I

Continuous Signals and Spectra

2. Continuous signals

2.1 Continuous and Discrete Signals. What distinguishes "continuous" signals is that they are defined for all times in some interval. They need not be continuous functions in the mathematical sense of the word (although real signals usually are). A discrete signal (regarded as a sampled signal for example) is defined only at certain instants separated by finite intervals. It is not zero between sampled values, it is simply undefined. For example the continuous signal f t A t t( ) sin( ) [ , )= ∈ ∞ω : 0 has well defined values of all times from zero to infinity, whereas the discrete signal which results from sampling at intervals T, f n A n T n( ) sin( ) , , , ,= = ⋅ ⋅ ⋅ω ; 0 1 2 3 , is (strictly speaking) undefined between sampling instants. Notwithstanding this distinction, discrete time signal processing and continuous signal processing share many methods and it is convenient to review some of the basics of continuous signal analysis before dealing more directly with discrete time signals.

2.2 Energy and Power of Signals. For voltage (or current) signals, the instantaneous power delivered to a load is (generally speaking) proportional to the square of the signal amplitude ( RIRVP 22 == ). The total energy in the signal (assume that the signal is defined for all time) is then proportional to

or ∫∞

∞−

dttV )(2 ∫∞

∞−

dttI )(2

This relation is used as a convenient definition for all signals, regardless of their physical nature. Thus the total energy of a signal, f(t), is defined to be

∫∞

∞−

= dttf 2)(E (the magnitude allows for the case in which f(t) is not real)

and the average power in a signal is defined by

∫−∞→

=T

TTdttf

TP 2)(

21lim

2.3 Finite Energy Signals For a finite energy signal, E is finite, ie.

f t dt2 ( ) < ∞−∞

∞

∫

A bounded finite duration pulse is an example of a finite energy signal. For example the unit height rectangular pulse of half-width T, which we will write as P , is defined by

tT ( )

⎩⎨⎧

>≤

=TtTt

tPT : 0 : 1

)(

For this signal

and the pulse is clearly finite energy. E= = = =−−∞

∞

−∞

∞

∫∫∫ f t dt P t dt dt TTT

T2 2 2( ) ( )

-T T

1

t

P (t)T

< ∞

Not all finite energy signals are of "compact support" (ie. zero for all t outside some finite interval or set of intervals). An important example is the "causal exponential"

1

t

U(t)e-t

f tt

e t T( ) /=<≥

⎧⎨⎩

−

0 0 : : t 0

This signal has finite values for all t > 0, but the energy is finite :

[ ]E= = = − =− − ∞∞

−∞

∞

∫∫ f t dt e dt T e Tt T t T2 2 20

0 2 2( ) / /

< ∞

As a final example consider the Gaussian pulse ( )f t e t T( ) /= − 12

2

. The indefinite integral of the Gaussian cannot be written in terms of elementary functions, but it is

not too difficult to show that e dtt−

−∞

∞

=∫12

2

2π , from which is should be

straightforward to show that the Gaussain pulse is also finite energy.

2.4 Finite Power Signals For a finite power signal the average power is finite, ie.

T T

T

Tf t dt

→∞ −

< ∞∫lim ( )12

2

In reality all physical signals must be finite energy, but many signals are conveniently idealized by functions which do not have the finite energy property. For example ∞<<∞−= tttf ; ) sin()( ω is clearly not finite energy since

E T

T

T

T

T

T

T

T

T

f t dt t dt t dt t t T T= = = − = −⎡

⎣⎢⎤⎦⎥

= −−−− −∫∫∫ 2 2 1

2121 2 2

22

2( ) sin ( ) ( cos( )) sin( ) sin( )

ω ωω

ωω

ω and this becomes infinite as T approaches infinity. The signal is finite power, however, since

PTT

T= =→∞

lim E

212

< ∞

In fact any bounded periodic function is finite power. The converse is not true, not all finite power signals are periodic. Noise is an important example of a finite power signal which is not periodic.

2.5 Periodic Signals A periodic signal repeats itself exactly after a finite interval. If T is the period of the signal then f t f t T t( ) ( )= + for all

Note that this immediately implies that f t f t nT n( ) ( ) , , ,= + = ± ± ⋅⋅ ⋅ ; 0 1 2 .

For example f t t t( ) sin( )= − ∞ < < ∞ω ; is periodic with period T =2πω

since

f t t t t f t( ) sin( ( )) sin( ) sin( )+ = + ( )= + = =2 2 2π ω ω π ω ω π ω .

Whereas the superficially similar signal is not strictly

periodic since, in general,

f tt

t( )

sin( )=

<≥

⎧⎨⎩

0 0 : : t 0

ω

sin( ( ))ω for t T T t+ ≠ − < <0 0 . Periodic signals are useful idealizations, real signals can only be approximately periodic. The consequences of this for signal analysis are discussed in detail later in the course.

2.6 Impulsive Signals The impulse of a signal is defined in analogy with the impulse due to a force in physics

∫∞

∞−

= dttfI )(

An impulsive (or impulse) signal is an idealization of a pulse that has a finite "impulse" but is of too short a duration for its exact shape to be of importance. Impulse-type signals also have a number of useful mathematical properties that will be exploited in subsequent lectures. The basic impulse signal is the Dirac delta function δ( )t . Very loosely speaking this is a pulse with the following properties :

δ : :

( )ttt

=≠

∞ =⎧⎨⎩

0 00

and δ for all and otherwise

( )t dta b

a

b

=< >⎧

⎨⎩

∫1 00

0

ie. the Dirac delta function is a "unit impulse" pulse of vanishing duration. It is often convenient to think of a delta function as the limiting form of a unit area pulse as the pulse width goes to zero. For example if we use a rectangular pulse

∆T TtT

P t( ) ( )=1

2,

then clearly

∆T tt T

Tt T

( ) =>

<

⎧⎨⎪

⎩⎪

01

2

:

: and ∆T

a

b

t dta T bb T a T

( )∫ =< − >< − >

⎧⎨⎩

10 : and : or

-T T

1/2T

t

T∆

T

and in the limit as T approaches zero these formulas reduce to the informal defining properties of a delta function. In fact any reasonable unit area pulse sharply peaked about t = 0 will produce the same results in the limit as the pulse width vanishes.

