ĐỀ TÀI THIẾT KẾ BỘ LỌC FIR THÔNG CAO BẰNG PHƯƠNG PHÁP LẤY MẪU TẦN SỐ

7/30/2019 TI THIT K B LC FIR THNG CAO BNG PHNG PHP LY MU TN S

1/22

I HC QUY NHN--- oOo ---

BI TP LNMN HC: X L S TN HIU

TI: THIT K B LC FIR THNG CAOBNG PHNG PHP LY MU TN S

Ngi hng dn :

Sinh vin thc hin :

Lp :

Quy Nhn , thng 5/2011


2/22

Thit k lc FIR thng cao bng phng php ly mu tn s Trang 2

LI NI U

Vi xu hng s ha cc h thng thng tin hin nay,vic x l tn hiu s ngycng tr nn quan trng vi kh nng x l thng tin mt cc u vit.

c th tip cn c lnh vc ny, chng ta cn c nhng kin thc c bn vtn hiu s v cc phng php x l. Mt trong nhng kin thc quan trng lthit k b lc s- h thng c th lm thay i tn hiu ph hp vi mc ch cacon ngi.

Trong x l s tn hiu, tn ti nhiu b lc s khc nhau nh: b lc thng thp,b lc thng di, b lc vi phn, thit k cc b lc thch hp, trc ht phi xcnh yu cu thc t da trn cc ch tiu k thut cho trc, trn c s nh hnhcu trc b lc v phng php thit k ti u. Cu trc b lc c th l: cu trc FIR(b lc s c p ng xung chiu di xc nh) hoc cu trc IIR (b lc s c png xung chiu di khng xc nh). Phng php thit k c th l: phng php cas, phng php ly mu tn s, hoc phng php xp x ti u,

c s phn cng ca thy gio , trn c s nhng kin thc hc, ti tmhiu b lc FIR theo phng php ly mu tn s.

Ti xin chn thnh cm n thy gio, bn b cng lp tn tnh hng dn ti c th hon thnh ti ny. Chc chn ti s khng trnh khi nhng thiu strt mong c s gp ca qu thy c v cc bn.

Xin chn thnh cm n!

Quy nhn, thng 5 nm 2011Ngi thc hin


3/22


MC LC

LI NI U.............................................................................................................2

MC LC...................................................................................................................3 Phn 1. C S L THUYT ......................................................................................4 1.1. Dn nhp...........................................................................................................4 1.2. Cu trc ca b lc FIR.....................................................................................6

a. Cu trc dng trc tip .....................................................................................6 b. Cu trc dng ghp tng:..................................................................................7 c. Cu trc dng pha tuyn tnh: ...........................................................................7

1.3. Cc c tnh ca b lc FIR pha tuyn tnh .......................................................8 a. p ng xung h(n): ..........................................................................................9 b. p ng tn s H(ej):....................................................................................11

1.4 Phng php thit k ly mu tn s : ..............................................................14

a. Phng php thit k n gin .......................................................................15b. Phng php thit k ti u............................................................................15 Phn 2. THIT K LC FIR THNG CAO .............................................................16

2.1. Bi ton thit k ..............................................................................................16 2.2. Phng php thit k.......................................................................................16 2.3. Thut ton v chng trnh Matlab..................................................................17

a. Lu thut ton: .......................................................................................... 17b) Chng trnh..................................................................................................19 c) Kt qu ..........................................................................................................21

TI LIU THAM KHO..........................................................................................22


4/22


5/22


Trong :

Band [0, wp]c gi l di thng, v 1 l dung sai (gn sng) c chpnhn trong p ng di thng l tng.

Band [ws, ]c gi l di chn, v 2 l dung sai di chn. Band [wp, ws]c gi l di chuyn tip, v khng c rng buc no v p

ng bin trong di ny

Cc ch tiu tng i gm c:

Rp: gn sng trong di thng tnh theo dB. As : Suy hao trong di chn tnh theo dB.

Quan h gia cc ch tiu ny nh sau:

011 11

10log20

pR (0) (1.1)

011

210log20

sA (>>1) (1.2)

Cc ch tiu trn c a ra i vi b lc FIR thng thp, v tt nhin i vi cc

b lc khc nh thng cao HPF (High Pass Filter), thng di BPF (Band Pass Filter)

u c th c nh ngha tng t. Tuy nhin, cc tham s thit k quan trng nht

l cc dung sai di tn v cc tn s cnh-di (tolerance or ripples and band-edge

frequencies). Bi vy, trong phn 1 v c s l thuyt ny chng ta ch tp trung vo

b lc FIR thng thp. Vic thit k c th cho b lc FIR thng di bng k thut ca

s s c pht trin trn c slc thng thp v s c m t chi tit trong phn 2.

