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Init high pass report matlab part, Fix gain

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Apostolos Fanakis 6 years ago
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  1. 9382
      High Pass Butterworth/Adobe Illustrator circuit layouts/high_pass_butterworth_circuit_layout.ai
  2. 4205
      High Pass Butterworth/Adobe Illustrator circuit layouts/high_pass_butterworth_circuit_layout.pdf
  3. 10
      High Pass Butterworth/Adobe Illustrator circuit layouts/high_pass_butterworth_circuit_layout.svg
  4. BIN
      High Pass Butterworth/Multisim/high_pass_butterworth.ms14
  5. 9
      High Pass Butterworth/high_pass_design.m
  6. 52
      report/4_high_pass/4_high_pass.pug
  7. 375
      report/4_high_pass/4_high_pass_design.pug
  8. 2
      report/4_high_pass/assets/diagrams/high_pass_butterworth_units_diagram.svg
  9. 270
      report/4_high_pass/assets/diagrams/high_pass_general_transfer_function_plot.svg
  10. 145
      report/4_high_pass/assets/diagrams/matlab_high_pass_butterworth_zero_pole.svg
  11. 354
      report/4_high_pass/assets/diagrams/multisim_high_pass_butterworth_circuit_layout_only_filter.svg
  12. 5
      report/report.pug

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High Pass Butterworth/Adobe Illustrator circuit layouts/high_pass_butterworth_circuit_layout.ai

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High Pass Butterworth/Adobe Illustrator circuit layouts/high_pass_butterworth_circuit_layout.pdf

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High Pass Butterworth/Adobe Illustrator circuit layouts/high_pass_butterworth_circuit_layout.svg

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High Pass Butterworth/Multisim/high_pass_butterworth.ms14

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9
High Pass Butterworth/high_pass_design.m

@ -83,7 +83,7 @@ low_pass_prototype_half_power_radial_frequency = 1/ ...
(1/(2*design_filter_order)); % rad/s
% Transforms the result to get the corresponding frequency for the high
% pass using the eq. 12-12
% pass using the eq. 12-3
design_half_power_radial_frequency = specification_pass_radial_frequency/ ...
low_pass_prototype_half_power_radial_frequency; % rad/s
@ -284,7 +284,6 @@ end
% Clears unneeded variables from workspace
clear i
clear -regexp ^geffe_
clear -regexp ^transformation_
% ========== POLES DE-NORMALIZATION END ==========
@ -384,7 +383,7 @@ clear clearVars
%% ========== GAIN ADJUSTMENT START ==========
total_gain_high = units_k(1,1)*units_k(1,2);
unit_adjustment_gain = 1/total_gain_high;
unit_adjustment_gain = (10^(10/20))/total_gain_high;
% We arbitrarily choose to use a 10KOhm series resistor in the adjustment
% unit
unit_adjustment_feedback_resistor = 10*10^3*unit_adjustment_gain;
@ -427,7 +426,7 @@ ltiview('bodemag', total_transfer_function);
ltiview('bodemag', units_transfer_functions(1,1), units_transfer_functions(1,2), ...
total_transfer_function);
%}
%{
hold off
sampling_time_seconds = 60; % s
@ -473,7 +472,7 @@ Pyy = system_output_fft.*conj(system_output_fft)/sampling_length_L;
figure(3)
semilogx(frequency_vector,Pyy(1:sampling_length_L/2+1))
grid on
%}
% Clears unneeded variable from workspace
clearVars = {'high_frequency', 'total_transfer_function', 'Pyy', ...
'frequency_vector', 'system_output', 'system_output_fft'};

