sachtructuyen.info_bai tap su ly tin hieu so tren matlab

8/3/2019 Sachtructuyen.info_bai Tap Su Ly Tin Hieu So Tren Matlab

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BI 1: LM QUEN VI MATLABI- L thuyt v thc hnh

1- m 1 hm m-file v thc hin yu cu sau:a- to ma trn c chiu di n x m: b- cng tr nhn chia hai ma trn va to( a ra kt qu)

c- trch ra ng cho ca 2 ma trn ban u sau ghp li thnh 1 ma trn mid- trch 2 dng u ca ma trn 1 v 2 dng cui ca ma trn 2. sau ghp chngthnh ma trn mi

e- trch ct cui ca ma trn th nht v ct u ca ma trn 2. sau ghp vi mtrn 1 to ma trn mi

% chng trnh thc hin>> a=[2 4 6 0 ;3 5 7 7;10 4 5 4];>> b=[2 4 6 1; 4 9 2 4 ;1 5 6 7 ];>> x=a+bx =

4 8 12 17 14 9 1111 9 11 11

>> y=a-by =

0 0 0 -1-1 -4 5 39 -1 -1 -3

>> z=a.*bz =

4 16 36 012 45 14 2810 20 30 28

>> t=a./b

t =

1.0000 1.0000 1.0000 00.7500 0.5556 3.5000 1.750010.0000 0.8000 0.8333 0.5714

>> diag(a); % ham co cong dung trich duong cheo cua ma tran>> diag(b);>> x1=[diag(a) diag(b)]

x1 =

2 25 9


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5 6>> a1=a(1:2,:);>> b1=a(2:3,:);>> x2=[a1;b1]

x2 =2 4 6 03 5 7 73 5 7 710 4 5 4>> a2=a(:,4);>> b2=b(:,1);>> x3=[a2 b2]

x3 =

0 27 44 1

2- ha2.1- trong khng gian 2D:m 1 hm m-file v v th ca cc hm sau(v trn cng 1 ca s)hm f(x) : f(x)= -x.sin(x)o hm ca f(x) : f(x)=-x.cosx-sinxo hm xp x : f 12=diff(f(x)/x(2)-x(1))sai s lin quan: f 22=(f 12-f(x)(1:999))/norm(f 12)% chng trnhx=linspace(-20,20,1000); %chon khoang lay mau%ham f(x);y=-x.*sin(x);subplot(2,2,1); % chia o trong do thi plot(x,y) %ve do thi trong khong gian 2dgridtitle('ham sin'); %tao tieu de cho do thixlabel('truc x' ); %tao nhan cho truc xylabel('truc y'); %tao nhan cho truc y%ham dao ham;y1=-x.*cos(x)-sin(x);subplot(2,2,2); plot(x,y1)grid %ve luoi trong do thititle('ham dao ham');xlabel('truc x');


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ylabel('truc y');%ham xap xiy2=diff(y/x(2)-x(1));subplot(2,2,3); plot(x(1:999),y2);

gridtitle('ham xap xi');xlabel('truc x');ylabel('truc y');y3=(y2-y1(1:999))/norm(y2);subplot(2,2,4); plot(x(1:999),y3);gridtitle('ham sai so lien quan');xlabel('truc x');ylabel('truc y');

%ket qua chuong trinh

2.2-trong khng gian 3D:m 1 hm m-file v v th ca cc hm sau(v trn tng ca s)a- z1=f(x,y)=sinx.siny, vi x,y=[0, ] b- z2=f(x,y)=x - x3 + y2 + 1, vi x,y=[-3,3]


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c- z3=f(x,y)=2 2

2 2

sin( )( ).

x y

x y x

++

, vi x,y=[-8,8]

% do thi 1x=(0:0.05:pi)

y=(0:0.05:pi)[x,y]=meshgrid(x,y); % tao ma tran he thong trong do thi 3dz1=sin(x).*sin(y);figure; %tao them 1 do thi moimesh(x,y,z1); %ve do thi 3dxlabel('Truc x');ylabel('Truc y');zlabel('Truc z1');title(' Do thi ham so Z= sinx*siny');% do thi 2[x,y]=meshgrid(-3:0.5:3);

z2=x-x.^3+y.^2+1;figure;mesh(x,y,z2);xlabel('Truc x');ylabel('Truc y');zlabel('Truc z2');title(' Do thi ham so z2 = x-x^3+y^2+1');% do thi 3[x,y]=meshgrid(-8:0.5:8);z3=sin (sqrt(x.^2+y.^2))./sqrt((x.^2+y.^2).*x);figure;

mesh(x,y,z3);xlabel('Truc x');ylabel('Truc y');zlabel('Trucz3');title(' Do thi ham so z3 = sin((sqrt(x.^2+y.^2)./sqrt((x.^2+y.^2).*x))');% ket qua chuong trinh


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II- NHN XT V KT LUNBi thc hnh gip ta bc u lm quen vi matlab, nh m ta co th d dng tcc ma trn cc hm va thc hin tinh ton n, v cc th n gin trongkhng gian 2d v 3dBi thc hnh cn gip ta hiu r chc nng ca cc lnh trong matlab

BI 2:M PHNG V TO TN HIU1-Tm tt l thuyt

Mt tn hiu thi gian ri rc c biu din nh 1 dy s hay cn gi l 1 dymu,c k hiu l {x[n]}; trong i s l nhng s nguyn chy t - n+, c trng cho thi gian. Gi tr ca dy mu ti thi im n l x[n]. v th tin li,tn hiu thi gian ri rc bt k thng c k hiu l x[n]

tn hiu thi gian ri rc c th l 1 dy mu c chiu di v hn hoc hhn. dy c chiu dy hu hn l dy c gi tr khc 0 trong 1 khong thigian hu hn t thi im N1 n N2: N1 n N2. Vi N2 N1. Dy nyc chiu di N=N2-N1+1 mu

dy tha mn iu kin[ ] [ ] x n x n kN = +% % c gi l dy tun hon vichu k c bn N l 1 s nguyn dng v k l 1 s nguyn bt k

nng lng ca dy x[n] dc xc nh bng cng thc 2| [ ] |n

X n

=

=

nng lng ca dy trong 1 khong xc nh t -K n K c xc nh bng biu thc 2| [ ] |

k

n k

X n =

= cng sut trung bnh ca 1 dy khng tun hon x[n] c xc nh bng

cng thc 21 1lim lim | [ ] |

1 1

K

K av K K n K

p X n K K

=

= =+ +


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Cng sut trung bnh ca 1 dy tun hon[ ] x n% vi chu k N c cho

bi cng thc 20

1 | [ ] | N

avn

p X n N =

= %

Dy xung n v c k hiu bng[ ]n v c xc nh t biu thc:

{1, 0

0, 0[ ]khin

khinn =

= Dy nhy bc n v k hiu u[n] dc xc dnh t biu thc :

