Pengolahan Sinyal Digital B 2011 2012 12 (Bab 7 & 8)

7/28/2019 Pengolahan Sinyal Digital B 2011 2012 12 (Bab 7 & 8)

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TUGAS

PENGOLAHANSINYAL DIGITAL

KELOMPOK 12

AGGOTA :

AHADI ANGGUN RAHARJO H1C009029

WISNU HERDANANTO H1C009032

HERU BUDIANTO H1C009033

KEMENTERIAN PENDIDIKAN DAN KEBUDAYAAN

UNIVERSITAS JENDERAL SOEDIRMAN

FAKULTAS SAINS DAN TEKNIK

PROGRAM STUDI TEKNIK ELEKTRO

PURBALINGGA

2012


2/28

Frekuensi cutoff yang ternormalisasi :

c=2 fcTs=21500

8000

c=0.375

saat 2M + 1 = 3

h(0)=c

n, n1

h(n)=(sin c n)(n )

=(sin0.375n)

(n)

Koefisien filter yang dihitung melalui ekspresi sebelumnya terdaftar sebagai:

h(0)=0.375

=0.375

h(1)=sin(0.3751)(10.375)

=0.2942

menggunakan symetri mengarah ke :

h(1)=h (1)=0.2942

sehingga menunda h(n) oleh M = 1 sampel menggunakan persamaan 7.2 yg diberikan :

b(0)h(01)=h (1)=0.2942b(1)=b (11)=h (0)=0.375b(2)=b(21)=h(1)=0.2942

fungsi jendela rectangular :

w rec(n)=1,MnMw rec(0)=1w rec(1)=1


3/28

respone impluse dengan jendela dihitung sebagai :hw (0)=h(0)wrec(0)=0.3751=0.375hw (1)=h(1)wrec (1)=0.29421=0.2942hw (1)=hw(1)=0.2942

dengan demikian menunda hw(n) oleh M = 1 diperoleh sampel :

b(0)=b(2)=0.2942,b(1)=0.375fungsi alih yang diperoleh sebagai berikut :

H(Z)=0.2942+ 0.375z1+ 0.2942z2

Y(Z)X(Z)

=H(Z)=0.2942+ 0.375z1+ 0.2942 z2

Y(Z)=0.2942X(Z)+ 0.375z1X(Z)+ 0.2942z2 X(Z)

Menerapkan z-transform balik di kedua sisi, persamaan perbedaan dihasilkan sebagai

y (n)=0.2942x (n)+ 0.375x (n)+ 0.2942x (n2)

magnitud frekuensi respone :

H(e j)=0.2942+ 0.375 e(j)+ 0.2942 e(j2)

H e( j)=e(j)[0.2942e(j)+ 0.375+ 0.2942 e(j)]

H e( j)=e(j)(0.375+ 0.5584 cos())

maka magnitud respon dan frekuensi respon diperoleh :

| H e( j)

| = | e(j)(0.375+ 0.5584cos()) |

and sudut

H e( j)=, if0.375+ 0.5884cos, n> 0

H e( j)=+ , if0.375+ 0.5884cos, n> 0

b. jendel hamming :

whm (0)=0.54+ 0.46cos (01

)=1.0

whm (1)=0.54+ 0.46cos(1

1)=0.08

wham (-1) = wham(1) = 0.54


4/28

The windowed impulse response is calculated as

hw (0)=h(0)wham(0)=0.951=0.95hw (1)=h(1)wham(1)=0.29420.08=0.0235hw (1)=hw(1)=0.0235

transfer function is achieved as:

H(z)=0.0235+ 0.95Z1+ 0.0235z2

c. magnitud frekuensi dan respone frekuensi :

H(e j)=0.0235+ 0.95 e(j)+ 0.0235 e(j2)

H e( j)=e(j)[0.0235 e(j)+ 0.375+ 0.0235 e(j)]

H e( j)=e(j)(0.375+ 0.047cos())


| H e( j)

| = | e(j)(0.375+ 0.047cos()) |

and sudut

H e( j)=, if0.375+ 0.047cos, n> 0

H e( j)=+ , if0.375+ 0.047cos, n< 0


5/28

7.2 Desain a 3-tap FIR highpass filter with a cutoff frequency of 1600 Hz and a sampling rate of 8000

Hz using :

a. rectangular window function.b. Hamming window function.

