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Lecture IX: Fourier transform

Maxim Raginsky

BME 171: Signals and Systems

Duke University

October 8, 2008

Maxim RaginskyLecture IX: Fourier transform

This lecture

Plan for the lecture:

1Recap: Fourier series representation of periodic signals

2Frequency content of aperiodic signals: the Fourier transform

3The inverse Fourier transform

4Properties of the Fourier transform

5Generalized Fourier transform

6Bandlimited and timelimited signals

7Frequency response of LTI systems

Maxim RaginskyLecture IX: Fourier transform

Recap: Fourier series

Recall from the last lecture that any sufficiently regularT-periodic continuous-time signalx(t)can be expanded, e.g., in a complex exponential Fourier series: x(t) =∞? k=-∞c kejkω0t, whereω0= 2π/Tis the fundamental frequency, and the Fourier coefficients{ck}are given by c k=1 T? T/2 -T/2x(t)e-jkω0tdt, k=...,-2,-1,0,1,2,... The Fourier coefficients{ck}tell us about thefrequency content(or spectral content) ofx(t).

Maxim RaginskyLecture IX: Fourier transform

Spectral content of aperiodic signals: the Fourier transform

What aboutaperiodicsignals?

Any continuous-time signalx(t)that has finite "energy", i.e., x2(t)dt <+∞, can be represented in the frequency domain via theFourier transform:

X(ω) =?

x(t)e-jωtdt

We will also write

X(ω) =F[x(t)]

to denote the fact thatX(ω)is the Fourier transform ofx(t).

Maxim RaginskyLecture IX: Fourier transform

Example: rectangular pulse

Consider the rectangular pulse

p

0,|t|> τ/2

F[pτ(t)] =?

p

τ(t)e-jωtdt

τ/2

-τ/2e-jωtdt -1 jωe-jωt?

τ/2

-τ/2 ejωτ/2-e-jωτ/2 jω

2sin(ωτ/2)

=τsinc?τω

2π?

Maxim RaginskyLecture IX: Fourier transform

Inverse Fourier transform

The signalx(t)can be recovered from its Fourier transform X(ω) =F[x(t)]using theinverse Fourier transformformula x(t) =F-1[X(ω)] =12π?

X(ω)ejωtdω

Note: There is a factor of1/2πin front of the integral. The integration is with respect toω, for a fixed value oft.

We will also write

x(t)↔X(ω) and say thatx(t)[time domain] andX(ω)[freq. domain] are aFourier transform pair.

Maxim RaginskyLecture IX: Fourier transform

Proof:

1

2π?

X(ω)ejωtdω=12π?

x(t?)e-jωt?dt?? e jωtdω x(t?)?1

2π?

ej(t-t?)ωdω? dt 1

2π?

-πΩej(t-t?)ωdω = lim

Ω→∞1

2πj(t-t?)Ω?

ej(t-t?)ω?πΩ = lim

Ω→∞sin(πΩ(t-t?))

πΩ(t-t?)

= lim

Ω→∞sinc(Ω(t-t?))

=δ(t-t?)

Hence,

1

2π?

X(ω)ejωtdω=?

x(t?)δ(t-t?)dt?=x(t)

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Properties of the Fourier transform

The Fourier transform has many useful properties that make calculations easier and also help thinking about the structure of signalsand the action of systems on signals. The properties are listed in any textbook on signals and systems. We will look at and prove a few of them.

Maxim RaginskyLecture IX: Fourier transform

Linearity

The Fourier transform islinear: if

x

1(t)↔X1(ω) andx2(t)↔X2(ω),

then c

1x1(t) +c2x2(t)↔c1X1(ω) +c2X2(ω)

for any two numbersc1andc2.

Proof:obvious -

F[c1x1(t) +c2x2(t)] =?

[c1x1(t) +c2x2(t)]e-jωtdt =c1? x

1(t)e-jωtdt+c2?

x

2(t)e-jωtdt

=c1X1(ω) +c2X2(ω)

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Time shift

Ifx(t)↔X(ω), thenx(t-c)↔X(ω)e-jωcfor any constantc.

Proof:

F[x(t-c)] =?

x(t-c)e-jωtdt x(t)e-jω(t+c)dt =e-jωc?∞ x(t)e-jωtdt =X(ω)e-jωc.

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Multiplication by a complex exponential

Ifx(t)↔X(ω), thenx(t)ejω0t↔X(ω-ω0)for anyrealω0.

Proof:

F ?x(t)ejω0t?=? x(t)ejω0te-jωtdt x(t)e-j(ω-ω0)tdt =X(ω-ω0).

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Multiplication by a cosine

Ifx(t)↔X(ω), thenx(t)cos(ω0t)↔12[X(ω+ω0) +X(ω-ω0)].

Proof:use linearity and the last property to get

F[x(t)cos(ω0t)] =F?1

2x(t)?ejω0t+e-jω0t??

1

2F?x(t)ejω0t?+12F?x(t)e-jω0t?

1

2[X(ω-ω0) +X(ω+ω0)].