{1, 00, 0[ ] khinkhinu n


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%day xung don vi chieu dai N tre M maum=4;x1=[zeros(1,m) 1 zeros(1,n-m-1)]; %zeros :ma tran dong nhatkhongsubplot(5,1,2);

stem(x1)title('day xung don vi chieu dai N tre M mau')%tao day nhay bac dai N mauu=[ones(1,n)];subplot(5,1,3);stem(u)title('day nhay bac dai N mau')%tao day nhau bac dai N bi tre M maum=4;u1=[ones(1,n-m)];subplot(5,1,4);

stem(u1)title('day nhau bac dai N bi tre M mau')%chuong trinh phat day xungn=-10:20;y=[zeros(1,10) 1 zeros(1,20)];subplot(5,1,5);stem(n,y)xlabel(' chi so thoi gian')ylabel('bien do')title('day xung don vi')axis([10 20 0 1.5]) %tao kich thuoc truc do thi

% ket qua chuong trinh


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Nhn xt: nhn vo th ta thy cc tn hiu c biu din di dng ri rnc n v (bin =1), th th hin r cc c im ca dy xung n v tnh cht tr , dy nhy bc n v ph hp vi l thuyt hc

2- biu din tn hiu sin phc v sin thcdy sin phc m t bng phng trnh:0 0 0 0( )

0 0| | | | cos( ) | | sin( )n j w n n n A e A e w n j A e w n + + = + + +

dy sin thc [ ] cos( )n x n A w n= + trong A , w0, l nhng s thc c

gi l bin , tn s gc v pha ban u ca dy s sin x[n], cn00 2w

f

= l

tn s% chuong trinht=-(1/12)+(pi/6).*j;k=2;

n=0:40;x=k*exp(t*n);subplot(2,1,1);stem(n,real(x));xlabel('chi so thoi gian');ylabel('bien do')title('phan thuc')subplot(2,1,2);


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stem(n,imag(x));xlabel('chi so thoi gian');ylabel('bien do');title('phan ao');


Nhn xt : kt qu thu c la th biu din tn hiu phn thc v phn di dng ri rc nc n v trn trc thc v trc o nh hm stem. Tn hi biu din dn tin v khng

3- cc tn hiu thi gian ri rc sin thc to tn hiu sin thc trong matlab ta s dng cc hm sin v cos%chuong trinh tin hieu thoi gian roi rac sin thucn=0:40;f=0.1; phase=0;A=1.5;arg=2*pi*f*n-phase;x=A*sin(arg);stem(n,x);axis([0 40 -2 2]);grid;title(' day tin hieu sin');xlabel(' chi so thoi gian n');ylabel(' bien do');



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Nhn xt : kt qu thu c cng l th dng ri rc hnh sin c bin l1,5, chu k l 10 bin thin lin tc theo thi gian4- cc tn hiu ngu nhin ri rctn hiu ngu nhin thi gian ri rc c chiu di n mu phn b u n tronkhong [0,1] c to bi lnh trong matlab l x=rand(1,N)to 1 tn hiu ngu nhin dng gauss c gi tr trung bnh bng 0 v phngsai bng n v , dng lnh x=rand(1,N)%chuong trinhn=40;x=randn(1,n);stem(x);


III- p dng


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1- biu din tn hiu sin thc c tn s 0,9 v 1,1%chuong trinh biu din tn hiu sin thc c tn s 0,9n=0:40;f=0.9; phase=0;

A=1.5;arg=2*pi*f*n-phase;x=A*sin(arg);stem(n,x);axis([0 40 -2 2]);grid;title(' day tin hieu sin');xlabel(' chi so thoi gian n');ylabel(' bien do');


%chuong trinh biu din tn hiu sin thc c tn s 1,1n=0:40;f=1,1; phase=0;A=1.5;arg=2*pi*f*n-phase;x=A*sin(arg);stem(n,x);axis([0 40 -2 2]);grid;title(' day tin hieu sin');xlabel(' chi so thoi gian n');ylabel(' bien do');



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So snh 2 dy ny vi dy trong chng trnh 2.3 (f=0,1) ta nhn thy : tns l 0,9 th chu k s l 9 dng xung ra c m phng gn nh tn s l 0,1- tn s 1,1 ta thy so vi f=0,1 th th l dng ng thng do gi tr maca hm sin l 1 v day gi tr l 1,1 nn th s c dng dng thng ( dgi tr =0)

2- biu din tn hiu sin thc c chiu di 50 mu, tn s 0,08, bin 2,5 vc dch pha 900

-thay lnh stem thnh lnh plot v lnh stairs%chuong trinh

n=0:50;f=0.08; phase=90;A=2.5;arg=2*pi*f*n-phase;x=A*sin(arg);subplot(3,1,1);stem(n,x);axis([0 40 -3 3]);title(' day tin hieu sin bd bang lenh stem');xlabel(' chi so thoi gian n');ylabel(' bien do');subplot(3,1,2); plot(n,x);title(' day tin hieu sin bd bang lenh plot');xlabel(' chi so thoi gian n');ylabel(' bien do');subplot(3,1,3);


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stairs(x);gridtitle('tin hieu dang luy thua bd bang lenh stairs')xlabel('chi so thoi gian')ylabel(' bien do')


Nhn xt: tn hiu ly tha c biu din bng 3 lnh plot, stem v stairs chra 3 dng sng khc nhau ri rc, lin tc, bc thang. Ta nhn thyTn hiudc biu din i t 0 v bt u tng dn ln khi gn ti 20 v t ti gi trnh l >40 ph hp vi l thuyt hc

IV- NHN XT V KT LUNBi thc hnh ny gip ta nm r cc vn v m phng v to tn hiu: cng dng cc hm trong tng bi, cch s dng cc lnh , cch v th , hiu c th

BI 3:H THNG LTII- L THUYT

H thng thi gian ri rc thc hin php nh x tn hiu li vo vi cc gi trx[n] thnh tn hiu ri rc li ra vi nhng tnh cht mong mun bng cch pdng nhng thut ton cho trc . bi thc hnh ny l tin hnh m phng 1 sh thng ri rc tuyn tnh v bt bin vi thi gian n gin v nghin cu cctnh cht ca chng trn lnh vc thi gian. Cc h thng ny c c trng bp ng xung h[n] v c m hnh ha trn hnh 3.1


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hnh 3.1 Quan h vo / ra ca h thng LTI c xax1 nh bi tng nhn chp

sau: [ ] [ ] [ ] [ ] [ ]r k

y n h k x n k x k h n k

= =

= = . v c k hiu bng:y[n]=x[n]*h[n]

2 h thng LTI c p ng xung ln lt l h1[n] v h2[n] ghp ni tipvi nhau th h thng tng th c p ng xung: h[n]=h1[n]*h2[n] . nu 2h thng ghp ni tip nhau sao cho :1[ ]* 2[ ] [ ]h n h n n = th h thng LTIc p ng xung h2[n] c gi l nghch o ca h thng LTI c p

ng xung h1[n] v ngc li 1 h thng LTI c gi l n nh theo ngha BIBO nu p ng xung

xa n tha mn iu kin: 2| [ ] |n

h k

=

< 1 h thng LTI c gi l nhn qu khi v ch khi p ng xung ca n

tha mn iu kin h[n]=0 khi n0 . vy h thng FIR nhn qu c phng

trnh0 0

[ ] [ ]M

m

m

b y n x n m

a==

H thng thi gian ri rc gi l tuyn tnh nu p ng ln 1 tng bng tng cc p ng , cch khc nu y1[n] v y2[n] l p ng ln cctnh hiu li vo x1[n] v x2[n] th i vi li vo tng : x[n]=ax1[n]+bx2[n] s c p ng l y[n]=ay1[n]+by2[n]. phng trnh ny ng vicc hng s a,b bt k v i vi tt c c gi tr kh d ca tn hiu li vx1[n] v x2[n] , nu phng trnh khng nghim ng vi t nht 1 gi trkhc khng ca a hoc b , ca x1[n], x2[n] th h thng l phi tuyn

H thng gi l bt bin i vi thi gian nu y1[n] l p ng i vtn hiu li vo x1[n], th p ng i vi phin bn tr ca tn hiu li vx[n]=x1[n-n0] s l y[n]=y1[n-n0], y n0 l s nguyn dng hoc m.


