EE602 CIRCUIT ANALYSIS

The Fourier Series

=
+ + =
1
0
) sin cos ( ) (
n
n n
t n b t n a a t f e e
In other words, any periodic function can be
resolved as a summation of
constant value and cosine and sine functions:

=
+ + =
1
0
) sin cos ( ) (
n
n n
t n b t n a a t f e e
) sin cos (
1 1
t b t a e e + +
0
a =
) 2 sin 2 cos (
2 2
t b t a e e + +
) 3 sin 3 cos (
3 3
t b t a e e + + +
dc
ac
The computation and study of Fourier
series is known as harmonic analysis and
is extremely useful as a way to break up
an arbitrary periodic function into a set of
simple terms that can be plugged in,
solved individually, and then recombined
to obtain the solution to the original
problem or an approximation to it to
whatever accuracy is desired or practical.
=
+ +
+ + +
Periodic Function
0
a
t a e cos
1
t a e 2 cos
2
t b e sin
1
t b e 2 sin
2
f(t)
t

=
+ + =
1
0 0 0
) sin cos ( ) (
n
n n
t n b t n a a t f e e
where
frequency l Fundementa
2
0
= =
T
t
e
}
+
=
T t
t
n
dt t n t f
T
a
0
0
0
cos ) (
2
e
}
+
=
T t
t
dt t f
T
a
0
0
) (
1
0
}
+
=
T t
t
n
dt t n t f
T
b
0
0
0
sin ) (
2
e
Notice that
component dc
value Average
period a over graph below Area
) (
1 0
0
0
=
=
=
=
}
+
T
dt t f
T
a
T t
t
[*The right sign of the area below graph must be
obeyed. +ve sign for area above x-axis & ve sign
for area below x-axis]
Symmetry Considerations
Symmetry functions:
(i) even symmetry
(ii) odd symmetry
Even symmetry
Any function f (t) is even if its plot is
symmetrical about the vertical axis, i.e.

) ( ) ( t f t f =
Even symmetry (cont.)
The examples of even functions are:
2
) ( t t f =
t t
t
| | ) ( t t f =
t t f cos ) ( =
Even symmetry (cont.)
The integral of an even function from A to
+A is twice the integral from 0 to +A
t
} }
+ +

=
A A
A
dt t f dt t f
0
even even
) ( 2 ) (
A +A
) (
even
t f
Odd symmetry
Any function f (t) is odd if its plot is
antisymmetrical about the vertical axis, i.e.

) ( ) ( t f t f =
Odd symmetry (cont.)
The examples of odd functions are:
3
) ( t t f =
t
t
t
t t f = ) (
t t f sin ) ( =
Odd symmetry (cont.)
The integral of an odd function from A to
+A is zero
t
0 ) (
odd
=
}
+

A
A
dt t f
A +A
) (
odd
t f
Even and odd functions
(even) (even) = (even)
(odd) (odd) = (even)
(even) (odd) = (odd)
(odd) (even) = (odd)

The product properties of even and odd
functions are:

Symmetry consideration
From the properties of even and odd
functions, we can show that:
for even periodic function;
}
=
2 /
0
cos ) (
4
T
n
tdt n t f
T
a e
0 =
n
b
for odd periodic function;
}
=
2 /
0
sin ) (
4
T
n
tdt n t f
T
b e
0
0
= =
n
a a
How?? [Even function]
2
T

2
T
} }
= =

2 /
0
2 /
2 /
cos ) (
4
cos ) (
2
T T
T
n
tdt n t f
T
tdt n t f
T
a e e
(even) (even)
| |
(even)
0 sin ) (
2
2 /
2 /
= =
}

T
T
n
tdt n t f
T
b e
(even) (odd)
| |
(odd)
) (t f
t
How?? [Odd function]
2
T

2
T
} }
= =

2 /
0
2 /
2 /
sin ) (
4
sin ) (
2
T T
T
n
tdt n t f
T
tdt n t f
T
b e e
(odd) (odd)
| |
(even)
0 cos ) (
2
2 /
2 /
= =
}

T
T
n
tdt n t f
T
a e
(odd) (even)
| |
(odd)
) (t f
t
0 ) (
2
2 /
2 /
0
= =
}

T
T
dt t f
T
a
(odd)
The amplitude-phase form

is known as the sine-cosine form
We can also express the Fourier series in
the cosine form only, that is


This form is called as the amplitude-phase
form

=
+ + =
1
0 0 0
) sin cos ( ) (
n
n n
t n b t n a a t f e e

=
+ + =
1
0 0
) cos( ) (
n
n n
t n A a t f | e
From this form, we can plot the amplitude
spectrum, vs. n and the phase
spectrum, vs. n.
It can be shown that the combination of
cosine and sine function can be expressed
as a cosine function only:



