HW03 Solutions boser

7/22/2019 HW03 Solutions boser

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B. E. Boser Page 1 of 20

Last modified 9/14/2011 4:08:00PM

HOMEWORK #3

1. 2ndorder band-pass filter.a) Find the sensitivity of the to (i.e. find ).

( )Substituting the expression for :

This architecture has a sensitivity that is independent of the !Notice also that:

b) Size the components for . How large is the component spread of the resistors?

Solving the equation for for in terms of :

Choosing and substituting :

Notice that for a given , the component spread of the resistors is fixed to , which is 400 inthis problem! Even a moderate can result in unmanageable resistor component spreads.2. Design a 4thorder Butterworth low-pass filter.

a) Synthesize the filter in MATLAB and determine the nominal component values (resistorsand capacitors) that meet the desired specifications.

The transfer function for our filter is this:

( )( )

This requires two Sallen-Key stages. Using the equations from chapter 2, slide 25 and

setting G = 1,

We can solve for C1and C2using these equations:


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By inspection, for a fixed p and Qp, increasing R1 or R2will decrease both C1 and C2.

Therefore, we set both resistors to their maximum values.

For the first stage, For the second stage,

The MATLAB code to determine the transfer function, plot the function, and then solve for

the values is included below.

clear all; close all;w_edge=2*pi*2e6;ripple=3;N=4;

% Synthesize the filter[z,p,k]=butter(N,wp,'s');sys=zpk(z,p,k);set(sys, 'DisplayFormat', 'frequency');sys


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%% Generate the magnitude and phase plot with annotationf=logspace(5,7,2000);[mag,phase]=bode(sys,2*pi*f);db = 20*log10(squeeze(mag));angle = squeeze(phase);mindB = -80; maxdB = 5; % Set vertical range of plot

figure(1)subplot(2,1,1),set(gca, 'fontsize', 14);semilogx(f, db, 'linewidth', 2), hold on;xlabel('Frequency [Hz]'),ylabel('Magnitude [dB]')axis([min(f) max(f) mindB maxdB]),grid;ind = find (db > -ripple, 1, 'last');H = line( [min(f) f(ind)], [-ripple -ripple], ...

'LineStyle', ':','LineWidth', 2, 'Color', 'r');H = line( [f(ind) f(ind)], [-ripple mindB], ...

'LineStyle', ':','LineWidth', 2, 'Color', 'r');H = line( [min(f) max(f)], [0 0], ...

'LineStyle', ':','LineWidth', 2, 'Color', 'r');

subplot(2,1,2),set(gca, 'fontsize', 14);semilogx(f, angle, 'linewidth', 2), hold on;xlabel('Frequency [Hz]'),ylabel('Phase [degrees]')axis([min(f) max(f) -400 100]),grid;

%% Calculate element values for the Sallen-Key Bi-quads%First Stagewp1 = sqrt(p(1)*p(2));Qp1 = wp1/(-2*real(p(1)));s1r1 = 100e3;s1r2 = 100e3;s1c1 = Qp1/wp1*(s1r1+s1r2)/s1r1/s1r2s1c2 = 1/Qp1/wp1/(s1r1+s1r2)

%Second Stagewp2 = sqrt(p(3)*p(4));Qp2 = wp2/(-2*real(p(3)));s2r1 = 100e3;s2r2 = 100e3;s2c1 = Qp2/wp2*(s2r1+s2r2)/s2r1/s2r2s2c2 = 1/Qp2/wp2/(s2r1+s2r2)

b)Re-scale the components such that under worst case conditions, the filter still meets thepassband specification under the worst case variations in the resistors and capacitors. What

is the minimum frequency (fsmin) at which the filter has a guaranteed stopband attenuation

of 45 dB?Recall from slide 25, with G=1:

Because there is variation in the resistors and capacitors, we subsitute:


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Note that the Q of the filter is unaffected. Therefore, the passband frequency will decreasethe most when both R and C increase, which is our worst-case variation. So we can pre-

scale the R and C in the decreasing direction, in order to just meet the spec in the worst-

case. This can be expresses as:

Note that the only thing matters is scaling down the RC product, so in this case, we choose

to keep R at its maximum value of 100k, in order to minimize total capacitance. Also notethat by doing this we still maintain the Butterworth shape of our filter in the nominal case,

since Q is unaffected.