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Nu h thc khng tha mn th h thng thay i i vi thi gian. Hthng va tha mn tnh tuyn tnh va bt bin vi thi gian gi l hthng LTI

Trong matlab d m phng cc h thng thi gian ri rc LTI nhn qu c phng trnh sai phn trn ta c th dng lnh filter y=filter(num,den,x)

Tn hiu li ra y[n] ca h thng LTI c p ng xung n v h[n] vi livo x[n] cng c xc nh bng lnh conv(h,x) p ng xung n v h[n] ca 1 h thng dc xc dnh bng lnh

h=inpz(num,den,N+1)

II- THC HNH1- cc h thng tuyn tnh v phi tuyn tnh Nghin cu tnh cht tuyn tnh ca h thng

y[n]-0,4y[n-1]+0,75y[n-2]=2,2403x[n]+2,4908x[n-1]+2,2403x[n-2]chng trnh di y m phng h thng ny vi 3 tn hiu vo l x1[n],x2[n],x[n]=ax1[n]+bx2[n]. tnh v v th ca tn hiu li ra y1[n],y2[n] v y[n] vix1=cos(0,2 n ),x2=cos(0,8 n ),x[n]=2x1[n]-3x2[n]%chuong trinhn=0:40;a=2; b=-3;x1=cos(2*pi*0.1*n);x2=cos(2*pi*0.2*n);x=a*x1+b*x2;

num=[2.2403 2.4908 2.2403]; %tao ma tran tu soden=[ 1 -0.4 0.75]; % tao ma tran mau soy1=filter(num,den,x1); % ham filter thuc hien bo loc matran tu, mau cua ham x1y2=filter(num,den,x2); % thuc hien bo loc matran tu, mau cua ham x2y=filter(num,den,x);yt=a*y1+b*y2;d=y-ytsubplot(3,1,1);stem(n,y);title('tin hieu loi ra theo x')axis([0 40 -50 50]) %chia truc cua do thi truc x tu 0-40, truc y tu -50-50subplot(3,1,2);stem(n,yt);title('tin hieu loi ra theo y')axis([0 40 -50 50])

subplot(3,1,3);stem(n,d);title('tin hieu sai so d')%ket qua chay chuong trinh


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Nhn xt: nhn vo d th tha thy tn hiu li ra theo x(y[n]), tn hiu li ra theoy(yt[n]) l ging nhau v th hai h thng ny l tuyn tnh v y[n]=yt[n]2- xc nh p ng xung n v ca h thng LTIPhp tnh v v p ng xung ca phng trnh sai phn sau:y[n]-0,4y[n-1]+0,75y[n-2]=2,2403x[n]+2,4908x[n-1]+2,2403x[n-2]

%tinh va ve dap ung xungn=40;num=[2.2403 2.4908 2.2403 ];

den=[1 -0.4 0.75];h=impz(num,den,n); %hm impz : xac dinh dap ung xung don vi cua he thongstem(h);title('dap ung xung cua he thong');


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Nhn xt: kt qu thu c l dang tn hiu c biu din di dng ri rc v tin v 03- cc h thng bt bin vi thi gianM phng h thng c phng trnh sai phn sau

y[n]= 2,2403x[n]+2,4908x[n-1]+2,2403x[n-2]+ 0,4y[n-1]+0,75y[n-2]

mc ch ca chng trnh ny l tm tn hiu li ra y[n] i vi 2 tn hiu li vox[n] v x[n-n0] v hiu ca 2 tn hiu n=0:40;n0=10;a=0.3; b=-2;xn=a*cos(2*pi*0.1*n)+b*cos(2*pi*0.4*n);xn0=[zeros(1,n0) xn];num=[2.243 2.4908 2.2403];den=[1 -0.4 0.7];yn=filter(num,den,xn);

yn0=filter(num,den,xn0);dn=yn-yn0(1+n0:41+n0);subplot(3,1,1)stem(n,yn);title('tin hieu loi ra trheo y[n]');subplot(3,1,2);stem(n,yn0(1:41));title('tin hieu loi ra tre n0 mau');subplot(3,1,3);stem(n,dn);title('tin hieu hieu')


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Nhn xt: tn hiu li ra y[n] gim dn, tn hiu li ra theo y[n-n0] b tr i 10 muvi tn hiu y[n] . hai tn hiu ny c bn l ging nhau nhng thi im bt u ctn hiu l khc nhauV tn hiu ra l 0 nn khng p ng vi tn hiu li vo nn h thng khng bt bivi thi gian

4- ghp ni tip cc h thngGhp ni 2 h thng bc 2 c phng trnh sai phn sau:y1[n]+0,9y1[n-1]+0,8y1[n-2]=0,3x[n]-0,3x[n-1]+0,4x[n-2]

vy2[n]+0,7y2[n-1]+0,85y2[n-2]=0,2y1[n]-0,5y1[n-1]+0,3y1[n-2]

thu c h thng bc 4 c phng trnh sai phn sauy[n]+1,6y[n-1]+2,28y[n-2]+1,325y[n-3]+0,68y[n-4]=0,06x[n]-0,19x[n-

1]+0,27x[n-2]-0,26x[n-3]+0,12x[n-4]%chuong trinhx=[1 zeros(1,40)];%tao tin hieu vao x[n]

n=0:40;%cac he so cua he thong bac 4den=[1 1.6 2.28 1.325 0.68];num=[0.06 -0.19 0.27 -0.26 0.12];%tinh tin hieu loi ra cua he thong bac 4y=filter(num,den,x);%cac he so cua hai he thong bac hainum1= [0.3 -0.2 0.4];den1=[1 0.9 0.8];num2=[0.2 -0.5 0.3];den2=[1 0.7 0.85];%tin hieu ra y[n]y1=filter(num1,den1,x);%tin hieu ra y[2]y2=filter(num2,den2,y1);%hieud[n]=y[n]-y2[n]d=y-y2;%ve cac tin hieusubplot(3,1,1);stem(n,y);ylabel('bien do')title('tin hieu ra cua he thong bac 4');grid;subplot(3,1,2);stem(n,y2);ylabel('bien do')title('loi ra cua he thong noi tiep');grid;subplot(3,1,3);stem(n,d);


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xlabel('chi so thoi gian n');ylabel('bien do');title('tin hieu sai so');grid;

Nhn xt:da vo th ta nhn thy tn hiu li ra y[n] ging tn hiu li ra y2[nn dy y[n] ging dy y2[n]

