M. J. Roberts - 2/18/07
Web Appendix F - Derivations of the
Properties of the Continuous-Time
Fourier Series
F.1 Numerical Computation of the CTFS
The harmonic function of a periodic signal with period TF is
X k =
x (t ) e
1
TF
j 2 kf F t
TF
dt .
Since the starting point of the integral is arbitrary, for convenience set it to t = 0
1
X k =
TF
TF
x (t ) e
j 2 kf F t
dt .
0
()
Suppose we don’t know the function x t but we have a set of N F samples over one
period starting at t = 0 , the time between samples is Ts = TF / N F . Then we can
approximate the integral by the sum of several integrals, each covering a time of length
Ts
1
X k TF
N F 1
n=0
( n+1)Ts
x nTs e j 2 kf F nTs dt nTs
( )
(Figure F-1).
x(t)
...
T0
x(t)
...
t
Ts
...
...
t
F-1
(F.1)
M. J. Roberts - 2/18/07
Figure F-1 Sampling the arbitrary periodic signal to estimate its CTFS harmonic function
(In Figure F-1, the samples extend over one fundamental period but they could extend
over any period and the analysis would still be correct.) If the samples are close enough
()
together x t does not change much between samples and the integral (F.1) becomes a
good approximation. We can now complete the integration.
1
X k TF
1
X k TF
N F 1
N F 1
( n+1)Ts
1
x nTs e j 2 kf F t dt =
TF
nTs
( )
n=0
N F 1
( )
n=0
e j 2 kf F nTs e j 2 kf F ( n+1)Ts 1 e j 2 kf F Ts
x nTs =
j2 kf F
j2 kf F TF
( )
n=0
Using Ts = TF / N F ,
j 2 k / N F N F 1
1 e
X k j2 k
x ( nT ) e
j 2 nk / N F
=e
s
j k / N F
n=0
X k e
e
j k / N F
j k / N F
(
( n+1)Ts
e j 2 kf F t x nTs j2 kf F nT
s
N F 1
x ( nT ) e
j 2 kf F nTs
s
n=0
j 2 sin ( k / N F )
j k / N F
j k / N F N F 1
e
e
j 2 nk / N F
x nTs e
j2 k
n=0
( )
)
N 1
1 sin k / N F F
x nTs e j 2 nk / N F
NF k / NF
n=0
sinc ( k / N F )
(
sinc k / N F
)
( )
N F 1
x ( nT ) e
j 2 nk / N F
s
NF
n=0
For harmonic numbers k << N F we can further approximate the harmonic function as
1
X k NF
N F 1
x ( nT ) e
j 2 nk / N F
s
.
n=0
F.2 Linearity
()
()
()
Let z t = x t + y t . If TF = mT0 x = qT0 y , where m and q are integers, then
()
the CTFS harmonic function of z t is
Z k =
1
TF
z (t ) e
TF
j 2 kf F t
dt =
1
TF
F-2
TF
()
()
x t + y t e j 2 kf F t dt
M. J. Roberts - 2/18/07
and
() z t =
n= Z k e
j 2 kf F t
( X k + Y k ) e
=
j 2 kf F t
n= and
Z k = X k + Y k .
So the linearity property is
TF = mT0 x = qT0 y
()
()
FS
x t + y t X k + Y k .
F.3 Time Shifting
() (
)
Let z t = x t t0 and let TF = mT0 x = mT0 z , where m is an integer. Then
Z k =
1
TF
z (t ) e
jk F t
TF
dt =
1
TF
x (t t ) e
jk F t
0
TF
dt .
Making the change of variable = t t0 d = dt ,
Z k =
1
TF
x ( ) e
(
jk F +t0
)
TF
d = e
jk F t0
()
1
jk x e F d
T TF
F = X k Z k = e
jk F t0
X k .
So the time-shifting property is
TF = mT0
( )
x ( t t ) e
FS
x t t0 e
j 2 kf F t0
FS
0
F-3
jk F t0
X k .
X k M. J. Roberts - 2/18/07
F.4 Frequency Shifting
()
Let z t = e
j 2 k0 f F t
()
x t , with k0 being an integer and TF = mT0 x = mT0 z , where
m is an integer. Then
Z k =
1
TF
z (t ) e
j 2 kf F t
TF
Z k =
1
TF
dt =
x (t ) e
1
TF
(
e
TF
)
j 2 k k0 f F t
TF
j 2 k0 f F t
()
x t e
j 2 kf F t
dt
dt = X k k0 .
