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Add a resistor to invert a sine wave??

Discussion in 'Electronics Homework Help' started by Greg J., Sep 7, 2015.

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  1. Greg J.

    Greg J.

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    0
    Oct 8, 2013
    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
     
  2. Alec_t

    Alec_t

    2,865
    768
    Jul 7, 2015
    The current through R2 and R3 is the same, but they are on opposite sides of the ground point.
     
  3. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    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 ;)
     
    Harald Kapp likes this.
  4. Greg J.

    Greg J.

    24
    0
    Oct 8, 2013
    Indeed, thank you. Alec_t's answer reminded me that V1- is not the reference point.
     
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