vi
R/K
QR
C
R
VBPVLP
-VLP
C r
r
45By Edgar Sánchez-Sinencio
3,2,1i,s
GBALet i
i
Furthermore assume the range of interest .1Q,1GB
and A
GB
i
o
oi
i
Then D(s) becomes:
3
o
2
o
1
o
o
2oa
a
oa2
2o
2
oo2
2
o
1
o3
3
o
2
o
1
o
o
GB2
GBGB
ss
QssD
sGB
1Q
sGBGB
21sGB
2GBGB
1)s(D
Thus
GB2
3
GB2
1
GB
then,1 Q and 1for
or
1
o
2
o
1
o
a
o
ooa
ooa
GB2GB1GB
46By Edgar Sánchez-Sinencio
o
o
ooa
oa
321
3
o
2
o
1
o
a
4
GBQ
or
1GBQ4
filter stablea for that Note
1GB
Q4for ,GB
Q41QQ
GBQ41
GBGBGB equalFor
GB2
GBGBQ1
1
Q
Q
47By Edgar Sánchez-Sinencio
ECEN 622 (ESS)
TAMUKEY FILTER PARAMETERS IN ACTIVE-RC FILTERS
• Dynamic Range
• Signal-To-Noise Ratio
• Total Output Noise
• Noise Power Spectral Density
• Total Area
Resistor and Capacitors can be expressed as:
resistor R of terminals theinput to thefrom function transfer theis fH Where
1 r
fH
R2
V
R2
HVfP
dsinput yiel sinusoidala for ndissipatiopower resistor The
ues.filter val normalized theare c and r where
CcC,RrR
i
2i
2i
2ii
R
Reference. L. oth et all, “General Results for Resistive Noise in Active RC and MOSFET-C Filters”, IEEE Trans on Circuits
and Systems II, Vol. 42, No. 12, pp. 785-793, December 1995.
48By Edgar Sánchez-Sinencio
Focusing on the noise resistor, the power spectral density is given by
(2) fHrkTR4fHkTR4fS2
o2
oR
The definition of fH o is pictorially shown below:
Thus, the total output noise (mean squared value) due to the resistors become
In practice the upper limit of the integration is limited to a useful practical value.
o RR dffSN
+
-Vi
VR
+
-
fHV o
49By Edgar Sánchez-Sinencio
The signal-to-ratio for a given Vi and frequency f is given by
2cc
12c
1
BP
BP
2c
c2
c
BP
2
R
f2
maxi
max,R
max,RRf
R
2i
fjffQjf
jffQfH
yieldssH above thenotation following theUsing
sQ
s
sQ
sH
examplefiler order BP-seconda consider usLet
(4) N2
fHmaxVDR
Then resistors. thein ndissipatiopower specified maximum theis P where
PfPmax
and
(3) N2
fHVSNR
50By Edgar Sánchez-Sinencio
12
icR Qaa
R
VfP
For the biquad shown below
and
a
a/Q1
C
kT2NR
OBPR VaNa limited by linearity and by resistor power dissipation which is
proportional to (a)2.
R
CC
RQR/a
QR/a 1/a
VOBP
-1Vi
51By Edgar Sánchez-Sinencio
Fully Differential Fully Balanced Circuits
What is the problem with single-input / single-output?
A
o
o
FZ1Z
iVnV
oVo
A
0
A
VVV
ZZ
ZVV
on1
F1
1on
)VV(Z
Z1V
VVVFor
icmid1
Fo
icmidi
No elimination of
common-mode signal.
How to solve this problem?
o
FZ1Z
oVo
1Z
o
2V
1V
FZ
)VV(Z
ZV 21
1
Fo
No common-mode output.
2
)VV()VV(VVVFor 21
21icmidi
52By Edgar Sánchez-Sinencio
How to obtain a fully differential circuit?
