fourier decomposition example
7 mins read ·
Fourier decomposition example: Three terms – constant, cosine, sine.
integral identities x
∫0Lsin(mL2πx)sin(nL2πx) dx={0,21L,m=n,m=n.(1)
∫0Lcos(mL2πx)cos(nL2πx) dx={0,21L,m=n,m=n.(2)
∫0Lcos(mL2πx)sin(nL2πx) dt=0.(3)
∫0Lcos(mL2πx) dt=0.(3b)
∫0Lsin(nL2πx) dt=0.(3c)
coefficients x
f(x)=m=1∑∞amcos[m(L2π)x]+m=1∑∞bmsin[m(L2π)x]+c0(4)
am=L2∫0Lf(x) cos[m(L2π)x] dx(5)
bm=L2∫0Lf(x) sin[m(L2π)x] dx(6)
c0=L1∫0Lf(x) dx.(7)
km=mL2π, m=1,2,3,..(8)
integral identities t
∫0Tsin(mT2πt)sin(nT2πt) dt={0,21T,m=n,m=n.(9)
∫0Tcos(mT2πt)cos(nT2πt) dt={0,21T,m=n,m=n.(10)
∫0Tcos(mT2πt)sin(nT2πt) dt=0.(11)
∫0Tcos(mT2πt) dt=0.(11b)
∫0Tsin(nT2πt) dt=0.(11c)
coefficients t
f(t)=m=1∑∞amcos[m(T2π)t]+m=1∑∞bmsin[m(T2π)t]+c0(12)
am=T2∫0Tf(t) cos[m(T2π)t] dt(13)
bm=T2∫0Tf(t) sin[m(T2π)t] dt(14)
c0=T1∫0Tf(t) dt.(15)
ωm=mT2π, m=1,2,3,..(16)
example
f(t)=2+5cos20πt+3sin8πt.(17)
f(t)==2+5cos(0.12πt)+3sin(0.252πt)2+5cos(1012πt)+3sin(412πt)(18)
consine terms coefficients
- Use Eqns (13), (10), (11) on (18).
am=++=T2∫0T2 cos[m(T2π)t] dtT2∫0T5cos(1012πt) cos[m(12π)t] dtT2∫0T3sin(412πt) cos[m(12π)t] dt0+T221T5δ10,m+0 = 5δ10,m.(19)
am=0,a10=5,am=0,0<m<10,10<m.(19b)
sine terms coefficients
- Use Eqn (14), (9), (11) on (18).
bm=++=T2∫0T2 sin[m(T2π)t] dtT2∫0T5cos(1012πt) sin[m(12π)t] dtT2∫0T3sin(412πt) sin[m(12π)t] dt0+0+T221T3δ4,m = 3δ4,m.(20)
bm=0,b4=3,bm=0,0<m<4,4<m.(20b)
constant term
- Use Eqn (15), (11b), (11c) on (18).
c0=++=T1∫0T2 dtT1∫0T5cos(1012πt) dtT1∫0T3sin(412πt) dt2+0+0 = 2.(21)
function f(t)
f(t)=m=1∑∞amcos[m(12π)t]+m=1∑∞bmsin[m(12π)t]+c0(12)
f(t)==a10cos[10(12π)t]+b4sin[4(12π)t]+c05cos[10(12π)t]+3sin[4(12π)t]+2.(22)
f(t)=2+5cos20πt+3sin8πt.(17)
f1=2 Hz
f(t)=2+5cos20πt+3sin8πt.(17)
f(t)==2+5cos(0.12πt)+3sin(0.252πt)2+5cos(50.52πt)+3sin(20.52πt)(23)
consine terms coefficients
- Use Eqns (13), (10), (11) on (23).
am=++=T1∫0T2 cos[m(0.52π)t] dtT2∫0T5cos(50.52πt) cos[m(0.52π)t] dtT2∫0T3sin(20.52πt) cos[m(0.52π)t] dt0+T221T5δ10,m+0 = 5δ5,m.(24)
am=0,a5=5,am=0,0<m<5,5<m.(24b)
sine terms coefficients
- Use Eqn (14), (9), (11) on (23).
bm=++=T1∫0T2 sin[m(0.52π)t] dtT2∫0T5cos(50.52πt) sin[m(0.52π)t] dtT2∫0T3sin(20.52πt) sin[m(0.52π)t] dt0+0+T221T3δ4,m = 3δ2,m.(25)
bm=0,b2=3,bm=0,0<m<2,2<m.(25b)
constant term
- Use Eqn (15), (11b), (11c) on (23).
c0=++=T1∫0T2 dtT2∫0T5cos(50.52πt) dtT2∫0T3sin(20.52πt) dt2+0+0 = 2.(26)
function f(t)
f(t)=m=1∑∞amcos[m(0.52π)t]+m=1∑∞bmsin[m(0.52π)t]+c0(12)
f(t)==a5cos[5(0.52π)t]+b2sin[2(0.52π)t]+c05cos[5(0.52π)t]+3sin[2(0.52π)t]+2.(27)
f(t)=2+5cos20πt+3sin8πt.(17)
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