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Unformatted text preview: CHAPTER 10  SINUSOIDAL STEADYSTATE ANALYSIS List of topics for this chapter :
Nodal Analysis
Mesh Analysis
Superposition Theorem
Source Transformation Thevenin and Norton Equivalent Circuits
AC Op Amp Circuits NODAL ANALYSIS Problem 10.1 Given the circuit in Figure 10.1 and i(t) = 5 Sin(1000t) amps, ﬁnd v0 (t)
using nodal analysis. 109 109 Figure 10.1 > Carefully DEFINE the problem.
Each component is labeled, indicating the value and polarity. The problem is clear. > PRESENT everything you know about the problem.
The goal of the problem is to ﬁnd V() (t) , which is clearly labeled in Figure 10.1, using nodal analysis. Thus, we need to label the nodes and ground. To ﬁnd Vo (t) without using derivatives and integrals, we must transform the circuit to the frequency domain. This allows 'us to ﬁnd the answer using algebra with complex numbers.
We can transform the circuit to the frequency domain after setting a reference value. Let us use a reference of A Sin(1000t + 4)) . ' In transforming to the frequency domain, remember that X L = jcoL and X C =1/(jcoC) .
Hence, the inductor becomes jcoL = j(103 )(10 x 103) = j10 and the capacitor becomes . l/jcoC = l/[j(103)(50x10‘6 )] = ~j20. 167 Let us draw the circuit after the transformation into the frequency domain and labeling the nodes
and ground. ' 10 Q V2 10 Q > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success.
The problem clearly states that the problem be solved using nodal analysis. Thus, the
technique to solve the problem is set. There is no reason to look at an alternative at this point. > ATTEMPT a problem solution. Let us begin by writing the node equations.
At node 1 : At node 2 :
VlO Vl—V2 Vz—Vl V2—0 V2—0 + =0 + + =
30 10 10 j20 10+j10 0 —5+ Simplifying,
V2‘V1+ﬂz_+V2(1*J)=0 v 3v —3V =150
1+ 1 2 10 20 20 ‘4V1’3V2 =150 (2)(V2_V1)+jV2+V2(l"j):O
2V1+3V2 = 0 Thus, the system of simultaneous equations is l4 3livll_F150l [2 allvzﬂol
Fvll 1 [3 3i
iV2l=(12—6)i2 4i [—_l
O
l_—_l
ll
(3va—
l—‘—I
W A
O U]
0 O
l_____l
ll
F—'—l
U‘I \]
l O L11
;_1 So,
V1 = 7540° or Vl (t) = 75 sin(1000t) volts V2 = 50100 or v2(t) = SOsin(1000t) volts Clearly, and I  ——“— — ——— — —— = (2.5)(1 — = Z  45°) amps 168 Hence, . V0 : j10 10 = (10290°)(2.5)(J§ 2 —45°) = 2542" 245° = 35.36245o volts Therefore,
Vo (t) = 35.36 sin(1000t + 45°) volts > EVALUATE the solution and check for accuracy.
Solving the problem with an alternate method, such as mesh analysis in this case, would show that the results of the problem solution are correct. Let us draw the circuit deﬁning the loop currents for mesh analysis. 109 109 Write the loop equations. Loop 1 : I1 = 540° amps Loop2: 30(I2 —Il)+1012—j20(12——I3)=0
. Loop3: —j20(I3—12)+IOI3+j10I3:0 Simplifying the equations for loops 2 and 3, Loop2: —30I1+(40—j20)I2 +j2013 = 0 Loop3: j2012 +(lO—Ij10)I3 = 0 Solve the third loop equation for I3 in terms of 12.
__  j20
”‘ 10 — j 10 13 I2 = (1—j)12 =1.4142245°12 Now, substitute I1 and I3 into the second loop equation and simplify to ﬁnd 12.
(—30)(540°) + (40 — j20) I2 + (j20)(l — j) I2 = 0
(30)(540°) + (40 — j20)I2 + (20 + j20) I2 = 0
6012 = (30)(540°)
_ (30)(5£0°) I2 60 = 2.510° amps Now, ﬁnd 13.
