SOLUTIONS
1.1a: HOMEWORK Proof that f 1 [
B ] =
f 1 [B ].
Proof. Assume x f 1 [ B ] then by denition there is a y B such that y = f (x) . This y must be in at least one of the sets, B , call this set B0 . By denition, x f 1 [B0 ] and thus x f 1 [B ] , therefore f 1 [ B ] f 1 [B ] . Conversely, assume that x f 1 [B ] , then there is some (call it 0 ) for which x f 1 [B0 ] . This means that there is a y = f (x) B0 . Because it is in one set, this y must be also be in the union of all the sets B . Therefore, f (x) = y B . By denition, then x f 1 [ B ] . This shows that f 1 [B ] f 1 [ B ] . These two sets can be subsets of each other if and only if they are equal. 1.1b: Proof that f 1 [
B ] =
f 1 [B ].
Proof. Assume x f 1 [ B ] , then there must be a y B such that f (x) = y. To be in the intersection, this y must be in all of the sets B . Therefore, x f 1 [B ] for all of the sets B and thus x f 1 [B ] . We have thus far shown that f 1 [ B ] f 1 [B ] . To show the converse, assume x f 1 [B ] this means that x is in all of the sets f 1 [B ] , but this implies that there is a y = f (x) that is in all of the sets B and is therefore in the intersection of these sets so y B . Because y = f (x) this implies that x f 1 [ B ] by denition of inverse images. Consequently, f 1 [B ] f 1 [ B ] and the two sets are equal. 2.2a: Proof that in a Euclidean space (f, g ) = (f, g ) Proof. The requirements of the inner product provide that (f, g ) = (g, f ) = { (g, f )} = (g, f ) = (f, g ) .
2.2b: Proof of the Schwartz inequality: Proof. If (f, g ) = 0, then the inequality is trivially satised by the requirements of the inner product. Therefore, assume that (f, g ) = 0. Let t be any scalar such that t = t (i.e. a real number). The inner product of f + t (f, g ) g with itself must be positive thus, (f + t (f, g ) g, f + t (f, g ) g ) 0 expanding this equation using the linearity of the inner product gives (f, f ) + 2t (f, g ) (g, f ) + t2 (f, g ) (g, f ) (g, g ) 0 This is a quadratic equation in t which is non-negative for all t (graphically its a smiling parabola that lies either entirely above the t axis, or just touches the t axis at one point). Therefore, the quadratic does not have two distinct real roots. Thus, the roots of this quadratic must be either complex or a single real number. In either case, the discriminate of the quadratic (b2 4ac) must be non-positive: 4 |(f, g )| 4 (f, f ) (g, g ) |(f, g )| 0. Dividing by 4 |(f, g )| gives the result |(f, g )| (f, f ) (g, g ) 0.
1 2 2 4 2
2
HOMEWORK SOLUTIONS
2.3: Show that the induced norm is a well-dened norm. Clearly (f, f ) 0 and (f, f ) = 0 implies that (f, f ) = 0 which from the denition of the inner product means that f = 0. f = (f, f ) = (f, f ) = || (f, f ) = || f 2 2 f + g = (f + g, f + g ) = (f, f ) + 2 {(f, g )} + (g, g ) and ( f + g ) = (f, f ) + 2 (f, f ) (g, g ) + 2 (g, g ) but, 2 {(f, g )} 2 |(f, g )| 2 (f, f ) (g, g ) by the Cauchy-Schwartz inequality, thus f + g 2 ( f + g ) and the triangle inequality is satised. 2.5: Prove the Parseval equality and generalized Parseval equality Proof. We know that in a separable Hilbert space
f=
i=1
(f, vi ) vi
where the {vi } form an orthonormal set. Thus by the linearity of the inner product,
(f, h) =
i=1
(f, vi ) vi , h
=
i=1
(f, vi ) (vi , h) =
i=1
(f, vi ) (h, vi ) .
This is the generalized Parseval equality, The Parseval equality s found by taking h = f so that
f
2
= (f, f ) =
i=1
(f, vi ) (f, vi ) =
i=1
|(f, vi )| .
2
3.1: Prove that for a linear operator, A: N (A) = R (A ) ,
N (A ) = R (A) .