Comparing both sides of eqn:


n
A
n
|
x A x A
x x A
x A x b x a
sin ) sin ( cos ) cos (
) sin sin cos (cos
) cos( sin cos
| |
| |
|
=
=
+ = +
.....(1) cos a A = | .....(2) sin b A = |
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2 2
) sin (cos
sin cos
b a A b a A
b a A
b a A A
+ = + =
+ = +
+ = +
| |
| |
|
.
|

\
|

a
b
a
b
a
b
A
A
1
tan tan
cos
sin
| |
|
|
: ) 2 ( ) 1 (
2 2
+
:
) 1 (
) 2 (
Hence
) cos( sin cos
0 0 0 n n n n
t n A t n b t n a | e e e + = +
where
2 2
amplitude,
n n n
b a A + =
|
|
.
|

\
|

=

n
n
n
a
b
1
tan phase,|
Or in phasor/complex form:
|
|
.
|

\
|

Z + = = Z

n
n
n n n n n n
a
b
b a jb a A
1 2 2
tan |
0 0
a A =
Example 1
Determine the Fourier series of the following
waveform. Obtain the amplitude and phase
spectra.
Solution
First, determine the period & describe the one period
of the function:
T = 2

< <
< <
=
2 1 , 0
1 0 , 1
) (
t
t
t f ) ( ) 2 ( t f t f = +
t
t
e = =
T
2
0
We find that
Then, obtain the coefficients a
0
, a
n
and b
n
:
2
1
) 0 1 (
2
1
0 1
2
1
) (
2
1
) (
1
2
1
1
0
2
0
0
0
= =
(

+ = =
=
} } }
}
dt dt dt t f
dt t f
T
a
T
Or
2
1
2
1 1 graph below Area
0
=

= =
T
a
t
t
t
t
t
e
n
n
n
t n
dt tdt n
tdt n t f
T
a
n
sin sin
0 cos 1
cos ) (
2
1
0
2
1
1
0
2
0
0
=
(

= + =
=
} }
}
Notice that n is integer which leads ,
since
0 sin = t n
0 3 sin 2 sin sin = = = = t t t
Therefore, .
0 =
n
a
t
t
t
t
t
e
n
n
n
t n
dt tdt n
tdt n t f
T
b
n
cos 1 cos
0 sin 1
sin ) (
2
1
0
2
1
1
0
2
0
0

=
(

= + =
=
} }
}
1 5 cos 3 cos cos = = = = t t t
1 6 cos 4 cos 2 cos = = = = t t t
Notice that
Therefore,

=

=
even , 0
odd , / 2
) 1 ( 1
n
n n
n
b
n
n
t
t
or
n
n ) 1 ( cos = t
+ + + + =
|
.
|

\
|
+ =
(

+ =
+ + =

=
t t t
t n
n
t n
n
t n b t n a a t f
n
n
n
n
n
n n
t
t
t
t
t
t
t
t
t
t
e e
5 sin
5
2
3 sin
3
2
sin
2
2
1
sin
2
2
1
sin
) 1 ( 1
2
1
) sin cos ( ) (
odd
1
1
1
0 0 0
Finally,
+ +
+ + =

|
.
|

\
|
+ =
|
.
|

\
|
+ =

=
) 90 5 cos(
5
2
) 90 3 cos(
3
2
) 90 cos(
2
2
1
) 90 cos(
2
2
1
sin
2
2
1
) (
odd
1
odd
1
t
t t
t n
n
t n
n
t f
n
n
n
n
t
t
t
t
t
t
t
t
t
t
In amplitude-phase form,
Amplitude spectrum: Phase spectrum:
Some helpful identities
For n integers,
n
n ) 1 ( cos = t
0 sin = t n
0 2 sin = t n 1 2 cos = t n
x x sin ) sin( =
x x cos ) cos( =
) 90 cos( sin = x x
) 90 cos( sin + = x x
) 180 cos( cos = x x
The sum of the Fourier series terms can
evolve (progress) into the original
waveform
From Example 1, we obtain
+ + + + = t t t t f t
t
t
t
t
t
5 sin
5
2
3 sin
3
2
sin
2
2
1
) (
It can be demonstrated that the sum will
lead to the square wave:
Notes:
t t t t t
t
t
t
t
t
t
t
7 sin
7
2
5 sin
5
2
3 sin
3
2
sin
2
+ + + t t t t
t
t
t
t
t
5 sin
5
2
3 sin
3
2
sin
2
+ +
t t t
t
t
t
3 sin
3
2
sin
2
+ t t
t
sin
2
(a) (b)
(c) (d)
t t t t t t
t
t
t
t
t
t
t
t
t
9 sin
9
2
7 sin
7
2
5 sin
5
2
3 sin
3
2
sin
2
+ + + +
t t t t
t
t
t
t
t
23 sin
23
2
3 sin
3
2
sin
2
2
1
+ + + +
(e)
(f)
Example 2
Given
, ) ( t t f = 1 1 s s t
) ( ) 2 ( t f t f = +
Sketch the graph of f (t) such that . 3 3 s s t
Then compute the Fourier series expansion of f (t).
Plot the amplitude and phase spectra until the forth
harmonic.
Solution
The function is described by the following graph:
T = 2
t
t
e = =
T
2
0
We find that
t t t
t
t
t
t
t
t
t
t
t
t
t
t e
n n n
n
n
n
n
t n
n
n
dt
n
t n
n
t n t
tdt n t tdt n t f
T
b
n n
n
1
2 2
1
0
2 2
1
0
1
0
1
0
1
0
0
) 1 ( 2
0
) 1 ( 2 sin 2 cos 2
sin
2
cos 2
cos
2
cos
2
sin
2
4
sin ) (
4
+

