ich rechne hier mal gerade etwas über rfid
ich habe eine Paralellschaltung aus eine Stromquelle, einem Kondensator, einer Spule, und einem Wiederstand.
Dann habe ich einen Zweiten Schwingkeis der genauso aufgebaut ist wie der erste, jedoch keine Stromquelle enthällt und ggfs. andere
Werte für Wiederstand, Kapazität und Spule hat.
Die Schwingkeise seinen über die magnetfelder ihrenen Spulen gekoppelt.
Beide Schwingkreise haben für sich betrachtet je eine Resonanzfrequenz. Diese beiden Frequenzen sollen nicht alzu stark abweichen.
Jetzt soll die Frequenz der Stromquelle in der nähe der Resonanzfrequenz varriert werden. Gesucht ist die Stromamplitute im zweiten Schwingkreis.
Was wissen wir?
Widerstand:
U
=
R
⋅
I
{\displaystyle U=R\cdot I}
Kapazität
U
=
Q
C
{\displaystyle U={\frac {Q}{C}}}
Induktion
U
=
A
⋅
n
⋅
cos
(
β
)
d
B
d
t
{\displaystyle U=A\cdot n\cdot \cos(\beta ){\frac {\mathrm {d} B}{\mathrm {d} t}}}
Magnetfeld eine Spule wir sind im Fernfeld, daher quadratischer Abfall, aber Konstante $\kappa$ unbekannt.
B
=
κ
n
I
r
2
{\displaystyle B={\frac {\kappa nI}{r^{2}}}}
Selbstinduktion
U
=
L
d
I
d
t
{\displaystyle U=L{\frac {\mathrm {d} I}{\mathrm {d} t}}}
Es kommt hier zusätzlich zum Selbstinduktionsterm ein Kopplungsterm hinzu.
U
L
1
=
L
d
I
L
1
d
t
+
A
1
n
1
cos
(
β
)
κ
n
2
r
2
d
I
L
2
d
t
{\displaystyle U_{\mathrm {L1} }=L{\frac {\mathrm {d} I_{\mathrm {L1} }}{\mathrm {d} t}}+A_{1}n_{1}\cos(\beta ){\frac {\kappa n_{2}}{r^{2}}}{\frac {\mathrm {d} I_{\mathrm {L2} }}{\mathrm {d} t}}}
Die konstanten fassen wir zur vereinfachung der Schreibweise zusammen
ψ
1
=
A
1
n
1
cos
(
β
)
κ
n
2
r
2
{\displaystyle \psi _{1}=A_{1}n_{1}\cos(\beta ){\frac {\kappa n_{2}}{r^{2}}}}
Also
U
L
1
=
L
d
I
L
1
d
t
+
ψ
1
d
I
L
2
d
t
{\displaystyle U_{\mathrm {L1} }=L{\frac {\mathrm {d} I_{\mathrm {L1} }}{\mathrm {d} t}}+\psi _{1}{\frac {\mathrm {d} I_{\mathrm {L2} }}{\mathrm {d} t}}}
Einbau in die Gesamtkonstuktion:
I
g
e
s
1
=
I
C
1
+
I
R
1
+
I
L
1
=
I
1
s
⋅
exp
(
i
ω
t
)
U
g
e
s
1
=
U
C
1
=
U
R
1
=
U
L
1
U
g
e
s
1
=
Q
C
1
1
C
=
R
1
⋅
I
R
1
=
L
1
d
I
L
1
d
t
+
ψ
1
d
I
L
2
d
t
d
U
g
e
s
d
t
=
I
C
1
1
C
1
=
R
1
⋅
d
I
R
1
d
t
=
L
1
d
2
I
L
d
t
2
+
ψ
1
d
2
I
L
2
d
t
2
I
1
s
⋅
sin
(
ω
t
)
=
C
1
