Because 0 . 1 2 5 ~ n/8, and = cos x(n) is periodic with period N = 16.

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Here we have the sum of two periodic signals,

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with the period of the first signal being equal to Nl = 24, and the period of the second, N2 = 36. Therefore, the period of the sum is N1N2 - (24)(36) - (24)(36) - 72 N = gcd(N1, N2) gcd(24,36) 12

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CHAP. 11

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SIGNALS AND SYSTEMS

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(c) In order for this sequence to be periodic, we must be able to find a value for N such that

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+ 0 . h ) = sin(a + 0.2 ( n + N ) )

The sine function is periodic with a period of 2 ~ Therefore, 0.2N must be an integer multiple of 271.. However, . because H is an irrational number, no integer value of N exists that will make the equality true. Thus, this sequence is aperiodic.

(4 Here we have the product of two periodic sequences with periods N I = 32 and N2 = 34. Therefore, the fundamental veriod is

Find the even and odd parts of the following signals:

The even part of a signal x ( n ) is given by x,(n) = f [ x ( n ) x(-n)l With x ( n ) = u(n), we have

which may be written concisely as

x,(n) = f fS(n) Therefore, the even part of the unit step is a sequence that has a constant value of it has a value of 1. The odd part of a signal x ( n ) is given by the difference xo(n) = i [ x ( n )- x(-n)] With x ( n ) = u(n), this becomes n>O

4 for all n except at n = 0,where

where sgn(n) is the signum function. With x ( n ) = a n u ( n ) ,the even part is

or The odd part, on the other hand, is

x,(n) = fa'"'

+ f&n)

If X I ( n ) is even and x2(n) is odd, what is y(n) = xl ( n ) . x2(n) If y ( n ) = x d n ) . x d n ) , y(-n) =X I ( - n ).xz(-n) Because x , ( n ) is even, x l ( n ) = x l ( - n ) , and because xz(n) is odd, x2(n) = - x z ( - n ) . Therefore,

and it follows that y ( n ) is odd.

SIGNALS AND SYSTEMS

[CHAP. 1

If x ( n ) = 0 for n < 0, derive an expression for x ( n ) in terms of its even part, xe(n), and, using this expression, find x ( n ) when x e ( n ) = (0.9)lnlu(n). Determine whether or not it is possible to derive a similar expression for x ( n ) in terms of its odd part. Because

x d n ) = f [ x ( n ) x(-n)l

and note that when x ( n ) = 0 for n

x,(n) = i [ x ( n ) - x ( - n ) ]

-= 0 ,

xe(n) = i x ( n )

n >0

n =0

xe(n) = x ( n )

Therefore, x ( n ) may be recovered from its even part as follows:

For example, with xe(n) = (0.9)lnlu(n), have we

Unlike the case when only the even part of a sequence is known, if only the odd part is given, it is not possible to recover x(n). The problem is in recovering the value of x(0). Because x,(O) is always equal to zero, there is no information in the odd part of x ( n ) about the value of x(0). However, if we were given x ( 0 ) along with the odd part, then, x ( n ) could be recovered for all n .

If xe(n) is the conjugate symmetric part of a sequence x(n), what symmetries do the real and imaginary parts of xe(n) possess The conjugate symmetric part of x ( n ) is

x&) = $ [ x ( n ) x*(-n)]

Expressing x ( n ) in terms of its real and imaginary parts, we have

Therefore, the real part of x,(n) is even, and the imaginary part is odd.

Find the conjugate symmetric part of the sequence

x ( n ) = je jnn/4

The conjugate symmetric part of x ( n ) is

xe(n) = i [ x ( n ) + x*(-n)] =

2 J'ejn"/4-

jejnnI4] = 0

Thus, this sequence is conjugate antisymmetric.

Given the sequence x ( n ) = ( 6 - n ) [ u ( n )- u ( n - 6 ) ] ,make a sketch of

(4 y l ( n ) = x ( 4 - n )

(c) y d n ) = x ( 8 - 3 n )

( h ) y2(n) = x(2n - 3)