5- tnh n dnh ca h thng LTIy[n]-x[n]-0,8x[n-1]-1,5y[n-1]-0,9y[n-2]%chuong trinhnum=[1 -0.8];den=[1 1.5 0.9]; N=250;h=impz(num,den,N+1);sum=0; %thuc hien phep lapfor k=1:N+1;

sum=sum + abs(h(k));if abs(h(k))


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title('dap ung xung don vi');%In gia tri tuyet doi cua h(k)td=abs(h(k))grid

Nhn xt: da vo th ta nhn thy h thng l n nh v n tin dn v 0III- P DNG

1-cc h thng tuyn tnh v phi tuyn tnhy[n]-0,5y[n-1]+0,25y[n-2]=x[n]+2x[n-1]+x[n-3]

vi x1[n]=cos(0,5n + 3 )

x2[n]=sin(0,2n )x[n]=3x1[n]+2x2[n]tnh v v tn hiu li ra y1[n],y2[n],y[n]. xt xem h thng c tuyn tnh hay khng? Vsao?%chuong trinh nghien cuu ve tinh chat tuyen tinh va khong tuyen tinhn=0:40;a=3; b=2;x1=cos(0.5*pi*n+pi/3);x2=sin(0.2*pi*n);x=a*x1+b*x2;num=[1 2 1];den=[1 -0.5 0.25];y1=filter(num,den,x1);y2=filter(num,den,x2);y=filter(num,den,x);yt=a*y1+b*y2;


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d=y-ytsubplot(3,1,1);stem(n,y);title('tin hieu loi ra theo x')axis([0 40 -50 50])

subplot(3,1,2);stem(n,yt);title('tin hieu loi ra theo y')axis([0 40 -50 50])subplot(3,1,3);stem(n,d);title('tin hieu sai so d')

V tn hiu li ra theo x ging tn hiu li ra theo y nn h thng l tuyn tnh2- xc nh p ng xung n v ca h thng LTI. Tnh n dnh ca h thngCho h thng LTI c phng trnh sai phna- y[n]-0,5y[n-1]+0,25y[n-2]=x[n]+2x[n-1]+x[n-3]%chuong trinhnum=[1 2 0 1];den=[1 -0.5 0.25 0]; N=100;h=impz(num,den,N+1);

sum=0;for k=1:N+1;sum=sum + abs(h(k));if abs(h(k))


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n=0:N;stem(n,h);xlabel('chi so thoi gian n');ylabel('bien do');title('dap ung xung don vi');

%In gia tri tuyet doi cua h(k)td=abs(h(k))%ket qua chay chuong trinh

b-y[n]=x[n]-4x[n-1]+3x[n-2]+1,7y[n-1]-y[n-2]%chuong trinhnum=[1 4 3];den=[1 -1.7 1]; N=100;h=impz(num,den,N+1);sum=0;for k=1:N+1;

sum=sum + abs(h(k));if abs(h(k))


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Nhn xt: da vo th ta thy h thng khng n nh v n bin thin lin tuc theodng sng sin3-cc h thng bt bin vi thi gianCho h thng LTI c phng trnh sai phny[n]-0,5y[n-1]+0,25y[n-2]=x[n]+2x[n-1]+x[n-3]vit chng trnh v tn hiu li ra y[n] v y[n-n0] vi n0=5, tn hiu li vo l

x[n]=3cos(0,5n +3

)+2sin(0,2 n ). H thng c bt bin vi thi gian khng?

%he thong dap ung xung bat bien theo thoi giann=0:40;n0=5;

a=3; b=2;xn=a*cos(0.5*pi*n+pi/3)+b*sin(0.2*pi*n);xn0=[zeros(1,n0) xn];num=[1 2 0 1];den=[1 -0.5 0.25 0];yn=filter(num,den,xn);yn0=filter(num,den,xn0);dn=yn-yn0(1+n0:41+n0);subplot(3,1,1)stem(n,yn);

title('tin hieu loi ra y[n]');subplot(3,1,2);stem(n,yn0(1:41));title('tin hieu loi ra tre n0 mau');subplot(3,1,3);stem(n,dn);title('tin hieu hieu');


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Nhn xt : da vo th ta nhn thy h thng khng bt bin vi thi gian

4-Ghp ni tip cc h thng LTIVit chng trnh tnh v v cc tn hiu li ra y1[n] v y2[n] tha mn phng trnh sa phn sau:y1[n]=0,5x[n]+ 0,27x[n-1]+0,77x[n-2]v y2[n]=0,45x[n]+0,5x[n-1]+0,45x[n-2]+0,53y[n-1]-0,46y[n-2]vi tn hi li vo x[n=cos(20n /256)+cos(200n /256) vi 0n29

%chuong trinhx= cos(20*pi*n/256)+cos(200*pi*n/256)%tao tin hieu vao x[n]n=0:29;%tinh tin hieu loi ra cua he thong bac 4y=filter(num,den,x);%cac he so cua hai he thong bac hainum1= [0.5 0.27 0.77];den1=[1 0 0 0];num2=[0.45 0.5 0.45];den2=[1 0.53 -0.46];%tin hieu ra y[n]y1=filter(num1,den1,x);%tin hieu ra y[2]y2=filter(num2,den2,y1);%hieud[n]=y[n]-y2[n]d=y-y2;%ve cac tin hieusubplot(3,1,1);


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stem(n,y);ylabel('bien do')title('tin hieu ra cua he thong bac 4');grid;subplot(3,1,2);

stem(n,y2);ylabel('bien do')title('loi ra cua he thong noi tiep');grid;subplot(3,1,3);stem(n,d);xlabel('chi so thoi gian n');ylabel('bien do');title('tin hieu sai so');grid;% kt qu

IV- NHN XT V KT LUNBi thc hnh gip ta m phng 1 s h thng ri rc tuyn tnh v bt bin vi thgian . n con gip ta hiu dc cc tnh cht ca h thng, cc thut ton v cc hmtrong tng bi ton, cc hng d gii quyt bi ton Tn hiu ra phn nh r tnh cht v vn ca bi ton ph hp vi l thuyt h

BI 4:PH TN S CA TN HIU-BIN I FORIER THI GIAN RI RCI-L THUYT1-Bin i forier thi gian ri rc2- Bin i forier thi gian ri rc nghch o (IDTFT)3-cc tnh cht ca DTFT

Tnh cht dch thi gian Dch v tn s Tnh cht nhn chp


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Tnh cht iu ch Tnh cht ngc thi gian

II-THC HNH1- tnh ton DTFT2- tnh DTFT X(e-jw) ca dy x[n] dng

X(e-jw) = 21 0,6 jw

jwe

e

+

+%chuong trinh%Tinh DTFT%Tinh cac mau tan so cua DTFTw=-4*pi:8*pi/511:4*pi;num=[2 1];den=[1 -0.6];h=freqz(num,den,w);% do thi cua DTFTfiguresubplot(2,1,1) plot(w/pi,real(h));grid;title('phan thuc bien doi DTFT cua x[n]');xlabel('omega/pi');ylabel('bien do');subplot(2,1,2) plot(w/pi,imag(h));grid;title('phan ao bien doi DTFT cua x[n]');xlabel('omega/pi');ylabel('bien do');figuresubplot(2,1,1) plot(w/pi,abs(h));grid;title('pho bien do cua tin hieu x[n]');xlabel('omega/pi)');ylabel('bien do');subplot(2,1,2) plot(w/pi,angle(h));grid;title('pho pha cua tin hieu x[n]');xlabel('omega/pi');ylabel('pha do bang radians')grid


