10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
10
53
Sinusoidal Steady-State Analysis-AC Analysis
Here we consider the forcing function v(t) = Vm sin(ωt), where the radian frequency in radians per second ω = 2πf
and the frequency in Hertz f = T1 and T is the period of the sine wave.
Consider an RL circuit in series with a voltage source v(t) = Vm sin(ωt). Find the current i(t) =?
R
i L
+ vL −
+
Vm cos(ωt)
−
RL-AC.m4
KVL: − Vm cos(ωt) + Ri + L
di
=0
dt
−P t
+ e−P t
Recall that the first-order ODE, dy
dt + P y = Q has the solution y(t) = Ae
Vm
R
Here we have P = L , Q = L cos(ωt), y = i. Substitute we get
Z t
x
Vm
− τt
− τt
+e
cos(ωx)e τ dx
i(t) = Ke
0 L
R
Qe+P t dt.
From Calculus,
Z
cos(ωt)et/τ dt =
τ et/τ
[cos(ωt) + (ωτ ) sin(ωt)]
1 + ω2τ 2
substitute in the relation for i(t) to get
i(t) = Ke−t/τ +
Vm R
Vm ωL
cos(ωt) + 2
sin(ωt)
R 2 + ω 2 L2
R + ω 2 L2
the natural response Ke−t/τ will die out after 5τ seconds and the forced response will be
i(t) =
R2
Vm R
Vm ωL
cos(ωt) + 2
sin(ωt)
2
2
+ω L
R + ω 2 L2
R
, then sin(θ) = √R2ωL
since cos2 θ + sin2 θ = 1, and letting Im =
If we let cos(θ) = √R2 +ω
2 L2
+ω 2 L2
current i(t) will be
i(t) = Im cos(θ) cos(ωt) + Im sin(θ) sin(ωt) = Im cos(ωt − θ)
Decaying Exponential:
2
t
1
2
3
4
5
6
7
8
9
10
e−t
0.3679
0.1353
0.0498
0.0183
0.0067
0.0025
0.0009
0.0003
0.0001
0.0000
1
e−t
0
-1
0
1
2
3
4
5
Vm
R2 +ω 2 L2
then the
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
54
The following is a Maxima session that verifies the previously obtained results:
(%i1) batch("C:/_teach/_eee202/_doc/RL-AC.mac");
batching #pC:/_teach/_eee202/_doc/RL-AC.mac
(%i2) assume(w>0)
(%o2) [w>0]
(%i3) v(t):=Vm*cos(w*t)
(%o3) v(t):=Vm*cos(w*t)
(%i4) ev(kcl:L*(’diff(i(t),t,1))+R*i(t)-v(t)=0)
(%o4) i(t)*R+(’diff(i(t),t,1))*L-Vm*cos(t*w)=0
(%i5) soln:desolve(kcl,i(t))
(%o5) i(t)=((i(0)*L*R^2-Vm*L*R+i(0)*w^2*L^3)*%e^(-(t*R)/L))/(L*(R^2+w^2*L^2))
+(Vm*cos(t*w)*R)/(R^2+w^2*L^2)+(Vm*w*sin(t*w)*L)/(R^2+w^2*L^2)
10.1
Leading and Lagging
In an inductive circuit, the voltage leads the current, while in a capacitive circuit the current leads the voltage.
ELI the ICE man.
We start from left-to-right, both sinusoids are rising with phase difference less than π = 180 degrees, the first we meet
is leading.
cos(t) leads cos(t − π/2) as shown
1
cos(t)
cos(t−pi/2)
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−3
−2
−1
0
1
2
3
The figure was obtained from Matlab via the commands:
>> fplot(’[cos(x)]’,[-pi pi -1 1],[’-’]); hold on
>> fplot(’[cos(x-pi/2)]’,[-pi pi -1 1],[’:’]); legend(’cos(t)’,’cos(t-pi/2)’); grid
>> print -deps cosleading.eps
Useful relations:
• sin(a ± b) = sin a cos b ± cos a sin b
• cos(a ± b) = cos a cos b ∓ sin a sin b
• sin x = cos(x − 90o )
• − sin x = cos(x + 90o )
• sin(x + 90o ) = cos x
• sin(x − 90o ) = cos(x + 180o)
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
10.2
55
The Rotational Operator j
Any point in a two-dimensional space can be reached given its Cartesian coordinates, the pair of numbers (x, y). Let j
be an operator that when multiplied by a vector, the vector will be rotated 90o (π/2 radians) in the counter clockwise
direction (Anti clockwise), hence a unit vector in the x−direction when multiplied by j it will become a unit vector
in y−direction.
Multiplying a unit vector in the x−direction by j twice will produce a vector in the negative x−axis direction, i.e. it
changes sign, while when multiplied four times it will be back to the original direction. We can conclude from this
that j 4 = 1 and j 2 = −1, and from which we get
√
j = −1
10.3
Euler’s Formula
ejθ = cos θ + j sin θ
since cos is an even function (cos(−x) = cos(x)) and sin is an odd function sin(−x) = − sin(x), we write
e−jθ = cos θ − j sin θ
and from which we get
cos θ =
ejθ + e−jθ
ejθ − e−jθ
and
2
2j
Note that multiplying by ejθ is equivalent to rotation in the (anti) counter clock-wise direction by an angle θ.
If θ = π/2, then ejθ = ejπ/2 = cos π/2 + j sin π/2 = 0 + j = j which is as we started by defining j as a rotational
vector by π/2 in CCW direction.
