Tài liệu Digital Signal Processing P2 doc

This sequence is shown in the figure below: 1.9 The power in a real-valued signal xn is defined as the sum of the squares of the sequence values: Suppose that a sequence xn has an even

CHAP 11 SIGNALS AND SYSTEMS 2

(a) The sequence x(n), illustrated in Fig 1-8(a), is a linearly decreasing sequence that begins at index n = 0 and ends at index n = 5 The first sequence that is to be sketched, yl(n) = x(4 - n), is found by shifting x(n) by four and time-reversing Observe that at index n = 4, yl(n) is equal to x(0) Therefore, yl(n) has a value of 6

at n = 4 and decreases linearly to the left (decreasing values of n) until n = - 1, beyond which y (n) = 0 The sequence y (n) is shown in Fig 1-8(b)

Fig 1-8 Performing signal manipulations

(b) The second sequence, y2(n) = x(2n - 3), is formed through the combination of time-shifting and down- sampling Therefore, y&~) may be plotted by first shifting x(n) to the right by three (delay) as shown in

Fig 1-8(c) The sequence y2(n) is then formed by down-sampling by a factor of 2 (i.e., keeping only the even index terms as indicated by the solid circles in Fig 1-8(c)) A sketch of yn(n) is shown in Fig I-8(d) (c) The third sequence, y3(n) = x(8 - 3n), is formed through a combination of tirne-shifting, down-sampling, and time-reversal To sketch y3(n) we begin by plotting x(8 - n), which is formed by shifting x(n) to the left by

eight (advance) and reversing in time as shown in Fig 1 -8(e) Then, y3(n) is found by extracting every third

sample of x(8 - n), as indicated by the solid circles, which is plotted in Fig 1-8( f )

(4 Finally, y4(n) = x(n2 - 2n + 1) is formed by a nonlinear transformation of the time variable n This sequence may be easily sketched by listing how the index n is mapped First, note that if n 2 4 or n 5 -2, then

n2 - 2n + 1 2 9 and, therefore, y4(n) = 0 For - I 5 n 5 3 we have

The sequence y4(n) is sketched in Fig 1-8(g)

1.8 The notation ~ ( ( n ) ) ~ is used to define the sequence that is formed as follows:

~ ( ( n ) ) ~ = x(n modulo N) where (n modulo N) is the positive integer in the range [0, N - 11 that remains after dividing n by N For example, ((3))g = 3, ((12))g = 4, and ((-6))d = 2 If x(n) = (i)%in(nn/2)u(n), make a sketch of

(a) x((n))3 and ( b ) x((n - 2))3

(a) We begin by noting that ((n))3, for any value of n, is always an integer in the range [ O , 21 In fact, because ((n))3 = ((n + 3k)h for any k ,

Therefore, x((n))3 is periodic with a period N = 3 It thus follows t h a t ~ ( ( n ) ) ~ is formed by periodically repeating the first three values of x(n) as illustrated in the figure below:

(b) The sequence x((n - 2))3 is also periodic with a period N = 3, except that the signal is shifted to the right by

no = 2 compared to the periodic sequence in part (a) This sequence is shown in the figure below:

1.9 The power in a real-valued signal x(n) is defined as the sum of the squares of the sequence values:

Suppose that a sequence x(n) has an even part x,(n) equal to

CHAP I] SIGNALS AND SYSTEMS 23

If the power in x(n) is P = 5, find the power in the odd part, x,(n), of x(n)

This problem requires finding the relationship between the power in x ( n ) and the power in the even and odd parts

By definition, x ( n ) = x , ( n ) + x,(n) Therefore,

Note that x,(n)x,(n) is the product of an even sequence and an odd sequence and, therefore, the product is odd Because the sum for all n of an odd sequence is equal to zero,

