Lecture IX: Fourier transform - University Of...

Lecture IX: Fourier transform

Maxim Raginsky

BME 171: Signals and Systems

Duke University

October 8, 2008

Maxim Raginsky Lecture IX: Fourier transform

This lecture

Plan for the lecture:

1 Recap: Fourier series representation of periodic signals

2 Frequency content of aperiodic signals: the Fourier transform

3 The inverse Fourier transform

4 Properties of the Fourier transform

5 Generalized Fourier transform

6 Bandlimited and timelimited signals

7 Frequency response of LTI systems


Recap: Fourier series

Recall from the last lecture that any sufficiently regular T -periodiccontinuous-time signal x(t) can be expanded, e.g., in a complexexponential Fourier series:

x(t) =∞∑

k=−∞

ckejkω0t,

where ω0 = 2π/T is the fundamental frequency, and the Fouriercoefficients ck are given by

ck =1

T

∫ T/2

−T/2

x(t)e−jkω0tdt, k = . . . ,−2,−1, 0, 1, 2, . . .

The Fourier coefficients ck tell us about the frequency content (or

spectral content) of x(t).


Spectral content of aperiodic signals: the Fourier transform

What about aperiodic signals?

Any continuous-time signal x(t) that has finite “energy”, i.e.,

∫ ∞

−∞

x2(t)dt < +∞,

can be represented in the frequency domain via the Fourier transform:

X(ω) =

∫∞

−∞

x(t)e−jωtdt

We will also writeX(ω) = F [x(t)]

to denote the fact that X(ω) is the Fourier transform of x(t).


Example: rectangular pulse

Consider the rectangular pulse

pτ (t) =

1, |t| ≤ τ/20, |t| > τ/2

F [pτ (t)] =

∫ ∞

−∞

pτ (t)e−jωtdt

=

∫ τ/2

−τ/2

e−jωtdt

=

[

−1

jωe−jωt

]τ/2

−τ/2

=ejωτ/2 − e−jωτ/2

jω

=2 sin(ωτ/2)

ω

= τsinc(τω

2π

)

.


Inverse Fourier transform

The signal x(t) can be recovered from its Fourier transformX(ω) = F [x(t)] using the inverse Fourier transform formula

x(t) = F−1[X(ω)] =1

2π

∫ ∞

−∞

X(ω)ejωtdω

Note:

There is a factor of 1/2π in front of the integral.

The integration is with respect to ω, for a fixed value of t.

We will also writex(t) ↔ X(ω)

and say that x(t) [time domain] and X(ω) [freq. domain] are a Fourier

transform pair.


Proof:

1

2π

∫∞

−∞

X(ω)ejωtdω =1

2π

∫∞

−∞

(∫∞

−∞

x(t′)e−jωt′dt′)

ejωtdω

=

∫∞

−∞

x(t′)

(1

2π

∫∞

−∞

ej(t−t′)ωdω

)

dt′.

1

2π

∫∞

−∞

ej(t−t′)ωdω =1

2πlim

Ω→∞

∫ πΩ

−πΩ

ej(t−t′)ωdω

= limΩ→∞

1

2πj(t − t′)Ω

[

ej(t−t′)ω]πΩ

−πΩ

= limΩ→∞

sin(πΩ(t − t′))

πΩ(t − t′)

= limΩ→∞

sinc(Ω(t − t′))

= δ(t − t′)

Hence,

1

2π

∫ ∞

−∞

X(ω)ejωtdω =

∫ ∞

−∞

x(t′)δ(t − t′)dt′ = x(t)

Q.E.D.


Properties of the Fourier transform

The Fourier transform has many useful properties that make calculationseasier and also help thinking about the structure of signals and the actionof systems on signals.

The properties are listed in any textbook on signals and systems. We willlook at and prove a few of them.


Linearity

The Fourier transform is linear: if

x1(t) ↔ X1(ω) and x2(t) ↔ X2(ω),

thenc1x1(t) + c2x2(t) ↔ c1X1(ω) + c2X2(ω)

for any two numbers c1 and c2.