10.3.1
Useful Tricks
ejx = cos x + j sin x
jejx = j cos x + j 2 sin x = j cos x − sin x
• cos x = Re{ejx }
• sin x = Re{−jejx }
• sin x = cos(x − π/2) = 16 (−90o)
• cos x = − sin(x − π/2)
10.4
Eigenfunction
Eignefunction f (t) is defined as the forcing function (input to the system) that when applied to the system, its response (output) will be Kf (t) where K is a constant that is specific to the system architecture and (components and
configuration.)
For linear time-invariant systems (Electric circuits that we are addressing in this course are considered linear systems),
ejωt is an eigenfunction of the system. Hence the response will be Aejωt .
ejωt
jωt
Linear SystemAe
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
56
Re-consider the RL circuit discussed in the previous section and let the forcing function be Vm ejωt and its response
(current in this case) i1 (t), KVL for the circuit, we get
−Vm ejωt + Ri1 + L
di1
=0
dt
Since the forcing function is of the form ejωt , hence the response i1 (t) = Aejωt .
Substituting into the KVL (differential equation) we get:
−Vm ejωt + RAejωt + L(Aejωt) )(jω) = ejωt (−Vm + RA + LA(jω)) = 0
from which we get −Vm + RA + LA(jω) = 0 and the constant A will be given by
A=
R−jωL
R−jωL to
ωL
R2 +ω 2 L2 we
R
R2 +ω 2 L2 and sin θ
V
m
√
, we can write
R2 +ω 2 L2
=
Vm (R−jωL)
R2 +ω 2 L2 .
√ θ−j sin θ ).
write A = Vm ( cos
R2 +ω 2 L2
get A =
We can multiply the right-hand side by
If we let cos θ =
Vm
R + jωL
can
Let Im =
A = Im (cos θ − j sin θ). Using Euler’s formula, e−jθ = cos θ − j sin θ we can write
A = Im e−jθ
Substituting for A in the current i1 (t) = Aejωt to get
i1 (t) = Im ej(ωt−θ)
If the forcing function is Vm cos(ωt) which is the real part of Vm ejωt , then the response will be real(i1 (t)) = Im cos(ωt−θ)
which is the same as obtained previously.
10.4.1
Integration and Differentiation of ejωt
1 jωt
e , while differentiation of ejωt = jωejωt .
Integration of ejωt = jω
1
We can replace integration with jω
and differentiation with jω.
10.4.2
Matlab and the Complex quantities
Useful functions: Let Z = R + jX = A6 (θ), then
• R=real(Z)
• X=imag(Z)
• A=abs(Z)
• θ = angle(Z)
• Z = A ∗ exp(j ∗ θ)
• Z =R+j∗X
If Z is an array of complex numbers, then:
P = angle(Z) returns the phase angles, in radians, for each element of complex array Z. The angles lie between ±π.
For complex Z, the magnitude R and phase angle θ are given by
R = abs(Z)
theta = angle(Z) and the statement
Z = R. ∗ exp(i ∗ theta) converts back to the original complex Z.
Measurements are carried out with reference to the real axis.
eωt = cos(ωt) + j sin(ωt) ⇒ cos(ωt) = real(eωt ) = 16 0 and
−jeωt = −j cos(ωt) + sin(ωt) ⇒ sin(ωt) = real(−jeωt ) = 16 (−π/2).
j ↔ 16 π/2 and
−j = 1j ↔ 16 −π/2
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
10.5
57
Using PSPICE
Here is how to describe an AC source and some needed dot commands:
• source + - ac magnitude phase(in degrees)
vs 1 0 ac 1.5 58.2
• analysis-type sweep-type number-of-points initial-value final-value
for AC analysis, vary frequency.
Here is an example for one (1) AC frequency analysis (from 50Hz to 50Hz)
.ac lin 1 50 50
• .print analysis-type current-into-voltage-source voltage-at-node
.print ac im(v30) ip(v30) ir(v30) ii(v30)
.print ac vm(1) vp(1)
vm(1)=magnitude of voltage at node 1
vp(1)=phase (degrees) of voltage at node 1
ir(v30)=real part of the current flowing into voltage source v30
ii(v30)=imaginary part of the current flowing into voltage source v30
Example: For the AC circuit shown, find iL using PSPICE.
100Ω
iL
0.3H
+
0.2iL
cos 500t
+
−
−
hk7p9.m4
HK7p9 page 242
* vs(t)=cos(500t), w=500 rad/sec, f=500/(2 pi)=79.5775Hz
*vs
+
ac
mag
phase(degrees)
vs
1
0
ac
1
0
r12
1
2
100
f02
0
2
v30
0.2
l23
2
3
0.3h
v30
3
0
ac
0
0
;inserted to measure iL
.ac
lin
1
79.5775 79.5775
.print ac
im(v30) ip(v30) ir(v30) ii(v30)
.print ac
vm(1)
vp(1)
.end
****
NODE
(1)
SMALL SIGNAL BIAS SOLUTION
VOLTAGE
0.0000
NODE
(2)
VOLTAGE
0.0000
TEMPERATURE =
NODE
(3)
27.000 DEG C
VOLTAGE
0.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
vs
0.000E+00
v30
0.000E+00
TOTAL POWER DISSIPATION
0.00E+00 WATTS
**** 12/16/106 07:54:46 ******** NT Evaluation PSpice (July 1997) ************
HK7p9 page 242
****
AC ANALYSIS
TEMPERATURE =
27.000 DEG C
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
FREQ
7.958E+01
IM(v30)
5.882E-03
IP(v30)
-6.193E+01
FREQ
VM(1)
7.958E+01
1.000E+00
JOB CONCLUDED
TOTAL JOB TIME
10.5.1
IR(v30)
2.768E-03
58
II(v30)
-5.190E-03
VP(1)
0.000E+00
.05
Example to be solved using time domain technique, then by using phasors (frequency domain)
Find i(t) in the circuit shown below.
