ECE 3610 Spring 2004 Project: LC Passive Filter Statistical Modeling and Analysis 1 Introduction In this final team project you will be using probability and statistics to determine yield expectations on one or more lumped element radio frequency (RF) bandpass filter designs. The candidate filters will be required to have a prescribed amplitude response versus frequency characteristic, which must be met or exceeded. Filters not meeting the requirements will be considered failed subassemblies. The parts chosen for the filter design will be parts having prescribed component tolerances and catalog stock design center values of inductance and capacitance. A repeated trials simulation of many filters, 1000 per design, will be tested using MATLAB based modeling. Independent random variable distributions will be used to assign component values. Each filter trial will thus have realistic variations in component value, reflecting what one would expect from a production run of the filter subassemblies. To assist the teams some of the MATLAB code (functions) needed in constructing the simula- tion are included in the Appendix and the source code itself will be on the class web site. The project teams will be limited to at most three members. I encourage you to work together, hence do not work alone unless you feel this is your only option. On the other hand since each team member receives the same project grade, a group of three should attempt to give all team members equal responsibility. The due date for the completed project will be 4:00 pm, Wednesday, May 12, 2004. 2 Filter Design Requirements For this project the filters of interest are bandpass filters based on third-order lowpass prototypes. The frequency response of an ideal bandpass filter is shown in Figure 1. The filter frequency f f 1 f 2 f o Hf () 1 0 Figure 1: Ideal bandpass filter. response magnitude, | H ( f )|, is unity over a band of frequencies running from f 1 to f 2 and zero otherwise. A realizable filter will not be this perfect, having gain less than unity in portions of the pass band, a transition band either side of the passband, and a stopband gain at most only approaching zero as the distance from the center frequency increases.
Transcript
ECE 3610 Spring 2004 Project: LC PassiveFilter Statistical Modeling and Analysis
1 Introduction
In this final team project you will be using probability and statistics to determine yield expectationson one or more lumped element radio frequency (RF) bandpass filter designs. The candidatefilters will be required to have a prescribed amplitude response versus frequency characteristic,which must be met or exceeded. Filters not meeting the requirements will be considered failedsubassemblies. The parts chosen for the filter design will be parts having prescribed componenttolerances and catalog stock design center values of inductance and capacitance.
A repeated trials simulation of many filters, 1000 per design, will be tested using MATLABbased modeling. Independent random variable distributions will be used to assign componentvalues. Each filter trial will thus have realistic variations in component value, reflecting what onewould expect from a production run of the filter subassemblies.
To assist the teams some of the MATLAB code (functions) needed in constructing the simula-tion are included in the Appendix and the source code itself will be on the class web site.
The project teams will be limited to at most three members. I encourage you to work together,hence do not work alone unless you feel this is your only option. On the other hand since each teammember receives the same project grade, a group of three should attempt to give all team membersequal responsibility. The due date for the completed project will be 4:00 pm, Wednesday, May 12,2004.
2 Filter Design Requirements
For this project the filters of interest are bandpass filters based on third-order lowpass prototypes.The frequency response of an ideal bandpass filter is shown in Figure 1. The filter frequency
ff1 f2fo
H f( )
1
0
Figure 1: Ideal bandpass filter.
response magnitude, |H( f )|, is unity over a band of frequencies running from f1 to f2 and zerootherwise. A realizable filter will not be this perfect, having gain less than unity in portions ofthe pass band, a transition band either side of the passband, and a stopband gain at most onlyapproaching zero as the distance from the center frequency increases.
2 FILTER DESIGN REQUIREMENTS 2
Hf() dB
0ε– dB
As–
f1fs1 f2 fs2
f
Passband
StopBand
TransitionBand
TransitionBand
StopBand
PassbandDesignTolerance
Realizable AmplitudeResponse in dB
f0 f1f2=
Figure 2: Realizable bandpass filter frequency response in dB, i.e., 20 log10 |H( f )|.
The magnitude response in decibels (dB) of a realizable bandpass filter is shown in Figure 2.For lumped element filters considered in this project, the design equations yield a filter centerfrequency, f), that is the geometric mean of the band edge frequencies, i.e.,
fo = √f1 f2 (1)
and a fractional bandwidth
w = f2 − f1
f0(2)
Due to random component variations, the realizable filter is only expected to fall within anfrequency response performance mask such as shown in Figure 3. A filter that does not fall withinthe given performance mask is unacceptable, and fails to meet the desired design specifications.