(
L
1
d
2
I
L
1
d
t
2
+
ψ
1
d
2
I
L
2
d
t
2
)
+
1
R
1
(
L
1
d
I
L
1
d
t
+
ψ
1
d
I
L
2
d
t
)
+
I
L
1
{\displaystyle {\begin{matrix}I_{\mathrm {ges1} }&=&I_{\mathrm {C1} }+I_{\mathrm {R1} }+I_{\mathrm {L1} }=I_{1s}\cdot \exp(i\omega t)\\U_{\mathrm {ges1} }&=&U_{\mathrm {C1} }=U_{\mathrm {R1} }=U_{\mathrm {L1} }\\U_{\mathrm {ges1} }&=&Q_{\mathrm {C1} }{\frac {1}{C}}=R_{1}\cdot I_{\mathrm {R1} }=L_{1}{\frac {\mathrm {d} I_{\mathrm {L1} }}{\mathrm {d} t}}+\psi _{1}{\frac {\mathrm {d} I_{\mathrm {L_{2}} }}{\mathrm {d} t}}\\{\frac {\mathrm {d} U_{\mathrm {ges} }}{\mathrm {d} t}}&=&I_{\mathrm {C1} }{\frac {1}{C_{1}}}=R_{1}\cdot {\frac {\mathrm {d} I_{\mathrm {R1} }}{\mathrm {d} t}}=L_{1}{\frac {\mathrm {d^{2}} I_{\mathrm {L} }}{\mathrm {d} t^{2}}}+\psi _{1}{\frac {\mathrm {d^{2}} I_{\mathrm {L_{2}} }}{\mathrm {d} t^{2}}}\\I_{1s}\cdot \sin(\omega t)&=&C_{1}\left(L_{1}{\frac {\mathrm {d^{2}} I_{\mathrm {L_{1}} }}{\mathrm {d} t^{2}}}+\psi _{1}{\frac {\mathrm {d^{2}} I_{\mathrm {L_{2}} }}{\mathrm {d} t^{2}}}\right)+{\frac {1}{R_{1}}}\left(L_{1}{\frac {\mathrm {d} I_{\mathrm {L_{1}} }}{\mathrm {d} t}}+\psi _{1}{\frac {\mathrm {d} I_{\mathrm {L_{2}} }}{\mathrm {d} t}}\right)+I_{\mathrm {L_{1}} }\end{matrix}}}
Analog erhalten wir
0
=
C
2
(
L
2
d
2
I
L
2
d
t
2
+
ψ
2
d
2
I
L
1
d
t
2
)
+
1
R
2
(
L
2
d
I
L
2
d
t
+
ψ
2
d
I
L
1
d
t
)
+
I
L
2
{\displaystyle 0=C_{2}\left(L_{2}{\frac {\mathrm {d^{2}} I_{\mathrm {L_{2}} }}{\mathrm {d} t^{2}}}+\psi _{2}{\frac {\mathrm {d^{2}} I_{\mathrm {L_{1}} }}{\mathrm {d} t^{2}}}\right)+{\frac {1}{R_{2}}}\left(L_{2}{\frac {\mathrm {d} I_{\mathrm {L_{2}} }}{\mathrm {d} t}}+\psi _{2}{\frac {\mathrm {d} I_{\mathrm {L_{1}} }}{\mathrm {d} t}}\right)+I_{\mathrm {L_{2}} }}
Ansatz
I
L
1
=
I
1
exp
(
i
ω
t
)
I
L
2
=
I
2
exp
(
i
ω
t
)
{\displaystyle {\begin{matrix}I_{\mathrm {L_{1}} }&=&I_{1}\exp(i\omega t)I_{\mathrm {L_{2}} }&=&I_{2}\exp(i\omega t)\\\end{matrix}}}
Einsetzen
I
1
s
=
−
C
1
(
L
1
ω
2
I
1
+
ψ
1
ω
2
I
2
)
+
i
1
R
1
(
L
1
ω
I
1
+
ψ
1
ω
I
2
)
+
I
1
{\displaystyle I_{1s}=-C_{1}\left(L_{1}\omega ^{2}I_{1}+\psi _{1}\omega ^{2}I_{2}\right)+i{\frac {1}{R_{1}}}\left(L_{1}\omega I_{1}+\psi _{1}\omega I_{2}\right)+I_{1}}
und
0
=
−
C
2
(
L
2
ω
2
I
2
+
ψ
2
ω
2
I
1
