241/341 Luftstromeigenschaften von RLC-Glidern

Needs["ErrorBarPlots`"]

Aufgabe 3

Auswertung

Zunächst übernehmen wir die Daten aus dem Heft um daraus das entsprechende Diagramm zur Phase zu generieren. Wir verzichten auf übernahme der gesamten Daten, sondern tragen nur Frequenz und Phase ein. {f,ϕ,Δϕ}

(Subscript[data, 3Hochpass])=
{{1,-18},
{2,-32},
{3,-44},
{4,-51},
{5,-58},
{6,-62},
{7,-65},
{8,-69},
{9,-70},
{10,-72}}

(Subscript[Δdata, 3Hochpass])={0.4,1,2,2,2,3,3,3,3,4}

(Subscript[data, 3Tiefpass])=
{{1,71},
{2,55},
{3,45},
{4,37},
{5,31},
{6,27},
{7,24},
{8,22},
{9,18},
{10,18}}

(Subscript[Δdata, 3Tiefpass])={1,1,2,2,2,2,1,1,2,2}

Nun tragen wir diese Daten in einem logarithmischen Plot auf.

JetPlot1=ListLogLinearPlot[(Subscript[data, 3Hochpass]),Joined->True];

ΔJetPlot1=ListLogLinearPlot[Transpose[{(Subscript[data, 3Hochpass])[[All,1]],(Subscript[data, 3Hochpass])[[All,2]]-(Subscript[Δdata, 3Hochpass])}],Joined->True,PlotStyle->Red];

JetPlot2=ListLogLinearPlot[(Subscript[data, 3Tiefpass]),Joined->True];

ΔJetPlot2=ListLogLinearPlot[Transpose[{(Subscript[data, 3Tiefpass])[[All,1]],(Subscript[data, 3Tiefpass])[[All,2]]-(Subscript[Δdata, 3Tiefpass])}],Joined->True,PlotStyle->Red];

Show[JetPlot1,Graphics[{Red,Dashed,Line[{{2.92//Log,-20},{2.92//Log,-80}}]}],Graphics[{Black,Line[{{3.11//Log,-20},{3.11//Log,-80}}]}],ΔJetPlot1,Plot[-45,{x,-1,10}],ImageSize->Full]

Show[JetPlot2,Graphics[{Red,Dashed,Line[{{2.79//Log,0},{2.79//Log,80}}]}],Graphics[{Black,Line[{{3//Log,0},{3//Log,80}}]}],ΔJetPlot2,Plot[+45,{x,-1,10}],ImageSize->Full]

Aufgabe 4 - Frequenzgang eines Serienschwingkreises

Wir berechnen die Induktivität mit der Formel:

(Subscript[L, F])[C_,ω_]:=(1/(4*π^2*ω^2*C))

(Subscript[ΔL, F])[C_,ω_,Δω_]:=Sqrt[(((Δω)/(4*π^2*ω^3*C)))^2]

(Subscript[ω, R])=Quantity[((4.06+3.89+3.87)/3),"Kilohertz"]

(Subscript[Δω, R])=Quantity[Sqrt[(((4.06-3.94)^2+(3.89-3.94)^2+(3.87-3.94)^2)/2)],"Kilohertz"]

Cr=Quantity[47,"Nanofarads"]

(Subscript[L, 1])=(Subscript[L, F])[Cr,(Subscript[ω, R])]//UnitConvert[#,"Henries"]&//N

(Subscript[ΔL, 1])=(Subscript[ΔL, F])[Cr,(Subscript[ω, R]),(Subscript[Δω, R])]//UnitConvert[#,"Henries"]&//N

(Subscript[L, 1])±(Subscript[ΔL, 1])

Nun berechnen wir den Gesamtwiderstand R+(Subscript[R, v]).

(Subscript[R, G])[Δf_,Λ_]:=2*π*Δf*Λ

(Subscript[ΔR, G])[Δf_,ΔΔf_,Λ_,ΔΛ_]:=Sqrt[(2*π*Λ*ΔΔf)^2+(2*π*Δf*ΔΛ)^2]

(Subscript[R, G,1000Ω])=(Subscript[R, G])[Quantity[5.43,"Kilohertz"],(Subscript[L, 1])]//UnitConvert[#,"Ohms"]&

(Subscript[ΔR, G,1000Ω])=(Subscript[ΔR, G])[Quantity[5.43,"Kilohertz"],Quantity[0.03,"Kilohertz"],(Subscript[L, 1]),(Subscript[ΔL, 1])]//UnitConvert[#,"Ohms"]&

(Subscript[R, G,220Ω])=(Subscript[R, G])[Quantity[1.33,"Kilohertz"],(Subscript[L, 1])]//UnitConvert[#,"Ohms"]&

