elna-filter3

Elektronika Analog 2006, TE - UGM 1

Classification of FiltersClassification of Filters

Signal Filter

Analog Filter Digital Filter

Element Type Frequency Band

Active Passive Low-Pass

High-Pass

Band-Pass

Band-Reject

All-Pass


Filter

1. Dalam pengolahan sinyal, fungsi tapis adalah untuk menghilangkan bagian sinyal yang tidak diiginkan

misalnya derau random,

2. Memilah bagian sinyal yang berguna, misal bagian yang berada di daerah frekuensi tertentu.


Analog Filters

Tapis analog menggunakan untai analog ( resistor, Kapasitor dan op-amp) untuk menghasilkan proses penapisan.

Aplikasi : pengurangan derau, perbaikan sinyal video, graphic equalisers in hi-fi systems, dll.

Sudah banyak teknik baku untuk merancang tapis analog sesuai dengan unjuk kerja yang diinginkan.

Sinyal yang akan di tapis berupa tegangan ataupun arus listrik yang mewakili suatu besaran fisis analog.


Digital Filters A digital filter uses a digital processor to perform numerical

calculations on sampled values of the signal.

The analog input signal is sampled and digitised using an ADC (analog to digital converter).

The resulting binary numbers, representing the input signal, are transferred to the processor, which carries out numerical calculations on them.

Note that in a digital filter, the signal is represented by a sequence of numbers, rather than a voltage or current.


Analog Versus Digital Filters A digital filter is programmable, i.e. its operation is

determined by a program stored in the processor's memory.

An analog filter can only be changed by redesigning the filter circuit.

Digital filters are easily designed, tested and implemented on a general-purpose computer or workstation.

The characteristics of analog filter circuits (particularly those containing active components) are subject to drift and are dependent on temperature.


Analog Versus Digital Filters

Digital filters are very much more versatile in their ability to process signals in a variety of ways;

Digital filters do not suffer from these problems, and so are extremely stable with respect both to time and temperature.

Unlike their analog counterparts, digital filters can handle low frequency signals accurately.

Fast DSP processors can handle complex combinations of filters in parallel or cascade (series), making the hardware requirements relatively simple and compact in comparison with the equivalent analog circuitry.


Ideal Filter Design (or Goals)

The passband amplitude response is continuously flat at a value of 1. The frequencies which are allowed to pass through the filter do so completely undistorted.

The stopband amplitude response is continuously flat at a value of 0. The frequencies which are allowed to pass through the filter do so completely undistorted.

The transition between the passband and the stopband happens instantaneously.

Stopband

fc cutoff frequency


Phase Delay

Goal To have the time delay

to be independent of the frequency, hence each frequency component is modified exactly the same way. (see figure)

Why this is important?

Ideal


Time Offset

The delay through the filter circuit of each component frequency that makes up the desired signal may be different. This shifting distorts the captured signal's time characteristics. For example, a signal used to modulate a carrier may be phase aligned to the carrier before passing through the filter but have an unwanted time offset afterward.


Basic Structure for FIR

Z-1 Z-1 Z-1 Z-1 Z-1

x(n) x(n-1) x(n-2) x(n-3) x(n-N+1)

Xh0

Xh1

Xh2

Xh3

XhN-2

XhN-1

x(n-N+2)

SUM

y(n)

Tapped-delay line (N-1) delaysN multipliers1 adder (N inputs)

∑1-N

=k

*

0

k)- x(nh(k)=x(n)h(n)=y(n)


TAPIS (FILTER)

Tapis frekuensi rendah (LPF : Low pass filter) : melewatkan frekuensi rendah, sampai

dengan cut off frequency) Tapis frekuensi tinggi (HPF : High pass

filter) : melewatkan frekuensi tinggi, mulai dari

cut off frequency)


LPF

PASIF : hanya komponen RC

AKTIF : menggunakan op-amp atau transistor

R

CVin Vout

R

CVin

Karakteristik terpengaruh beban

Karakteristiktak terpengaruh beban

Vout


HPF

aktif

R

C

Vin Vout

pasif

1RCs

RCs

V

V

in

out

+

R

C

Vin Vout


Passive Low-Pass Filter

The pass-band is from 0 to some frequency wp.

Its stop-band extends form some frequency ws, to infinity.

In practical circuit design, engineers often choose amplitude gain of 0.95 for passive RC filters:

p s

)( jH

Vout

Vin

C

R

VVoutoutVVinin RL

-3dB


( )

2

1

( ) ( )

1( )

1

( ) tan ( )

jH A e

ARC

RC


Passive High-Pass FilterPassive High-Pass Filter

Its stop-band is form 0 to some frequency s

The pass-band is from some frequency p to infinity.