In this project filters are modeling in MATLAB, the frequency response in dB is obtained overa specific frequency band, and then the filter is compared to mask as shown in Figure 3 to see if itpasses.
2 FILTER DESIGN REQUIREMENTS 3
Hf() dB
0ε– dB
As–
f1fs1 f2 fs2
f
Actual response must not enter shaded regions
TheoreticalResponse
Figure 3: A possible filter frequency response acceptance mask in dB, i.e., we have plotted20 log10 |H( f )| versus frequency.
3 LUMPED ELEMENT BANDPASS FILTER DESIGN 4
3 Lumped Element Bandpass Filter Design
One approach to passive filter design is by the so-called insertion loss method [1]. The steps in thedesign process are shown in Figure 4. The filter design specifications are typically given in terms
Filter Specifications
LowpassPrototype
Design
Scaling andConversion Implementation
Figure 4: Filter design by the insertion loss method.
a desired amplitude response, see for example Figure 2 where the passband amplitude toleranceis −εdB ≤ |H( f )|dB ≤ 0 for f1 ≤ f ≤ f2 and the stopband tolerance is |H( f )|dB ≤ −As forf ≤ fs1 and f ≥ fs2.
3.1 Lowpass Prototype
From these filter design specifications comes a passive inductor (L) capacitor (C) lowpass filterprototype. The filter is called a prototype because it generally will have a lowpass cutoff frequencyof 1 radian/second and a normalized impedance level of one ohm. Lumped lowpass prototypesrelevant to this project are of the form of a ladder network as shown in Figure 5, where the filterorder is given by N . The element values are ohms, henries, or farads as given by the element
. . .
. . .
0g g g g
g g g g g
g g g
g g g g
g
g
2 4 N
1 3 N + 1 N N + 1
N
N + 1N4
3
2
10
N + 1
N even o r N odd
Figure 5: Lumped element lowpass prototype ladder networks; two equivalent forms.
type. The element g0 represents the source resistance and the element gN+1 represents the loadresistance. Filters of this type are said to be doubly terminated since they require both a sourceresistance and load resistance for proper operation.
3 LUMPED ELEMENT BANDPASS FILTER DESIGN 5
The element values g1 to gN are chosen to give a desired shape to the lowpass frequencyresponse. Two popular shapes are Butterworth which has a maximally flat passband and Chebyshevwhich has an equal-ripple passband. The Chebyshev has the advantage of the Butterworth is thatby allowing passband ripple the transition from passband to stopband is smaller for the same filterorder N . See Appendix A for more detailed design formulas, not required to work this project.
3.2 Scaling and Conversion
The third block of Figure 4 is scaling and conversion. In the lowpass prototype the impedancelevel is 1 ohm. To scale the impedance level to say R ohms, multiply all resistors and inductors byR and divide all capacitors by R. To summarize we remap the resistors, inductors, and capacitorsas follows:
Rk → R · Rk Lk → R · Lk, Ck → Ck/R (3)
Under the bandpass transformation inductors become series LC circuits and capacitors becomeparallel LC circuits. Assuming impedance scaling has already taken place, the only variablesneeded besides L and C are the fractional bandwidth, w, and the center frequency, f0. The trans-formation equations and the circuit element replacements are shown in Figure 6.
L
C
L
L
C
C
s
p
s
p
LsL
2πf0w---------------=
Csw
2πf0L--------------=
CpC
2πf0w---------------=
Lpw
2πf0C---------------=
Lowpass Bandpass
Figure 6: Element transformations from lowpass to bandpass.
The general topology of an N = 3 bandpass filter is shown in Figure 7. The filter topology of
C1
C2
C3L1
L2
L3
Figure 7: A general bandpass filter topology based on an N = 3 lowpass prototype, less the sourceand load terminations.
3 LUMPED ELEMENT BANDPASS FILTER DESIGN 6
Figure 7 is the focus of this simulation project.MATLAB functions have been written to obtain bandpass LC values for Butterworth and
Chebyshev lowpass prototypes. The function descriptions are given below. A complete listingof the function source code can be found in Appendix C.