2.7 Properties of the Dirac Delta Function The Dirac delta function must be used with some care, and many of its properties are only rigorously true when discussing integrals involving delta functions. The following properties can be derived from the "defining" property, and tend to be reasonably easy to see from our informal discussion above.

Defining : for all functions f(t) continuous at t = 0. δ ( ) ( ) ( )t f t dt f−∞

∞

∫ = 0

Shift : δ ( ) ( ) (t T f t dt f T− =−∞

∞

∫ )

Symmetry: δ δ ( ) ( )t t= − Multiplication : f t t T f T t T( ) ( ) ( ) ( )δ δ − = −

Scaling : δ δ ( ) ( ) : ata

t a= ≠1 0

Note that the strict definition of the delta function as a "generalized function" is in terms of the defining property above. It is easy to establish this property from our informal definition using (say) the ∆T t( ) functions: the integral is equal to the average value of f(t) over a small interval (-T,T) around the origin, which approaches f(0) arbitrarily closely as provided f(t) is continuous (in the mathematical sense) at t = 0.

T → 0

Other properties of the delta function can be defined (even differentiation), the most important of these is the "convolution" property. The convolution of two functions f(t) and g(t) is defined by the integral

f g t f t g t t dt∗ = ′ − ′ ′−∞

∞

∫( ) ( ) ( )

(The importance of integrals of this type will become evident in later lectures.) Note that convolution is commutative : f g t g f t∗ = ∗( ) ( ) The convolution property of the delta function is f t t T f t T( ) ( ) ( )∗ − = −δ ie. convolving a function with a delta function simply shifts the function.

2.8 Harmonic Functions and Complex Exponentials. Harmonic functions are of the form f t A t( ) cos( )= +ω φ , which can be written in the alternative form f t a t b t( ) cos( ) sin( )= +ω ω where a A b A= = −cos sinφ φ and

The quantity ω ππ

= =2 2 fT

is the frequency of the signal in rad/sec (or simply s-1),

f is the frequency in cycles per second or Hz, T is the period of the signal in seconds.

The choice of the cosine is arbitrary, we could equally well have used f t A t( ) sin( ) /= + = +ω ϕ ϕ φ π where 2 .

Harmonic signals at integer multiples of the frequency of a given signal are called the harmonics of that signal, and the given signal is called the fundamental. ( The same terms are also used to refer to the frequencies of the signals.) So, if f t t1( ) cos( )= + 1ω φ is the fundamental then f t n tn n( ) cos( )= +ω φ is the n-th harmonic. The relationships between harmonic signals and complex exponentials all follow from deMoivre's theorem: e jjθ θ θ= +cos sin From which

( ) ( )cos sinθ θθ θ θ θ= + = −− −12

12

e ej

e ej j j j and

Re

Im

θ

z = e j θ

cos

sin

θ

θ With we can relate harmonic signals to complex exponentials and vv.

θ ω φ= + t

For θ real, the magnitude of e is one, since : jθ

e e ej j jθ θ θ θ θ= + = +(Re ) (Im ) cos sin2 2 2 2 1=

The complex conjugate of e jθ is ( )e e jj jθ θ θ θ

∗ −= = −cos sin

2.9 Problems 1. Which of the following functions are finite energy, which are finite power and

which are neither ? (i). (ii). sin t e t−

(iii). t (iv). 112t +

(v). e tt− 2

sin (Note that it may not be necessary to explicitly calculate the integrals involved in order to answer this question.)

2. What is the period of the full wave rectified cosine wave f t t( ) cos( )= π 2 ?

What is the period of the half-wave rectified cosine

f tt t

t( )

cos( ) cos( )cos( )

=≥<

⎧⎨⎩

π ππ

2 22

: if 0 : if 0

0

t t

)

?

3. Use the properties of the delta function to evaluate or simplify the following

expressions:

(i). (ii).

(iii). (iv). e t

e t dt4 δ( )−∞

∞

∫ e t dt4 0 25δ( . )−−∞

∞

∫δ( )7t t4 0 25∗ −δ( .

4. By using the ∆T tT

P t( ) ( )=1

2 T function as an approximation to the delta

function, show that the defining property of the delta function is true in the limit as provided the function f(t) is continuous at the origin. T → 0

5. Use deMoivre’s theorem to prove the identity 2 2sin cos sinθ θ θ= .

3. Fourier analysis of continuous time signals I Fourier series

3.1 Fourier Series Periodic signals (which in addition satisfy some not-very-restrictive extra conditions) can be approximated arbitrarily closely by a sum of the harmonics of a fundamental with the same frequency as the signal. This result, first published by Fourier in 1812, is a cornerstone of system analysis and signal processing. Thus if f(t) is a periodic signal with period T, it has a Fourier Series expansion of the form

f t a a nT

t b nT

tn nn

( ) cos sin= + ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

=

∞

∑01

2 2π π

The quantities an and bn are called Fourier coefficients, their calculation is described in the next section. The approximation gets better as the number of terms included in the summation increases. There are (at least) two other versions of this result, which can be derived from the

properties of harmonic functions. Thus, using ω ππ

1 12 2= = f

T for ease of notation,

we have an expansion entirely in terms of cosine functions

( )f t a A n tn nn

( ) cos= + +=

∞

∑0 11

ω φ

and an expansion in terms of complex exponentials

f t F enjn t

n

( ) ==−∞

∞

∑ ω1

If f(t) is a real function then an, bn, An and φn are all real quantities. In general the quantities Fn are complex numbers. Relations between the Fourier coefficients of the three forms are easily found , for example, for real functions :

A a b ban n n n

n

n

= + = −2 2 and tan φ

( ) ( )F a F a jb F a jb nn n n n n n0 012

12 1 2 3= = − = + =−, , and for ⋅ ⋅ ⋅, ,

In what follows we will usually use the complex exponential form of the Fourier expansion.