)
+
i
1
R
2
(
L
2
ω
I
2
+
ψ
2
ω
I
1
)
+
I
2
{\displaystyle 0=-C_{2}\left(L_{2}\omega ^{2}I_{2}+\psi _{2}\omega ^{2}I_{1}\right)+i{\frac {1}{R_{2}}}\left(L_{2}\omega I_{2}+\psi _{2}\omega I_{1}\right)+I_{2}}
Ausrechnen:
0
=
−
C
2
L
2
ω
2
I
2
−
C
2
ψ
2
ω
2
I
1
+
i
1
R
2
L
2
ω
I
2
+
i
1
R
2
ψ
2
ω
I
1
+
I
2
{\displaystyle 0=-C_{2}L_{2}\omega ^{2}I_{2}-C_{2}\psi _{2}\omega ^{2}I_{1}+i{\frac {1}{R_{2}}}L_{2}\omega I_{2}+i{\frac {1}{R_{2}}}\psi _{2}\omega I_{1}+I_{2}}
0
=
(
−
C
2
L
2
ω
2
+
i
1
R
2
L
2
ω
+
1
)
I
2
+
(
−
C
2
ψ
2
ω
2
+
i
1
R
2
ψ
2
ω
)
I
1
{\displaystyle 0=\left(-C_{2}L_{2}\omega ^{2}+i{\frac {1}{R_{2}}}L_{2}\omega +1\right)I_{2}+\left(-C_{2}\psi _{2}\omega ^{2}+i{\frac {1}{R_{2}}}\psi _{2}\omega \right)I_{1}}
γ
:=
I
2
I
1
=
−
−
C
2
ψ
2
ω
2
+
i
1
R
2
ψ
2
ω
−
C
2
L
2
ω
2
+
i
1
R
2
L
2
ω
+
1
{\displaystyle \gamma :={\frac {I_{2}}{I_{1}}}=-{\frac {-C_{2}\psi _{2}\omega ^{2}+i{\frac {1}{R_{2}}}\psi _{2}\omega }{-C_{2}L_{2}\omega ^{2}+i{\frac {1}{R_{2}}}L_{2}\omega +1}}}
Einsetzen
I
1
s
I
1
=
−
C
1
(
L
1
ω
2
+
ψ
1
ω
2
γ
)
+
i
1
R
1
(
L
1
ω
+
ψ
1
ω
γ
)
+
1
{\displaystyle {\frac {I_{1s}}{I_{1}}}=-C_{1}\left(L_{1}\omega ^{2}+\psi _{1}\omega ^{2}\gamma \right)+i{\frac {1}{R_{1}}}\left(L_{1}\omega +\psi _{1}\omega \gamma \right)+1}
Ausrechnen:
I
1
s
=
I
1
(
−
C
1
L
1
ω
2
+
−
C
1
ψ
1
ω
2
γ
+
i
1
R
1
L
1
ω
+
i
1
R
1
ψ
1
ω
γ
+
1
)
{\displaystyle I_{1s}=I_{1}\left(-C_{1}L_{1}\omega ^{2}+-C_{1}\psi _{1}\omega ^{2}\gamma +i{\frac {1}{R_{1}}}L_{1}\omega +i{\frac {1}{R_{1}}}\psi _{1}\omega \gamma +1\right)}
I
1
I
1
s
=
1
−
C
1
L
1
ω
2
+
−
C
1
ψ
1
ω
2
γ
+
i
1
R
1
L
1
ω
+
i
1
R
1
ψ
1
ω
γ
+
1
{\displaystyle {\frac {I_{1}}{I_{1s}}}={\frac {1}{-C_{1}L_{1}\omega ^{2}+-C_{1}\psi _{1}\omega ^{2}\gamma +i{\frac {1}{R_{1}}}L_{1}\omega +i{\frac {1}{R_{1}}}\psi _{1}\omega \gamma +1}}}
I
2
I
1
s
=
I
2
I
1
I
1
I
1
s
=
I
1
I
1
s
=
γ
−
C
1
L
1
ω
2
+
−
C
1
ψ
1
ω
2
γ
+
i
1
R
1
L
1
ω
+
i
1
R
1
ψ
1
ω
γ
+
1
{\displaystyle {\frac {I_{2}}{I_{1s}}}={\frac {I_{2}}{I_{1}}}{\frac {I_{1}}{I_{1s}}}={\frac {I_{1}}{I_{1s}}}={\frac {\gamma }{-C_{1}L_{1}\omega ^{2}+-C_{1}\psi _{1}\omega ^{2}\gamma +i{\frac {1}{R_{1}}}L_{1}\omega +i{\frac {1}{R_{1}}}\psi _{1}\omega \gamma +1}}}
Kehrwert vereinfacht das Leben.
