SIGNALS AND SYSTEMS [CHAP. 1 Because a linear shift-invariant system will be stable only if

we see that this system is not stable.

1.22 Determine which of the following systems are invertible:

To test for invertibility, we may show that a system is invertible by designing an inverse system that uniquely recovers the input from the output, or we may show that a system is not invertible by finding two different inputs that produce the same output. Each system defined above will be tested for invertibility using one of these two methods.

( a ) This system is clearly invertible because, given the output y ( n ) , we may recover the input using x ( n ) = 0 . 5 y ( n ) . ( h ) This system is not invertible, because the value of x ( n ) at 11 = 0 cannot be recovered from y ( n ) . For example,

the response of the system to X ( R ) and to x l ( n ) = x ( n )

+

a & n ) will be the same for any a .

(c) Due to the differencing between two successive input values, this system will not be invertible. For example, note that the inputs x ( n ) and x ( n )

+

c will produce the same output for any value of c.

(6) This system corresponds to an integrator and is an invertible system. To show that i t is invertible, we may construct the inverse system, which is

.u(n) = y ( n ) - y ( n - I) To show that this is the inverse system, note that

n - l

(e) Invertibility must hold for complex as well as real-valued signals. Therefore, this system is noninvertible because it discards the imaginary part 01' x ( n ) . One could state, however, that this system is invertible over the set of

( a ) Linearity, shift-invariance, stability, and causality are easily shown to be preserved in a cascade. For example, the response of S I to the input nxl ( n )

+

h x z ( n ) will be a w l ( n )

+

b w 2 ( n ) due to the linearity of S,. With this as the input to S2, the response will be, again by linearity, a y , ( n )

+

hy7(n). Therefore, if both S I and S2 are linear, the cascade will be linear.

CHAP. 11 SIGNALS AND SYSTEMS 35

( b ) If SI and S2 are nonlinear, it is not necessarily true that the cascade will be nonlinear because the second system may undo the nonlinearity of the first. For example, with

although both SI and Sz are nonlinear, the cascade is the identity system and, therefore, is linear.

(c) As in ( b ) , if S I and S2 are shift-varying, it is not necessarily true that the cascade will be shift-varying. For example. if the first system is a modulator.

and the second is a demodulator,

y ( n ) = w ( n ) . e-Inq

the cascade is shift-invariant, even though a modulator and a demodulator are shift-varying. Another example is when S l is an up-sampler

and S2 is a down-sampler

y ( n ) = w ( 2 n )

In this case, the cascade is shift-invariant, and y ( n ) = x ( n ) . However, if the order of the systems is reversed, the cascade will no longer be shift-invariant. Also, if a linear shift-invariant system, such as a unit delay, is inserted between the up-sampler and the down-sampler, the cascade of the three systems will, in general, be shift-varying. and what is its value? What about the last nonzero value?

Because we are convolving two finite-length sequences, the index of the first nonzero value in the convolution is equal to the sum of the indices of the first nonzero values of the two sequences that are being convolved. In this case, the index is n =

-

12, and the value is

y ( - 1 2 ) = x 2 ( - 6 ) = 9 Similarly, the index of the last nonzero value is at n = 48 and the value is

1.25

The convolution of two finite-length sequences will be finite in length. 1s it true that the convolution of a finite-length sequence with an infinite-length sequence will be infinite in length?

SIGNALS AND SYSTEMS [CHAP. I It is not necessarily true that the convolution of an infinite-length sequence with a finite-length sequence will be infinite in length. It may be either. Clearly, if x ( n ) = 6 ( n ) and h ( n ) = (OS)"u(n), the convolution will be an infinite-length sequence. However, it is possible for the finite-length sequence to remove the infinite-length tail of an infinite-length sequence. For example, note that

Therefore, the convolution of x ( n ) = 6 ( n ) - f S ( n - I) with h ( n ) = (OS)"u(n) will be finite in length:

1.26

Find the convolution of the two finite-length sequences:

Shown in the figure below are the sequences x ( k ) and h ( k ) .