The nominal component values we get are:



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The minimum frequency that the filter can guarantee stopband attenutation to be 45 dB is

when both R and C decrease. From the plot, we find fsminto be: The MATLAB code to rescale the component value and to generate the plot is shown

below.

%% Rescale R's and C's to account for the worst-case, in which R and C% are both too larges1c1 = s1c1/1.1/1.15s1c2 = s1c2/1.1/1.15s2c1 = s2c1/1.1/1.15s2c2 = s2c2/1.1/1.15

figure(2)hold on;i = 2;%Worst-case variation%when i = 1, both R and C decrease%when i = 2, nominal case%when i = 3,fori=[1:3]

rvar=[0.85 1 1.15];cvar=[0.9 1 1.1];

s1r1var = rvar(i)*s1r1;s1r2var = rvar(i)*s1r2;


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s1c1var = cvar(i)*s1c1;s1c2var = cvar(i)*s1c2;s2r1var = rvar(i)*s2r1;s2r2var = rvar(i)*s2r2;s2c1var = cvar(i)*s2c1;s2c2var = cvar(i)*s2c2;

wp1 = 1/sqrt(s1r1var*s1c1var*s1r2var*s1c2var);Qp1 = wp1/(1/(s1r1var*s1c1var)+1/(s1r2var*s1c1var));h1 = tf([1], [(1/wp1)^2 1/(wp1*Qp1) 1]);wp2 = 1/sqrt(s2r1var*s2c1var*s2r2var*s2c2var);Qp2 = wp1/(1/(s2r1var*s2c1var)+1/(s2r2var*s2c1var));h2 = tf([1], [(1/wp2)^2 1/(wp2*Qp2) 1]);sys = zpk(h1*h2);

f = logspace(5, 7, 2000);w = 2*pi*f;[mag, phase] = bode(sys, w);db = 20*log10(squeeze(mag));angle = squeeze(phase);

%Generate the magnitude plot wizth annotation, phasesubplot(2,1,1),set(gca, 'fontsize', 14);semilogx(f, db, 'linewidth', 2), hold on;xlabel('Frequency [Hz]'),ylabel('Magnitude [dB]')axis([min(f) max(f) mindB maxdB]);

%both R and C decrease, draw 45dB lineif(i == 1)

ind = find (db < -45, 1, 'first');H = line( [min(f) f(ind)], ...

[-45 -45], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');H = line( [f(ind) f(ind)], ...

[-45 mindB], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');disp('fsmin='), disp(f(ind));

%both R and C increaseelseif(i == 3)

ind = find (db > -ripple, 1, 'last');H = line( [min(f) f(ind)], ...

[-ripple -ripple], 'LineStyle', ':','LineWidth', 2, 'Color','r');

H = line( [f(ind) f(ind)], ...[-ripple mindB], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');

H = line( [min(f) max(f)], ...[0 0], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');

end

subplot(2,1,2),set(gca, 'fontsize', 14);semilogx(f, angle, 'linewidth', 2), hold on;xlabel('Frequency [Hz]'),ylabel('Phase [degrees]')axis([min(f) max(f) -400 100])

end

subplot(2,1,1); grid;subplot(2,1,2); grid;


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c) Enter the final circuit schematic in Virtuoso (or any other tool you like) and perform acsimulations for the nominal component sizes you re-scaled in part b. Also perform two more

simulations using the worst-case variations of R and C that result in the minimum and

maximum passband frequency. Submit a plot showing the filters magnitude response for allthree cases in one diagram. Clearly annotate the passband specifications as well as fsmin.