>> help lc_butter
[Rg,L,C,RL] = lc_butter(N,f1,f2,R): lumped elementbandpass filter design using a maximally flat or Butterworthlowpass prototype. The LC ladder network is assumed to beginwith a parallel LC resonator circuit followed by a series LC resonator,followed by a another parallel LC resonator circuit, etc.The filter is assumed to be doubly terminated, that isa resistive source impedance is required as well asa resistive load termination.
N = lowpass filter order (bandpass design is of order 2Nf1 = lower 3dB bandedge frequency in Hzf2 = upper 3dB bandedge frequency in HzR = impedance scaling factor in ohms
Rg = scaled source termination resistance in ohmsL = vector of inductance values in Henries [parallel series parallel ...]C = vector of capacitance values in Farads [parallel series parallel ...]
RL = scaled load resistance in ohms (same as Rg for Butterworth)
[Rg,L,C,RL] = lc_cheby(N,RdB,f1,f2,R): lumped elementbandpass filter design using a Chebyshevlowpass prototype. The LC ladder network is assumed to beginwith a parallel LC resonator circuit followed by a series LC resonator,followed by a another parallel LC resonator circuit, etc.The filter is assumed to be doubly terminated, that isa resistive source impedance is required as well asa resistive load termination.
3 LUMPED ELEMENT BANDPASS FILTER DESIGN 7
N = lowpass filter order (bandpass design is of order 2NRdB = passband ripple in dBf1 = lower 3dB bandedge frequency in Hzf2 = upper 3dB bandedge frequency in HzR = impedance scaling factor in ohms
Rg = scaled source termination resistance in ohmsL = vector of inductance values in Henries [parallel series parallel ...]C = vector of capacitance values in Farads [parallel series parallel ...]
RL = scaled load resistance in ohms (same as Rg only for N odd)
These functions return the series and parallel L and C values in henries and farads respectively.The source and load resistances are also returned. The inputs to the functions are: (1) filter orderN, (2) for the Chebyshev the ripple value in dB RdB, which is also the passband attenuation at thecutoff frequencies (the Butterworth is always 3 dB down), (3) the lower passband cutoff frequencyin Hz f1, (4) the upper passband cutoff frequency in Hz f2, (5) the filter impedance level in ohmsR. Note that in place of f1 and f2 you insert the filter center frequency in Hz f0 and the fractionalbandwidth w. In practice the filter order N would be determined using all of the amplitude responsevalues of Figure 2, e.g., εdB, f1, f2, As, fs1, fs2. Since we are only going to study an N = 3 designin this project, we will always force N to be three and take what ever stopband performance comesas a result of constraining the passband.
As an example suppose we want to design a Butterworth bandpass filter having f1 = 9 MHz,f2 = 11 MHz, and scaled to a 50 ohm impedance level.
>> [Rg,L,C,RL] = lc_butter(3,9e6,11e6,50)***** f0 = 9.95e+006 Hz and w = 0.201 *****
Suppose we want to design a Chebyshev bandpass filter with 2 dB ripple a center frequency of10 MHz, a 20% fractional bandwidth, and a 50 ohm impedance level.
>> [Rg,L,C,RL] = lc_cheby(3,2.0,10e6,0.20,50)***** f1 = 9.05e+006 Hz and f2 = 1.1e+007 Hz *****
What is missing is a means of taking the filter description in terms of element values and obtainingthe frequency response i.e., |H( f )|dB versus frequency f in Hz. From circuit theory we know thata lumped element two-port network has a transfer function that relates the voltage at the output tothe voltage at the input. In circuit theory we first learn about frequency response when studyingsinusoidal steady-state analysis. For a two-port network as shown in Figure 8 the transfer functionis ordinarily the ratio VL( jω)/Vg( jω). In this case we modify the formula slightly since when
Two-PortNetwork
Rg 50=
RL 50=vg vL
+
-
+
-
Figure 8: A two-port network showing source and load voltages for transfer function calculation.
the two-port is removed we are left with a simple voltage divider having a gain of 1/2 for allfrequencies. A doubly terminated network, such as a transmission line environment where an LC
3 LUMPED ELEMENT BANDPASS FILTER DESIGN 9
filter may occur with source and load matching, always has this 1/2 factor in the voltage gain.We define the transfer function as the gain relative to the same circuit with the two-port removed,hence we include a factor of two in our definition of transfer function, i.e.,
H( j2π f ) = 2VL( j2π f )
Vg( j2π f )(4)
We also learn in circuit theory that a Laplace transform domain transfer function is related to thefrequency response by letting s = jω = j2π f .