We will discuss two potential approaches
Approach 1
2PV
FR
o2V o
1R
1R 1oV
FR
o1V o
1R
1R 2oV
FR
2nV
2P1P VV
1PV1nV
Remark:More robust to reject
common-mode signals
Approach 2
FR
o
1V o
1R
)VV(R
R
)VV(R
RV
211
F
121
F1o
1PV
1nV
FR
o
o
2V o
o
)VV(R
RV 12
1
F2o
2PV
2nV
FR
o
o
)VVVV(R
RVV 1221
1
F2o1o
)VV(R
R2V 21
1
FoD
conditions 1P1n2P2n VV;VV
Remark: sensitive to CM signals
FR
53By Edgar Sánchez-Sinencio
First-Order FB Low Pass with Op Amp
*
.subckt opamp non inv out
rin non inv 100K
egain 1 0 (non, inv) 200K
ropen 1 2 2K
copen 2 0 15.9155u
eout 3 0 (2, 0) 1
rout 3 out 50
.ends
*vin 3 31 ac 1.0
vin 31 0 ac 1.0
x1 4 1 2 opamp
x2 4 11 22 opamp
R1 3 1 1K
R11 3 4 1K
R1B 31 4 1K
R1BB 31 11 1K
RF1 2 1 1K
RF1B 22 11 1K
RF11 4 0 1K
RF11B 4 0 1K
C1 2 1 0.159155u
C1B 22 11 0.159155u
C1A 4 0 0.159155u
C11B 4 0 0.159155u
rdummy 3 31 1
.ac dec 10 10Hz 10KHz
.probe
.end
1C
1FR
oiV o
1R
11R11FR
A1C
B11C
o
iV o
B1R
BB1R
B1C
B1FR
oV
B11FR
oV
22
1131
4
23 1
54By Edgar Sánchez-Sinencio
Fully Balanced T-T Active-RC Implementation
1C
2C
1oRR
R
2oR
o oV
QR
o
iV
o
R
KR
KR1oR
QR
2oR
1C
QR
2oR
1C
R
R
2C
2C
1oR
R
o
o
iV
o R
KR
KR
oV
1oR1C
QR
R
2C
2oR
55By Edgar Sánchez-Sinencio
Introduction to Matlab and
Simulink For Filter Design
622 Active Filters
By Edgar Sánchez-Sinencio
Texas A&M University
56By Edgar Sánchez-Sinencio
Example 1: Ideal Integrator
57
R = 1K C = 0.159mF
By Edgar Sánchez-Sinencio
Bode Plot: Ideal Integrator (Matlab)
s=tf(‘s’);
R=1e3; %Resistor Value
C=0.159e-3; %Capacitor Value
hs=1/(R*C*s); %hs= Vo(s)/Vi(s)
figure(1)
bode(hs) %Create Bode Plot
grid minor %Add grid to plot
H= gcr; %change X-axis
units
h.AxesGrid.Xunits = ‘Hz’; %Set units to
Hz
pole(hs); %calculates hs poles
zero(hs); %calculates hs zeros
58
-20
-10
0
10
20
30
Magnitu
de (
dB
)
10-1
100
101
-91
-90.5
-90
-89.5
-89
Phase (
deg)
Bode Diagram
Frequency (Hz)
By Edgar Sánchez-Sinencio
59
2) Go to: Tools => Control Design=> Linear Analysis 3) Then press: Linearize model
1) Create Model using Gain, Integrator, and In/Out blocks
Bode Plot: Ideal Integrator (Simulink)
By Edgar Sánchez-Sinencio
Tow-Thomas Biquad (Simulink)
60By Edgar Sánchez-Sinencio
Output Waveform (Scope)
61By Edgar Sánchez-Sinencio
Integrator Non-ideal amplifier
62
clear clcs=tf('s'); R=1; %Resistor ValueC=0.159e-3; %Capacitor Valuehs1=-1/(R*C*s); %hs= Vo(s)/Vi(s)figure(1) bodemag(hs1)hold onf=1e3;for i=1:5;GBW=2*pi*f;A=GBW/s;Beta=R/(R+1/(s*C));hs2=-1/(R*C*s)*1/(1+1/(A*Beta));hold onbodemag(hs2,{2*pi*1,2*pi*1e5}) f=10*f;endgrid minor %Add grid to ploth= gcr; %change X-axis unitsh.AxesGrid.Xunits = 'Hz'; %Set units to Hzlegend('ideal', 'GBW=1kHz','GBW=10kHz', 'GBW=100kHz','GBW=1MHz', 'GBW=10MHz',1)
100
101
102
103
104
105
-80
-60
-40
-20
0
20
40
60
80
Magnitu
de (
dB
)
Bode Diagram
Frequency (Hz)
ideal
GBW=1kHz
GBW=10kHz
GBW=100kHz
GBW=1MHz
GBW=10MHz
By Edgar Sánchez-Sinencio
Filter Approximation: Low-Pass Butterworth
