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1 - Sendo f(t)=sem(wt), encontrar a transfomada Z[f[n]]

Z[f[n]]= z [ ejwn-e-jwn2j ]
= 12 [ejwn-e-jwnj].Z
Z[n] = n=o∞ejwn . z-n
Z[n] = ejw0 . z0 + ejw1 . z-1 + ejw2 . z-2 + ejw3 . z-3 ...
Z[n] = (ejw)0 . ( z-1)0 + (ejw)1 . ( z-1)1 + (ejw)2 . ( z-1)2 + (ejw)3 . ( z-1)3 ...
= a1-r a=1, r=ejw . z-1, = 11- ejw . z-1 = 11-ejwnz = 1z- ejwnz = zz-ejwn

Z[n] = n=o -∞e-jwn . z-n
Z[n] = n=o - ∞e-jwn . z-n
Z[n] =(-e-jw0 . z0 )+(- e-jw1 . z-1 )+(- e-jw2 . z-2 )+(- e-jw3 . z-3 ) ...
Z[n] = (-1)(e-jw)0 . ( z-1)0 + (-1)(ejw)1 . ( z-1)1 + (-1)(ejw)2 . ( z-1)2 +
(-1)(-ejw)3 . (z-1)3 ...

a1-r a=-1, r=-e-jw . z-1, = -11- ejw . z-1 = -11-e-jwnz = -1z- e-jwnz = -zz- e-jwn

Z[n] =12j [ zz- ejwn - -zz-e-jwn ]
Z[n] =12j [z-e-jw- (- z-ejw)z2-ze-jw-ze-jw+e-jw.ejw]
Z[n] =12j[ z(ejw-e-jw)z2-2zejw+e-jw+1]
Z[n] =12j[zsen(jw)z2-2zcosjw+1]

2 - Sendo f(t)=cos(wt), encontrar a transfomada Z[f[n]]
Z[f[n]]= z [ ejwn+e-jwn2 ]
= 12 [ejwn+e-jwn].Z
Z[n] = n=o∞ejwn . z-n
Z[n] = ejw0 . z0 + ejw1 . z-1 + ejw2 . z-2 + ejw3 . z-3 ...
Z[n] = (ejw)0 . ( z-1)0 + (ejw)1 . ( z-1)1 +(ejw)2 . ( z-1)2 + (ejw)3 . ( z-1)3 ...
= a1-r a=1, r=ejw . z-1, = 11- ejw . z-1 = 11-ejwnz = 1z- ejwnz = zz- ejwn

Z[n] = n=o ∞e-jwn . z-n
Z[n] = n=o ∞e-jwn . z-n
Z[n] =e-jw0 . z0 + e-jw1 . z-1 + e-jw2 . z-2 + e-jw3 . z-3 ...
Z[n] = (e-jw)0 . ( z-1)0 + (ejw)1 . ( z-1)1 + (ejw)2 . ( z-1)2 +(-ejw)3 . (z-1)3 ...

a1-r a=1, r=e-jw . z-1, = 11-ejw . z-1 = 11-e-jwnz = 1z- e-jwnz = zz- e-jwn

Z[n] =12 [ zz- ejwn + zz- e-jwn ]
Z[n] =12 [ze-jw+ z(ejw)z2-ze-jw-zejw+e-jw.ejw]
Z[n] =12[ z(ejw+e-jw)z2-2zejw+e-jw+1]
Z[n] =12[zcos(jw)z2-2zcosjw+1]

01 y’’(t) + 2y’(t) + 4y(t) = 0 y(0)=1 , y’(0)=1
Y =-b±b2-4ac2a = -2±22-4.1.42.2 = -2±4-164 = x’=-1-3 , x’’=-1-3
Suloção geral.
Y=e(-1∓3)
Y= e-x(c1cos3.x+(c2sen3.x))Y(0) = 1 = e0(c1cos3.0+c2sen3.0)
1= 1(c1 +0)
C1=1
Y= e-x(c1cos3.x+(c2sen3.x))
Y’= -e-x(1cos3.x+c2sen3.x+e-x-3 sen3.x+(c23.xcos(3.x)))
Y(0) = 1 = -e-0(cos (0)+0)+ e-0(0+c23))
Y(0) = 1 = -1 + c23
c23=2
c2=23
Yg(t)= e-x(cos3.x+23sen3.x

fplot('(1*exp(-x))*(cos(x*sqrt(3))+((2/sqrt(3))*sin(x*sqrt(3))) )',[-5,2])
grid
xlabel('tempo')
ylabel('amplitude')title('exercicio 01')

05 y’’(t) + 4y’(t) + 5y(t) = 0 , y(0)=1 , y’(0)=0
Y=e(-2∓1i)
Y(t)= e-2x(c1cosx+c2senx)
Y(0) = 1 = e0(c1cos0+c2sen0)
C1= 1
Y(t)= e-2x(c1cosx+c2senx)
Y’(0)= -2e-2x1cosx+c2senx+e-2x-c1senx+c2cosx
Y’(0)= 0 = -2e0(1)+ 1 (c2)
C2= 2
Yg(t)= e-2x(cosx+2senx

fplot('(1*exp(-2*x)*(cos(x)+2*sin(x)))' ,[0,4])
grid
xlabel('tempo')
ylabel('amplitude')
title('exercicio05')


02 6y’’(t) + 5y’(t) + 5y(t) = 0 , y(0)=4 , y’(0)=0
Y=e(-512∓9512)
Y(t)= e-512x(c1cos9512x+c2sen9512x)
Y(0)= 4 =e0 (c1cos0+c2sen0)
Y(0)= 4 = 1(c1 + 0 )
C1 = 4
Y’(t)=-512e-512x(4cos9512x+c2sen9512x) +
e-512x (-49512sen9512x+9512c2cos9512x)
Y(0)=0 = -5124+ 1 (c29512x)
C2 =20129512 = 2012 . 1295
C2 = 2095

Yg(t)= e-512x(4cos9512x+2095sen9512x)fplot('((1*exp(-(5/12)*x)*((4*cos((sqrt(95)/12)*x)+sin((sqrt(95)/12)*x)))))' ,[0,15])
grid
xlabel('tempo')
ylabel('amplitude')
title('exercicio 02')

Questoes aula dia 22/05
Encontre y[n] para a equaçao diferencial
* Md2y(t)dt2 + Ddy(t)dt +kyt=f(t)

Ms2ys+ Dsys+ ky(s)=f(s)
yzMz2+ Dz+ k=fzyzfz =1Mz2+ Dz+ k
s=(z-1ts)
ts=1
yzfz =1M(z-1ts)2+ D(z-1ts)+ k = 1M(z2-2z+1)+ D(z-1)+ k = 1Mz2-2Mz+1M+ Dz-1D+ k =1Mz2+z(2M+D)+ M-D+ k = 1Mz2+z(-2M+D)+ M-D+ k .M = 1/Mz2+z(-2M+D)M+ KM-DM+ 1 . z-2z-2
1M.z-2z2.z-2+z-2M+DM.z-2+ KM-DM+ 1.z-2 = 1M.z-21+z-1(-2M+D)M+ KM-DM+ 1.z-2
1M.z-21+z-1(-2+DM)+ KM-DM+ 1.z-2
yzz-2 KM-DM+ 1+ z-1-2+DM+1= 1M.z-2f(z)
z-1{z-2 KM-DM+ 1yz+...
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