Proof. Suppose A is an operator from X to Y , and thus A is the adjoint operator from Y to X . Let f N (A) then Af = 0. Consider any element f2 R (A ) , then there is some element g2 such that A g = f2 , The inner product between (f, f2 ) is (f, f2 ) = (f, A g2 ) = (Af, g2 ) = 0, thus f R (A ) . Now, suppose that f is in the orthogonal complement of the range space of A then for every f2 R (A ) corresponding to some g2 with A g2 = f2 we have (f, f2 ) = 0 or (f, A g2 ) = (Af, g2 ) = 0. Because this is true for every g2 it must be that Af = 0. Thus, N (A) = R (A ) . This same argument holds by replacing A with A and because (A ) = A we have, N (A ) = R (A) . 3.2: Prove the projection operator is bounded with unit norm. Proof. The projection operator is one which satises P = P and P 2 = P. Thus Pf
2
= (P f, P f ) = (f, P P f ) = (f, P f ) (P f, P f ) Pf f
(f, f ) (P f, P f )
where the last result is true via the Cauchy-Schwartz inequality, thus Pf = (f, f ) = f
which shows that the operator is bounded. The norm is the supremum of
over all f. From above this relation is always 1 and therefore, the norm is 1. Now, from the submultiplicative property we get P = PP P P , and thus the norm of P is 1. These two facts together show that P = 1. To prove the submultiplicative property, note that ABx A B x A B x
HOMEWORK SOLUTIONS
3
by denition of the norms of A and B as the supremum of the ratios ABx / B x and B x / x . Thus, ABx / x A B . Therefore, the supremum (least upper bound) of ABx / x must be less than A B as well. 3.5: Prove that the bandlimiting operator is a projection operator. Proof. Let P be bandlimiting the operator for the band B , then P 2 f is P 2f = F 1 B ( ) (P f ) ( ) = F 1 {B ( ) B ( ) f ( )} = F 1 {B ( ) f ( )} = Pf therefore P 2 = P . To nd P we note that 1 B ( ) f ( ) ej x g (x) d dx, (P f, g ) = q (2 ) 1 = B ( ) f (x ) ej (xx ) g (x) d dx dx, q (2 ) 1 = f (x) ( ) ej x g ( ) d dx q B (2 ) = (f, P g ) . Thus, because the adjoint is unique and satises (P f, g ) = (f, P g ) we must have P = P. 4.1: The operator satises the ODE y + 2y + 5 = x for Ax = y together with initial rest The system (transfer) function after partial fraction expansion is 1 1 1 1 H (s) = 2 = = + . s + 2s + 5 (s + 1 + 2j ) (s + 1 2j ) 4j (s + 1 + 2j ) 4j (s + 1 2j ) Initial rest means that for zero-input the output must be zero. Therefore, the system must be causal and have a right-sided impulse response. This means that the region of convergence of the system function must be right sided. The ROC must also be bounded by the poles. Thus, the ROC of H (s) is s > 1. Consequently the inverse (bilateral) Laplace transform gives the impulse response which is 1 (1+2j )t e e(12j )t u (t) . h (t) = 4j Simplifying this gives 1 h (t) = et sin (2t) u (t) . 2 To nd Ax we must either convolve h (t) with x (t) or multiply the Laplace transform of x (t) with H (s) and nd another inverse Laplace transform. The former solution gives y (t) = = = In an integral table we nd that
t
h ( ) x (t ) d u (t) t e sin (2 ) e4(t ) d 2 0 u (t) 4t t 3 e e sin (2 ) d. 2 0 [a sin (bx) b cos (bx)] a2 + b2 x=0 a sin (bt) b cos (bt) b = et +2 . a2 + b2 a + b2
t
ea sin (b ) d
0
= ex
4
HOMEWORK SOLUTIONS
Therefore,
t
e3 sin (2 ) d = e3t
0
2 3 sin (2t) 2 cos (2t) +. 13 13
y (t)
= =
3 sin (2t) 2 cos (2t) 1 4t e + et u (t) 13 26 1 4t 13 t 2 e + e sin 2t tan1 u (t) . 13 26 3
Solving using the Laplace transform gives Y (s) = = 1 (s + 1 + 2j ) (s + 1 2j ) (s + 4) (4j ) (3 2j ) (s + 1 + 2j )
1 1
+
(4j ) (3 + 2j ) (s + 1 2j )
1
1
+
(16 8 + 5) (s + 4)
1
,
with an ROC that is still s > 1. Thus y (t) is y (t) = = (3 + 2j ) (12j )t 3 2j (1+2j )t 1 e + e + e4t u (t) 13 4j 13 4j 13 3 sin (2t) 2 cos (2t) 1 4t e + et u (t) . 13 26
as before. 4.3: Compute (f, h) exactly (f, h) = = = = gauss (x, 3) gauss (x, 4) dx e 18 e 32 dx e( 18 + 32 )x dx
1 1 2 x2 x2
25 1 1 1 1832 2 where 2c = 18 + 32 = 18+32 = 1832 = 352 = 11.52 thus (f, g ) = 11.52 . Using a band-limited 2 50 approximation we can compute the inner product from samples of f and h as
2 2c
(f, h) (f, h)est =
n=
f
n n h
with =
5 3.