3.2 Fourier Analysis The problem here is to find expressions for the Fourier coefficients given the form of the signal. We shall concentrate on the complex exponential form, since the others can all be calculated once these are known. The simplest way to derive expressions for the Fourier coefficients is to use the

orthogonality relations for complex exponentials with ω ππ

1 12 2= = f

T :

e e dtn m n

T n mjn t jm t

T−

+

=≠

=⎧⎨⎩

∫ ω ω

τ

τ1 1

0 if and are integers and if

m

By multiplying both sides of the complex exponential Fourier expansion by e and integrating over one cycle of the signal we get

jm t− ω1

f t e dt F e e dt TFT

jm tn

jn t jm tT

nm( )

τ

τω ω ω

τ

τ+− −

+

=−∞

∞

∫ ∫∑= =1 1 1

The expression for the complex exponential Fourier coefficients is thus

FT

f t e dtm

Tjm t=

+−∫

11( )

τ

τω

The choice of the value of τ in this formula is a matter of convenience; typically

. τ τ= = −0 2 or T /

3.3 Parseval's Relation and Power Spectra The equivalence of a signal and its Fourier series expansion is not quite straightforward, although for most signals it reduces to equality in the usual sense.

If f(t) is the signal and represents its partial Fourier expansion,

then the mean square error is defined by

~ ( )f t F eN njn t

n N

N

==−∑ ω1

[ ]∫+τ

τ

−=T

NN dttftfT

M2

)(~)(1

The equality of a signal and its Fourier expansion really only means that M NN → → ∞0 as .

This has two consequences : (a). The Fourier series expansion of a signal is equal to the original signal (in the usual sense) only where f(t) is continuous. At discontinuities the series is equal to the average of the values on either side of the discontinuity.

(b). There is a simple relationship between the power in a signal and its Fourier coefficients. This is really just the equation M∞ = 0. Using the definitions of M and of the Fourier series, together with the orthogonality of the complex exponentials, we get : 1 2 2

Tf t dt Fn

n

T

( ) ==−∞

∞+

∑∫τ

τ

This is Parseval's relation. Its physical interpretation is that the power in a signal is equal to the sum of the powers in each harmonic (it is not too difficult to show that Fn

2 is the power in the term ). This may seem obvious, but it is usually not true that the power in the sum of two or more signals is equal to the sum of the powers of the component signals.

F enjn tω1

The quantities S F S F F nn n n0 0

2 2 2 1 2 3= = + =− , ; , , ,⋅ ⋅ ⋅ are proportional to the power contained in the component with frequency ω ωn n= 1 , and thus give the distribution of power with frequency. The set of such quantities is the power spectrum of the signal. The information in many signals is more easily extracted from its power spectrum than from the original signal (speech signals are an example - the human ear appears to be insensitive to the phases of the harmonics). Note, however, that the power spectrum is not a complete description of the signal, information about the relative phases of the harmonics is lost in forming the power spectrum.

3.4 Example of Fourier Analysis of a Simple Periodic Signal The Rectangular Pulse Train:

f t P t kT Tk

( ) ( )= − <=−∞

∞

∑ τ τ ; 0 <2

T-T −τ τ t

f

1

Clearly this is periodic with period T.

So we can write

f t T P t T kT P t k T P t k T f tk k k

( ) ( ) ( ( ) ) ( ) (+ = + − = − − = − ′ ==−∞

∞

=−∞

∞

′=−∞

∞

∑ ∑ ∑τ τ τ1 )

[ ]

f t F e

FT

P t e dtT

e dtjn

e

njn T t

n

njn T t jn T t jn T

T

T

( )

( )

( / )

( / ) ( / ) ( / )

=

= = =−

=−∞

∞

−

−

−

−−

∑

∫ ∫

2

2 21 1 12

2

2

π

τπ π

τ

τπ

τ

τ

π

2

and hence

[ ]Fjn

e e n Tnn

jn T jn T=−

− =−12

22 2

ππτπ

π τ π τ( / ) ( / ) sin( / )

It is convenient to tidy this up a little by defining a function

sinc( ) sinx xx

= so that FT

n Tn =2 2τ

πτsinc( / )

- 2 0 - 1 0 0 1 0 2 0- 0 . 4

- 0 . 2

0

0 . 2

0 . 4

0 . 6

0 . 8

1

x

sinc

(x)

In this example the Fourier coefficients are all real, and since sinc( ) sinc( )− =x x , the coefficients are symmetric about n = 0; F Fn n− = . The rate at which the coefficients fall of with increasing order is not particularly fast, and a relatively large number are required to produce a good approximation to the rectangular pulse train.

The case of the square wave ( ) is of special interest.

τ = T / 4

F nnn m

mn m

n m= =

==

−+

= +

⎧

⎨⎪⎪

⎩⎪⎪

12

20

0 21

2 12 1

12

sinc( / )( )

( )

π

π

: :

:

Thus all the even harmonics (except the DC term (n = 0)) are zero.