Determine the transfer function and difference equation of the designed FIR system, andcompute and plot the magnitude frequency response for =0, /4, /2, 3/4, and radians.

requency cutoff ternormalisasi :

c = 2 fc Ts =2x 1600

8000=0,4 radian

2M + 1 = 3 M = 1

h(0) =c

=0.4

=0,6

h(1) =sin c n

n=sin 0,4

=sin 72

3,14=0,303

Applying Hamming :

ham(0) = 0,540,46cos nM

=0,540,46cos0=1

ham(1) = 0,540,46cos n

M

=0,540,46cos=0,08

simetry ham(1) = ham(-1) = 0,08

the windowed impuls respons is calculated as :

h(0) = h(0) ham(0) = 0,6 x 1 = 0,6

h(1) = h(1) ham(1) = - 0,303 x 0,08 = - 0,02424

h(-1) = h(-1) ham(-1) = - 0,303 x 0,08 = - 0,02424

delay

b0 = h(n M) = h(0 -1) = h(-1)b1 = h(n M) = h(1 -1) = h(0)

b2 = h(n M) = h(2 -1) = h(1)

jadi h(1) = h(-1)

H(z) = - 0,02424 + 0,6 z-1 0,02424 z-2


6/28

YzXz

=Hz

Y(z) = -0,02424 X(z) + 0,6 z-1 X(z) 0,02424 z-2 X(z)

y(n) = - 0,02424x(n) + 0,6 x(n -1) 0,02424x(n -2)

H(z) = - 0,02424 + 0,6 z-1 0,02424 z-2

H(ej) = - 0,02424 + 0,6 e-j 0,02424 e-j2

= e-j (-0,02424 ej + 0,6 0,02424e-j )= e-j (0,6 0,4848 cos )

|H(ej)| = |0,6 0,04848 cos |

0

- + if 0,6 0,04848 cos < 0

f 0,6 0,04848 cos |H(ej)| |H(ej)| dB


7/28

L=2 fL T = 21600

8000= 0,4 rad

L=2 fHT = 21800

8000 = 0,45 rad

2M1=5

h n=. HL

n=0

sin Hn

n-

sin L nn

n=0 - n2

When , n=0 --> h(0) =HL

=

0,450,4

= 0,05

h(1) =sin 0,451

1-

sin 0,411

= 0.1166

h(2) =sin 0,452

2-

sin 0,422

= -0,43789

h(-1)= 0,1166

h(-2) = -0,43789

sec= rec n=1 , MnM

ham= n=0,546046cos nM

, MnM

a rec 0=1


8/28

rec 1=1 rec 2=1 rec 1=1 rec 2=1

h0 =h 0 rec 0=0,051=0,01

h1=h 1=0,11661=0,1166

h2 =h2 =0,437891=0,43789

b0=b

4=0,43789, b

1=b

3=0,1166danb

2=0,05

Hz=0,437890,1166z10,05770,1166z30,43789z4

b. ham0=0,540,46cos

2=1

ham1 =0,540,46cos

12

=0,54

ham2=0,540,46cos22

=0,08

ham1 =ham1 =0,54

ham2=ham2=0,08

h

0

=h0

.ham

0

=0,051=0,01

h1=h1=0,11660,54=0,06296

h2 =h2 =0,437890,08=0,03503

HZ=0,035030,06296z10,05z20,06296z

30,03503z4


9/28

Frekuensi cutoff yang ternormalisasi :

L=2 fLT=21600

8000=0.4

H=2 fH T=21800

8000=0.45

Since 2M + 1 = 5 in this case, using the equation in Table 7.1 yields

when n = 0, we have

h(0)= H+ L

=0.45+ 0.4=0.95

The other computed filter coefficients via the previous expression are

listed as

h(1)=sin(0.41)