Using only |n| 10 we get an error of approximation of only (f, h) (f, h)est = 1.5277 1013 .
C 4.4: Find F (x) = (2)2 H (1 , 2 ) ej (x1 1 +x2 2 ) d1 d2 where H is the smallest regular hexagon inscribing the circle of radius . Symmetry allows us to consider only integration over the rst quadrant so that
F (x) =
4C (2 )
2 0
0
H (1 , 2 ) cos (x1 1 ) cos (x2 2 ) d1 d2 .
The characteristic function of the hexagon suggests that the integral be broken up into two pieces: 2 (F1 + F2 ) F = 23 where
2 3 2
F1 =
0
d1 cos (x1 1 )
0
d2 cos (x2 2 )
HOMEWORK SOLUTIONS
5
and
3(1 )
F2 =
2
d1 cos (x1 1 )
0
d2 cos (x2 2 ) .
Simplifying these expressions gives
2
F1
=
0
1 3 d1 cos (x1 1 ) sin x2 x2 2
3 2 x2
sin = and F2 = =
sin x1 2
x1 x2
.
= =
= Thus,
1 sin x2 3 ( 1 ) x2 /2 cos x1 x2 3 + x2 3 cos x1 x2 3 + x2 3 2 2x2 x1 x2 3 cos x1 + x2 3 x2 3 cos x1 + x2 3 x2 3 2 + , 2x2 x1 + x2 3 cos x1 + x2 3 cos (x1 ) cos (x1 ) cos x1 x2 3 2 2 + 2x2 x1 x2 3 2x2 x1 + x2 3 cos x1 + x2 3 cos x1 x2 3 cos (x1 ) 3 2 2 x2 3x2 2x2 x1 x2 3 2x2 x1 + x2 3 1 2 3 3 3 cos x1 2 cos x2 2 3 cos (x1 ) sin 2 x2 sin x1 2 x2 3x2 x2 (x2 3x2 ) /x1 1 2 1 2
d1 cos (x1 1 )
F1 + F2 = sin + =
3 x2 2
sin
3 cos x1 2 3 x2 2
1 x1 x1 x2 x2 (x2 3x2 ) 1 2 3 cos x2 2 3 cos (x1 ) . x2 3x2 1 2 x1 2
x1 2
sin +
sin
3 cos x1 2
x2 3x2 x2 1 2 1 x1 x2 (x2 3x2 ) 1 2 cos x2 23 3 cos (x1 ) x2 3x2 1 2
3 2 x2 3x2 1 2
3 cos x1 cos x2 2
3 cos (x1 )
3x2 sin
3 2 x2 sin x1 (x2 3x2 ) 1 2 3 2
x1 2
F (x1 , x2 )
=
2 x1 cos x1 2 cos x2 2
cos (x1 ) x2 3 sin 23 x2 sin x1 2
x1 (x2 3x2 ) 1 2
.
4.5: Prove that the null space of a convolution operator is orthogonal to the range space of the operator.
6
HOMEWORK SOLUTIONS
Proof. Let A be a convolution operator (with kernel, or impulse response, hA ). Let f N (A) and let g R (A) We know that Af = 0, therefore using the spectral respresentation of the operator we know that 1 hA ( ) f ( ) ej x d = 0 q (2 ) but this requires that hA ( ) f ( ) = 0. A is 0 or f is 0. If hA is zero then h is zero also and therefore, for all . This means that for every either h
A
h ( ) f ( ) = 0. A Because g is in the range of A it can be written as g = Af2 for some f2 thus, 1 g (x) = hA ( ) f2 ( ) ej x d , q (2 ) or in other words g ( ) = hA ( ) f2 ( ) Now by the generalized Parseval equality, 1 (f, g ) = f ( ) g ( ) d , q (2 ) but g ( ) = h ( ) f2 ( ) , A so 1 (f, g ) = f ( ) h ( ) f2 ( ) d q A (2 ) 1 = 0 f2 ( ) d q (2 ) = 0. Because f and g were arbitrary elements of the null and range space, we have that that the null and range space of a convolution operator are orthogonal.