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Home , Fourier transform, Sign function

EE 102 spring 2001-2002 Handout #23

Lecture 11 The transform

• definition • examples • the of a unit step • the Fourier transform of a periodic signal • properties • the inverse Fourier transform

11–1 The Fourier transform we’ll be interested in signals defined for all t the Fourier transform of a signal f is the

 ∞ F (ω)= f(t)e−jωtdt −∞

• F is a function of a real variable ω;thefunction value F (ω) is (in general) a

 ∞  ∞ F (ω)= f(t)cosωt dt − j f(t)sinωt dt −∞ −∞

•|F (ω)| is called the spectrum of f;  F (ω) is the spectrum of f • notation: F = F(f) means F is the Fourier transform of f;asfor Laplace transforms we usually use uppercase letters for the transforms (e.g., x(t) and X(ω), h(t) and H(ω),etc.)

The Fourier transform 11–2 Fourier transform and

Laplace transform of f

 ∞ F (s)= f(t)e−st dt 0

Fourier transform of f

 ∞ G(ω)= f(t)e−jωt dt −∞ very similar definitions, with two differences:

• Laplace transform is over 0 ≤ t<∞;Fouriertransform integral is over −∞

The Fourier transform 11–3 therefore,

• if f(t)=0for t<0, – if the imaginary axis lies in the ROC of L(f),then

G(ω)=F (jω),

i.e.,theFourier transform is the Laplace transform evaluated on the imaginary axis – if the imaginary axis is not in the ROC of L(f),thenthe Fourier transform doesn’t exist, but the Laplace transform does (at least, for all s in the ROC)

• if f(t) =0for t<0,thenthe Fourier and Laplace transforms can be very different

The Fourier transform 11–4 examples • one-sided decaying exponential  0 t<0 f(t)= e−t t ≥ 0

Laplace transform: F (s)=1/(s +1)with ROC {s |s>−1} Fourier transform is  ∞ 1 f(t)e−jωt dt = = F (jω) −∞ jω +1

• one-sided growing exponential  0 t<0 f(t)= et t ≥ 0

Laplace transform: F (s)=1/(s − 1) with ROC {s |s>1} f doesn’t have a Fourier transform

The Fourier transform 11–5 Examples double-sided exponential: f(t)=e−a|t| (with a>0)

 ∞  0  ∞ F (ω)= e−a|t|e−jωt dt = eate−jωt dt + e−ate−jωt dt −∞ −∞ 0 1 1 = + a − jω a + jω 2 = a a2 + ω2

1 2/a ) ) t ω ( ( f F

0 0 −a a −1/a t 1/a ω

The Fourier transform 11–6  1 −T ≤ t ≤ T rectangular pulse: f(t)= 0 |t| >T  T −1   2sinωT F (ω)= e−jωt dt = e−jωT − ejωT = −T jω ω

1 2T ) ) t ω ( ( f F

0 0 −T T −π/T π/T t ω unit impulse: f(t)=δ(t)

 ∞ F (ω)= δ(t)e−jωt dt =1 −∞

The Fourier transform 11–7  11− T ≤ t ≤ 1+T shifted rectangular pulse: f(t)= 0 t<1 − T or t>1+T  1+T −1   F (ω)= e−jωt dt = e−jω(1+T ) − e−jω(1−T ) 1−T jω −e−jω   = e−jωT − ejωT jω 2sinωT = e−jω ω (phase assuming T =1) 2T −π | ) ) ω ω ( 0 ( F F  |

0 −π −π/T π/T −π/T π/T ω ω

The Fourier transform 11–8 Step functions and signals by allowing impulses in F(f) we can define the Fourier transform of a or a constant signal unit step what is the Fourier transform of  0 t<0 f(t)= ? 1 t ≥ 0 the Laplace transform is 1/s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1/jω in fact, the integral

 ∞  ∞  ∞  ∞ f(t)e−jωt dt = e−jωt dt = cos ωt dt − j sin ωt dt −∞ 0 0 0 is not defined

The Fourier transform 11–9 however, we can interpret f as the limit for α → 0 of a one-sided decaying exponential  e−αt t ≥ 0 gα(t)= 0 t<0,

(α>0), which has Fourier transform

1 a − jω a jω Gα(ω)= = = − a + jω a2 + ω2 a2 + ω2 a2 + ω2 as α → 0, a jω 1 → πδ(ω), − → a2 + ω2 a2 + ω2 jω let’s therefore define the Fourier transform of the unit step as  ∞ 1 F (ω)= e−jωt dt = πδ(ω)+ 0 jω