1

sin(0.451)1

=0.302880.31455=0.01167

h(2)=sin(0.42)

2

sin(0.452)2

=0.0935960.04921=0.0444

Using the symmetry leads to


10/28

h(-1) = h(1) = -0.01167h(-2) = h(2) = 0.0444

dikarenakan penjendelaan rectangular memiliki fungsi

w rec(n)=1,MnMw rec(0)=1w rec(1)=1w rec(2)=1w rec(3)=1

respone impluse dengan jendela dihitung sebagai :hw (0)=h(0)wrec(0)=0.951=0.95hw (1)=h(1)wrec (1)=0.011671=0.01167hw (2)=h(2)wrec (2)=0.04441=0.0444h(1)=h(1)=0.01167h(2)=h(2)=0.0444

thus, delaying h(n) by M = 2 sample gives :b0=b4=0.0444, b1=b3=0.01167b2=0.95

the transfer function is achieved as

H(z)=0.04440.01167z1+ 0.95z20.01167z3+ 0.0444z4

The magnitude frequency response and phase response can be obtained

using the technique introduced in Chapter 6. Substituting z=e j intoH(z), it follows that

H(e j)=0.04440.01167(ej)+ 0.95 (e j2)0.01167(ej3)+ 0.0444(ej4)

b. dengan menggunakan penjendelaan hamming

whm (0)=0.54+ 0.46cos(0

2)=1

whm (1)=0.54+ 0.46cos(12

)=0.54

whm (1)=0.54+ 0.46cos( 22 )=0.08

whm (-1) = whm (1) = 0.54whm (-2) = whm (2) = 0.08

The windowed impulse response is calculated as

hw (0)=h(0)wham(0)=0.951=0.95hw (1)=h(1)wham(1)=0.011670.54=0.006


11/28

hw (2)=h(2)wham (2)=0.04440.08=0.0035hw(-1) = hw(1) = -0.006hw(-2) = hw(2) = 0.0035

thus, delying hw(n) by M = 2 sample gives

bo = b4 = 0.006 , b1=b3 0.0035 b2 = 0.95The transfer function is achieved as

H(z)=0.006+ 0.0035z1+ 0.95z2+ 0.0035z30.006z4

dengan demikian menunda hw(n) oleh M = 1 diperoleh sampel :

b(0)=b(2)=0.2942,b(1)=0.375fungsi alih yang diperoleh sebagai berikut :

H(Z)=0.2942+ 0.375z1+ 0.2942z2

Y(Z)X(Z)

=H(Z)=0.2942+ 0.375z1+ 0.2942 z2

Y(Z)=0.2942X(Z)+ 0.375z1X(Z)+ 0.2942z2 X(Z)

Menerapkan z-transform balik di kedua sisi, persamaan perbedaan dihasilkan sebagaiy (n)=0.2942x (n)+ 0.375x (n)+ 0.2942x (n2)magnitud frekuensi respone :

H(e j)=0.2942+ 0.375 e(j)+ 0.2942 e(j2)

H e( j )=e(j)[0.2942e(j)+ 0.375+ 0.2942 e(j)]

H e( j)=e(j)(0.375+ 0.5584 cos())


| H e( j)

| = | e(j)(0.375+ 0.5584cos()) |

and sudut

H e( j)=, if0.375+ 0.5884cos, n> 0

H e( j)=+ , if0.375+ 0.5884cos, n> 0


12/28

7.5 Given an FIR lowpass filter design with the following spesifications

Passband = 0 800 Hz

Stopband = 1200 4000 Hz

Passband ripple = 0.1 dBStopband attenuation = 40 dB

Sampling rate = 8000 Hz

determine the following :a. window method

b. length of the FIT filter

c. cutoff frequency for the design equation.

a. stopband atenuation = 40 dB jadi menggunakan metode Hanning

passband ripple 0,0546

b.

