Dirac Delta Function Convolution

7/27/2019 Dirac Delta Function Convolution

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Topic 3

The -function & convolution.Impulse response & Transfer function

In this lecture we will described the mathematic operation of the convolution oftwo continuous functions. As the name suggests, two functions are blended orfolded together.

We will then discuss the impulse responseof a system, and show how it is relatedto the transfer function of the system.

First though we will define a special function called the -function or unit impulse.It is, like the Heaviside step function u(t), a generalized function or distributionand is best defined by considering another function in conjunction with it.

3.1 The -function

Consider a function

g(t) =

1/w 0< t < w

0 otherwise

One thing of note about g(t) is that

w

0

g(t)d t= 1.

The lower limit 0 is a infinitesimally small amount less than zero. Now, supposethat the widthwgets very small, indeed as small at 0+, an number an infinitesimalamount bigger than zero. At that point, g(t) has become like the function, avery thin, very high spike at zero, such that

(t)dt=

0+0

(t)dt= 1

1


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w1/

w0

g(t)

0

(t)

(a) (b)

Figure 3.1: As w becomes very small the function g(t) turns into a -function (t) indicated by

the arrowed spike.

In some sense it is akin to the derivative of the Heaviside unit step

t

(t)dt=u(t) .

More formally the delta function is defined in association with any arbitrary functionf(t), as

The delta function ...

f(t)(t)dt=f(0) .

Picking out values of a function in this way is called sifting of f(t) by(t). Wecan also see that

f(t)(t )dt=f() ,

a result that we will return to.

0

(t)f(t)

0

f(t)

(t)

t

(t)

0

t

A

(a) (b) (c)

Figure 3.2: (a,b) Sifting. (c) With an amplitude A.

Although the -function is infinitely high, very often you will see a described asthe unit-function, or see a-function spike with an amplitude A by it. This is todenote a delta-function where

(t)dt= 1 or

A(t)dt=A.


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3.2 Properties of the -function

Fourier transform of the delta function:

FT [(t)] = 1

Proof: Use the definition of the -function and sift the function f(t) = eit:

(t)eitdt= ei0 = 1 .

Symmetry: The -function has even symmetry.

(t) =(t)

Parameter Scaling:

(at) = 1

|a|(t)

Proof: To prove this return to the fundamental definition,

f(t)(t)dt=f(0)Ifa 0, substitute (at) for t(no swap in limits)

f(at)(at)d(at) = a

f(at)(at)dt = f(0)

But

f(at)(t)dt=f(0) a (at) =(t).

Now ifa


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3.3 Fourier Transforms that involve the-function

Fourier Transform ofei0t

FTei0t=

ei(0)tdt= 2( 0) .

Fourier Transform of1

FT[1] = 2() .

You could obtain this either by putting 0 = 0 just above, or by using the dualproperty,FT[1] = 2(), then the even symmetry property () =().

Fourier Transform ofcos 0t

F T [cos 0t] = FT

1

2

ei0t + ei0t

= (( 0) +(+0)) .

Fourier Transform ofsin 0t

FT [cos 0t] = FT

1

2i ei0t ei0t

= i (( 0) (+0)) .

Fourier Transform of Complex Fourier Series yes, this can be useful!

FT

n=n=

Cnein0t

= 2

n=n=

Cn( n0) .

3.4 Convolution

We turn now to a very important technique is signal analysis and processing. Theconvolution of two functions f(t) and g(t) is denoted by f g. The convolutionis defined by an integral over the dummy variable .

The convolution integral. The value off gat t is

(f g)(t) =

f()g(t )d


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The process is commutative, which means that

(f g)(t) (g f)(t) or

f()g(t )d

f(t )g()d

3.4.1 Example 1It is easier to see what is going on when convolving a signal fwith a functiongof even or odd symmetry. However, to get into a strict routine, it is best to startwith an example with no symmetry.

[Q] Find and sketch the convolution of f(t) = u(t)eat with g(t) = u(t)ebt,where botha and bare positive.

[A] Using the first form of the convolution integral, the short answer must bethe unintelligible

f g=

u()eau(t )eb(t)d .