I3 = (1.41424 —45°) I2 = (1.41424 45°)(2.540°) = 3.5361  45° amps 169 Clearly, I0 :13 = 3.5364—45° amps. Evidently,
V0 =j101o = (10490°)(3.536z  45°) : 35.36445o volts and Von) = 35.36sin(1000t + 45°) volts This answer is the same as the answer obtained using nodal analysis. Our check for accuracy was
successful. ' > Has the problem been solved SATISFACTORILY? If so, present the solution; if not,
then return to “ALTERNATIVE solutions” and continue through the process again.
This problem has been solved satisfactorily. vo (t) = 35.36sin(1000t + 45°) volts Problem 10.2 [10. 5] Given the circuit in Figure 10.2 and vi (t) = 1000800000 volts, use nodal analysis to ﬁnd vo (t) . 20S) Figure 10.2 ' Let us start by building the ac circuit. The voltage source, vi (t) = 1000500000 volts, becomes
1010°volts, with 0) = 1000. The inductive reactance becomes jooL = j10 . The capacitive reactance becomes 1/ jcoC = 1/[j(1000)(50 X10'6)] = j20 . Now, draw the ac circuit. 209 V1 —j20§2 V2 j10§2 309 1040° V O 170 At node 1 : At node 2 : . 10V1 V1+V1—V2 Vl—V2 4V1 V2
=~— . . =——+ .
20 20 120 . 120 20 30+110
10—V1=V1+j(V1—V2) (4+j)V1=(0.6+j0.8)V2 . . 0.6+‘O.8 10=(2+j)V1—J2+JV2 Vl=—4:j—V2 Note that I0 = V1 / 20 was substituted when writing the equation for node 2. Substituting the equation for node 2 into the equation for node 1, 2+' 0.6+0.8‘
1OZ< 1x J)V_ 'V
_4+j 2 J 2
V _ 170 
or ' 2 ‘ 0.6— 3262
Clearly,
30 3 170
:——"V=_" —=.14,olt
V° 30+le 2 [3+jl[0.6—j26.2] 6 5 47026 v0 5 With a reference of A cos(1000t + (1)) ,
v0 (t) = 6.154 cos(1000t + 70.26°) volts Problem 10.3 Given the circuit in Figure 10.3 and V(t) = 20 cos(1000t) volts, ﬁnd i0 (t)
using nodal analysis. 6 Q 250 HF 10 Q
io(t)
V(t) a 8 Q E 10 mH
Figure 10.3 io (t) = 632.5 cos(1000t  18.44°) milliamps 171 MESH ANALYSIS Problem 10.4 Given the circuit in Figure 10.4 and i(t) = 2005(1000t) amps, ﬁnd io (t) using mesh analysis.
10 Q
io(t)
i(t) a 20 mH Figure 10.4 H" First, we transform the circuit to the frequency domain using a reference of A cos(1000t + and deﬁne the mesh currents as seen in the following circuit. Remember that XL = jcoL and XC = — ij '
10 Q
210° A 0 E j20 Q There is only one unknown loop current, Io.
Writing the loop equation, 10(I0 —— 2) +101o +j201o = O (20 + j20)Io = 20 20 1 110°
I  ——'— ~ — — m = 0.7071145° amps ° ‘20+j20‘1+j‘,/§445o Therefore,
io (t) = 707.1 cos(1000t — 45°) milliamps 172 . Problem 10.5 Given the circuit in Figure 10.5 and i(t) = 5 sin(1000t) amps, ﬁnd io (t)
using mesh analysis. 109 109 io(t)
10 mH Figure 10.5 Transform the circuit to the frequency domain using a reference of A sin(1000t + (1)) and deﬁne
the mesh currents. 109 100 Clearly, IO :12. Use mesh analysis to ﬁnd I1 and 12.
Forloopl: 30(11——5)+1011—j20(11—IZ)=0
Forloop2: j20(IZ—Il)+1012+j1012=0 Simplifying,
(40— j20)I1 + j2OI2 = 150 j2011+(10—j10)12 =0 Simplifying further,
(2—j)11+jI2 = 7.5
j211+(l—j)I2 = 0 'Find II in terms of I2 for the second loop equation.
l+j l ,1 1 ,
.11=( j2 )12 =[§+JE]12 =[3)(1+J)12 . . Substituting this equation into the ﬁrst loop equation we get, 173 (2 ')[——1+j)1+'1 75
‘J 2 J 2 : ~ 
2 .