f =1200800

8000=

400

8000=0,05

N =3,1

f=

3,1

0,05=62

c.

fc=fpassfstop

2

=8001200

2

=1000Hz


13/28

FIR High pass filter

Pass band = 0-1500 HzStep = 2000-4000 Hz

Pass Band ripple = 0,02 dB hamming : pass band ripple= 0,0194 & topband attenuation= 53 dB

stopband attenuation = 60 dB

sampling rate = 8000 Hz

f = |2000-1500|/8000=0,0625

N = 3,3/Af

= 3,3/0,625

= 52,8pilih angka yang ganjil, maka N=53

cutt off frequency= (1500+2000)/2 = 1750 Hz

c=17502

8000=0,4375 rad


14/28

7.7 Diberikan design FIR bandpass filter dengan spesifikasi sebagai berikut:

frekuensi cutoff bawah = 1500 Hz

Lebar transisi bawah = 600 Hz

Frekuensi cutoff atas = 2300Hz

Lebar transisi atas = 600 HzPassband ripple = 0,1 dB

Attenuasi Stopband = 50 dB

Kecepatan laju sampling = 8000 Hz

Tentukan:

a. Metode window

b. Panjang dari FIR filter

c. Frekuensi cutoff

a) Berdasarkan pada spesifikasi diatas, jendela Hamming adalah metode jendela yang tepat untuk

permasalahan ini, karena memiliki passband ripple 0,0194 dB dan attenuasi stopband sebesar 53 dB

b) Menghitung lebar transisi normal:

f1=600

8000=0.075 dan f2=

600

8000=0.075

dan panjang filternya yang ditentukan menggunakan jendela Hamming adalah:

N1=3,3

f=3,3

0,075=44

N2=3,3

f=

3,3

0,075=44

kita pilih agka ganjil untuk nilai N, maka N = 45.

c) Frekuensi cutoff lower dan upper dapat dihitung:

L=21500

8000=0,375 radian

H=223008000=0,575 radian

N= 2M+1= 45


15/28

7.8 Given an FIR bandstop filter design with the following spesifications :

Lower passband n= 0 120 Hz

Stopband = 1600 2000 Hz

Upper passband = 2400 4000 HzPassband ripple = 0,05 dB

Stopband attenuation = 60 dB

Sanpling rate = 8000 Hzdetermine the following :

a. window method

b. length of the FIR filterc. cutoff frequency for the design equation

answer:

a.Stopband attenuation 60 dB blackman passband ripple = 0,0017

b. f1=16001200

8000=0,05 f2=

240020008000

=0,05

N=5,5

f=

5,5

0,05=110

c. fclow=fpassfstop

2=

120016002

=1400 Hz

frequency cutoff ternormalisasi

L =1400 x 2

2=0,35 radians

fch=fpass fstop

2=

400016002

=5600

2=2800Hz

frequency cutoff ternomalisasi

H =2800x 2

8000=0,7 radians


16/28

a. y(n)=0.25x(n)-0.5x(n-1)+0.25x(n-2)

b. y n=0,25x n0,5x n10,2x n20,5x n30,2x n4


17/28

8.1 Given an analog filter with the transfer function

H(s) =1000

s1000,

convert it to the digital filter transfer function and difference equation using theBLT if the DSP system has a sampling period of T = 0.001 second.

Answer :

Menggunakan BLT

H(z) = H(s) | s =2z1Tz1

=1000

s1000| s =

2z1Tz1

T = 0.001 s

H(z) =

1000

2z1Tz1

1000=

1000

2z10.001z1

1000=

1000

2000z1z1

1000

=

0.5

z1T10.5

=0.5z1

z10.5z1=

0.5z0.51.5z0.5

jadi :

H(z) =0.5z0.5/1.5z 1.5z0.5/1.5z

=0.330.33z1

10.33z1

maka :y(n) = 0.33x(n) + 0.33 x(n-1) + 0.33 y(n-1)