Proof: obvious –

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

∫ ∞

−∞

[c1x1(t) + c2x2(t)] e−jωtdt

= c1

∫∞

−∞

x1(t)e−jωtdt + c2

∫∞

−∞

x2(t)e−jωtdt

= c1X1(ω) + c2X2(ω)

Q.E.D.


Time shift

If x(t) ↔ X(ω), then x(t − c) ↔ X(ω)e−jωc for any constant c.

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.


Multiplication by a complex exponential

If x(t) ↔ X(ω), then x(t)ejω0t ↔ X(ω − ω0) for any real ω0.

Proof:

F[x(t)ejω0t

]=

∫∞

−∞

x(t)ejω0te−jωtdt

=

∫ ∞

−∞

x(t)e−j(ω−ω0)tdt

= X(ω − ω0).

Q.E.D.


Multiplication by a cosine

If x(t) ↔ X(ω), then x(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

]+

1

2F

[x(t)e−jω0t

]

=1

2[X(ω − ω0) + X(ω + ω0)] .

Q.E.D.


Convolution in time domain

If x(t) ↔ X(ω) and v(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 Raginsky Lecture IX: Fourier transform

Parseval’s theorem

Let x(t) and v(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, since x(t) is real,

X(ω) =

∫ ∞

−∞

x(t)ejωtdt = X(−ω).

Q.E.D.


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 of 1/2π).


Duality

If x(t) ↔ X(ω), then X(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.


Duality: an example

Let

x(t) = τ sinc

(τt

2π

)

.

Then by duality we have

X(ω) = 2πpτ (ω).

In more detail:pτ (t) ↔ τ sinc

(τω

2π

)

Thus, by duality,

τ sinc

(τt

2π

)

↔ 2πτ (ω).


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 impulse δ(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 have1 ↔ 2πδ(−ω) = 2πδ(ω).


Fourier transform of the cosine

The cosine signal x(t) = cos(ω0t) does not have the Fourier transform inthe ordinary sense. It does, however, have a generalized Fouriertransform:

F [cos(ω0t)] = F

[1

2(ejω0t + e−jω0t)

]

=1

2F

[1 · ejω0t

]+

1

2F

[1 · e−jω0t

]

=1

2[2πδ(ω − ω0) + 2πδ(ω + ω0)]

= πδ(ω − ω0) + πδ(ω + ω0).

Q.E.D.


Fourier transform of a periodic signal

Using the generalized Fourier transform, we can analyze periodic signalsthat do not have a Fourier transform in the ordinary sense. Thus, if x(t)is a T -periodic signal, we can expand it in a complex exponential Fourierseries as

x(t) =

∞∑

k=−∞

ckejkω0t.

X(ω) = F

[∞∑

k=−∞

ckejkω0t

]

=

∞∑

k=−∞

ckF[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 frequency ω0.


Bandlimited and timelimited signals

A signal x(t) is called:

bandlimited if there exists a number B > 0 (called thebandwidth), such that

X(ω) = 0, for all |ω| ≥ B.

timelimited if there exists a number T > 0, such that

x(t) = 0, for all |t| ≥ T.

It can be proved that a bandlimited signal cannot be timelimited, andvice versa. We’ve seen an example of this with the transform pairs

pτ (t) ↔ τ sinc(τω

2π

)

and τ sinc

(τt

2π

)

↔ 2πpτ (ω)

However, a signal can be approximately timelimited and bandlimited —that is, there exist numbers B > 0 and T > 0, such that |x(t)| is smallfor |t| ≥ T and |X(ω)| is small for |ω| ≥ B.


Frequency response of LTI systems

Consider an LTI system with the impulse response h(t). Then the outputof the system due to input x(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 asfollows:

Y (ω) = H(ω)X(ω).

This is a consequence of the fact that convolution in time domaincorresponds to multiplication in frequency domain.

The Fourier transform H(ω) of the impulse response h(t) is called the

frequency response of the system.


Lecture IX: Fourier transform - University Of...

Documents

Transcript of Lecture IX: Fourier transform - University Of...