Now, for a lumped element two-port containing 2N reactive elements (L’s and C’s) the s-domain transfer can be written as a ratio of polynomials in s where the highest positive power ofs is 2N . The bandpass filter containing 2N reactive elements has an s-domain transfer function ofthe form
H(s) = 2VL(s)
Vg(s)= b2N s2N + b2N−1s2N−1 + · · · + b1s + b0
a2N s2N + a2N−1s2N−1 + · · · + a1s + a0(5)
where a 2 has been included in the bn coefficients to properly scale the to unity in the passband.For the N = 3 lowpass prototype bandpass filter the highest degree in s is just 2 · 3 = 6.
Circuit analysis of the N = 3 topology in Figure 7 yields the following transfer function
H(s) = 21 + s2 C2 L2
s C2 Rg
[− 1+
(1 + s2 C2 L2
)2(
s C1 + 1s L1
+ s C21+s2 C2 L2
+ 1Rg
) (s C3 + s C2
1+s2 C2 L2+ 1
s L3+ 1
RL
)s2 C2
2
]−1
(6)
See Appendix B for the circuit analysis details. Multiplying out the numerator and denominator,and collecting like terms with like powers of s, yields the factored form
The MATLAB function lc filter(), described below, takes the element values of the N =3 network and generates the bn and an coefficients (here s6 is the highest power in H(s)).
>> help lc_filter6
[b,a] = lc_filter6(Rg,L,C,RL): A function for creating
3 LUMPED ELEMENT BANDPASS FILTER DESIGN 10
the s-domain polynomial description of a 6-pole lumped elementLC bandpass filter in terms of the filter L and C element values.
Lg = source resistance in ohms, a scalarL = is a row vector holding the inductances [L1 L2 L3] in henriesC = is a row vector holding the capacitors [C1 C2 C3] in faradsRL = load resistance in ohms, a scalar
For further simulation we form the system function H(s) = b(s)/a(s) sowe can use the MATLAB function freqs() for frequency domain analysis ofthe filter. To account for the source/load termination we gain scale thenumerator coefficients by 2 so the filter has a nominal passband gain ofunity.
b = coefficients of the numerator polynomial (descending powers of S)a = coefficients of the denominator polynomial
Mark Wickert April 2001
Notice that this function takes as input the exact variables that are returned by the filter designfunctions lc butter and lc cheby. The function then returns vectors b and a containingthe polynomial coefficients corresponding to numerator and denominator polynomials of H(s)respectively. For statistical filter analysis the L and C vectors will be tweaked or modified byrandom numbers representing component errors before being inserted in lc filter6(). Thecoefficients in b and a will be shifted from their ideal values, thus leading to a distorted frequencyresponse when compared with the theoretical expectations.
To finally get the frequency response from the element values, we need to have a means foreasily evaluating H(s = j2π f ) over a specified range of frequency values. In MATLAB’s signalprocessing toolbox there is a defined function named freqs() that does exactly this.
>> help freqs
FREQS Laplace-transform (s-domain) frequency response.H = FREQS(B,A,W) returns the complex frequency response vector Hof the filter B/A:
given the numerator and denominator coefficients in vectors B and A.The frequency response is evaluated at the points specified invector W (in rad/s). The magnitude and phase can be graphed bycalling FREQS(B,A,W) with no output arguments.
[H,W] = FREQS(B,A) automatically picks a set of 200 frequencies W onwhich the frequency response is computed. FREQS(B,A,N) picks Nfrequencies.
See also LOGSPACE, POLYVAL, INVFREQS, and FREQZ.
To demonstrate how to obtain a frequency response as designed in Section 3.2, we will nowplot the magnitude response in dB the two 10 MHz bandpass filters designed earlier.
Figure 9: Frequency response in dB of Butterworth design example.
4 PROJECT TASKS 12
• The Chebyshev design:
>> [b,a] = lc_filter6(Rg,L,C,RL);>> f = 5e6:10e6/400:15e6;>> Hc = freqs(b,a,2*pi*f);>> plot(f/1e6,20*log10(abs(Hc)))>> grid>> axis([5 15 -60 0])>> print -tiff -deps fig10.eps
5 6 7 8 9 10 11 12 13 14 15−60
−50
−40
−30
−20
−10
0
f in MHz
H f( ) dB
Figure 10: Frequency response in dB of Chebyshev design example.
The tools are in place for performing statistical analysis on N = 3 LC bandpass filters so whatremains is to assign specific project tasks that need to be carried out.
4 Project Tasks
1. Begin by designing an N = 3 Chebyshev filter having the following specifications:
N = 3, f1 = 25 MHz, f2 = 35 MHz, εdB = 1.0 dB, R = 50 ohms
(a) List the ideal LC filter element values in µh and pf respectively.
(b) Plot the frequency response (theoretical response), |H( f )|dB for 15 ≤ f ≤ 50 MHz,using these exact component values. A good range for the y-axis is -40 to 0 dB.
(c) In a real design implementation catalog values of L and C must be used. Choose partsfrom the inductor/capacitor catalog given in Appendix D. You may form series andparallel combinations of inductors and capacitors (three at most in any one position).Keeping the parts count down keeps the cost down. Overlay the magnitude response of
4 PROJECT TASKS 13
the actual component values filter with the part (b) plot. What is the total per unit costfor the inductors and capacitors used in your design?
2. Using the filter designed in Problem 1, statistically analyze the frequency response of thefilter by using independent random number generators to choose LC component values.
(a) Referring to the parts catalog assume that each inductor value is randomly chosen froma uniform distribution whose mean is the true mean and the PDF range (support inter-val) is ±10% of the mean. For the capacitors also use a uniform PDF except now thesupport interval swings ±5% about the true value. Even though we are taking the tol-erance from the parts catalog assume for now that the mean value of each L and C arethe ideal values as calculated in 1a. Overlay plot 100 trials of the frequency response,choosing randomly new L and C values for each plot. As a final overlay plot the idealfrequency response in a different line type/color/width so the deviation about the truefilter response is clearly evident (color printing would be nice here, but with a grayscaling printer discrimination should still be possible).
(b) Using 100 trials, plot in dB a histogram of the frequency response values occurring atf = 25 MHz (use 24 MHz to move below the band edge if you wish). The amplitudedistribution is a function of 6 random variables. According to the central limit theoremit may be reasonable to assume that the distribution is approximately Gaussian. Whatdo you think?
(c) Repeat 2a using the actual component values you picked in 1c. When inductors andcapacitors are put in parallel and series you will need to generate additional randomcapacitor values and then combine using the rules for parallel and series connectionsof elements.
(d) Develop a program that will determine the number of failed filters based the responseacceptance mask shown in Figure 11. Note that since the filter is passive, including
f MHz,0
01.0–
2.0–
2.0
Filte
r G
ain
in d
B
30–
18 26 34 47
Figure 11: Frequency response acceptance mask for Problem 2.
the source and load which have only positive values of resistance, the maximum filtergain is always less than or equal to one (zero dB). Your program does not need to plot
REFERENCES 14
anything, it simply needs to indicate how many filters failed out of the total simulationrun. A failure occurs on the frequency response interval (here 15 to 50 MHz) if theshaded region of Figure 11 is entered. Run the simulation for 1000 trials using thetheoretical L and C values.
(e) Repeat 2d using the catalog based values of L and C .
(f) Repeat 2e except now increase the precision of the inductors from ±10% to ±5%. Arethe results what you would expect?
3. In this problem you will compare a Butterworth and a Chebyshev with random componenterrors. Design a Butterworth bandpass filter to have the following specifications:
N = 3, f1 = 25 MHz, f2 = 35 MHz, R = 50 ohms
and then a Chebyshev bandpass filter to have a similar passband by virtue of εdB = 3 dB:
N = 3, f1 = 25 MHz, f2 = 35 MHz, εdB = 3.0 dB, R = 50 ohms
The Chebyshev filter will have a more rapid roll-off into the stopband, but how easy is it tobuild this filter. Using the ideal component values, but with catalog tolerance values of ±10%for inductors and ±5% for capacitors, compare the yield for each filter on a run of 1000.To accomplish this comparison modify the mask of Figure 11 by eliminating the stopbandrequirement (e.g., raise the amplitude threshold from -30 dB up to -10 dB) and lower thepassband lower threshold from -2 dB to -4 dB. You should find that the Butterworth designis more robust to random component variations.
References
[1] David M. Pozar, Microwave and RF Design of Wireless Systems, John Wiley, New York,2001.
[2] Robert E. Collin, Foundations for Microwave Engineering, second edition, McGraw-Hill,New York , 1992.
A Butterworth and Chebyshev Prototype Design
A.1 Butterworth
In a Butterworth lumped element design the normalized gk values are given by [2]
g0 = gN+1 = 1 ohm (8)
gk = 2 sin
[(2k − 1)π
2N
], k = 1, 2, . . . , N (9)
B DETAILED CIRCUIT ANALYSIS 15
A.2 Chebyshev
In a Chebyshev lumped element design the normalized gk values are given by [2]
g0 = 1 ohm (10)
gN+1 ={
1, N odd
1 + 2ε2 + 2ε√
1 + ε2, N even(11)
gk = 4ak−1ak
bk−1gk−1, k = 2, 3, . . . , N (12)
where
ak = sin2k − 1
2Nπ (13)
bk = sinh2 β
2N+ sin2 kπ
N(14)
β = ln
√1 + ε2 + 1√1 + ε2 − 1
(15)
g1 = 2a1
sinh β/(2N )(16)
andε =
√10εdB/10 − 1 (17)
B Detailed Circuit Analysis
To obtain the transfer function for the N = 3 bandpass filter we use nodal analysis of the circuitin Figure 12. Begin by writing a node voltage equation in the s-domain at vi by summing currents
��
��
����
��
��
��
��
��
��
��
Figure 12: Nodal analysis of the N = 3 bandpass filter to obtain H(s).
away from the node to zero
Vi − Vg
Rg+ Vi
L1s+ Vi C1s + Vi − VL
L2s + 1C2s
= 0 (18)
C MATLAB FUNCTION LISTINGS 16
or solving for Vg we have
Vg
Rg= Vi
[1
Rg+ 1
L1s+ C1s + C2s
1 + L2C2s2
]− VL
C2s
1 + L2C2s2(19)
Secondly we write a nodal equation at vL
VL
RL+ VL
L3s+ VLC3s + VL − vi
L2s + 1C2s
= 0 (20)
Next we solve (20) for Vi
Vi = 1 + L2C2s2
C2s
[1
RL+ 1
L3s+ C3s + C2s
1 + L2C2s2
]VL (21)
Now we can use (21) to eliminate Vi in (19) and ultimately solve for the ratio VL/Vg
Vg1 + L2C2s2
RgC2s=
(1 + L2C2s2
C2s
)2 [1
RL+ 1
L3s+ C3s + C2s
1 + L2C2s2
]
·[
1
Rg+ 1
1s+ C1s + C2s
1 + L2C2s2
]VL − VL (22)
Rearrange by pulling 2VL/Vg = H(s) alone on the left results in
H(s) = 21 + s2 C2 L2
s C2 Rg
[− 1+
(1 + s2 C2 L2
)2(
s C1 + 1s L1
+ s C21+s2 C2 L2
+ 1Rg
) (s C3 + s C2
1+s2 C2 L2+ 1
s L3+ 1
RL
)s2 C2
2
]−1
(23)
To factor into the ratio of polynomials the symbolic math capability of Mathematica is used.
C MATLAB Function Listings
C.1 The Butterworth filter design m-file, lc butter.m
function [Rg,L,C,RL] = lc_butter(N,f1,f2,R)% [Rg,L,C,RL] = lc_butter(N,f1,f2,R): lumped element% bandpass filter design using a maximally flat or Butterworth% lowpass prototype. The LC ladder network is assumed to begin% with a parallel LC resonator circuit followed by a series LC resonator,% followed by a another parallel LC resonator circuit, etc.% The filter is assumed to be doubly terminated, that is% a resistive source impednace is required as well as% a resistive load termination.%% N = lowpass filter order (bandpass design is of order 2N
C MATLAB FUNCTION LISTINGS 17
% f1 = lower 3dB bandedge frequency in Hz% f2 = upper 3dB bandedge frequency in Hz% R = impedance scaling factor in ohms%% Rg = scaled source termination resistance in ohms% L = vector of inductance values in Henries [parallel series parallel ...]% C = vector of capacitance values in Farads [parallel series parallel ...]% RL = scaled load resistance in ohms (same as Rg for Butterworth)%% The filter topology is the following:%% o---Rg---o----|---o----L2----o----C2-----o---|----o--- - +% | | |% + | | |% ----o---- ----o---- |% | | | | |% Input | | | | cont. |% Source L1 C1 L3 C3 ... RL Output% | | | | |% | | | | N > 3 |% - ----o---- ----o---- |% | | |% 0-------------|------------------------------|----o--- | -% Ground% Mark Wickert April 2001
Rg = R;RL = R;
%Normalized (R=1) lowpass g values for Butterworth prototypek = 1:N;g = 2*sin((2*k-1)/(2*N)*pi);
%Scale impedance level of all g valuesg(1:2:end) = g(1:2:end)/R; % Capacitor scalingg(2:2:end) = g(2:2:end)*R; % Inductor scaling
%Lowpass to bandpass element transformationL = zeros(1,N);C = zeros(1,N);
elsef0 = sqrt(f1*f2); % Center frequency in Hzw = (f2 - f1)/f0; % Fractional bandwidths = sprintf(’***** f0 = %10.3g Hz and w = %5.3f *****\n’,f0,w);disp(s);
C MATLAB FUNCTION LISTINGS 18
end
%C to parallel LC resonatorsL(1:2:end) = w./(2*pi*f0*g(1:2:end));C(1:2:end) = g(1:2:end)/(2*pi*f0*w);%L to series LC resonatorsL(2:2:end) = g(2:2:end)/(2*pi*f0*w);C(2:2:end) = w./(2*pi*f0*g(2:2:end));
C.2 The Chebyshev filter design m-file, lc cheby.m
function [Rg,L,C,RL] = lc_cheby(N,RdB,f1,f2,R)% [Rg,L,C,RL] = lc_cheby(N,RdB,f1,f2,R): lumped element% bandpass filter design using a Chebyshev% lowpass prototype. The LC ladder network is assumed to begin% with a parallel LC resonator circuit followed by a series LC resonator,% followed by a another parallel LC resonator circuit, etc.% The filter is assumed to be doubly terminated, that is% a resistive source impedance is required as well as% a resistive load termination.%% N = lowpass filter order (bandpass design is of order 2N% RdB = passband ripple in dB% f1 = lower 3dB bandedge frequency in Hz% f2 = upper 3dB bandedge frequency in Hz% R = impedance scaling factor in ohms%% Rg = scaled source termination resistance in ohms% L = vector of inductance values in Henries [parallel series parallel ...]% C = vector of capacitance values in Farads [parallel series parallel ...]% RL = scaled load resistance in ohms (same as Rg only for N odd)%% The filter topology is the following:%% o---Rg---o----|---o----L2----o----C2-----o---|----o--- - +% | | |% + | | |% ----o---- ----o---- |% | | | | |% Input | | | | cont. |% Source L1 C1 L3 C3 ... RL Output% | | | | |% | | | | N > 3 |% - ----o---- ----o---- |% | | |% 0-------------|------------------------------|----o--- | -% Ground% Mark Wickert April 2001
%Convert Ripple in dB (RdB) to eps2 = epsilonˆ2eps2 = 10ˆ(RdB/10) - 1;
Rg = R;
C MATLAB FUNCTION LISTINGS 19
if fix(N/2)*2 == N, % If N is even RL is not equal to RRL = R*(1 + 2*eps2 + 2*sqrt(eps2*(1 + eps2)));
elsef0 = sqrt(f1*f2); % Center frequency in Hzw = (f2 - f1)/f0; % Fractional bandwidths = sprintf(’***** f0 = %10.3g Hz and w = %5.3f *****\n’,f0,w);disp(s);
end