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Convolution in time domain

Ifx(t)↔X(ω)andv(t)↔V(ω), then

x(t)? v(t)↔X(ω)V(ω)

Proof:

F[x(t)? v(t)] =?

[x(t)? v(t)]e-jωtdt x(λ)v(t-λ)dλ? e -jωtdt x(λ)? v(t-λ)e-jωtdt?

F[v(t-λ)]dλ

x(λ)V(ω)e-jωλdλ =V(ω)? x(λ)e-jωλdλ =X(ω)V(ω).

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Parseval"s theorem

Letx(t)andv(t)be real-valued signals. Then?∞

x(t)v(t)dt=1

2π?

-∞X(ω)V(ω)dω

Proof:

x(t)v(t)dt=? x(t)?1

2π?

V(ω)ejωtdω)?

dt 1

2π?

V(ω)?

x(t)ejωtdt? dω 1

2π?

V(ω)X(-ω)dω

1

2π?

-∞X(ω)V(ω)dω, where we used the fact that, sincex(t)is real,

X(ω) =?

x(t)ejωtdt=X(-ω).

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Parseval"s theorem: cont"d

An important consequence of Parseval"s theorem is that x2(t)dt=1

2π?

|X(ω)|2dω. In other words, signal energy can be computed both in time domain and in frequency domain (up to a factor of1/2π).

Maxim RaginskyLecture IX: Fourier transform

Duality

Ifx(t)↔X(ω), thenX(t)↔2πx(-ω).

Proof:

F[X(t)] =?

X(t)e-jωtdt

= 2π·1

2π?

X(t)e-jωtdt

= 2π·1

2π?

X(ω?)e-jωω?dω?

=F-1[X(ω)](-ω) = 2π·1

2πx(-ω)

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Duality: an example

Let x(t) =τsinc?τt

2π?

Then by duality we have

X(ω) = 2πpτ(ω).

In more detail:

p

τ(t)↔τsinc?τω

2π?

Thus, by duality,

τsinc?τt

2π?

↔2πτ(ω).

Maxim RaginskyLecture IX: Fourier transform

Generalized Fourier transform

The Fourier transform is defined only for signals with finite energy. However, we can extend its scope by allowing singularity functions. We begin by computing the Fourier transform of the unit impulseδ(t).

F[δ(t)] =?

δ(t)e-jωtdt

δ(t)dt

= 1, where we used the sifting property of the unit impulse.

By duality, we have

1↔2πδ(-ω) = 2πδ(ω).

Maxim RaginskyLecture IX: Fourier transform

Fourier transform of the cosine

The cosine signalx(t) = cos(ω0t)does not have the Fourier transform in the ordinary sense. It does, however, have a generalized Fourier transform:

F[cos(ω0t)] =F?1

2(ejω0t+e-jω0t)?

1

2F?1·ejω0t?+12F?1·e-jω0t?

1

2[2πδ(ω-ω0) + 2πδ(ω+ω0)]

=πδ(ω-ω0) +πδ(ω+ω0).

Q.E.D.

Maxim RaginskyLecture IX: Fourier transform

Fourier transform of a periodic signal

Using the generalized Fourier transform, we can analyze periodic signals that do not have a Fourier transform in the ordinary sense. Thus, ifx(t) is aT-periodic signal, we can expand it in a complex exponential Fourier series as x(t) =∞? k=-∞c kejkω0t.

X(ω) =F?

k=-∞c kejkω0t? k=-∞c kF?ejkω0t? k=-∞2πckδ(ω-kω0). Thus, the (generalized) Fourier transform of a periodic signal is a train of impulses located at integer multiples of the fundamental frequencyω0.

Maxim RaginskyLecture IX: Fourier transform

Bandlimited and timelimited signals

A signalx(t)is called:

bandlimitedif there exists a numberB >0(called the bandwidth), such that

X(ω) = 0,for all|ω| ≥B.

timelimitedif there exists a numberT >0, such that x(t) = 0,for all|t| ≥T. It can be proved that a bandlimited signal cannot be timelimited, and vice versa. We"ve seen an example of this with the transform pairs p

τ(t)↔τsinc?τω

2π?

andτsinc?τt2π? ↔2πpτ(ω) However, a signal can be approximately timelimited and bandlimited - that is, there exist numbersB >0andT >0, such that|x(t)|is small for|t| ≥Tand|X(ω)|is small for|ω| ≥B.

Maxim RaginskyLecture IX: Fourier transform

Frequency response of LTI systems

Consider an LTI system with the impulse responseh(t). Then the output of the system due to inputx(t)is given by the convolution integral, y(t) =x(t)? h(t) =? x(λ)h(t-λ)dλ. In frequency domain, the action of the system can be described as follows:

Y(ω) =H(ω)X(ω).

This is a consequence of the fact that convolution in time domain corresponds to multiplication in frequency domain. The Fourier transformH(ω)of the impulse responseh(t)is called the frequency responseof the system.

Maxim RaginskyLecture IX: Fourier transform

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