1kΩ
i(t)
1.5kΩ
+
vs (t)
40 sin 3000t
−
+
vL
−
iL
1
3H
+
vC
−
iC
1
6 µF
hk8dp4.m4
1. Using time-domain technique:
vs:40*sin(3000*t); R1:1500; R2:1000; L:1/3; C:(1e-6)/6;
KVL1: -vs+R1*i(t)+L*’diff(iL(t),t,1)=0;
KVL2: -L*’diff(iL(t),t,1)+R2*C*’diff(vC(t),t,1)+vC(t)=0;
KCL2: i(t)-iL(t)-C*’diff(vC(t),t,1)=0;
atvalue(iL(t),t=0,0);
atvalue(vC(t),t=0,0);
sol:desolve([KVL1,KVL2,KCL2],[i(t),iL(t),vC(t)]);
Solution from Maxima
i (t) = e−2100 t
√
√
!
62 sin 300 71 t
6 cos 300 71 t
8 sin (3000 t) 6 cos (3000 t)
√
+
−
+
625
625
625
625 71
iL (t) = e−2100 t
√
√
!
sin 300 71 t
13 cos 300 71 t
9 sin (3000 t) 13 cos (3000 t)
√
+
−
+
625
625
625
625 71
vC (t) = e
−2100 t
√
√
!
1232 sin 300 71 t
16 cos 300 71 t
112 sin (3000 t) 16 cos (3000 t)
√
−
−
+
+
5
5
5
5 71
It is clear that the transient terms will die out very quickly since they are multiplied by e−2100t and
iL (t) =
1
(9 sin(3000t) − 13 cos(3000t))
625
2. Using frequency-domain technique
First convert all time-domain specs into phasor form
vs = 40 sin 3000t = 40{Re(−jej3000t )} from which we write Vs = 406 (−90o ). Zin as seen by the voltage source
can be written as
Zin = 1500 +
j( 31 (3000)(1000 −
j
)
(3000)( 16 )(e−6)
j1000 + 1000 − j2000
then
I=
= 2000 + j1500 = 25006 36.9o
Y
406 −90o
=
= 166 (−126.9o)mA
Zin
25006 36.9o
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
59
and in time-domain, i(t) = 16 cos(3000t − 126.9o).
Or we can write the mesh equations for the two meshes 1 and 2. I is the current in the first mesh while I2 is the
second mesh current.
KVL1: −406 −90 + 1500I + j1000(I − I2) = 0
KVL2: j1000(I2 − I) + 1000I2 − j2000I2 = 0
rewrite the two equations:
(1500 + j1000)I − j1000I2 = 406 −90
−j1000I + (1000 − j1000)I2 = 0
using Matlab:
[1500 + j ∗ 1000, −j ∗ 1000; −j ∗ 1000, 1000 − j ∗ 1000] ∗ [I; I2] = [40 ∗ exp(−j ∗ pi/2); 0];
~ , solving
AI~ = V
I=inv(A)*V;
abs(I) gives [0.0160; 0.0113] an angle(I)*180/pi gives [−126.9; 8.13] degrees.
10.5.2
Lead and Lag in the Phasor domain
V1
φ1 − φ2
V1 leads V2
V2
ωt + φ1
ωt + φ2
Real Axis
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
10.6
60
Time-domain ↔ Phasor domain
• time domain variables are written as lowercase, e.g. v1, i5, · · ·
• phasor domain variables are written as uppercase, e.g. V 1, I5, · · ·
• usually angles are written in degrees in time-domain while they are written in radians in phasor domain.
• All sinusoids must be in cos(ωt ± θ) form. No sin functions and No cos(−ωt ± θ)
•
d
dt
•
R
↔ jω
(.)dt ↔
1
jω
• sin(ωt) ↔ 16 −π/2 = −j,
to make j coincides with the +ve real axis, we need to multiply it with
or sin(ωt) = Re{−jejωt }
1
j
= −j,
• cos(ωt) ↔ 16 0 = 1,
(cos represents the real part of ejωt = cos(ωt) + j sin(ωt) and sin represents the imaginary part (90o )
di
• Inductor voltage L dt
↔ jωLI
• Capacitor current C dvdtC ↔ jωCVC
• Reactance of an L = XL = ωL
• Reactance of a C = XC =
−1
ωC
• Resistance R in series with an inductor L have an impedance Z = R + jXL = R + jωL
1
• Resistance R in series with a capacitor C have an impedance Z = R + jXC = R − j ωC
1
• R, L, C series circuit have an impedance Z = R + jXL + jXC = R + jωL − j ωC
= R + j(ωL −
Time domain
Begin
Phasor domain
Begin
Circuit in the
sinusoidal
steady state
Circuit
in the
phasor
Forced response of
circuit differtial eqn
Algebraic
solution
techniques
Sinusoidal
response
waveforms
phasor
responses
End
End
flow.diagram.for.phasor.circuit.analysis.m4
1
ωC )
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
10.7
61
Demo questions
• Complex Arithmetic
Cartesian form c = a + jb, Polar form c = |r|6 θ
|r|= magnitude/absolute value
θ=angle
a= real part, b= imaginary part
a − jb= complex conjugate of a + jb.