(2—j)(1+j)12+j212=15
(3+j3)12 =15
15 5 540° Hence,
I0 = 3.5364 — 45° amps Therefore, .i0 (t) = 3.536 sin(1000t — 45°) amps Wm. SUPERPOSITION THEOREM The superposition theorem applies to ac circuits the same as it does for do circuits. This
theorem is important if the circuit has sources operating at different frequencies. Since the
impedances depend on frequency, we must have a different frequencydomain circuit for each.
source. The total response is obtained by adding the individual responses in the time domain. Problem 10.6 Given the circuit in Figure 10.6, i(t) = 200$(1000t) amps and
v(t) = lOsin(4000/ 3t) volts, ﬁnd iC (t) . 200 Figure 10.6 Because the two sources have different frequencies, we need to use superposition to solve this
problem. Thus, iC (t) = iCl (t) + icz (t). Start with the current source and a reference of A cos(1000t + ¢) . 240° A 20 Q 174 Using current division, 20 2 2 1+'
IC1 =( )(2) = 1———,=(—)(—J—)= 1+j = @249: 1.4142245o amps Hence, iCI (t) = 1.4142 COS(1 OOOt + 45°) amps Next, use the voltage source with a reference of A sin(4000 / 3t + (1)) . 20 Q
Icz
—j15 Q 1040° V
Clearly,
' 2 2 4+ '3 2 5436.87°
IC2 = = = 956$: = lag—“J = 0.4L36.87° amps
Hence, iC2 (t) = O.4sin(4000/3t + 36.87°) amps Recall that iC (t) = iCl (t) + iC2 (t) . Therefore,
iC (t) = [1.4142 c0s(1000t) + 0.4 sin(4000/ 3t + 36.87°)] amps Problem 10.7 [10.33] Given the circuit in Figure 10.7 and vi (t) = l2cos(3t) volts, solve for V() (t) using the principle of superposition. 10V Figure 10.7 175 Let V0 : Vl + V2 + V3 where V1, V2 , and V3 are respectively the voltages produced by the 10 volt dc source, the ac current source, and the ac voltage source acting independently. For V1, .
consider the circuit shown in Figure (a). 69 2H The capacitor is an open circuit to do while the inductor is a short circuit. Hence, V1 =10 volts and V1 (t) = 10 volts. For V2 , consider the circuit in Figure (b). to = 2 , so the inductor becomes jcoL = j4.  Likewise, the capacitor becomes 1/ ij =  j / (2 / 12) = 1'6. 69 Applying nodal analysis,
4 YA+£+XL£1+A .1},
‘6 j6 j4‘6 J6 J4 2
which leads to = 21.45.426.56o volts o (4)02)
2  J V2: With a reference of A Sin(2t + ,
V2 (t) = 21.45 sin(2t + 26.56°) volts 176 For V3, consider the circuit in Figure (0). 0) = 3 which leads to ij = j6 for the inductor and
1/ jcoC =  (3/12) = —j4 for the capacitor. At the nonreference node,
12 — V3 V3 + V3
6  j4 j6 which leads to (2x12) _ o 1
V3 = . — 1073312656 volts
2+] With a reference of A cos(3t + (1)) ,
v3 (t) = 10.733 cos(3t — 26.56°) volts Recall that vo (t) = Vl (t) + v2 (t) + v3 (t). Therefore,
V() (t) = [10 + 21.45 sin(2t + 25.56°) + 10.733 cos(3t ~ 25.56°)] volts Problem 10.8 Given the circuit in Figure 10.8 and vs (t) = cos(l 0000 volts, ﬁnd v(t) . 20 Q 50 uF 20 mH O vs(t) Figure 10.8 "H v(t) = 10 + 20 c0s(1000t — 90°) volts
or v(t) = 10 + 20 sin(1000t) volts 177 SOURCE TRANSFORMATION Source transformation in the frequency domain involves transforming a voltage source in