52
report/4_high_pass/4_high_pass.pug

@ -0,0 +1,52 @@
div(style="page-break-before:always")
// Chapter title
h2 Ανωδιαβατό φίλτρο Butterworth
br/
// Chapter description/introduction
p.
Με βάση τον αριθμό ΑΕΜ (8261), υπολογίστηκαν οι προδιαγραφές του ανωδιαβατού φίλτρου προς σχεδίαση:
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th Προδιαγραφή
th Τιμή
tbody
tr
td Pass frequency (f#[sub p])
td 5000 Hz
tr
td Pass radial frequency (ω#[sub p])
td 31415.927 rad/s
tr
td Stop frequency (f#[sub s])
td 1923.077 Hz
tr
td Stop radial frequency (ω#[sub s])
td 12083.049 rad/s
tr
td Min stop attenuation (a#[sub min])
td 24.667 dB
tr
td Max pass attenuation (a#[sub max])
td 0.667 dB
figcaption
.reference #[span.table-count]
.caption.
Προδιαγραφές σχεδίασης ανωδιαβατού φίλτρου
figure.block-center.width-15cm
img(src="4_high_pass/assets/diagrams/high_pass_general_transfer_function_plot.svg").width-15cm
figcaption
.reference #[span.plot-count]
.caption.title.
Ποιοτικό γράφημα συνάρτησης μεταφοράς ανωδιαβατού Butterworth φίλτρου.
.caption.
Στο γράφημα φαίνονται οι συχνότητες που ορίζουν τη ζώνη αποκοπής (f#[sub s]/ω#[sub s]) και τη ζώνη διόδου (f#[sub p]/ω#[sub p]), καθώς και οι προδιαγραφές α#[sub min] και α#[sub max].
// Sub-Chapters
include 4_high_pass_design
//- include 1_low_pass_transfer_function_matlab
//- include 1_low_pass_transfer_function_multisim

