Tài liệu Signals And Systems With Matlab Applications P2 doc

Chapter 1 Elementary Signals1-20 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications 1.9 Exercises 1.. Chapter 1 Elementary Signals1-22 Signals and Systems

Trang 1

Signals and Systems with MATLAB Applications, Second Edition 1-19

Orchard Publications

Summary

int(d) % Integrate the delta function

ans =

Heaviside(t-a)*k

1.8 Summary

• The unit step function that is defined as

• The unit step function offers a convenient method of describing the sudden application of a volt-age or current source

• The unit ramp function, denoted as , is defined as

•The unit impulse or delta function, denoted as , is the derivative of the unit step It is also defined as

and

•The sampling property of the delta function states that

•The sifting property of the delta function states that

•The sampling property of the doublet function states that

u 0( )t

u 0( )t 0 t<0

1 t>0

⎩

⎨

⎧

=

u 1( )t

u 1( )t u 0( ) τ τ d

∞ –

t

∫

=

δ τ ( ) τd

∞ –

t

∫ = u 0( )t

δ t( ) = 0 for all t≠0

f t ( )δ t a( – ) = f a ( )δ t( )

a = 0

f t ( )δ t( ) = f 0 ( )δ t( )

f t ( )δ t α( – ) t d

∞ –

∞

δ' t( )

f t ( )δ' t a( – ) = f a ( )δ' t a( – ) f– ' a ( )δ t a( – )

Trang 2

Chapter 1 Elementary Signals

1-20 Signals and Systems with MATLAB Applications, Second Edition

1.9 Exercises

1 Evaluate the following functions:

a

b

c

d

e

f

2

a Express the voltage waveform shown in Figure 1.24, as a sum of unit step functions for

b Using the result of part (a), compute the derivative of , and sketch its waveform

Figure 1.24 Waveform for Exercise 2

tδ

sin t π

6

-–

⎝ ⎠

⎛ ⎞

2tδ

4

-–

⎝ ⎠

⎛ ⎞

t

2 δ t π

2

-–

⎝ ⎠

⎛ ⎞

cos

2tδ

8

-–

⎝ ⎠

⎛ ⎞

t 2 e–t δ t 2( – ) t d

∞

–

∞

∫

t

2δ1

t π

2

-–

⎝ ⎠

⎛ ⎞

sin

v t( )

0< <t 7 s

v t( )

−10

−20

10 20

0

v t( )

t s( )

e–2t

V

( )

v t( )

Trang 3

Solutions to Exercises 1.10 Solutions to Exercises

Dear Reader:

The remaining pages on this chapter contain the solutions to the exercises

You must, for your benefit, make an honest effort to solve the problems without first looking at the solutions that follow It is recommended that first you go through and solve those you feel that you know For the exercises that you are uncertain, review this chapter and try again If your results do not agree with those provided, look over your procedures for inconsistencies and computational errors Refer to the solutions as a last resort and rework those problems at a later date

You should follow this practice with the exercises on all chapters of this book

Trang 4

1 We apply the sampling property of the function for all expressions except (e) where we apply the sifting property For part (f) we apply the sampling property of the doublet

a

b

c

d

e

We recall that the sampling property for the doublet states that

Thus,

f

2

a

or

δ t( )

f t ( )δ t a( – ) = f a ( )δ t a( – )

tδ

sin t π

6

-–

⎝ ⎠

t=π 6⁄ δ t π

6

-–

⎝ ⎠

⎛ ⎞

6

-δ t π 6

-–

⎝ ⎠

⎛ ⎞

sin 0.5 δ t π

6

-–

⎝ ⎠

⎛ ⎞

2tδ

4

-–

⎝ ⎠

t=π 4⁄ δ t π

4

-–

⎝ ⎠

⎛ ⎞

2

-δ t π

4

-–

⎝ ⎠

⎛ ⎞

t

2 δ t π

2

-–

⎝ ⎠

⎛ ⎞

2

- 1( + cos 2t)