= +

= + =
(

=
+
(

=
= =
}
} }
Then we compute the coefficients:
0
0
= =
n
a a
since f(t) is an odd function.

+ + + +
+ + =
+ + =

=
+ + =

=
+

=
) 90 4 cos(
2
1
) 90 3 cos(
3
2
) 90 2 cos(
1
) 90 cos(
2
4 sin
2
1
3 sin
3
2
2 sin
1
sin
2
sin
) 1 ( 2
) sin cos ( ) (
1
1
1
0 0 0
t t
t t
t t t t
t n
n
t n b t n a a t f
n
n
n
n n
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
e e
Hence,
Amplitude spectrum: Phase spectrum:
0.64
0.32
0.21
0.16
Example 3
Given

< <
< <
=
4 2 , 0
2 0 , 2
) (
t
t t
t v
) ( ) 4 ( t v t v = +
(i) Sketch the graph of v (t) such that
. 12 0 s st
(ii) Compute the trigonometric Fourier series of v (t).
(iii) Express the Fourier series of v (t) in the
amplitude-phase form. Then plot the amplitude
and phase spectra until the forth harmonic.
Solution
(i) The function is described by the following graph:
T = 4
2
2
0
t t
e = =
T
We find that
0 2 4 6 8 10 12
t
v (t)
2
(ii) Then we compute the coefficients:
2
1
2
2
4
1
) 2 (
4
1
0 ) 2 (
4
1
) (
1
2
0
2
2
0
4
2
2
0
4
0
0
=
(

= =
)
`

+ = =
}
} } }
t
t dt t
dt dt t dt t v
T
a
Or
2
1
4
2 2
2
1
0
=

= a

=

=

=
(

+ =
+
(

=
+ = =
}
} } }
odd , / 4
even , 0
] ) 1 ( 1 [ 2 ) cos 1 ( 2
2
2 cos 1 cos
2
1
0
sin
2
1 sin ) 2 (
2
1
0 cos ) 2 (
2
1
cos ) (
2
2 2 2 2 2 2
2
0
2
0
2
0
2
0
2
0
2
0
0
0
2
0
0
0
4
2
2
0
0
4
0
0
n n
n
n n
n
n
n
n
t n
dt
n
t n
n
t n t
tdt n t tdt n t v
T
a
n
n
t t t
t
e
e
e
e
e
e
e
e
e e
t e e
e
e
e
e
e
e
e
e
e
e e
n n n
n
n
n
t n
n
dt
n
t n
n
t n t
tdt n t tdt n t v
T
b
n
2 1
2
2 sin 1
sin
2
1 1
cos
2
1 cos ) 2 (
2
1
0 sin ) 2 (
2
1
sin ) (
2
0
2
0
2
0
0
2
0
2
0
2
0
0
2
0
0
0
2
0
0
0
4
2
2
0
0
4
0
0
= = =
(

=
+ = =
}
} } }
since
0 sin 2 sin
0
= = t e n n

=
+ + =
)
`

|
.
|

\
|
+
|
.
|

\
|

+ =
+ + =
1
0 0
1
2 2
1
0 0 0
) cos(
2
sin
2
2
cos
] ) 1 ( 1 [ 2
2
1
) sin cos ( ) (
n
n n
n
n
n
n n
t n A a
t n
n
t n
n
t n b t n a a t v
| e
t
t
t
t
e e
Hence,

|
.
|

\
|
Z
|
|
.
|

\
|
+
Z
=

|
.
|

\
|

=
= Z

odd ,
2
tan 1
4 2
even , 90
2
odd ,
2 2
even ,
2
1
2 2
n
n
n n
n
n
n j
n n
n
n
j
jb a A
n n n n
t
t t
t
t t
t
|
(iii) Where
Z = Z 5 . 57 75 . 0
1 1
| A
Z = Z 0 . 78 22 . 0
3 3
| A
Z = Z 90 32 . 0
2 2
| A
Z = Z 90 16 . 0
4 4
| A
5 . 0
0 0
= = a A
We obtained that
( )
( ) + +
|
.
|