I
1
s
I
2
=
−
C
1
L
1
ω
2
γ
−
1
+
−
C
1
ψ
1
ω
2
+
i
1
R
1
L
1
ω
γ
−
1
+
i
1
R
1
ψ
1
ω
+
γ
−
1
{\displaystyle {\frac {I_{1s}}{I_{2}}}=-C_{1}L_{1}\omega ^{2}\gamma ^{-1}+-C_{1}\psi _{1}\omega ^{2}+i{\frac {1}{R_{1}}}L_{1}\omega \gamma ^{-1}+i{\frac {1}{R_{1}}}\psi _{1}\omega +\gamma ^{-1}}
Es gilt:
γ
−
1
=
−
−
C
2
L
2
ω
2
+
i
1
R
2
L
2
ω
+
1
−
C
2
ψ
2
ω
2
+
i
1
R
2
ψ
2
ω
=
−
(
−
C
2
L
2
ω
2
+
i
1
R
2
L
2
ω
+
1
)
(
−
C
2
ψ
2
ω
2
−
i
1
R
2
ψ
2
ω
)
C
2
2
ψ
4
ω
4
+
1
R
2
2
ψ
2
2
ω
2
{\displaystyle \gamma ^{-1}=-{\frac {-C_{2}L_{2}\omega ^{2}+i{\frac {1}{R_{2}}}L_{2}\omega +1}{-C_{2}\psi _{2}\omega ^{2}+i{\frac {1}{R_{2}}}\psi _{2}\omega }}=-{\frac {\left(-C_{2}L_{2}\omega ^{2}+i{\frac {1}{R_{2}}}L_{2}\omega +1\right)\left(-C_{2}\psi _{2}\omega ^{2}-i{\frac {1}{R_{2}}}\psi _{2}\omega \right)}{C_{2}^{2}\psi _{4}\omega ^{4}+{\frac {1}{R_{2}^{2}}}\psi _{2}^{2}\omega ^{2}}}}
γ
−
1
=
−
C
2
2
ψ
2
L
2
ω
4
+
C
2
ψ
2
ω
2
−
1
R
2
2
ψ
2
L
2
ω
2
+
i
C
2
L
2
ω
3
1
R
2
ψ
2
+
i
1
R
2
ψ
2
ω
−
i
C
2
ψ
2
ω
3
1
R
2
L
2
C
2
2
ψ
2
2
ω
4
+
1
R
2
2
ψ
2
2
ω
2
{\displaystyle \gamma ^{-1}={\frac {-C_{2}^{2}\psi _{2}L_{2}\omega ^{4}+C_{2}\psi _{2}\omega ^{2}-{\frac {1}{R_{2}^{2}}}\psi _{2}L_{2}\omega ^{2}+iC_{2}L_{2}\omega ^{3}{\frac {1}{R_{2}}}\psi _{2}+i{\frac {1}{R_{2}}}\psi _{2}\omega -iC_{2}\psi _{2}\omega ^{3}{\frac {1}{R_{2}}}L_{2}}{C_{2}^{2}\psi _{2}^{2}\omega ^{4}+{\frac {1}{R_{2}^{2}}}\psi _{2}^{2}\omega ^{2}}}}
a
(
ω
)
:=
−
C
2
2
ψ
2
L
2
ω
4
+
C
2
ψ
2
ω
2
−
1
R
2
2
ψ
2
L
2
ω
2
{\displaystyle a(\omega ):=-C_{2}^{2}\psi _{2}L_{2}\omega ^{4}+C_{2}\psi _{2}\omega ^{2}-{\frac {1}{R_{2}^{2}}}\psi _{2}L_{2}\omega ^{2}}
b
(
ω
)
:=
+
C
2
L
2
ω
3
1
R
2
ψ
2
+
1
R
2
ψ
2
ω
−
C
2
ψ
2
ω
3
1
R
2
L
2