In Matlab: c=3+j*4;
r=abs(c); theta=angle(c); cstar=conj(c);
√ a=real(c); b=imag(c);
−1 b
2
2
6
a + jb = r θ, |r| = a + b , θ = tan ( a )
a + jb = r 6 θ, a = |r| cos(θ), b = |r| sin(θ)
1
1
= a+jb
? a−jb
= aa−jb
2 +b2
a+jb
a−jb
When adding/subtracting complex numbers use Cartesian form while when multiplying/dividing
use polar form.
c1 ± c2 = (a1 + jb1 ) ± (a2 + jb2 ) = (a1 ± a2 ) + j(b1 ± b2 )
|r1 |6 θ1
|r1 | 6
c1
= |r
(θ1 − θ2 )
= |r
c2
2|
2 | 6 θ2
6
c1 ? c2 = |r1 | θ1 ? |r2 |6 θ2 = |r1 r2 |6 (θ1 + θ2 )
• Series RLC circuits. The phasor domain voltages are as follows:
Observer
jωL
VL leads IL , IC leads VC
ωt
RI
I
jωC
• Zin for R series with R//L:
500Ω
Zin ⇒
RsRpL.m4
1mH
100Ω
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
Zin = 500 + 100//(jω/1000) = 500 +
⇒ Zin = 500 +
ω 2 (e−4)
1e4+ω 2 (e−6)
+
0.1ω
100+jω/1000
j10ω
1e4+ω 2 (e−6)
?
100−jω/1000
100−jω/1000
62
= 500 +
0.1jω(100−jω/1000
10000+ω 2 /(1e6)
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
63
TR5eEx8.6 The circuit in the figure below is operating in the sinusoidal steady state with vs (t) = 35 cos(1000t)V .
25mH
i(t) 50Ω
10µF
+
vs (t) = 35 cos 1000t
−
tr5eex8.6.m4
a. Transform the circuit into the phasor domain.
ZR = R = 50Ω
ZL = jωL = j1000 × 25 × (1e − 3) = j25Ω
1
1
ZC = jωC
= j1000×(10e−6)
= −j100Ω
Using these results, we obtain the phasor-domain circuit as shown below:
+
−
I 50Ω
+V −
R
−j100Ω
+ −
VC
j25Ω
+ V −
L
Vs = 356 0o
tr5eex8.6.phasor.m4
b. Solve for the phasor current I.
KVL: −Vs + VR + VC + VL = 0 ⇒ −356 0 + 50 × I + (−j100) × I + (j25) × I = 0
6 0
356 0
356 0
o
6
= 50−j75
= 90.135
⇒ I = 50−j100+j25
6 −56.3o = 0.388 56.3
c. Solve for the phasor voltage across each element.
VR = ZR × I = 50 × 0.3886 56.3o = 19.46 56.3o , VR in phase with I
VL = ZL × I = j25 × 0.3886 56.3o = 9.706 146.3o, VL leads I by 90o
VC = ZC × I = −j100 × 0.3886 56.3o = 38.86 −33.7o , VC lags the current by 90o
d. Construct the waveforms corresponding to the phasor found in (b) and (c).
o
i(t) = Re{0.388ej56.3 ej1000t } = 0.388 cos(1000t + 56.3o )
o
vR (t) = Re{19.4ej56.3 ej1000t } = 19.4 cos(1000t + 56.3o )
o
vC (t) = Re{9.70ej146.3 ej1000t } = 9.7 cos(1000t + 146.3o )
o
vL (t) = Re{38.8e−j33.7 ej1000t } = 38.8 cos(1000t − 33.7o )
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
64
• Example 10.12 HKD7e
In the circuit shown, given that the voltage V = 16 0o construct a phasor diagram showing
IR , IL , and IC . Combining these currents, determine the angle by which Is leads IR , IC , and Ix .
Ix
IR
IL
IC
j0.3S
Is
−j0.1S
+
0.2S V
−
hkd7eex10.12.m4
IR = (0.2)(16 0) = 0.26 0
IL = (−j0.1)(16 0) = 0.16 −90o
IC = (j0.3)(16 0) = 0.36 90o
and the phasor diagram
Is=Ix+IC
IC
IR
IL
Ix=IL+IR
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
65
e
100mH
c
+
100∠0o −
10Ω
b
a
10Ω
d
Skilling2ep164.m4
10µF
Skilling2e Page 164: Find Zae , currents and voltages.
Using Maxima:
/* Skilling2ep164.mac */
restart; kill(all); globalsolve:true;
ZpZ(Z1,Z2):=Z1*Z2/(Z1+Z2);
R1:10; R2:10; C:10e-6; L:100e-3; Vs:100*%e^(%i*0);
Zab:R1;
Zbde:1/(%i*w*C);
Zbce:R2+%i*w*L;
Zae:Zab+ZpZ(Zbde,Zbce);
f:60;
w:2*%pi*f;
zae:ev(Zae); rectform(%); polarform(%);
/* zae=43.3486628577 %i + 23.5616138933= 49.3381821798 %e^(1.07292844092 %i) */
float(1.0729*180/%pi); /* 61.472 */
Iab:polarform(float(Vs/zae)); rectform(%);
/* 2.02682781533 %e^(- 1.07292844092 %i)=0.9679184011 - 1.780776505 %i
*/
Ibce:float(ev(Iab*Zbde/(Zbde+Zbce)));
rectform(Ibce); polarform(Ibce);
/* Ibce= 1.0350520 - 2.1212779 %i; 2.3603290 %e^(- 1.1168448 %i); */
Ibde:float(ev(Iab*Zbce/(Zbde+Zbce))); rectform(Ibde); polarform(Ibde);
/* Ibde= 0.340501 %i - 0.0671336; 0.347056 %e^(1.7654608 %i); */
/* as a check: Iab=Ibce+Ibde */
Isum: float(ev(Ibce+Ibde)); rectform(Isum); polarform(Isum);
/* find the voltages Vab, Vbe and check their sum must equal Vs */
Vab: Vs*Zab/Zae;
vab:float(ev(Vab)); rectform(vab); polarform(vab);
/* vab=9.679184011 - 17.80776505 %i= 20.2682781533 %e^(- 1.07292844092 %i) */
Vbe: Vs*ZpZ(Zbde,Zbce)/Zae;
vbe:float(ev(Vbe)); rectform(vbe); polarform(vbe);
/* vbe=17.80776505 %i + 90.3208159881= 92.0595801474 %e^(0.194664509 %i) */
vsum: vab+vbe; rectform(vsum); polarform(vsum); /* answer: vsum=100 */
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
66
TR5eEx8.10 Find the steady-state currents i(t), iC (t), and iR (t) in the circuit shown below for vs (t) =
100 cos(2000t)V . L = 250mH, C = 0.5µF, and R = 3kΩ.
