F CTFSDerivations

M. J. Roberts - 1/3/11

F-1

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 T is

cx k =

1T

x t( )e j2 kt /T dtT .

Since the starting point of the integral is arbitrary, for convenience set it to t = 0

cx k =

1T

x t( )e j2 kt /T dt0

T

. Suppose we dont know the function

x t( ) but we have a set of N samples over one

period starting at t = 0 , the time between samples is Ts = T / N . Then we can approximate the integral by the sum of several integrals, each covering a time of length Ts

cx k 1T

x nTs( )e j2 knTs /T dtnTs

n+1( )Ts

n=0

N 1

(F.1)

(Figure F-1).


F-2

t

x(t)

t

x(t)

T0

Ts

...

...

...

...

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.

cx k 1T

x nTs( ) e j2 kt /T dtnTs

n+1( )Ts

n=0

N 1

= 1T x nTs( )e j2 kt /T

j2k / T

nTs

n+1( )Ts

n=0

N 1

cx k

1T

x nTs( ) e j2 knTs /T e j2 k n+1( )Ts /T

j2k / T

n=0

N 1

= 1 e j2 kTs /T

j2kx nTs( )e j2 knTs /T

n=0

N 1

Using Ts = T / N ,

cx k

1 e j2 k / N

j2kx nTs( )e j2nk / N

n=0

N 1

= e j k / N ej k / N e j k / N

j2 sin k / N( )

j2kx nTs( )e j2nk / N

n=0

N 1

cx k e j k / N 1

Nsin k / N( )k / Nsinc k / N( )

x nTs( )e j2nk / N

n=0

N 1

e j k / N / N( )sinc k / N( ) x nTs( )e j2nk / Nn=0

N 1


F-3

For harmonic numbers k


F-4

So the time-shifting property is

T = mT0x t t0( ) FST e j2 kt0 /T cx k

.

F.4 Frequency Shifting (Harmonic Number Shifting) Let

z t( ) = e j2 k0t /T x t( ) , with k0 being an integer and T = mT0x = mT0z , where m

is an integer. Then

cz k =

1T

z t( )e j2 kt /T dtT =

1T

e j2 k0t /T x t( )e j2 kt /T dtT

cz k =

1T

x t( )e j2 k k0( )t /T dtT = cx k k0 .

So the frequency shifting property is

T = mT0e j2 k0t /T x t( ) FST cx k k0

.

F.5 Time Reversal

Let z t( ) = x t( ) and let T = mT0x = mT0z , where m is an integer. If

x t( ) = cx k e j2 kt /T

k=

then

x t( ) = cx k e j2 k t( )/T

k=

= cx k e j2 k( )t /Tk=

. Let q = k , then

x t( ) = cx q e j2qt /T

q=

and, since changing the order of summation does not change the sum,

x t( ) = cx q e j2qt /T

q=

.


F-5

Therefore, since

z t( ) = cz k e j2 kt /T

k=

we can say that

cz k e

j2 kt /T

k=

= cx k e j2 kt /Tk=

and

cz k = cx k . So the time reversal property is

T = mT0x t( ) FST cx k

.

F.6 Time Scaling

Let z t( ) = x at( ) , a > 0 and let T = mT0x , m an integer. (Figure F-2).

x(t)

t

t

z(t)T0x

T0xa 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 T0x that

z t( ) is periodic with fundamental period T0z = T0x / a .


F-6

Case 1. z t( ) represented by a CTFS over a period of z t( ) , T / a .

The CTFS harmonic function will be

cz k =

aT

z t( )e j2 kt / T / a( ) dtt0

t0 +T / a

=aT

x at( )e j2 kat /T dtt0

t0 +T / a

. We can make the change of variable = at d = adt yielding

cz k =

aT

1a

x ( )e j2 k /T dat0

at0 +T

=1T

x ( )e j2 k /T dat0

at0 +T

. Since the starting point t0 is arbitrary

cz k =

1T

x ( )e j2 k /T dT = cx k

and the CTFS harmonic function describing

z t( ) over the period T / a is the same as the

CTFS harmonic function describing x t( ) over the period T .