8.2 Hps=

1

s1 dengan frekuensi cut off = 30 Hz dan sampling rate of 200 Hz

a.) first we obtain the digital frekuensi as

wd=2 f=230=60 rad/sec T=1

fs=

1

200=0,005sec

wa=2

Ttan

wd T

2=

2

0,005tan

60 x 0,0052

=400tan 27=203,81 rad/s


18/28

lalu perform the prototype transformation

Hs=Hps =wa

swa=

203,81

s203,81

apply the BLT

Hs=Hps =wa

swa=

203,81

s203,81dengan s=

2z1Tz1

Hz=203,81

400z1z1

203,81=

203,81

400

z1z1

203,81

400

=0,509

z1z1

0,509

lalu kalikan dengan (z+1)

0,509z1

z10,509z1=

0,509z0,509z10,509z0,509

=0,509z0,5091,509z0,491

terakhir bagi pembilang dan penyebut dengan 1,509z

Hz=0,3370,377 z1

10,325 z1

b) program matlab


19/28

% Plot the magnitude and phase responsesfs = 200;% Sampling rate (Hz)[B, A] = lp2lp([1],[1 1],203.81);[b, a] = bilinear(B, A, fs)%b = [0.337 0.337] numerator coefficients of the digital filter from MATLAB%a = [1 -0.325]denominator coefficients of the digital filter from MATLAB[hz, f] = freqz([0.337 0.337],[1-0.325],512,fs);%the frequency responsephi = 180*unwrap(angle(hz))/pi;

subplot(2,1,1), plot(f, abs(hz)),grid;axis([0 fs/2 0 1]);xlabel('Frequency (Hz)'); ylabel('Magnitude Response')subplot(2,1,2), plot(f, phi); grid;axis([0 fs/2 100 0]);xlabel('Frequency (Hz)'); ylabel('Phase (degrees)')

8.3. The normalized lowpass filter with a cutoff frequency of 1 rad/sec is given as

Hps=1

s1a. Use Hp(s) and the BLT to obtain a corresponding IIR digital highpass filter with a cutofffrequency of 30 Hz, assuming a sampling rate of 200 Hz.b. Use MATLAB to plot the magnitude and phase frequency responses of H(z).

Answer :a. First, we obtain the digital frequency as d=2 f=230=60 rad/sec ;T=1/ fs=1/200sec

We then follow the design procedure:1. First calculate the prewarped analog frequency as

a= 2Ttan dT

2= 2

1/200tan 60/200

2

that is, a=400x tan 27o=400x 0,51=203,81 rad/sec

2. Then perform the prototype transformation (lowpass to lowpass) as follows:

HS=Hpss=

1

a

=1

as1

=s

as

which yields an analog filter:


20/28

Hs=s

203,81s3. Apply the BLT, which yields

Hz=[ s203,81s ]s=2z1Tz1

We simplify the algebra by dividing both the numerator and the denominator by

180:

Hz=180 x

z1z1

203,81180xz1z1

Hz=

z1z1

203,81

180

z1z1

Hz=

z1z1

1,13z1z1

Then we multiply both numerator and denominator by (z + 1) to obtain

Hz= z1z1 x z1

1,13z1z1 x z1Hz=

z11,13z1z1

Hz=z1

1,13z1,13z1

Hz=z1

0,13z0,13Finally, we divide both numerator and denominator by 0.13z to getnthe transferfunction in the standard format:

Hz=z1/0,13z

0,13z0,13/0,13z

Hz=7.697.69z1

1z1

b. The corresponding MATLAB design is listed in Program 8.2. Figure 8.12 shows themagnitude and phase frequency responses.