i(t) L
vs (t)
C
iC (t) iR (t)
R
tr5eex8.10.m4
a. Transform the circuit into the phasor domain.
ZR = R = 3000Ω
ZL = jωL = j2000 × 250 × (1e − 3) = j500Ω
1
1
= j2000×(0.5e−6)
= −j1000Ω
ZC = jωC
Vs = 1006 0
Using these results, we obtain the phasor-domain circuit as shown below:
I
j500Ω
+
Vs = 1006 0o
−
IC
−j1000Ω
IR
3000Ω
tr5eex8.10.phasor.m4
b. Solve for the phasor currents I, IC , IL .
KCL: I − IC − IR = 0 ⇒ −(V2 − Vs )/ZL − V2 /ZC − V2 /ZR = 0
1
⇒ −(V2 − 100)/(j500) − V2 /(−j1000) − V2 /3000 = 0 ⇒ 100 = V2 ( j500
−
100
1
1
1
= V2 ( j500 + −j1000 + 3000 )
j500
⇒ 100 = V2 (1 − 0.5 + j0.1666) ⇒ V2 =
=
100
0.52706 18.4o
189.7446 −18.4o
= 0.06326 −18.4o
3000
6 −18.4o
V2
= 189.744
= 0.18976 90 − 18.4o = 0.18976 71.6o
−j1000
−j1000
100−189.76 −18.4o
= −80+j59.878
= 0.1198 + j0.16 = 0.26 53.2o
j500
j500
IR =
IC =
I=
100
0.5+j0.1666
V2
R
1
−j1000
−
= 189.7446 −18.4o
=
c. Construct the waveforms corresponding to the phasors found in (b).
o
iR (t) = Re{0.0632e−j18.4 ej2000t } = 0.0632 cos(2000t − 18.4o )
o
iC (t) = Re{0.1897ej71.6 ej2000t } = 0.19 cos(2000t + 71.6o )
o
i(t) = Re{0.2ej53.2 ej2000t } = 0.2 cos(2000t + 53.2o )
1
)
3000
⇒
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
67
HKD7ep10.78 Use ω = 1 rad/s, and find the Norton equivalent circuit as seen by a load connected between a, b.
1F
+
+
o
16 0 −
−
+ VL −
2H
0.25VL
a
⇐ Norton
b
hkd7ep10.78.m4
OC condition: Voc = Va = VT h = 16 0o
SC condition: kcl 2:
6 0o
− V2 −1
− 0.25VL − Isc = 0 ⇒ −jV2 + j − 0.25(j2 × Isc ) − Isc = 0
−j
and V2 = VL = Isc (j2)
substitute in kcl 2 ⇒ −j(j2 ×Isc ) + j −0.5jIsc −Isc = 0 ⇒ Isc (1 −j0.5) = −j ⇒ Isc =
j
= 0.896 −63.43o
1.126 153.43o
6
o
0
o
oc
6
ZT h = VIsc
= 0.896 1 −63.43
o = 1.12 63.43
YN = ZT1 h = 0.896 −63.43o
IN = Isc = 0.896 −63.43o
j
−1+j0.5
=
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
68
HKD7ep10.81 Find the Thevenin equivalent circuit as seen by a load connected between a, b.
−j300Ω
200Ω
+
+
o
1006 0 −
−
j100Ω
a
+
o
6
− 100 −90
⇐ T hevenin
b
hkd7ep10.81.m4
OC condition: Voc = Va
6 90o
a−100)
kcl a: − (V−j300
− V a−100
= 0 ⇒ Va − 100 − 3Va + 3006 90o ⇒ 2Va = 3006 90o − 100 =
j100
j300 − 100 ⇒ Va = j150 − 50
SC condition:
6 90o
− (0−100)
− 0−100
− Isc = 0 ⇒ Isc = 3j − j 6 90o ⇒ Isc = 1 − 3j
−j300
j100
oc
= j150−50
= j150Ω
= j3 150+j50
ZT h = VIsc
3+j1
1+ j
3
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
69
KD7eEx10.5 Find Zin .
−j1
200mF 2H
Zin ⇒ 10Ω
6Ω
j10
Zin ⇒ 10Ω
500nF
6Ω
−j0.4
hkd7eex10.5.pasor.m4
hkd7eex10.5.m4, ω = 5 rad/sec
) = 10//(−j1 + j10 + 0.0265 −
Zin = 10//(−j1 + j10 + (6//(−j0.4))) = 10//(−j1 + j10 + 6(−j0.4)
6−j0.4
j0.398) = 10//(0.0265 + j8.602) =
(10)(0.0265+j8.602)
(10+0.0265+j8.602)
= 4.255 + j4.929 = 6.5116 49.2oΩ
KD7eEx10.6 Find the time domain node current i(t) in the circuit shown below.