Vic thit k v thc hin lc FIR c nhng thun li sau y:

p ng pha l tuyn tnh. D thit k do khng gp cc vn n nh (lc FIR lun n nh). Vic thc hin rt hiu qu. C th s dng DFT thc hin

p ng pha l tuyn tnh (linear phase response) mang li nhng thun li sau:

Bi ton thit k ch gm cc php tnh s hc thc ch khng cn php tnhs hc phc.

B lc pha tuyn tnh khng c mo tr nhm v ch b tr mt khong khngi.


6/22


i vi b lc c chiu di M (hoc bc M-1) s php ton c bc M/2 nh kho st trong thc hin b lc c pha tuyn tnh.

1.2. Cu trc ca b lc FIR

Mt b lc p ng xung hu hn vi hm h thng c dng:

1M

0n

nn

M11M

110 zbzbzbb)z(H (1.3)

Nh vy p ng xung h(n) l:

else

Mnbnh

n

0

10)( (1.4)

V phng trnh sai phn l:

)1Mn(xb)1n(xb)n(xb)n(y 1M10 (1.5)y chnh l tch chp tuyn tnh ca cc dy hu hn.

Bc ca b lc l M-1, trong khi chiu di ca b lc l M (bng vi s lng cc

h s). Cc cu trc b lc FIR lun lun n nh, v tng i n gin hn so vi

cc cu trc b lc IIR. Hn th na, cc b lc FIR c th c thit k c mt

p ng pha tuyn tnh v l iu cn thit trong mt s ng dng.

Chng ta s xem xt ln lt cc cu trc ca b lc FIR sau y:

a. Cu trc dng trc tipPhng trnh sai phn c thc hin bi mt dy lin tip cc b tr do khng c

ng phn hi:

)1Mn(xb)1n(xb)n(xb)n(y 1M10 (1.6)

Do mu thc bng n v nn ta ch c mt cu trc dng trc tip duy nht. Cu

trc dng trc tip c cho trong hnh 1.2 vi M = 5:

b0z-1 b1

z-1 b2z-1 b3

z-1 b4y(n)

x(n)

Hnh 1.2 Cu trc lc FIR dng trc tip


7/22


b. Cu trc dng ghp tng:Hm h thng H(z) c bin i thnh cc tch ca cc khu bc 2 vi cc h s

thc. Cc khu ny c thc hin dng trc tip v b lc tng th c dng ghp

tng ca cc khu bc 2.

M1

0

1M1

0

10

M11M

110 z

bbz

bb1bzbzbb)z(H

K

1k

22,k

11,k0 )zBzB1(b (1.7)

trong

2M

K , Bk,1 v Bk,2 l cc s thc i din cho cc h s ca cc khu bc

2. Cu trc dng ghp tng c cho trong hnh 1.3 vi M = 7:

c. Cu trc dng pha tuyn tnh:i vi cc b lc chn tn, ngi ta mong mun c p ng pha l hm tuyn tnh

theo tn s, ngha l:

)e(H j (1.8)

trong 0 hoc2

v l mt hng s.

i vi b lc FIR nhn qu c p ng xung trong khong [0, M-1], th cc iu

kin tuyn tnh l:

1Mn0,0);n1M(h)n(h (1.9)

1Mn0,2/);n1M(h)n(h (1.10)

Xt phng trnh sai phn c cho trong phng trnh (1.5) vi p ng xung i

xng trong phng trnh (1.9), ta c:

)1Mn(xb)2Mn(xb)1n(xb)n(xb)n(y 0110

)]2Mn(x)1n(x[b)]1Mn(x)n(x[b 10

B1,1z-1 z-1 z-1

y(n)x(n)

B2,1 B3,1

b0

B1,2z-1 z-1 z-1

B2,2 B3,2

Hnh 1.3 Cu trc lc FIR dng ghp tng


8/22


S khi thc hin phng trnh sai phn trn c m t trong hnh 1.4 di y

i vi c M l v M chn:

i vi M l: M = 7, cn i vi M chn: M = 6

R rng, vi cng mt bc ca b lc (cng M) cu trc pha tuyn tnh s tit kim

c 50% cc b nhn so vi cu trc dng trc tip.

