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Ex7041: FDT for RLC circuit

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Ex7041: FDT for RLC circuit
Ex7041: FDT for RLC circuit
Submitted by: Afik Shachar
The problem:
An electrical circuit has in series components with capacitance C, inductance L, resistance R and
a voltage source V0 cos ωt with frequency ω.
(a) Identify the responsefunction αQ (ω) = hQ(ω)i/( 21 V0 ) . Use this to write the energy dissipation
rate.
(b) Use the fluctuation dissipation relation to identify the Fourier transform ΦQ (ω) of the charge
correlation function. Evaluate hQ2 (t)i and compare with the result from equipartition.
(c) Evaluate the current fluctuations hI 2 (t)i and compare with the result from equipartition.
BT
Under what conditions does one get Nyquist’s result hI 2 iω1 ↔ω2 = 2kπR
(ω2 − ω1 ) ?
R∞
R
2
∞
ω dω/2π
dω/2π
1
1
Hint: −∞ (ω2 −ω
2 )2 +γ 2 ω 2 = 2γω 2 ,
−∞ (ω 2 −ω 2 )2 +γ 2 ω 2 = 2γ .
0
0
0
The solution:
(a)
this RLC circuit is discribed by the equation:
LhQ̈i + RhQ̇i +
1
hQi = V0 cos (ωt)
C
applying Fourier transform:
r 1
π
−LΩ hQ(Ω)i − iΩhQ(Ω)i + hQ(Ω)i = V0
δ(Ω − ω) + δ(Ω + ω)
C
2
p V0 π2 δ(Ω − ω) + δ(Ω + ω)
hQ(Ω)i = χ(Ω)F T [V (t)] =
1
2 − iΩR
−
LΩ
C
r
r
π
1
π C1 − Lω 2 + iωR
hQ(ω)i = V0
= V0
2 C1 − Lω 2 − iωR
2 1 − Lω 2 2 + ω 2 R2
C
1
2 + iωR
hQ(ω)i √
C − Lω
αQ (ω) = 1
= 2π
1
2 2 + ω 2 R2
2 V0
C − Lω
1
− Lω 2 + iωR
αQ (ω)
C
χ(ω) = √
=
1
2 2 + ω 2 R2
2π
C − Lω
2
and the energy dissipation rate (from lecture notes, sec. 12.2, eq. 619):
1
1
Im [χ(ω)]
1
Ẇ = η(ω)V02 ω 2 = ω 2 V02
= V02
2
2
ω
2
we denote γ =
Ẇ =
R
L
and ω02 =
1
CL
1
γV02 ω 2
2L ω − ω 2 2 + γ 2 ω 2
0
1
1
C
ω2R
2
− Lω 2 + ω 2 R2
(b)
ΦQ (ω) = F T [hQ(0)Q(τ )i]
from FDT:
1
Im [χ(ω)]
=
tanh
ω
~ω
~ω
2kB T
ΦQ (ω)
for ~ω << kB T :
Im [χ(ω)]
1
=
ΦQ (ω)
ω
2kB T
ΦQ (ω) =
again γ =
R
L
1
L
1
CL
2kB T R
L
2
2
2
−ω
+ ω2 R
L2
and ω02 =
ΦQ (ω) =
1
CL
1
2kB T γ
L ω2 − ω2 2 + ω2γ 2
0
and we recall that:
Z
∞
hQ(0)Q(τ )i =
ΦQ (ω) exp−iωτ
−∞
dω
2π
therefore:
Z
∞
dω
2kB T γ
hQ (t)i =
ΦQ (ω)
=
2π
L
−∞
2
Z
∞
−∞
ω2
dω/2π
2kB T γ
=
2
2
2
2
2γω02 L
− ω0 + ω γ
hQ2 (t)i = CkB T
from equipartition we have
1
1 Q2
1
CV 2 =
= kB T
2
2 C
2
we then get the same result:
2 Q (t) = CkB T
(c)
I = Q̇
I(ω) = −iωQ(ω)
therefore
1
2kB T γω 2
L ω2 − ω2 2 + ω2γ 2
0
Z ∞
Z
dω
2kB T γ ∞
ω 2 dω/2π
2kB T γ
2
hI (t)i =
ΦI (ω)
=
=
2
2π
L
2γL
−∞
−∞ ω 2 − ω02
+ ω2γ 2
ΦI (ω) = ω 2 ΦQ (ω) =
2
kB T
L
from equipartition we have
1 2
1
LI
= kB T
2
2
hI 2 (t)i =
we then get the same result:
kB T
I 2 (t) =
L
to get Nyquist’s result, we look again at ΦI (ω)
ΦI (ω) =
1
L
ω2
2kB T γ
ω2 2
1 − ω02 + γ 2
it can be seen that for ω 1 −
ΦI (ω) =
ω02
ω2
<<
R
L,
this becomes
2kB T
R
which in turn gives
2
I ω1 ↔ω2 = 2
Z
ω2
ω1
2kB T
dω
=
ΦI (ω)
2π
πR
Z
ω2
dω
ω1
2
2kB T
I ω1 ↔ω2 =
(ω2 − ω1 )
πR
3
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