So the frequency shifting property is
TF = mT0
e
()
x ( t ) X k k j 2 k0 f F t
e
jk0 F t
FS
x t X k k0 .
FS
0
F.5 Time Reversal
() ( )
Let z t = x t and let TF = mT0 x = mT0 z , where m is an integer. If
( ) X k e
x t =
j 2 kf F t
k = then
( ) x t =
k = X k e
( )
j 2 kf F t
=
X k e
(
)
j 2 kf F t
.
k = Let q = k , then
( ) X q e
x t =
j 2 qf F t
q=
and, since changing the order of summation does not change the sum,
( ) X q e
x t =
j 2 qf F t
q = Therefore, since
( ) Z k e
z t =
k = we can say that
F-4
j 2 kf F t
.
M. J. Roberts - 2/18/07
k = Z k e
j 2 kf F t
=
X k e
j 2 kf F t
k = and Z k = X k . So the time reversal property is
( )
TF = mT0
.
FS
x t X k F.6 Time Scaling
() ( )
Let z t = x at , a > 0 and let TF = mT0 x , m an integer. (Figure F-2).
x(t)
t
T0x
z(t)
t
T0x
a
()
()
Figure F-2 A signal x t and a time-scaled version z t of that signal
()
The first thing to realize is that if x t is periodic with fundamental period T0 x that
()
z t is periodic with fundamental period T0 z = T0 x / a and fundamental frequency af0 x .
()
()
Case 1. z t represented by a CTFS over a period of z t , TF / a .
The CTFS harmonic function will be
a
Z k =
TF
t0 +TF / a
t0
()
z t e
j 2 kaf F t
a
dt =
TF
t0 +TF / a
( )
x at e
t0
We can make the change of variable = at d = adt yielding
F-5
j 2 kaf F t
dt .
M. J. Roberts - 2/18/07
a 1
Z k =
TF a
at0 +TF
x ( ) e
j 2 kaf F / a
at0
1
d =
TF
at0 +TF
x ( ) e
j 2 kf F d .
at0
Since the starting point t0 is arbitrary
Z k =
1
TF
x ( ) e
j 2 kf F TF
d = X k ()
and the CTFS harmonic function describing z t over the period TF / a is the same as
()
the CTFS harmonic function describing x t over the period TF .
() ( )
z t = x at
TF = mT0 x TF / a = mT0 x /a
Z k = X k ()
()
Even though the CTFS harmonic functions of x t and z t are the same, the
CTFS representations themselves are not because the fundamental frequencies are
different. The representations are
() x t =
k = X k e
j 2 kf F t
()
( ) Z k e
z t = x at =
and
j 2 kaf F t
.
k = ()
()
Case 2. z t represented by a CTFS over a period of x t , TF
The CTFS harmonic function will be
1
Z k =
TF
t0 +TF
z (t ) e
j 2 kf F t
t0
1
dt =
TF
t0 +TF
x ( at ) e
j 2 kf F t
dt .
t0
Let = at d = adt . Then
1
Z k =
aTF
at0 + aTF
()
x e
j 2 kf F / a
d .
at0
If a is not an integer, the relationship between the two harmonic functions Z k and
X k cannot be simplified further.
F-6
M. J. Roberts - 2/18/07
()
Let a be a non-zero integer. The signal x t is made up of frequency components
at integer multiples of f F .
()
Therefore for ratios k / a that are not integers, x and
j 2 kf F / a
are orthogonal on the interval at0 < < at0 + aTF and Z k = 0 . For ratios
k / a that are integers, the integral over a periods is a times the integral over one period
and
e
1
Z k = a aTF
at0 +TF
()
x e
(
)
j 2 k / a f F at0
d = X k / a , k / a an integer .
So the time scaling property, for this kind of time scaling, is
() ( )
TF = mT0 x , z t = x at , a a non-zero integer
X k / a , k / a an integer
Z k = , otherwise
0
.
F.7 Change of Period
()
()
If the CTFS harmonic function of x t over any period TF is X k , we can find
the CTFS harmonic function X q k of x t over a time qTF where q is a positive
integer. The new fundamental CTFS frequency is then f F / q and
X q k =
1
qTF
qTF
()
x t e
(
)
j 2 kf F / q t
dt
This is exactly the same as the result for time scaling by a positive integer in the previous
section and the result is
X k / q , k / q an integer
TF qTF X q k = .