1.3. Cc c tnh ca b lc FIR pha tuyn tnh

Trong phn ny chng ta s tho lun v hnh dng ca p ng xung, p ng tn

s trong hm h thng ca cc b lc FIR pha tuyn tnh.

Cho h(n), trong 0 n M 1, l p ng xung c chiu di M th hm truyn h

thng l:

1M

0n

n1M)1M(1M

0n

n z)n(hzz)n(h)z(H (1.11)

c (M-1) im cc gc (trivial poles) v M-1 im khng nm v tr bt k trn

mt phng z. p ng tn s l:

,e)n(h)e(H1M

0n

njj (1.12)

b0

z-1

b1

z-1

b2

x(n)

z-1 z-1

z-1

y(n)

b0

z-1

b1

z-1

b2 b3 y(n)

x(n)

z-1 z-1

z-1

z-1M=7

M=6

Hnh 1.4 Cu trc lc FIR pha tuyn tnh vi cc h s M chn v l


9/22


a. p ng xung h(n):Chng ta c th a ra rng buc pha tuyn tnh:

,)e(H j (1.13)

trong : l mt hng s tr pha. Ta bit rng h(n) phi i xng, ngha l:

21M,1Mn0),n1M(h)n(h (1.14)

Do h(n) l i xng theo , l ch s i xng. C hai kiu i xng:

M l: Trong trng hp ny,2

1M l mt s nguyn. p ng xung

c m t trong hnh 1.5 di y:

M chn: Trong trng hp ny,2

1M khng phi l mt s nguyn. p

ng xung c m t bng hnh 1.6 di y:

Hnh 1.5 p ng xung i xng, M l

Hnh 1.6 p ng xung i xng, M chn


10/22


Ta cng c b lc FIR pha tuyn tnh loi hai nu ta yu cu p ng pha jeH

tho mn iu kin:

)e(H j vi (1.15)

p ng pha l ng thng nhng khng i qua gc. Trong trng hp ny khng phi l hng s tr pha, nhng:

d)e(Hd j

(1.16)

l hng s, chnh l tr nhm ( l mt hng s tr nhm). Trong trng hp ny, cc

tn s c lm tr vi mt tc khng i. Nhng mt s tn s c th c l m tr

vi tc ln hn hoc nh hn.

i vi kiu pha tuyn tnh ny, c th thy rng:

1Mn0),n1M(h)n(h v2

,2

1M

(1.17)

iu ny c ngha rng p ng xung h(n) l phn i xng (antisymmetric). Ch

s i xng vn l2

1M . Mt ln na chng ta li c 2 kiu, cho M l v M chn.

M l: Trong trng hp ny,2

1M l mt s nguyn. p ng xung

c m t bng hnh 1.7 di y:

Lu rng mu h() ti2

1M phi bng 0, ngha l, 0

21M

h

.

Hnh 1.7 p ng xung phn i xng, M l


11/22


M chn: Trong trng hp ny,2

1M khng phi l mt s nguyn. p

ng xung c m t trong hnh 1.8.

b. p ng tn s H(ej):Khi t hp hai loi i xng v phn i xng vi M chn v M l, ta c bn kiu

lc FIR pha tuyn tnh. p ng tn s ca mi kiu c biu thc v hnh dng ring.

nghin cu cc p ng pha ca cc kiu ny, ta vit biu thc ca H(ej) nh sau:

21M

,2

;e)e(H)e(H )(jjrj

(1.18)

trong Hr(ej) l hm p ng ln ch khng phi l hm p ng bin . p

ng ln l mt hm thc, c th va dng va m, khng ging p ng bin

lun lun dng. p ng pha kt hp vi p ng bin l mt hm khng lin tc,

trong khi kt hp vi p ng ln l mt hm tuyn tnh lin tc.

