Download The Complex Form

The Complex Form
23.6
Introduction
In this Section we show √
how a Fourier series can be expressed more concisely if we introduce
the complex number i = −1. By utilising the Euler formula:
eiθ = cos θ + i sin θ
we can replace the trigonometric functions by complex exponential functions. By also combining
the Fourier coefficients an and bn into a complex coefficient cn through
1
cn = (an − ibn )
2
we find that, for a given periodic signal, both sets of constants can be found in one operation.
We also obtain Parseval’s Theorem which has important applications in electrical engineering.
The complex formulation of a Fourier series is an important pre-cursor of the Fourier transform
which attempts to Fourier analyse non-periodic functions.
'
$
① know how to obtain a Fourier series
Prerequisites
Before starting this Section you should . . .
&
Learning Outcomes
After completing this Section you should be
able to . . .
② be familiar with the complex numbers
③ be familiar with the relation between the
exponential function and the trigonometric
functions
✓ express a periodic function in terms of its
Fourier series in complex form
✓ understand Parseval’s theorem
%
1. Complex Exponential Form of a Fourier Series
So far we have discussed the trigonometric form of a Fourier Series i.e. we have represented
functions of period T in the terms of sinusoids, and possibly a constant term, using
∞ a0 2nπt
2nπt
f (t) =
+
+ bn sin
.
an cos
2
T
T
n=1
If we use the angular frequency
ω0 =
2π
T
We obtain the more concise form
a0 (an cos nω0 t + bn sin nω0 t).
+
f (t) =
2
n=1
∞
We have seen that the Fourier coefficients are calculated using the following integrals:
T
an = T2 −2T f (t) cos nω0 t dt
n = 0, 1, 2, . . .
(1)
2
bn =
2
T
T2
− T2
f (t) sin nω0 t dt
n = 1, 2, . . .
(2)
An alternative, more concise form, of a Fourier Series is available using complex quantities. This
form is quite widely used by Engineers, for example in Circuit Theory and Control Theory, and,
in this course, will lead naturally into a later Section on the Fourier Transform.
2. Revision of the exponential form of a complex number
Recall that a complex number in Cartesian form
z = a + ib,
where a and b are real numbers and i2 = −1, can be written in polar form
z = r(cos θ + i sin θ)
where r = |z| =
√
a2 + b2 and θ, the argument or phase of z, is such that
a = r cos θ
b = r sin θ.
A more concise version of the polar form of z can be obtained by defining a complex exponential quantity eiθ by
eiθ = cos θ + i sin θ
(This is sometimes known as Euler’s relation.) The polar angle θ is normally expressed in
radians.
Replacing i by −i we obtain
e−iθ = cos θ − i sin θ
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
2
Write down in cosine, sine form (i) eiπ/6
Cartesian form.
(ii) e−iπ/6 . Then write each in
Your solution
e−iπ/6 = cos
π √3
π 1
− i sin
=
−i
6
6
2
2
6
We have, by definition,
π √3
π 1
+ i sin
=
+i
6
6
2
2
π
eiπ/6 = cos
Write down (i) cos
(ii) sin
π
6
in terms of eiπ/6 and e−iπ/6 .
Your solution
(Don’t forget the factor i in this latter case)
eiπ/6 − e−iπ/6 = 2i sin
π 6
or
π
1
=
eiπ/6 − e−iπ/6
6
2i
sin
π 1 =
eiπ/6 + e−iπ/6
6
2
cos
Similarly, subtracting the two results,
eiπ/6 + e−iπ/6 = 2 cos
π 6
or
We have, adding the two results from the previous guided exercise
Clearly similar calculations could be carried out for any angle θ and the results are summarised
in the following Key Point.
3
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
Key Point
cos θ =
1 iθ
e + e−iθ
2
sin θ =
1 iθ
e − e−iθ
2i
Using these results we can redraft an expression of the form
an cos nθ + bn sin nθ
in terms of complex exponentials.
(This expression, with θ = ω0 t, is of course the n’th harmonic of a trigonometric Fourier Series.)
Using the results from the last Key Point (with nθ instead of θ) rewrite
an cos nθ + bn sin nθ
in complex exponential form.
First do direct substitution
Your solution
an cos nθ + bn sin nθ =
so
an cos nθ =
bn inθ
a n
einθ + e−inθ +
e − e−inθ
2
2i
an inθ
e + e−inθ
2
bn sin nθ =
bn inθ
e − e−inθ
2i
We have
Now collect the terms in einθ and in e−inθ
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
4
Your solution
or, since
1
1
(an − ibn )einθ + (an + ibn )e−inθ .