Submit a plot of the complete filter schematic.Note:

i. Dynamic range scaling or Biquad ordering is not required for this problem.ii. In Virtuoso, you may use design variables to scale component values more easily.

The nominal component values are below:


The lowest passband-frequency case (increase R by 15%, increase C by 10%)The largest fsmincase (decrease R by 15%, decrease C by 10%)

The plot of all three cases is shown below. Two things to note here:

i. The variation of the component values are accomplished by a parametric set sweepin Cadence ADE. To learn how to do this, refer to the Virtuoso Analog Design

Environment User Guide (p.420).

ii. In this case, the Spectre simulation result is imported into MATLAB using theSpectreRF MATLAB Toolbox. The plot is then created in MATLAB. The

MATLAB code to for this is also shown. Of course, the curves can also be plotted

in Cadence.


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%% Grabbing data from Cadence simulation directoryclear all; close all;

simDir =

;s=cds_srr(simDir,'sweep1_ac-sweep','out');

f = s.value(1).YVec.value.YVec.freq;magdb(1,:) = 20*log10(abs(s.value(1).YVec.value.YVec.V));magdb(2,:) = 20*log10(abs(s.value(2).YVec.value.YVec.V));magdb(3,:) = 20*log10(abs(s.value(3).YVec.value.YVec.V));

%% Plottingfigure(1);mindB = -80; maxdB = 5;ripple = 0.5;fori=1:3

db = magdb(i,:);%Generate the magnitude plot wizth annotationset(gca, 'fontsize', 14);semilogx(f, db, 'linewidth', 2), hold on;xlabel('Frequency [Hz]'),ylabel('Magnitude [dB]')axis([min(f) max(f) mindB maxdB]);

%both R and C decrease, draw 45dB lineif(i == 1)

ind = find (db < -45, 1, 'first');H = line( [min(f) f(ind)], ...


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[-45 -45], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');H = line( [f(ind) f(ind)], ...

[-45 mindB], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');disp('fsmin='), disp(f(ind));

%both R and C increaseelseif(i == 3)

ind = find (db > -ripple, 1, 'last');H = line( [min(f) f(ind)], ...[-ripple -ripple], 'LineStyle', ':','LineWidth', 2, 'Color',

'r');H = line( [f(ind) f(ind)], ...

[-ripple mindB], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');H = line( [min(f) max(f)], ...

[0 0], 'LineStyle', ':','LineWidth', 2, 'Color', 'r');end

end

outin out

I2

gain_ideal

G: 1

in out

I0

gain_ideal

G: 1

s2c1

C3

s2c2

C2

s1c2

C1

s1c1

C0

R3

s2r1

R2

s2r2

R1

s1r2

R0

s1r1

V0

acm:1vdc:0

3. Design a filter with the following response:

a) What kind of a filter is this (low-pass, band-pass, etc.)? Solve for the poles and zeroes(i.e. find the real and imaginary component of all poles and zeroes). Then find the Q of

each complex conjugate pole. Plot the pole-zero diagram.

This filter is a high-pass filter. You can see this from viewing the magnitude responsecurve, but also from the pole-zero plot as well.


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The pole zero plot shows there is one zero at the origin and no zeroes or poles at infinity.

This would seem to indicate a high pass filter.

Poles Zeros

-1.726e7 0

-2.502e6 i7.997e6 i4.6648e6

-3.57e5 i6.281e6 i3.2296e6

b) Plot the magnitude response using MATLAB. What are the filters passband edgefrequency, passband ripple, and stopband attenuation?

passband edge = 1 MHz

passband ripple = 1 dB

stopband attenuation = 45 dB


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MATLAB code is shown here:clear all; close all;s = zpk('s')sys = s*(s^2+2.176e13)*(s^2+1.043e13) ...