B lc FIR pha tuyn tnh Loi-1 (Type 1): p ng xung i xng, M l

Trong trng hp ny 0 , 2

1M

l mt bin nguyn, v n1Mhnh ,1Mn0 , th ta c th chng t rng:

2/1Mj2/1M

0n

j encosna)e(H

(1.19)

trong :

Hnh 1.8 p ng xung phn i xng, M chn


12/22


2

1Mh0a vi mu chnh gia (1.20)

n2

1Mh2na vi

23M

n1

B lc FIR pha tuyn tnh Loi-2 (Type 2): p ng xung i xng, M chnTrong trng hp ny 0 , n1Mhnh , 1Mn0 , nhng

2

1M

khng phi l mt bin nguyn, th ta c th chng t rng:

2/1Mj2/M

1n

j e2

1ncosnb)e(H

(1.21)

trong :

n2M

h2nb vi2

M,...,2,1n (1.22)

So snh (1.21) v (1.18), ta c:

2/M

1nr

21

ncosnb)(H (1.23)

Lu : Ti , ta c 0

2

1ncosnb)(H

2/M

1n

r

m khng cn quan tm

n b(n) hoc h(n). Do chng ta khng th s dng loi ny (h(n) i xng, M

chn) i vi b lc thng cao hoc b lc chn di.

Lc FIR pha tuyn tnh Loi-3 (Type 3): p ng xung phn i xng, M lTrong trng hp ny ta c

2

,2

1M l mt bin nguyn,

n1Mhnh , 1Mn0 , v 021M

h

th ta c th chng t:

2

1M

2j2/1M

0n

j ensinnc)e(H (1.24)

trong

n2

1Mh2nc vi

2M

,...,2,1n (1.25)


13/22


So snh (1.24) v (1.18), ta c:

2/1M

0nr nsinncH (1.26)

Lu: Ti 0 v , ta c 0H r m khng cn quan tm c(n) hoc h(n).

Hn th na, je 2j

, iu c ngha l rjH l thun o. Do , loi b lc ny

khng thch hp i vi vic thit k b lc thng thp hoc thng cao. Tuy nhin,

iu ny thch hp i vi vic xp x cc b vi phn v b bin i Hilbert s l

tng.

Lc FIR pha tuyn tnh Loi-4(Type 4):p ng xung phn i xng, M chnTrong trng hp ny 2

, n1Mhnh , 10 Mn , nhng 21M

khng phi l mt bin nguyn, th ta c th chng t rng:

21M

2j2/M

1n

j e21

nsinnd)e(H (1.27)

trong :

n2M

h2nd vi2M

,...,2,1n (1.28)

So snh (1.27) v (1.18), ta c:

2/M

1nr

21

nsinnd)(H (1.29)

Lu : Ti , 0)0(H r v je 2j

. Do vy, loi ny cng thch hp cho vic

thit k cc b vi phn s v b bin i Hilbert s.

Bng sau y m t kh nng thch hp trong vic thit k cc b lc v cc b bin

i Hilbert s, b vi phn s ca 4 loi lc FIR pha tuyn tnh nu:

Type LPF HPF BPF SBF Hilbert Differentiator

FIR Type 1 FIR Type 2 FIR Type 3 FIR Type 4


14/22


1.4 Phng php thit k ly mu tn s :

Theo phng php ly mu tn s, p ng tn s yu cu Hd(ejw) trc tin c

ly mu u M im cch u nhau gia 0 v 2pi :

H(k)= Hd(ej2k/M) k=0,1,........M-1

Cc mu tn s ny to thnh DFT M im m bin i nghch l b lc FIR c bc

M-1:

1

0

/2)(1

)(M

k

MnkjekHM

nh 10 Mn

p ng thc t l ni suy ca cc mu c cho bi:

1

0/21

1

0 1

)(1)()(

M

kMnkj

MM

n

n

ez

kH

M

zznhzH

p ng pha i vi kiu 1 v 2 :

p ng pha i vi kiu 3 v 4

Hnh 1.9 m t k thut ly mu tn s :

T hnh trn ta nhn thy :

Li xp xl hiu ca p ng l tng v p ng thc t bng khng ti cc tn s

c ly mu.

0 1 2 3 4 5 6 7

1 .

0 1 2 3 4 5 6 7 8

.1 . .

. . ..

. . .

1,,1

21

,)(2

21

21,,0,2

21

)(M

Mk

M

kMM

Mk

M

kM

kH

1,,12

1,

)(22

12

21

,,0,2

21

2)(M

Mk

M

kMM

Mk

M

kM

kH


15/22


Li xp x tt c cc tn s khc ph thuc vo hnh dng ca p ng tn s l

tng; ngha l, p ng tn s l tng cng sc ntth li xp x cng ln.