375
report/4_high_pass/4_high_pass_design.pug

@ -0,0 +1,375 @@
h3 Σχεδίαση φίλτρου
p.
Για τη σχεδίαση του φίλτρου ακολουθήθηκε η διαδικασία που περιγράφεται στο κεφάλαιο 12 των σημειώσεων του μαθήματος:
figure.block-center.width-15cm
div.ui.list.ordered.celled.striped
div.item Υπολογισμός των προδιαγραφών ενός πρωτότυπου κατωδιαβατού Butterworth φίλτρου, μέσω των προδιαγραφών του επιθυμητού ανωδιαβατού φίλτρου.
div.item Υπολογισμός της τάξης και της συχνότητας ημίσειας ισχύος του πρωτότυπου φίλτρου.
div.item Υπολογισμός των πόλων του πρότυπου φίλτρου Butterworth.
div.item Υπολογισμός των πόλων και μηδενικών του ανωδιαβατού φίλτρου μέσω του ευθύ μετασχηματισμού LP → HP.
div.item Υλοποίηση των ανωδιαβατών μονάδων με χρήση φίλτρων Sallen-Key με βάση τα κυκλώματα του κεφαλαίου 7.
div.item Κλιμακοποίηση του κυκλώματος με στόχο τη μεταφορά στις πραγματικές συχνότητες και σε στοιχεία με πρακτικές (υλοποιήσιμες) τιμές.
div.item Έλεγχος των κερδών των μονάδων και ρύθμιση κέρδους με επιβολή απόσβεσης ή ενίσχυσης.
div(style="page-break-before:always")
h4 Υπολογισμός συνάρτησης μεταφοράς
p.
Αρχικά σχεδιάζεται ένα πρωτότυπο κατωδιαβατό Butterworth φίλτρο, το οποίο αργότερα θα μετατραπεί στο επιθυμητό ανωδιαβατό Butterworth.
p Υπολογίζονται οι προδιαγραφές του πρωτότυπου κατωδιαβατού χρησιμοποιώντας τις εξισώσεις #[span.course-notes-equation 12-4]:
p.latex-equation.
$$\Omega_p = 1\frac{rad}{s}$$
και $$\Omega_S = \frac{\omega_p}{\omega_s} = \frac{31415.927}{12083.049} = 2.6\frac{rad}{s}$$
p Οι προδιαγραφές απόσβεσης παραμένουν ίδιες.
p Υπολογίζεται η τάξη του φίλτρου χρησιμοποιώντας την εξίσωση #[span.course-notes-equation 9-52]:
p.latex-equation.
$$n = \left \lceil \frac{\log\bigg(\frac{10^{\frac{a_{min}}{10}}-1}{10^{\frac{a_{max}}{10}}-1}\bigg)}{2\log(\frac{\Omega_s}{\Omega_p})} \right \rceil = \left \lceil \frac{\log\bigg(\frac{10^{2.4667}-1}{10^{0.0667}-1}\bigg)}{2\log(2.6)} \right \rceil = \left \lceil \frac{3.2453}{0.8299} \right \rceil = \left \lceil 3.91 \right \rceil = 4$$
p.
Από τον παραπάνω τύπο φαίνεται ότι κατά τον υπολογισμό της τάξης του φίλτρου γίνεται στρογγυλοποίηση της τάξης προς τον επόμενο #[strong μεγαλύτερο] ακέραιο. Αυτό γίνεται επειδή δεν είναι δυνατή η υλοποίηση ενός φίλτρου ρητής τάξεως, έτσι είναι απαραίτητο η τάξη να στρογγυλοποιηθεί. Η στρογγυλοποίηση είναι σημαντικό να γίνει προς τα επάνω (ceiling) ώστε να επιτευχθούν οι προδιαγραφές του φίλτρου. Μία πιθανή στρογγυλοποίηση προς τα κάτω θα είχε ως αποτέλεσμα την αποτυχία στη σχεδίαση.
p.
Λόγω της στρογγυλοποίησης αυτής, αναμένεται μάλιστα να υπάρχει υπερκάλυψη σε τουλάχιστον μία από τις προδιαγραφές. Στη συγκεκριμένη περίπτωση, λόγω της επιλογής κανονικοποίησης ως προς το pass band, αναμένεται να υπάρχει υπερκάλυψη της προδιαγραφής a#[sub max].