\
|
+
+
|
.
|

\
|
+ =
+ + =

=
90 2 cos 16 . 0 0 . 78
2
3
cos 22 . 0
90 cos 32 . 0 5 . 57
2
cos 75 . 0 5 . 0
) cos( ) (
1
0 0
t
t
t
t
t n A a t v
n
n n
t
t
t
t
| e
Amplitude spectrum: Phase spectrum:
0.5
0.75
0.32
0.22
0.16
0
/2 3/2 2
/2 3/2 2
57.5
78.0
90 90
Example 4
Given

< <
< <
< <
=
2 1 , 1
1 1 ,
1 2 , 1
) (
t
t t
t
t f
) ( ) 4 ( t f t f = +
Sketch the graph of f (t) such that . 6 6 s s t
Then compute the Fourier series expansion of f (t).
Solution
The function is described by the following graph:
T = 4
2
2 t t
e = =
T
We find that
0 4 6 2 4 6
t
f (t)
2
1
1
Then we compute the coefficients. Since f (t) is
an odd function, then
0 ) (
2
2
2
0
= =
}

dt t f
T
a
0 cos ) (
2
2
2
= =
}

tdt n t f
T
a
n
e
and
2 2 2 2
1
0
2 2
2
1
1
0
1
0
2
1
1
0
2
0
2
2
) 2 / sin( 4 cos 2 sin 2 cos
cos 2 cos sin cos
cos cos cos
sin 1 sin
4
4
sin ) (
4
sin ) (
2
t
t
t
t
e
e
e
e
e
e e
e
e
e
e
e
e
e
e
e
e
e e
e e
n
n
n
n
n
n
n
n
n
n n
n
t n
n
n
n
t n
dt
n
t n
n
t n t
tdt n tdt n t
tdt n t f
T
tdt n t f
T
b
n
+ = + =

+ =
(

+ +
(

=
(

+ =
= =
}
} }
} }

since
0 sin 2 sin = = t e n n

=
+

=
|
.
|

\
|
=
+ + =
1
1
1
1
0
2
sin
) 1 (
2
2
sin
cos 2
) sin cos (
2
) (
n
n
n
n
n n
t n
n
t n
n
n
t n b t n a
a
t f
t
t
t
t
t
e e
Finally,

Example 5
Compute the Fourier series expansion of f (t).
Solution
The function is described by
T = 3
3
2 2
0
t t
e = =
T
and

< <
< <
< <
=
3 2 , 1
2 1 , 2
1 0 , 1
) (
t
t
t
t f
) ( ) 3 ( t f t f = +
T = 3
Then we compute the coefficients.
3
8
1
2
3
2 ) 0 1 (
3
4
2 1
3
4
) (
4
) (
2
2 / 3
1
1
0
2 / 3
0
3
0
0
=
(

|
.
|

\
|
+ =
(

+ = = =
} } } }
dt dt dt t f
T
dt t f
T
a
| |
3
8
) 2 3 ( ) 1 2 ( 2 ) 0 1 (
3
2
1 2 1
3
2
) (
2
3
2
2
1
1
0
3
0
0
= + + =
(

+ + = =
} } } }
dt dt dt dt t f
T
a
Or, since f (t) is an even function, then
Or, simply
3
8
4
3
2
period a in
graph below area Total
2
) (
2
3
0
0
= =
|
|
.
|

\
|
= =
}
T
dt t f
T
a
3
2
sin
2
3
2
sin sin 2
2
sin
2
3
sin 2
3
4
sin
2
3
sin 2 sin
3
4
sin 2
3
4 sin
3
4
cos 2 cos 1
3
4
cos ) (
4
cos ) (
2
2 / 3
1
1
0
2 / 3
1
1
0
2 / 3
0
3
0
t
t
t
t
t
e
e
e
e
e
e
e
e
e
e
e
e e
e e
n
n
n
n
n
n
n
n
n
n
n
n
n
t n
n
t n
tdt n tdt n
tdt n t f
T
tdt n t f
T
a
n
=
|
.
|