Li cng ln khi gn cnh di v cng b khi bn trong di.

- C hai cch tip cn thit k :

a. Phng php thit k n gin : s dng tng c bn v khng a ra mt

rng buc no v li xp x, ngha l chp nhn li sinh ra do thit k.

Trong phng php ny ta t H(k)= Hd(ej2k/M) k=0,1,........M-1 v s dng cc

cng thc thu c p ng xung h(n). Phng php ny t c s dng trong

thc t.

b. Phng php thit k ti u : c gng ti thiu ha li trong di chn bng cch

thay i cc gi tr ca mu trong di chuyn tip.

Trong phng php ny, chng ta phi tng M to ra cc mu t do trong dichuyn tip ngha l chng ta thay i cc gi tr ca chng thu c h s suy

gim ln nht i vi M v rng di chuyn tip cho. y l mt bi ton ti u

ha v c gii quyt bng k thut quy hoch tuyn tnh.

Trong thc t rng di chuyn tip ni chung kh b, ch cha c mt hoc hai

mu. Do chng ta cn ti u ha tt nht hai mu thu c h s suy gim di

chn ln nht.


16/22


Phn 2. THIT K B LC FIRTHNG CAO

2.1. Bi ton thit k

Hy thit k b lc FIR thng cao pha tuyn tnh theo phng php ly mu tn s,

vi cc ch tiu b lc cn thit k c cho nh sau:

Cnh di chn: ws Cnh di thng: wp gn sng trong di thng: Rp Suy hao trong di chn: As

Cc i lng ny c th c m t trn hnh 2.1 nh sau:

iu kin: ws< wp


17/22


trong di chuyn tip c thc hin lp xc inh c b lc c Rp v As l tt

nht.

Bc 3. Tm p ng xung ca b lc thng cao cn thit k

p ng xung ca b lc thng cao c th tm c bng php bin i DFT

ngc cc mu hd(n) :

1

0

/2)(1

)(M

k

MnkjekHM

nh 10 Mn

2.3. Thut ton v chng trnh Matlab

Trong phn ny s thc hin chng trnh thit k b lc thng cao bng phng

php ly mu tn s. Chng trnh s nhn cc ch tiu yu cu ca b lc cn thit k,

sau thc hin cc bc thit k tm c p ng xung h(n).

kho st b lc va thit k, chng trnh cng s thc hin tnh ton v v png bin - tn s ca b lc theo dB, cng nh v cc p ng xung l tng hd(n),

hm ca s w(n) v p ng xung b lc thc t h(n).

Chng trnh c vit v chy trn nn Matlab, vi vic s dng mt s hm h

trc sn ca Matlab cho x l tn hiu s, v mt s hm vit thm c tham kho

t ti liu [1] (cc hm di dng cc file .m).

a. Lu thut ton:


18/22


BEGIN

Nhp cc ch tiuws, wpAs, Rp

Ch tiu chp lkhng?

No

Tnh s mu M

Tm ga tr ca T1, T2 ttnht ( ng vi Rpd v

Asd tt nht )

Tnh hd(n)

Yes

V hd(n), h(n) v png bin (dB) ca b

lc thit k.

END

Rpd v Asd ctha mn yucu khng?

Yes

No


19/22


b) Chng trnhfunction [ws wp As Rp]=loc_FIR_thongcao()fprintf('\n');fprintf('Nhap vao cac thong so cua bo loc thong cao :');fprintf('\n');ws1=input('Nhap 0ws ,021, As = ');

endRp1=input('Nhap 0 < Rp < 1, Rp = ');while((Rp1=1))

Rp1=input('Nhap sai, nhap lai 0 < Rp < 1, Rp = ');endws=ws1;wp=wp1;As=As1;Rp=Rp1;M=round(1/(wp-ws))*6+1;alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l;

%Tim cac gia tri toi uuMaxAs=As;MinRp=Rp;

N1=ceil(ws*alpha)+1;N2=ceil((2-wp)*alpha)+1;for T1 = 0.05:0.01:0.5

for T2 = 1:-0.01:0.5Hrs = [zeros(1,N1),T1,T2,ones(1,N2-N1-2),T2,T1,zeros(1,M-N2-2)];Hdr = [0,0,1,1]; wdl = [0,ws,wp,1];k1 = 0:floor((M-1)/2); k2 = floor((M-1)/2)+1:M-1;angH = [-alpha*(2*pi)/M*k1, alpha*(2*pi)/M*(M-k2)];