Thus, the power in x ( n ) is

which says that the power in x ( n ) is equal to the sum of the powers in its even and odd parts Evaluating the power

in the even part of x ( n ) , we find

2n

PC = ): (ynl = - I + 2 ): ( f ) = f

Therefore, with P = 5 we have

10

P, = 5 - P, = T

1.10 Consider the sequence

Find the numerical value of

Compute the power in x(n),

W

If x(n) is input to a time-varying system defined by y(n) = nx(n), find the power in the output signal (i.e., evaluate the sum)

This is a direct application of the geometric series

With the substitution of -n for n we have

Therefore, it follows from the geometric series that

( h ) To find the power in x ( n ) we must evaluate the sum

Replacing n by -n and using the geometric series, this sum becomes

(c) Finally, to find the power in y ( n ) = n x ( n ) we must evaluate the sum

In Table 1- I there is a closed-form expression for the sum

but not for C:,n2an However, we may derive a closed-form expression for this sum as follow^.^ Differenti- ating both sides of Eq (1.19) with respect to a , we have

Therefore we have the sum

Using this expression to evaluate Eq (1.18) we find

1.11 Express the sequence

I 1 n = O

2 n = l .r(n) =

3 n = 2

0 else

as a sum of scaled and shifted unit steps

In this problem, we would like to perform a signal decomposition, expressing x ( n ) as a sum of scaled and shifted

unit steps There are several ways to derive this decomposition One way is to express x ( n ) as a sum of weighted

and shifted unit samples,

x ( n ) = S(n) + 2S(n - I) + 3S(n - 2 )

and use the fact that a unit sample may be written as the difference of two steps as follows:

Therefore, x ( n ) = u ( n ) - u(n - I ) + 2[u(n - I) - u(n - 2)] + 3[u(n - 2 ) - u(n - 3)]

which gives the desired decomposition:

"his method is very useful and should be remembered

CHAP I] SIGNALS AND SYSTEMS 25

Another way to derive this decomposition more directly is as follows First, we note that the decomposition should begin with a unit step, which generates a value of I at index n = 0 Because x(n) increases to a value of 2 at n = 1,

we must add a delayed unit step u(n - 1) At n = 2, x(n) again increases in amplitude by 1, so we add the delayed unit step u(n - 2) At this point, we have

Thus, all that remains is to bring the sequence back to zero for n > 3 This may be done by subtracting the delayed unit step 3u(n - 3), which produces the same decomposition as before

Discrete-Time Systems

1.12 For each of the systems below, x ( n ) is the input and y(n) is the output Determine which systems are homogeneous, which systems are additive, and which are linear

(a) If the system is homogeneous,

y(n) = T[cx(n)] = cT[x(n)]

for any input x(n) and for all complex constants c The system y(n) = log(x(n)) is not homogeneous because the response of the system to xl(n) = cx(n) is

which is not equal to c log(x(n)) For the system to be additive, if yl(n) and y2(n) are the responses to the inputs and xz(n), respectively, the response to x(n) = x l ( n ) + x2(n) must be y(n) = yl(n) + y2(n) For this system we have

T[xl(n) + x h N = log[x~(n) + x2(n)l # log[x~(n)l+ log[x2(n)l Therefore, the system is not additive Finally, because the system is neither additive nor homogeneous, the system is nonlinear

(b) Note that if y(n) is the response to x(n)

the response to xl(n) = cx(n) is

which is not the same as y1 (n) Therefore, this system is not homogeneous Similarly, note that the response to x(n) = x,(n) + x2(n) is

which is not equal to yl(n) + y2(n) Therefore, this system is not additive and, as a result, is nonlinear

This system is homogeneous, because the response of the system to xl(n) = cx(n) is

The system is clearly, however, not additive and therefore is nonlinear

Let y,(n) and yz(n) be the responses of the system to the inputs x,(n) and x2(n), respectively The response to the input

x(n) = axl(n) + bxz(n) y(n) = x(n) sin ( y ) = [axl (n) + bx2(n)] sin r; - 1

Thus, it follows that this system is linear and, therefore, additive and homogeneous

Because the real part of the sum of two numbers is the sum of the real parts, if y,(n) is the response of the system toxl(n), and yz(n) is the response to x2(n), the response to y(n) = yl(n) + yz(n) is

