Tài liệu Signals and Systems P2 pdf

Sketch and label each of the following signals... 1-22, represent each of the following signals by a graph and by a sequence of numbers... Let x r be the complex exponential signal w

SIGNALS AND SYSTEMS [CHAP 1

( a ) x(r - 2 ) is sketched in Fig 1-18(a)

( 6 ) x ( 2 0 is sketched in Fig 1-18(b)

( c ) x(t/2) is sketched in Fig 1-18(c)

( d ) X ( - t ) is sketched in Fig 1-1Nd)

( c )

Fig 1-18

1.2 A discrete-time signal x [ n ] is shown in Fig 1-19 Sketch and label each of the following signals

( a ) x [ n - 21; ( b ) x [ 2 n ] ; ( c ) x [ - n ] ; ( d ) x [ - n + 21

Fig 1-19

CHAP 11 SIGNALS AND SYSTEMS

( a ) x[n - 21 is sketched in Fig 1-20(a)

( b ) x[2n] is sketched in Fig 1-20(b)

( c ) x [ - n ] is sketched in Fig 1-2Nc)

( d ) x [ - n + 21 is sketched in Fig 1-2Nd)

(4

Fig 1-20

13 Given the continuous-time signal specified by

x ( t ) = (A - It' otherwise - 1 l f l l

determine the resultant discrete-time sequence obtained by uniform sampling of x(t)

with a sampling interval of ( a ) 0.25 s, ( b ) 0.5 s, and (c) 1.0 s

It is easier to take the graphical approach for this problem The signal x(t) is plotted in Fig 1-21(a) Figures 1-21(b) to ( d l give plots of the resultant sampled sequences obtained for the three specified sampling intervals

( a ) T, = 0.25 s From Fig 1-21(b) we obtain

x [ n ] = ( ,0,0.25,0.5,0.75,1,0.75,0.5,0.25,0, .)

T

( b ) T, = 0.5 s From Fig 1-21(c) we obtain

x [ n ] = { , 0 , 0 5 , 1 , 0 5 , 0 , I

T

SIGNALS AND SYSTEMS [CHAP 1

(4

Fig 1-21

( c ) T, = 1 s From Fig 1-21(d) we obtain

x [ n ] = ( , O , 1,O .) = S[nl

1.4 Using the discrete-time signals x , [ n ] and x , [ n ] shown in Fig 1-22, represent each of

the following signals by a graph and by a sequence of numbers

( a ) y J n 1 = x , [ n l + x , [ n l ; ( b ) y , [ n I = 2 x , [ n l ; ( c ) y,[nI = x J n I x J n l

Fig 1-22

CHAP 11 SIGNALS AND SYSTEMS

( a ) y , [ n ] is sketched in Fig 1-23(a) From Fig 1-23(a) we obtain

(b) y 2 [ n ] is sketched in Fig 1-23(b) From Fig 1-23(b) we obtain

(c) y J n ] is sketched in Fig 1-23(c) From Fig 1-23(c) we obtain

(d

Fig 1-23

Using Eqs (1.5) and (1.6), the even and odd components of the signals shown in Fig 1-24 are sketched in Fig 1-25

SIGNALS AND SYSTEMS [CHAP 1

(4

Fig 1-24

1.6 Find the even and odd components of x ( r ) = e J '

Let x , ( r ) and x , ( I ) be the even and odd components of ei', respectively

eJ' = x , ( I ) + x , ( I )

From Eqs ( 1 5 ) and ( 1 6 ) and using Euler's formula, we obtain

x,( I ) = $ ( e J r + e - J ' ) = cos I

x , , ( I ) = f ( e i ' - e - j ' ) = j s i n t

Show that the product of two even signals o r of two odd signals is an even signal and that the product of an even and an odd signaI is an odd signal

Let x ( t ) = x l ( t ) x 2 ( t ) If X J I ) and x 2 ( l ) are both even, then

x ( - l ) = x , ( - I ) X , ( - t ) = x I ( I ) x 2 ( t ) = x ( t )

and x ( t ) is even If x , ( t ) and x 2 ( t ) are both odd, then

x ( - I ) = x , ( - I ) x , ( - I ) = - x , ( t ) [ - x 2 ( t ) ] = x 1 ( t ) x 2 ( t ) = x ( t )

and x ( t ) is even If x , ( t ) is even and x 2 ( f ) is odd, then

and X ( I ) is odd Note that in the above proof, variable I represents either a continuous or a discrete variable

