ee247 administrative - eecs instructional support group...
TRANSCRIPT
![Page 1: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/1.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 1
EE247Administrative
• Final exam group has been changed to group 3– Final exam new date/time Dec. 13th, 5pm to 8pm
• Homework #1 has been posted on course website and is due Sept 11th (next Thurs.)– Submissions can be on paper or via email– Paper submission in class or during same day office
hours – Please show your derivations and explain your work
EECS 247 Lecture 3: Filters © 2008 H.K. Page 2
EE247 Lecture 3
• Active Filters– Active biquads
• Sallen- Key & Tow-Thomas• Integrator-based filters
– Signal flowgraph concept– First order integrator-based filter– Second order integrator-based filter & biquads
– High order & high Q filters• Cascaded biquads & first order filters
– Cascaded biquad sensitivity to component mismatch• Ladder type filters
![Page 2: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/2.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 3
Filters2nd Order Transfer Functions (Biquads)
• Biquadratic (2nd order) transfer function:
PQjH
jH
jH
P=
=
=
=
∞→
=
ωω
ω
ω
ω
ω
ω
)(
0)(
1)(0
2
2P P P
2 22
2P PP
P 2P
P
1P 2
1H( s )
s s1
Q
1H( j )
1Q
Biquad poles @: s 1 1 4Q2Q
Note: for Q poles are real , complex otherwise
ω ω
ωω ωω ω
ω
=
+ +
=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
⎛ ⎞= − ± −⎜ ⎟⎝ ⎠
≤
EECS 247 Lecture 3: Filters © 2008 H.K. Page 4
s-Plane12
2
Complex conjugate poles:
s 1 4 12
P
PP
P
Q
j QQ
ω
> →
⎛ ⎞= − ± −⎜ ⎟⎝ ⎠
P2Q- part real Pω=
PQ21arccos
poles
ωj
σ
Pω radius =
![Page 3: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/3.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 5
Implementation of Biquads• Passive RC: only real poles can’t implement complex conjugate
poles
• Terminated LC– Low power, since it is passive– Only fundamental noise sources load and source resistance– As previously analyzed, not feasible in the monolithic form for f
<a few 100s of MHz
• Active Biquads– Many topologies can be found in filter textbooks! – Widely used topologies:
• Single-opamp biquad: Sallen-Key• Multi-opamp biquad: Tow-Thomas• Integrator based biquads
EECS 247 Lecture 3: Filters © 2008 H.K. Page 6
Active Biquad Sallen-Key Low-Pass Filter
• Single gain element• Can be implemented both in discrete & monolithic form• “Parasitic sensitive”• Versions for LPF, HPF, BP, …
Advantage: Only one opamp used to obtain 2polesDisadvantage: Sensitive to parasitic – all pole no zeros
outV
C1
inV
R2R1
C2
G
2
2
1 1 2 2
1 1 2 1 2 2
( )1
1
1 1 1
P P P
P
PP
GH ss sQ
R C R C
Q GR C R C R C
ω ω
ω
ω
=+ +
=
= −+ +
![Page 4: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/4.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 7
Addition of Imaginary Axis Zeros
• Sharpen transition band• Can “notch out” interference• High-pass filter (HPF)• Band-reject filter
Note: Always represent transfer functions as a product of a gain term,poles, and zeros (pairs if complex). Then all coefficients have a physical meaning, and readily identifiable units.
