R(T) = 1 To = 5 0.2 -1 dt = 0.2 √ {2+5 cos [10xt+tan~¹ (4/3)]} {2+5 cos [10z (t + 7) + tan`¹ (4/3)]} -0.2 = 5 √.⁰². 0 + To = 5 0.2 + x(t)x(t+t) dt 125 2 -0.2 +50 of cos [10x (1 + r) + tan¯¹ (4/3)] dt +0.2 $ 1.0². = 4+ 4dt + 50 4+1050 8 (+ + +) + +25 cos [10´t + tan¯¹ (4/3)] cos [10n (t + t) + tan−¹ (4/3)] 0.2 1.⁰² 0 125 1.03 -0.2 0 4dt + 0 + 0 + COS [10nt + tan-¹ (4/3)] dt cos (10лT) dt + +25 cos cos (10лT) 125 2 +0.2 6.0² 125 2 -1 cos (10лt) dt COS : [20ñt + 10πt̃ +2tan¯¹ (4/3)] dt (4/3)] } 0.2 cos [20nt + 10πT + 2 tan`¹ (4/3)] dr dt (2.148)
R(T) = 1 To = 5 0.2 -1 dt = 0.2 √ {2+5 cos [10xt+tan~¹ (4/3)]} {2+5 cos [10z (t + 7) + tan`¹ (4/3)]} -0.2 = 5 √.⁰². 0 + To = 5 0.2 + x(t)x(t+t) dt 125 2 -0.2 +50 of cos [10x (1 + r) + tan¯¹ (4/3)] dt +0.2 $ 1.0². = 4+ 4dt + 50 4+1050 8 (+ + +) + +25 cos [10´t + tan¯¹ (4/3)] cos [10n (t + t) + tan−¹ (4/3)] 0.2 1.⁰² 0 125 1.03 -0.2 0 4dt + 0 + 0 + COS [10nt + tan-¹ (4/3)] dt cos (10лT) dt + +25 cos cos (10лT) 125 2 +0.2 6.0² 125 2 -1 cos (10лt) dt COS : [20ñt + 10πt̃ +2tan¯¹ (4/3)] dt (4/3)] } 0.2 cos [20nt + 10πT + 2 tan`¹ (4/3)] dr dt (2.148)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Question
I don't know how to integral this answer
![R(t) =
=
1/0 1 1 0
To
-0.2
1
√ {2+5 cos [107r+ tan~¹ (4/3)]} {2+5 cos [10,7 (1 + r) + tan−¹ (4/3)]} dt
0.2
·0.2
0
0.2
= 560²
+50
+
0.2
= 56,0²
=
125
= 4+
x(t)x(t + t) dt
+ 125/0²
4 + 10 cos [10лt + tan−¹ (4/3)] + 10 cos [10ñ (t + t) + tan−¹ (4/3)] dt
+25 cos [10лt + tan¯¹ (4/3)] cos [10π (t + 7) + tan−¹ (4/3)]
25
4dt + 50
0.2
cos [107 (t + 7) + tan¯¹ (4/3)] dt
0.2
S 6.⁰²
0.2
[0².
cos (10лT) dt +
4dt +0+0+
COS
COS [10nt + tan¯¹ (4/3)] dt
cos (10лT)
125
2
125/08
-0.2
10²
-1
COS [20лt + 10 +2 tan¯¹ (4/3)] dt
cos (10лT) dt
[20лt + 10л + 2 tan-¹ (4/3)] dt
(2.148)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e554a09-0f9c-418e-bf2f-f101c15adca6%2F74c33b36-57dc-460d-a3b6-e4615db34405%2F7apvkdf_processed.png&w=3840&q=75)
Transcribed Image Text:R(t) =
=
1/0 1 1 0
To
-0.2
1
√ {2+5 cos [107r+ tan~¹ (4/3)]} {2+5 cos [10,7 (1 + r) + tan−¹ (4/3)]} dt
0.2
·0.2
0
0.2
= 560²
+50
+
0.2
= 56,0²
=
125
= 4+
x(t)x(t + t) dt
+ 125/0²
4 + 10 cos [10лt + tan−¹ (4/3)] + 10 cos [10ñ (t + t) + tan−¹ (4/3)] dt
+25 cos [10лt + tan¯¹ (4/3)] cos [10π (t + 7) + tan−¹ (4/3)]
25
4dt + 50
0.2
cos [107 (t + 7) + tan¯¹ (4/3)] dt
0.2
S 6.⁰²
0.2
[0².
cos (10лT) dt +
4dt +0+0+
COS
COS [10nt + tan¯¹ (4/3)] dt
cos (10лT)
125
2
125/08
-0.2
10²
-1
COS [20лt + 10 +2 tan¯¹ (4/3)] dt
cos (10лT) dt
[20лt + 10л + 2 tan-¹ (4/3)] dt
(2.148)
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Follow-up Question
![.. R(T) = 5 √ {4+
·+10 cos [1077t + Tan" (413)] +10 COS [COTT (t+7) + Tan`¹ (413)]
0.2
5. √ [4+10 cas [10πt + Tan (43)] +10 cos [10TT (t+7) + Tan` (4/3)]
= 5
let W₁ = 107T →⇒ T₁ =
002
it.
0
+ 25 Cos (LOTTT) + 25 cos [2017t + 10157+ 2 Tan ( 13
[²= 2 COSA COSB = cos [A-B) + Cos (A+B)]
21T
W₁
[cox(iont +0) dt =0
=
+25 cos [10πt+Tan (413)] (08 [1077(+7) + Tan
(Kr)= "720dt +0 + 0 +
R(T)
2TT = 0.2 sec
LOTT
ran (4/3)] }dt
+o) dt =0, ie. any sinusoidal signal integration
over one complete period = =0
an ²¹(+13)] } dt)
0.2
125 cos (1017) √dt + 125
2
D
(
√ cos(2017+ + a) dt
let 0₁ = 101TT + 2 Tant (4/3)
O](https://content.bartleby.com/qna-images/question/1e554a09-0f9c-418e-bf2f-f101c15adca6/f7d944b8-ff75-4e5a-8c24-275e773c3a4f/j98mhf8_processed.jpeg)
Transcribed Image Text:.. R(T) = 5 √ {4+
·+10 cos [1077t + Tan" (413)] +10 COS [COTT (t+7) + Tan`¹ (413)]
0.2
5. √ [4+10 cas [10πt + Tan (43)] +10 cos [10TT (t+7) + Tan` (4/3)]
= 5
let W₁ = 107T →⇒ T₁ =
002
it.
0
+ 25 Cos (LOTTT) + 25 cos [2017t + 10157+ 2 Tan ( 13
[²= 2 COSA COSB = cos [A-B) + Cos (A+B)]
21T
W₁
[cox(iont +0) dt =0
=
+25 cos [10πt+Tan (413)] (08 [1077(+7) + Tan
(Kr)= "720dt +0 + 0 +
R(T)
2TT = 0.2 sec
LOTT
ran (4/3)] }dt
+o) dt =0, ie. any sinusoidal signal integration
over one complete period = =0
an ²¹(+13)] } dt)
0.2
125 cos (1017) √dt + 125
2
D
(
√ cos(2017+ + a) dt
let 0₁ = 101TT + 2 Tant (4/3)
O
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