63
The squared magnitude of a low-pass butterworth filter is given by:
By Edgar Sánchez-Sinencio
Pole-zero plot
64
Pole-Zero Map
Real Axis
Imagin
ary
Axis
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
10.080.170.280.380.50.64
0.8
0.94
0.080.170.280.380.50.64
0.8
0.94
0.2
0.4
0.6
0.8
1
0.2
0.4
0.6
0.8
1
By Edgar Sánchez-Sinencio
Bode Plot
65
-120
-100
-80
-60
-40
-20
0M
agnitu
de (
dB
)
10-2
10-1
100
101
102
-270
-180
-90
0
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
By Edgar Sánchez-Sinencio
Low-pass Chebyshev Filter
• Use the Matlab cheb1ap function to design a second order
Type I Chebyshev low-pass filter with 3dB ripple in the pass
band
w=0:0.05:400; % Define range to plot
[z,p,k]=cheb1ap(2,3);
[b,a]=zp2tf(z,p,k); % Convert zeros and poles of G(s) to polynomial form
bode(b,a)
grid minor;
66By Edgar Sánchez-Sinencio
Low-pass Chebyshev Filter
67
-80
-60
-40
-20
0
Magnitu
de (
dB
)
10-2
10-1
100
101
102
-180
-135
-90
-45
0
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
By Edgar Sánchez-Sinencio
Low-pass Chebyshev Filter
w=0:0.01:10;
[z,p,k]=cheb1ap(2,3);
[b,a]=zp2tf(z,p,k);
Gs=freqs(b,a,w);
xlabel('Frequency in rad/s');
ylabel('Magnitude of G(s)');
semilogx(w,abs(Gs));
title('Type 1 Chebyshev Low-Pass Filter');
Grid;
68
% Another way to write the code!
By Edgar Sánchez-Sinencio
Low-pass Chebyshev Filter
69
10-2
10-1
100
101
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Type 1 Chebyshev Low-Pass Filter
By Edgar Sánchez-Sinencio
Inverse Chebyshev• Using the Matlab cheb2ap function, design a third order
Type II Chebyshev analog filter with 3dB ripple in the stop band.
w=0:0.01:1000;
[z,p,k]=cheb2ap(3,3);
[b,a]=zp2tf(z,p,k); Gs=freqs(b,a,w);
semilogx(w,abs(Gs));
xlabel('Frequency in rad/sec');
ylabel('Magnitude of G(s)');
title('Type 2 Chebyshev Low-Pass Filter, k=3, 3 dB ripple in stop band');
grid
70By Edgar Sánchez-Sinencio
Inverse Chebyshev
71
10-2
10-1
100
101
102
103
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in rad/sec
Magnitude o
f G
(s)
Type 2 Chebyshev Low-Pass Filter, k=3, 3 dB ripple in stop band
By Edgar Sánchez-Sinencio
Elliptic Low-Pass Filter
• Use Matlab to design a four pole elliptic analog low-pass filterwith 0.5dB maximum ripple in the pass-band and 20dBminimum attenuation in the stop-band with cutoff frequency at200 rad/s.
w=0: 0.05: 500;
[z,p,k]=ellip(4, 0.5, 20, 200, 's');
[b,a]=zp2tf(z,p,k);
Gs=freqs(b,a,w);
plot(w,abs(Gs))
title('4-pole Elliptic Low Pass Filter');
grid
72By Edgar Sánchez-Sinencio
Elliptic Low-Pass Filter
73
0 50 100 150 200 250 300 350 400 450 5000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
14-pole Elliptic Low Pass Filter
By Edgar Sánchez-Sinencio
Transformation Methods
• Transformation methods have been developed where a low
pass filter can be converted to another type of filter by simply
transforming the complex variable s.