2
Z2
P P P
02
P
Z
s1
H( s ) Ks s
1Q
H( j ) K
H( j ) K
ω
ω
ω
ω ω
ω
ωωω
→
→∞
⎛ ⎞+ ⎜ ⎟⎜ ⎟⎝ ⎠=
⎛ ⎞+ + ⎜ ⎟⎜ ⎟
⎝ ⎠
=
⎛ ⎞= ⎜ ⎟⎜ ⎟
⎝ ⎠
EECS 247 Lecture 3: Filters © 2008 H.K. Page 8
Imaginary Zeros• Zeros substantially sharpen transition band• At the expense of reduced stop-band attenuation at
high frequencyPZ
P
P
ffQ
kHzf
32100
===
104 105 106 107-50
-40
-30
-20
-10
0
10
Frequency [Hz]
Mag
nitu
de [
dB]
With zerosNo zeros
Real Axis
Imag
Axi
s
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2x 106
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2x 10
6
Pole-Zero Map
![Page 5: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/5.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 9
Moving the Zeros
104 105 106 107-50
-40
-30
-20
-10
0
10
20
Frequency [Hz]
Mag
nitu
de[d
B]
Pole-Zero Map
Real AxisIm
ag A
xis
-6 -4 -2 0 2 4 6-6
-4
-2
0
2
4
6
x105
x105
filterrejectBandffQ
kHzf
PZ
P
P
−→===
2100
EECS 247 Lecture 3: Filters © 2008 H.K. Page 10
Tow-Thomas Active Biquad
Ref: P. E. Fleischer and J. Tow, “Design Formulas for biquad active filters using three operational amplifiers,” Proc. IEEE, vol. 61, pp. 662-3, May 1973.
• Parasitic insensitive• Multiple outputs
![Page 6: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/6.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 11
Frequency Response
( ) ( )
( ) ( )01
21001020
01
3
012
012
22
012
0021122
1
1asas
babasabbakV
Vasasbsbsb
VV
asasbabsbabk
VV
in
o
in
o
in
o
++−+−−=
++++=
++−+−−=
• Vo2 implements a general biquad section with arbitrary poles and zeros
• Vo1 and Vo3 realize the same poles but are limited to at most one finite zero
EECS 247 Lecture 3: Filters © 2008 H.K. Page 12
Component Values
8
72
173
2821
111
21732
80
6
82
74
81
6
8
111
21753
80
1
1
RRk
CRRCRRk
CRa
CCRRRRa
RRb
RRRR
RR
CRb
CCRRRRb
=
=
=
=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
=
827
2
86
20
015
112124
10213
20
12
111
111
11
1
RkRbRR
Cbak
R
CbbakR
CakkR
CakR
CaR
=
=
=
−=
=
=
=
thatfollowsit
11
21732
8
CRQCCRRR
R
PP
P
ω
ω
=
=
zerospolesdesiredthefromexampleforRCCkba iii / and , ,,,given 821
![Page 7: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/7.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 13
Higher-Order Filters in the Integrated Form
• One way of building higher-order filters (n>2) is via cascade of 2nd
order biquads, e.g. Sallen-Key,or Tow-Thomas
2nd orderFilter ………
Nx 2nd order sections Filter order: n=2N
1 2 Ν
Cascade of 2nd order biquads:☺ Easy to implement
Highly sensitive to component mismatch -good for low Q filters only
Good alternative: Integrator-based ladder type filters
2nd orderFilter
2nd orderFilter
EECS 247 Lecture 3: Filters © 2008 H.K. Page 14
Integrator Based Filters• Main building block for this category of filters
Integrator• By using signal flowgraph techniques
Conventional RLC filter topologies can be converted to integrator based type filters
• Next few pages:– Introduction to signal flowgraph techniques– 1st order integrator based filter– 2nd order integrator based filter– High order and high Q filters
![Page 8: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/8.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 15
What is a Signal Flowgraph (SFG)?
• SFG Topological network representation consisting of nodes & branches- used to convert one form of network to a more suitable form (e.g. passive RLC filters to integrator based filters)
• Any network described by a set of linear differential equations can be expressed in SFG form
• For a given network, many different SFGs exists
• Choice of a particular SFG is based on practical considerations such as type of available components
*Ref: W.Heinlein & W. Holmes, “Active Filters for Integrated Circuits”, Prentice Hall, Chap. 8, 1974.