%Example 8.6% Plot the magnitude and phase responsesfs = 200;% Sampling rate (Hz)[B, A] =lp2lp([1],[1 1],203.81);[b, a] =bilinear(B, A, fs)%b = [0.3660 0.3660] numerator coefficients of the digital filter from MATLAB%a =[1 _0:2679]denominator coefficients of the digital filter from MATLAB


21/28

[hz, f] = freqz([7.69 7.69],[1-0.13],512,fs);%the frequency responsephi =180*unwrap(angle(hz))/pi;subplot(2,1,1), plot(f, abs(hz)),grid;axis([0 fs/2 0 50]);xlabel('Frequency (Hz)'); ylabel('Magnitude Response')subplot(2,1,2), plot(f, phi); grid;axis([0 fs/2 -100 0]);xlabel('Frequency (Hz)'); ylabel('Phase (degrees)')

8.4 Consider the normalized lowpass filter with a cutoff frequency of 1 rad/ sec:

Hp(s) =1

s1

a. Use Hp(s) and the BLT to design a corresponding IIR digital notch (bandstop) filter

with a lower cutoff frequency of 20 Hz, an upper cutoff frequency of 40 Hz, and a

sampling rate of 120 Hz.

b. Use MATLAB to plot the magnitude and phase frequency

responses of H(z).

Answer :

A. wh = 2 fh = 2 (40) = 80 rad/secwl = 2 fl = 2 (20) = 40 rad/sec

T = 1/fs = 1/120 sec

wah = 2/T tanwh T

2= 240 x tan

80/1202

= 4.156 x 102 rad/sec


22/28

wal = 240 tanwl T

2= 240 x tan

40/1202

= 1.385 x 102 rad/sec

W = wah wal = 415.6 138.5 = 277.03 rad/sec

wo2 = wah x wal = 5.75606 x 104

Hp(s) =1

s1

H(s) = Hp(s)|s

2wo2

sW=

Ws

s2wsw02

=277.03s

s2277,03s5.75606x104

H(z) =277.03s

s2277,03s5.75606x104

| s = 240 (z-1)/(z+1)

B. dengan menggunakan MATLAB

% Design of the digital bandpass Butterworth filterformat longfs = 120;[B A] = lp2bp([1],[1 1],sqrt(5.75606*10^4),277.03)% Complete step 2[b a] = bilinear(B,A,fs) % Complete step 3% Plot the magnitude and phase responses%b = [0:0730 - 0:0730]; numerator coefficients from MATLAB%a = [1 0:7117 0:8541]; denominator coefficients from MATLABfreqz(b, a,512,fs);axis([0 fs/2-40 10])

Tampilan pada common

B =

1.0e+002 *

2.77030000000000 0.00000000000008

A =

1.0e+004 *

0.00010000000000 0.02770300000000 5.75606000000000

b =

0.36602259977825 0.00000000000000 -0.36602259977825


23/28

a =

1.00000000000000 -0.00043380652009 0.26795480044350

Grafik :