• Matlab lp2lp, lp2hp, lp2bp, and lp2bs functions can be used to
transform a low pass filter with normalized cutoff frequency,
to another low-pass filter with any other specified frequency,
or to a high pass filter, or to a band-pass filter, or to a band
elimination filter, respectively.
74By Edgar Sánchez-Sinencio
LPF with normalized cutoff frequency, to
another LPF with any other specified frequency
• Use the MATLAB buttap and lp2lp functions to find the transfer function of a third-order Butterworth low-pass filter with cutoff frequency fc=2kHz.
% Design 3 pole Butterworth low-pass filter (wcn=1 rad/s)
[z,p,k]=buttap(3);
[b,a]=zp2tf(z,p,k); % Compute num, den coefficients of this filter (wcn=1rad/s)
f=1000:1500/50:10000; % Define frequency range to plot
w=2*pi*f; % Convert to rads/sec
fc=2000; % Define actual cutoff frequency at 2 KHz
wc=2*pi*fc; % Convert desired cutoff frequency to rads/sec
[bn,an]=lp2lp(b,a,wc); % Compute num, den of filter with fc = 2 kHz
Gsn=freqs(bn,an,w); % Compute transfer function of filter with fc = 2 kHz
semilogx(w,abs(Gsn));
grid;
xlabel('Radian Frequency w (rad/sec)')
ylabel('Magnitude of Transfer Function')
title('3-pole Butterworth low-pass filter with fc=2 kHz or wc = 12.57 kr/s')
75By Edgar Sánchez-Sinencio
LPF with normalized cutoff frequency, to
another LPF with any other specified frequency
76
103
104
105
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Radian Frequency w (rad/sec)
Magnitude o
f T
ransfe
r F
unction
3-pole Butterworth low-pass filter with fc=2 kHz or wc = 12.57 kr/s
By Edgar Sánchez-Sinencio
• Use the MATLAB commands cheb1ap and lp2hp to find the transfer function ofa 3-pole Chebyshev high-pass analog filter with cutoff frequency fc = 5KHz.
% Design 3 pole Type 1 Chebyshev low-pass filter, wcn=1 rad/s
[z,p,k]=cheb1ap(3,3);
[b,a]=zp2tf(z,p,k); % Compute num, den coef. with wcn=1 rad/s
f=1000:100:100000; % Define frequency range to plot
fc=5000; % Define actual cutoff frequency at 5 KHz
wc=2*pi*fc; % Convert desired cutoff frequency to rads/sec
[bn,an]=lp2hp(b,a,wc); % Compute num, den of high-pass filter with fc =5KHz
Gsn=freqs(bn,an,2*pi*f); % Compute and plot transfer function of filter with fc = 5 KHz
semilogx(f,abs(Gsn));
grid;
xlabel('Frequency (Hz)');
ylabel('Magnitude of Transfer Function')
title('3-pole Type 1 Chebyshev high-pass filter with fc=5 KHz ')
77
High-Pass Filter
By Edgar Sánchez-Sinencio
High-Pass Filter
78
103
104
105
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency (Hz)
Magnitude o
f T
ransfe
r F
unction
3-pole Chebyshev high-pass filter with fc=5 KHz
By Edgar Sánchez-Sinencio
Band-Pass Filter
• Use the MATLAB functions buttap and lp2bp to find the transfer function of a 3-pole Butterworth analog band-pass filter with the pass band frequency centered at fo = 4kHz , and bandwidth BW =2KHz.