EECS 247 Lecture 3: Filters © 2008 H.K. Page 16
What is a Signal Flowgraph (SFG)?• Signal flowgraph technique consist of nodes & branches:
– Nodes represent variables (V & I in our case) – Branches represent transfer functions (we will call the
transfer function branch multiplication factor or BMF)• To convert a network to its SFG form, KCL & KVL is used to
derive state space description• Simple example:
Circuit State-space description SFG
ZZ
VoinIinI
VoI Z Vin o× =
![Page 9: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/9.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 17
Signal Flowgraph (SFG)Examples
1SL
Circuit State-space description SFG
R
L
R
oV
C 1SC
Vin
Vo
oI
VoinI
inI
inI
inI Vo
Vin oI
I R Vin o
1V Iin o
SL
1I Vin o
SC
× =
× =
× =
EECS 247 Lecture 3: Filters © 2008 H.K. Page 18
Useful Signal Flowgraph (SFG) Rules
1Va
2Vb
a+b1V
2V
a.b1V
2V3Va b1V 2V
a.V1+b.V1=V2 (a+b).V1=V2
a.V1=V3 (1)b.V3=V2 (2)Substituting for V3 from (1) in (2) (a.b).V1=V2
• Two parallel branches can be replaced by a single branch with overall BMF equal to sum of two BMFs
• A node with only one incoming branch & one outgoing branch can be eliminated & replaced by a single branch with BMF equal to the product of the two BMFs
![Page 10: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/10.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 19
Useful Signal Flowgraph (SFG) Rules
• An intermediate node can be multiplied by a factor (k). BMFs for incomingbranches have to be multiplied by k and outgoing branches divided by k
3V
a b1V 2V k.a b/k
1V 2V
3.Vk
a.V1=V3 (1)b.V3=V2 (2)
Multiply both sides of (1) by k(a.k) . V1= k.V3 (1)
Divide & multiply left side of (2) by k(b/k) . k.V3 = V2 (2)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 20
Useful Signal Flowgraph (SFG) Rules
hiV2V a
oV
b
g-b
3VhiV
2V a/(1+b)oV
b
g
3V
3VciV
2V aoV
b
d-bciV2V a
oV-1 -b
1 d
3V
• Simplifications can often be achieved by shifting or eliminating nodes• Example: eliminating node V4
• A self-loop branch with BMF y can be eliminated by multiplying the BMFof incoming branches by 1/(1-y)
V4
![Page 11: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/11.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 21
Integrator Based Filters1st Order LPF
• Conversion of simple lowpass RC filter to integrator-based type by using signal flowgraph techniques
in
V 1os CV 1 R
=+
oV
RsCinV
EECS 247 Lecture 3: Filters © 2008 H.K. Page 22
What is an Integrator?Example: Single-Ended Opamp-RC Integrator
∫inV
o in o in1 1V V , V V dt
sRC RC= − × = − ∫
oV
C
inV
-
+R -
Note: Practical integrator in CMOS technology has input & output both in the form of voltage and not current Consideration for SFG derivation
oV
RCτ =
∫
![Page 12: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/12.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 23
Integrator Based Filters1st Order LPF
1. Start from circuit prototype-Name voltages & currents for all components
2. Use KCL & KVL to derive state space description in such a way to have BMFs in the integrator form:
Capacitor voltage expressed as function of its current VCap.=f(ICap.)Inductor current as a function of its voltage IInd.=f(VInd.)