i(t)1.5kΩ
+
+
vs (t) −
−
1kΩ
1
H
3
I 1.5kΩ
1
µFVs
6
+
+
= 406 −90 −
−
1kΩ
j1k
−j2k
vs (t) = 40 sin 3000t
hkd7eex10.6.phasor.m4
hkd7eex10.6.m4
In the phasor domain:
kcl2: −(V2 − Vs )/(1.5k) − V2 /(j1k) − V2 /(1k − j2k) = 0
Using Matlab:
Vs=40*exp(j*(-90)*pi/180)
den=(1/(1.5e3)+1/(i*1e3)+1/(1e3-i*2e3))
V2=Vs/(1.5e3)/den
I=(Vs-V2)/(1.5e3)
abs(I)
angle(I)*180/pi
Using Maxima
Vs:40*kcl: -(V2-Vs)/(1.5e3)-V2/(den:(1/(1.5e3)+1/(V2:Vs/(1.5e3)/den; float(polarform(V2));
float(rectform(V2)); I:float(polarform((Vs-V2)/(1.5e3)));
or we can find Zin then I = ZVins as follows:
= 1.5k + j1k+2k
= 1.5k + 2+j1
? 103 =
Zin = 1.5k + (j1k)//(1k − j2k) = 1.5k + (j1k)(1k−j2k)
j1k+1k−j2k
1−j1
1−j1
6
2.2361 0.4636
o
3
3
3
6
6
(1.5 + 1.4142
6 −0.7854 ) ? 10 = (1.5 + 0.5 + j1.5) ? 10 = (2 + j1.5) ∗ 10 = 2.5 0.6435k = 2.5 36.87
6
o
40 −90
o
6
I = ZVins = 2500
6 36.87o = 16 −126.9 mA.
⇒ i(t) = 16 cos(3000t − 126.9o ) mA.
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
70
KD6eEx10.7 Find the time domain node voltages v1 (t) and v2 (t) in the circuit shown below.
−j5Ω
V1
16 0o
V2
5Ω
j10Ω
−j10Ω
j5Ω
10Ω
−.56 −90o
hkd6efig10.21.m4
KCL at nodes 1 and 2 and using Maxima we get:
/* hkd6eex10.7.mac */
globalsolve:true;
restart; kill(all);
I1:1*exp(%i*0); /* left current source */
I2:0.5*exp(%i*(-90)*%pi/180); /* right current surce */
KCL1: I1-V1/5-V1/(-%i*10)-(V1-V2)/(%i*10)-(V1-V2)/(-%i*5)=0;
KCL2:-(V2-V1)/(-%i*5)-(V2-V1)/(%i*10)-V2/(%i*5)-V2/(10)-I2;
soln:solve([KCL1,KCL2],[V1,V2]);
soln1:rectform(soln); /* put the answer in rectangular form */
float(cabs(V1)); float(carg(V1)*180/%pi); /* magnitude and angle in degrees */
float(cabs(V2)); float(carg(V2)*180/%pi); /* magnitude and angle in degrees */
/* time domain solution */
v1t:float(cabs(V1))*cos(w*t+float(carg(V1)*180/%pi));
v2t:float(cabs(V2))*cos(w*t+float(carg(V2)*180/%pi));
Maxima answers:
(%o5)
(%o8)
(%o9)
(%o10)
(%o11)
(%o12)
(%o13)
[[V1 : 1 - 2 %i, V2 : 4 %i - 2]]
2.23606797749979
- 63.43494882292201
4.47213595499958
116.565051177078
2.23606797749979 cos(t w - 63.43494882292201)
4.47213595499958 cos(t w + 116.565051177078)
D6eEX10.11 (Circuit with Two different frequencies) Determine the power dissipated by the 10Ω resistor in the circuit shown below.
10Ω
2 cos(3t)A
0.2F
hkd6efig10.31.m4
0.5F
2 cos 5tA
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
71
Note that there are two different frequencies in the circuit, hence apply one source at a time,
i.e. use superposition.
I1 10Ω
−jΩ
−j0.4Ω
I2 10Ω
26 0o A
−j1.667Ω
56 0A
hkd6efig10.31b.m4
−j0.667Ω
hkd6efig10.31c.m4
in Maxima,
kill(all); restart; globalsolve:true;
R:10; w1:3; w2:5;
KCL1: -V1/(10-%i)-V1/(-0.4*%i)-2=0;
KCL2: 5-V2/(-1.66*%i)-V2/(10-0.667*%i)=0;
linsolve(KCL1,V1);
I1:-V1/(10-%i);
float(rectform(I1));
i1t:float(cabs(I1)*cos(w1*t+carg(I1)*180/%pi));
linsolve(KCL2,V2);
I2:V2/(10-0.6667*%i);
float(rectform(I2));
i2t:float(cabs(I2)*cos(w2*t+carg(I2)*180/%pi));
p:R*(i1t+i2t)^2;
Results from Maxima:
I1: 0.010984699882307 - 0.078462142016477 %i
i1t : 0.079227339733946 cos(3.0 t - 82.03038960567865)
I2: 0.18319662747181 - 0.78737179752408 %i
i2t : 0.80840296378313 cos(5.0 t - 76.90210746936657)
p : 10 (0.80840296378313 cos(5.0 t - 76.90210746936657)+0.079227339733946 cos(3.0 t - 82.
Q25 Using phasors, determine i(t) in the following equations:
di
1. 2 dt
+ 3i(t) = 4 cos(2t − 45o )
2. 10 idt +
R
di
dt
+ 6i(t) = 5 cos(5t + 22o )
Rewrite in phasor form: 2jωI + 3I = 46 (−π/4),
6 −π/4
46 (−π/4)
4 6
6
I = 43+j2ω
= 3.6056
6 0.5880 = 3.6056 (−pi/4 − 0.5880) = 1.1094 (−1.3734).
