Question of Phase Shift

Discussion in 'General Electronics Chat' started by Mazaag, Apr 26, 2008.

  1. Mazaag

    Thread Starter Senior Member

    Oct 23, 2004
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    0
    hey guys..

    suppose we have a RC low Pass filter... the phase response of Vo/Vin of the circuit goes from 0 to -90 in the stopband...

    my question is as follows... if we have an input sinwave Asin(wt) with (phase 0 deg) of a particular frequency ( assume it is exactly at the cut-off frequency of the lowpass), so we would have a phase of -45..

    the output would be A'sin(wt +phi) where A' is the scaled amplitude and phi is the phase shift....

    would the output, in this case, be A'sin(wt-45) ? if that is the case, doesn't that mean that the output is actually leading the input by 45 degrees (which obviously doesn't make sense? where did i go wrong in my analysis...

    Thanks guys
     
  2. Caveman

    Active Member

    Apr 15, 2008
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    No, it is right. First of all remember that phi is not in degrees, it is in radians, so -45 deg = PI/4 radians.

    If t = 0, then input = 0, output = -0.7*A'
    If t = PI/(4*w), then input = 0.7*A', output = 0
    See how input got to 0 before output returned to zero. Of course, this is assuming continuous sine waves.
     
  3. Mazaag

    Thread Starter Senior Member

    Oct 23, 2004
    255
    0
    Alright thanks Caveman..

    Another question..

    could someone explain to me (fundamentally) why there is a phase changE? what is the significance of this phase change.. like..why does it occur ? where does this delay come from? (like .. what happens to the charge? etc..)
     
  4. Caveman

    Active Member

    Apr 15, 2008
    471
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    Okay, a phase change is just a frequency version of time delay. The same time delay for different frequencies looks like different phase changes. For example 1ms delay 100Hz is 36 degrees delay and is 360 degrees for 1kHz.

    Remember that a capacitor has to have current to charge up:
    dV/dt = I/C
    But the resistor limits the current to the capacitor. At low frequencies, the rate of the change in voltage of the input is slow enough that the capacitor charges up quickly enough that there is no significant phase delay. But as the frequency goes up, the charge up delay becomes a more significant phase shift.

    I never thought about why it maxes out at 90 degrees. I'll have to think about it...
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Because when it gets to 90° to the primary signal (at maximum), it's at 0v.

    Think of a sinewave. It starts out at 0v, at 0°
    By the time it gets to 90°, it's at it's apogee.
    By the time it gets to 180°, it's at zero again.

    So, if a signal is at 90° at it's apogee, 90° later, it's at zero.
     
  6. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    Well, my way of stating it is that at low frequencies, the resistor doesn't have much current, so the difference between input and output is very little as is the phase. But at large frequencies, the capacitor resists the quick voltage change by drawing more current. The current is 90 degrees ahead in phase of the voltage change in the capacitor. So the voltage across the resistor is 90 degrees ahead. But the output voltage is Vin - Vr = Vout. So a 0 degrees input - 90 degrees positive phase results in a 90 degrees negative phase output.
     
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