Goed - stel dat de huidige Falstad simulatie in essentie correct is. Dan laat zich nu een vergelijking afleiden die wiskundig uitdrukt hoe het ingangs- en uitgangssignaal van de Ladik integrator met elkaar samenhangen:
Uit bovenstaand schema zien we dat:
[imath] I_1 = \frac{U_{in}}{ (1-\alpha)\mathrm{R}_1 + \mathrm{R}_2 } \,\,\,\,\,\, (1) [/imath]
[imath] I_2 = \mathrm{C}_1 \frac{ \mathrm{d} (- U_1) }{ \mathrm{d} t} [/imath]
[imath] I_2 = - \mathrm{C}_1 \frac{ \mathrm{d} U_1 }{ \mathrm{d} t} \,\,\,\,\, (2) [/imath]
[imath] I_3 = \frac{- U_{out} }{ (1-\alpha)R_5 + R_4 } \,\,\,\,\,\ (3) [/imath]
[imath] I_1 = I_2 + I_3 \,\,\,\,\,\,\, (4) [/imath]
[imath] I_4 = \frac{ U_{out} }{ \mathrm{R}_3 } \,\,\,\,\,\ (5) [/imath]
[imath] I_4 = \mathrm{C}_2 \cdot \frac{ \mathrm{d} }{ \mathrm{d} t } (-U_1 - U_{out} ) [/imath]
[imath] I_4 = \mathrm{C}_2 \cdot \left ( \frac{ \mathrm{d} (-U_1) }{ \mathrm{d} t } + \frac{ \mathrm{d} ( - U_{out} )}{ \mathrm{d} t } \right ) [/imath]
[imath] I_4 = \mathrm{C}_2 \cdot \left ( - \frac{ \mathrm{d} U_1 }{ \mathrm{d} t } - \frac{ \mathrm{d} U_{out} }{ \mathrm{d} t } \right ) [/imath]
[imath] I_4 = - \mathrm{C}_2 \cdot \frac{ \mathrm{d} U_1 }{ \mathrm{d} t } - \mathrm{C}_2 \cdot \frac{ \mathrm{d} U_{out} }{ \mathrm{d} t } \,\,\,\,\, (6) [/imath]
Uit (1), (2), (3) en (4) volgt:
[imath] \frac{U_{in}}{ (1-\alpha)\mathrm{R}_1 + \mathrm{R}_2 } \, = \, - \mathrm{C}_1 \cdot \frac{ \mathrm{d} U_1}{ \mathrm{d} t} \, + \, \frac{- U_{out} }{ (1-\alpha)\mathrm{R}_5 + \mathrm{R}_4 } [/imath]
[imath] \frac{U_{in}}{ ((1-\alpha)\mathrm{R}_1 + \mathrm{R}_2) \mathrm{C}_1 } \, = \, - \frac{ \mathrm{d} U_1}{ \mathrm{d} t} \, + \, \frac{- U_{out} }{ ((1-\alpha)\mathrm{R}_5 + \mathrm{R}_4) C_1 } \,\,\,\,\, (7) [/imath]
Uit (5) en (6) volgt:
[imath] \frac{ U_{out} }{ \mathrm{R}_3 } = - \mathrm{C}_2 \cdot \frac{\mathrm{d} U_1}{\mathrm{d} t} - \mathrm{C}_2 \cdot \frac{\mathrm{d} U_{out}}{\mathrm{d} t}[/imath]
[imath] \frac{ U_{out} }{ \mathrm{R}_3 \mathrm{C}_2} = -\frac{\mathrm{d} U_1}{\mathrm{d} t} - \frac{\mathrm{d} U_{out}}{\mathrm{d} t}[/imath]
[imath] \frac{ U_{out} }{ \mathrm{R}_3 \mathrm{C}_2} + \frac{\mathrm{d} U_{out}}{\mathrm{d} t} = - \frac{\mathrm{d} U_1}{\mathrm{d} t} \,\,\,\,\,\, (8) [/imath]
En uit (7) en (8 ) volgt dan nog dat:
[imath] \frac{U_{in}}{ ((1-\alpha)\mathrm{R}_1 + \mathrm{R}_2) C_1 } \, = \, \frac{ U_{out} }{ \mathrm{R}_3 \mathrm{C}_2} + \frac{\mathrm{d} U_{out}}{ \mathrm{d} t} \, + \, \frac{- U_{out} }{ ((1-\alpha)\mathrm{R}_5 + \mathrm{R}_4) \mathrm{C}_1 } [/imath]
[imath] \frac{U_{in}}{ ((1-\alpha)\mathrm{R}_1 + \mathrm{R}_2) C_1 } \, = \, \frac{\mathrm{d} U_{out}}{ \mathrm{d} t} + \frac{ U_{out} }{ \mathrm{R}_3 \mathrm{C}_2} + \frac{- U_{out} }{ ((1-\alpha)\mathrm{R}_5 + \mathrm{R}_4) \mathrm{C}_1 } [/imath]
[imath] \frac{ U_{in}}{ ((1-\alpha)\mathrm{R}_1 + \mathrm{R}_2) C_1 } \, = \, \frac{\mathrm{d} U_{out}}{ \mathrm{d} t} + \left ( \frac{ 1 }{ \mathrm{R}_3 \mathrm{C}_2} - \frac{ 1 }{ ((1-\alpha)\mathrm{R}_5 + \mathrm{R}_4) \mathrm{C}_1 } \right ) \cdot U_{out} \,\,\,\,\,\, (9) [/imath]
Schrijf voor het gemak:
[imath] \mathrm{a} = \frac{1}{\mathrm{R}_3 \mathrm{C}_2} - \frac{1}{((1-\alpha)\mathrm{R}_5 + \mathrm{R}_4) \mathrm{C}_1} \,\,\, \& \,\,\, \mathrm{b} = ((1-\alpha)\mathrm{R}_1 + \mathrm{R}_2) \mathrm{C}_1 \,\,\,\,\,\,\ (10) [/imath]
Dan vinden we als vergelijking voor (de simulatie van) de Ladik integrator:
[imath] \frac{U_{in}}{ \mathrm{b}} \, = \, \frac{\mathrm{d} U_{out}}{ \mathrm{d} t} + \mathrm{a} \cdot U_{out} \,\,\,\,\,\, (11) [/imath]