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Nhn xt: bi ton ny hin th DTFT ca dy X(e-jw) da vo hm freqz v v thca nHm abs ly bin Hm angle ly pha

3- tnh cht ca DTFTa-tnh cht dch chuyn thi gian%chuong trinhw=-pi:2*pi/225:pi;wo=0.4*pi;D=10;x=[1 2 3 4 5 6 7 8 9];


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h1=freqz(x,1,w);h2=freqz([zeros(1,D) x],1,w);subplot(2,2,1) plot(w/pi,abs(h1));grid;

title('pho bien do day goc')xlabel('omega/pi');ylabel('bien do');subplot(2,2,2) plot(w/pi,abs(h2));gridtitle('pho bien do cua day bi dich')xlabel('omega/pi');ylabel('bien do');subplot(2,2,3) plot(w/pi,angle(h1));

gridtitle('pho pha cua day goc')xlabel('omega/pi');ylabel('bien do');subplot(2,2,4) plot(w/pi,angle(h2));gridtitle('pho pha cua day bi dich')xlabel('omega/pi');ylabel('bien do');


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Thng s c trng cho s dch chuyn thi gian l h1 ,h2( xc nh p ng tn sca h thng) b-tnh cht dch chuyn v tn s%chuong trinhw=-pi:2*pi/225:pi;

wo=0.4*pi;x1=[1 3 5 7 9 11 13 15 17];L=length(x1);h1=freqz(x1,1,w)n=0:L-1;x2=exp(wo*i*n).*x1;h2=freqz(x2,1,w);subplot(2,2,1) plot(w/pi,abs(h2));grid;title('pho bien do cua day goc')

xlabel('omega/pi');ylabel('bin do');subplot(2,2,2) plot(w/pi,abs(h2));grid;title('pho bien do cua day bi dich tan so')xlabel('omega/pi');ylabel('bin do');subplot(2,2,3); plot(w/pi,angle(h1));title('pho pha cua day goc');xlabel('omega/pi');ylabel('bin do');subplot(2,2,4); plot(w/pi,angle(h2));grid;title('pho pha cua bi dich tan so');xlabel('omega/pi');ylabel('bin do');


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Nhn xt: thng s c trng cho s dch chuyn tn s l h1,h2,x1,x2, trong x2=exp(wo*i*n).*x1;h2=freqz(x2,1,w);l lnh c trng nht (xc nh p ng tn s ca hm e m :x2)

d- tnh cht iu ch%chuong trinhw=-pi:2*pi/225:pi;x1=[1 3 5 7 9 11 13 15 17];x2=[1 -1 1 -1 1 -1 1 -1 1];y=x1.*x2;h1=freqz(x1,1,w)h2=freqz(x2,1,w)

h3=freqz(y,1,w)subplot(3,1,1) plot(w/pi,abs(h2));grid;title('pho bien do cua day x1')xlabel('omega/pi');ylabel('bien do');subplot(3,1,2)


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plot(w/pi,abs(h2));grid;title('pho bien do cua day x2')xlabel('omega/pi');ylabel('bien do');

subplot(3,1,3); plot(w/pi,angle(h1));title('pho bien do cua day tich');xlabel('omega/pi');ylabel('bien do');

Nhn xt: thng s c trng cho tnh cht diu ch l: h1,h2,h3.d-tnh cht ngc thi gian%chuong trinhw=-pi:2*pi/225:pi;x=[1 2 3 4];

L=length(x)-1;h1=freqz(x,1,w)h2=freqz(fliplr(x),1,w);h3=exp(w*L*i).*h2;subplot(2,2,1) plot(w/pi,abs(h2));grid;title('pho bien do cua day goc')


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xlabel('omega/pi');ylabel('bien do');subplot(2,2,2) plot(w/pi,abs(h2));grid;

title('pho bien do cua day nguoc thoi gian')xlabel('omega/pi');ylabel('bien do');subplot(2,2,3); plot(w/pi,angle(h1));title('pho pha cua day goc');xlabel('omega/pi');ylabel('bien do');subplot(2,2,4); plot(w/pi,angle(h3));title(' pho pha cua day nguoc thoi gian');

xlabel('omega/pi');grid;

Nhn xt : lnh cho tnh cht ngc thi gian l h1,h2,h3 trong h2=freqz(fliplr(x),1,w) l lnh c trng nht ( fliplr l hm ta ma trn ngc vi mtrn x): xc nh p ng tn s ca ma trn ngc vi x

III-P DNG


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1-tnh ton DTFT%Tinh cac mau tan so cua DTFTw=0:8*pi/511:pi;num=[0.9 0.7 -0.5 0.3 1];den=[1 0.3 -0.5 0.7 0.9];

h=freqz(num,den,w);% do thi cua DTFTfiguresubplot(2,1,1) plot(w/pi,real(h));grid;title('phan thuc bien doi DTFT cua x[n]');xlabel('omega/pi');ylabel('bien do');subplot(2,1,2) plot(w/pi,imag(h));

grid;title('phan ao bien doi DTFT cua x[n]');xlabel('omega/pi');ylabel('bien do');figuresubplot(2,1,1) plot(w/pi,abs(h));grid;title('pho bien do cua tin hieu x[n]');xlabel('omega/pi)');ylabel('bien do');subplot(2,1,2) plot(w/pi,angle(h));grid;title('pho pha cua tin hieu x[n]');xlabel('omega/pi');ylabel('pha do bang radians')grid


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NX: im nhy trong ph pha chnh l khong thi gian tn hiu nhy sangtrng thi khc

2-kho st tnh cht ca DTFT


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a-dch chuyn v thi gian vi hai dy c chiu di thay i v hai dch thi khcnhau%bai 4.2 Tinh chat dich chuyen thoi gian%chuong trinhw=-pi:2*pi/225:pi;

wo=0.4*pi;d=5;x1=[1 3 5 7 9 ]D=10;x=[1 2 3 4 5 6 7 8 9];h1=freqz(x,1,w);h2=freqz([zeros(1,d) x],1,w);h3=freqz(x1,1,w);h4=freqz([zeros(1,D) x1],1,w);subplot(4,2,1) plot(w/pi,abs(h1));

grid;title('pho bien do day goc')subplot(4,2,2) plot(w/pi,abs(h2));grid;title('pho bien do cua day bi dich')subplot(4,2,3) plot(w/pi,angle(h1));gridtitle('pho pha cua day goc')subplot(4,2,4) plot(w/pi,angle(h2));grid;title('pho pha cua day bi dich')subplot(4,2,5) plot(w/pi,abs(h3));grid;title('pho bien do day goc')subplot(4,2,6) plot(w/pi,abs(h4));grid;title('pho bien do cua day bi dich')subplot(4,2,7) plot(w/pi,angle(h3));gridtitle('pho pha cua day goc')subplot(4,2,8) plot(w/pi,angle(h4));grid;title('pho pha cua day bi dich')