Back to time-domain: i(t) = 1.1094 cos(2t − 1.3734 ∗ π/180) (angle in degrees). Using Matlab
>> n=4*exp(-j*pi/4);
>> d=3+j*2;
>> I=n/d
I =
0.2176 - 1.0879i
>> abs(I)
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
72
ans =
1.1094
>> angle(I)
ans =
-1.3734 in radians
For the second part in phasor form:
1
10( jω
)I + (jω)I + 6I = 56 0.3840 with ω = 5 rad/sec.
10
I + j5I + 6I = 56 0.3840
j5
−j2I + j5I + 6I = 56 0.3840
6 0.3840
56 0.3840
6
= 6.7082
(6 + j3)I = 56 0.3840, and I = 5 6+j3
6 0.4636 = 0.7454 −0.0796
and in time domain: i(t) = 0.7454 cos(5t − 0.0796 ∗ 180/π) = 0.7454 cos(5t − 4.5607o)
Q54 In the circuit of the Figure, find Vs if Io = 26 0 A.
Write the mesh equations (by inspection)
−j2Ω
I1
j4Ω
2Ω
Vs
+−
−j1Ω
I3
I2
Io
j2Ω
1Ω
ch9p54.m4





0
I1
2 − j2 + j4
−j4
0





−j4
j4 + j2
−j2
  I2  =  −Vs 

0
I3
0
−j2
1 + j2 − j1
with Io = I3 = 26 0.
Using the third equation, −j2I2 + (1 + j1)I3 = 0, we get I2 =
1+j1
.
j2
j4(I2 )
j4
= 1+j1
2+j2
j2 2+j2
Using the first equation, (2 + j2)I1 − j4I2 = 0, we get I1 =
= 1.
Finally, using the second
(−j4)I1 + (j6)I2 − (j2)I3 = −Vs to get Vs = j4(1) − 3(1 +
√ equation,
o
6
j1) + j2 = −3 + j3 = 3 3 −45
Vs = 6 − j ∗ 6 = 8.48536 (−0.7854)
Solution using Maxima
eq1:(%I*2+2)*I1-%I*4*I2=0; (2*%I+2)*I1-4*%I*I2=0;
eq2:-%I*2*I3+%I*6*I2-%I*4*I1=-Vs; 6*%I*I2-4*%I*I1-4*%I=-Vs;
eq3:(%I+1)*I3-%I*2*I2=0; 2*(%I+1)-2*%I*I2=0;
I3:0*%I+2;
soln:solve([eq1,eq2,eq3],[I1,I2,Vs]);
[[I1=2,I2=(%I+1)/%I,Vs=6*%I-6]];
Q70 Find the equivalent impedance of the circuit in the figure below
Insert a 1-Volt source at the input, write the mesh equations (by inspection):
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
73
10Ω
+ j15Ω
+
16 0 −
−
I1
2Ω
I2
−j10Ω
5Ω
8Ω
I3
−j5Ω
ch9p70.m4
(12+j15)I1 −(10+j15)I2 −2I3 = 1, −(10+j15)I1 +(15+j5)I2 −5I3 = 0, −2I1 −5I2 +(15−j5)I3 =
0 and Z = 1/I1 .
Using maxima
globalsolve:true;
eq1:-2*I3-(%i*15+10)*I2+(%i*15+12)*I1=1;
eq2:-5*I3+(%i*5+15)*I2-(%i*15+10)*I1=0;
eq3:(15-%i*5)*I3-5*I2-2*I1=0;
soln:solve([eq1,eq2,eq3],[I1,I2,I3]);
[[I1:(3726*%i+5067)/97673,I2:(39163*%i+11971)/488365,I3:(3239*%i+394)/97673]];
RTh:1/I1
97673/(3726*%i+5067);
float(rectform(RTh)) 12.511111111111111-9.1999999999999993*%i
float(polarform(RTh)) 15.529581489356623/2.7182818284590451^(%i*atan(414/563))
angle:float((atan2(imagpart(RTh),realpart(RTh))*180)/%pi)
-36.328768708782754 degrees
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
74
Q74 Design an RL circuit to provide a 90o leading phase shift.
Use the circuit
I1 R
R
+
vi
L
−
I+2
L vo
−
CH9P54AS.m4
If vi = 16 0, write the two mesh equations:
−1+(R+%i∗XL)∗I1−%i∗XL = 0; −%i∗XL∗I1+(2∗R+%i∗XL)∗I2
= 0. vo = I2∗%i∗XL.
√
2
Solve for imagpart(vo)=0 to get XL = 2 ∗ R − 3 ∗ R.
In Maxima
/* ch9.p74.mac */
globalsolve:true;
eq1:-1+(R+%i*XL)*I1-%i*XL=0;
eq2:-%i*XL*I1+(2*R+%i*XL)*I2=0;
eq3: vo=I2*%i*XL;
soln:solve([eq1,eq2,eq3],[I1,I2,vo]);
[[I1:(XL-%i)/(XL-%i*R),I2:(XL^2-%i*XL)/(XL^2-3*%i*R*XL-2*R^2),
vo:(%i*XL^3+XL^2)/(XL^2-3*%i*R*XL-2*R^2)]]
solve(imagpart(vo)=0,XL);
[XL=-sqrt(2*R^2-3*R),XL=sqrt(2*R^2-3*R),XL=0]
√
and the acceptable answer is XL = 2R2 − 3R.
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
75
Q85 The ac bridge circuit of the figure below is called a Wien bridge. It is used for measuring the
frequency of a source. Show that when the bridge is balanced,
1
f= √
2π R2 R4 C2 C4
Write KVL for the three meshes and solve to find I1, I2, I3.