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( ) = cx k e j2 kt /T

k=

and z t( ) = x at( ) = cz k e j2 kat /T

k=

.

Case 2. z t( ) represented by a CTFS over a period of x t( ) , T

The CTFS harmonic function will be

cz k =

1T

z t( )e j2 kt /T dtt0

t0 +T

=1T

x at( )e j2 kt /T dtt0

t0 +T

. Let = at d = adt . Then

cz k =

1aT

x ( )e j2 k / a( )/T dat0

at0 +aT

.


F-7

If a is not an integer, the relationship between the two harmonic functions

cz k and

cx k cannot be simplified further.

Let a be a non-zero integer. The signal x t( ) is made up of frequency components

at integer multiples of 1 / T . Therefore for ratios k / a that are not integers, x ( ) and e

j2 k / a( )/T are orthogonal on the interval at0 < < at0 + aT and cz k = 0 . For ratios k / a that are integers, the integral over a periods is a times the integral over one period and

cz k = a

1aT

x ( )e j2 k / a( ) /T dat0

at0 +T

= cx k / a , k / a an integer .

So the time scaling property, for this kind of time scaling, is

T = mT0x , z t( ) = x at( ) , a a non-zero integercz k =

cx k / a , k / a an integer

0 , otherwise

.

F.7 Change of Period If the CTFS harmonic function of

x t( ) over any period T is cx k , we can find

the CTFS harmonic function cx q k of x t( ) over a time qT where q is a positive

integer.

cx q k =

1qT

x t( )e j2 kt / qT dtqT

This is exactly the same as the result for time scaling by a positive integer in the previous section and the result is

T qT cx q k =

cx k / q , k / q an integer

0 , otherwise

.


F-8

F.8 Time Differentiation Let

z t( ) = ddt x t( )( ) and let T = mT0x = mT0z , where m is an integer. Then we can

represent z t( ) by

z t( ) = ddt x t( )( ) =

ddt

cx k ej2 kt /T

n=

= j2k / T( )cx k e j2 kt /T

n=

. Then, if

z t( ) = cz k e j2 kt /T

k=

it follows that

cz k e

j2 kt /T =k=

j2k / T( ) cx k e j2 kt /Tk=

,

and cz k = j2k / T( )cx k .

So the differentiation property is

T = mT0ddt

x t( )( ) FST j2k / T( )cx k .

F.9 Time Integration Let

z t( ) = x ( )d

t

and let T = mT0x , where m is an integer. We must consider two cases separately,

cx 0 = 0 and

cx 0 0 . If cx 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-9

......t

x(t)

......t

......t

x(t)

t

cx 0[ ] = 0 cx 0[ ] & 0

x o( )do


F-10

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

e j2q /T d

t

which is

e j2q /T

j2q / T

t

.

Evaluation at the upper limit is not a problem but the lower limit does present a problem. Since the complex sinusoid e j2q /T 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 magnitude of e j2q /T is one but its phase could be any value in a range of 2 radians. But, it turns out that this wont matter. Call the indeterminate value at the lower limit C. Then

e j2q /T

j2q / T

t

=e j2qt /T

j2q / T C

and

cz k =1T

cx q e j2qt /T

j2q / T C

q=

e j2 kt /T dtT

=cx q

Te j2 q k( )t /T

j2q / T Ce j2 kt /T

dtT

q=

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 an integer number of periods of a complex sinusoid. So, in the end,

cz k =

cx k T

1j2k / T

dtT =

cx k j2k / T

, k 0

and the integration property is

T = mT0

If cx 0 = 0 , x ( )d

t

FST cx k

j2k / T , k 0

.


F-11

F.10 Multiplication-Convolution Duality

Let z t( ) = x t( )y t( ) and let T = mT0x = qT0 y , where m and q are integers. Then

cz k =

1T

z t( )e j2 kt /T dtT =

1T

x t( )y t( )e j2 kt /T dtT .

Then, using

y t( ) = cy k e j2 kt /T

k=

= cy q e j2qt /Tq=

we get

cz k =

1T

x t( ) cy q e j2qt /Tq=

e j2 kt /T dt

T . Reversing the order of integration and summation,

cz k =

1T

cy q x t( )e j2qt /T e j2 kt /T dtTq=

or

cz k = cy q 1T

x t( )e j2 kq( )t /T dtT

=cx kq

q=

.

