introduction to signals and systems lecture #6 - frequency...

Introduction to Signals and Systems Lecture #6 - Frequency-Domain Representation of Signals

Guillaume Drion Academic year 2017-2018

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The complex exponential

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Transmission of complex exponentials through LTI systems

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where is the transfer function of the LTI system.

LTI system

Continuous case:


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Continuous case:

x(t) = a1es1t + a2e

s2t + a3es3t

y(t) = a1H(s1)es1t + a2H(s2)e

s2t + a3H(s3)es3t

would totally define the system input-output properties if

x(t) =X

k

akeskt

H(s)


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where is the transfer function of the LTI system.

Discrete case:

LTI system


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Discrete case:

would totally define the system input-output properties if

x[n] = a1zn1 + a2z

n2 + a3z

n3

y[n] = a1H(z1)zn1 + a2H(z2)z

n2 + a3H(z3)z

n3

H(z)

x[n] =X

k

akznk

Outline

Frequency-domain representation of periodic signals: Fourier series.

Frequency-domain representation of aperiodic signals: the Fourier transform

Convergence of the Fourier transform

Properties of the Fourier transform

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Continuous-time periodic signals

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x(t) = e

j!0t = e

j 2⇡T t

The complex exponential of period has a set of harmonically related complex exponentials: that are all periodic with period .

T

�k(t) = ejk!0t = ejk2⇡T t, k = 0,±1,±2, . . .

T

So the signalis also periodic with period .

x(t) =k=1X

k=�1ake

jk!0t =k=1X

k=�1ake

jk 2⇡T t

T

Fourier series of a periodic continuous-time signal

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x(t) =k=1X

k=�1ake

jk!0t =k=1X

k=�1ake

jk 2⇡T t

ak =1

T

Z

Tx(t)e�jk!0t

dt

=1

T

Z

Tx(t)e�jk 2⇡

T tdt

Fourier series coefficients

Example: is the average value over one period.a0 =1

T

Z

Tx(t)dt

Two classes of conditions that a periodic signal can satisfy to guarantee that it can be represented by a Fourier series

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(i) Signals having finite energy over a period, i.e. are representable through the Fourier series.

Z

T|x(t)|2dt < 1.

When this condition is satisfied, we are guaranteed that the coefficients are finite. Furthermore, we are guaranteed that the energy in the approximation error of converges to 0 as :

ENxN (t) N ! 1

e(t) = x(t)�1X

k=�1ake

jk!0t !Z

T|e(t)|2dt = 0.


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(ii) The Dirichlet conditions to ensure that a signal equals its Fourier series representation, except at isolated values of time for which the signal is discontinuous: 1. Over a period, must be absolutely integrable, that is, 2. is of bounded variations, i.e. no more than a finite number of maxima and minima during a single period. 3. There are only a finite number of discontinuities, and each of these discontinuities is finite.

x(t)Z

T|x(t)|dt < 1

x(t)


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(ii) The Dirichlet conditions ensures that

8t 2 [O, T ) : limN!1

xN (t) = x(t)

1X

k=�1ake

jk!0t = x(t) for all where is continuous.t x(t)

How does the Fourier series converges for a periodic signal with discontinuities?

Discrete-time periodic signals

A signal is periodic with a period if .

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x[n] = x[n+N ] 8nN

with .!0 =2⇡

N

�k[n] = ejk!0n = ejk2⇡N n, k = 0,±1,±2, . . .

The complex exponential of period has a set of harmonically related complex exponentials:

N

In discrete time, there are only N dinstinct harmonic components:

�k[n] = �k+rN [n]

Fourier series of a periodic discrete-time signal

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Fourier series coefficients

Finite serie!

x[n] =X

k=<N>

akejk!0n =

X

k=<N>

akejk 2⇡

N n

ak =1

N

X

n=<N>

x[n]e�jk!0n

=1

N

X

n=<N>

x[n]e�jk 2⇡N n

Outline





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We start by deriving the Fourier series representation of the continuous-time periodic square wave defined by

Can we derive a frequency representation of aperiodic signals?

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x(t) =

⇢1, |t| < T1

0, T1 < |t| < T/2

over one period and periodically repeating with period .T

We start by deriving the Fourier series representation of the continuous-time periodic square wave of period .


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T

Fourier series coefficients: .ak =2 sin(k!0T1)

k!0T

The Fourier series coefficients can be rewritten as:

Tak =2 sin(!T1)

!

��!=k!0

where the function represents the envelope of , and the coefficients are simply equally spaced samples of this envelope. This envelope is independent of .

(2 sin!T1)/! Tak

T


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Tak =2 sin(!T1)

!

��!=k!0

T

T = 4T1

T = 8T1

T = 16T1


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Tak =2 sin(!T1)

!

��!=k!0

As the period increases, the envelope is sampled with closer and closer spacing. As :

The periodic square-wave approaches a rectangular pulse (aperiodic).

The set of Fourier series coefficients approaches the envelope function.

TT ! 1

We think of an aperiodic signal (a) as the limit of the periodic signal (b) as the period becomes arbitrarily large, and we examine the limiting behaviour of the Fourier series representation for this signal.

Frequency representation of an aperiodic signal

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On the interval , we have:


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�T/2 t T/2

x̃(t) =+1X

k=�1ake

jk!0t : periodic signalx̃(t)

: aperiodic signalx(t)

ak =1

T

Z T/2

�T/2x̃(t)e�jk!0t

dt

=1

T

Z T/2

�T/2x(t)e�jk!0t

dt

=1

T

Z +1

�1x(t)e�jk!0t

dt

The envelope of is defined as


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X(j!) Tak

X(j!) =

Z +1

�1x(t)e�j!t

dt

with ak =1

TX(jk!0)

This gives the Fourier series representation of the periodic signal

x̃(t) =+1X

k=�1

1

T

X(jk!0)ejk!0t

=+1X

k=�1

1

2⇡X(jk!0)e

jk!0t!0

As :

the periodic signal approaches the aperiodic signal: .

the fundamental frequency approaches 0: .