-20 -10 0 10 20 -0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

freq

Four

ier C

oeffi

cien

t

The average signal power in the square wave is

PT

f t dtT

P t dtT

dtTT

T

T

T

= = = =− − −∫ ∫ ∫

1 1 1 22

2

2

22

2

2

( ) ( )/

/

/

/

ττ

τ τ=

1

The power in the DC term of the expansion is

S FT0 0

222 1

4= = ⎛

⎝⎜⎞⎠⎟

=τ

The power in the n-th harmonic (counting both positive and negative n) is

S Fn

n n= =⎧⎨⎪

⎩⎪2

02

2 2

: n even2 : n oddπ

-20 -10 0 10 20 0

0.05

0.1

0.15

0.2

0.25

freq

Pow

er

The fact that the square wave contains considerable high frequency power is shown by the fact that 2% of total power (and 4% of non-DC power) is contained in harmonics higher than the tenth. The relatively slow convergence for the square wave is shown in the progressive synthesis of the signal from its Fourier components. The peculiar "ringing" around the edges of the partially reconstructed pulses is also a feature ("Gibbs phenomenon") of the behaviour of Fourier series near discontinuities.

-10 -5 0 5 10-2

0

2

-10 -5 0 5 10-2

0

2

-10 -5 0 5 10-2

0

2

-10 -5 0 5 10-2

0

2

-10 -5 0 5 10-2

0

2

An intriguing purely mathematical result also follows from Parseval’s relation for the square wave:

12 1 82

0

2

( )nn +=

=

∞

∑ π

3.5 Properties of Fourier Coefficients 3.5.1 Convergence: If a periodic function satisfies some conditions, then the Fourier series converges to the function at any time t at which the function is continuous. A sufficient set of conditions is the Dirichlet conditions : i. The function can have at most a finite number of discontinuities in any cycle. ii. The function can have at most a finite number of maxima and minima in any cycle. iii. The function must be absolutely integrable over any cycle.

ie. f t dtT

( )τ

τ+

∫ < ∞

3.5.2 Linearity: If f(t) and g(t) are two functions with the same period, then the Fourier coefficients of the sum of the two functions is the sum of the Fourier coefficients of the individual signals. Slightly more generally, h t f t g t H F Gn n( ) ( ) ( )= + ⇒ = n+α β α β where α are arbitrary constants.

β and

3.5.3 Symmetries: Suppose f(t) is a real function of time then If f(t) is real and even ( )

F Fn n−∗=

f t f t( ) ( )= − then Im Fn = 0 If f(t) is real and odd ( )f t f t( ) ( )= − − then Re Fn = 0 If g(t) = f(-t) then G Fn n= −

3.5.4 Shift: If then g t f t( ) ( )= − τ G e Fn

jnn= − ω τ1

3.6 Problems 1. Show that any two distinct complex exponentials taken from the set

; ,..., , , , ,..., e nj n t T2 1 0 1 2π = −∞ − ∞ are orthogonal on the interval [0,T].

2. For the square wave shown below, calculate

(i). total average power in the signal (ii). power in the DC (zero frequency) term (iii). power in the first harmonic.

4 8 12-4-8-12 0

1

t

f

3. Find the complex Fourier series coefficients for the full wave rectified cosine

f t t( ) cos( / )= π 2 .

4. A idealised, simple signal detector consists of a full wave rectifier followed by

a filter that passes only the DC component of the rectified signal. For a cosine input signal f t t( ) cos( )= π 2

estimate the maximum power detected as a fraction of the total power in the original signal. (Hint : use your solution to question 3.)

4. Fourier analysis of continuous time signals II Fourier transforms

4.1 The Fourier Transform Fourier analysis via the Fourier series is limited to periodic signals. Most signals are not periodic and it is important to extend the idea of Fourier analysis to such signals. If the signals are finite energy this can be done in a straightforward manner using the Fourier transform. Consider a fixed pulse (f(t)) for which f t t( ) = >0 : τ . Suppose this pulse is repeated at intervals T to generate a periodic function.

T-T 0

t

f

τ−τ

∼

The fundamental frequency is ω π

12

=T

and the frequency of the n-th harmonic is

ωπ

ωn nT

n= =2

1 . The signal can be expanded as a Fourier series

~( ) ~

~ ~( ) ( )

f t F e

FT

f t e dtT

f t e dt

njn t

n

njn t j t

T

T

n

=

= =

=−∞

∞

−

−

−

−

∑

∫ ∫

ω

ω ω

τ

τ

1

1

2

21 1

-5 0 5 -0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

frequency (Hz)

Four

ier c

oeffi

cien

t

T = 2.0 s

-5 0 5

-0.2 -0.1

0 0.1 0.2 0.3 0.4 0.5

frequency (Hz)

Four

ier c

oeffi

cien

t

T = 1.0 s

-5 0 5

-0.2 -0.1

0 0.1

0.2 0.3

0.4 0.5

frequency

Four

ier

ff

T = 3.0

-5 0 5 -0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

frequency (Hz)

Four

ier c

oeffi

cien

t

T = 4.0 s

As T the spacing between the harmonics decreases and the number of harmonics in a given frequency interval increases. In the limit as T it is convenient to treat the frequency as a continuous variable.

→ ∞→ ∞

Note that the quantity

F TF f t eTn

nj t

n

( ) ~ ( )limω

ω ω

ω

τ

τ

= =→∞→∞

→

−

−∫ dt

is independent of T , and depends only on the form of the pulse f(t). Also we can let provided we understand that Tτ → ∞ → ∞ first.

The function

F f t e j t( ) ( )ω ω= −

−∞

∞

∫ dt

F

is the Fourier Transform of the signal f(t). It exists whenever the integral in the definition converges.