/ ((s+1.726e7)*(s^2+5.005e6*s+7.021e13)*(s^2+7.148e5*s+3.958e13));

[z,p,k] = zpkdata(sys);z = z{1};p = p{1};figure(1);w_limit_x = 2e7;w_limit_y = 2e7;set(gca, 'fontsize', 14);

%plot pole-zero diagramplot(p,'x','linewidth',2),hold on,plot(z,'o','linewidth',2);axis([-w_limit_x w_limit_x -w_limit_y w_limit_y]), axis square;xlabel('\sigma'),ylabel('j\omega'),title('Complex s-plane')

grid on%find Q values of each pole pairQp1 = 1/(2*(-real(p(2))/abs(p(2))))Qp2 = 1/(2*(-real(p(4))/abs(p(4))))

f = logspace(5, 7, 3000);[mag, phase] = bode(sys, 2*pi*f);db = 20*log10(reshape(mag, 1, length(f)));angle = squeeze(phase);

%Plot the magnitude response to find the -3dB frequency and


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%stop band edge%Generate the magnitude plot with annotation, phase,%and group delay plotsmindB = -80; maxdB = 5;figure(2);set(gca, 'fontsize', 14);

semilogx(f, db, 'linewidth', 2), hold on;xlabel('Frequency [Hz]'),ylabel('Magnitude [dB]')axis([min(f) max(f) mindB maxdB]),grid;

%draws vertical line intersecting 3-dB pointind = find (db > -3, 1, 'first');H = line( [f(ind) f(ind)], [mindB -3], ...

'LineStyle', ':','LineWidth', 2, 'Color', 'r');H = line( [min(f) max(f)], [0 0], ...


%draws horizontal line showing passband ripple%estimate of passband frequency, doesn't need to be perfect%but should be outside of the transition bandf_passband = 1e6;%find max ripple for frequencies greater than 1.03e6ripple = -min(db(find(f>f_passband)))H = line( [f_passband max(f)], [-ripple -ripple], ...


%draw horizontal line showing stopband attenuation%estimate of stopband frequency, doesn't need to be perfect but should %beoutside of the transition bandf_stopband = 7.667e5attenuation = -max(db(find(f


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MATLAB code is here:%% try changing the high-Q poles[z,p,k] = zpkdata(sys);z = z{1};p = p{1};

p(4) = 1.05*p(4)p(5) = 1.05*p(5)sys_newhiQ = zpk(z,p,k);

f = logspace(5, 7, 3000);[mag_newhiQ, phase_newhiQ] = bode(sys_newhiQ, 2*pi*f);db_newhiQ = 20*log10(reshape(mag_newhiQ, 1, length(f)));

%Plot the magnitude response to find the -3dB frequency and stop band edge%Generate the magnitude plot with annotation, original curve will be dashedmindB = -80; maxdB = 5;figure(3)set(gca, 'fontsize', 14);

semilogx(f, db_newhiQ, f, db, '--', 'linewidth', 2);xlabel('Frequency [Hz]'),ylabel('Magnitude [dB]');title('Varying the high-Q poles');legend('Original: dashed','Location','SouthEast');axis([min(f) max(f) mindB maxdB]),grid;

%% now change the low-Q pole pair[z,p,k] = zpkdata(sys);z = z{1};p = p{1};p(2) = 1.05*p(2);p(3) = 1.05*p(3);sys_newloQ = zpk(z,p,k);

f = logspace(5, 7, 3000);[mag_newloQ, phase_newloQ] = bode(sys_newloQ, 2*pi*f);db_newloQ = 20*log10(reshape(mag_newloQ, 1, length(f)));

