enter image description here why we put negative sign in the phase shift can someone explain this to me ? any tutorial or lecture ? looking forward 4 your suggestion

  • \$\begingroup\$ If inductor voltage leads current then current phase is neg. If Capacitor current leads voltage, then voltage phase is negative for a sine input. if comparing visa-versa, then phase is positive \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 4 '16 at 15:19
  • \$\begingroup\$ @TonyStewart.EEsince'75 over my head \$\endgroup\$ – Hassnain Abbas Nov 4 '16 at 15:23

For an inductor, the basic relationship between voltage and current is: -

V = L di/dt

Or, put another way, the rate of change of current (with respect to time) determines how big the voltage will be.

We can mathematically integrate both sides and we see that the time-integral of the applied voltage is proportional to current. So, if the input voltage is a sinewave, it's time integral is a negative cosine wave: -

enter image description here

For the low pass filter, as frequency rises, the inductive reactance becomes more dominant (despite the presence of the resistor) and the output magnitude begins to noticeably fall as the frequency gets close to and advances past the -3 dB point: -

enter image description here

Ultimately, at a high enough frequency, the phase angle of the current through the resistor is dominated by the inductive reactance. In other words, XL has become very large with respect to R. This means that the phase angle of the output voltage starts at zero degrees (low frequencies) and ends up at -90 degrees (high frequencies).

E.g. if XL has risen to 1000 and R is 1 then arctan of -1000 is -89.94 degrees.

  • \$\begingroup\$ any references ? need more detail \$\endgroup\$ – Hassnain Abbas Nov 5 '16 at 15:43
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    \$\begingroup\$ Which part do you need references and/or more detail. I've started with the extremely fundamental relationship between inductor voltage and current and extrapolated this to a phase angle between 0 and -90 for an RL low pass filter. I'm not going to give references how to integrate a sine into a -cosine because that is also fundamental and a math law. Inductive reactance rises with frequency as examinable by that inductor fundamental equation; because of this there is a 3 dB point when configured with a resistor (again a fundamental and mathematical certainty). So, what are you looking for? \$\endgroup\$ – Andy aka Nov 5 '16 at 15:53
  • \$\begingroup\$ well I'm a newbie and don't know all that stuff. please excuse me if you faced trouble Regards \$\endgroup\$ – Hassnain Abbas Nov 6 '16 at 13:23

The (parallel with the output, to ground) capacitor responds to the integral of the input current, so the result will be a lag, which is represented as a negative number for the phase (the output lags the input). The lag is small for a very low frequency and is close to 90° for a very high frequency where there almost no output voltage.

The complex impedance of a capacitor is \$X_C=\frac{1}{j\omega C} = -\frac{j}{\omega C}\$
(multiply numerator and denominator by \$j\$)

If you use the voltage divider equation to find the output you can find your output magnitude and phase with a tiny bit of complex algebra.

Vo/Vin = Xc/(Xc+R). Find the real and imaginary parts and do a rectangular to polar conversion to find the magnitude and phase angle. Don't forget to multiply the denominator by the complex conjugate to make it real so you can separate the two parts.


Simple answer: The transfer function of an L-R lowpass is


The phase shift of such a ratio (Numerator N)/(Denominator D) is

Phi(total)=Phi(N) - Phi(D), with N=R and D=(R+jwL)

Because the numerator N is constant we have Phi(N)=0

and therefore:

Phi(total)=0 - Phi(D)= - arctan[Im(D)/Re(D)]= -arctan (wL/R).


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