Then

cz k = cy q cx k q

q=

and the multiplication property is

T = mT0x = qT0 y

x t( )y t( ) FST cy q cx k q q=

= cx k cy k .

This result

cy q cx k q q=

is a convolution sum. So the product of periodic continuous-time signals with a common period T corresponds to the convolution sum of their CTFS harmonic functions, both of which are based on that common period. Now let

cz k = cx k cy k and let T = mT0x = qT0 y , where m and q are

integers. Then


F-12

z t( ) = cx k cy k e j2 kt /T

k=

z t( ) = 1T x ( )e j2 k /T d

T

cy k e

j2 kt /T

k=

= 1T x ( )d cy k ej2 k t( )/T

k=

y t( )

T

or

z t( ) = 1T x ( )y t ( )dT .

This integral looks just like a convolution integral except that it covers the range t0 < t0 + T instead of < < . This integral operation is called periodic convolution and is indicated by the notation

x t( ) y t( ) = x ( )y t ( )d

T . Therefore

z t( ) = 1 / T( )x t( ) y t( ) .

Since

x t( ) is periodic it can be expressed as the periodic extension of an

aperiodic function xap t( )

x t( ) = xap t qT( )

q=

= xap t( )T t( ) .

(The function xap t( ) is not unique. It can be any function that satisfies this equation.)

Then

x t( ) y t( ) = xap qT( )

q=

y t ( )dT

x t( ) y t( ) = xap qT( )y t ( )d

t0

t0 +T

q=

. Let = qT . Then d = d and

x t( ) y t( ) = xap ( )y t + qT( )( )d

t0 +qT

t0 + q+1( )T

q=

.


F-13

Since

y t( ) is periodic, with period T ,

y t + qT( )( ) = y t qT ( ) = y t ( )

and the summation of integrals

t0 +qTt0 + q+1( )Tq=

is equivalent to the single integral over

infinite limits

we conclude that

x t( ) y t( ) = xap ( )y t ( )d

= xap t( ) y t( ) .

So the periodic convolution of two functions x t( ) and y t( ) with a common period T

can be expressed as an aperiodic convolution of y t( ) with a function xap t( ) which,

when periodically repeated with the same period T equals x t( ) . The periodic convolution of two periodic functions corresponds to the product of their CTFS harmonic function representations and the period T and the convolution property is

T = mT0x = qT0 yx t( ) y t( ) FST T cx k cy k

.

F.11 Conjugation Let

z t( ) = x* t( ) and let T = mT0x = mT0z , where m is an integer. Then

cz k ej2 kt /T

k=

= cx k e j2 kt /Tk=

*

= cx* k e

j2 kt /T

k=

= cx* k e j2 kt /Tk=

and, since changing the order of summation does not change the sum,

cz k e

j2 kt /T

k=

= cx* k e j2 kt /Tk=

,


F-14

cz k = cx

* k and the conjugation property is

T = mT0x* t( ) FST cx* k

.

F.12 Parseval's Theorem The signal energy in any period T = mT0x , where m is an integer, of any periodic

signal x t( ) is

Ex ,T = x t( )

2dt

T = cx k ej2 kt /T

k=

2

dtT

Ex ,T = cx k e

j2 kt /T

k=

cx q ej2qt /T

q=

*

dtT

Ex ,T = cx k e

j2 kt /T cx* q e

j2qt /T

q=

k=

dt

T

Ex ,T = cx k cx

* q ej2 kq( )t /T

q=

k=

dt

T

Ex ,T = cx k cx* k

k=

+ cx k cx* q e j2 kq( )t /Tq=

k=kq

dtT

Ex ,T = cx k 2

k=

dtT + cx k cx* q e

j2 kq( )t /Tq=

k=kq

dtT

=0 , kq

Ex ,T = T cx k

2

k=


F-15

Therefore, for any periodic signal

x t( ) the average signal power can be calculated either

of two ways.

T = mT01T

x t( ) 2 dtT = cx k

2

k=

.

F CTFSDerivations

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