2
2
i
1
= 2 = −i
i
i
1
2
We get
an +
bn
i
einθ +
1
2
an −
bn
i
e−inθ
We can now write this expression in more concise form by defining
1
cn = (an − ibn )
2
which has complex conjugate
1
c∗n = (an + ibn ).
2
Hence, we have finally,
Your solution
an cos nθ + bn sin nθ = cn einθ + cn∗ e−inθ
Clearly, we can now rewrite the trigonometric Fourier Series
a0 (an cos nω0 t + bn sin nω0 t)
+
2
n=1
∞
as
a0 inω0 t
+ c∗n e−inω0 t
cn e
+
2
n=1
∞
(3)
A neater, and particularly concise, form of this expression can be obtained as follows:
Firstly write a20 = c0 (which is consistent with the general definition of cn since b0 = 0)
The second term in the summation
∞
c∗n e−inω0 t = c∗1 e−iω0 t + c∗2 e−2iω0 t + . . .
n=1
can be written, if we define c−n = c∗n = 12 (an + ibn ), as
−iω0 t
c−1 e
−2iω0 t
+ c−2 e
−3iω0 t
+ c−3 e
+ ... =
−∞
cn einω0 t
n=−1
5
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
Hence (3) can be written
c0 +
∞
inω0 t
cn e
+
n=1
−∞
cn einω0 t
n=−1
or, finally, in the extremely concise form
∞
cn einω0 t .
n=−∞
The complex Fourier coefficients cn can be readily obtained as follows using (1) and (2) for
an , b n .
Firstly
T
2
a0
1
c0 =
f (t)dt
(4)
=
2
T − T2
For n = 1, 2, 3, . . . we have
1
1
cn = (an − ibn ) =
2
T
cn =
i.e.
1
T
T2
− T2
T
2
− T2
f (t)(cos nω0 t − i sin nω0 t)dt
f (t)e−inω0 t dt
(5)
Also for n = 1, 2, 3, . . . we have
c−n =
1
1
= (an + ibn ) =
2
T
c∗n
T
2
f (t)einω0 t dt
− T2
This last expression is equivalent to stating that for n = −1, −2, −3, . . .
T
cn = T1 −2T f (t)e−inω0 t dt
(6)
2
The 3 equations (4), (5), (6) can thus all be contained in the one expression
T
2
1
cn =
f (t)e−inω0 t dt
for n = 0, ±1, ±2, ±3, . . .
T − T2
The results of this discussion are summarised in the following Key Point.
Key Point
Fourier Series in Complex Form
A function f (t) of period T has a complex Fourier series
f (t) =
∞
inω0 t
cn e
where
n=−∞
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
1
cn =
T
T
2
f (t)e−inω0 t dt
− T2
6
For the special case T = 2π so that ω0 = 1 these formulae become particularly simple
π
∞
1
int
cn e
cn =
f (t)e−int dt.
f (t) =
2π
−π
n=−∞
Note the extremely concise form of these results.
3. Properties of the complex Fourier coefficients
Using properties of the trigonometric Fourier coefficients an , bn we can readily deduce the
following results for the cn ’s:
a0
2
1. c0 =
is always real
2. Suppose the periodic function f (t) is even so that all bn are zero. Then since, in the
complex form, the bn arise as the imaginary part of cn , it follows that for f (t) even the
coefficients cn (n = ±1, ±2, . . .) are wholly real.
If f (t) is odd, what can you deduce about the Fourier coefficients cn ?
Your solution
Since, for an odd periodic function the Fourier coefficients an (which constitute the real part
of cn ) are zero, then in this case the complex coefficients cn are wholly imaginary.
Further important points about complex Fourier coefficients
1. Since
1
cn =
T
T
2
f (t)e−inω0 t dt
− T2
then if f (t) is even, so cn will be real, we have two possible methods for evaluating cn :
(a) evaluate the integral above as it stands i.e. over the full range − T2 , T2 : note
carefully that the second term in the integrand is neither an even nor an odd function
so the integrand itself is
(even function) × (neither even nor odd function) = neither even nor odd function.
Thus we cannot write
2
cn =
T
7
T /2
f (t)e−inω0 t dt
0
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
(b) Put e−inω0 t = cos nω0 t − i sin nω0 t so
f (t)e−inω0 t = f (t) cos nω0 t − if (t) sin nω0 t
= (even)(even) − i(even)(odd)
= (even) − i(odd).
Hence
2
cn =
T
T
2
f (t) cos nω0 t dt =
0
an
.