p.
Υπολογίζεται η κανονικοποιημένη συχνότητα ημίσειας ισχύος χρησιμοποιώντας την εξίσωση #[span.course-notes-equation 9-48]:
p.latex-equation.
$$\Omega_{hp} = \frac{\Omega_p}{\big [ 10^{\frac{a_{max}}{10}}-1 \big ]^{\frac{1}{2n}}} = \frac{1}{\big [ 10^{0.0667}-1 \big ]^{\frac{1}{2n}}} = 1.2517\frac{rad}{s}$$
p και στη συνέχεια μεταφέρεται στη πραγματική συχνότητα χρησιμοποιώντας την εξίσωση #[span.course-notes-equation 12-3]:
p.latex-equation.
$$\omega_{hp} = \frac{\omega_p}{\Omega_{hp}} = \frac{31415.927}{1.2517} = 25097.784\frac{\text{rad}}{\text{s}}$$
p.
Οι γωνίες Butterworth μπορούν να υπολογιστούν με βάση τον αλγόριθμο Guillemin ή να βρεθούν απευθείας από γνωστούς πίνακες γωνιών Butterworth. Για φίλτρο τέταρτης τάξης, οι γωνίες είναι:
p.latex-equation.
$$\pm 22.5^\circ$$ και $$\pm 67.5^\circ$$
p.
Με βάση τις εξισώσεις #[span.course-notes-equation 9-31] και #[span.course-notes-equation 9-38]:
p.latex-equation.
$$\sigma_k = -\cos(\psi_k)$$
$$\Omega_k = \sin(\psi_k)$$
p.latex-equation.
$$Q_k = \frac{1}{2\cos(\psi_k)}$$
p προκύπτουν οι πόλοι του #[strong κατωδιαβατού Butterworth]:
p.latex-equation.
$$\text{Pole 1: } -0.924\pm0.383\mathrm{i}\text{, }\hspace{2mm}\Omega_0 = 1\text{, }\hspace{2mm}Q=0.541$$
$$\text{Pole 2: } -0.383\pm0.924\mathrm{i}\text{, }\hspace{2mm}\Omega_0 = 1\text{, }\hspace{2mm}Q=1.307$$
p.
Οι πόλοι μετασχηματίζονται χρησιμοποιώντας τον ευθύ μετασχηματισμό #[span.course-notes-equation 12-6]:
p.latex-equation.
$$S = \frac{1}{s}$$
p.
Σύμφωνα με τις σημειώσεις του μαθήματος (κεφάλαιο 12, σελίδα 5, τέλος σελίδας) και τη θεωρία ο ευθύς μετασχηματισμός LP → HP φίλτρων Butterworth αφήνει τους πόλους αναλοίωτους. Έτσι οι πόλοι του #[strong ανωδιαβατού] Butterworth είναι οι ίδιοι με αυτούς του #[strong κατωδιαβατού] που υπολογίστηκαν. Ωστόσο, στη συνάρτηση μεταφοράς εμφανίζονται τέσσερα μηδενικά στο μηδέν.
p.latex-equation.
$$T_{LP}(S) = \frac{1}{(S^2+1.848S+1)(S^2+0.765S+1)}$$
$$T_{HP}(s) = \frac{s^4}{(s^2+1.848s+1)(s^2+0.765s+1)} $$
p.
Στην παραπάνω σχέση, η συχνότητα ημίσειας ισχύος για το ανωδιαβατό φίλτρο είναι ω#[sub 0]=1, δηλαδή έχουμε ένα κανονικοποιημένο ανωδιαβατό φίλτρο. Στην πραγματικότητα όμως η συχνότητα 3dB του ανωδιαβατού φίλτρου προκύπτει ω#[sub 0]=25097.784 (rad/s), διότι Ω#[sub 0]≠1 και ω#[sub p]≠1. Επομένως, οι πόλοι της T#[sub HP](s) κείνται πάνω σε ένα κύκλο με ακτίνα ω#[sub 0]. Οι πόλοι της ανωδιαβατής συνάρτησης δίνονται από τον παρακάτω πίνακα, όπου αντί για Ω#[sub 0k]=1 θεωρούμε μέτρα ω#[sub 0], δηλαδή:
p.latex-equation.
$$s_k = \omega_0(-\cos{\psi_k}+\mathrm{i}\sin{\psi_k})$$
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th p#[sub k]
th ψ#[sub k]
th σ#[sub k]±jω#[sub k]
th Ω#[sub 0k]
th Q#[sub k]
tbody
tr
td p#[sub 1]
td ±22.5°