(Subscript[ΔR, G,220Ω])=(Subscript[ΔR, G])[Quantity[1.33,"Kilohertz"],Quantity[0.03,"Kilohertz"],(Subscript[L, 1]),(Subscript[ΔL, 1])]//UnitConvert[#,"Ohms"]&

(Subscript[R, G,47Ω])=(Subscript[R, G])[Quantity[0.68,"Kilohertz"],(Subscript[L, 1])]//UnitConvert[#,"Ohms"]&

(Subscript[ΔR, G,47Ω])=(Subscript[ΔR, G])[Quantity[0.68,"Kilohertz"],Quantity[0.03,"Kilohertz"],(Subscript[L, 1]),(Subscript[ΔL, 1])]//UnitConvert[#,"Ohms"]&

Nun berechnen wir die Verlustwiderstände aus (Subscript[U, A ])und (Subscript[U, E]).

(Subscript[R, V])[R_,UE_,UA_]:=((UE*R)/UA)-R

(Subscript[ΔR, V])[R_,UE_,ΔUE_,UA_,ΔUA_]:=Sqrt[(R*((ΔUE)/UA))^2+(UE*R*((ΔUA)/(UA^2)))^2]

(Subscript[R, V,1000Ω])=(Subscript[R, V])[Quantity[1000,"Ohms"],Quantity[0.97,"Volts"],Quantity[0.93,"Volts"]]

(Subscript[ΔR, V,1000Ω])=(Subscript[ΔR, V])[Quantity[1000,"Ohms"],Quantity[0.97,"Volts"],Quantity[0.01,"Volts"],Quantity[0.93,"Volts"],Quantity[0.01,"Volts"]]

(Subscript[R, V,220Ω])=(Subscript[R, V])[Quantity[220,"Ohms"],Quantity[0.98,"Volts"],Quantity[0.76,"Volts"]]

(Subscript[ΔR, V,1000Ω])=(Subscript[ΔR, V])[Quantity[220,"Ohms"],Quantity[0.98,"Volts"],Quantity[0.01,"Volts"],Quantity[0.76,"Volts"],Quantity[0.01,"Volts"]]

(Subscript[R, V,1000Ω])=(Subscript[R, V])[Quantity[47,"Ohms"],Quantity[0.98,"Volts"],Quantity[0.36,"Volts"]]

(Subscript[ΔR, V,1000Ω])=(Subscript[ΔR, V])[Quantity[47,"Ohms"],Quantity[0.98,"Volts"],Quantity[0.01,"Volts"],Quantity[0.36,"Volts"],Quantity[0.01,"Volts"]]

5 Bestimmung der Dämpfungskonstanten eines freien, gedämpften Schwingkreises

Zunächst berechnen wir das logarithmische Dekrement. Wir mitteln über alle Nachbarn.

Λ[B_,C_]:=Log[(B/C)]

Λ12=Λ[Quantity[2.25,"Volts"],Quantity[1.42,"Volts"]]

Λ23 =Λ[Quantity[1.42,"Volts"],Quantity[0.91,"Volts"]]

Λ34=Λ[Quantity[0.91,"Volts"],Quantity[0.60,"Volts"]]

Λ45=Λ[Quantity[0.60,"Volts"],Quantity[0.39,"Volts"]]

ΛM=((Λ12+Λ23+Λ34+Λ45)/4)

ΔΛM=Sqrt[(1/3)*((Λ12-ΛM)^2+(Λ23-ΛM)^2+(Λ34-ΛM)^2+(Λ45-ΛM)^2)]

Für die Berechnung der Dämpfungskonstante ist es nötig auch die Periode zu messen. Wir mitteln über alle unsere Werte und multiplizieren sie mit 2.

T=2*{128,126,127,128,124,126}

TM=Median[T]//N

ΔTM=StandardDeviation[T]//N

Jetzt ergibt sich die Dämpfungskonstante

δ=Quantity[((ΛM)/TM),(1/("Microseconds"))]

Δδ=Quantity[Sqrt[(((ΔΛM)/TM))^2+(((ΔTM*ΛM)/(TM^2)))^2],(1/("Microseconds"))]

Nun berechnen wir aus (Subscript[L, 1])und δ den Gesamtwiderstand.

(Subscript[R, GG])=δ*2*(Subscript[L, 1])//UnitConvert[#,"Ohms"]&

(Subscript[ΔR, GG])=Sqrt[((Subscript[ΔL, 1])*2*δ)^2+((Subscript[L, 1])*2*Δδ)^2]//UnitConvert[#,"Ohms"]&

Wir berechnen nun mit (Subscript[L, 1 ])die Resonanzfrequenz.

Resonanz=(1/(2*π*Sqrt[(Subscript[L, 1])*Cr]))//UnitConvert[#,"Hertz"]&

ΔResonanz=Sqrt[((((Subscript[ΔL, 1])*Cr)/(2*((Subscript[L, 1])*Cr)^(3/2))))^2]//UnitConvert[#,"Hertz"]&