t=π 2⁄

δ t π 2

-–

⎝ ⎠

2

- 1( + cosπ)δ t π

2

-–

⎝ ⎠

2

- 1( –1 )δ t π

2

-–

⎝ ⎠

2tδ

8

-–

⎝ ⎠

t=π 8⁄ δ

8

-–

⎝ ⎠

4

-δ t π 8

-–

⎝ ⎠

⎛ ⎞

8

-–

⎝ ⎠

⎛ ⎞

f t ( )δ t α( – ) t d

∞ –

∞

t 2 e–t δ t 2( – ) t d

∞

–

∞

∫ = t 2 e–t t=2 = 4e–2 = 0.54

f t ( )δ' t a( – ) = f a ( )δ' t a( – ) f– ' a ( )δ t a( – )

t

2 δ1

t π

2

-–

⎝ ⎠

⎛ ⎞

sin 2 t t=π 2⁄ δ1

t π

2

-–

⎝ ⎠

⎛ ⎞

dt

- 2 t t=π 2⁄ δ t π

2

-–

⎝ ⎠

⎛ ⎞

sin

–

=

1 2

- 1( –cos 2t)

t=π 2⁄ δ1

t π

2

-–

⎝ ⎠

t=π 2⁄ δ t π

2

-–

⎝ ⎠

⎛ ⎞

sin

–

=

1 2 - 1( +1)δ1

t π

2

-–

⎝ ⎠

⎛ ⎞ πδ t π

2

-–

⎝ ⎠

⎛ ⎞

sin

t π

2

-–

⎝ ⎠

⎛ ⎞

=

v t( ) = e–2t[u 0 ( ) u t – 0(t–2) ] + (10t–30 ) u[ 0(t–2 ) u– 0(t–3) ]

+ (– t 50 10 + ) u[ 0(t–3 ) u– 0(t–5) ] + (10t–70 ) u[ 0(t–5 ) u– 0(t–7) ]

Trang 5

Solutions to Exercises

b

(1)

Referring to the given waveform we observe that discontinuities occur only at , ,

function

and with these simplifications (1) above reduces to

The waveform for is shown below

v t( ) e–2t u 0 ( ) e t –2t

u 0(t–2 ) 10tu+ 0(t–2 ) 30u– 0(t–2 ) 10tu– 0(t–3 ) 30u+ 0(t–3) –

=

10tu 0(t–3) – +50u 0(t–3 ) 10tu+ 0(t–5 ) 50u– 0(t–5 ) 10tu+ 0(t–5)

70u 0(t–5 ) 10tu– 0(t–7 ) 70u+ 0(t–7) –

e–2t u 0( )t + ( –e–2t+10t–30 )u 0(t–2) + ( –20t+80 )u 0(t–3) + (20t–120 )u 0(t–5)

=

+ ( –10t+70 )u 0(t–7)

dv

dt

- = –2e–2t u 0 ( ) e t + –2t δ t( ) + (2e–2t+10 )u 0(t–2) + ( –e–2t+10t–30 )δ t 2( – )

20u 0(t–3) + ( –20t+80 )δ t 3( – ) 20u+ 0(t–5) + (20t–120 )δ t 5( – )

–

10u 0(t–7) + ( –10t+70 )δ t 7( – )

–

t = 2 t = 3

t = 5 δ t( ) = 0 δ t 7( – ) = 0

e–2t+10t–30

–

( )δ t 2( – ) ( –e–2t+10t–30)

t=2 δ t 2( – )

20t

– +80

( )δ t 3( – ) = ( –20t+80) t=3 δ t 3( – ) = 20 δ t 3( – )

20t–120

( )δ t 5( – ) = (20t–120) t=5 δ t 5( – ) = – δ t 5 20 ( – )

dv dt⁄ 2e–2t u 0 ( ) 2e t –2t

u 0(t–2 ) 10u 0(t–2 ) 10δ t 2– ( – )

–

=

20u 0(t–3 ) 20δ t 3+ ( – ) 20u+ 0(t–5 ) 20δ t 5– ( – ) 10u– 0(t–7) –

2e–2t[u 0 ( ) u t – 0(t–2)] 10δ t 2– ( – ) 10 u+ [ 0(t–2 ) u– 0(t–3)] 20δ t 3+ ( – ) –

=

10 u[ 0(t–3 ) u– 0(t–5) ] – –20 δ t 5( – ) 10 u+ [ 0(t–5 ) u– 0(t–7) ]

dv dt⁄

20 10

V s⁄ ( )

t s( )

20

–

10

10 δ t 2( – ) –

20 δ t( –3)