First, make sketches of the functions f() andg(t ) as varies. Functionf()looks just like f(t) of course. But g(t ) is a reflected (time reversed) andshifted version of g(t). (The reflection is easy enough. To check that the shiftis correct, ask yourself where does the function g(p) drop? The answer is atp= 0. So g(t ) must drop when t = 0, that is when =t.)

0

f( ) )g(

0

)g(

0 0

g(t )

t

(a) (b) (c) (d)

Figure 3.3:

We now multiply the two functions, BUT we must worry about the fact that t isa variable. In this case there are two different regimes, one when t < 0 and theother whent 0. Figure 3.4 shows the result.

So now to the integration. For t < 0, the function on the bottom left of Figure3.4 is everywhere zero, and the result is zero. For t 0

u()eau(t)eb(t)d = ebt t

0

e(ba)d = ebt

b a

e(ba)t 1


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f( )g(t )

0

g(t ) f( )

t

t0

t

0

t0

Figure 3.4:

So

f g(t) =

eat ebt

/(b a) for t 0

0 for t 0 part of the convolution for b= 2and a= 1.

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6

(exp(-x)-exp(-2*x))

Figure 3.5: f g(t) plotted for t >0 when b= 2 and a= 1.


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3.4.2 Example 2

[Q] Derive an expression for the convolution of an arbitrary signal f(t) with thefunction g(t) shown in the figure. Determine the convolution when f(t) = A, aconstant, and when f(t) =A + (B A)u(t).

[A] Follow the routine. Function f() looks exactly like f(t), but g(t ) isreflected and shifted. Multiply and integrate over from to . Because gonly has finite range, we can pinch in the limits of integration, and the convolutionbecomes

f g=

tta

f()d+

t+at

f()d

f(t)

t

g(t)

taa

1

1

g(t )f(t)

t

1

1

t+a

ta t

Figure 3.6:

Whenf(t) =A, a constant, it is obvious by inspection that the convolution is zerofor all t. When f(t) = A + (B A)u(t), we have to be more careful becausethere is a discontinuity in the function.

From Figure 3.7(a):

The convolution is zero for all t < a and all t > a (Diagram positions1,2,5).

The maximum value is when t= 0 (Position 4). By inspection, or using theintegrals above, (f g)(t= 0) =a(B A).

For a < t


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g(t )

4: t=0

f( ) f( ) g(t )

A

B

aaa2a 00 0

1: 2:t=a a


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Now suppose that h iscausal. If > tthenh(t ) = 0, and therefore, providedh(t) is causal,

the time response y(t) to an input x(t) is

y(t) =

x()h(t )d = x h .

The output is the input CONVOLVED with the Impulse Response

Function.

3.6 How do we connect this up ...

Can we reconcile the following things you now know about systems and signals?

1. The temporal output is the temporal input CONVOLVED with the ImpulseResponse Function.

2. The frequency domain output is the frequency domain input MULTIPLIEDby the Transfer Function.

3. The frequency domain signal is the Fourier Transform of the temporal signal

Mathematically, it must be that the FT of a convolution is a product.

y(t) = x(t) h(t)

F T F T

? ?

Y() = X()H()

3.7 The Fourier Transform of a Convolution

To prove this we need to develop the integration for the Fourier Transform of aconvolution. Now x h is a perfectly respectable function oft, so

FT [x h] =

t=

[(x h)(t)] eitdt

=

t=

=

x()h(t )deitdt


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For absolute clarity, lets switch the order of integration, then write t=+ p. Inthe inner integral is a constant, so that dt= dp.

FT [x h] =

=

t=

x()h(t )eitdtd

=

=

p=

x()h(p)eieipdpd

=

=

x()eid

p=

h(p)eipdp

= FT [x] FT [h]

= X() H()

So, the first important thing we discover is

The time-convolution/frequency-modulation property of Fourier

transform

f(t) g(t) F()G()

or FT [f(t) g(t)] = F T [f(t)] F T [g(t)] =F()G()

3.8 Impulse response and Transfer Function

The second important connection is that

The Fourier Transform of the Impulse Response Function is the Trans-

fer Function

h(t) H()

or FT [h(t)] = H()


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*h(t)x(t) y(t)

X( Y() ) )H(

Figure 3.9:

3.9 The time-modulation/frequency-convolution property

There is a further result involving convolution that we state now.

The modulation/convolution property of Fourier transform

f(t)g(t) 1

2F() G()

3.10 Example1

[Q1] Show that (t) f(t) =f(t).