In practical circuit design, engineers choose amplitude gain of 0.95 for passive CR filters:

s p

)( jH

Vout

Vin

R

C

VVoutoutVVinin


Design of Passive FiltersDesign of Passive Filters

1

1

jRCjH

1

1

RCssH

Transfer Function

21

1

RCV

V

in

out

The amplitude response:

2

1

2

13

RCf dB

The 3dB break-point is at:

LF

L

ZZ

ZG

The amplitude gain:

C

R

VVoutoutVVinin RL


Guideline of Pass Filter DesignGuideline of Pass Filter Design

R

1

1

ssH

Transfer Function

C VVoutoutVVinin RL

RC

Time Constant

Select resistor based on amplitude gain:

95.0

LF

L

ZZ

ZG

LLF RZRZ 053.095.0

05.0

Select capacitor based on cut-off freq:

dBRfRC

32

1


Higher Higher Order FiltersOrder Filters

C

R

VVoutoutVVinin

First Order RC Low Pass Second Order RC Low Pass

C2 VVoutoutVVininC1

R1 R2

The higher the order of the filter, the closer it approaches ideal characteristics.


Active FiltersActive Filters

Active filters employ Op-Amps to attenuate select frequencies and amplify signal during filtering process.

Q factor of a filter is defined as the ratio of the center frequency fc to the bandwidth fH - fL :

LH

Cff

fQ


Bandpass Filter Parameters


Proportional- or Variable-Q


Constant-Q


Example:Design a low pass filter with cut-off frequency of 5kHz, and DC gain of 10:

Two equations, three unknowns

Design of Low Pass Active FiltersDesign of Low Pass Active Filters

-

+

Vin

Vout

R1

RF

A

B

C2

Transfer Function: Transfer Function:

0

0..

s

KFT LP

221

CRfF

H

The -3 dB cut-off frequency:

1RRK F

LP

The DC gain:


Design of High Pass Active FiltersDesign of High Pass Active Filters

Vout

-

+

Vin

R1

RF

A

B

C1

The -3 dB cut-off frequency:

The DC gain:

Two equations, three unknowns

Select one component based on other conditions, and determine the values of the other two components.

1121

CRfH

1RRK F

HP

Transfer Function: Transfer Function:

0

..

s

sKFT HP


Filter ClassFilter Class

o A filter of a given order can be made to approximate to ideal characteristics in a number of ways, depending on the values of the filter components (or say: depending on the filter class.

o Two useful classes are Butterworth (maximally flat) and Chebyshev (equal-ripple) filters (n is the filter order)

n

C

in

out

ffV

V2

1

1

Butterworth Filter

Chebyshev Filter

Cn

in

out

ffCEV

V

221

1


ORDE LPF

1RCs

1

V

V

in

out

12

1

V

V222

in

out

RCssCR

133

1

V

V222333

in

out

RCssCRsCR

Orde satu :

Orde dua :

Orde tiga :


Karakteristik LPF

0 50 100 150 200 250 300 350 400 450 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1bentuk tanggapan

frekuensi (Hz)

magnitude

Orde 1

Orde 3

Orde 2


Butterworth ApproximationButterworth Approximation

Typical magnitude responses with 1c

0 1 2 30

0.2

0.4

0.6

0.8

1

Mag

nitu

de

Butterworth Filter

N = 2N = 4N = 10


Chebyshev ApproximationChebyshev Approximation

Typical magnitude response plots of the analog lowpass Type 2 Chebyshev filter are shown below

0 1 2 30

0.2

0.4

0.6

0.8

1

Mag

nitu

de

Type 2 Chebyshev Filter

N = 3N = 5N = 7


Elliptic ApproximationElliptic Approximation

Typical magnitude response plots with are shown below 1 p

0 1 2 30

0.2

0.4

0.6

0.8

1

Mag

nitu

de

Elliptic Filter

N = 3N = 4


Higher Order Active FiltersHigher Order Active Filters

Vout

-

+Vin

R2

Rb

C1

R1

Ra

C2

Gain=K

Filter Class R1 R2 C1 C2 K

Buterworth3.01 dB at H

1.00 1.001.001.00 1.59

Chebyshev1 dB ripple

1.00 1.000.94 0.97 2.00

The above list gives the gain and component valves for one of themany choices for H=1. You may find more combinations from filter design handbook(s).


313142

42

31)(GGsGGCsCC

GGsH

Given transfer function:

012

2

)(bsbsb

asH

Find circuit parameters:

4321 ,,, CGCG

Conductance:i

i RG

1


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH


4321 ,,, CGCG1. Let the resistors be 1 (i.e., 1S)2. Solve for the capacitors: - Make sure b0 = 1 (divide it out if necessary) - 2xC4 = b1 ---- C4 = b1/2 F - C2C4 = b2 ---- C2 = b2/C4 F3. Scale components - multiply resistors by K, divide capacitors by K - e.g., let K = 500,000


423142

31

231)(

GGsCCGsCC

sCCsH


012

2

22)(

bsbsb

sbsH


4321 ,,, GCGC1. Let the Capacitors be 1F2. Solve for the Resistors: - Make sure b2 = 1 (divide it out if necessary) - 2xG4 = b1 ---- G4 = b1/2 S - G2G4 = b2 ---- G2 = b2/G4 F3. Scale components - multiply resistors by K, divide capacitors by K - e.g., let K = 500,000