\
|
=
|
.
|

\
|
=
(

|
.
|

\
|
+ =
(

+
(

=
(

+ =
= =
} }
} }
;
3
2t
e =

=
|
.
|

\
|
=
|
.
|

\
|
+ =
+ + =
1
1
1
0
3
2
cos
3
2
sin
1 2
3
4
3
2
cos
3
2
sin
2
3
4
) sin cos (
2
) (
n
n
n
n n
t n n
n
t n n
n
t n b t n a
a
t f
t t
t
t t
t
e e
Finally,

and
0 =
n
b
since f (t) is an even function.
Parsevals Theorem
Parservals theorem states that the
average power in a periodic signal is equal
to the sum of the average power in its DC
component and the average powers in its
harmonics
=
+ +
+ + +
t a e cos
1
t a e 2 cos
2
t b e sin
1
t b e 2 sin
2
f(t)
t
P
avg
P
dc
P
a1
P
b1
P
a2
P
b2
0
a
For sinusoidal (cosine or sine) signal,
R
V
R
V
R
V
P
2
peak
2
peak
2
rms
2
1 2
=
|
.
|

\
|
= =
For simplicity, we often assume R = 1,
which yields
2
peak
2
rms
2
1
V V P = =
And since ,
Therefore, the total power of all harmonics
is

+ + + + + =
+ + + + + =
2
2
2
2
2
1
2
1
2
0
dc avg
2
1
2
1
2
1
2
1
2 2 1 1
b a b a a
P P P P P P
b a b a


=

=
+ = + + =
1
2 2
0
1
2 2 2
0 avg
2
1
) (
2
1
n
n
n
n n
A a b a a P
2
rms
V P =


=

=
+ = + + =
1
2 2
0
1
2 2 2
0 rms
2
1
) (
2
1
n
n
n
n n
A a b a a V
Circuit applications
Steps for applying Fourier series:
1. Express the excitation as a Fourier
series
2. Transform the circuit from the time
domain to the frequency domain
3. Find the response of the dc and ac
components in the Fourier series
4. Add the individual dc and ac responses
using the superposition principle
The signal in Figure 6.1 is applied to the circuit in
Figure 6.2.

(i) Find the Fourier series of i
s
(t).
(ii) Plot the amplitude spectrum for the dc
component and the first three non-zero
harmonics of i
s
(t).
(iii) Find the load current i
L
(t).
(iv) Find the average power dissipated in the
resistor R
L
.
Example 6
) (t i
s
) (t i
s
H 1
O 1
O = 1
L
R
) (t i
L
Figure 6.2
Figure 6.1
Solution
(i) The current source is described by
and

< <
< <
=
< < =
0 ,
0 ,
, | | ) (
t t
t t
t t t i
s
t
t
t t
1 , 2
0
= = e t T
Compute the Fourier series. Since i
s
(t) is an even
function,
or
2 2
1
2
2
) (
2
0
2
0
2 /
0
0
t
t t
t
t
=
(

= = =
} }
t
tdt dt t i
T
a
T
s
2 2
1 1
0 over
graph below Area
1
0
t
t t
t t t
=
|
.
|

\
|
=
|
|
.
|

\
|
< <
=
t
a

=

=

=
(

+ =

= =
=
} }
}
even , 0
odd , / 4 ] 1 ) 1 [( 2
) 1 (cos 2
cos 2 sin 2
sin 2 sin 2
cos
2
4
cos ) (
4
2
2
2
0
2
0
0
0
2 /
0
0
n
n n
n
n
n
n
nt
n
n
dt
n
nt
n
nt t
ntdt t
tdt n t i
T
a
n
T
s n
t
t
t
t
t
t t
t
t t t
e
t
t
t
t
Therefore, the Fourier series of i
s
(t) is
0 =
n
b
+ + +
+ + + + =
+
|
.
|

\
|
+ =
|
.
|

\
|
+ =

=
) 180 5 cos(
25
4
) 180 3 cos(
9
4
) 180 cos(
4
2
) 180 cos(
4
2
cos
4
2
) (
odd
1
2
odd
1
2
t
t t
nt
n
nt
n
t i
n
n
n
n
s
t
t t
t
t
t
t
t
n
s
I
n
s
I Z
n n
1.571
1.273
0.141
0.051
(ii) Hence,
, 5 , 3 , 1 , 180
4
2
= Z = n
n
I
n
s
t 2
0
t
=
s
I
(iii) To compute i
L
(t), separate to dc and ac analysis:
dc analysis (n = 0)
S/c the inductor, hence
0
s
I
O 1
O 1
4 2
2 /
2
0
0
t t
= = =
s
L
I
I
|
.
|