20 δ t( –5) –

2e–2t

–

Trang 6

NOTES

Trang 7

Chapter 2

The Laplace Transformation

his chapter begins with an introduction to the Laplace transformation, definitions, and proper-ties of the Laplace transformation The initial value and final value theorems are also discussed and proved It concludes with the derivation of the Laplace transform of common functions

of time, and the Laplace transforms of common waveforms

2.1 Definition of the Laplace Transformation

The two-sided or bilateral Laplace Transform pair is defined as

(2.1)

(2.2)

Inverse Laplace transform, and is a complex variable whose real part is , and imaginary part ,

In most problems, we are concerned with values of time greater than some reference time, say

, and since the initial conditions are generally known, the two-sided Laplace transform pair of (2.1) and (2.2) simplifies to the unilateralor one-sided Laplace transform defined as

(2.3)

(2.4)

The Laplace Transform of (2.3) has meaning only if the integral converges (reaches a limit), that is, if

(2.5)

To determine the conditions that will ensure us that the integral of (2.3) converges, we rewrite (2.5)

T

L f t{ ( )} F s= ( ) f t( )

∞ –

∞

∫ e–st dt

=

L–1{F s( )} f t= ( ) 1

2 πj

- F s( )

σ jω–

σ jω+

=

F s( )

s = σ jω+

t

t = t 0 = 0

L f t{ ( )} F= ( )s f t( )

t 0

∞

∫ e–st dt f t( )

0

∞

∫ e–st dt

L–1{F s( )} f= ( )t 1

2 πj

- F s( )

σ jω–

σ jω+

=

f t( )

0

∞

∫ e–st dt < ∞

Trang 8

Chapter 2 The Laplace Transformation

as

(2.6)

The term in the integral of (2.6) has magnitude of unity, i.e., , and thus the condition for convergence becomes

(2.7)

Fortunately, in most engineering applications the functions are of exponential order* Then, we can express (2.7) as,

(2.8)

and we see that the integral on the right side of the inequality sign in (2.8), converges if Therefore, we conclude that if is of exponential order, exists if

(2.9) where denotes the real part of the complex variable

Evaluation of the integral of (2.4) involves contour integration in the complex plane, and thus, it will not be attempted in this chapter We will see, in the next chapter, that many Laplace transforms can

be inverted with the use of a few standard pairs, and therefore, there is no need to use (2.4) to obtain the Inverse Laplace transform

In our subsequent discussion, we will denote transformation from the time domain to the complex frequency domain, and vice versa, as

(2.10)

2.2 Properties of the Laplace Transform

1 Linearity Property

The linearity property states that if

have Laplace transforms

* A function is said to be of exponential order if

f t ( )e–σt

0

∞

∫ e–j ωt dt < ∞

f t ( )e–σt

0

∞

∫ dt < ∞

f t( )

f t( ) f t( ) keσ0 t

for all t≥0

<

f t ( )e–σt

0

∞

∫ dt keσ0 t e–σt

0

∞

<

σ σ > 0

Re s{ } = σ σ > 0

f t( ) ⇔F s( )

f 1 ( ) f t , 2 ( ) … f t , , n( )t

Trang 9

Properties of the Laplace Transform

respectively, and

are arbitrary constants, then,

(2.11)

Proof:

Note 1:

It is desirable to multiply by to eliminate any unwanted non-zero values of for

2 Time Shifting Property

The time shifting property states that a right shift in the time domain by units, corresponds to mul-tiplication by in the complex frequency domain Thus,

(2.12)

Proof:

(2.13)

on the right side of (2.13) becomes

3 Frequency Shifting Property

The frequency shifting property states that if we multiply some time domain function by an exponential function where a is an arbitrary positive constant, this multiplication will produce a shift of the s variable in the complex frequency domain by units Thus,

F 1 ( ) F s , 2 ( ) … F s , , n( )s

c 1, ,c 2 … c, n

c 1 f 1 ( ) c t 2 f 2 ( ) … c t n f

n( )t + + + c 1 F 1 ( ) c s 2 F 2 ( ) … c s n F

n( )s

⇔

L c{ 1 f 1 ( ) c t + 2 f 2 ( ) … c t + + n f n( )t } [c 1 f 1 ( ) c t + 2 f 2 ( ) … c t + + n f n( )t ]

t 0

∞

=

c 1 f 1( )t

t 0

∞

∫ e–st dt c 2 f 2( )t

t 0

∞

∫ e–st dt … + c n f n( )t

t 0

∞

∫ e–st dt

=

c 1 F 1 ( ) c s + 2 F 2 ( ) … c s + + n F n( )s

=

a

e–as

f t( –a )u 0(t–a) ⇔e–as F s( )