4.2 The Inverse Fourier Transform Writing the Fourier series in terms of F Tn n( ) ~

ω =~( ) ( )f t F e

Tnj t

n

n==−∞

∞

∑ ω ω 1

and noting that ∆ωn n n T= − =+ω ω

π1

2 , leads to

~( ) ( )f t F enj t

nn

n==−∞

∞

∑12π

ω ω ∆ω

In the limit as T the sum on the RHS becomes a (Reimannian) integral, and, for finite t, the function on the LHS is f(t). thus

→ ∞

f t F e dj t( ) ( )=−∞

∞

∫1

2πω ωω

4.3 Plancherel's Theorem and Parseval's Relation and Energy Spectra. The derivation of the Fourier transform and its inverse given above are not quite rigorous, but the results do hold, in much the same sense that the Fourier series formulae hold, provided suitable conditions are applied to the functions. If the function f(t) is absolutely integrable

f t dt( )−∞

∞

∫ < ∞

then, in general, the Fourier transform exists. Less generally, Plancherel's theorem assures us that if f(t) is finite energy then exists and f(t) is given in terms of

F( )ωF( )ω by the inverse transform. Also is

square integrable and we have F( )ω

f t dt F d( ) ( )2 212−∞

∞

−∞

∞

∫ ∫=π

ω ω

This last relation is Parseval's relation for the Fourier transform. Physically this means that the total energy in the signal is equal to the sum of the energies over all the frequencies. We can identify the quantity S F( ) ( )ω ω= 2 with as the "Energy spectral density" of the signal - playing much the same role as the Power spectrum for Fourier series. S d( )ω ω is proportional to the energy in the signal in the frequency range ( , )ω ω ω+ d . As for Fourier Series, the energy spectral density is often characteristic of the signal, but is not equivalent to it; information about the phase of the signal is lost in forming the spectrum.

ω

S(ω)

S(ω)dω

4.4 Examples of Common Transform Pairs. 4.4.1 The Rectangular Pulse: f t P tT( ) ( )=

F f t e dt P t ej tT

j t( ) ( ) ( )ω ω ω= =−

−∞

∞−

−∞

∞

∫ ∫ dt

so

F e dt ej

Tj t

T

T j t

T

T

( ) sin( )ω

ωω

ωω

ω

= =−

⎡

⎣⎢

⎤

⎦⎥ =−

−

−

−∫ 2

This is very similar to the Fourier coefficients of the rectangular pulse train, and it is also convenient to write this in terms of the sinc() function : P t T TT ( ) sinc( )↔ 2 ω The double headed arrow notation means that the function on the right is the Fourier transform of the function on the left.

-6 -4 -2 0 2 4 6 -0.5

0

0.5

1

1.5

2

frequency (Hz)

Four

ier T

rans

form

T = 1 s

4.4.2. The Triangular Pulse

Q ttT

t T

t TT ( ) =

−⎛⎝⎜

⎞⎠⎟ ≤

>

⎧

⎨⎪

⎩⎪

1

0

:

:

F Q t e dt t T e dt

T TT

t t dt

Tj t j t

T

T

T

( ) ( ) ( / )

sinc( ) cos( )

ω

ω ω

ω ω= = −

= −

−

−∞

∞−

−∫ ∫

∫

1

2 2

0

t

f

T-T

1

where we have used the fact that |t| is an even function of t. The second term can be

integrated by parts to yield : 2 2 2 1

02T

t t dt T T TT

T

cos( ) sinc( ) ( cos( )ω ω

ωω∫ = −

− )

Substituting, and using the half-angle formula for leads eventually to

cos( ) sin ( / )ω ωT T= −1 2 22

Q t T TT ( ) sinc ( / )↔ 2 2ω

-6 -4 -2 0 2 4 6 0

0.2

0.4

0.6

0.8

1

frequency (Hz)

Four

ier T

rans

form

T = 1 s

4.4.3. The Causal Exponential :

f t U t et

e tt T

t T( ) ( )= =<≥

⎧⎨⎩

−−

0 00

: :

[ ]F U t e e dt e dt

eT j

t T j t T j tT j t

( ) ( ) ( )( )

ωω

ω ωω

= = = −+

− −

−∞

∞− +

∞ − + ∞

∫ ∫ 1

0

10

1

which leads to U t e Tj T

t T( ) /− ↔+1 ω

-6 -4 -2 0 2 4 6 0

0.2

0.4

0.6

0.8

1

frequency (Hz)

Rea

l par

t of F

ourie

r Tra

nsfo

rm T = 1 s

-6 -4 -2 0 2 4 6 -0.6

-0.4

-0.2

0

0.2

0.4

0.6

frequency (Hz)

Imag

inar

y pa

rt of

Fou

rier T

rans

form

T = 1 s

4.4.4. The Gaussian :

( )f t e t T( ) = − 12

2

It can be shown that the result e dtt−

−∞

∞

=∫12

2

2π leads to an expression for the Fourier

transform of the Gaussian

( )e Tet T T− −↔12

2 12

2

2π ω( ) ie. the Fourier transform of a Gaussian is a Gaussisn.

4.5 Amplitude and Phase The Fourier transform is usually a complex function (ie has both real and imaginary parts) and it is frequently more useful to describe the transform in terms of its amplitude and phase. Thus : F A e j( ) ( ) ( )ω ω φ ω= where the amplitude (which is the related to the energy spectral density) is A F F F F F( ) ( ) ( ) ( ) (Re ) (Im )ω ω ω ω= = = +∗ 2 2 and the phase is given by

tan ( ) Im ( )Re ( )

φ ωωω

=FF

together with the usual conventions for determining which quadrant the angle is in. (Note that the phase is only defined up to an integer multiple of 2π and for display purposes it is often convenient to restrict φ to (say) the interval −π π, .) For example the Fourier transform of the causal exponential can be written in terms of reals and imaginary parts :

F Tj T

Tj T

j Tj T

TT

j TT

( )( ) ( )

ωω ω

ωω ω

ωω

=+

=+

−−

=+

−+1 1

11 1 12

2

2

Re ( )( )

Im ( )( )

F TT

F TT

ωω

ωω

ω=

+= −

+1 12

2

2 ;

or in terms of amplitude and phase

A F F Tj T

Tj T

TT

( )( )

ωω ω ω

= =−

⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟ =

+∗

1 1 1 2

tan ImRe

φ ω π φ= = −FF

T ; - < <12

12 π

(The restricted range for φ follows from the fact that Re ( )F ω > 0 for all finite frequencies.)