3. Use state space description to draw signal flowgraph (SFG) (see next page)
1IoV
RsCinV
2I
1V+ −
CV
+
−
EECS 247 Lecture 3: Filters © 2008 H.K. Page 24
Integrator Based FiltersFirst Order LPF
1IoV
1
RsCinV
2I
1V+ −
1Rs
1sC
2I1I
CVinV 1−1 1V
• All voltages & currents nodes of SFG
• Voltage nodes on top, corresponding current nodes below each voltage node
SFG
CV
+
−
oV1
Integrator formV V V1 in C
1V IC 2 sCV Vo C
1I V1 1 RsI I2 1
= −
= ×
=
= ×
=
![Page 13: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/13.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 25
Normalize• Since integrators are the main building blocks require in & out signals
in the form of voltage (not current)
Convert all currents to voltages by multiplying current nodes by a scaling resistance R*
Corresponding BMFs should then be scaled accordingly
1 in o
11
2o
2 1
V V VV
IRsI
VsC
I I
=
=
= −
=
1 in o*
*1 1
*2
o *
* *2 1
V V V
RI R V
Rs
I RV
sC R
I R I R
=
=
= −
=
* 'x xI R V=
1 in o*
'1 1
'2
o *
' '2 1
V V V
RV V
Rs
VV
sC R
V V
=
=
= −
=
EECS 247 Lecture 3: Filters © 2008 H.K. Page 26
1st Order Lowpass Filter SGFNormalize
'2V
*1
sCR
oVinV 1−1 1V
1
*RRs
1Rs
1sC
2I1I
oVinV 1−1
1
1V
'1V
oVinV 1−1 1V
*RRs
1I R ∗× 2I R ∗×
*1
sCR
*
*R
R
![Page 14: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/14.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 27
1st Order Lowpass Filter SGFSynthesis
'1V
1*1
sC R
oVinV 1−1 1V
'2V1
*RRs
* * CR Rs , Rτ= = ×
'1V
1sτ ×
oVinV 1−1 1V
'2V1
1sτ ×
oVinV 1−1 1V
'2V
1
Consolidate two branches
*Choosing
R Rs=
EECS 247 Lecture 3: Filters © 2008 H.K. Page 28
First Order Integrator Based Filter
oV
inV
-+ ( )1
H ssτ
=×
∫+
1sτ ×
oVinV 1−1 1V
'2V
1
![Page 15: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/15.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 29
1st Order Filter Built with Opamp-RC Integrator
oV
inV
-+
∫+
• Single-ended Opamp-RC integrator has a sign inversion from input to output
Convert SFG accordingly by modifying BMF
oV
inV-
∫+
-
oV
'in inV V= −
+
∫+
-
EECS 247 Lecture 3: Filters © 2008 H.K. Page 30
1st Order Filter Built with Opamp-RC Integrator
• To avoid requiring an additional opamp to perform summation at the input node:
oV
'in inV V= −
+
∫+
-
oV
'inV
∫--
![Page 16: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/16.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 31
1st Order Filter Built with Opamp-RC Integrator (continued)
o'
in
V 11 sRCV
= −+
oV
C
in'V
-
+R
R
--
oV'
inV
∫
EECS 247 Lecture 3: Filters © 2008 H.K. Page 32
Opamp-RC 1st Order Filter Noise
2n1v
( )
k 22o mm0m 1
thi
2 21 2 2
2 2n1 n2
2o
v S ( f ) dfH ( f )
S ( f ) Noise spectral densi ty of m noise source1H ( f ) H ( f )
1 2 fRC
v v 4KTR f
k Tv 2 C2
π
α
∞=
=
→
= =+
= = Δ
=
=
∑ ∫
oV
C
-
+R
R
Typically, α increases as filter order increases
Identify noise sources (here it is resistors & opamp)Find transfer function from each noise sourceto the output (opamp noise next page)
2n2v
α
![Page 17: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/17.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 33
Opamp-RC Filter NoiseOpamp Contribution
2n1v
2opampv oV
C
-
+R
R
• So far only the fundamental noise sources are considered
• In reality, noise associated with the opamp increases the overall noise