series with an impedance to a current source in parallel with an impedance or vice versa. We
must keep the following relationship in mind when performing source transformations. Vs vS = 2515 (=9 IS = 2—
S Problem 10.9 Given the circuit in Figure 10.9 and V(t) = 20 cos(1000t) volts, ﬁnd v0 (t)
using source transformations. 6 Q 250 HF 10 Q
8 Q 10 mH v(t) Figure 10.9 Transform the circuit to the frequency domain using a reference of Acos(1000t + . 6Q —j4Q 109 2040° V O 8 Q Reduce the circuit using source transformations. Begin with the 2040°V source in series with
69. which becomes a 10/340°A source in parallel with 69. j10 r2 —j4 Q 10 Q 10/340° A 8 Q j10 (2 NH 178 Now combine 69  89 = 24/79. The 10/310°A source in parallel with 24/79 becomes an
80/740°V source in series with 24/79. 24/7 9 —j4 9 10 9 80/740° V 0 To perform the next source transformation, we need to ﬁnd the parallel equivalent of the resistor
and capacitor in series. We know that two series impedances and two parallel impedances are l l 1
= Z + Z d = —— + —
51 52 an Z ZPl ZP2 I Peq j10 o "H Z Seq To ﬁnd the parallel equivalence of two series impedances, let ZSeq = ZPeq . Then, __1___ ;+_1_
ZSI +ZS2 ZPl ZPZ It can be shown that 1 1 1
24/7— j4 ‘ 8.095 + — j6.939 Thus, the 80/740°V source in series with 24/7 —j49 becomes a 2.169449.4°A source in parallel
with 8.095 ~j6.9399. 2.169449.4°A j10 9 Combining 8.0959  209 = 5.7639. We now need to ﬁnd the series equivalent of the resistor
in parallel with the capacitor. It can be shown that 1 1 1 5.763 + ~j6.939 z 3.411—j2.833 Thus, the 2.169.449.40A source in parallel with 5.763 — j6.9399 becomes a 9.61849.69°V source
in series with 3.411 ~j2.8339. Before redrawing the circuit, let’s combine the series resistances of3.4119 +109 = 13.4119. 179 13.411 Q —j2.833 Q . 9.61849.69°V Now that we have found a less complex circuit via source transformations, it is simple to solve for v0 (t) . 9.61819.69° 9.61849.69° 9.61849.69°
I — W w — — = 0.63254 18.43° " 13.411ej2.833+j10 =13.411+j7.167 _15.206428.12° and
V0 =j101= (10190°)(0.63251 18.43°) = 6.325471.57° Recall that we used a reference of A cos(1000t + (1)) . Therefore,
v°‘(t) = 6.325 cos(1000t + 71.57 °) volts THEVENIN AND NORTON EQUIVALENT CIRCUITS Problem 10.10 Given the circuit in Figure 10.10 and V(t) = 100 cos(1000t) volts, ﬁnd ' V111 , IN , and Zeq looking into terminals a and b. Figure 10.10 Because the circuit in Figure 10.10 has a capacitor, we can only determine VTh , IN , and Zeq for
an ac circuit in the frequency domain. Hence, the circuit becomes, 180 10040° V 3 Begin by ﬁnding the opencircuit voltage, V0c . Note that there is no current ﬂowing through the capacitor. Thus, Voc = V1. Using nodal analysis, V,—100 Vl—O
5 + 5 ~3I=O
100—V1
whereI= .
5
So,
Vl—IOO Vl—O [100~V1)
5 + 5 “(3) 5 =
. (VI—100)+V1—(3)(100—V1)=0
5V1=400
V1=80volts
Hence, VCC = 8OZO° volts Now, ﬁnd the short—circuit current, Isc . 100£0° V C Use nodal analysis to ﬁnd V1 and then 181 Writing the nodal equation, V1—100 Vl—O Vl—O
+ — — 31+ . = 0
5 5  _]
100 — V1
where I = —— .