L f t{ ( –a )u 0(t–a) } 0

0

a

∫ e–st dt f t( –a)

a

∞

∫ e–st dt

+

=

t–a = τ t = τ a+ dt = dτ

f( ) τ

0

∞

∫ e–s(τ a+ )dτ e–as f( ) τ

0

∞

∫ e–sτdτ e–as F s( )

f t( )

e–at

a

Trang 10

(2.14)

Proof:

Note 2:

A change of scale is represented by multiplication of the time variable by a positive scaling factor Thus, the function after scaling the time axis, becomes

4 Scaling Property

Let be an arbitrary positive constant; then, the scaling property states that

(2.15)

Proof:

and letting , we get

Note 3:

Generally, the initial value of is taken at to include any discontinuity that may be present

at If it is known that no such discontinuity exists at , we simply interpret as

5 Differentiation in Time Domain

The differentiation in time domain property states that differentiation in the time domain corresponds

to multiplication by in the complex frequency domain, minus the initial value of at Thus,

(2.16)

Proof:

e–at f t( ) ⇔F s( +a)

L e{ –at f t( ) } e–at f t( )

0

∞

∫ e–st dt f t( )

0

∞

∫ e–(s+a )t dt F s( +a)

t

a

f at( ) 1

a

-F s a

-⎝ ⎠

⎛ ⎞

⇔

L f at{ ( ) } f at( )

0

∞

∫ e–st dt

=

t = τ a⁄

L f at{ ( ) } f( ) τ

0

∞

∫ e–s(τ a⁄ )d τ

a

-⎝ ⎠

⎛ ⎞ 1 a

- f( ) τ

0

∞

∫ e–(s a⁄ ) τd( ) τ 1

a

-F s a

-⎝ ⎠

⎛ ⎞

f t( ) t = 0−

f ' t( ) d

dt - f t( )

= ⇔sF s ( ) f 0– ( )−

L f{ ' t( ) } f ' t( )

0

∞

∫ e–st dt

=

Trang 11

Properties of the Laplace Transform

Using integration by parts where

(2.17)

The time differentiation property can be extended to show that

(2.18)

(2.19)

and in general

(2.20)

To prove (2.18), we let

and as we found above,

Then,

Relations (2.19) and (2.20) can be proved by similar procedures

We must remember that the terms , and so on, represent the initial conditions Therefore, when all initial conditions are zero, and we differentiate a time function times, this corresponds to multiplied by to the power

v u d

∫ = uv–∫u v d

du = f ' t( ) v = e–st u = f t( ) dv = –se–st

L f{ ' t( ) } f t ( )e–st

0−

∞

s f t( )

0−

∞

∫ e–st dt

+ f t ( )e–st

0−

a

e–sa f a ( ) f 0– ( )−

a lim→ ∞ +sF s( ) = 0–f 0( )− +sF s( )

=

d 2

dt 2 - f t( ) ⇔s 2 F s ( ) sf 0– ( ) − –f ' 0( )−

d 3

dt 3 - f t( ) ⇔s 3 F s ( ) s– 2 f 0( )− –sf ' 0( )− –f '' 0( )−

d n

dt n - f t( ) s n F s ( ) s n–1

f 0( )− – –s n–2 f ' 0( )− … f– n–1

0−

( ) –

⇔

g t( ) f ' t( ) d

dt - f t( )

L g{ ' t( ) } = sL g t{ ( )} g 0– ( ) −

L f{ '' t( ) } = sL f{ ' t( )} f– ' 0( )− = s sL f t[ [ ( )] f 0– ( ) − ] f– ' 0( )−

s 2 F s ( ) sf 0– ( )− –f ' 0( )−

=

f 0 ( ) f− , ' 0 ( ) f− , '' 0( )−

f t ( ) n

Trang 12

6 Differentiation in Complex Frequency Domain

This property states that differentiation in complex frequency domain and multiplication by minus one, corresponds to multiplication of by in the time domain In other words,