-6 -4 -2 0 2 4 6 0

0.2

0.4

0.6

0.8

1

frequency (Hz)

Am

plitu

de o

f Fou

rier T

rans

form

T = 1 s

-6 -4 -2 0 2 4 6 -100

-50

0

50

100

frequency (Hz)

Pha

se o

f Fou

rier T

rans

form

(deg

) T = 1 s

4.6 Properties of the Fourier Transform. The Fourier transform has a number of properties, similar in kind to those of the Fourier series, which are frequently used in practice. We shall use the double-headed arrow notation ( f t F( ) ( )↔ ω ) to denote Fourier transform pairs 4.6.1. Linearity: If and then

f t F g t Gf t g t F G( ) ( ) ( ) ( )( ) ( ) ( ) ( )

↔ ↔+ ↔ +

ω ωα β α ω β ω

where α and β are arbitrary constants Thus the Fourier transform of a sum is the sum of the Fourier transforms. One consequence of this is that, if the spectra of signals do not overlap, then we can transmit both signals simultaneously over the same medium and still be able to separate the signals at the receiver. This is the principle behind frequency division multiplexing used extensively in communications.

4.6.2. Symmetries: Most of the symmetry properties of the Fourier transform can be traced to the results If then and

f t Ff t Ff t F

( ) ( )( ) ( )( ) ( )

↔− ↔ −

↔ −∗ ∗

ωω

ω

If the signal is real then and we get the "Hermitian property" of the Fourier transform

f t f t∗ =( ) ( )

f t f t F F( ) ( ) ( ) ( )= ⇒ = −∗ ∗ ω ω

Thus, for real signals, the negative frequency parts of the transform can be obtained from the positive frequency components; they carry no extra information. If, in addition to being real the signal is also even, f t f t( ) ( )− = , then and the Fourier transform is real. By the Hermitian property it is then also even ,

F F( ) ( )ω ω= ∗

F F( ) (− =ω ω) If the signal is real and odd, f t f t( ) ( )− = − , then − = and the Fourier transform is pure imaginary. The Hermitian property then shows that the transform is also odd,

∗F F( ) ( )ω ω

)F F( ) (− = −ω ω . In general a signal can be written as a sum of an odd and an even signal,

( ) ( )f t f t f t f t f t f t f te o( ) ( ) ( ) ( ) ( ) ( ) ( )= + − + − − = +12

12

By linearity, F F Fe o( ) ( ) ( )ω ω= + ω And if the signal f(t) is real, the symmetry properties ensure that Fe( )ω is real and even and Fo ( )ω is imaginary and odd. Thus the real part of the Fourier transform of a signal is the Fourier transform of the even part of the signal, and the imaginary part of the Fourier transform is proportional to the Fourier transform of the odd part of the signal. f t F f t j Fe o( ) Re ( ) ( ) Im ( )↔ ↔ω ω ,

4.6.3. Shifting and Modulation The shifting property of the Fourier series generalizes to the Fourier transform. f t F f t T e Fj T( ) ( ) ( ) ( )↔ ⇒ − ↔ −ω ωω

-1 0 10

0.5

1

t

f(t)

-0.5 0 0.5 0

5

10

15

freq

Re(

F)

-0.5 0 0.5 -10

0

10

20

freq

Re(

F)

-1 0 10

0.5

1

t

f(t-T

)

Thus shifting a signal in time affects only the phase of the Fourier transform, in effect adding a term to the phase which is linear in the frequency. There is also a similar result which holds if the transform is shifted in frequency

f t F e f t Fj t( ) ( ) ( ) ( )↔ ⇒ ↔ −ω ωΩ Ω Thus to shift a signal to higher or lower frequencies one need only multiply the signal by a complex exponential of the appropriate frequency. In practice the multiplying function is a real harmonic signal, eg cos . But this can be written as a sum of two complex exponentials, and we get the "Modulation property" of the Fourier transform

( )Ωt

( )f t F t f t F F( ) ( ) cos( ) ( ) ( ) ( )↔ ⇒ ↔ − + +ω ω ωΩ Ω Ω1

2

-0.5 0 0.5 0

5

10

15

freq

Re(

F)

-0.5 0 0.5 0

2

4

6

8

freq

Re(

F)

-1 0 1-1

-0.5

0

0.5

1

t

f(t)c

os(a

t)

-1 0 10

0.5

1

t

f(t)

The most obvious application of this is in communications; amplitude modulation is one way of shifting the frequency ranges of different signals so that they do not overlap, allowing Frequency division multiplexing. The principle also finds application in many types of measurement involving narrow band high frequency signals, eg NMR spectroscopy. 4.6.4 Scaling A scaled version of a signal is just a signal with the same "shape" but evolving faster or slower than the original signal. Mathematically if f(t) is the original signal, then f(at) is a scaled version of the signal, speeded up (shorter duration) for |a| > 1, and slowed down for 0<|a|<1. The scaling property is

f t F f ata

F a( ) ( ) ( ) ( )↔ ⇒ ↔ω ω1

Thus shortening the duration of a pulse (a > 1) broadens its spectrum (1 a < 1) and lengthening a pulse narrows its spectrum. High frequencies are associated with rapid transitions in signals and low frequencies with slow changes

-0.5 0 0.5 0

5

10

15

freq

Re(

F)

-1 0 1 0

0.5

1

t

f(10t

)

-1 0 1 0

0.5

1

t

f(t)

-0.5 0 0.5 0

1

2

3

freq

Re(

F)

4.6.5. Duality. The form of the inverse transform equation is very similar to that of the transform, this similarity can be used to derive the Duality property of Fourier transform pairs. f t F F t f( ) ( ) ( ) ( )↔ ⇒ ↔ −ω π ω2