{\displaystyle b(\omega ):=+C_{2}L_{2}\omega ^{3}{\frac {1}{R_{2}}}\psi _{2}+{\frac {1}{R_{2}}}\psi _{2}\omega -C_{2}\psi _{2}\omega ^{3}{\frac {1}{R_{2}}}L_{2}}
c
(
ω
)
:=
C
2
2
ψ
2
2
ω
4
+
1
R
2
2
ψ
2
2
ω
2
{\displaystyle c(\omega ):=C_{2}^{2}\psi _{2}^{2}\omega ^{4}+{\frac {1}{R_{2}^{2}}}\psi _{2}^{2}\omega ^{2}}
γ
(
ω
)
−
1
=
a
(
ω
)
+
i
b
(
ω
)
c
(
ω
)
{\displaystyle \gamma (\omega )^{-1}={\frac {a(\omega )+ib(\omega )}{c(\omega )}}}
I
1
s
I
2
=
−
C
1
L
1
c
(
ω
)
ω
2
a
(
ω
)
−
i
C
1
L
1
c
(
ω
)
ω
2
b
(
ω
)
−
C
1
ψ
1
ω
2
+
i
a
(
ω
)
1
R
1
c
(
ω
)
L
1
ω
−
b
(
ω
)
1
R
1
c
(
ω
)
L
1
ω
+
i
1
R
1
ψ
1
ω
+
a
(
ω
)
+
i
b
(
ω
)
c
(
ω
)
{\displaystyle {\begin{matrix}{\frac {I_{1s}}{I_{2}}}&=&-{\frac {C_{1}L_{1}}{c(\omega )}}\omega ^{2}a(\omega )-i{\frac {C_{1}L_{1}}{c(\omega )}}\omega ^{2}b(\omega )\\&&-C_{1}\psi _{1}\omega ^{2}+ia(\omega ){\frac {1}{R_{1}c(\omega )}}L_{1}\omega -b(\omega ){\frac {1}{R_{1}c(\omega )}}L_{1}\omega \\&&+i{\frac {1}{R_{1}}}\psi _{1}\omega +{\frac {a(\omega )+ib(\omega )}{c(\omega )}}\end{matrix}}}
Interessant ist jedoch nur der Kehrwert des Betrags:
x
:=
a
(
ω
)
c
(
ω
)
−
C
1
L
1
c
(
ω
)
ω
2
a
(
ω
)
−
C
1
ψ
1
ω
2
{\displaystyle x:={\frac {a(\omega )}{c(\omega )}}-{\frac {C_{1}L_{1}}{c(\omega )}}\omega ^{2}a(\omega )-C_{1}\psi _{1}\omega ^{2}}
y
:=
b
(
ω
)
c
(
ω
)
+
1
R
1
ψ
1
ω
+
a
(
ω
)
1
R
1
c
(
ω
)
L
1
ω
−
C
1
L
1
c
(
ω
)
ω
2
b
(
ω
)
{\displaystyle y:={\frac {b(\omega )}{c(\omega )}}+{\frac {1}{R_{1}}}\psi _{1}\omega +a(\omega ){\frac {1}{R_{1}c(\omega )}}L_{1}\omega -{\frac {C_{1}L_{1}}{c(\omega )}}\omega ^{2}b(\omega )}
Es gilt:
I
2
I
1
s
=
1
x
2
+
y
2
{\displaystyle {\frac {I_{2}}{I_{1s}}}={\frac {1}{\sqrt {x^{2}+y^{2}}}}}
Vereinfachte Betrachtung
Bearbeiten
