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uncertainty principle

The Uncertainty Principle

Bob Williamson

Australian National University

November18,2000

The uncertainty principle shows that one can not jointly localize a signal in time and frequency arbitrarily well;either one has poor frequency localization or poor time localiza-tion.The degree of localization is measured in the theorem below by the quantitities d and D;these are like the standard deviation of a probability distribution.

The proof we present was originally due to Weyl[H.Weyl,Theory of groups and Quan-tum Mechanics,Dover,NY,(1950);Appendix A]

Theorem1(Uncertainty Principle)Suppose f(t)is a?nite energy signal with Fourier

transform F(ω).Let

E:= ∞?∞|f(t)|2dt=12π ∞

?∞

|F(ω)|2dω

d2:=1

E

?∞

t2|f(t)|2dt

D2:=

1

2πE

?∞

ω2|F(ω)|2dω

If |t|f(t)→0as|t|→∞,then

Dd≥1 2

and equality holds only if f(t)has the form

f(t)=Ce?αt2.

In order to prove the theorem we need the following Lemma which is very useful in many other situations.

Lemma2(Cauchy-Schwarz Inequality)For any square integrable functions z(x)and w(x)de?ned on the interval[a,b],

b

a

z(x)w(x)dx

2≤

b

a

|z(x)|2dx b a|w(x)|2dx(1)

and equality holds if and only if z(x)is proportional to w?(x)(almost everywhere on [a,b]).

Proof Assume z(x)and w(x)are real(the extension to complex-valued functions is straight-forward).Let

I(y)= b a[z(x)?yw(x)]2dx

= b a z2(x)dx

A ?2y b a z(x)w(x)dx

B

+y2 b a w2(x)dx

C

=A?2yB+y2C

Clearly I(y)≥0for all y∈R.But if I(y)=A?2yB+y C≥0for all y∈R then B2?Ac≤0.If B2?AC=0,then I(y)has a double real root:?k:I(k)=0for y=k.Thus(1)holds and if it is an equality,then I(y)has a real root which implies ?k:I(k)= b a[z(x)?kw(x)]2dx=0.But this can only occur if the integrand is identi-cally zero;thus z(x)=kw(x)for all x.

Proof(Theorem)Assume f(t)is real.Lemma2implies

?∞

tf

d f

dt

dt 2≤

?∞

t2f2dt ∞?∞

d f dt

2dt.(2)

Let

A:= ∞?∞tf d f dt dt

= t d(f2/2)dt dt (by the chain rule for differentiation)

=t f2

2

?∞

α? ∞?∞f22dt β

using integration by parts.By assuption |t|f→0?|t|f2→0?tf2→0.Thus α=0.Furthermoreβ=E/2and so

A=?E

2

(3)

Recalling that d

dt

f(t)?jωF(ω),by Parseval’s theorem we have

∞?∞

d f dt

2dt=12π

?∞

ω2|F(ω)|dω(4)

Substituting(3)and(4)into(2)we obtain

?E2

2=

?∞

tf

d f

dt

dt 2≤

?∞

t2f2dt

Ed2

×

1

?∞

ω2|F(ω)|2dω

ED2

(5)

?dD≥1

2

(6)

If(6)is an equality,then(2)must be also which is possible only if

d

dt

f(t)=ktf(t)

?f(t)=Ce?αt2

(note e?π(bt)2?e?π(f/b)2—the Fourier transform of a Gaussian is a Gaussian).