Because h ( n ) is equal to zero outside the interval [-3, 31, and x ( n ) is zero outside the interval [ l , 51, the convolution y ( n ) = x ( n )

*

^{h ( n )} is zero outside the interval 1-2, 81.

One way to perform the convolution is to use the slide rule approach. Listing x ( k ) and h ( - k ) across two pieces of paper, aligning them at k = 0 , we have the picture as shown below (the sequence h ( - k ) is in front).

Forming the sum of the products x(k)h(-k), we obtain the value of y ( n ) at time n = 0 , which is y ( 0 ) = 2 . Shifting h(-k) to the left by one, multiplying and adding, we obtain the value of y ( n ) at n = -1, which is y ( - I ) = 2.

Shifting one more time to the left, forming the sum of products, we find y ( - 2 ) = 1, which is the last nonzero value of y ( n ) for n < 0 . Repeating the process by shifting h ( - k ) to the right, we obtain the values of y ( n ) for n > 0 , which are

Another way to perform the convolution is to use the fact that

x ( n )

*

S(n - no) = x ( n - n o ) Writing h ( n ) as

CHAP. I ] SIGNALS AND SYSTEMS we may evaluate y ( n ) as follows

y ( n ) = 2 x ( n

+

^{3 ) - 2 x ( n}

+

^{1 )}

+

2 x ( n - I) - 2 x ( n - 3 ) Making a table of these shifted sequences.

and adding down the columns, we obtain the sequence y ( n ).

1.27 Derive a closed-form expression for the convolution of

x ( n )

and

h ( n )

where

I N-6

x ( n )

= ( 6 )

u ( n ) h ( n )

⁼

( f ) " u ( n

-

3)

Because both sequences are infinite In length. it is easier to evaluate the convolution sum directly:

Note that because x ( n ) = 0 for n < 0 and h ( n ) = 0 for n < 3 , y ( n ) will be equal to zero for n < 3. Substituting x ( n ) and h ( n ) into the convolution sum, we have

Due to the step u ( k ) , the lower limit on the sum may be changed to k = 0, and because u ( n - k - 3) is zero for k > n - 3 , the upper limit may be changed to k = n - 3 . Thus. for n 2 3 the convolution sum becomes

Using the geometric series to evaluate the sum, we have

1.28 A linear shift-invariant system has

a

unit sample response

Find the output if the input is

x ( n )

=

-n3"u(-n)

Shown below are the sequences x ( n ) and h ( n ) .

3 8

SIGNALS AND SYSTEMS [CHAP. I Because x ( n ) is zero for n > -1, and h ( n ) is equal to zero for n > - 1 , the convolution will be equal to zero for n z - 2 . Evaluating the convolution sum directly, we have

Because u ( - k ) = 0 fork > 0 and u ( - ( n - k ) - 1) = 0 fork < n

+

I, the convolution sum becomes

With the change of variables m = -k, and using the series formulas given in Table I -I, we have

Let us check this answer for a few values of n using graphical convolution. Time-reversing x ( k ) , we see that h ( k ) and x ( - k ) do not overlap for any k and, thus, y ( 0 ) = 0 . In fact, it is not until we shift x ( - k ) to the left by two that there is any overlap. With x ( - 2 - k ) and h ( k ) overlapping at one point, and the product being equal to

i,

it follows that y ( - 2 ) =

4.

Evaluating the expression above for y ( n ) above at index n = -2, we obtain the same result. For n = -3, the sequences x ( - 3 - k ) and h ( k ) overlap at two points, and the sum of the products gives y ( - 3 ) =

f + $

⁼

$,

which, again, is the same as the expression above.

1.29

If the response of a linear shift-invariant system to a unit step (i.e., the step response) is

Dans le document Schaum's Outline of Theory and Problems of Digital Signal Processing (Page 44-48)