, otherwise
0
F.8 Time Differentiation
( ) dtd ( x (t )) and let T
can represent z ( t ) by
Let z t =
F
= mT0 x = mT0 z , where m is an integer. Then we
F-7
M. J. Roberts - 2/18/07
()
z t =
d
d jk t jk t
x t = X k e F = jk F X k e F .
dt
dt n= n= ( ( ))
Then, if
( ) Z k e
z t =
jk F t
k = it follows that
k = Z k e
jk F t
= jk F
X k e
jk F t
,
k = Z k = jk F X k .
So the differentiation property is
TF = mT0
( ( ))
( ( ))
.
d
d
FS
FS
x t j2 kf F X k ,
x t jk F X k dt
dt
F.9 Time Integration
( ) x ( ) d
Let z t =
t
and let TF = mT0 x , where m is an integer. We must
consider two cases separately, X 0 = 0 and X 0 0 . If X 0 0 then, even though
x t is periodic, z t is not and we cannot represent it exactly for all time with a CTFS
()
()
(Figure F-3).
F-8
M. J. Roberts - 2/18/07
X[0] = 0
X[0] = 0
x(t)
x(t)
...
...
...
...
t
t
t
tt
∫x(λ)dλ
x(τ)dτ
∫−∞
∫ x(τ)dτ
−∞
...
...
t
t
Figure F-3 Effect of a non-zero average value on the integral of a periodic function
()
If X 0 = 0 then we can represent z t by
t
( ) x ( ) d
z t =
and its CTFS harmonic function is
Z k =
1
TF
x ( ) d e
t
TF
jk F t
dt .
For k = 0 ,
Z 0 =
1
TF
x ( ) d dt
t
TF
()
which is the average value of the integral of x t over one fundamental period. We
()
know that the average value of x t is zero but, without some other information, we
()
don’t know what the average value of the integral of x t is and cannot determine the
value of Z 0 . However we can determine all the other values of Z k .
t
t
( ) x ( ) d = X k e
d = X k e
= k
k = z t =
jk F t
jk F d
()
x 1
Z k =
T
F
jk F t
t
jq F e
X
q
d
dt
e
TF q = F-9
(F.2)
M. J. Roberts - 2/18/07
In (F.2) k has been replaced by q to avoid confusion with k which is independent of q. To
finish the integration we must evaluate the integral
t
e
jq F d which is
t
e jq F .
jq F Evaluation at the upper limit is not a problem but the lower limit does present a problem.
jq Since the complex sinusoid e F is periodic it is impossible to define what its value is at
the lower limit of negative infinity because negative infinity is a limit, not a number. The
jq magnitude of e F is one but its phase could be any value in a range of 2 radians.
But, it turns out that this won’t matter. Call the indeterminate value at the lower limit C.
Then
t
jq t
e jq F e F
C
=
jq F jq F
and
jk t
e jq F t
X
q
C
e F dt
TF q = jq
F
j
q
k
t
X q e ( ) F
jk t
= Ce F dt
TF
q = TF
jq F
1
Z k =
TF
For each k 0 the computation involves a summation of terms involving q. But all of
those terms are zero unless q = k because the integration is over a integer number of
periods of a complex sinusoid. So, in the end,
Z k =
X k TF
X k X k 1 dt
=
T
=
, k0
TF jk jk F TF F
jk F
F and the integration property is
TF = mT0
t
()
x d FS
X k j2 kf F
t
,
()
FS
x d F-10
X k jk F
, k 0 , if X 0 = 0
.
M. J. Roberts - 2/18/07
F.10 Multiplication-Convolution Duality
() () ()
Let z t = x t y t and let TF = mT0 x = qT0 y , where m and q are integers. Then
Z k =
1
TF
z (t ) e
j 2 kf F t
TF
dt =
x (t ) y (t ) e
1
TF
j 2 kf F t
TF
dt .
