Bilinear switched capacitor resistor equivalent resistance

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tenso

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I am trying to find the equivalent resistance of the following circuit in baker's book.



How do you get the factor of 4 in the denominator? I am trying to get the answer by using the principle of transfer of charge and conservation of charge like in the case below:
https://en.wikipedia.org/wiki/Switched_capacitor#The_switched-capacitor_resistor


we can assume that the voltage on the left side of cap is Vin and on the right Vout.

during phi1 voltage across the cap is V1-V2 and during phi2 voltage across cap is V2-V1. So shouldn't delta q be 2C(V1-V2)?

so shouldn't the answer be 1/2Cf ?
 

The capacitor is charging from -ΔV to +ΔV in phase 1 and back in phase 2, total q = 4CΔV.
 

FvM said:
The capacitor is charging from -ΔV to +ΔV in phase 1 and back in phase 2, total q = 4CΔV.
Click to expand...

This is not clear.

Let say the voltage at left side is V1 and on the right V2.

When s1 is closed the voltage across C is V1-V2. The charge of C is C(V1-V2). Now s1 is opened and s2 is closed. The voltage across C is V2-V1 and the charge on it is C(V2-V1). So the charge transferred out of the capacitor should be C (V1-V2) - C (V2-V1) is 2C(V1-V2). Why is this quantity 2C(V1-V2) multiplied by 2 again?
 

Charge is transferred to the output also when s1 is closed. What you describe happens 2 times per clock cycle.
 
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    tenso

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So I had one more question, what would be the expression output equivalent resistance for the following switched capacitor circuit. Here V2 is connected to gnd during phi2.

 

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