\
|
+ Z
+
+
=
|
.
|

\
|
Z
|
|
.
|

\
|
+
+
=
= =
+ +
+
=

180
2
tan tan
4
1 4
180
4
2
1
,
) (
1 1
2
2
2
2
0
n
n
n
n
n
n jn
jn
n n I
L j R R
L j R
I
n s
n L
n
L
n n
t
t
e e
e
e
s
L
L
I
L j R R
L j R
I
) ( e
e
+ +
+
=
ac analysis (n 1)
or
+
+ + + +
+ + =
) 49 . 190 5 cos( 048 . 0 ) 26 . 195 3 cos( 124 . 0
) 43 . 198 cos( 805 . 0 785 . 0 ) (
t t
t t i
L
Hence,
Z =
Z = Z =
=
49 . 190 048 . 0
, 26 . 195 124 . 0 , 43 . 198 805 . 0
, 785 . 0
5
3 1
0
L
L L
L
I
I I
I
W 949 . 0 ) 048 . 0 124 . 0 805 . 0 (
2
1
785 . 0
2
1
2 2 2 2
1
2 2
2
) (
0
= + + + ~
|
.
|

\
|
+ =
=

=
L
n
L L
L rms L
R I I
R I P
n
(iv) Average power absorbed by R
l
is
The signal in Figure 7.1 is applied to the circuit in
Figure 7.2.
(i) Find the Fourier series of v
s
(t).
(ii) Find the output voltage v
o
(t).
(iii) Plot the amplitude spectrum for the dc
component and the first three non-zero harmonics
of the output voltage v
o
(t).
(iv) Find the r.m.s value of v
o
(t) .
Example 7
) (t v
s
F 1
O 1
) (t v
o
Figure 7.2
Figure 7.1
O 1
) (t v
s
+

Solution
(i) The voltage source is described by
and

< <
< <
=
0 1 , 1
1 0 , 1
) (
t
t
t v
s
t e = =
0
, 2 T
Compute the Fourier series. Since v
s
(t) is an odd
function,
0
0
= =
n
a a

=

=

=
(

= =
=
}
}
even , 0
odd , / 4
] ) 1 ( 1 [ 2
) cos 1 ( 2
cos
2 sin
2
4
sin ) (
4
1
0
1
0
2 /
0
0
n
n n
n
n
n
n
t n
tdt n
tdt n t v
T
b
n
T
s n
t
t
t
t
t
t
t
e
Therefore, the Fourier series of v
s
(t) is
+ +
+ =

|
.
|

\
|
=
|
.
|

\
|
=


=

=
) 90 5 cos(
5
4
) 90 3 cos(
3
4
) 90 cos(
4
) 90 cos(
4
sin
4
) (
odd
1
odd
1
t
t t
t n
n
t n
n
t v
n
n
n
n
s
t
t
t
t
t
t
t
t
t
Hence,
, 5 , 3 , 1 , 90
4
= Z = n
n
V
n
s
t
0
0
=
s
V
(ii) To compute v
o
(t), separate to dc and ac analysis:
dc analysis (n = 0)
O/c the capacitor, hence
0
0
=
s
V
O 1
0
0
=
o
V
O 1
+

0 = I
( ) Z
+
+
=
|
.
|

\
|
Z
|
|
.
|

\
|
+
+
=
= =
+
+
=

90 2 tan tan
4 1
1 4
90
4
2 1
1
,
2 1
1
1 1
2 2
2 2
0
t t
t
t
t
t t
t
t e e
e
e
n n
n
n
n
n n j
jn
n n V
RC j
RC j
V
n s
n
n
o
n n
s s o
V
RC j
RC j
V
C j R R
C j R
V
e
e
e
e
2 1
1
)] / 1 ( [
) / 1 (
+
+
=
+ +
+
=
ac analysis (n 1)
or
+ +
+ =
) 82 . 91 5 cos( 128 . 0
) 02 . 93 3 cos( 213 . 0 ) 61 . 98 cos( 660 . 0 ) (
t
t t t v
o
Hence,
Z =
Z = Z =
=
82 . 91 128 . 0
, 02 . 93 213 . 0 , 61 . 98 660 . 0
, 0
5
3 1
0
o
o o
o
V
V V
V
n
o
V
n
o
V Z
n
n
0.660
0.213
0.128
98.61
(iii)
93.02
91.82
(iv)
V 705 . 0 ) 128 . 0 213 . 0 660 . 0 (
2
1
2
1
2 2 2
1
2 2
) rms (
0
= + + ~
+ =