td -23187.329±j9604.506
td 25097.784
td 0.541
tr
td p#[sub 2]
td ±67.5°
td -9604.506±j23187.329
td 25097.784
td 1.307
figcaption
.reference #[span.table-count]
.caption.
Πόλοι ανωδιαβατού Butterworth φίλτρου
p.
Με άλλα λόγια η ω#[sub 0] δίνει την συνολική κλιμακοποίηση συχνότητας, αφ'ενός μεν λόγω της αρχικής κλιμακοποίησης (ω#[sub p]) και στην συνέχεια λόγω κλιμακοποίησης του Ω#[sub hp]: ω#[sub hp]=31415.927/1.2517=25097.784.
figure.block-center.width-15cm
img(src="4_high_pass/assets/diagrams/matlab_high_pass_butterworth_zero_pole.svg").width-15cm
figcaption
.reference #[span.plot-count]
.caption.
Πόλοι και μηδενικά του Butterworth.
p.
Κάθε ζεύγος μιγαδικών πόλων υλοποιείται από ένα κύκλωμα Sallen-Key, προκύπτουν έτσι οι παρακάτω ανωδιαβατές μονάδες προς υλοποίηση:
figure.block-center.width-15cm
img(src="4_high_pass/assets/diagrams/high_pass_butterworth_units_diagram.svg").width-12cm.block-center
figcaption
.reference #[span.plot-count]
.caption.
Ανωδιαβατές μονάδες Sallen-Key προς υλοποίηση
h4 Υλοποίηση συνάρτησης μεταφοράς
p.
Από τον αριθμό ΑΕΜ (8261) υποδεικνύεται η χρήση των κυκλωμάτων Sallen-Key του κεφαλαίου 6, με χρήση της πρώτης στατηγικής σχεδίασης (Στρατηγική 1).
h5 Μονάδα 1
p Η πρώτη μονάδα Sallen-Key πρέπει να υλοποιεί:
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th Προδιαγραφή
th Τιμή
tbody
tr
td ω#[sub 01]
td 25097.784
tr
td Q
td 0.541
figcaption
.reference #[span.table-count]
.caption.
Προδιαγραφές πρώτης ανωδιαβατής μονάδας Sallen-Key
p.
Υπολογίζονται τα στοιχεία του κυκλώματος του φίλτρου, με χρήση της μεθοδολογίας που περιγράφεται στο κεφάλαιο 6.2.1 (σελίδα 27) και των εξισώσεων #[span.course-notes-equation 7-74] και #[span.course-notes-equation 7-75]:
p.latex-equation.
$$R_1 = R_2 = 1\text{ Ohm}$$
$$C_1 = C_2 = 1\text{ F}$$
$$k = 3-\frac{1}{Q} = 3-\frac{1}{0.541} = 1.1522\hspace{1cm}\text{(Gain at high frequencies)}$$
$$r_1 = 1$$
$$r_2 = 2-\frac{1}{Q} = 2-\frac{1}{0.541} = 0.1522$$
p #[strong Κλιμακοποίηση]
p.
Γίνεται κλιμακoποίηση των στοιχείων της μονάδας για να μεταφερθούν οι συχνότητες στις πραγματικές τιμές. Επιλέγεται:
p.latex-equation.
$$k_{f} = \omega_{01} = 25097.784$$
p.
Με βάση τον αριθμό ΑΕΜ (8261) επιλέγεται κατάλληλος συντελεστής κλιμακοποίησης πλάτους ώστε να επιτευχθεί τιμή πυκνωτών ίση με 0.1μF, γίνεται χρήση του τύπου #[span.course-notes-equation 6-33]:
p.latex-equation.
$$k_{m} = \frac{C_{old}}{k_fC_{new}} = \frac{1}{25097.784*0.1*10^{-6}} = 398.4415$$
p Οι τελικές τιμές των στοιχείων φαίνονται στον παρακάτω πίνακα:
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th Στοιχείο/Κέρδος
th(colspan="2") Τιμή
tbody
tr
td C#[sub 1] = C#[sub 2]
td(colspan="2") 0,1 μF
tr
td R#[sub 1]
td(colspan="2") 398.4415 Ohm
tr
td R#[sub 2]
td(colspan="2") 398.4415 Ohm