For example, we know that the Fourier transform of a rectangular pulse is a sinc function, duality implies that the Fourier transform of a sinc function is a rectangular pulse. More precisely P t T T T tT PT T( ) sinc( ) sinc( ) ( )↔ ⇒ ↔ −2 2 2ω π ω Or, on changing notation slightly, and using the fact that the rectangular pulse is even,

sinc( ) ( )ΩΩ Ωt P↔π

ω

-1 0 10

0.5

1

t

f(t)

-1 0 1-20

0

20

40

t

F(t)

-0.5 0 0.5 -20

0

20

40

freq

Re(

F)

-1 0 1 0

0.5

1

freq

f(fre

q)

(This result is used in Magnetic Resonance Imaging to ensure that only a thin slice of an object being examined is excited by the RF pulse.) 4.6.6. Transforms of Derivatives A result of great importance in the application of Fourier analysis to the study of the behaviour of linear systems, and of considerable use in extending tables of transform pairs.

f t F d f tdt

j Fn

nn( ) ( ) ( ) ( ) (↔ ⇒ ↔ω ω )ω

For example dedt

te te j et

t t−

− − −= − ⇒ ↔ −12

212

2 12

2 12

2

2ω π ω

A similar result holds for derivatives of transforms

f t F jt f t d Fd

nn

n( ) ( ) ( ) ( ) ( )↔ ⇒ − ↔ω

ωω

(Note that the conditions for the existence of the Fourier transform of f(t) do not guarantee the existance of the Fourier transform of its derivatives, so that the results above should be used with some care.)

4.6.7. The Convolution Theorem Perhaps the most important property of the Fourier transform as far as the description of filters and related systems is concerned. If and then

f t F g t Gf g t F G

( ) ( ) ( ) ( )( ) ( ) ( )↔ ↔

∗ ↔ω ω

ω ω

ie. the Fourier transform of a convolution is the product of the Fourier transforms. A similar result holds for the Fourier transform of a product of two signals:

f t g t F G F G d( ) ( ) ( ) ( ) ( )↔ ∗ = ′ − ′ ′−∞

∞

∫1

21

2πω

πω ω ω ω

For example, it is not too difficult to show that the convolution of a rectangular pulse with itself is a triangular pulse of twice the width. P P t TQ tT T T∗ =( ) ( )2 2 and thus

, which can be verified using the explicit expression for the Fourier transform of a triangular pulse. 2 2 2 42

2 2TQ t T T T T T TT ( ) sinc( ) sinc( ) sinc ( )↔ =ω ω ω

4.7 Problems 1. Show that the Fourier transform of the unit height triangular pulse of width 2T

shown below

-T T

1

t

QT(t)

is given by F T T( ) ( )ω ω sinc = 2 2 .

2. Find the Fourier transforms of the following signals

(i). f tt

t t( )

cos( )=

>≤

⎧⎨⎩

0 12 1

: : π

(ii). f tt

t e tt( )sin( )

=<≥

⎧⎨⎩

−

0 02 0

: : π

3. Find the amplitudes of the Fourier transforms calculated in question 2. In what

way does the spectrum of the decaying oscillation differ from that of a pure oscillation ?

4. Given that

P t T T Q t T T U t ejT T

t( ) ( ) ( ) ( ) ( )↔ ↔+

−2 2 11

2sinc and sinc and ω ωω

↔

Use the properties of the Fourier transform to find the Fourier transforms of the following functions : (i). P t TT ( − 2) (ii)

(iii) (iv)

U t e t( ) −4

sinc2( )t 11 2+ t

(Hint : consider

Re 11+

⎛⎝⎜

⎞⎠⎟

jω.)

5. The Fourier transforms of which of the following signals are real, which are imaginary and which are complex ? (Note : there is no need to do the transformations to determine this.)

(i). 11 2+ t

(ii). sin( (iii) )t e t− 2

cos( )t e t−

(iv).

f

t (v)

f

t1

-1

6. Starting with the definition of the Fourier transform, prove the scaling

theorem,

ie that f t F f ata

F a a( ) ( ) ( ) ( )↔ ⇒ ↔ ≠ω ω1 0 for

7. A signal has a Fourier transform of the form

F P P( ) ( ) ( )ω ω ω ω ω ω= − + + >>Ω Ω Ω0 0 0 where . This signal is multiplied by a signal of the form cos( )ω1t . Sketch the forms of the spectrum of the original signal and the modulated signal for the cases (i) and (ii) ω ω ω ω1 0 1 0< > .

5. Fourier Transforms of Impulsive and Periodic Signals

Strictly speaking periodic signals do not have Fourier transforms in the simple sense. If the integral used to define the transform is, however, interpreted in such a way as to allow results which are "generalized functions" , eg. delta functions, then it is possible to define the Fourier transform of periodic signals.

5.1 Fourier Transform of a Delta Function For convenience of notation we introduce a second shorthand notation for Fourier transforms:

)]([)( tfF ωω F= or more usually )]([)( tfF F=ω

Then ie. ∫∞

∞−

−− === 1)()]([ 0 ωωδδ jtj edtettF δ ( )t ↔ 1

Thus the Fourier transform of a delta function at the origin is a constant function of frequency.

t

f(t) = (t)δ (ω)

ω

F = 1

If the delta function is shifted away from the origin a similar calculation yields

∫∞

∞−

−− =−=− Tjtj edteTtTt ωωδδ )()]([F ie δ ω ( )t T e j T− ↔ −

so that, in general, the Fourier transform of a delta function is a complex exponential, in effect a function with constant amplitude and phase linearly proportional to frequency.

