Math 1. Review 1.1a: Proof that f 1 [
B ] =
f 1 [B ].
Proof. Assume x f 1 [ B ] then by denition f (x) B such that y = f (x) . This value of f (x) must therefore be in at least one of the sets, B , call this set B0 . Thus, x f 1 [B0 ] and therefore, x 1 [B ] . We have just shown that f 1 [ B ] f 1 [B ] . Conversely, assume that x f 1 [B ] , f then there is some (call it 0 ) for which x f 1 [B0 ] . This means that f (x) B0 . Because it is in one set, this value must be also be in the union of all the sets B . Therefore, f (x) B . By denition, 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 it must be that f (x) B . To be in the intersection, this value must be in all of the sets B . Therefore, x f 1 [B ] for every set 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 f (x) is in all of the sets B and is therefore in the intersection of these sets so f (x) B . 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.
2 2 4 2
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
1
2
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
2. Linear Operators 1.1: 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) . 1.4: Prove that the space-limiting operator is a projection operator. Proof. Let P be the space-limiting operator for the region D, then P 2 f is P 2f = D (x) D (x) f (x) = D (x) f (x) = Pf
therefore P 2 = P . To nd P we note that (P f, g ) = = = D (x) f (x) g (x) dx, f (x) [D (x) g (x)] dx, (f, P g ) .
Therefore, because the adjoint is unique and satises (P f, g ) = (f, P g ) it must be that P = P. 2.1: 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 H (s) = 1 1 1 1 = = + . s2 + 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 h (t) = Simplifying this gives h (t) = 1 t e sin (2t) u (t) . 2 1 (1+2j )t e e(12j )t u (t) . 4j
3
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 b t a sin (bt) b cos (bt) =e +2 . a2 + b2 a + b2 = ex 3 sin (2t) 2 cos (2t) 2 +. 13 13
t
ea sin (b ) d
0
Therefore,
t
e3 sin (2 ) d = e3t
0
y (t)
= =
1 4t 3 sin (2t) 2 cos (2t) 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) = = 1 (3 + 2j ) (12j )t 3 2j (1+2j )t e + e + e4t u (t) 13 4j 13 4j 13 1 4t 3 sin (2t) 2 cos (2t) e + et u (t) . 13 26
as before. 2.4: 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 1832 1 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= 5 3.
f
n n h
with = Using only |n| 10 we get an error of approximation of only (f, h) (f, h)est = 1.5277 1013 . 2.6: Prove that the null space of a convolution operator is orthogonal to the range space of the operator.
4
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 q (2 ) but this requires that hA ( ) f ( ) = 0. for all . This means that for every either hA is 0 or f is 0. If hA is zero then h is zero also and therefore, A h ( ) f ( ) = 0. A Because g is in the range of A it can be written as g = Af2 for some f2 thus, g (x) = or in other words g ( ) = hA ( ) f2 ( ) Now by the generalized Parseval equality, (f, g ) = but g ( ) = h ( ) f2 ( ) , A so (f, g ) 1 q (2 ) 1 = q (2 ) = 0. = f ( ) h ( ) f2 ( ) d A 0 f2 ( ) d 1 q ) (2 f ( ) g ( ) d , 1 q (2 ) hA ( ) f2 ( ) ej x d , hA ( ) f ( ) ej x d = 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. 3.4: Compute discretization of system g (t) = f (t) over domain [0, 1] . To use the sampled impulse response method we need the solution to this system which is readily obtained as
t
g (t) = g (0) +
0
f ( ) d.
For the case of g (0) = 0 we have the linear system
t
g (t)
=
0 1
f ( ) d u (t ) f ( ) d.
0
= Sampling this system with t = 1/N gives g (nt)
m
u (nt mt) f (mt) t
or gn =
m
Anm fm .