CHAP 11 SIGNALS AND SYSTEMS

(4

Fig 1-25

SIGNALS AND SYSTEMS

1.8 Show that

[CHAP 1

( a ) If x(t) and x [ n ] are even, then

( b ) If x(t) and x[n] are odd, then

x(0) = 0 and x[O] = O

k

/ a ~ ( r ) dr = 0 and x x [ n ] = O

( a ) We can write

Letting t = - A in the first integral on the right-hand side, we get

Since x ( t ) is even, that is, x ( - A ) = x ( A ) , we have

Hence,

Similarly,

Letting n = - m in the first term on the right-hand side, we get

Since x [ n ] is even, that is, x [ - m ] = x [ m ] , we have

Hence,

(1.75a)

( I 75b)

( 6 ) Since x ( t ) and x [ n ] are odd, that is, x( - t ) = - x ( t ) and x [ - n ] = - x [ n ] , we have

X ( - 0 ) = - x ( O ) and x [ - 0 1 = - x [ O ]

CHAP 11 SIGNALS AND SYSTEMS

Hence,

Similarly,

and

in view of Eq (1.76)

( t ) = , j @ d

is periodic and that its fundamental period is 27r/00

By Eq (1.7), x(t) will be periodic if

e i @ d t + TI = e i w d

Since

e i w ~ ( r + T ) = e i q r e i q , T

we must have

eimoT = 1 (1.78)

If w, = 0, then x(t) = 1, which is periodic for any value of T If o0 # 0, Eq (1.78) holds if

27T

o o T = m 2 r or T = m - m = positive integer

a 0

Thus, the fundamental period To, the smallest positive T, of x(t) is given by 2 r / o o

1.10 Show that the sinusoidal signal

x ( t ) = cos(w,t + 8 )

is periodic and that its fundamental period is 27r/wo

The sinusoidal signal x(l) will be periodic if

cos[o,(t + T) + 81 = w s ( o o t + 8)

We note that

cos[w,(t + T) + 81 = cos[oot + 8 + woT] = cos(oot + 8 )

SIGNALS A N D SYSTEMS [CHAP 1

2 7 w0T=m2.rr or T = m - m = positive integer

*o Thus the fundamental period To of x ( r ) is given by 2.rr/wo

x [ n ] = e ~ " ~ "

is periodic only if fl0/2.rr is a rational number

By Eq ( 1 9 ) , x[n] will be periodic if

,iflo(" + N l = , i n , , n , i ~ h p = , i n o n

or

e i n ~ N = 1

Equation ( 1 7 9 ) holds only if

f l o N = m 2 ~ m = positive integer

o r

2.rr N Thus, x[n] is periodic only if R0/27r is a rational number

1.12 Let x ( r ) be the complex exponential signal

with radian frequency wo and fundamental period To = 2.rr/oo Consider the discrete-time sequence x [ n ] obtained by uniform sampling of x ( t ) with sampling interval Ts That is,

x [ n ] = x ( n T , ) = e J " u n T

Find the condition on the value of T, so that x [ n ] is periodic

If x[n] is periodic with fundamental period N,,, then

, i o u ( n + N , , ) T , = , i w ~ n T , , i w u N , J ' , = ejwun-l;

Thus, we must have

T, m

- = - - - rational number

To No

Thus x [ n ] is periodic if the ratio T,/T,, of the sampling interval and the fundamental period of

x ( t ) is a rational number

Note that the above condition is also true for sinusoidal signals x ( t ) = cos(o,,t + 8 )

CHAP 11 SIGNALS AND SYSTEMS

1.13 Consider the sinusoidal signal

x ( t ) = cos 15t

Find the value of sampling interval T, such that x [ n ] = x ( n T , ) is a periodic sequence

Find the fundamental period of x [ n ] = x ( n T , ) if TT = 0 1 ~ seconds

The fundamental period of x ( t ) is To = 2*rr/wo = 2 7 / 1 5 By Eq (1.81), x [ n ] = x ( n T s ) is periodic if

where m and No are positive integers Thus, the required value of T, is given by

Substituting T, = 0 1 ~ = ~ / 1 0 in Eq (1.821, we have

Thus, x [ n ] = x ( n T , ) is periodic By Eq (1.82)

The smallest positive integer No is obtained with m = 3 Thus, the fundamental period of

x [ n l = x ( 0 l ~ n ) is N , = 4

.4 Let x , ( t ) and x , ( t ) be periodic signals with fundamental periods T, and T 2 , respec- tively Under what conditions is the sum x ( t ) = x , ( t ) + x 2 ( t ) periodic, and what is the fundamental period of x( t ) if it is periodic?