5.2 Fourier Transform of a Constant.. A constant function does not satisfy the conditions for the existence of the Fourier integral unless some limiting procedure is agreed upon to handle the improper integral. This is, in effect, what is done when we use the Duality property of the transform to calculate the transform of a constant function :

f t F F t ft

( ) ( ) ( ) ( )( ) ( )

↔ ⇒ ↔ −↔ ⇒ ↔ −

ω π ωδ π

21 1 2

δ ω

δ ω )

or, since the delta function is an even function, 1 2↔ πδ ω ( ) Physically, this can be interpreted to mean that all the energy in a constant (ie DC signal) is contained in the zero frequency component.

5.3 Fourier Transform of Harmonic Signals The Fourier transform of harmonic signals follows a similar argument: use duality on the transform of a shifted delta function. δ πω ( ) (t T e e Tj T jtT− ↔ ⇒ ↔ − −− − 2 or, on changing notation slightly e j tΩ Ω ↔ −2πδ ω( ) Thus all the energy in a single (complex exponential) harmonic term is concentrated at one frequency. The general harmonic function can be transformed by using the usual relations between sines and cosines and complex exponentials. The result is

( )cos( ) ( ) ( )Ω Ω t e ej j+ ↔ − + +−φ π δ ω δ ωφ φ Ω And by taking φ φ π= =0 2 and − we find, respectively

( )cos( ) ( ) ( )Ω Ω Ω t ↔ − + +π δ ω δ ω ( )sin( ) ( ) ( )Ω Ω Ω t j↔ − − − +π δ ω δ ω

-0.5 0 0.5

-1

0

1

t

cos(

ft)

-0.5 0 0.5

-1

0

1

t

sin(

ft)

-50 0 50

-200

0

200

freq

Re(

F[si

n(ft)

])

-50 0 50 0

100

200

300

freq

Re(

F[co

s(ft)

])

Note that the transform of cos( ) is real and even and the transform of sin( ) is pure imaginary and odd.

5.4 Fourier Transforms of Periodic Signals. The results above can be used to find expressions, in terms of delta functions, for the Fourier transforms of general periodic functions. Not surprisingly they resemble the Fourier series. In fact it is easiest to start with a Fourier series expansion and then transform this.

f t F e F nTn

j n T t

nn

n

( ) ( / )= ↔ −⎛⎝⎜

⎞⎠⎟=−∞

∞

=−∞

∞

∑ ∑2 2 2π π δ ωπ

The Fourier coefficients are given by the usual expression, but is convenient to define a new function, which is just one cycle of the periodic function:

f tf t T t T

t Tr ( )( )

=− ≤ << − ≥

⎧⎨⎩

: : or t

12

12

12

120 T

Then FT

f t e dtT

F n Tnj n T

T

T

r= =−

−∫

1 1 22

12

12

( ) ( )( / )π π , where f t Fr r( ) ( )↔ ω , and we can

write

f tT

F n T nTr

n

( ) ( )↔ −⎛⎝⎜

⎞⎠⎟=−∞

∞

∑2 2 2ππ δ ω

π

The multiplication property of the Fourier transform allows us to write

F n T nT

F nTr r( ) ( )2 2 2

π δ ωπ

ω δ ωπ

−⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟

,

so that, finally,

f t FT

nTr

n

( ) ( )↔ −⎛⎝⎜

⎞⎠⎟=−∞

∞

∑ωπ

δ ωπ2 2

Thus the Fourier transform of a periodic signal is a sequence of delta functions located at the harmonics of the fundamental frequency modulated by , the Fourier transform of the "restricted function" .

Fr ( )ω

-100 0 100 -100

-50

0

50

100

150

200

250

300

freq

Re(

F)

-0.5 0 0.5 0

0.5

1

1.5

t

f(t)

5.5 The Dirac Comb and its Fourier Transform The result for the Fourier transform of a periodic signal can be concisely written in terms of a special function, the "Dirac Comb", which is a net of uniformly spaced delta functions of uniform impulse.

S x x nXXn

( ) ( )= −=−∞

∞

∑δ

This functions is the idealization of an infinite sequence of uniform short pulses. In terms of this we can write

)(2)()( 2 ωπ

ω↔ π Tr ST

Ftf

The Fourier transform of a Dirac comb can be found by an instructive trick. Using the convolution property of delta functions, we can write, for a periodic function f(t),

f t f t nT f t t nT f t t nT f t S trn

rn

rn

r T( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )= − = ∗ − = ∗ − = ∗=−∞

∞

=−∞

∞

=−∞

∞

∑ ∑ ∑δ δ

The Convolution theorem can then be used to transform this

)]([)()( tSFtf Tr Fω↔ Comparing this with the other result for the Fourier transform of a periodic signal, we see that

S tT

ST T( ) ( )↔2

2π

ωπ

Thus the Fourier transform of a Dirac comb is a Dirac comb. The spacing between the delta functions in the transform is proportional to the reciprocal of the spacing in the time domain. This result is physically familiar from the theory of the diffraction grating, the spacing of lines in the spectrum is inversely proportional to the spacing of the lines in the grating. In fact the pattern produced at large distances when coherent light is passed through an aperture of any shape (the Fraunhofer diffraction pattern) is proportional to the Fourier transform of the aperture. This result has actually been used in the past to evaluate the Fourier transforms.

-100 0 100 0

20

40

freq

Re(

F)

-0.5 0 0.50

0.5

1

1.5

t

S(t;

2T)

-100 0 100 0

20

40

freq

Re(

F)

-0.5 0 0.50

0.5

1

1.5

t

S(t;

T)

5.6 Problems

1. Show that the Gaussian ( )f tT

e t T( ) = −12

12

2

π

behaves like a delta function as T → 0. 2. Find the Fourier transform of the following periodic signals

(i). (ii). (iii). A train of unit height triangular pulses of half width 1 and period 2.

f t t t( ) sin cos= + f t t( ) cos= 2

3. A finite duration segment of a cosine signal can be written

where 2T is the duration of the segment. Find the Fourier transform of this signal. In what way does it differ from the Fourier transform of a true cosine signal ?

f t P t tT( ) ( ) cos=