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% ket qua

b-dch chuyn v tn s vi hai dy c chiu di thy i v hai dch tn khc nh% tinh chat dich chuyen tan so%chuong trinhw=-pi:2*pi/225:pi;wo=0.4*pi;x1=[1 3 5 7 9 11 13 15 17];L=length(x1);x2=[1 2 3 4 5 6];l=length(x2);h1=freqz(x1,1,w);h2=freqz(x2,1,w);n=0:L-1;m=0:l-1;

x3=exp(wo*i*n).*x1;h3=freqz(x3,1,w);x4=exp(wo*i*m).*x2;h4=freqz(x4,1,w);subplot(4,2,1) plot(w/pi,abs(h1));grid;title('pho bien do cua day goc')


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subplot(4,2,2) plot(w/pi,abs(h3));grid;title('pho bien do cua day bi dich tan so')subplot(4,2,3);

plot(w/pi,angle(h1));title('pho pha cua day goc');subplot(4,2,4); plot(w/pi,angle(h3));grid;title('pho pha cua bi dich tan so');subplot(4,2,5) plot(w/pi,abs(h2));grid;title('pho bien do cua day goc')subplot(4,2,6)

plot(w/pi,abs(h4));grid;title('pho bien do cua day bi dich tan so')subplot(4,2,7); plot(w/pi,angle(h2));title('pho pha cua day goc');subplot(4,2,8); plot(w/pi,angle(h4));grid;title('pho pha cua bi dich tan so');%kt qu


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c-ngc thi gian vi hai dy c chiu di thay i%tinh chat nguoc thoi gian%chuong trinhw=-pi:2*pi/225:pi;x=[1 2 3 4];x1=[1 4 6 8 3 2 9 1 2];L=length(x)-1;l=length(x1)-1;h1=freqz(x,1,w)h2=freqz(fliplr(x),1,w);h3=exp(w*L*i).*h2;h4=freqz(x1,1,w)h5=freqz(fliplr(x1),1,w);h6=exp(w*L*i).*h5;subplot(4,2,1) plot(w/pi,abs(h1));grid;title('pho bien do cua day goc')xlabel('omega/pi');ylabel('bien do');subplot(4,2,2) plot(w/pi,abs(h3));grid;title('pho bien do cua day nguoc thoi gian')xlabel('omega/pi');


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ylabel('bien do');subplot(4,2,3); plot(w/pi,angle(h1));title('pho pha cua day goc');xlabel('omega/pi');

ylabel('bien do');subplot(4,2,4); plot(w/pi,angle(h3));title(' pho pha cua day nguoc thoi gian');xlabel('omega/pi');grid;subplot(4,2,5) plot(w/pi,abs(h4));grid;title('pho bien do cua day goc')xlabel('omega/pi');

ylabel('bien do');subplot(4,2,6) plot(w/pi,abs(h6));grid;title('pho bien do cua day nguoc thoi gian')xlabel('omega/pi');ylabel('bien do');subplot(4,2,7); plot(w/pi,angle(h4));title('pho pha cua day goc');xlabel('omega/pi');ylabel('bien do');subplot(4,2,8); plot(w/pi,angle(h6));title(' pho pha cua day nguoc thoi gian');xlabel('omega/pi');grid;% ket qua


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IV-NHN XT V KT LUNBi thc hnh ny gip ta tm hiu v bin i DTFT (bin i fourier thi gian rirc) v cc tnh cht ca DTFT ,cc thut ton v cc hm, ngha ca cc hm

BIN I Z

I-L THUYT1-trong phn ny s dng MATLAB nghin cu bin i z ca dy x[n].Trc tinghin cu bin i z ca dy x[n] c bin din bng phn thc hu t ca hai a

thc vi bin s phc z-1 dng:

1 20 1 2

1 20 1 2

...( )...

M M

M M

b b z b z b z X z

a a z a z a z

+ + + +=+ + + +

Hoc di dng khai trin:

0 0

0

1

( )( )

( )

M

mn M m

N

k k

z z b

X z z a z p

=

=

=

Trong Pk l cc im cc,cn Zml cc im khng.Ngoi ra cn c thm (N-M)im khng ti gc ta z=0 nu N>M hoc M-N cc im cc cng ti z=0 nu N>M Nu nh gi X(z) trn vng trn n v z=e jw th s thu c X(e jw) l bin iFOURIER ri rc ca dy x[n] .Trong MATLAB nh gi bin i z trn vngtrn n v s dng hm FREQZ2-Phn tch bin i z


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Bin i z ca 1 dy thi gian ri rc x[n] c nh ngha theo cng thc:

{ }( ) [ ] [ ] nn

X z Z x n x n z

=

= = trong z l 1 bin s phc l 1 php nh x 1 dy thi gian ri rc thnh1 hm bin s phc X(z).Nu bini z tn ti c ngha l hm s phc X(z) c xc nh t tch phn CAUCHY

{ }1 11( ) [ ] ( )2n

C

x z Z X n X z z dz j

= = Trong C l vng kn bao quanh gc ta nm trong min hi t ROC ca X(z)Trong trng hp h thng LTI thi gian ri rc c hm chuyn v trong DSP th phn ln cc h thng u c M


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0.1147 + 0.2627i0.1147 - 0.2627i

k =

0.4000

sos =

0.4000 0.8000 1.2000 1.0000 9.2293 2.43441.0000 0.5000 0.5000 1.0000 -0.2293 0.0822

Nx: hm zp2sos c chc nng phn tch h thng cho ra ma trn c c LX6Hm tf2zp : xc nh im cc v im khng ca hm truyn2-bin i z ngc% chuong trinh tim bien doi z nguoc cua ham truyen co dang phuong trinh(*)num=[2 5 9 5 3];den=[5 45 2 1 1];[x,t]=impz(num,den)%ket qua chuong trinhx =

1.0e+007 *

0.0000-0.00000.0000-0.00000.0002


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-0.00180.0161-0.14391.2887

t =

01234567

8 Nx: hm impz: xc nh p ng xung ca h thng3-biu din hm truyn theo bin s trng thi% tim ma tran trang thai cua ham truyennum=[-11/6 3/2 -7/12 1/12];den=[3 1.5 1 0.5];[A B C D]=tf2ss(num,den)% ket qua thuc hienA =

-0.5000 -0.3333 -0.16671.0000 0 0

0 1.0000 0

B =

100

C =

0.8056 0.0093 0.1296 Nx: hm tf2ss: tm cc ma trn trng thi

III-P DNG1-vit chng trnh matlab tnh v hin th cc im cc v im khng, v gin im cc v im khng ca 1 bin i z c dng phn thc hu t ca z-1.dng


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chng trnh phn tch bin i z ca h thng c hm truyn sau:1 2 3 4

1 2 3 4

0,9 0,7 0,5 0,3( )1 0,3 0,5 0,7 0,9

z z z z H z

z z z z

+ + +=+ + +

% chuong trinhnum=[0.9 0.7 -0.5 0.3 1]den=[1 0.3 -0.5 0.7 0.9]zplane(num,den)[z,p,k]=tf2zp(num,den);sos=zp2sos(z,p,k)%ket quanum =