Since x , ( t ) and x , ( t ) are periodic with fundamental periods T I and T,, respectively, we have

x l ( t ) = x , ( t + T I ) = x , ( t + m T , ) m = positive integer

x 2 ( t ) = x 2 ( t + T 2 ) = x 2 ( f + k T 2 ) k = positive integer Thus,

In order for x ( t ) to be periodic with period T , one needs

Thus, we must have

m T , = kT2 = T

T I k

- - - - = rational number

T2 m

In other words, the sum of two periodic signals is periodic only if the ratio of their respective periods can be expressed as a rational number Then the fundamental period is the least

SIGNALS AND SYSTEMS [CHAP 1

common multiple of T, and T2, and it is given by Eq (1.84) if the integers m and k are relative prime If the ratio T,/T, is an irrational number, then the signals x,(t) and x,(t) do not have a common period and x ( t ) cannot be periodic

1.15 Let x,[n] and x2[n] be periodic sequences with fundamental periods N , and N2, respectively Under what conditions is the sum x[n] =x,[n] +x2[n] periodic, and what

is the fundamental period of x[n] if it is periodic?

Since x,[n] and x2[n] are periodic with fundamental periods N, and N2, respectively, we have

x I [ n ] = x I [ n + N,] = x , [ n + m N , ] m = positive integer x2[n] =x,[n + N,] =x,[n + kN,] k = positive integer Thus,

~ [ n ] = x , [ n + m N , ] + x 2 [ n + kN,]

In order for x[n] to be periodic with period N, one needs

x[n + N ] = x , [ n + N ] + x 2 [ n + N ] = x , [ n + mN,] +x,[n + kN2]

Thus, we must have

mN, = kN2 = N Since we can always find integers m and k to satisfy Eq (1.861, it follows that the sum of two periodic sequences is also periodic and its fundamental period is the least common multiple of

N, and N,

1.16 Determine whether or not each of the following signals is periodic If a signal is periodic, determine its fundamental period

2TT

( a ) x ( t ) = cos ( b ) x ( t ) = s i n p t

3

( c ) x ( t ) = c o s - I +sin -t

( g ) x[n] = ej("/4)" ( h ) x [ n ] = c o s f n

( i ) x[n] = cos -n + sin -n

8

x ( t ) is periodic with fundamental period T , = 27r/w0 = 27r

x(r) is periodic with fundamental period TO = 27r/o,, = 3

( c ) x ( t ) = cos I + sin -t = x , ( t ) + x 2 ( t )

where x,(t) = cos(7r/3)r = cos w,t is periodic with T, = 27r/w, = 6 and x 2 ( t ) =

s i n ( ~ / 4 ) t = sin w2t is periodic with T2 = 21r/w2 = 8 Since T,/T, = = is a rational number, x ( t ) is periodic with fundamental period To = 4T, = 3T2 = 24

CHAP 11 SIGNALS AND SYSTEMS

( d l x(t) = cos r + sin f i r = x , ( r ) +x2(r)

where x,(t) = cos r = cos o , t is periodic with TI = 27r/01 = 27r and x2(t) = sin f i t = sin w2t is periodic with T2 = 27r/02 = fir Since T,/T2 = fi is an irrational number,

x ( t ) is nonperiodic

(e) Using the trigonometric identity sin2 0 = t(l - cos 201, we can write

I

where x,(t) = $ is a dc signal with an arbitrary period and x2(t) = - $ cos2r = - cos 0 2 t

is periodic with T2 = 2n/w2 = 7 Thus, x(t) is periodic with fundamental period To = T

7T

( f ) x ( t ) = e j t ( r / 2 ) r - 11 = e - j e j ( r / 2 ) r = -I 'e j w d , Wo = Ir

L

x(t) is periodic with fundamental period To = 27r/w0 = 4

Since R0/27r = $ is a rational number, x[nl is periodic, and by Eq (1.55) the fundamen- tal period is No = 8

x[n] = cos f n = cos n o n , R o = $

Since n0/27r = 1 / 8 ~ is not a rational number, x[n] is nonperiodic

x[n] = cos -n + sin -n = x,[n] + x2[n 1

where

x2[n] = sin -n = cos f12n + 0, = -

Since R , / 2 ~ r = (= rational number), xl[n] is periodic with fundamental period N, = 6,

and since R2/27r = $ ( = rational number), x2[n] is periodic with fundamental period N2 = 8 Thus, from the result of Prob 1.15, x[n] is periodic and its fundamental period is given by the least common multiple of 6 and 8, that is, No = 24

Using the trigonometric identity cos2 8 = i ( l + cos28), we can write

x [ n ] = cost -n = - + - cos -n = x , [ n ] + x 2 [ n ]

where x,[n] = $ = $(l)" is periodic with fundamental period Nl = 1 and x2[n] =

1

cos(a/4)n = cos R 2 n , Q 2 = ~ / 4 Since R2/27r = ( = rational number), x2[n] is periodic with fundamental period N2 = 8 Thus, x[n] is periodic with fundamental period

No = 8 (the least common multiple of N, and N,)

1.17 Show that if x ( t + T ) = x ( t ) , then

for any real a, p, and a

SIGNALS A N D SYSTEMS [CHAP 1

If x(t + T ) = x ( t ) , then letting t = 7 - T , we have

X ( T - T + T ) = x ( r ) = x ( T - T )

and

Next, the right-hand side of Eq ( 1 8 8 ) can be written as

By E q ( 1 8 7 ) we have

( t ) d = / x ( t ) dt

a + T

Thus

1.18 Show that if x ( t ) is periodic with fundamental period T o , then the normalized average

power P of x ( t ) defined by Eq ( 1 1 5 ) is the same as the average power of x ( 0 over any interval of length T , , that is,

By Eq ( 1.15)

1

P = lim - /T'2 1 x ( t ) 1' dt

T-.r: T - 7 / 2

Allowing the limit to be taken in a manner such that T is an integral multiple of the fundamental period, T = kT,, the total normalized energy content of x ( t ) over an interval of

length T is k times the normalized energy content over one period Then

1.19 The following equalities are used on many occasions in this text Prove their validity

Then

SIGNALS AND SYSTEMS

N- 1

a S = a C a n = a + a Z + a " + - + a N

n = O

Subtracting Eq (1.95) from Eq (1.941, we obtain

Hence if a # 1, we have

If a = 1, then by Eq (1.94)

( 6 ) For la1 < 1, lim a N = 0 Then by Eq (1.96) we obtain

N - m

- -

x a n = lim x a n = lim - -

(c) Using Eq (1.911, we obtain

(d) Taking the derivative of both sides of Eq (1.91) with respect to a, we have

and

Hence,

( a ) x ( t ) = e - " ' u ( t ) , a > O ( b ) x ( t ) = A c o s ( w , t + 8 )

(c) x ( t ) = t u ( t ) ( d l x [ n ] = ( - 0.5)"u[n]

( e ) x [ n l = u[nl (f x [ n ] = 2ej3"

SIGNALS AND SYSTEMS [CHAP 1

Thus, x ( t ) is an energy signal

( b ) The sinusoidal signal x ( t ) is periodic with To = 2 7 r / o o Then by the result from Prob 1.18, the average power of x ( t ) is

Thus, x ( t ) is a power signal Note that periodic signals are, in general, power signals

Thus, x ( t ) is neither an energy signal nor a power signal

( d ) By definition ( 1 1 6 ) and using Eq (1.91), we obtain

Thus, x [ n l is an energy signal

( e ) By definition ( 1 1 7 )

P = lim - C l x b 1 I 2

N + % 2 N + 1 , = - N

N + C = 2 N + 1 ,,=,, ~ + m2 N + 1 2

Thus, x [ n ] is a power signal

( f Since I x [ n ] l = I2eiJnI = 2IeJ3"l = 2 ,

N-+= 2 N + l n = - N

2'

N - m 2 N + 1 .= - N

1

= lim - 4 ( 2 N + 1 ) = 4 < m

~ + m 2 N + 1 Thus, x [ n ] is a power signal

BASIC SIGNALS

1.21 Show that

CHAP 11 SIGNALS AND SYSTEMS

Let T = - t Then by definition ( 1 1 8 )

Since T > 0 and 7 < 0 imply, respectively, that t < 0 and r > 0, we obtain

which is shown in Fig 1-26

Fig 1-26

1.22 A continuous-time signal A t ) is shown in Fig 1-27 Sketch and label each of the following signals

( a ) x ( t ) u ( l - t ) ; ( b ) x ( t ) [ u ( t ) - u ( t - I)]; (c) x ( t ) H t -

Fig 1-27

(a) By definition ( 1 1 9 )

and x(r)u(l - t ) is sketched in Fig 1-28(a)

( 6 ) By definitions (1.18) and (1.19)

O < t l l

u - ( t - 1 =

otherwise and x ( t ) [ u ( r ) - u(t - I ) ] is sketched in Fig 1-28(b)