[A1] One could do this the long way by writing

f()(t )d, etc, but aquicker way is to argue that

F T [(t) f(t)] = FT [(t)] F T [f(t)] = 1 . F()

FT1 [FT [(t) f(t)]] = FT1 [F()]

(t) f(t) = f(t)

We could also show that f(t) (t ) =f(t ) .

Both of these are manifestations of the

Sifting property of-functions

f(t) (t ) =f(t )


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Previously this was expressed as an integral but we now recognize that integralas the convolution integral!


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3.11 More examples

[Q2] Convolve the two top hat functions f(t) and g(t) shown in below, and plotthe resulting convolved signal r(t). Find its Fourier Transform R().

1

a a

f(t)

2

a/2a/2

g(t)

Figure 3.10:

[A2] Plotted below as functions of aref(), the reversed and shifted g(t )fort= 3a/2, 3a/2< t < a/2 andt= a/2, and the resultingf()g(t).

g(t)

1

a

f()

t a

g(t)

1

a a

f()

t

g(t)

1

a a

f()

t

a t+a/2

0a

fg

fg

fg

Figure 3.11:

From the plots it is obvious that

for all t < 3a/2 f g is zero, so the convolution integral is zero.

For 3a/2< t < a/2 the integral is t+a/2a

2d= 2

t+

3a

2

.


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Att= a/2 there is complete overlap, at which point the integral is 2a.

It will continue constant at 2a until t= 0.

The convolution has even symmetry.

The result is given in Figure 3.12.

2a

03a/2 3a/2a/2a/2

g(t)f(t) *

Figure 3.12:

To find the Fourier transform of this convolution, we can use f g F()G().

Function f(t) is a top hat of height 1, but its width is 2a. We can use theParameter Scaling Property from Lecture 2 that says if

If f(t) F() then FT [f(t)] = 1

||F

We have to be VERY careful. It is tempting to say that because our function is 2

times wider than the standard top hat, the scaling must be 2. WRONG!The scaling is = 1/2. For the unit top-hat of half-width a/2

() = 2

sin

a2

= a sinc

a2

.

So for our parameter scaled version

F() = 1

|(1/2)|

(1/2)

= 2

2

2sin

2a

2

=

2

sin (a) = 2asinc

a

.

Functiong(t) is a top hat of height 2 and half-width a/2. From Lecture 2, Section2.4,

G() = 2() = 4

sin

a2

= 2a sinc

a2

.

So the Fourier Transform of the convolution is

FT [f g] = 8

2sin(a)sin

a2

= 4a2 sinc

a2

sinc

a

.


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[Q3] A signal f(t) is amplitude modulated by cos 0 into a signal y(t). Find theresulting Fourier Transform Y() in terms ofF().

[A3] We could use the method given in Lecture 2 Amplitude modulation by acosine. Instead, lets use the property at Section 3.6 of this Lecture.

f(t)g(t) 12

F() G()

where our g(t) = cos 0t. By writing

cos 0t=1

2

ei0t + ei0t

one can determine that

FT [cos 0t] = ((+0) +( 0)) .

(0)(+

0)

0 0

Figure 3.13: Frequency spectrum of cos 0tHence

FT [f(t)cos 0t] = 1

2F() ((+0) +( 0))

= 1

2(F(+0) +F( 0))

So suppose f(t) had a amplitude spectrum as sketched below. After amplitudemodulation, the amplitude spectrum would split into two parts.

CosineModulation

F( )

0

F( )

00

0

Figure 3.14: An amplitude spectrum split into two by modulation with a cosine wave.


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3.12 Summary

We started by developing the definition of the unit delta function (t). Althoughthis is an infinitely high spike, it is also infinitely narrow, so that the integral overit is unity.

The delta function was more formally defined via its sifting properties.We looked at properties of the delta function and at Fourier transforms that involvethe delta function

We then defined the convolution integral, and looked at how to lay out theintegral graphically.

We defined the Impulse Response function and showed that the temporal outputof a system is the temporal input convolved with the input.

Finally, by showing that the FT of a convolution of two temporal function is the

product of their individual FTs, we found that our old friend the Transfer Functionis the Fourier Transform of the Impulse Response.

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Dirac Delta Function Convolution