5 So, V1 «100 V1~0 [ICC—VI] . 5 + 5 ~(3) 5 +JV1 — 0
(V1 —— 100) +Vl — (3)(100—V1)+ j5V1 = O
(5 + j5) VI = 400
400 ’ 80 (80)(1 — j)
V : ———— = ————. : ———— : —— t
1 5+j5 1+] 2 (40)“ J)
Thus,
V 40 1— ' ISO = = = (40)(1+ j) = 4045449 amps Finally,
°q ’ Isc — 40J§z45° _ 0 ms Therefore, VT11 = V0c = 8010° or VT11 (t) = 80 c0s(1000t) volts IN = Isc = 40ﬁz45°_ or iN (t) = 56.57 c0s(1000t + 45°) amps zeq = Jig45° or zeq = (l—j)ohms
Problem 10.11 [10. 43] Find the Thevenin and Norton equivalent circuits for the circuit shown in Figure 10.11. 601120° V 5 Q —j10 Q 2 o
j20 Q Figure 10.11 182 To ﬁnd ZTh , consider the circuit shown in Figure (a). SS) —j10§2 2E2 j20 Q «— (a)
zeq = 1'20” (5 —j10)+2 = (16—j12)+2 =18—j12 = 21.634—33.69° ohms To obtain VTh , consider the circuit in Figure (b). 5 Q —j10 s2 2 Q
+
604120° V j20 o VDC
(1))
1'20 j4
= —————— 604120° = 420°
V°° 5—j10+j20 1+j260 Voc =(l.7889£26.57°)(601120°)=107.334146.57° volts 07.334146.57°
= V—‘m— — —1—————— = 49614180260 amps “I
where ISC is the current ﬂowing downward through a short across the terminals. Vm=V oc’ Recall that ZTh = ZN = Z and IN = Isc. eq’ Therefore, zTh = 2N = (18— j12) ohms
VTh = 107.334146.57° volts
IN = 49614180260 amps 183 Problem 10.12 Given the circuit in Figure 10.12 and V(t) = 100 cos(1000t) volts, ﬁnd
i(t) , the current through R, for R = 09, IQ, 109, 1009, and 10009. 10 mH SQ 109 Figure 10.12 R I i(t)
r 0 £2 56574  45° A 5.657 c0s(1000t — 45°) A
1 9 5.4391  45°A 5.439 cos(1000t  45°) A
10 (2 4.0411  45° A 4.041 cos(1000t  45°) A
100 Q 1.13144  45° A 1.1314 c0s(1000t  45°) A
1000 Q 0.1381  45°A 138 cos(1000t  45°) mA
AC OP AMP CIRCUITS Given the ac circuit in Figure 10.13, ﬁnd Vout as a function of Vin . Problem 10.13 —j1000 Q Figure 10.13 184 Using nodal analysis at node a, Va  Vin Va — Vout
————+—————~=0 103  le3
where Va = Vb = 0.
So,
 Vin + — Vout __
103 ﬂ03 —
 jVout : Vin
V.
Vow = "T = jVin
' J
Therefore, Vout : Vin 490° The output is equal to the input except for a phase shift of 90°. Problem 10.14 Given the circuit in Figure 10.14 and Vin (t) = lOsin((ot) volts, ﬁnd vout (t)
for a) = l, 10, 100, 1000 rad/sec. lOuF Figure 10.14 185 Use the following ac circuit, with a reference of A Sin(0)t + , to ﬁnd V in terms of CD. out —j105/(o Q 1040° Using nodal analysis at node a,
Va ~ Va — Vout Va — Vout _ .——‘+ . —0
104 + 1105/0) 3100)
where Va = Vb = 0.
So,
'10 'Vout Vout
—— + =0
104+j105/03 lem
. le4
100—_]0)V0m+ a) Vow—0
104
[— j0)+O‘)“]Vout = 100
100 i 100 100
V ——————~—=L= w 490° Now, substitute the values for co into the equation for Vow. 100
At (D :1 rad/sec, ‘70ut E _104 490° :: 90° vent (t) = 10 sin(t ~ 90°) mV 03 103 At a) = 10 rad/sec, Vout 5 W490 :  9900 v0“t (t) = 101.01 sin(10t — 90°) mV 186 490°: 0.101014 90° 1 10
At a) = 100 rad/sec, Vout E m190° = ‘0—490° This corresponds. to the case where the LC combination forms a parallel resonant circuit and the
output goes to inﬁnity. 05 105 1
At 0) = 1000 rad/sec, Vout E Wig)" = WZ90° = (1101014900 vow (t) = 101.01 sin(1000t + 90°) mV In conclusion, the output, Vout (t) , has a ~90° phase shift for all values of 00 less than 100 and has
a 90° phase shift for values of co greater than 100. 187 ...
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This note was uploaded on 09/06/2010 for the course EE 50145 taught by Professor Chi during the Spring '10 term at UCSF.
 Spring '10
 Chi

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