0.9000 0.7000 -0.5000 0.3000 1.0000

den =

1.0000 0.3000 -0.5000 0.7000 0.9000

sos =

0.9000 1.8655 1.0900 1.0000 1.7115 0.82571.0000 -1.2950 0.9175 1.0000 -1.4115 1.0900

2-nu cc im cc v im khng ca bin i z cho th c th tm li c bithc ca bin i z bng cch dng lnh sau :[num,den]=zp2tf(z,p,k)


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Hy vit chng trnh matlab tnh v hin th bin i z ca cc im 0: z1=0,3;z2=2,5; z3=-0,2+j0,4; z4=-0,2-j0,4V cc im cc : p1=0,5; p2=-0,75; p3=0,6+j0,7; p4=0,6-j0,7Vi k=3,9% (ap dung)chuong trinh tim bien doi z tu diem cuc va diem khonh

z=[0.3;2.5;-0.2+j*0.4;-0.2-j*0.4] p=[0.5;-0.75;0.6+j*0.7;0.6-j*0.7]k=3.9[num,den]=zp2tf(z,p,k)%ket qua chuong trinhz =

0.30002.5000-0.2000 + 0.4000i-0.2000 - 0.4000i

p =

0.5000-0.75000.6000 + 0.7000i0.6000 - 0.7000i

k =

3.9000

num =

3.9000 -9.3600 -0.6630 -1.0140 0.5850

den =

1.0000 -0.9500 0.1750 0.6625 -0.31873-tm cc ma trn trng thi ca cc hm truyn sau:

a/1 2

1 2 3

1 4,2 0,8( )1 2,5 3

z z H z

z z z

+ += +

%(ap dung) tim ma tran trang thai cua ham truyennum=[1 4.2 0.8 0];den=[1 -2.5 3 -1];[A B C D]=tf2ss(num,den)


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% ket qua chuong trinhA =

2.5000 -3.0000 1.00001.0000 0 0

0 1.0000 0

B =

100

C =

6.7000 -2.2000 1.0000

D =

1

b/3 2

3 2

8 16 12 9( )1,6 1,1 0,3

z z z H z

z z z

+ + +=+ + +

%(ap dung) tim ma tran trang thai cua ham truyennum=[9 12 16 8];den=[3 1.1 1.6 1];[A B C D]=tf2ss(num,den)% ket qua thuc hienA =

-0.3667 -0.5333 -0.33331.0000 0 0

0 1.0000 0

B =100

C =


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2.9000 3.7333 1.6667

D =

3V- NHN XT V KT LUNBi thc hnh nghin cu v bin di z cua dy x[n], v v xc dnh cc im ccim khng hiu c cc hm s dng trong chng trnh d gii quyt bi ton

BI 6: THIT K CC B LCI-L THUYT1- Cc ch tiu thit k b lc2-thit k b lc IIR 3-Thit k b lc FIR

II-THC HNH1-thit k b lc IIR Thit k mch lc elliptic va butterworth thng thp vi cc quy nh sau:

Tn s mp ca di thng f p=800hz, mp ca di chn f s=1khz , mp m ca dithng 0,5Db, suy gim cc tiu ca di chn 40dB v tn s ly mu F=4khz

%thiet ke bo loc elliptic[N,Wn]=ellipord(0.4,0.5,.05,40);[b,a]=ellip(N,0.5,40,Wn);[h,omega]=freqz(b,a,256);subplot(1,2,1);

plot(omega/pi,20*log10(abs(h)));grid;xlabel('omega/pi');ylabel('bien do(db)');title('mach loc IIR elliptic');%thiet ke mach loc[N,Wn]=buttord(0.4,0.5,.05,40);[b,a]=butter(N,Wn);[h,omega]=freqz(b,a,256);subplot(1,2,2); plot(omega/pi,20*log10(abs(h)));

grid;xlabel('omega/pi');ylabel('bien do(db)');title('mach loc IIR butterworth');kt qu chng trnh


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Nx: thit k b lc IIR ta can c lng bc ca b lc , xc nh bc v xc nh ng tn s ca b lc

thit k b lc thng di s dng php bin i song song tuyn tnh thik b lc chebychev loi 1 bc 5, di thng 0,1-0,5hz tn s ly mu 2hz. C

tn s ct l1 5

= 2 =

%tk bo loc thong dai su dung phep bien doi song song tuyen tinh[z,p,k]=cheb1ap(5,3);[A,B,C,D]=zp2ss(z,p,k);fs=2;u1=2*fs*tan(0.1*(2*pi/fs)/2);u2=2*fs*tan(0.5*(2*pi/fs)/2);Bw=u2-u1;W0=sqrt(u1*u2);[At,Bt,Ct,Dt]=lp2bp(A,B,C,D,W0,Bw);[Ad,Bd,Cd,Dd]=bilinear(At,Bt,Ct,Dt,2,0.1);[bz,az]=ss2tf(Ad,Bd,Cd,Dd);[h,f]=freqz(bz,az,256,2);semilogy(f,abs(h));grid;xlabel('frequency');title('dap ung bien do');kt qu chay chng trnh


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2-thit k b lc FIR thit k b lc FIR bng phng php ca s%thiet ke bo loc fir wp=0.2*pi;ws=0.3*pi;As=50; bw=ws-wp; N=ceil((As-7.95)/(2.285*bw))+1;n=0:(N-1); beta=0.5842*(50-21)^0.4+0.07886*(50-21);wc=(ws+wp)/2;h=(wc/pi)*sinc(wc*(n-N/2)).*besseli(0,beta*sqrt(1-4*((n-N/2)/(N/2)).^2))/besseli(0,beta);stem(n,h)title('dap ung xung don vi cua bo loc FIR cua so kaiser'); pause b=h;a=[1];freqz(b,a,500,1000);title('dap ung tan so cua bo loc FIR cua so kaiser') pauset=0:0.001:0.1;x=sin(2*pi*30*t)+sin(2*pi*450*t);y=filter(b,a,x);subplot(2,1,1); plot(t,x);title('tin hieu gom 2 tan so:30hz va 450hz');subplot(2,1,2); plot(t,y);title('tin hieu da loc');xlabel('thoi gian(s)');


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thit k b lc nhiu di tn:%tk bo loc FIR nhieu dai tann=129;f=[0 0.3 0.5 0.7 0.9 1];a=[0 0.5 0 1 0];up=[0.00 0.51 0.03 1.02 0.05];lo=[-0.005 0.49 -0.03 0.98 -0.05]; b=fircls(n,f,a,up,lo);[hh,ff]=freqz(b,1,512,2); plot(ff,abs(hh),'b-');grid;xlabel('tan so chuan hoa');ylabel('bien do');


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thit k b lc FIR c p ng tn s ty chn%tk bo loc FIR co dap ung tan so tuy chon b=cfirpm(38,[-1 -0.5 -0.4 0.3 0.4 0.8],{'multiband',[5 1 2 2 2 1]},[1 10 5]);

[hh,ff]=freqz(b,1,512,2,'whole'); plot(ff,abs(hh),'b-');grid;xlabel('tan so chuan hoa');ylabel('bien do');

