How did he get the value inside the circle? Can you explain step by step and a clear line? 2 { 30²5 [0² (-na) + 3 sin(-na) -1]} jna ¹ = tosd +jsing]
How did he get the value inside the circle? Can you explain step by step and a clear line? 2 { 30²5 [0² (-na) + 3 sin(-na) -1]} jna ¹ = tosd +jsing]
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 19E
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![solution:
Here
f(t) = 24 30<+<1
Tos
0 14+42
Here period of f is
2L = 2 L=1
". The Complex Fourier Series of f(t) is
jnat
f(t)
Σ Cne
= Co + Σ Cn
Y=L
where
Co = 2/₁ √ f(t) df
S
2.1
O
= ½ 5²4d²
C₂ = 2.1 ( f(t) @ shat of
1
einat df
-
.4. So
12/22
= 2 [-²]!
jna
е
eo
-
= 2 [
+
उनम
Jna
2{-}* (@³²-1)}
jna
How did he get the value inside
the circle? Can you explain step by
step and a clear line?
=2
j²
१ उनम
[ cos (-nx) +j sin(-na)
F
=
2 J
· [ los (na) - jsin (na) - 1]
मंत्र
". The required
is f(t)
na
ز2
= 2 +
[Dos na -1] [: Sinna =0]
complex Fourier series of f(t)
2√² (Cosna-1) @√nät
37
N=18
inat
nto
= 2 [+] ! = 2 [1-0) = 2
nto
2= 658 +jsing]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F43c7c067-6ecb-4f5e-9b68-b0237412635d%2F67d1b8ef-bdb5-49fd-b10d-46cad3eafbc6%2Flfbbkz9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:solution:
Here
f(t) = 24 30<+<1
Tos
0 14+42
Here period of f is
2L = 2 L=1
". The Complex Fourier Series of f(t) is
jnat
f(t)
Σ Cne
= Co + Σ Cn
Y=L
where
Co = 2/₁ √ f(t) df
S
2.1
O
= ½ 5²4d²
C₂ = 2.1 ( f(t) @ shat of
1
einat df
-
.4. So
12/22
= 2 [-²]!
jna
е
eo
-
= 2 [
+
उनम
Jna
2{-}* (@³²-1)}
jna
How did he get the value inside
the circle? Can you explain step by
step and a clear line?
=2
j²
१ उनम
[ cos (-nx) +j sin(-na)
F
=
2 J
· [ los (na) - jsin (na) - 1]
मंत्र
". The required
is f(t)
na
ز2
= 2 +
[Dos na -1] [: Sinna =0]
complex Fourier series of f(t)
2√² (Cosna-1) @√nät
37
N=18
inat
nto
= 2 [+] ! = 2 [1-0) = 2
nto
2= 658 +jsing]
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