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# Integral Transforms (laplace

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Category: Lab Report

Subcategory: Mathematics

Pages: 1

Words: 275

According to formula (L9), and (L10)
L2y”+y’=2Ly”+Ly’=2s2Ys-sy0-y’0+sYs-y0=0Ys2s2+s-2s-1=0Ys=2s+1s(2s+1)=1sAccording to formula (L3)
yt=1
According to formula (L9), and (L10)
L3y”+y-1=3Ly”+Ly-L1=3s2Ys-sy0-y’0+Ys-1s=0Ys3s2+1-3s-1s=0Ys=3s+1s3s2+1=1sAccording to formula (L3)
yt=1

u=Ri+Li’Us=RIs+LsIs-i0=(R+Ls)I(s)I=1R+LsUs=12+3sss2+16=-3/742+3s+(s+24)/74s2+16it=-174e-23t+174cos4t+674sin⁡(4t)

Hs=LsI(s)U(s)=LsR+Ls=3s2+3s

It is 1st order high-pass filter.
The curve is from matlab.
Matlab Command:
bode(tf([3 0],[3,2]))

Hs=3s2+3sht=L-13s2+3s=L-11-22+3s=δt-23e-23t
yt=L-132+3s=e-23t

c0=5 ⇒c0=5, argc0=0c1=2-2j ⇒c1=22,argc1=-π4c2=-3-3j ⇒c2=32,argc2=-3π4c3=4j ⇒c3=4,argc3=π2
Matlab code:
subplot(1,2,1)
bar(0:3,[5 8^.5 18^.5 4], 0.1)
gridtitle(‘magnitude’)
subplot(1,2,2)
bar(0:3,[0 -pi/4 -3*pi/4 pi/2], 0.1)
gridtitle(‘phase’)

Pav=1T0Tf(t)2dt=k=-∞∞ck2=c02+2k=1∞ck2=25+28+18+16=109c02=25The asked ratio is c02Pav=0.2294Hence, 22.94%

2Rec1ejωbt=4cosπt+4sinπt2Rec2ej2ωbt=-6cos2πt+6sin2πt2Rec3ej3ωbt=-8sin3πtIt follows the formula (2.4) that
ft=c0+2Rek=13ckejkωbt=5+4cosπt+4sinπt-6cos2πt+6sin2πt-8sin3πt
Matlab code:
t = 2:0.01:4
plot(t, 5+4*cos(pi*t)+4*sin(pi*t)-6*cos(2*pi*t)+6*sin(2*pi*t)-8*sin(3*pi*t))

gcd(5, 9) = 1
hence, the period of u(t) is 2π.

Hs=4+4s2(1+s)=2Hence the output is
yt=2ut=6+4sin5t-6cos⁡(9t)
c0=6 ⇒c0=6, argc0=0c5=42j=-2j ⇒c5=2,argc5=-π2c9=-62=-3 ⇒c9=3,argc9=π
Matlab code:
subplot(1,2,1)
bar(0:9,[6 0 0 0 0 2 0 0 0 3], 0.1)
gridtitle(‘magnitude’)
subplot(1,2,2)
bar(0:9,[0 0 0 0 0 -pi/2 0 0 0 pi], 0.1)
gridtitle(‘phase’)