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T ! 1x̃(t) ! x(t)

!0 = 2⇡/T ! 0

x̃(t) =+1X

k=�1

1

2⇡X(jk!0)e

jk!0t!0

x(t) =1

2⇡

Z +1

�1X(j!)ej!t

d!

The continuous-time Fourier transform

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x(t) =1

2⇡

Z +1

�1X(j!)ej!t

d!

X(j!) =

Z +1

�1x(t)e�j!t

dt

Inverse Fourier transform

Fourier transform

is called the frequency spectrum of .

It tells you how to describe as a linear combination of sinusoïdal signals at different frequencies (i.e. what frequencies are “present” in the signal).

X(j!) x(t)

x(t)

The continuous-time Fourier transform: example

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Fourier series of periodic signals vs Fourier transform of aperiodic signals

To represent a periodic signals, we use a set of complex exponentials that are harmonically related:

To represent an aperiodic signals, we use a set of complex exponentials that are infinitesimally close in frequency:

x(t) =1

2⇡

Z +1

�1X(j!)ej!t

d!

x̃(t) =+1X

k=�1

1

2⇡X(jk!0)e

jk!0t!0

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Fourier transform of periodic signals

We can construct the Fourier transform of a periodic signal directly from its Fourier series representation.

The resulting transform is a train of impulses in the frequency domain, with the areas of the impulses proportional to the Fourier series coefficients.

Indeed, the Fourier transform corresponds to the signal

X(j!) = 2⇡�(! � !0)

x(t) =1

2⇡

Z 1

�12⇡�(! � !0)e

j!td!

= e

j!0t

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Fourier transform of periodic signals

We can construct the Fourier transform of a periodic signal directly from its Fourier series representation.

The resulting transform is a train of impulses in the frequency domain, with the areas of the impulses proportional to the Fourier series coefficients.

If we now take a linear combination of impulses equally spaced in frequency it gives the signal

x(t) =+1X

k=�1ake

jk!0t

X(j!) =+1X

k=�12⇡ak�(! � k!0)

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Fourier transform of periodic signals: example

Fourier transform of a periodic square wave.

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Fourier transform of Dirac delta function :

Examples of Fourier transforms

x(t) = �(t)

X(j!) =

Z +1

�1�(t)ej!tdt = 1

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x(t) = cos!0t

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x(t) = sin!0t

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Outline





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Convergence of Fourier transforms

The Fourier transform remain valid for an extremely broad class of signals of infinite duration. The convergence criteria are similar to the Fourier series:

There is no energy in the difference between a signal and the reconstruction of a signal from its Fourier representation if the signal has finite energy:

Z +1

�1|x(t)|2dt < +1

The Dirichlet conditions ensure that a signal equals its Fourier representation, except at isolated values of time for which the signal is discontinuous (see Fourier series).

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Outline





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Properties of the Fourier transform: linearity

x(t)F ! X(j!)

y(t)F ! Y (j!)

ax(t) + by(t)F ! aX(j!) + bY (j!)

If

and

then

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Properties of the Fourier transform: time shifting

x(t)F ! X(j!)

If

then

x(t� t0)F ! e

�j!t0X(j!)

It is very useful when taking time delays into account in systems.

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Properties of the Fourier transform: differentiation and integration

dx(t)

dt

F ! j!X(j!)

Z t

�1x(⌧)d⌧

F ! 1

j!

X(j!) + ⇡X(0)�(!)

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Properties of the Fourier transform: time and frequency scaling

x(t)F ! X(j!)

If

then

Example: playing an audio signal faster sounds higher in frequency.

x(at)F ! 1

|a|X(j!

a

)

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Properties of the Fourier transform: time and frequency scaling

x(t)F ! X(j!)If then x(at)

F ! 1

|a|X(j!

a

)

A signal that is localized in time is not localized in frequency, and conversely!

Time domain Frequency domain

Example: incertitude principle in physics.

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Properties of the Fourier transform: the convolution property

y(t) = x(t) ⇤ h(t) F ! Y (j!) = X(j!)H(j!)

If

then

x(t)F ! X(j!)

y(t)F ! Y (j!)

h(t)F ! H(j!)

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The discrete-time Fourier transform

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Synthesis equation

Analysis equation

x[n] =1

2⇡

Z

2⇡X(ej!)ej!n

d!

X(ej!) =+1X

n=�1x[n]e�j!n

The discrete-time Fourier transform: example

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where is the Fourier transform of the impulse response of the system.

LTI system

u(t) = ej!t

y(t) = h(t) ⇤ u(t)

=

Z +1

�1h(⌧)ej!(t�⌧)d⌧

= ej!t

Z +1

�1h(⌧)e�j!⌧d⌧

= ej!tH(j!)

H(j!) =

Z +1

�1h(⌧)e�j!⌧d⌧

u(t) = ej!t

y(t) = ej!tH(j!)


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where is the discrete-time Fourier transform of the impulse response of the system.

LTI system

u[n] = ej!n

y[n] = h[n] ⇤ u[n]

=X

k2Zh[k]ej!(n�k)

= ej!nX

k2Zh[k]e�j!k

= ej!nH(ej!)

H(ej!) =X

k2Zh[k]e�j!k

u[n] = ej!n

y[n] = ej!nH(ej!)

introduction to signals and systems lecture #6 - frequency...

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