In the following circuit: If vs (t) is given as a square waves as shown: If L=2H and R=492, Find i(t)? Vs (t)⭑ 8 Solution: Vs (t) 3 5 i(t) R Example 1: In the following circuit: If v,(t) is given as a square waves, If L=2H and R=492, Find i(t)? Vs (1)4 10 -π/2 π/2 3/2 2π 5/2 t t=0 i(t) R Vs (t) + Solution: Since T= then 10 = 2π/T = 2 rad/s 2/דר 30= dt = = 10 at 10 dt = 5 an an πο f(t) cos(noot)dt /2 10 cos(2nt)dt = 0 元 T f(t) sin(noot)dt bn = /2 20 bn 10 sin(2nt)dt for odd n пл So, Fourier Transformation of V, (t): 00 20 Vs (t)=5+Σ sin(2nt) ηπ odd n Now, we need to find i(t) for: 1- V, (t) = 5V using Laplace Analysis: V(s) 5/s 5 5/2 11(s)- = Z(s) (SL+R) s(2s+4) s(s+2) 5/4 5/4 = 11(s) S (s+2) 11(t)=(5/4) -(5/4) e 2-Vs (t)=- 20 -21 sin(2nt) using frequency analysis: nπ 20 V= ηπ & 0-2n Z=R+joL=4+ j(2n)(2) = 4+ j4n=4(1+jn) V 5 5 I= = /-tan 'n Z nл(1+jn) n√√1+n² 5 i2(t) sin(2nt-tan h) n√1+n² The complete solution is: 00 i(t) = i(t) + Σ i2(t) odd n
In the following circuit: If vs (t) is given as a square waves as shown: If L=2H and R=492, Find i(t)? Vs (t)⭑ 8 Solution: Vs (t) 3 5 i(t) R Example 1: In the following circuit: If v,(t) is given as a square waves, If L=2H and R=492, Find i(t)? Vs (1)4 10 -π/2 π/2 3/2 2π 5/2 t t=0 i(t) R Vs (t) + Solution: Since T= then 10 = 2π/T = 2 rad/s 2/דר 30= dt = = 10 at 10 dt = 5 an an πο f(t) cos(noot)dt /2 10 cos(2nt)dt = 0 元 T f(t) sin(noot)dt bn = /2 20 bn 10 sin(2nt)dt for odd n пл So, Fourier Transformation of V, (t): 00 20 Vs (t)=5+Σ sin(2nt) ηπ odd n Now, we need to find i(t) for: 1- V, (t) = 5V using Laplace Analysis: V(s) 5/s 5 5/2 11(s)- = Z(s) (SL+R) s(2s+4) s(s+2) 5/4 5/4 = 11(s) S (s+2) 11(t)=(5/4) -(5/4) e 2-Vs (t)=- 20 -21 sin(2nt) using frequency analysis: nπ 20 V= ηπ & 0-2n Z=R+joL=4+ j(2n)(2) = 4+ j4n=4(1+jn) V 5 5 I= = /-tan 'n Z nл(1+jn) n√√1+n² 5 i2(t) sin(2nt-tan h) n√1+n² The complete solution is: 00 i(t) = i(t) + Σ i2(t) odd n
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Question
I need help with this problem and an explanation for the solution described below. (Electric Circuits 2: Fourier Circuit Analysis)

Transcribed Image Text:In the following circuit:
If vs (t) is given as a square waves as shown:
If L=2H and R=492, Find i(t)?
Vs (t)⭑
8
Solution:
Vs (t)
3
5
i(t)
R

Transcribed Image Text:Example 1:
In the following circuit:
If v,(t) is given as a square waves,
If L=2H and R=492, Find i(t)?
Vs (1)4
10
-π/2
π/2
3/2
2π
5/2 t
t=0
i(t)
R
Vs (t) +
Solution:
Since T= then 10 = 2π/T = 2 rad/s
2/דר
30= dt =
= 10 at 10 dt = 5
an
an
πο
f(t) cos(noot)dt
/2
10 cos(2nt)dt = 0
元
T
f(t) sin(noot)dt
bn
=
/2
20
bn
10 sin(2nt)dt
for odd n
пл
So, Fourier Transformation of V, (t):
00 20
Vs (t)=5+Σ sin(2nt)
ηπ
odd n
Now, we need to find i(t) for:
1- V, (t) = 5V using Laplace Analysis:
V(s)
5/s
5
5/2
11(s)-
=
Z(s)
(SL+R)
s(2s+4)
s(s+2)
5/4 5/4
=
11(s) S (s+2)
11(t)=(5/4) -(5/4) e
2-Vs (t)=-
20
-21
sin(2nt) using frequency analysis:
nπ
20
V=
ηπ
& 0-2n
Z=R+joL=4+ j(2n)(2) = 4+ j4n=4(1+jn)
V
5
5
I=
=
/-tan 'n
Z
nл(1+jn) n√√1+n²
5
i2(t)
sin(2nt-tan h)
n√1+n²
The complete solution is:
00
i(t) = i(t) + Σ i2(t)
odd n
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