Für unendlichenen Wiederstand und gleiche Kreise
γ
−
1
=
−
C
2
L
2
ω
2
+
1
C
2
ψ
2
ω
2
{\displaystyle \gamma ^{-1}={\frac {-C_{2}L_{2}\omega ^{2}+1}{C_{2}\psi _{2}\omega ^{2}}}}
I
1
s
I
2
=
−
C
1
L
1
ω
2
γ
−
1
+
−
C
1
ψ
1
ω
2
+
γ
−
1
{\displaystyle {\frac {I_{1s}}{I_{2}}}=-C_{1}L_{1}\omega ^{2}\gamma ^{-1}+-C_{1}\psi _{1}\omega ^{2}+\gamma ^{-1}}
Also
I
1
s
I
2
=
L
−
C
L
ω
2
+
1
ψ
−
C
ψ
ω
2
+
−
C
L
ω
2
+
1
C
ψ
ω
2
{\displaystyle {\frac {I_{1s}}{I_{2}}}=L{\frac {-CL\omega ^{2}+1}{\psi }}-C\psi \omega ^{2}+{\frac {-CL\omega ^{2}+1}{C\psi \omega ^{2}}}}
Die kann man vereinfachen zu
I
1
s
I
2
=
−
C
2
ω
4
(
L
2
+
ψ
2
)
+
1
C
ψ
ω
2
{\displaystyle {\frac {I_{1s}}{I_{2}}}={\frac {-C^{2}\omega ^{4}\left(L^{2}+\psi ^{2}\right)+1}{C\psi \omega ^{2}}}}
I
1
s
I
2
=
−
ω
4
(
1
+
k
)
+
ω
0
4
ψ
ω
2
ω
0
2
L
{\displaystyle {\frac {I_{1s}}{I_{2}}}={\frac {-\omega ^{4}\left(1+k\right)+\omega _{0}^{4}}{\frac {\psi \omega ^{2}\omega _{0}^{2}}{L}}}}
from math import *
L=1
C=1
R=30
A=1
wr=sqrt(1/(L*C)-1/(2*R*R*C*C)) #Resonanz
L1=L
L2=L
C1=C
C2=C
R1=R
R2=R
A1=A
A2=A
psi=0.1
psi1=psi
psi2=psi
def a(o):return C2*C2+psi2*L2*pow(o,4)+C2*psi2*pow(o,2)-(1/(R2*R2)*psi2*L2*o*o)
def b(o):return C2*L2*pow(o,3)*psi2/R2+psi2*o/R2-C2*psi2*pow(o,3)*L2/R2
def c(o):return C2*C2*psi2*psi2*pow(o,4)+psi2*psi2*o*o/(R2*R2)
def x(o):return a(o)/c(o)-C1*L1*o*o*a(o)/c(o)-C1*psi1*o*o
def y(o):return b(o)/c(o)+psi1*o/R1+a(o)*L1*o/(R1*c(o))-C1*L1*o*o*b(o)/c(o)
def s(o):return 1.0/sqrt(x(o)*x(o)+y(o)*y(o))
max=1000
delta=1.0
for i in range(max):
dw=(i-max/2.0)*delta*wr/max
w=wr+dw
print w,s(w)
Der Erwartete Doppelpeak konnte nicht beobachtet werden, es ergab sich etwas was einer gewönlichen Resonanzkurve sehr ähnlich sah.