8.5 Hps=1

s1dengan frekuensi lowpass cut off = 15 Hz , upper cutoff = 25

Hz dan sampling rate of 200 Hz

a) untuk yg lowpass = 15 Hz


fs=

1

120=0,0083sec

wa=2

Ttan

wd T

2=

2

0,0083tan

30x 0,00832

=240,96tan 22,41=99,36 rad/s


Hs=Hps =wa

swa=

99,36

s99,36


24/28

apply the BLT

Hs=Hps =wa

swa=

99,36

s99,36dengan s=

2z1Tz1

Hz= 99,36240,96z1z1

99,36=

99,36

240,96z1z1

99,96

240,96

= 0,412z1z1

0,412


0,412z1z10,412z1

=0,412z0,412

z10,412z0,412=

0,412z0,4121,412z0,588


Hz=0,2920,292z1

10,42z1

untuk yang upper cutoff = 25 Hz


fs=

1

120=0,0083sec

wa=2

Ttan

wd T

2=

2

0,0083tan

50x 0,00832

=240,96tan 37,35=183,89 rad/s


Hs=Hps =wa

swa=

183,89

s183,89

apply the BLT

Hs=Hps =wa

swa=

183,89

s183,89dengan s=

2z1Tz1

Hz=183,89

240,96z1z1

183,89=

183,89

240,96

z1z1

183,89

240,96

=0,763

z1z1

0,763



25/28

0,763z1z10,763z1

=0,763z0,763

z10,763z0,763=

0,763z0,7631,763z0,237


Hz=0,4330,433z1

10,134z1

b) program matlab untuk frekuensi cutt off = 15 Hz

% Plot the magnitude and phase responsesfs = 120;% Sampling rate (Hz)[B, A] = lp2lp([1],[1 1],99.36);[b, a] = bilinear(B, A, fs)%b = [0.292 0.292] numerator coefficients of the digital filter from MATLAB%a = [1 -0.42]denominator coefficients of the digital filter from MATLAB[hz, f] = freqz([0.292 0.292],[1-0.42],512,fs);%the frequency responsephi = 180*unwrap(angle(hz))/pi;subplot(2,1,1), plot(f, abs(hz)),grid;axis([0 fs/2 0 1]);xlabel('Frequency (Hz)'); ylabel('Magnitude Response')

subplot(2,1,2), plot(f, phi); grid;axis([0 fs/2 100 0]);xlabel('Frequency (Hz)'); ylabel('Phase (degrees)')

untuk yang frekuensi cuttoff = 25 Hz


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% Plot the magnitude and phase responsesfs = 120;% Sampling rate (Hz)[B, A] = lp2lp([1],[1 1],183.89);[b, a] = bilinear(B, A, fs)%b = [0.433 0.433] numerator coefficients of the digital filter from MATLAB%a = [1 -0.134]denominator coefficients of the digital filter from MATLAB[hz, f] = freqz([0.433 0.433],[1-0.134],512,fs);%the frequency responsephi = 180*unwrap(angle(hz))/pi;subplot(2,1,1), plot(f, abs(hz)),grid;axis([0 fs/2 0 1]);xlabel('Frequency (Hz)'); ylabel('Magnitude Response')subplot(2,1,2), plot(f, phi); grid;axis([0 fs/2 100 0]);

xlabel('Frequency (Hz)'); ylabel('Phase (degrees)')

8.6. Design a first-order digital lowpass Butterworth filter with a cutoff frequency of 1.5kHz and a passband ripple of 3 dB at a sampling frequency of 8,000 Hz.a. Determine the transfer function and difference equation.b. Use MATLAB to plot the magnitude and phase frequency responses.

Answera.

d=21500=3000 rad/sec dan T= 1fs= 1

8000sec

1. a 2/Ttan d T2=1600tan

3000/80002

= 1,069104 rad/sec

2. Hp(s)=1

s1


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H(s)=Hp(s)|as

=

1

as1

=s

s1.069x104 rad/sec

3. dengan BLT :

H(z) =s

s1.069x104

rad/sec

|s=16000(z-1)/(z+1)

H(z)=

16000z1z1

16000z1z1

1.069x104

z1z1

=16000z1

16000z11.069x104z1

= 16000 z116000 z11.069x10 4z1

=16000z16000

16000z1600010690z10690

=16000z1600026690z5310

=0.559z0.599

z0.1989

pembilang dan penyebut dibagi dengan z:

H(z)=0.5990.599z1

10.1989z1

b. program MATLAB:

%Soal 8.6

% Design of the digital lowpass Butterworth filterformat longfs = 8000; % Sampling rate[B A] = lp2lp([1],[0 1 1], 1.0690*10 ^ 4) % Complete step 2

[b a] = bilinear(B,A,fs)% Complete step 3% Plot the magnitude and phase responses%b = [0.7157 1.4315 0.7157]; numerator coefficients from MATLAB%a = [1 1.3490 0.5140]; denominator coefficients from MATLABfreqz(b,a,512,fs);axis([0 fs/2 -40 10])

Setelah di-run:


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B =

10690

A =

1 10690

b =

0.40052454102660 0.40052454102660

a =

1.00000000000000 -0.19895091794680

Pengolahan Sinyal Digital B 2011 2012 12 (Bab 7 & 8)

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Transcript of Pengolahan Sinyal Digital B 2011 2012 12 (Bab 7 & 8)