• For a well-designed filter opamp is designed such that noise contribution of opamp is negligible compared to other noise sources
• The bandwidth of the opamp affects the opamp noise contribution to the total noise
2n2v
EECS 247 Lecture 3: Filters © 2008 H.K. Page 34
Integrator Based Filter2nd Order RLC Filter
oV
1
1sL
R CinI•State space description:
R L C o
CC
RR
LL
C in R L
V V V VI
VsC
VI
RV
IsL
I I I I
=
=
= = =
=
= − −
• Draw signal flowgraph (SFG)
SFG
L
1R
1sC
CIRI
CV
inI
1−1
RV 1
1−
LV
LI
CV
LI
+
- CILV
+
-
Integrator form
+
-RV
RI
![Page 18: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/18.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 35
1 1
*RR *
1sCR
'1V
2V
inV
1−1
1V
*RsL
1−
oV
'3V
2nd Order RLC Filter SGFNormalize
1
1R
1sC
CIRI
CV
inI
1−1
RV 1
1−
LV
LI
1sL
• Convert currents to voltages by multiplying all current nodes by the scaling resistance R*
* 'x xI R V=
'2V
EECS 247 Lecture 3: Filters © 2008 H.K. Page 36
inV
1*R R 21
sτ11
sτ1 1
*RR *
1sCR
'1V
2V
inV
1−1
1V
*RsL
1−
oV
'3V
2nd Order RLC Filter SGFSynthesis
∑
oV
--
* *1 2R C L Rτ τ= × =
![Page 19: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/19.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 37
-20
-15
-10
-5
0
0.1 1 10
Second Order Integrator Based Filter
Normalized Frequency [Hz]
Mag
nitu
de (d
B)
inV
1*R R 21
sτ11
sτ
BPV
-- LPV
HPV
∑
Filter Magnitude Response
EECS 247 Lecture 3: Filters © 2008 H.K. Page 38
Second Order Integrator Based Filter
inV
1*R R 21
sτ11
sτ
BPV
-- LPV
HPV
∑
BP 22in 1 2 2
LP2in 1 2 2
2HP 1 2
2in 1 2 2* *
1 2*
0 1 2
1 2
1 2 *
V sV s s 1V 1V s s 1V sV s s 1
R C L R
R R11
Q 1
From matching pointof v iew desirable :RQ
R
L C
β
β
β
β
β
ττ τ τ
τ τ ττ τ
τ τ τ
ω τ τ
τ τ
τ τ
τ τ
=+ +
=+ +
=+ +
= × =
=
= =
= ×
= → =
![Page 20: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/20.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 39
Second Order Bandpass Filter Noise
2 2n1 n2
2o
v v 4KTRdf
k Tv 2 Q C
= =
=
inV
1*R R2
1sτ
11
sτ
BPV
--
2n1v
2n2v
• Find transfer function of each noise source to the output
• Integrate contribution of all noise sources
• Here it is assumed that opamps are noise free (not usually the case!)
k 22o mm0m 1
v S ( f ) dfH ( f )∞=
= ∑ ∫
Typically, α increases as filter order increasesNote the noise power is directly proportion to Qα
∑
EECS 247 Lecture 3: Filters © 2008 H.K. Page 40
Second Order Integrator Based FilterBiquad
inV
1*R R 21
sτ11
sτ
BPV
--
21 1 2 2 2 30
2in 1 2 2
a s a s aVV s s 1β
τ τ ττ τ τ
+ +=+ +
∑oV
∑
a1 a2 a3
HPV
LPV
• By combining outputs can generate general biquad function:
s-planejω
σ
![Page 21: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/21.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 41
SummaryIntegrator Based Monolithic Filters
• Signal flowgraph techniques utilized to convert RLC networks to integrator based active filters
• Each reactive element (L& C) replaced by an integrator• Fundamental noise limitation determined by integrating capacitor
value:
– For lowpass filter:
– Bandpass filter:
where α is a function of filter order and topology
2o
k Tv Q Cα=
2o
k Tv Cα=
EECS 247 Lecture 3: Filters © 2008 H.K. Page 42
Higher Order Filters• How do we build higher order filters?