High Interchange Capacitors and Resistors ii RsC


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH




313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH



8073.1484.2

8073.12 ss


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH



18073.1484.2

8073.11

12 ss


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH


4321 ,,, CGCG1. Let the resistors be 1 (i.e., 1S)2. Solve for the capacitors: - Make sure b0 = 1 (divide it out if necessary) - 2xC4 = b1 ---- C4 = b1/2 F - C2C4 = b2 ---- C2 = b2/C4 F3. Scale components - multiply resistors by K, divide capacitors by K - e.g., let K = 500,000 Dallas Semiconductor, Application Note 1795: Dec 03, 2002

6872.0

28073.1

484.2

4 C

18073.1484.2

8073.11

12 ss


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH



8052.0

6872.08073.1

1

2 C

18073.1484.2

8073.11

12 ss


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH



8052.02 C

6872.04 C

11 G

13 G

18073.1484.2

8073.11

12 ss


313142

42

31)(GGsGGCsCC

GGsH


012

2

0)(bsbsb

bsH


4321 ,,, CGCG1. Let the resistors be 1 (i.e., 1S)2. Solve for the capacitors: - Make sure b0 = 1 (divide it out if necessary) - 2xC4 = b1 ---- C4 = b1/2 F - C2C4 = b2 ---- C2 = b2/C4 F3. Scale components - multiply resistors by K, divide capacitors by K - e.g., let K = 500,000 Dallas Semiconductor, Application Note 1795: Dec 03, 2002

FC 6104.12

FC 3744.14

KR 5001

KR 5003

18073.1484.2

8073.11

12 ss


FC 6104.12

FC 3744.14 KR 5001

KR 5003

First Second Order Section


Program Matlab

r=1000;c=4*10^-12;f=1:5:10^4;w=2*pi*f;s=j*w;z=(r*c*s)./(r*c*s+l);zmag=abs(z);semilogx(f,zmag,'b',f)xlabel('frekuensi, hz')ylabel('besar z,ohm')


Gambar hasil program

100

101

102

103

104

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

frekuensi, hz

besa

r z,

ohm


PERSAMAAN FILTER

22

)(

nn s

Qs

HsN

22

)(

nn s

Qs

HssN

22

2

)(

nn s

Qs

HssN

LOW PASS

BAND PASS

HIGH PASS


FILTER PASIF

CsLsRCsLs

sN1//

1//)(

LCs

RCs

sRCL

sN11

)(2

C

R

VVoutoutVVinin L

22

)(

nn s

Qs

HssN

LCn1

L

CRQ

RC

LH


LCn1 Cf

L

221

L

CRQ C

LQR


BANDPASS FILTER

1. C=100*10^-6;2. q=25.0; faktor kualitas3. f0=50.0; frekuensi center4. n=2*pi*f0;5. L=1/((n^2)*C)6. R=q*sqrt(L/C)7. f=10:0.1:100;8. w=2*pi*f;9. s=j*w;10. z=((L/R*C)*s)./(s.^2+(1/(R*C))*s+(1/(L*C)));11. zmag=abs(z);12. plot(f,zmag,'b')

Elektronika Analog 2006, TE - UGM 5010 20 30 40 50 60 70 80 90 100

0

0.2

0.4

0.6

0.8

1

1.2x 10

-9


313142

42

31

1111

11)(

RRsRRCsCCRRsH


Salens-Key Gain satu

1)1(

1)(

2220

smRCsCmnRs

V

V

i

R3=R

R1=mR

C4=C

C2=nC


Tugas 1

Buktikan rumus itu berdasar rangkaian yang ada.


20

02

)(

s

Qs

HsN

1)1(

1)(

2220

smRCsCmnRs

V

V

i

1

m

mnQ

mnRCf

2

10


PROSEDUR PERANCANGAN

Ditentukan dulu f0 dan Q,

kemudian :

1. Pilih m=1, R antara 10 K-Ohm s/d 100K-Ohm

RQfC

04

1

2. Hitung C :

3. Hitung : n = 4Q2



4. Tentukan nilai kapasitor yang standar untuk C dan nC n yang baru (nbaru > nlama)

22

Q

nk

2

42

kkm

5. Dengan n yang baru, hitung k sbb :

6. hitung m baru:



7. Hitung R

8. Tentukan mR dan R yang standar

9. Uji dulu secara simulatif dengan C dan R yang real. Apakah hasilnya = teoritis ?

nmCfR

02

1


Merancang filter low pass salen-key gain = 1dengan

f0=1KHz dan Q=2

2. Buatlah tanggapan frekuensi dengan simulasi program matlab.

3. Pertama dipilih R=33 KOhm. Hasil akhir di simulasikan.

TUGAS 2 dan 3


FILTER KRC Gain dapat diubah tanpa mengubah parameter

filter Menggunakan penguat non inverting

mnRCf

2

10

Vout

-

+Vin

R2

Rb

C1

R1

Ra

C2

Gain=K)1(1 Kmnm

mnQ


Butterworth

Dicapai dengan m=1 hambatan sama Dan n=2 kapasitor berbanding 2:1

2

1Q

elna-filter3

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