ĐỀ tÀi thiẾt kẾ bỘ lỌc fir thÔng cao bẰng phƯƠng phÁp lẤy mẪu tẦn sỐ
TRANSCRIPT
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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
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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
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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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 5
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.
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 6
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
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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
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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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 9
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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 10
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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 11
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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 12
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)
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 13
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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 14
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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 15
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.
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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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 17
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:
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 18
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
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Thit k lc FIR thng cao bng phng php ly mu tn s Trang 19
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
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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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[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.