[z,p,k]=buttap(3); % Design 3 pole Butterworth low-pass filter with wcn=1 rad/s
[b,a]=zp2tf(z,p,k); % Compute numerator and denominator coefficients for wcn=1 rad/s
f=100:100:100000; % Define frequency range to plot
f0=4000; % Define centered frequency at 4 KHz
W0=2*pi*f0; % Convert desired centered frequency to rads/s
fbw=2000; % Define bandwidth
Bw=2*pi*fbw; % Convert desired bandwidth to rads/s
[bn,an]=lp2bp(b,a,W0,Bw); % Compute num, den of band-pass filter
% Compute and plot the magnitude of the transfer function of the band-pass filter
Gsn=freqs(bn,an,2*pi*f);
semilogx(f,abs(Gsn));
grid;
xlabel('Frequency f (Hz)');
ylabel('Magnitude of Transfer Function');
title('3-pole Butterworth band-pass filter with f0 = 4 KHz, BW = 2KHz')
79By Edgar Sánchez-Sinencio
Band-Pass Filter
80
102
103
104
105
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Frequency f (Hz)
Magnitude o
f T
ransfe
r F
unction
3-pole Butterworth band-pass filter with f0 = 4 KHz, BW = 2KHz
By Edgar Sánchez-Sinencio
Band-Elimination (band-stop) Filter
• Use the MATLAB functions buttap and lp2bs to find the transfer function of a 3-pole Butterworth band-elimination (band-stop) filter with the stop band frequency centered at fo = 5 kHz , and bandwidth BW = 2kHz.
[z,p,k]=buttap(3); % Design 3-pole Butterworth low-pass filter, wcn = 1 r/s
[b,a]=zp2tf(z,p,k); % Compute num, den coefficients of this filter, wcn=1 r/s
f=100:100:100000; % Define frequency range to plot
f0=5000; % Define centered frequency at 5 kHz
W0=2*pi*f0; % Convert centered frequency to r/s
fbw=2000; % Define bandwidth
Bw=2*pi*fbw; % Convert bandwidth to r/s
% Compute numerator and denominator coefficients of desired band stop filter
[bn,an]=lp2bs(b,a,W0,Bw);
% Compute and plot magnitude of the transfer function of the band stop filter
Gsn=freqs(bn,an,2*pi*f);
semilogx(f,abs(Gsn));
grid;
xlabel('Frequency in Hz'); ylabel('Magnitude of Transfer Function');
title('3-pole Butterworth band-elimination filter with f0=5 KHz, BW = 2 KHz')
81By Edgar Sánchez-Sinencio
Band-Elimination (band-stop) Filter
82
102
103
104
105
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in Hz
Magnitude o
f T
ransfe
r F
unction
3-pole Butterworth band-elimination filter with f0=5 KHz, BW = 2 KHz
By Edgar Sánchez-Sinencio
How to find the minimum order to meet
the filter specifications ?
The following functions in Matlab can help you to find the
minimum order required to meet the filter specifications:
• Buttord for butterworth
• Cheb1ord for chebyshev
• Ellipord for elliptic
• Cheb2ord for inverse chebyshev
83By Edgar Sánchez-Sinencio
Calculating the order and cutoff frequency
of a inverse chebyshev filter
• Design a 4MHz Inverse Chebyshev approximation with Ap gain at passband corner.
The stop band is 5.75MHz with -50dB gain at stop band.
clear all;
Fp = 4e6; Wp=2*pi*Fp;
Fs=1.4375*Fp; Ws=2*pi*Fs;
Fplot = 20*Fs;
f = 1e6:Fplot/2e3:Fplot ;
w = 2*pi*f;
Ap = 1;
As = 50;
% Cheb2ord helps you find the order and wn (n and Wn) that
%you can pass to cheby2 command.
[n, Wn] = cheb2ord(Wp, Ws, Ap, As, 's');
[z, p, k] = cheby2(n, As, Wn, 'low', 's');
[num, den] = cheby2(n, As, Wn, 'low', 's');
bode(num, den)
84By Edgar Sánchez-Sinencio
Bode Plot
85
-200
-150
-100
-50
0
Magnitu
de (
dB
)
105
106
107
108
109
1010
540
720
900
1080
1260
Phase (
deg)
Bode Diagram
Frequency (rad/sec)
By Edgar Sánchez-Sinencio
References
[1] S. T. Karris, “Signals and Systems with Matlab
Computing and Simulink Modeling,” Fifth Edition.
Orchard Publications
[2] Matlab Help Files
86By Edgar Sánchez-Sinencio