On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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GEN E 123 (week 10)
Prof Catherine Gebotys, Department of Electrical and Computer Engineering, DC3514
h
I. AC CIRCUITS
o Previous circuits we analyzed: solution = natural solution + forced
solution (where in natural solution, we killed independent sources.
forced solution was due to independent sources. Note the forced
response was a constant or DC response due to DC source)
o Next we look at other input sources : SINUSOIDAL
o Important in communication systems, power, etc
o Important in representing ‘any’ signal (Fourier series)
o Sinusoidal input source or sinusoidal excitation or ac circuits
o We will only analyze the forced response (since this ac response remains
after natural response has passed)
o IN a Circuit, IF (a) source(s) are sinusoidal THEN all currents and
voltages are sinusoidal.
o IN a Circuit IF sources are exponential, then all currents and voltages
are sinusoidal.
o We’re not limited to 1st order circuits (with one storage element, ie.
C,L) higher order circuits can be handled with same method as used for
resistive circuits!
o So ac circuits can be analyzed with no more difficulty than dc circuits!
Thanks to phasor analysis .
o However before we discuss phasors… we start with sinusoidals.
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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-review of sinusoidals…ωφπθ∠
o v(t) = Vm sin ωt
where ω is the angular frequency (radian frequency) in radians/sec or rad/s.
where v(t) = v(t+T), periodic function with period = T = 2π/ω sec, and frequency
= 1/period = ω/ 2π in cycles per second or Hertz or Hz.
v(t)
Vm
t
π/ω 2π/ω
0
/
Below we have v(t) = Vm sin (ωt+φ), where v(t) leads Vmsinωt by φ rad.
φ>0 implies a Left shift (leads), φ<0 implies a Right shift (lags).
φ/ω displacement
o
o
o
o
o
o
v1= 4sin(2t+300) leads v2= 6sin(2t-120) by 420
sinωt = cos(ωt-π/2) or cosωt = sin(ωt+π/2)
sin(ωt ±π )= -sinωt and cos(ωt ±π )= -cosωt
sin(ωt ±2π )= sinωt and cos(ωt ±2π )= cosωt
A cosωt + B sinωt = (A2+B2)0.5 cos(ωt - θ ), θ=tan-1B/A
M cos(ωt - θ ) = Acosωt + B sinωt , A=Mcosθ, B=Msinθ {quadrature
representation of sinusoid}
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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o Next we need the use of Complex Numbers
o A complex number is : a + j b , where a = real part, b= imaginary part (note:
j2 = -1)
o Why do we use complex numbers? Consider a source in a circuit vg(t) and a
current somewhere else in the circuit i(t).
o vg(t) = f1(t), we get i(t) = i1(t)
o vg(t)=f2(t), we get i(t)=i2(t)
o vg(t)=f1(t) + j f2(t) we get i(t) = i1(t) + j i2(t)
o Next consider vg(t) = Vm ej(wt+φ) , the real part of vg(t) is Vm cos(ωt+φ), the
imaginary part of vg(t) is Vm sin(ωt+φ). So we solve our circuit with vg(t) =
Vm ej(wt+φ) as source then we take the real and imaginary parts of our
solution as the solution to Vm cos(ωt+φ) and Vm sin(ωt+φ) input
sources.
o Example
o We use as a source : vg(t) = Vm ej(wt+φ) ; then we solve for i(t) = S ej(ws
t + φs)
,
o So if our given source is vg(t)= Vm cos(ωt+φ), we get i(t)= S cos(ωs
t + φs)
o or if our given source is vg(t)= Vm sin(ωt+φ), we get i(t)= S sin(ωs t
+ φs)
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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EXAMPLE:
2H
where vg(t)=15cos2t, Find i=?
2Ω
vg(t)
+
0.1F
First replace source with 15ej2t V source.
Next write KVL:
Differentiate and divide by 2:
Substitute trial form i =A ej2t
Next take the real part of the solution since the source was ‘cos’.
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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NOTES:
Source
Complex Source
Source Phasor
j(wt+φ)
A cos(ωt+φ)
Ae
A ∠φ
A sin(ωt+φ)
A ej(wt+φ)
A∠φ-90o
Rectangular coordinates: a + j b
Polar coordinates: r∠ θ , r = (a2+b2)0.5 , θ = tan-1b/a
a + j b <=> r∠θ
a = r cosθ , b = r sinθ
V is a voltage phasor, v(t) = V ejwt = |V| ejθ ejwt = |V| ej(wt+θ)
Phasors are the complex #’s we multiply ejwt in the expressions for i(t),v(t).
In previous example, we could convert from rectangular coordinates to polar
coordinates to get a current phasor i = 6.7 ∠26.6 o
EXAMPLE:
4
36 cos(2t+30o)
+
3H
2 sin(2t-15o)
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
I-V laws for PHASORS
All currents and voltages are of the form A ejwt. We’re only interested in the forced
response.
We have the following PHASOR laws
(Recall: V is a voltage phasor, v(t) = V e jwt = |V| e jθ e jwt = |V| e j(wt+θ) ):
Ohm’s Law:
V=RI
Inductor’s:
V = jwL I
Capacitor’s:
V = 1/(jwC) I
Impedance (in ohm’s) :
(V/I )=Z
o impedance in a general circuit plays the role of resistance in a resistive
circuit. Impedance is a complex number (not a phasor). A constant that scales
one phasor to produce another phasor. Real component of impedance is
resistance (resistive component), imaginary component of impedance is
reactance (reactive component).
o Now we can use the impedance concept in analysis of RLC circuits where
o ZR = R “resistive”
§ XR = 0
o ZL = jwL = wL ∠90o “reactive”
§ ZL = jXL , XL = inductive reactance (+ve)
o ZC = 1/(jwC) = -j 1/(wC) = (1/wC) ∠–90o “reactive”
§ ZC = 1/(jwC) = jXC , XC = - 1/(wC) =capacitive reactance (-ve)
Example: What is impedance seen from terminal?
IR = VR/R , IL = VL/ZL , IR = IL
KVL: I R + jwL I = V or I (R+jwL) = V
Impedance = V/I = R+jwL = (R2+w2L2)0.5 tan-1 wL/R
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On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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KVL:
V1 + V2 + … + Vn = 0
KCL:
I1 + I2 + … + In = 0
KVL, KCL holds for phasors
EXAMPLE (see first example of this weeks notes, now rewritten with phasors)
: Solve for i
j4
150o
+
2
-j5
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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NOTE: phasors are useful for finding FORCED response. Typically though the
natural responses decay to zero as time goes to infinity, whereas the forced
response is the steady – state response. These circuits are called stable.
So only for stable circuits, can phasor analysis be used to find steady-state response.
(unstable circuits have a natural response that does not decay to zero)
Analysis of the circuit:
1. Create a phasor diagram of the circuit showing impedances (& sources as
phasors).
2. identify the method used to solve the circuit (ie. if you have all current sources
use nodal analysis , etc)
3. write out the equations as done before
4. convert your solution from rectangular coordinates (a+jb) to polar coordinates (
r∠θ )
5. from polar coordinates transform directly into cosine