2
2. If f (t + T2 ) = −f (t) then of course only odd harmonic coefficients cn (n = ±1, ±3, ±5, . . .)
will arise in the complex Fourier Series just as with trigonometric series.
Example Find the complex Fourier Series of the sawtooth wave shown:
f (t)
A
−T
2T
T
t
Solution
We have
At
0<t<T
T
f (t + T ) = f (t)
f (t) =
The period is T in this case so
2π
.
T
Looking at the graph of f (t) we can say immediately
ω0 =
i. the Fourier Series will contain a constant term c0
ii. if we imagine shifting the horizontal axis up to A2 the signal can be written
f (t) =
A
+ g(t),
2
where g(t) is an odd function with complex Fourier coefficients that are purely imaginary.
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
8
Solution (contd.)
Hence we expect the required complex Fourier series of f (t) to contain a constant term A2
and complex exponential terms with purely imaginary coefficients. We have, from the general
theory, and using 0 < t < T as the basic period for integrating.
1 T At −inω0 t
A T −inω0 t
cn =
dt = 2
te
dt
e
T 0 T
T 0
We can evaluate the integral using parts:
T
−inω0 t
te
dt =
0
=
But ω0 =
2π
T
te−inω0 t
(−inω0 )
T
0
1
+
inω0
T
e−inω0 t dt
0
−inω0 t T
1
T einω0 T
e
−
0
(−inω0 ) (inω0 )2
so
e−inω0 T = e−in2π = cos 2nπ − i sin 2nπ
= 1 − i0 = 1
Hence the integral becomes
−inω0 T
T
1
−
−
1
e
−inω0 (inω0 )2
Hence
A
cn = 2
T
Note that
c−n =
Also c0 =
1
T
T
0
At
dt
T
=
A
2
T
−inω0
=
iA
2πn
n = ±1, ±2, . . .
iA
−iA
=
= c∗n
2π(−n)
2πn
as it must
as expected.
Hence the required complex Fourier Series is
f (t) =
∞
A iA einω0 t
+
2
2π n=−∞ n
n=0
which could be written, showing only the constant and the first two harmonics as
A
ei2ω0 t
e−i2ω0 t
−iω0 t
iω0 t
f (t) =
+ π + ie
+i
... − i
− ie
+ ... .
2π
2
2
The corresponding trigonometric Fourier Series for the function can be readily obtained from
this complex series by combining the terms in ±n, n = 1, 2, 3, . . .
9
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
For example this first harmonic is
A −ie−iω0 t + ieiω0 t
2π
=
A
{−i(cos ω0 t − i sin ω0 t) + i(cos ω0 t + i sin ω0 t)}
2π
A
A
(−2 sin ω0 t) = − sin ω0 t
2π
π
Performing similar calculations on the other harmonics we obtain the trigonometric form of the
Fourier series
∞
A A sin nω0 t
f (t) = −
.
2
π n=1
n
=
Find the complex Fourier Series of the 2π−periodic function
−π <t<π
f (t) = et
f (t + 2π) = f (t)
f (t)
−π
3π
π
t
Firstly write down an integral expression for the Fourier coefficients cn :
Your solution
cn =
We have, since T = 2π so
ω0
1
2π
−π
π
et e−int dt
=1
Now combine the real exponential and the complex exponential as one term and carry out the
integration
Your solution
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
10
11
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
This is essentially a mathematical theorem but has, as we shall see, an important engineering
interpretation particularly in electrical engineering.
4. Parseval’s Theorem
e(1−in)π = eπ e−inπ = eπ (cos nπ − i sin nπ) = eπ cos nπ
e−(1−in)π = e−π einπ = e−π cos nπ
Hence
cn =
1
sinh π (1 + in)
1
(eπ − e−π ) cos nπ =
cos nπ
2π (1 − in)
π (1 + n2 )
Note that the coefficients cn n = ±1, ±2, . . . have both real and imaginary parts in this case as
the function being expanded is neither even nor odd.
Also
c−n =
sinh π (1 − in)
sinh π (1 − in)
cos(−nπ) =
cos nπ = cn∗ as required.
π (1 + (−n)2 )
π (1 + n2 )
This includes the constant term c0 =
sinh π
.
π
Hence the required Fourier Series is
∞
sinh π (1 + in) int
f (t) =
(−1)n
e
π n=−∞
(1 + n2 )
since
cos nπ = (−1)n .
