18 Exercise Solutions
Chapter 1 Chapter 2
Exercise 2-1. Addition of two complex-valued signals is illustrated below:
Re{x( t )} x( t ) y( t ) x( t ) + y( t ) Im{x( t )} Re{y( t )} Im{y( t )} Re{x( t ) + y( t )} Im{x( t ) + y( t )}
and multiplication below:
x( t ) y( t ) x( t )y( t ) Re{x( t )} Im{x( t )} Im{y( t )} Re{y( t )} Im{x( t )y( t )} + Re{x( t )y( t )}
Complex addition is accomplished by two real additions, and complex multiplication by four real multiplications and two real additions. Exercise 2-2. A complex system with a real-valued input:
Re{h( t )} Im{h( t )} Re{y( t )} Im{y( t )} x( t ) h( t ) y( t )
Re{x( t )}
A real system with a complex-valued input:
Re{x( t )} Im{x( t )} Re{h( t )} Im{h( t )} Re{y( t )} Im{y( t )} x( t ) h( t ) y( t )
Exercise 2-3. We can treat the convolution x( t ) h( t ) just like complex multiplication, since the convolution operation is linear an integration. To check linearity, for a complex constant A and two input signals x1( t ) and x2( t ),
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(x1( t ) + A x2( t )) h( t ) = x1( t ) h( t ) + A (x2( t ) h( t )) following the rules of complex arithmetic. This establishes linearity. Exercise 2-4. Observe that so that Y( f ) =
j2ft
(18.1)
Y( f ) =
e j2ft = e j2fe j2f(t )
h()x(t )de x(t )e
dt . (18.2) = H( f )X( f ) , (18.3)
h()e
j2fd
j2f(t )dt
after a change of variables. Exercise 2-5. Take the Fourier transform of both sides of (2.2), getting X (f) = = =
m = xm(t mT)e
j2ftdt
m = xm (t mT)e j2ftdt m = xme j2fmT = X(e j2fT ) .
(18.4)
Exercise 2-6. The impulse response of the system is gk = g(kT), and hence (2.17) gives the frequency response directly, 1 G(e j2fT ) = --T
m = G(f m T ) .
for all |f|< 1 (2T) .
(18.5)
Exercise 2-7. Given X( f ) = 0 for all |f|> 1 (2T), (2.17) implies that 1 X(e j2fT) = --- X( f ) T (18.6)
To get x( t ) from xk, therefore, we can use, in Fig. 2-1, the following lter: F( f ) = T ; 0; |f|<1 (2T) otherwise (18.7)
Exercise 2-8. From (2.24), the modulus-squared of the complex envelope s ( t ) is | | s ( t ) 2 = (s2( t ) + s 2 ( t )) 2. (18.8)
Since the Hilbert transform is a phase-only lter, it doesnt change the energy, so the energy of s( t ) and its Hilbert transform s ( t ) are the same. Hence, the complex envelope energy is identical to that of s( t ). | Exercise 2-9. From (2.24), the real envelope is e( t ) = 2| s ( t ) = s2( t ) + s 2 ( t) 1 / 2 , which is the magnitude of the phase splitter output, before the complex exponential, and is clearly independent of the carrier frequency. Exercise 2-10. First show that S < implies BIBO. Suppose the input is bounded by xk L. Then
Exercise Solutions
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|yk| = |
m = hmxk m| L m = |hm| = LS < .
k xk = h * |hk|; 0; k such that hk 0 k such that hk = 0
(18.9)
Then show that if S = there exists a bounded input such that the output is unbounded. Such an input
is
Exercise 2-11.
(18.10)
(a) If the sequence is left-sided (2.37) becomes
k = |hk||z|k <
K
(18.11)
for some K. This sum can be rewritten |z|K
k = 0 |hKk||z|k
(18.12)
and we recognize that the |z|K cannot affect convergence except at |z|= 0 and |z|= . All the terms in the sum are positive powers of |z|, and hence if they converge for some |z|= R they must converge for all smaller |z| (except possibly |z|= 0). (b) If K > 0, the sequence is not anticausal, and there is a K-th order pole at z = 0, the ROC does not include z = 0. If K = 0, then H(0) = hK, the ROC includes z = 0. If K < 0, there is a K-th order zero at z = 0, which is therefore included in the ROC. Exercise 2-12. This follows directly from the observation that xk l for a xed integer l has Z transform zlX(z). Taking the Z transform of both sides of (2.41), we get the desired results. Exercise 2-13. This is a straightforward evaluation. For example,
y( t ), x( t ) = y( t )x*( t ) dt = ( x( t )y*( t ) dt)* = x( t ), y( t ) * .
(18.13)
Exercise 2-14. The inequality is obviously true (with equality) if X = 0 or Y = 0, so assume that X 0 and Y 0. Then we have the inequality 0 || X Y ||2 0 || X ||2 2Re{* X,Y } + ||2|| Y ||2 If we let (18.14) (18.15)
= X,Y || Y ||2,
then the previous inequality becomes 0 || X ||2 | X,Y |2 || Y ||2 , from which the Schwarz inequality follows immediately. (18.16)