%Plot the magnitude response to find the -3dB frequency and stop band edge%Generate the magnitude plot with annotation, original curve will be dashedmindB = -80; maxdB = 5;figure(4)set(gca, 'fontsize', 14);semilogx(f, db_newloQ, f, db, '--', 'linewidth', 2);xlabel('Frequency [Hz]'),ylabel('Magnitude [dB]');title('Varying the Low-Q poles');

legend('Original: dashed','Location','SouthEast');axis([min(f) max(f) mindB maxdB]),grid;

d) Realize the filter with a cascade of Tow-Thomas biquads and a single active first-ordersection that you will have to invent yourself. Use C=100 pF for all capacitors. Choose

k1=k2=1 and R8=10k. Choose the gains such that the peak magnitude response is one

(0 dB) for all amplifier outputs. Verify your result using Virtuoso/Spectre (or any other

tool you like) and compare with MATLAB by finding the passband edge, the passbandripple, and stopband attenuation. Build the Tow-Thomas stages such that the complex


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pole pair with the highest Q is paired with the zero pair nearest to it and order the filter

stages from lowest Q to highest Q. Submit a plot showing the overall magnitude

response as well as the responses at each stage output in the same plot. You will likely

want to write MATLAB scripts to find the required gain and create your plots.

Note: Use the same gain-ideal component as in Problem 1.

First Stage: an active first order high pass filter

s

-------------

(s+1.726e007)

Yes, there is an inversion but it is okay. If the students use a passive high pass filter with

a unity gain buffer, in order to get non-inverting gain, this is okay. However, it is not

ideal because the filter would be sensitive to parasitics.

For active 1st-order HPF, there are at least two configurations either by using 1 R and 2

Cs, or by using 2 R and 1C. They implement the same transfer function. But the 1R2C

circuit is preferred since capacitors have better matching characteristics.

Component Values

s1 =

sys: [1x1 zpk]

num: [1 0]

den: [1 17260000]

C: 1.0000e-010

R: 579.3743

For the 2nd

and 3rd

stage, DR scaling is required. This is done in MATLAB. Maximum

value in magnitude response of the combined 1stand 2

ndstage is found first, and the 2

nd

stage transfer function is then simply scaled down by this factor. Same for the 3rd

stage.

For the 2nd

and 3rd

stage, we also need to do DR scaling for the internal nodes. First using

k1=k2=1 to calculate the component values, then scale k1 and k2 based on the peak

frequency response of the internal nodes, and re-calculate all the component values.

Second Stage: Tow-thomas filter

(s^2 + 1.043e013)

------------------------------

(s^2 + 5.005e006s + 7.021e013)

Component Values (Solve for the component values using the equations on Chapter 2,

Slide 24, by initially using k1=k2=1. )

Frequency response of the internal nodes (1stand 3

rdamplifier) is shown:


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Final component values:

s2 =

sys: [1x1 zpk]

sys_drs: [1x1 zpk]

num: [1.0003 0 1.0433e+013]

den: [1 5.0050e+006 7.0210e+013]

k1: 0.6437

k2: 1.3149

C: 1.0000e-010

R1: 1.9980e+003

R2: 768.2464

R3: 1.4099e+003

R4: 1.5191e+003

R5: 5.1702e+003

R8: 10000

R6: 9.9974e+003

R7: 1.3149e+004


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Third Stage: Tow-thomas filter

(s^2 + 2.176e013)

------------------------------

(s^2 + 7.148e005s + 3.958e013)

Frequency response of the internal nodes (1

st

and 3

rd

amplifier) is shown below:

Rescale k1 and k2, then compute the final component values:

s3 =

sys: [1x1 zpk]

num: [0.9999 0 2.1758e+013]

den: [1 7.1480e+005 3.9580e+013]

sys_drs: [1x1 zpk]

k1: 0.9378

k2: 1.0565

C: 1.0000e-010

R1: 1.3990e+004R2: 1.4906e+003

R3: 1.6044e+003

R4: 1.3243e+004

R5: 2.7115e+003

R8: 10000

R6: 1.0001e+004

R7: 1.0565e+004


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MATLAB code is here:%% pole-zero grouping[z,p,k]=zpkdata(sys); z=z{1}; p=p{1};% Group p2,p3 with z4, z5 (high-Q pole with nearby zero)s1.sys=zpk(z(1), p(1),1);s2.sys=zpk(z(2:3),p(4:5),1); % Low Q stage first

