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[Moved]: Break circuit into two half circuits

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anhnha

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Could you prove this sentence from a paper in the image below:

"Since the circuit is fully symmetric, it can be broken into two half circuits, like the one shown in Figure 5."

9244564400_1449017216.png
 

Re: Break circuit into two half circuits

I think it'm simple that Vx=Vx/2-(-Vx/2). after they analyze the circuit in Figure 5.
 
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    sarge

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Re: Break circuit into two half circuits

You can break the Vin signal with 0V on the middle node. Using the KVL, the voltages at the input terminal are +Vin/2 and -Vin/2


 
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Re: Break circuit into two half circuits

Hi,
How can you know that the potential at the point between two voltage sources is 0V?
BTW, could you talk detail how would you apply KVL here?
I know KVL but I don't know how you use it here.
 

Re: Break circuit into two half circuits

The circuit is symmetric, so the average voltage between the two circuits will be zero. VL-0=Vx/2 gives VL=Vx/2 and 0-VR=Vx/2 gives VR=-Vx/2.
 

Re: Break circuit into two half circuits

Well, could you prove how the potential at the point between two voltage sources is 0V?
 

The middle node becomes a virtual ground (not actual ground) when left and right hand side of a differential circuit are same. Analyze the half circuit, and then multiply by 2 at the end to get the overall response.
 

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