– Cascade of biquads and 1st order sections• Each complex conjugate pole built with a biquad and real pole
with 1st order section• Easy to implement• In the case of high order high Q filters highly sensitive to
component mismatch– Direct conversion of high order ladder type RLC filters
• SFG techniques used to perform exact conversion of ladder type filters to integrator based filters
• More complicated conversion process• Much less sensitive to component mismatch compared to cascade
of biquads
![Page 22: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/22.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 43
Higher Order FiltersCascade of Biquads
Example: LPF filter for CDMA cell phone baseband receiver • LPF with
– fpass = 650 kHz Rpass = 0.2 dB– fstop = 750 kHz Rstop = 45 dB– Assumption: Can compensate for phase distortion in the digital domain
• Matlab used to find minimum order required 7th order Elliptic Filter
• Implementation with cascaded BiquadsGoal: Maximize dynamic range– Pair poles and zeros– In the cascade chain place lowest Q poles first and progress to higher Q
poles moving towards the output node
EECS 247 Lecture 3: Filters © 2008 H.K. Page 44
Overall Filter Frequency Response
Bode Diagram
Pha
se (d
eg)
Mag
nitu
de (d
B)
-80
-60
-40
-20
0
-540
-360
-180
0
Frequency [Hz]300kHz 1MHz
Mag
. (dB
)
-0.2
0
3MHz
![Page 23: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/23.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 45
Pole-Zero Map (pzmap in Matlab)
Qpole fpole [kHz]
16.7902 659.4963.6590 611.7441.1026 473.643
319.568
fzero [kHz]
1297.5836.6744.0
Pole-Zero Map
-2 -1.5 -1 -0.5 0-1
-0.5
0
0.5
1
Imag
Axi
s X1
07
Real Axis x107
s-Plane
EECS 247 Lecture 3: Filters © 2008 H.K. Page 46
CDMA FilterBuilt with Cascade of 1st and 2nd Order Sections
• 1st order filter implements the single real pole• Each biquad implements a pair of complex conjugate poles and a
pair of imaginary axis zeros
1st orderFilter Biquad2 Biquad4 Biquad3
![Page 24: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/24.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 47
Biquad Response
104
105
106
107
-40
-20
0
LPF1
-0.5 0
-0.5
0
0.5
1
104
105
106
107
-40
-20
0
Biquad 2
104
105
106
107
-40
-20
0
Biquad 3
104
105
106
107
-40
-20
0
Biquad 4
EECS 247 Lecture 3: Filters © 2008 H.K. Page 48
Individual Biquad Magnitude Response
Frequency [Hz]
Mag
nitu
de (d
B)
104 105 106 107-50
-40
-30
-20
-10
0
10
LPF1Biquad 2Biquad 3Biquad 4
-0.5 0
-0.5
0
0.5
1
![Page 25: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/25.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 49
Intermediate Outputs
Frequency [Hz]
Mag
nitu
de (d
B)
-80
-60
-40
-20
Mag
nitu
de (d
B)
LPF1 +Biquad 2
Mag
nitu
de (d
B)
Biquads 1, 2, 3, & 4
Mag
nitu
de (d
B)
-80
-60
-40
-20
0
LPF10
10kHz10
6
-80
-60
-40
-20
0
100kHz 1MHz 10MHz
5 6 7
-80
-60
-40
-20
0
4 5 6
Frequency [Hz]
10kHz10
100kHz 1MHz 10MHz
LPF1 +Biquads 2,3 LPF1 +Biquads 2,3,4
EECS 247 Lecture 3: Filters © 2008 H.K. Page 50
-10
Sensitivity to Relative Component MismatchComponent variation in Biquad 4 relative to the rest
(highest Q poles):
– Increase ωp4 by 1%
– Decrease ωz4 by 1%
High Q poles High sensitivityin Biquad realizations
Frequency [Hz]1MHz
Mag
nitu
de (d
B)
-30
-40
-20
0
200kHz
3dB
600kHz
-50
2.2dB
![Page 26: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/26.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 51
High Q & High Order Filters
• Cascade of biquads– Highly sensitive to component mismatch not suitable