%C to parallel LC resonatorsL(1:2:end) = w./(2*pi*f0*g(1:2:end));C(1:2:end) = g(1:2:end)/(2*pi*f0*w);%L to series LC resonatorsL(2:2:end) = g(2:2:end)/(2*pi*f0*w);C(2:2:end) = w./(2*pi*f0*g(2:2:end));
C.3 The m-file function responsible for converting filter element val-ues to polynomial coefficients, lc filt6.m
function [b,a] = lc_filter6(Rg,L,C,RL)% [b,a] = lc_filter6(Rg,L,C,RL): A function for creating
C MATLAB FUNCTION LISTINGS 20
% the s-domain polynomial description of a 6-pole lumped element% LC bandpass filter in terms of the filter L and C element values.%% The filter topology is the following:%% o---Rg---o-----|-----o----L2----o----C2-----o-----|-----o---- +% | | |% + | | |% ----o---- ----o---- |% | | | | |% Input | | | | |% Source L1 C1 L3 C3 RL Output% | | | | |% | | | | |% - ----o---- ----o---- |% | | |% 0--------------|----------------------------------|-----o---| -% Ground%% Lg = source resistance in ohms, a scalar% L = is a row vector holding the inductances [L1 L2 L3] in henries% C = is a row vector holding the capacitors [C1 C2 C3] in farads% RL = load resistance in ohms, a scalar%% For further simulation we form the system function H(s) = b(s)/a(s) so% we can use the MATLAB function freqs() for frequency domain analysis of% the filter. To account for the source/load termination we gain scale the% numerator coefficients by 2 so the filter has a nominal passband gain of% unity.% b = coefficients of the numerator polynomial (descending powers of S)% a = coefficients of the denominator polynomial%% Mark Wickert April 2001
% Define numerator polynomial:b = 2*[C(2)*L(1)*L(3)*RL 0 0 0]; % only sˆ3 term exits
Figure 13 and 14 lists small surface mount capacitor and inductor values useful for RF filter design.The surface mount inductors have a tolerance of ±10% or ±20% while the surface mount silvermica capacitors have a tolerance of ±5% for part numbers ending in J and ±0.5 pf for part numbersending in D.
Surface Mount MICA Capacitors
Size Temperature Capacit- Dimension - Inch (mm) Digi-Key Price Each Cornell DubilierCode Vdc Coefficient ppm/°C ance L W H I (Max) Part Number 1 10 100 Part Number
Cornell Dubilier mica capacitors are made of pure India Ruby muscovite mica. It isamong the most inert and stable materials in the universe. When used as the insulator(dielectric) in a capacitor, the result is an electronic component with the ultimate instability and reliability.The capacitance change from zero to full rated operating voltage is generally less than±0.1%. The series inductance of these capacitors is as little as eight nanohenrys (nH).The stability characteristics also apply to frequency which is why mica capacitors are souseful in the gigahertz range. Mica capacitors are so efficient that less that 0.05% of the applied volt-amperes is lostas heat. Dipped silver mica capacitors such as the Standard Types are manufactured byapplying pure silver to ultra-thin mica plates. The plates are stacked and secured by
brass clips with copper-clad steel leads welded to each clip. The extra mass of the metalacts as a heat sink and enables these capacitors to absorb voltage transients up to100,000 volts per microsecond.Type MC - High-Frequency, High Stability Chip for Instruments and RF• Low impedance to beyond 1 GHz, • Near-zero capacitance change, all environments
SPECIFICATIONS:Capacitance Tolerance:
• Manufacture’s part number suffix letter: D = ±0.5pF• Manufacture’s part number suffix letter: J = ±5%
Temperature Range: -55°C to +125°C
W
H
I I
L
† 1% Tolerance
NNEEWW!!
Tape and Reel packaging available with lead time.
Figure 13: A short surface mount capacitor catalog for RF filter design.
D INDUCTOR/CAPACITOR CATALOG 22
Surface Mount Chip Inductors
PM40 (1812) Series PM0805 Series PM1008 Series
1.78
2.41
1.78
1.781.02
0.76
1.02
2.79
2.92
2.29
2.541.02
1.27
1.02
Test R,dc I,dc Cut Tape Tape & Reel
L Tol. Q Freq. Max. Max. Digi-Key Price Each Pricing* J.W.Miller
(µH) % Min. MHz Ohms mA Part No. 1 10 100 500 Part No.