The current in the AC meter must be zero when the bridge is balanced, i.e. I2=I3. Solving for
ω we get ω = √R2 R14 C2 C4 .
1
I2
+
+
16 0 −
−
I1
R1
3
0 CH9P85.m4
Here is the Maxima code to do this:
R2
R3
µA
AC Meter
C2
2
I3
R4
4
C4
/* ch9.p85.mac */
kill(all); /* soft restart for Maxima */
globalsolve:true; /* make intermediate results as global */
/* the three mesh equations */
eq1:-1+(R1+R2+%i*(-1/(w*C2)))*I1-R1*I2-(R2-%i/(w*C2))*I3=0;
eq2:-R1*I1+(R1+R3)*I2=0;
eq3: -(R2-%i/(w*C2))*I1+(R2-%i/(w*C2)+R4*(-%i/(w*C4))/(R4-%i/(w*C4)))*I3;
/* call the Maxima solver */
soln:solve([eq1,eq2,eq3],[I1,I2,I3]);
/* we require that I2=I3, solve for w */
omega:solve(I2-I3=0,w);
/* w is real, hence imaginary part which must be zero, find R3 in terms of other componen
solve(imagpart(omega[2])=0,R3); /* R3:R4*(C2*R1-C4*R3)/(C2*R2); */
/* Algebraic systems such as Maxima, Maple, ... are not perfect
solvers when dealing with radicals ... hence take the square of w */
realomegasquare:(realpart(omega[2]))^2;
/* now substitute for the value of R3 that makes the imaginary part=0 */
realomegasquare:ratsubst(C2*R1*R4/(C4*R4+C2*R2), R3, realomegasquare);
The results are
R3=(C2*R1*R4)/(C4*R4+C2*R2);
w^2=((-C4^2*R3^2+2*C2*C4*R1*R3-C2^2*R1^2)*R4^2+(2*C2*C4*R2*R3^2+2*C2^2*R1*R2*R3)*R4-C2^2*
/(4*C2^2*C4^2*R2^2*R3^2*R4^2)
realomegasquare:ratsubst((C2*R1*R4)/(C2*R2+C4*R4),R3,realomegasquare);
w^2=1/(C2*C4*R2*R4)
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
• Find vx in the circuit
76
0.25F
1H
2Ω
+
2Ω
2 cos 4tA
3 cos(4t + 60o )V
vx (t)
−
+
−
schof6
ω = 4, Is = 26 0, Vs = 36 60o , ωL = 4(1) = 4, 1/(ωC) = 1/(4 ∗ 0.25) = 1, Z = jωL − j/(ωC) + 2 =
j4 − j1 + 2 = 2 + j3.
s +Vs
KCL@1 : +Is − Vx /2 − (Vx − Vs )/(Z) = 0 and solve for Vx = Z∗I
= 3.8235 + j1.4314 =
1+0.5Z
o
o
4.08276 20.524 , and finally vx (t) = 4.0827 cos(4t + 20.524 ).
Verify linearity in this AC circuit:
1.
2.
3.
4.
Is = 26 0, Vs = 0, Vxi = Is ∗ Z/(Z + 2) = 2.7200 + 0.9600i
Is = 0, Vs = 36 60o , Vxv = 2 ∗ Vs /(Z + 2) = 1.1035 + 0.4714i
Vx = Vxi + Vxv = 3.8235 + 1.4314i = 4.08276 20.524o
vx (t) = 4.0827 cos(4t + 20.524o)
• In a series circuit having vs (t) = 10 cos(10t) − 10 sin(10t) and impedance Z. If the current in
the loop i(t) = −1 sin(10t − 90o ), find the components of Z.
Recall that cos x = Re{ejx } = 16 0 and sin x = Re{−jejx } = −16 90o = 16 −90o .
Vs (ω) = 106 0 − 106 −900 = 10 − 10(−j) = 10 + j10, I(ω) = −16 −90 − 90 = −16 −180 =
−1(−1) = 1, Z = 10+j10
= 10 + j10 = R + jωL, from which we get R = 10, L = 10/10 = 1H.
1
• OPAMP in AC circuits: in the circuit shown, the output voltage Vo =
Cf
Rf
C
V0
+
−
+
+
16 0o V
−
wopr2c
I=
1
1
jωC
= jωC, Zf =
1
Rf ( jωC
)
f
1
Rf + jωC
=
f
Rf
1+jωCf Rf
KV L : I ∗ Zf + Vo = 0, we get
Rf
Vo = −IZf = −jωC
1 + jωCf Rf
!
=
−jωCRf
1 + jωCf Rf
• In the circuit shown, R = 4700Ω, C = 0.02µF . At which frequency will Vo be in phase with Vi ?
Zs = R −
j
,
ωC
Zp =
−j
R( ωC
)
j
R− ωC
1
. The series current Is = Vo R1 + Vo −j
, Vi = (Zs + Zp )Is
ωC
2
G=
2
2
Vi
w C R − 3j w C R + j2
=−
Vo
j wCR
10 SINUSOIDAL STEADY-STATE ANALYSIS-AC ANALYSIS
R
+ V
− i
rcsrcp
77
C
C
+
R Vo
−
for Vi , Vo to be in phase, then the angle of G must be zero, i.e. the angle of the numerator=angle
of denominator. It is clear that the angle of the denominator is 90o, hence the real part of the
numerator must be zero in order to have the numerator pure imaginary, i.e. ω 2 C 2 R2 − 1 = 0,
1
1
ω
ω = CR
= 4700∗0.02e−6
= 1.0638e + 004rad/sec, and f = 2π
= 1.6931e + 003 = 1693.1Hz.