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Add a resistor to invert a sine wave??

Greg J.

Oct 8, 2013
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Oct 8, 2013
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I must not understand what is going on at the negative side of the AC source vs. the ground in the attached diagram. Why is the signal at Node 2 ("n2") inverted?

Inv_Sine_Wave.png
 

Alec_t

Jul 7, 2015
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The current through R2 and R3 is the same, but they are on opposite sides of the ground point.
 

dorke

Jun 20, 2015
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A picture worth 1000 words...


Inv_Sine_Wave.png


The current in the loop is i.
Voltages on the resistors are:
VR1=i*R1
VR2=i*R2
VR3=i*R3
Voltages on the measurement points (vs.GND )are:
Vn1= VR2 = i*R2
Vn2= -VR3= -i*R3

It is not that "the signal is inverted",
It is just the way you measure and what is your reference point !
For instance, if you moved the reference point to be n2 itself ,
all voltages would measure the same "polarity".

Hope that helps ;)
 

Greg J.

Oct 8, 2013
24
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Oct 8, 2013
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24
Indeed, thank you. Alec_t's answer reminded me that V1- is not the reference point.
 
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