for implementation of high Q & high order filters– Cascade of biquads only used in cases where required
Q for all biquads <4 (e.g. filters for disk drives)
• Ladder type filters more appropriate for high Q & high order filters (next topic)– Will show later Less sensitive to component mismatch
EECS 247 Lecture 3: Filters © 2008 H.K. Page 52
Ladder Type Filters
• For simplicity, will start with all pole (no finite zero) ladder type filters
– Start with LC ladder type and find values for Ls & Cs– Convert to integrator based form – example shown
• Next will attend to high order ladder type filters incorporating zeros
• Implement the same 7th order elliptic filter in the form of ladder RLC with zeros
– Find level of sensitivity to component mismatch – Compare with cascade of biquads
• Convert to integrator based form utilizing SFG techniques– Effect of integrator non-Idealities on filter
frequency characteristics
![Page 27: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/27.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 53
LC Ladder Filters
• Made of resistors, inductors, and capacitors• Doubly terminated or singly terminated (with or w/o RL)
RsC1 C3
L2
C5
L4
inV RL
oV
Doubly terminated LC ladder filters Lowest sensitivity to component mismatch
EECS 247 Lecture 3: Filters © 2008 H.K. Page 54
LC Ladder Filters
• First step in the design process find values for Ls & Cs– Filter tables
• A. Zverev, Handbook of filter synthesis, Wiley, 1967.• A. B. Williams and F. J. Taylor, Electronic filter design, 3rd edition, McGraw-
Hill, 1995.– CAD tools
• Matlab• Spice
RsC1 C3
L2
C5
L4
inV RL
oV
![Page 28: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/28.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 55
LC Ladder Filter Design Example
Design a LPF with maximally flat passband:f-3dB = 10MHz, fstop = 20MHzRs >27dB @ fstop
• Maximally flat passband Butterworth
From: Williams and Taylor, p. 2-37
Stopband A
ttenuation
Νοrmalized ω
• Find minimum filter order :
• Here standard graphs from filter books are used
fstop / f-3dB = 2Rs >27dB
Minimum Filter Order5th order Butterworth
1
-3dB
2
-30dBP
assb
and
Atte
nuat
ion
EECS 247 Lecture 3: Filters © 2008 H.K. Page 56
LC Ladder Filter Design Example
From: Williams and Taylor, p. 11.3
Find values for L & C from Table:Note L &C values normalized to
ω-3dB =1
Denormalization:Multiply all LNorm, CNorm by:
Lr = R/ω-3dBCr = 1/(RXω-3dB )
R is the value of the source and termination resistor (choose both 1Ω for now)
Then: L= Lr xLNorm
C= Cr xCNorm
![Page 29: EE247 Administrative - EECS Instructional Support Group ...ee247/fa08/files07/lectures/L3_f08.pdf · EE247 Administrative • Final exam group has been changed to group 3 – Final](https://reader030.vdocuments.us/reader030/viewer/2022021801/5b3ce0937f8b9a28308b9c84/html5/thumbnails/29.jpg)
EECS 247 Lecture 3: Filters © 2008 H.K. Page 57
LC Ladder Filter Design Example
From: Williams and Taylor, p. 11.3
Find values for L & C from Table:Normalized values:C1Norm =C5Norm =0.618C3Norm = 2.0L2Norm = L4Norm =1.618
Denormalization:Since ω-3dB =2πx10MHz
R =1 Lr = R/ω-3dB = 15.9 nHCr = 1/(RXω-3dB )= 15.9 nF
C1=C5=9.836nF, C3=31.83nF
L2=L4=25.75nH
EECS 247 Lecture 3: Filters © 2008 H.K. Page 58
Magnitude Response Simulation
Frequency [MHz]
Mag
nitu
de (d
B)
0 10 20 30-50
-40
-30
-20
-10-50
-6 dB passband attenuationdue to double termination
30dB
Rs=1OhmC19.836nF
C331.83nF
L2=25.75nH
C59.836nF
L4=25.75nH
inV RL=1Ohm
oV
SPICE simulation Results