tr
td r#[sub 1]
td(colspan="2") 398.4415 Ohm
tr
td r#[sub 2]
td(colspan="2") 60.6591 Ohm
tr
td Κέρδος στις υψηλές συχνότητες
td 1.1522
td 1.23 dB
figcaption
.reference #[span.table-count]
.caption.
Οι τιμές των στοιχείων της πρώτης μονάδας και το κέρδος στις υψηλές συχνότητες
p.
Η συνάρτηση μεταφοράς της μονάδας υπολογίζεται χρησιμοποιώντας την εξίσωση #[span.course-notes-equation 6-68]:
p.latex-equation.
$$\begin{align*} T_{HP}^1(s) &= k\frac{s^2}{s^2+\big [ \frac{1}{R_2C_1}+\frac{1}{R_2C_2}+\frac{1-k}{R_1C_1} \big ] s+\frac{1}{R_1R_2C_1C_2}} \\[3.5mm] &= 1.1522\frac{s^2}{s^2+\big [ \frac{1}{398.4415*0.1*10^{-6}}+\frac{1}{398.4415*0.1*10^{-6}}+\frac{1-1.1522}{398.4415*0.1*10^{-6}} \big ] s+\frac{1}{398.4415^2*(0.1*10^{-6})^2}} \\[3.5mm] &=\frac{1.1522s^2}{s^2+46376s+6.299*10^8} \end{align*}$$
h5 Μονάδα 2
p Η δεύτερη μονάδα Sallen-Key πρέπει να υλοποιεί:
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th Προδιαγραφή
th Τιμή
tbody
tr
td ω#[sub 01]
td 25097.784
tr
td Q
td 1.307
figcaption
.reference #[span.table-count]
.caption.
Προδιαγραφές δεύτερης ανωδιαβατής μονάδας Sallen-Key
p.
Υπολογίζονται τα στοιχεία του κυκλώματος του φίλτρου, με χρήση της μεθοδολογίας που περιγράφεται στο κεφάλαιο 6.2.1 (σελίδα 27) και των εξισώσεων #[span.course-notes-equation 7-74] και #[span.course-notes-equation 7-75]:
p.latex-equation.
$$R_1 = R_2 = 1\text{ Ohm}$$
$$C_1 = C_2 = 1\text{ F}$$
$$k = 3-\frac{1}{Q} = 3-\frac{1}{1.307} = 2.2346\hspace{1cm}\text{(Gain at high frequencies)}$$
$$r_1 = 1$$
$$r_2 = 2-\frac{1}{Q} = 2-\frac{1}{1.307} = 1.2346$$
p #[strong Κλιμακοποίηση]
p.
Γίνεται κλιμακoποίηση των στοιχείων της μονάδας για να μεταφερθούν οι συχνότητες στις πραγματικές τιμές. Επιλέγεται:
p.latex-equation.
$$k_{f} = \omega_{01} = 25097.784$$
p.
Με βάση τον αριθμό ΑΕΜ (8261) επιλέγεται κατάλληλος συντελεστής κλιμακοποίησης πλάτους ώστε να επιτευχθεί τιμή πυκνωτών ίση με 0.1μF, γίνεται χρήση του τύπου #[span.course-notes-equation 6-33]:
p.latex-equation.
$$k_{m} = \frac{C_{old}}{k_fC_{new}} = \frac{1}{25097.784*0.1*10^{-6}} = 398.4415$$
p Οι τελικές τιμές των στοιχείων φαίνονται στον παρακάτω πίνακα:
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th Στοιχείο/Κέρδος
th(colspan="2") Τιμή
tbody
tr
td C#[sub 1] = C#[sub 2]
td(colspan="2") 0,1 μF
tr
td R#[sub 1]
td(colspan="2") 398.4415 Ohm
tr
td R#[sub 2]
td(colspan="2") 398.4415 Ohm
tr
td r#[sub 1]
td(colspan="2") 398.4415 Ohm
tr
td r#[sub 2]
td(colspan="2") 491.9291 Ohm
tr
td Κέρδος στις υψηλές συχνότητες
td 2.2346
td 6.98 dB
figcaption
.reference #[span.table-count]
.caption.
Οι τιμές των στοιχείων της δεύτερης μονάδας και το κέρδος στις υψηλές συχνότητες
p.
Η συνάρτηση μεταφοράς της μονάδας υπολογίζεται χρησιμοποιώντας την εξίσωση #[span.course-notes-equation 6-68]:
