Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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For this question i get all the way to the point where i have a + bj [ 0.0166 + 0.1502j ] however, calculating the phase of this specific question just does not match up with the answer provided by my university.
the formula that i am using is that the phase = 90-arctan(Imaginary/Real) which in this particular problem should add output a phase of -96,31 degrees. However it does only output 6,31 and the input voltage does not contain -90 so i dont get it.
Transcribed Image Text:11.24 (A)* Consider the high-pass, active filter shown in
the Figure P11.24. If v¡(t) = 12 cos(1000m. t) V,
R₁ = 160 kn, RF = 220 kn, and C₁=220 pF, find vo(t).
C₁
Vi
Sensor signal
R₁
WW
FIGURE P11.24 High-pass, active filter.
RE
W
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Follow-up Question
This is the solution provided by my school. Why is the phase calculated using -90-arctan? is this just incorrect??
Transcribed Image Text:24
A
Ci= 230.10³ f
A₂
10
11
=
11
j
W= 1000TT; Ri 160k. Rf=220k 2
Vo
=
Vi
jwRfG
H₂WRIC
2.1000 7.220000-220-10¹ -12
12
1+ J. 1000TT· 160000.220.10⁰
0,15205 j
1+0,11058j
0,152053+0,0168
10122288
1-0,110585
-0,11058j
=
0,15205 j
1 +0,11058j
0,15205-0,0168 )
1-0,01222883²
F
E
0,01661+ 0,1502
TAL CA= √2,01661²201502²) <-go-tan' (3715682
9,1502
0,01661
Vi (4) = 0,1511-12 Cos (1000 π € - 96,31°) V
1,81 Cos (1000TTH -96,319) V
Vim =
-
0,1511
2-96,31
0
Solution
by Bartleby Expert
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
This is the solution provided by my school. Why is the phase calculated using -90-arctan? is this just incorrect??
Transcribed Image Text:24
A
Ci= 230.10³ f
A₂
10
11
=
11
j
W= 1000TT; Ri 160k. Rf=220k 2
Vo
=
Vi
jwRfG
H₂WRIC
2.1000 7.220000-220-10¹ -12
12
1+ J. 1000TT· 160000.220.10⁰
0,15205 j
1+0,11058j
0,152053+0,0168
10122288
1-0,110585
-0,11058j
=
0,15205 j
1 +0,11058j
0,15205-0,0168 )
1-0,01222883²
F
E
0,01661+ 0,1502
TAL CA= √2,01661²201502²) <-go-tan' (3715682
9,1502
0,01661
Vi (4) = 0,1511-12 Cos (1000 π € - 96,31°) V
1,81 Cos (1000TTH -96,319) V
Vim =
-
0,1511
2-96,31
0
Solution
by Bartleby Expert
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