s3.sys=zpk(z(4:5),p(2:3),1);[s3.num, s3.den]=zp2tf(z(2:3),p(4:5),1);%% First stage, using 1R and 2C, C=100pF[num, den]=tfdata(s1.sys);s1.num = num{1};s1.den = den{1};s1.C=100e-12;s1.R=1/s1.den(2)/s1.C;%% 2nd stagesys12 = s1.sys * s2.sys;f = logspace(6,8,2000);[mag,phase]=bode(sys12,2*pi*f);maxmag = max(mag);s2.sys_drs = s2.sys/maxmag;[num, den]=tfdata(s2.sys_drs);s2.num = num{1};s2.den = den{1};a2=s2.den(1); a1=s2.den(2); a0=s2.den(3);b2=s2.num(1); b1=s2.num(2); b0=s2.num(3);k2=1/(10^(-2.378/20)); k1=10^(-3.826/20);s2.k1=k1; s2.k2=k2;s2.C=100e-12;s2.R1=1/a1/s2.C;s2.R2=k1/sqrt(a0)/s2.C;s2.R3=1/k1/k2/sqrt(a0)/s2.C;s2.R4=1/k2/(a1*b2-b1)/s2.C;s2.R5=k1*sqrt(a0)/b0/s2.C;s2.R8=10e3;

s2.R6=s2.R8/b2;s2.R7=k2*s2.R8;%% 3rd stagesys123 = s1.sys * s2.sys_drs * s3.sys;f = logspace(6,8,2000);[mag,phase]=bode(sys123,2*pi*f);maxmag = max(mag);s3.sys_drs = s3.sys/maxmag;[num, den]=tfdata(s3.sys_drs);s3.num = num{1};s3.den = den{1};a2=s3.den(1); a1=s3.den(2); a0=s3.den(3);b2=s3.num(1); b1=s3.num(2); b0=s3.num(3);k1=10^(-558e-3/20); k2=1/(10^(-477.1e-3/20));s3.k1 = k1; s3.k2=k2;s3.C=100e-12;s3.R1=1/a1/s3.C;s3.R2=k1/sqrt(a0)/s3.C;s3.R3=1/k1/k2/sqrt(a0)/s3.C;s3.R4=1/k2/(a1*b2-b1)/s3.C;s3.R5=k1*sqrt(a0)/b0/s3.C;s3.R8=10e3;s3.R6=s3.R8/b2;s3.R7=k2*s3.R8;s1, s2, s3


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Magnitude response of output of each stage (plotted using SpectreRF MATLAB Toolbox):

passband edge= 1 MHz

passband ripple = 1dB

stopband attenuation = 45 dB

Below is a plot of the scaled amplifier outputs of stage 2 and stage 3.


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PM

Below is the schematic:

Stage 1:

s1out

in out

I8

gain_ideal

G: -1e10

R16

s1r

s1c

C5

s1c

C4

V0

acm:1vdc:1

Stage2:

s2out

s1out

in out

I2

gain_ideal

G: -1e10

in out

I1

gain_ideal

G: -1e10

in out

I0

gain_ideal

G: -1e10

R7

s2r3

R6

s2r1

R5

s2r5

R4

s2r6

R3

s2r4

R2

s2r2

R1

s2r8

R0

s2r7

s2c2

C1

s2c1

C0

Stage 3:

s2out

s3out

in out

I7

gain_ideal

G: -1e10

in out

I6

gain_ideal

G: -1e10

in out

I4

gain_ideal

G: -1e10

R15

s3r7

R14

s3r8

R13

s3r2

R12

s3r6

R11

s3r5

R10

s3r1

R9

s3r3

R8

s3r4

s3c1

C3

s3c2

C2

HW03 Solutions boser

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