p.latex-equation.
$$\begin{align*} T_{HP}^2(s) &= k\frac{s^2}{s^2+\big [ \frac{1}{R_2C_1}+\frac{1}{R_2C_2}+\frac{1-k}{R_1C_1} \big ] s+\frac{1}{R_1R_2C_1C_2}} \\[3.5mm] &= 2.2346\frac{s^2}{s^2+\big [ \frac{1}{398.4415*0.1*10^{-6}}+\frac{1}{398.4415*0.1*10^{-6}}+\frac{1-2.2346}{398.4415*0.1*10^{-6}} \big ] s+\frac{1}{398.4415^2*(0.1*10^{-6})^2}} \\[3.5mm] &=\frac{2.2346s^2}{s^2+19210s+6.299*10^8} \end{align*}$$
h4 Ρύθμιση κέρδους
p.
Με βάση τον αριθμό ΑΕΜ (8261) πραγματοποιείται ρύθμιση κέρδους με στόχο την επίτευξη κέρδους 10 dB στη ζώνη διόδου.
p.
Κατά την υλοποίηση των μονάδων Sallen-Key διαπιστώθηκε ότι κάθε μονάδα εισάγει ένα κέρδος. Το συνολικό κέρδος που εισάγουν οι μονάδες είναι:
p.latex-equation.
$$k = k_1k_2 = 1.1522*2.2346 = 2.5747$$
p.
Για τη ρύθμιση του κέρδους χρησιμοποιείται μία αναστρέφουσα συνδεσμολογία με κέρδος:
p.latex-equation.
$${k}' = \frac{10^{\frac{10}{20}}}{k} = \frac{3.162}{2.5747} = 1.2281$$
p.
Επιλέγεται η χρήση αντίστασης εισόδου ίσης με r#[sub 1]=10 kOhm. Έτσι η αντίσταση ανατροφοδότησης υπολογίζεται:
p.latex-equation.
$$r_2 = r_1{k}' = 10*10^3*1.2281 = 12281\text{ Ohm}$$
h4 Συναρτήσεις μεταφοράς
p.
Στον παρακάτω πίνακα φαίνονται οι συναρτήσεις μεταφοράς των επιμέρους μονάδων που υπολογίστηκαν παραπάνω:
figure.block-center.width-15cm
table.ui.celled.table.teal.striped.center.aligned
thead
tr
th Μονάδα
th Συνάρτηση μεταφοράς
tbody
tr
td Πρώτη μονάδα (Unit 1)
td.
$$T_{HP}^1(s) = \frac{1.1522s^2}{s^2+46376s+6.299*10^8}$$
tr
td Δεύτερη μονάδα (Unit 2)
td.
$$T_{HP}^2(s) = \frac{2.2346s^2}{s^2+19210s+6.299*10^8}$$
figcaption
.reference #[span.table-count]
.caption.
Συναρτήσεις μεταφοράς των επιμέρους μονάδων
p.
Η συνολική συνάρτηση μεταφοράς του φίλτρου υπολογίζεται:
p.latex-equation.
$$\begin{align*} T_{HP(s)} &= {k}'T_{HP(s)}^1T_{HP(s)}^2 \\[3.5mm] &=\frac{3.1623s^4}{s^4+65584s^3+21.506*10^8s^2+4.1311*10^{13}s+39.677*10^{16}} \\[3.5mm] \end{align*}$$
p.
Στο παρακάτω σχήμα φαίνεται το τελικό κύκλωμα του φίλτρου:
figure.block-center.width-19cm
img(src="4_high_pass/assets/diagrams/multisim_high_pass_butterworth_circuit_layout_only_filter.svg").width-19cm
figcaption
.reference #[span.plot-count]
.caption.
Κύκλωμα ζωνοφρακτικού φίλτρου

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After

Width:  |  Height:  |  Size: 28 KiB

5
report/report.pug

@ -5,9 +5,10 @@
//- include 0_intro/0_intro
//- include 1_low_pass/1_low_pass
//- include 2_band_pass/2_band_pass
include 3_band_elimination/3_band_elimination
//- include 4_high_pass
//- include 3_band_elimination/3_band_elimination
include 4_high_pass/4_high_pass
//- Removes annoying date from top left corner
template#page-header
span \

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