The Fourier transform 11–10 negative time unit step  1 t ≤ 0 f(t)= 0 t>0

  0 ∞ 1 F (ω)= e−jωt dt = ejωt dt = πδ(ω) − −∞ 0 jω

constant signals: f(t)=1 f is the sum of a unit step and a negative time unit step:

 ∞  0  ∞ F (ω)= e−jωt dt = e−jωt dt + e−jωt dt =2πδ(ω) −∞ −∞ 0

The Fourier transform 11–11 Fourier transform of periodic signals similarly, by allowing impulses in F(f),wecandefinetheFourier transform of a periodic signal sinusoidal signals: Fourier transform of f(t)=cosω0t  1 ∞   F (ω)= ejω0t + e−jω0t e−jωt dt 2 −∞   1 ∞ 1 ∞ = e−j(ω−ω0)tdt + e−j(ω+ω0)tdt 2 −∞ 2 −∞

= πδ(ω − ω0)+πδ(ω + ω0)

F (ω)

π π

ω −ω0 ω0

The Fourier transform 11–12 Fourier transform of f(t)=sinω0t  1 ∞   F (ω)= ejω0t − e−jω0t e−jωt dt 2j −∞   1 ∞ 1 ∞ = e−j(ω−ω0)tdt + − e−j(ω0+ω)tdt 2j −∞ 2j −∞

= −jπδ(ω − ω0)+jπδ(ω + ω0)

F (ω)

jπ

ω 0 ω −ω0

−jπ

The Fourier transform 11–13 periodic signal f(t) with fundamental ω0 express f as

∞ jkω0t f(t)= ake k=−∞

∞  ∞ ∞ j(kω0−ω)t F (ω)= ak e dt =2π akδ(ω − kω0) k=−∞ −∞ k=−∞ F (ω)

2πa0

2πa−1 2πa1

2πa−2 2πa2 2πa−3 2πa3 ω −3ω0 −2ω0 −ω0 ω0 2ω0 3ω0

The Fourier transform 11–14 Properties of the Fourier transform

linearity af(t)+bg(t) aF (ω)+bG(ω)

1 ω time scaling f(at) |a|F ( a ) time shift f(t − T ) e−jωT F (ω)

df (t) differentiation dt jωF(ω)

k d f(t) ( )k ( ) dtk jω F ω  t ( ) F (ω) + (0) ( ) integration −∞ f τ dτ jω πF δ ω

k k ( ) k d F (ω) multiplication with ttf t j dωk  ∞ ( ) ( − ) ( ) ( ) −∞ f τ g t τ dτ F ω G ω  ( ) ( ) 1 ∞ ( ) ( − ) multiplication f t g t 2π −∞ F ω G ω ω dω

The Fourier transform 11–15 Examples  1 t ≥ 0 function: f(t)= −1 t<0 write f as f(t)=−1+2g(t),whereg is a unit step at t =0,andapply linearity 2 2 F (ω)=−2πδ(ω)+2πδ(ω)+ = jω jω sinusoidal signal: f(t)=cos(ω0t + φ) write f as f(t)=cos(ω0(t + φ/ω0)) and apply time shift property:

jωφ/ω0 F (ω)=πe (δ(ω − ω0)+δ(ω + ω0))

The Fourier transform 11–16  0 |t| > 10 pulsed cosine: f(t)= cos t −10 ≤ t ≤ 10

1

0.5 ) t

( 0 f

−0.5

−1 −10 0 10 t write f as a f(t)=g(t)cost where g is a rectangular pulse of width 20 (see page 12-7)

2sin10ω F(cos t)=πδ(ω − 1) + πδ(ω +1), F(g(t)) = ω

The Fourier transform 11–17 now apply multiplication property  ∞ sin 10ω F (jω)= (δ(ω − ω − 1) + δ(ω − ω +1))dω −∞ ω sin(10( − 1)) sin(10( +1)) = ω + ω ω − 1 ω +1

12

10

8

6 ) ω

( 4 F 2

0

−2

−4 −1 ω0 1

The Fourier transform 11–18 The inverse Fourier transform if F (ω) is the Fourier transform of f(t), i.e.,

 ∞ F (ω)= f(t)e−jωt dt −∞ then  1 ∞ f(t)= F (ω)ejωt dω 2π −∞ let’s check    1 ∞ 1 ∞ ∞ F (ω)ejωt dω = f(τ)e−jωτ ejωtdω 2π ω=−∞ 2π ω=−∞ τ=−∞   1 ∞ ∞ = f(τ) e−jω(τ−t)dω dτ 2π τ=−∞ ω=−∞  ∞ = f(τ)δ(τ − t)dτ −∞ = f(t)

The Fourier transform 11–19