= n
o o o
n
V V V
Exponential Fourier series
Recall that, from the Eulers identity,
x j x e
jx
sin cos =

yields
2
cos
jx jx
e e
x

+
=
2
sin
j
e e
x
jx jx

=
and
Then the Fourier series representation becomes

=
|
.
|

\
|
+
+
|
.
|

\
|

+ =
(

|
.
|

\
|
+
+
|
.
|

\
|

+ =
(

|
|
.
|

\
|

|
|
.
|

\
|
+
+ =
(

|
|
.
|

\
|

+
|
|
.
|

\
|
+
+ =
+ + =
1 1
0
1
0
1
0
1
0
1
0
2 2 2
2 2 2
2 2 2
2 2 2
) sin cos (
2
) (
n
t jn
n n
n
t jn
n n
n
t jn
n n
t jn
n n
n
t jn t jn
n
t jn t jn
n
n
t jn t jn
n
t jn t jn
n
n
n n
e
jb a
e
jb a a
e
jb a
e
jb a a
e e
jb
e e
a
a
j
e e
b
e e
a
a
t n b t n a
a
t f
e e
e e
e e e e
e e e e
e e
Here, let we name


=

=
|
.
|

\
|
+
+
|
.
|

\
|

+ =
1 1
0
2 2 2
) (
n
t jn
n n
n
t jn
n n
e
jb a
e
jb a a
t f
e e
2
n n
n
jb a
c

=
,
2
n n
n
jb a
c
+
=

Hence,

=

=

=
= + + =
+ + =
+ + =
n
t jn
n
n
t jn
n
n
t jn
n
n
t jn
n
n
t jn
n
n
t jn
n
n
t jn
n
e c e c c e c
e c e c c
e c e c c
e e e
e e
e e
1
0
1
1 1
0
1 1
0
and .
2
0
0
a
c =
c
0
c
n c
n
Then, the coefficient c
n
can be derived from
}
}
} }
} }

=
=
(

=
=

=
T
t jn
T
T T
T T
n n
n
dt e t f
T
dt t n j t n t f
T
tdt n t f j tdt n t f
T
tdt n t f
T
j
tdt n t f
T
jb a
c
0
0
0 0
0 0
) (
1
] sin )[cos (
1
sin ) ( cos ) (
1
sin ) (
2
2
cos ) (
2
2
1
2
e
e e
e e
e e
In fact, in many cases, the complex
Fourier series is easier to obtain rather
than the trigonometrical Fourier series
In summary, the relationship between the
complex and trigonometrical Fourier series
are:
n n
n n
n
A
jb a
c | Z =

=
2
1
2
2
n n
n
jb a
c
+
=

}
= =
T
dt t f
T
a c
0
0 0
) (
1
}

=
T
t jn
n
dt e t f
T
c
0
) (
1
e
-

=
n n
c c
or
2 2
2 2
n
n n
n
A
b a
c =
+
=
The average power and the rms value in
the term of Fourier complex coefficient are


=

=
O
+ = =
1
2
2
0
2
1
2
n
n
n
n
c c c P


=

=
+ = =
1
2
2
0
2
rms
2
n
n
n
n
c c c F
Example 8
For the given function,
(i) Obtain the complex Fourier series
(ii) Plot the amplitude and the phase spectra
of the complex Fourier series for 5 n 5
(iii) Calculate the average power and the rms value
of the signal
t 2 t 4 t 4 t 2 0
t 2
e
1
) (t v
t
(i) Since , . Hence
Solution
| |
t t
t
t
t
t
2
1
2
1
2
1
) (
1
2
2
0
2
0
0
0

= =
=
=
}
}
e
e
dt e
dt t v
T
c
t
t
T
1
0
= e
t 2 = T
) 1 ( 2
1
) 1 ( 2
1
) 1 ( 2
1
1 2
1
2
1
2
1
) (
1
2 2 2 ) 1 ( 2
2
0
) 1 (
2
0
) 1 (
2
0
0
0
jn
e
jn
e e
jn
e
jn
e
dt e dt e e
dt e t v
T
c
n j jn
t jn
t jn jnt t
T
t jn
n

=
(

=
= =
=

} }
}
t t t
t
t t
t t t t
t
t t
e
since
1 0 1 2 sin 2 cos
2
= = =

t t
t
n j n e
n j
jnt
n n
t jn
n
e
jn
e
e c t v


=

= =
) 1 ( 2
1
) (
2
0
t
t
e
Therefore, the complex Fourier series of v(t) is
0
2
0
2
0
2
1
) 1 ( 2
1
c
e
jn
e
c
n
n
n
=

=
=
=
t t
t t
*Notes: Even though c
0
can be found by substituting
c
n
with n = 0, sometimes it doesnt works (as shown
in the next example). Therefore, it is always better to
calculate c
0
alone.
n
n
n
n
e
c
n n
1
2
1
2
2
tan
1
85
1
tan 0
1 2
1