Your solution
Now simplify this as far as possible and write out the Fourier Series:
We have
cn =
1
2π
π
e(1−in)t dt =
−π
π
1 e(1−in)t
1
1
=
e(1−in)π − e−(1−in)π
2π (1 − in) −π 2π (1 − in)
Parseval’s Theorem states that if f (t) is a periodic function with period T and if cn (n =
0, ±1, ±2, . . .) denote the complex Fourier coefficients of f (t) then
1
T
T
2
∞
2
f (t)dt =
− T2
|cn |2 .
n=−∞
In words the theorem states that the mean square value of the signal f (t) over one period equals
the sum of the squared magnitudes of all the complex Fourier coefficients.
Proof of Parseval’s Theorem.
Assume f (t) has a complex Fourier Series of the usual form:
f (t) =
∞
inω0 t
cn e
n=−∞
where
1
cn =
T
T
2
2π
ω0 =
T
f (t)e−inω0 t dt
− T2
Then
f 2 (t) = f (t)f (t) = f (t)
cn einω0 t =
cn f (t)einω0 t dt
Hence
1
T
T
2
− T2
1
f (t)dt =
T
2
T
2
− T2
1
cn
=
T
=
cn c∗n
∞
=
cn f (t)einω0 t dt
T
2
f (t)einω0 t dt
− T2
|cn |2
n=−∞
which completes the proof.
Parseval’s Theorem can also be written in terms of the Fourier coefficients an , bn of the trigonometric Fourier Series.
Recall that
c0 =
so
a0
2
cn =
an − ibn
2
n = 1, 2, 3, . . .
a2n + b2n
|cn | =
4
2
cn =
an + ibn
2
n = −1, −2, −3, . . .
n = ±1, ±2, ±3, . . .
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
12
so
∞
∞
a20
a2n + b2n
|cn | =
+2
4
4
n=−∞
n=1
2
and hence Parseval’s Theorem becomes
T
∞
2
a20 1 2
1
2
f (t)dt =
(a + b2n )
+
T − T2
4
2 n=1 n
. . . (∗)
The engineering interpretation of this theorem is as follows:
suppose f (t) denotes an electrical signal (current or voltage): then from elementary circuit
theory f 2 (t) is the instantaneous power (in a 1 ohm resistor) so that
1
T
T
2
f 2 (t)dt
− T2
is the energy dissipated in the resistor during one period.
Now a sinusoid wave of the form
A cos ωt
(or A sin ωt)
2
2
has a mean square value A2 so a purely sinusoidal signal would dissipate a power A2 in a 1 ohm
resistor.
Hence Parseval’s Theorem in the form (∗) states that the average power dissipated over 1 period
equals the sum of the powers of the constant (or d.c.) components and of all the sinusoidal (or
alternating) components.
The triangular signal shown
f (t)
π
−π
π
t
has trigonometric Fourier Series (see Section 29.3, p10)
f (t) =
∞
π
4 cos nt
−
.
2 π n=1
n2
(odd n)
∞
π4
1
=
.
Use Parseval’s Theorem to show that
n4
96
n=1
(n odd)
13
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
Firstly, identify a0 , an and bn for this situation and write down the definition of f (t) for this
case.
Your solution
f (t + 2π) = f (t)
f (t) = |t|
−π <t<π
Also
bn = 0
an =
We have
a0
2
=
π
2

n = 1, 2, 3, 4, . . .
0

 − n42 π
n = 2, 4, 6, . . .
n = 1, 3, 5, . . .
Now evaluate the integral on the left hand side of Parseval’s Theorem and hence complete the
problem.
Your solution
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
14
HELM (VERSION 1: March 19, 2004): Workbook Level 2
23.6: The Complex Form
1. i
(−1)n
eint
n
15
(sum from −∞ to ∞ excluding n = 0).
1
(sum from −∞ to ∞ excluding n = 0).
eint
2. π + i
n
(1 + in) int
sinh π (sum from −∞ to ∞).
(−1)n
e
π
(1 + n2 )
3.
3. f (t) = et
−π ≤t≤π
1. f (t) = t
−π ≤t≤π
2. f (t) = t
0 ≤ t ≤ 2π
Obtain the complex Fourier series for each of the following functions of period 2π
Exercises
We have f 2 (t) = t2 so
1
T
T
2
f 2 (t)dt =
− T2
1
2π
π
t2 dt =
−π
π
1 t3
π2
=
2π 3 −π
3
The right hand side of Parseval’s Theorem is
∞
∞
a02 2
π 2 1 16
a
+
+
n =
4
4
2 n=1 n4 π 2
n=1
(n odd)
Hence
∞
π2
π2
8 1
=
+ 2
3
4
π n=1 n4
∞
8 1
π2
=
π 2 n=1 n4
12
∴
(n odd)
∴
∞
π4
1
= .
n4
96
n=1
(n odd)
(n odd)