H = Hrs.*exp(j*angH);h = real(ifft(H,M));[db,mag,pha,grd,w] = freqz_m(h,1);[Hr,ww,a,L] = Hr_Type1(h);delta_w=1/500;Asd = -max( db(1:ws/delta_w+1));Rpd = -min(db(wp/delta_w+1):501)if((Asd>=MaxAs)&(Rpd


20/22


MinRp=Rpd;T1op=T1;T2op=T2;

endend

endHrs = [zeros(1,N1),T1op,T2op,ones(1,N2-N1-2),T2op,T1op,zeros(1,M-N2-2)];Hdr = [0,0,1,1]; wdl = [0,ws,wp,1];k1 = 0:floor((M-1)/2); k2 = floor((M-1)/2)+1:M-1;angH = [-alpha*(2*pi)/M*k1, alpha*(2*pi)/M*(M-k2)];H = Hrs.*exp(j*angH);h = real(ifft(H,M));[db,mag,pha,grd,w] = freqz_m(h,1);[Hr,ww,a,L] = Hr_Type1(h);

subplot(1,1,1);subplot(2,2,1);plot(wl(1:alpha+1)/pi,Hrs(1:alpha+1),'o',wdl,Hdr);

axis([0,1,-0.1,1.1]); title('Frequency Samples at M=%2.4f',M);xlabel('Frequency in pi units'); ylabel('Hr(k)')set(gca,'XTickMode','manual','XTick',[0;ws;wp;1])set(gca,'XTickLabelMode','manual','XTickLabels',[0;ws;wp; 1])set(gca,'YTickMode','manual','YTick',[0,T1op,T2op,1]); gridsubplot(2,2,2); stem(l,h); axis([-1,M,-0.4,0.4])title('Impulse Response'); xlabel('n');ylabel('h(n)');text(M+1,-0.4,'n')subplot(2,2,3); plot(ww/pi,Hr,wl(1:alpha+1)/pi,Hrs(1:alpha+1),'o');axis([0,1,-0.1,1.1]); title('Amplitude Response')xlabel('Frequency in pi units'); ylabel('Hr(w)')set(gca,'XTickMode','manual','XTick',[0;ws;wp;1])

set(gca,'XTickLabelMode','manual','XTickLabels',[ 0;ws;wp;1])set(gca,'YTickMode','manual','YTick',[T1op,T2op]); gridsubplot(2,2,4);plot(w/pi,db); axis([0,1,-100,10]); gridtitle('Magnitude Response');xlabel('Frequency in pi units'); ylabel('Decibels');set(gca,'XTickMode','manual','XTick',[0;ws;wp;1])set(gca,'XTickLabelMode','manual','XTickLabels',[ 0;ws;wp; 1])set(gca,'YTickMode','Manual','YTick',[-As;0]);set(gca,'YTickLabelMode','manual','YTickLabels',[As; 0])


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c) Kt qu

Kt qu chng trnh c thc hin vi ws=0.4pi, wp=0.65pi, Rp=0.4, As=45


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TI LIU THAM KHO

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[1] Ng Vn S, ging X l Tn hiu s, i hc Bch Khoa Nng.

[2] Monson H. Hayes (Sch dch - Tng Vn On), L thuyt v Bi tp X l Tn

hiu s, Nh xut bn Lao ng x hi.

[3] Quch Tun Ngc, X l Tn hiu s, i hc Bch Khoa H Ni.

[4] Nguyn Quc Trung (2002), X l Tn hiu v Lc s, Nh xut bn Khoa hc

v K thut.

[5] Nguyn Phng Quang, Matlab & Simulink dnh cho k s iu khin t

ng, NXB Khoa Hc K Thut, 2004.

[6] Vinay K. Ingle & John G. Proalis, Digital Signal Procesing Using MATLAB,

Brooks/Cole Thomson Learning.

[7] H Vn Sung, Thc hnh X L S Tn Hiu Trn My Tnh PC Vi

Matlab, NXB Khoa Hc K Thut, 2005.

ĐỀ TÀI THIẾT KẾ BỘ LỌC FIR THÔNG CAO BẰNG PHƯƠNG PHÁP LẤY MẪU TẦN SỐ

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