Z
+
~
(

|
.
|

\
|

Z
+

= Z
t
u
t
(ii)
The complex frequency spectra are
kW 19 . 20
) 7 . 16 6 . 20 9 . 26 38 1 . 60 ( 2 85
2 2 2 2 2 2
2
1
=
+ + + + + ~
=

=
O
n
n
c P
(ii) The average power is (assume R = 1 )
and since
2
rms 1
V P =
O
V 09 . 142 k 19 . 20
rms
= = V
Example 9
Obtain the complex Fourier series of the function in
Example 1. Then plot the complex frequency spectra
for 4 n 4 and calculate the rms value of the signal.
Solution
2
1
1
2
1
) (
1
1
0 0
0
= = =
} }
dt dt t f
T
c
T
t e =
0
) 1 (
2 2
1
0 1
2
1
) (
1
1
0
2
1
1
0 0
=
(

=
+ = =

} } }
t
t
t e
t t
jn
t jn
t jn
T
t jn
n
e
n
j
jn
e
dt e dt e t f
T
c
) 1 (
2
=
t
t
jn
n
e
n
j
c
But
n jn
n n j n e ) 1 ( cos sin cos = = =

t t t
t
Thus,

= =
even , 0
odd , /
] 1 ) 1 [(
2 n
n n j
n
j
n
t
t
Therefore,


=
=

=
|
.
|

\
|
+ = =
odd
0
2
1
) (
0
n
n
n
t jn
n
t jn
n
e
n
j
e c t f
t e
t
*Here notice that .
0
0
c c
n
n
=
=
2
1
0
= c
= = 0 , 0
n n
c u
For n even,
For n odd,

<
>
= =
0 , 90
0 , 90
,
1
n
n
n
c
n n
u
t
0.5
692 . 0 ) 11 . 0 32 . 0 ( 2 5 . 0
2 2 2
2
rms
= + + ~
=

= n
n
c F
The average rms value is
Summary
Sine-cosine form



Amplitude-phase form


Exponential (or complex) form

=
+ + =
1
0 0 0
) sin cos ( ) (
n
n n
t n b t n a a t f e e
, cos ) (
2
0
0
}
=
T
n
tdt n t f
T
a e
}
=
T
n
tdt n t f
T
b
0
0
sin ) (
2
e
, ) (
1
0
0
}
=
T
dt t f
T
a
,
2
0
T
t
e =

=
+ + =
1
0 0
) cos( ) (
n
n n
t n A a t f | e

=
=
n
t jn
n
e c t f
0
) (
e
n n n n
jb a A = Z|
,
0 0
a c =
,
2
n n
n
jb a
c

=
*
n n
c c =

Prob.17.37, pg.803
If the sawtooth waveform in Fig.17.9 is the
voltage source v
s
(t) in the circuit of
Fig.17.22, find the response v
o
(t).
Prob.17.37, pg.803
If the periodic current waveform in
Fig.17.73(a) is applied to the circuit in
Fig.17.73(b), find v
o
.
+

3
) (t i
s
0 1 2 3
1
3
t
(a) (b)
Problem


1 2 3 0 4
1
5
2

3
6
O = 1
o
R
O = 1
s
R F 1 = C

+
) (t v
o
) (t v
s
) (t v
s
t
(i) Compute the output voltage v
o
(t).
(ii) Plot the amplitude and phase spectra of v
o
(t)
for the first three nonzero harmonics.
(ii) Calculate the average power dissipated in
the resistor R
o
.
QUIZ 1
) (t v
s

+
) (t v
o
H 1
H 1
O 1
O 1
Given the Fourier series expansion of the
voltage signal v
s
(t) is
Find v
o
(t) in the amplitude-phase form.

=
|
.
|

\
|
=
1
sin
10
5 ) (
n
s
nt
n
t v
t
Quiz1
Determine the Fourier series of the following
waveform.
) (t v
4
QUIZ 2
Find i(t) if v(t) is the signal as given in QUIZ 1.


=

=
+
|
.
|

\
|
+ =
|
.
|

\
|
+ =
odd
1
odd
1
) 90 cos(
2
2
1
sin
2
2
1
) (
n
n
n
n
t n
n
t n
n
t v t
t
t
t
QUIZ 3
Find the output voltage v
o
(t) if the current
source i
s
(t) is given by

=

|
.
|

\
|
+ =
1
) 90 cos(
2
1 ) (
n
s
nt
n
t i
t
) (t i
s

+
) (t v
o
O 5
F 5 . 0
O 3
QUIZ 4
Using time differentiation technique, find the
Fourier transform of f(t).