Then, using
() y t =
k = Y k e
j 2 kf F t
=
Y q e
j 2 qf F t
q = we get
Z k =
1
TF
j 2 kf t
j 2 qf F t
F
x
t
Y q e
dt .
e
TF q
= ()
Reversing the order of integration and summation,
Z k =
Y q x (t ) e
1
TF
j 2 qf F t j 2 kf F t
e
TF
q = dt
or
Y q T x (t ) e
Z k =
1
(
)
j 2 k q f F t
dt .
q = F
TF
= X k q Then
Z k =
Y q X k q q = and the multiplication property is
TF = mT0 x = qT0 y
() ()
x t y t FS
Y q X k q = X k Y k .
q = This result q = Y q X k q is a convolution sum. So the product of CT signals
corresponds to the convolution sum of their CTFS harmonic functions.
Now let Z k = X k Y k and let TF = mT0 x = qT0 y , where m and q are
integers. Then
F-11
M. J. Roberts - 2/18/07
( ) X k Y k e
z t =
jk F t
k = ()
z t =
1
T
k = F
()
TF
x e
jk F d Y k e
jk F t
=
1
TF
( ) Y k e
TF
x d
(
jk F t )
k = (
y t )
or
()
z t =
x ( ) y (t ) d
1
TF
TF
.
This integral looks just like a convolution integral except that it covers the range
t0 < t0 + TF instead of < < . This integral operation is called periodic
convolution and is indicated by the notation
()
( ) x ( ) y (t ) d
x t y t =
Therefore
TF
() (
) ()
.
()
z t = 1 / TF x t y t .
()
Since x t
is periodic it can be expressed as the periodic extension of an
()
aperiodic function x ap t
( ) x (t qT ) = x (t ) (t ) .
x t =
ap
q = F
ap
TF
()
(The function, x ap t , is not unique. It can be any function which satisfies this equation.)
Then
x t y t = x ap qTF y t d
TF
q = ()
()
()
t0 +TF
q = t0
() x t y t =
(
) (
(
) (
)
)
x ap qTF y t d .
Let = qTF . Then d = d and
()
(
t0 + q +1 TF
q = t0 + qTF
() x t y t =
)
( ) ( (
))
x ap y t + qTF d .
()
Since y t is periodic, with period TF ,
F-12
M. J. Roberts - 2/18/07
( (
)) = y (t qT
y t + qTF
and the summation of integrals
infinite limits
F
(
)
t0 + q +1 TF
q = t + qT
0
F
) (
= y t
)
is equivalent to the single integral over
we conclude that
()
( ) x ( ) y (t ) d = x (t ) y (t ) .
x t y t =
ap
ap
()
()
expressed as an aperiodic convolution of y ( t ) with a function x ( t ) which, when
periodically repeated with the same period T equals x ( t ) . The periodic convolution of
So the periodic convolution of two functions x t and y t each with period TF can be
ap
F
two periodic functions corresponds to the product of their CTFS harmonic function
representations and the period TF and the convolution property is
TF = mT0 x = qT0 y
()
()
FS
x t y t TF X k Y k .
F.11 Conjugation
()
()
Let z t = x * t and let TF = mT0 x = mT0 z , where m is an integer. Then
Z k e
j 2 kf F t
k = =
=
X k e
j 2 kf F t
k = X
*
k = k e
j 2 kf F t
*
=
X
k =
*
k e j 2 kf F t
and, since changing the order of summation does not change the sum,
k = Z k e
j 2 kf F t
=
X
k = *
k e j 2 kf F t ,
Z k = X * k and the conjugation property is
()
TF = mT0
FS
x * t X * k F-13
.
M. J. Roberts - 2/18/07
F.12 Parseval's Theorem
The signal energy in any period TF = mT0 x , where m is an integer, of any
()
periodic signal x t is
Ex,T =
F
Ex,T
F
F
Ex,T
F
Ex,T
F
x t
TF
=
TF Ex,T =
()
2
X k e
dt =
TF
2
j 2 kf F t
dt
k = *
X k e
j 2 kf F t
k = j 2 qf F t
dt
X q e
q = X k e j 2 kfF t X* q e j 2 qfF t dt
TF k
= q = j 2 k q f t = X k X * q e ( ) F dt
TF
k = q = j 2 ( k q ) f F t *
*
dt
=
X k X k + X k X q e
TF k = k = q = k q
Ex,T =
F
TF
k = 2
X k dt + TF
X k X
k = q = k q
*
q e
(
)
j 2 k q f F t
dt
=0 , k q
Ex,T = TF
F
X k k = 2
()
Therefore, for any periodic signal x t
TF = mT0
1
TF
()
TF
x t
2
dt =
k = F-14
X k 2
.