Therefore the system is additive It is not homogeneous, however, because

unless c is real Thus, this system is nonlinear

For an input x(n), this system produces an output that is the conjugate symmetric part of x(n) If c is a complex constant, and if the input to the system is xl(n) = cx(n), the output is

Therefore, this system is not homogeneous This system is, however, additive because

1.13 A linear system is one that is both homogeneous and additive

(a) Give an example of a system that is homogeneous but not additive

(b) Give an example of a system that is additive but not homogeneous

There are many different systems that are either homogeneous or additive but not both One example of a system that is homogeneous but not additive is the following:

x(n - I)x(n)

~ ( n ) =

x(n + I )

Specifically, note that if x(n) is multiplied by a complex constant c, the output will be

cx(n-l)cx(n) x ( n - I ) x ( n )

which is c times the response to x(n) Therefore, the system is homogeneous On the other hand, it should be clear that the system is not additive because, in general,

{xl(n - 1 ) + X Z ( ~ - l ) J ( x ~ ( n ) + xz(n)I x ~ ( n - l ) x ~ ( n ) xdn - l)xz(n)

x ~ ( n + 1) + x A n + I ) + x,(n + 1 ) +

xz(n + 1)

CHAP I ] SIGNALS AND SYSTEMS

An example of a system that is additive but not homogeneous is

Additivity follows from the fact that the imaginary part of a sum of complex numbers is equal to the sum of imaginary parts This system is not homogeneous, however, because

1.14 Determine whether or not each of the following systems is shift-invariant:

(a) Let y ( n ) be the response of the system to an arbitrary input x ( n ) To test for shift-invariance we want to compare the shifted response y ( n - n o ) with the response of the system to the shifted input r(n - nu) With

we have for the shifted response

Now, the response of the system to x l ( n ) = x ( n - n o ) is

Because y l ( n ) = y ( n - n o ) , the system is shifl-invariant

where f ( n ) is a shift-varying gain Systems of this form are always shift-varying provided f ( n ) is not a constant

To show this, assume that f ( n ) is not constant and let n I and nz be two indices for which f ( n , ) # f ( n z ) With

an input r l ( n ) = S(n - n l ) , note that the output y l ( n ) is

If, on the other hand, the input is x 2 ( n ) = 6 ( n - n 2 ) , the response is

Although .t.,(n) and x Z ( n ) differ only by a shift, the responses y l ( n ) and y 2 ( n ) differ by a shift and a change in amplitude 'Therefore, the systcm is shift-varying

(c) Let

be the response of the system to an arbitrary inpul r(n) The response of the system to the shifted input

r l ( n ) = x(n - no) is

Because this is equal to v ( n - n o ) , the system is shift-invariant

(d) This system is shift-varying, which may be shown with a simple counterexample Note that if x(n) = S(n), the

response will be y(n) = 6(n) However,ifxl(n) = 6(n-2) the response will be yl(n) = xl(n2) = 6(n2-2) = 0,

which is not equal to y(n - 2) Therefore, the system is shift-varying

(e) With y(n) the response to x(n), note that for the input xl(n) = x(n - N), the output is

which is the same as the response tox(n) Because yl (n) # y(n- N), ingeneral, this system isnot shift-invariant

(f) This system may easily be shown to be shift-varying with a counterexample However, suppose we use the direct approach and let x(n) be an input and y(n) = x(-n) be the response If we consider the shifted input,

x l (n) = x(n - no), we find that the response is

However, note that if we shift y(n) by no,

which is not equal to yl (n) Therefore, the system is shift-varying

1.15 A linear discrete-time system is characterized by its response h k ( n ) to a delayed unit sample S(n - k )

For each linear system defined below, determine whether or not the system is shift-invariant

( a ) hk(n) = ( n - k ) u ( n - k )

S(n - k - 1) k even

(a) Note that hk(n) is a function of n - k This suggests that the system is shift-invariant To verify this, let y(n)

be the response of the system to x(n):