Thus, A(1) = u ((n m) t) t. nm
5
There is some ambiguity in the denition of u (0) . We will choose u (0) = Fourier representation of u (t) . Thus, n<m 0 t A(1) = n=m . nm 2 t n>m gn gn1 = fn t
1 2
as that is the value it has in the
The nite dierence discretization matrix is obtained by converting the dierential equation directly:
(2)
for n = 1 . . . N. with boundary condition g0 = 0. We can nd Anm by solving this equation for fn = nm : A(2) An1m = tnm , nm with A0m = 0. Summing both sides from n = 1 to k gives
k (2)
Akm = t
n=1
(2)
nm =
0 t
k<m km
.
This is very similar to the discretization matrix as was found by sampling the impulse response except for at k = m where the sampled-Greens function has the value t/2. The nite-element discretization matrix is found by expanding g and f on an interpolating basis as before and then nding the coecients on g to make the error orthogonal to the subspace spanned by the interpolating basis. Thus ge (t) fe (t) =
m
gm m (t) fm m (t)
m
=
where gm and fm are sample values of g (t) and f (t) respectively at mt = m/N and m (t) are linearly interpolating basis functions. The equations used to solve for gm are for all n = 1 . . . N .
1
(ge (t) fe (t)) n (t) = 0.
0
This leads to gm (m , n ) =
m m 1
fm (m , n ) .
We need to compute (m , n ) = m (t) n (t) dt.
0 1 2.
If n = m 1 this is 1 . If n = m this inner product If |n m| 2, this is zero. If n = m + 1 this is 2 t is 0. From the notes we know that (m , n ) = 6 (4nm + nm1 + nm+1 ) . Thus the nite-element method leads to the dierence equation t 1 gm (nm1 nm+1 ) = (4fn + fn+1 + fn1 ) 2m 6 gn+1 gn1 = t (4fn + fn+1 + fn1 ) . 3
Because we have initial rest, lets rewrite this as t gn gn2 = (4fn1 + fn + fn2 ) . 3 To solve for samples of g at n = 1 . . . N note that we need g1 and f1 which we assume to be zero (initial (3) (3) (3) rest). To nd Anm we solve this system assuming fn = nm and initial rest (A0m = A1m = 0). Because we assume initial rest, we have a linear shift-invariant system and we can use the z-transform to nd gn . Specically, t Am (z ) z 2 Am (z ) = 4z 1 + 1 + z 2 z m . 3
6
or Am (z ) = This is Am (z ) = Therefore, A(3) nm
t m 1 + 4z 1 + z 2 . z 3 1 z 2
t m 1 3 . z 1 + 3 1 z 1 1 + z 1
t nm nm + 3u [n m] (1) u [n m] 3 0 n<m t n=m . = 3 t nm 3 (1) n>m =
3
2
The input is f (t) = exp 25(2t1) 8 g (t) =
. Samples are at t = n/N for n = 1 . . . N . The exact result is
t
exp
0
25 (2 1) 8
2
d + erf 5 22 .
=
2 erf 10
5t 5 2 22
Samples of the approximate result are computed by multiplication of the corresponding discretization matrix with the vector formed using samples of f. The results indicate that the sampled Greens function method performs better than the other methods again. It is interesting to note the importance of choosing u (0) = 1/2 when constructing the sampled Greens function as the choice u (0) = 1 would have led to nite-dierence discretization which performs worse. The poor performance of the nite-dierence discretization is actually overstated because most of the error comes from not correctly predicting the position of the rise in g (t) which occurs near t = 1 . If the nite-dierence 2 samples are interpreted as samples on a grid shifted by t/2, then this method performs comparably. In the gures below, di 2 represents the result of re-interpolating the nite-dierence result onto a grid shifted by one-half. When this is done, it results in the best discretization. 3. Vogel 2.4: Find the adjoint of the Fredholm rst kind integral operator: (Kf ) (x) =
k (x, y ) f (y ) dy.
The adjoint satises (Kf, g ) = (f, K g ) . But for this s...