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Background
As v cos(wt + f + 2p) = v cos(wt + f), restrict –p < f ≤ p
Engineers throw an interesting twist into this formulation
o The frequency term wt has units of radians
o The phase shift f has units of degrees: –180° < f ≤ 180°
Background
A positive phase shift causes the function to lead of f
For example, –sin(t) = cos(t + 90°) leads cos(t) by 90°
Background
0 45 90 135 180 225 270 315 360
-1
-0.5
0
0.5
1
time(t)
cos(t)
sin(t)
+90°
Compare cos(377t) and cos(377t + 45°)
Background
If the phase shift is 180°, the functions are out of phase
E.g., –cos(t) = cos(t – 180°) and cos(t) are out of phase
0 45 90 135 180 225 270 315 360
-1
-0.5
0
0.5
1
time(t)
cos(t)
sin(t)
-180°
• Compare cos(377t) and cos(377t – 180°)
Why we use Phasors?
Problem Solution
Complicated and difficult
solution process
Transformed
Problem
Transformed
Problem
Transformed
Solution
Transformed
Solution
Transform
Relatively simple
solution process, but
using complex numbers
Inverse
Transform
Solutions Using Transforms
Real, or time
domain
Complex or
transform domain
The idea of phasor representation is based on Euler’s identity.
In general,
we use this relation to express v(t). If v(t) defines as;
Phasors
Phasors
• If we use sine for the phasor instead of cosine,
• then v(t) = Vm sin (ωt + φ) = Im (Vm𝒆j(ωt + φ))
• and the corresponding phasor is the same as that
Phasors
• Differentiating a sinusoid:
This shows that the derivative v(t) is transformed to the phasor domain
as jωV
Phasors
• Integrating a sinusoid
• Similarly, the integral of v(t) is equivalent to dividing its corresponding
phasor by jω.
Find the sinusoids represent by phasors.
Solution:
Converting this to time domain gives
in time domain
Using the phasor approach, determine the curret i(t) in a
circuit described by the integrodiffential equation.
Solution:
We transform each term in the equation from time domain
to phasor domain.
İn time domain
İn phasor domain
ω=2 so;
İn time domain
 Begin with the resistor;
 İf the current trough a resistor R is
 With Ohm’s law voltage acrosss R;
 The phasor form of this voltage;
 İf the phasor form of current;
 For inductor L;
 The current trough L is
 The voltage acrosss L;
Transform to the phasor
 For capacitor C;
 The voltage across C is
 The current trough C is
Solution:
 İf Z=R+jX inductive
İf Z=R-jX capasitive
Phasor_Impedance_Admitance.ppt
Phasor_Impedance_Admitance.ppt

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Phasor_Impedance_Admitance.ppt

  • 1. Background As v cos(wt + f + 2p) = v cos(wt + f), restrict –p < f ≤ p Engineers throw an interesting twist into this formulation o The frequency term wt has units of radians o The phase shift f has units of degrees: –180° < f ≤ 180°
  • 3. A positive phase shift causes the function to lead of f For example, –sin(t) = cos(t + 90°) leads cos(t) by 90° Background 0 45 90 135 180 225 270 315 360 -1 -0.5 0 0.5 1 time(t) cos(t) sin(t) +90°
  • 4. Compare cos(377t) and cos(377t + 45°)
  • 5. Background If the phase shift is 180°, the functions are out of phase E.g., –cos(t) = cos(t – 180°) and cos(t) are out of phase 0 45 90 135 180 225 270 315 360 -1 -0.5 0 0.5 1 time(t) cos(t) sin(t) -180°
  • 6. • Compare cos(377t) and cos(377t – 180°)
  • 7. Why we use Phasors? Problem Solution Complicated and difficult solution process Transformed Problem Transformed Problem Transformed Solution Transformed Solution Transform Relatively simple solution process, but using complex numbers Inverse Transform Solutions Using Transforms Real, or time domain Complex or transform domain
  • 8. The idea of phasor representation is based on Euler’s identity. In general, we use this relation to express v(t). If v(t) defines as; Phasors
  • 9. Phasors • If we use sine for the phasor instead of cosine, • then v(t) = Vm sin (ωt + φ) = Im (Vm𝒆j(ωt + φ)) • and the corresponding phasor is the same as that
  • 10. Phasors • Differentiating a sinusoid: This shows that the derivative v(t) is transformed to the phasor domain as jωV
  • 11. Phasors • Integrating a sinusoid • Similarly, the integral of v(t) is equivalent to dividing its corresponding phasor by jω.
  • 12. Find the sinusoids represent by phasors. Solution:
  • 13. Converting this to time domain gives in time domain
  • 14. Using the phasor approach, determine the curret i(t) in a circuit described by the integrodiffential equation. Solution: We transform each term in the equation from time domain to phasor domain.
  • 15. İn time domain İn phasor domain ω=2 so;
  • 17.  Begin with the resistor;  İf the current trough a resistor R is  With Ohm’s law voltage acrosss R;  The phasor form of this voltage;
  • 18.  İf the phasor form of current;
  • 19.  For inductor L;  The current trough L is  The voltage acrosss L;
  • 21.
  • 22.  For capacitor C;  The voltage across C is  The current trough C is
  • 23.
  • 24.
  • 26.
  • 27.
  • 28.
  • 29.  İf Z=R+jX inductive İf Z=R-jX capasitive

Editor's Notes

  1. Y is the admittance, measured in siemens. G is the conductance, measured in siemens. B is the susceptance, measured in siemens.