The response to a shifted input, x(n - no), is

With the substitution 1 = k - no this becomes

From the expression for y(n) given in Eq ( 1 2 4 , we see that

which is the same as yl(n) Therefore, this system is shift-invariant

CHAP I] SIGNALS AND SYSTEMS 29

varying Let us see if we can tind an example that demonstrates that it is a shift-varying system For the input

~ ( 1 1 ) = 6(11), the response is

Because g l ( n ) # y(n - I ), the system is shift-varying

( c ) Finally, for the last system, we see that although h k ( n ) is a function of n - k fork even and a function of (n - k)

f o r k odd,

11k(n) # h k - ~ ( n - 1)

In other words, the response of the system to 6(n - k - 1) is not equal to the response of the system to 6(n - k) delayed by 1 Therefore this system is shift-varying

1.16 Let Tr.1 be a linear system, not necessarily shift-invariant, that has a response h k ( n ) to the input 6 ( n - k) Derive a test in terms of k k ( n ) that allows one to determine whether or not the system is stable and whether

or not the system is causal

(a) The response of a linear system to an input ~ ( n ) is

Therefore the output may be hounded as follows:

If x(n) is bounded, Ix(n)l 5 A < W ,

l y w i i A 2 I M ~ ) I

As a result if

the output will be bounded, and the system is stable Equation (1.23) is a necessary condition for stability

To establish the sufficiency of this condition, we will show that if this summation is not finite, we can find a bounded input that will produce an unbounded output Let us assume that h k ( n ) is bounded for all k and n

[otherwiue the system will be unstable because the response to the bounded input S(n - k) will be unbounded] With h i ( t l ) bounded for all k and n, suppose that the sum in Eq (1.23) is unbounded for some n, say n = no Let

x ( n ) = s g n ( h , ( n ~ ) l that is,

For this Input, the response at time n = no is

which, by assumption, is unbounded Therefore, the system is unstable and we have established the sufficiency

of the condition given in Eq (1.23)

( b ) Let us now consider causality For an input x ( n ) , the response is as given in Eq (1.22) In order for a system

to be causal, the output y ( n ) at time no cannot depend on the input x ( n ) for any n > no Therefore, Eq ( 1 2 2 )

must be of the form

,I

y ( n ) = x hk(n)x(k) k=-m

This, however, will be true for any x ( n ) if and only if

which is the desired test for causality

Determine whether o r not the systems defined in Prob 1 .I5 are (a) stable and (b) causal

(a) For the first system, h k ( n ) = ( n - k ) u ( n - k ) , note that h k ( n ) grows linearly with n Therefore, this system cannot be stable For example, note that if x ( n ) = S(n), the output will be

which is unbounded Alternatively, we may use the test derived in Prob 1 .I6 to check for stability Because

this system is unstable On the other hand, because h , ( n ) = 0 for n < k , this system is causal

( b ) For the second system, h k ( n ) = S(2n - k ) , note that h l ( n ) has, at most, one nonzero value, and this nonzero

value is equal to I Therefore,

for all n , and the system is stable However, the system is not causal To show this, note that if x ( n ) = &(n - 2 ) ,

the response is

y ( n ) = h 2 ( n ) = 6(2n - 2 ) = &(n - I)

Because the system produces a response before the input occurs, it is noncausal

(c) For the last system, note that

= x Su(n - k ) = 1 5

A= A=-.u

A add h odd

which is unbounded Therefore, this system is unstable Finally, because h k ( n ) = 0 for n < k , the system is

causal

Consider a linear system that has a response t o a delayed unit step given by

That is, s k ( n ) is the response of the linear system to the input x ( n ) = u ( n - k ) Find the response of this system to the input x ( n ) = 6 ( n - k ) , where k is a n arbitrary integer, and determine whether o r not this system is shift-invariant, stable, o r causal

Because this system is linear, we may find the response, h k ( n ) , to the input &(n - k ) as follows With &(n - k ) =

u ( n - k ) - u(n - k - I), using linearity it follows that

which is shown below: