eece 359 - signals and communications: part 1 spring 2014...

EECE 359 - Signals and Communications: Part 1 Spring 2014

Fourier transform representation of CT

aperiodic signals – Section 4.1

A large class of aperiodic CT signals can

be represented by the CT Fourier transform

(CTFT).

The (CT) Fourier transform (or spectrum) of

x(t) is

X(jω) =

∫ ∞

−∞

x(t)e−jωt dt.

x(t) can be reconstructed from its spectrum

using the inverse Fourier transform

x(t) =1

∫ ∞

−∞

X(jω)ejωt dω.

The above two equations are referred to as

the Fourier transform pair with the first one

SM 107

being the analysis equation and the second

being the synthesis equation.

Notation:

X(jω) = F{x(t)}x(t) = F−1{X(jω)}

x(t) and X(jω) form a Fourier transform

pair, denoted by

x(t)F←→ X(jω)

Often, we will use the simpler notation

x(t)←→ X(jω)

Example:

rect(t) or Π(t) =

1, |t| < 1/2

0, |t| > 1/2

1/2, |t| = 1/2

SM 108

Π t( )

2---– 0

rect(t)

SM 109

F Π t( ){ }

0 2π2– π4– π6– π 4π 6π

Fourier transform of rect(t)

Example: x(t) = e−atu(t), a > 0. We

want to show that

e−atu(t)←→ 1

a+ jω, a > 0 .

SM 110

Since the above FT is complex-valued, it is

customary to plot its magnitude and phase,

a+ jω| =

1√a2 + ω2

a+ jω= − tan−1

a jω+---------------

1 a⁄

0 aa–

Plot of spectrum magnitude

a jω+---------------∠

π 2⁄

0 aa–

π– 2⁄

Plot of spectrum phase

SM 111

Example: triangle function

∧(t) =

1− |t|, −1 ≤ t ≤ 10, elsewhere

Λ t( )

11– 0

triangle function

Exercise: Show that

∧(t)F←→ sinc2(

SM 112

Properties of CT Fourier transform –

Section 4.3

Table 4.1 on p. 328 summarizes many CTFT

properties.

1. Linearity – Section 4.3.1

Let x(t)←→ X(jω), y(t)←→ Y (jω).

z(t) = ax(t) + by(t)

←→ Z(jω) = aX(jω) + bY (jω) .

Proof:

SM 113

Example: What is the CTFT of the

signal, z(t), shown in the figure below?

z t( )

11– 0

0.50.5–

Signal z(t)

SM 114

2. Time shift – Section 4.3.2

If x(t)←→ X(jω), then

x(t− t0)←→ e−jωt0X(jω) .

Proof:

Note that when a signal is delayed by t0, the

spectrum amplitude is unchanged whereas

the spectrum phase is changed by −ωt0.

SM 115

Example: What are the CTFT’s of the

signals shown below?

Π t 0.5+( )

Π t 0.5–( )

Shifted rect signals

SM 116

3. Time scaling – Section 4.3.5

If x(t)↔ X(jω), then

x(at)↔1

where a is a non-zero real constant.

The time scaling property states that if a

signal is compressed in time by a factor a,

then its spectrum is expanded in frequency

by the same factor a and vice-versa.

This property is an example of the

inverse relationship between the time and

frequency domains.

Example: Since

Π(t)↔ sinc( ω

Π(2t)↔1

SM 117

Note also that if a = −1 (corresponding to

a time reversal) in the time scaling property,

we have

x(−t)↔ X(−jω)

i.e. the spectrum is also reversed.

Combining the time shift and scaling

properties

What is the Fourier transform of x(at− b)?

SM 118

4. Conjugation – Section 4.3.5

x∗(t)↔ X∗ (−jω) .

As a result, if x(t) is real, we have

X (−jω) = X∗ (jω)

i.e. the spectrum magnitude is an even

function of ω and the spectrum phase is an

odd function of ω.

More generally, we can summarize the

relationship between a signal and its

spectrum as follows:

SM 119

5. Convolution – Section 4.4y(t) = x(t) ∗ h(t)↔ Y (jω) = X (jω)H (jω) .

Proof:

Application: Since ∧(t) = Π(t) ∗Π(t),

F{∧(t)} = F{Π(t)}F{Π(t)}

= sinc2( ω

SM 120

6. Differentiation & Integration – Section 4.3.4

dtx(t)↔ jωX(jω) .

Proof:

This result indicates that differentiation

accentuates the high frequency components

in the signal.

Application: Consider y(t) as shown below

y t( )

1– 0

SM 121

Since y(t) = ddt∧ (t), we have

F{y(t)} = jω F{∧(t)}

= jω sinc2( ω

−∞

x(τ)dτ ↔1

jωX (jω) + πX(0)δ(ω) .

We see that in contrast to differentiation,

integration attenuates the high frequency

components in the signal.

7. Area property

∫ ∞

−∞

x(t)dt = X (0) .

SM 122

This result follows directly from the

definition of the Fourier transform, i.e.

X(jω) =

∫ ∞

−∞

x(t)e−jωt dt.

Also,∫ ∞

−∞

X(jω)dω = 2π x (0)

x(t) =1

∫ ∞

−∞

X(jω)ejωt dω.

Example: Evaluate∫∞

−∞sinc(x)dx.

This can be obtained as follows:

SM 123

8. Duality

Due to the similarity between the FT

analysis and synthesis equations, we have a

“duality” relationship.

X (jt)↔ 2πx(−ω) .

Proof:

SM 124

Example: Determine the inverse FT of

Π(ω).

Π(t)↔ sinc( ω

↔ 2πΠ(−ω) = 2πΠ(ω)

or equivalently

2πsinc

↔ Π(ω).

Duality can also be useful in suggesting

new properties of the FT.

SM 125

Example:

9. Parseval’s relation

∫ ∞

−∞

|x(t)|2 dt =1

∫ ∞

−∞

|X(jω)|2 dω .

Remarks:

(a) The LHS is the total energy in the signal

(b) |X(jω)|2 describes how the energy in

x(t) is distributed as a function of

frequency. It is commonly called the

energy density spectrum of x(t).

SM 126

Proof:

Example: Evaluate∫∞

−∞sinc2x dx.

SM 127

CT unit impulse function – pp. 32–38,

pp. 92-93

The CT unit impulse function is also known

as the Dirac delta function, δ(t). Contrast

with the Kronecker delta function, δij. The

introduction of δ(t) allows us to look at the

FT of periodic signals.

δ(t) is defined by the sifting property, namely

∫ ∞

−∞

δ(t)f(t) dt = f(0)

if f(t) is continuous at t = 0.

Properties of the Dirac delta function

∫ ∞

−∞

δ(t) dt = 1

SM 128

δ t( )represents the strength of

the impulse function

∫ ∞

−∞

δ(t− t0)f(t) dt =

∫ ∞

−∞

δ(τ)f(τ + t0) dτ

= f(t0)

The first line is obtained using τ = t− t0.

f(t) ∗ δ(t) =

∫ ∞

−∞

δ(τ)f(t− τ) dτ

= f(t)

The above is referred to as the replication

property of δ(t).

SM 129

δ(at) =1

|a|δ(t)

Proof:

5. What is the Fourier transform of δ(t)?

F{δ(t)} =

SM 130

6. What is the inverse FT of δ(ω)?

F−1{δ(ω)}∆=

∫ ∞

−∞

δ(ω)ejωt dω

Thus, 1↔ 2πδ(ω).

7. It follows from the last result that∫ ∞

−∞

e−jωt dt = 2πδ(ω)

Therefore,

2πδ(ω − ωc) =

∫ ∞

−∞

e−j(ω−ωc)t dt

∫ ∞

−∞

ejωcte−jωt dt

ejωct}

SM 131

We thus have ejωct ↔ 2πδ(ω − ωc). This

result is useful in examining the FT of a

periodic signal.

8. Using the last result, we can write

cosωct =1

ejωct + e−jωct]

2[2πδ(ω − ωc) + 2πδ(ω + ωc)]

= πδ(ω − ωc) + πδ(ω + ωc)

� ωctcos{ }

ω–c ω

SM 132

Similarly, we have

sinωct =1

ejωct − e−jωct]

j[δ(ω − ωc)− δ(ω + ωc)]

� ωc

sin t{ }

ω–c

j---–

SM 133

Fourier transform of periodic signals –

Section 4.2

Recall from our discussion of FS

representation: A periodic signal, x̃(t), with

fundamental period T can be represented by

x̃(t) =∞∑

k=−∞

ak ejk(2πT )t

where the (possibly complex) Fourier

coefficients {ak} are given by

∫ T2

x̃(t) e−jk(2πT )t dt,

k = 0,±1,±2, . . .

Let x(t) =

x̃(t), −T2 ≤ t ≤ T

0, otherwise.

SM 134

Thus, x(t) is simply one basic period of x̃(t).

Then, with ω0∆= 2π

T, we can write

∫ ∞

−∞

x(t) e−jkω0t dt

TX (jkω0) .

The last line follows since

X(jω) =

∫ ∞

−∞

x(t)e−jωt dt.

In other words, ak is equal to 1Tmultiplied by

the FT of x(t) evaluated at ω = kω0.

Therefore,

x̃(t) =

∞∑

k=−∞

TX (jkω0) ejkω0t

∞∑

k=−∞

X (jkω0) 2πδ (ω − kω0)

SM 135

In summary, the FT of a periodic signal

consists of a series of impulses located

at frequencies which are multiples of the

fundamental frequency ω0. The strength of

the impulse at the kth harmonic frequency

kω0 is 2πak.

Example: What is the FT of the impulse train

or comb function?

Recall that the comb function is given by

δT (t) =∞∑

k=−∞

δ(t− kT )

and is shown in the figure below.

δTt( )

T 2T 3TT–2T–3T–

SM 136

In this example, “x(t)” = δ(t) so that

X(jω) = 1.

Therefore,

δT (t)↔2π

∞∑

k=−∞

ω − k2π

T------

� δTt( ){ }

T------

T------2π

T------–

We will see that the comb function is very

useful in discussing sampling .

SM 137

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