(2.21)

Proof:

Differentiating with respect to s, and applying Leibnitz’s rule*for differentiation under the integral, we get

In general,

(2.22)

The proof for follows by taking the second and higher-order derivatives of with respect

to

7 Integration in Time Domain

This property states that integration in time domaincorresponds to divided by plus the initial value of at , also divided by That is,

(2.23)

* This rule states that if a function of a parameter is defined by the equation where f is some

known function of integration x and the parameter , a and b are constants independent of x and , and the

par-tial derivative exists and it is continuous, then

f t( ) t

tf t( ) d

ds

-– F s( )

⇔

L f t{ ( ) } F s( ) f t( )

0

∞

∫ e–st dt

a

b

∫

=

f

dα - ∂ (x, α )

α ( )

∂

- x d

a

b

∫

=

d ds

-F s( ) d

ds

- f t( )

0

∞

∫ e–st dt

s

∂∂

0

∞

∫ e–st f t ( )dt –t

0

∞

∫ e–st f t ( )dt

tf t( ) [ ]

0

∞

∫ e–st dt

– = –L tf t[ ( ) ]

=

t n f t( ) ( ) –1 n d n

ds n -F s( )

⇔

s

f( ) τ

∞ –

t

∫ dτ F s( )

s

- f 0

−

( )

s

-+

⇔

Trang 13

Properties of the Laplace Transform Proof:

We express the integral of (2.23) as two integrals, that is,

(2.24)

The first integral on the right side of (2.24), represents a constant value since neither the upper, nor the lower limits of integration are functions of time, and this constant is an initial condition denoted

as We will find the Laplace transform of this constant, the transform of the second integral

on the right side of (2.24), and will prove (2.23) by the linearity property Thus,

(2.25)

This is the value of the first integral in (2.24) Next, we will show that

We let

then,

and

Now,

(2.26)

and the proof of (2.23) follows from (2.25) and (2.26)

f( ) τ

∞ –

t

∫ dτ f( ) τ

∞ –

0

∫ dτ f( ) τ

0

t

+

=

f 0( )−

L f 0{ ( )− } f 0( )−

0

∞

∫ e–st dt f 0( )− e–st

0

∞

∫ dt f 0 ( )e− –st

s

–

-0

∞

f 0 ( ) 0− f 0( )−

s

-–

–

× f 0( )−

s

-=

=

f( ) τ

0

t

∫ dτ F s( )

s

-⇔

g t( ) f( ) τ

0

t

=

g' t( ) = f( ) τ

g 0( ) f( ) τ

0

L g' t{ ( ) } = G s( ) = sL g t{ ( )} g 0– ( )− = G s ( ) 0–

sL g t{ ( ) } = G s( )

L g t{ ( ) } G s( )

s

-=

L f( ) τ

0

t

⎧ ⎫ F s( )

s

-=

Trang 14

8 Integration in Complex Frequency Domain

This property states that integration in complex frequency domain with respect to corresponds to division of a time function by the variable , provided that the limit exists Thus,

(2.27)

Proof:

Integrating both sides from to , we get

Next, we interchange the order of integration, i.e.,

and performing the inner integration on the right side integral with respect to , we get

9 Time Periodicity

The time periodicityproperty states that a periodic function of time with period corresponds to the integral divided by in the complex frequency domain Thus, if we let

trans-form pair

(2.28)

s

t

-t lim→0

f t( )

t

- F s ( ) s d

s

∞

∫

⇔

F s( ) f t( )

0

∞

∫ e–st dt

=

F s ( ) s d

s

∞

0

∞

∫ e–st dt d s

s

∞

∫

=

F s ( ) s d

s

∞

s

∞

∫ d s f t ( ) t d

0

∞

∫

=

s

F s ( ) s d

s

∞

t

-– e–st s

∞

f t ( ) t d

0

∞

t

-e–st d t

0

∞

t

-⎩ ⎭

⎨ ⎬

⎧ ⎫

T

f t( )

0

T

T f t( ) = f t( +nT) n = 1 2 3, , , …

f t( +nT)

f t( )

0

T

∫ e–st dt

1–e–sT

-⇔

Tiêu đề	Signals and systems with matlab applications
Trường học	Orchard Publications
Thể loại	sách

Định dạng
Số trang	20
Dung lượng	545,38 KB