II-P DNG1- thit k b lc IIR Thit k b lc thng di dng hm cheby2 vi cc yu cu nh sau:-gii hn ca dy chn di :0,3 -gii hn ca dy chn trn : 0,6 - suy hao di chn :50dB- cc gii hn trn v di ca di thng: 0,4 v 0,5 - gn sng di thng :0,5dB[z,p,k]=cheb2ap(10,0.5);[A,B,C,D]=zp2ss(z,p,k);As=50;u1=2*As*tan(0.4*pi*(2*pi/As)/2);

u2=2*As*tan(0.5*pi*(2*pi/As)/2);Bw=u2-u1;W0=sqrt(u1*u2);[At,Bt,Ct,Dt]=lp2bp(A,B,C,D,W0,Bw);[Ad,Bd,Cd,Dd]=bilinear(At,Bt,Ct,Dt,2,0.1);[bz,az]=ss2tf(Ad,Bd,Cd,Dd);[h,f]=freqz(bz,az,256,2);semilogy(f,abs(h));


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grid;xlabel('frequency');title('dap ung bien do');%ket qua

2-thit k b lc FIR a-bng phng php ca sthit k b lc FIR s , thng thp pha tuyn tnh s dng phng php ca kaiser vi cc yu cu thit k : tn s ly mu 10khz, tn s gii hn giithng 1,5khz ,tn s gii hn gii chn 2khz , suy hao gii thng 0,1Db ,suyhao gii chn 80dB,. Tnh bc ca b lc%thiet ke bo loc fir

wp=1,5;ws=2;Ap=0.1;As=80; bw=ws-wp; N=ceil((As-7.95)/(2.285*bw))+1;n=0:(N-1); beta=0.5842*(50-21)^0.4+0.07886*(50-21);wc=(ws+wp)/2;h=(wc/pi)*sinc(wc*(n-N/2)).*besseli(0,beta*sqrt(1-4*((n-N/2)/(N/2)).^2))/besseli(0,beta);

stem(n,h)title('dap ung xung don vi cua bo loc FIR cua so kaiser'); pause b=h;a=[1];freqz(b,a,500,1000);title('dap ung tan so cua bo loc FIR cua so kaiser') pause


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t=0:0.001:0.1;x=sin(2*pi*30*t)+sin(2*pi*450*t);y=filter(b,a,x);subplot(2,1,1); plot(t,x);

title('tin hieu gom 2 tan so:30hz va 450hz');subplot(2,1,2); plot(t,y);title('tin hieu da loc');xlabel('thoi gian(s)');figure;

b- FIR nhiu di tnthit k b lc FIR c dp ng tn s c m t nh sau- di tn t -1 -0,8: bin gim t 5 xung 2 trng s bng 1- di tn t -0,70,5:bin bng 2 trng s bng 5

- di tn t -0,4

-0,1 : bin gim t 2 xung 1 trong s bng 1- di tn t 0,50,7 : bin bng 2 trng s bng 5- di tn t 0,81: bin tng t 2 ln 5, trng s bng 1

%tk bo loc FIR co dap ung tan so tuy chon b=cfirpm(38,[-1 -0.8 -0.7 0.5 -0.4 -0.1 0.1 0.4 0.5 0.7 0.8 1],{'multiband',[5 22 2 2 1 1 2 2 2 2 5]},[1 5 1 1 5 1]);[hh,ff]=freqz(b,1,512,2,'whole'); plot(ff,abs(hh),'b-');grid;xlabel('tan so chuan hoa');ylabel('bien do');

BI 7: M PHNG M HNH H THNG VIN THNGI-L THUYTBi th nghim ny tm hiu phng php m hnh ha mt h thng thng tinn gin nh hnh 7.1


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m ha nguno nn theo -lawo lng t hao m sa saio m hammingo m BCHo m tch chpo iu cho AM/FMo ASK o QPSK o

II-THC HNHTrong phn ny chng ta s thc hin m hnh 1 h thng truyn thng r

rc nh hnh 7.2

Truyn dn

Thng tintruyn

M hanun

M sasai iu ch

Knhtruyn

Thu nhnThng tinnhn

Gii mngun

Gii msa sai

Gii iuch


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Knhtruyn

(nhiugaussian)

iu chQASK

M haBCH

Tn hiudigital

Gii iuch QASK

Gii m

BCH

Tn hiudigital thu

t


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Hnh 7.2: m hnh h thng truyn tin ri rc%Tao tin hieu nguon

k=11msg=randint(k*2,1)subplot(4,1,1)stem(msg,'.')ylabel('digital message')%ma hoa bchn=15code=encode(msg,n,k,'bch')subplot(4,1,2)stem(code,'.')ylabel('bch')

%dieu che qask su dung bo dieu che m-qask m=16fd=1fc=10fs=30modu=dmod(code,fc,fd,fs,'qask',m)subplot(4,1,3) plot(modu)ylabel('qask')%kenh truyen co nhieustd_value=0.1modu_noise=modu+randn(length(modu),1)*std_value%giai dieu che qask demo=ddemod(modu_noise,fc,fd,fs,'qask',m)%giai ma bchmsg_r=decode(demo,n,k,'bch')subplot(4,1,4)stem(msg_r,'.')ylabel('received message')


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III-P DNGThc hin m hnh digital communications nh m hnh sau:


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Knhtruyn

(nhiugaussian)

iu chFSK

M haBCH

msg

Gii iuch FSK

Gii mBCHMsg thu

%Tao tin hieu nguonk=11;msg=randint(k*2,1);subplot(4,1,1);stem(msg,'.');ylabel('digital message');%ma hoa bchn=15;code=encode(msg,n,k,'bch');subplot(4,1,2);


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stem(code,'.');ylabel('bch');%dieu che fsk su dung bo dieu che m-fsk m=16;fd=1;

fc=10;fs=30;modu=fskmod(code,fc,fd,fs,'fsk',m);subplot(4,1,3); plot(modu);ylabel('qask');%kenh truyen co nhieustd_value=0.1;modu_noise=modu+randn(length(modu),1)*std_value;%giai dieu che fsk demo=fskdemod(modu_noise,fc,fd,fs,'fsk',m);

%giai ma bchmsg_r=decode(demo,n,k,'bch');subplot(4,1,4);stem(msg_r,'.');ylabel('received message');

IV-NHN XT V KT LUNBi thc hnh tm hiu v phng php m hnh ha 1 h thng thng tin.

Nguyn tc chung v th t ca m hnh l l to tn hiu ngun, m ha tn hiu, diuch tn hiu v gii iu ch, knh truyn nhiu cui cng l gii m ha thu c hiu

Bi thc hnh cn gip ta tm hiu c cc hm v nguyn tc s dng cc hm

KT LUN CHUNG: t thc tp ny gip ta bc u lm quen vi phnmm matlab , hiu c tm quan trng ca tm quan trng ca vic m phng ccchng trnh trn phn mm ny. Hiu c tm quan trng trong vic bin i tnhiu sang s( c rt nhiu tin ch nhng quan trng l c th kh nhiu v ti iuch c m n gin)

Cc kt qu thu c sau khi chy chng trnh m phng dng theo lthuyt hc

Kh khn: do bc du mi lm quen vi phn mm v cch m phng nnn c nhng vn vn cn khc mc nhng vi s gip tn tnh ca cc thyc em gii quyt c mt s vn

sachtructuyen.info_bai tap su ly tin hieu so tren matlab

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