40 From the finite to the infinite matrix elements. We have ac, + Bez = Ec, (for p = 1) Bc, + ac, + Bc, = Ec2 (for p = 2) Bc, + ac, + Bc, = Ec, (for p = 3) Bcj-1 + ac, + Bc,+1 = Ec, (for p = j) (3.5) BeN-2 + acN -1 + Bcx = EcN-1 (for p = N – 1) BCN-1 + acN = ECN (for p = N). There are N coupled linear equations here, and we have to solve them! But, it is not as bad as it looks. First of all we note that the equations have all the same form except for p = 1 and p = N, the end atoms. Al the equations can be simplified by first dividing by B. For example, for p =j we get Cj-1 - xc, + c,+1 = 0 (for p = j) (3.6) where x = (E - a)/B. (3.7) Let us try the solution c, = ev where i = (-1)2 and 0 is to be determined (do not confuse i in the exponential with the label of atom i). At this stage 0 may be real, imaginary, or complex. Inserting this guess into egn (3.6) we get elU- 1)0 - x ev + eU+1]e = 0 or x = (e" +e-i) = 2 cos 0. (3.8) If we replace 0 by -0 we get the same solution for x. Therefore we could have c, = A e + Be-, where A and B are arbitrary constants and still egn (3.8) would be satisfied. How do we determine A and B? Well, do not forget the end atoms. For p =1 we have c, = xc,. Therefore, A e210 + Be- 210 = 2 cos 8(A e" + Be-) (3.9) and this equation is satisfied by setting A = - B. Thus the equation c2 = xc, pins down A and B to the extent that c, = A(e - e-) where A is still an arbitrary constant and 0 satisfies eqn (3.8). We may rewrite c, as D sin je where D = 2iA, i.e. yet another arbitrary constant. The equation for the other end atom, cy-1= xCN, is used now to pin down the allowed values of 0 D sin(N – 1)8 = xD sin(Ne) = 2D cos 0 sin(N0) or sin(N – 1)8 = 2 cos e sin(NO) = sin(N8) cos 0 - cos(NO) sin 6 (3.10)
40 From the finite to the infinite matrix elements. We have ac, + Bez = Ec, (for p = 1) Bc, + ac, + Bc, = Ec2 (for p = 2) Bc, + ac, + Bc, = Ec, (for p = 3) Bcj-1 + ac, + Bc,+1 = Ec, (for p = j) (3.5) BeN-2 + acN -1 + Bcx = EcN-1 (for p = N – 1) BCN-1 + acN = ECN (for p = N). There are N coupled linear equations here, and we have to solve them! But, it is not as bad as it looks. First of all we note that the equations have all the same form except for p = 1 and p = N, the end atoms. Al the equations can be simplified by first dividing by B. For example, for p =j we get Cj-1 - xc, + c,+1 = 0 (for p = j) (3.6) where x = (E - a)/B. (3.7) Let us try the solution c, = ev where i = (-1)2 and 0 is to be determined (do not confuse i in the exponential with the label of atom i). At this stage 0 may be real, imaginary, or complex. Inserting this guess into egn (3.6) we get elU- 1)0 - x ev + eU+1]e = 0 or x = (e" +e-i) = 2 cos 0. (3.8) If we replace 0 by -0 we get the same solution for x. Therefore we could have c, = A e + Be-, where A and B are arbitrary constants and still egn (3.8) would be satisfied. How do we determine A and B? Well, do not forget the end atoms. For p =1 we have c, = xc,. Therefore, A e210 + Be- 210 = 2 cos 8(A e" + Be-) (3.9) and this equation is satisfied by setting A = - B. Thus the equation c2 = xc, pins down A and B to the extent that c, = A(e - e-) where A is still an arbitrary constant and 0 satisfies eqn (3.8). We may rewrite c, as D sin je where D = 2iA, i.e. yet another arbitrary constant. The equation for the other end atom, cy-1= xCN, is used now to pin down the allowed values of 0 D sin(N – 1)8 = xD sin(Ne) = 2D cos 0 sin(N0) or sin(N – 1)8 = 2 cos e sin(NO) = sin(N8) cos 0 - cos(NO) sin 6 (3.10)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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FROM BOOK SUTTON, ELECTRONIC STRUCTURE OF MATERIALS,
HOW TO GET 3.10
![40 From the finite to the infinite
matrix clements. We have
ac, + Bcz = Ec,
(for p = 1)
Bc, + ac2 + Bcz = Ec2
(for p= 2)
Bc2 + ac, + Bc, = Ec,
(for p = 3)
Bcj-1 + ac, + Bc,+1 = Ec,
(for p = j)
(3.5)
Bex-2 + aCN-1 + BCN = EcN-1
(for p = N - 1)
BCN-1+ acN = EcN
(for p = N).
There are N coupled linear equations here, and we have to solve them! But,
it is not as bad as it looks. First of all we note that the equations have all
the same form except for p = 1 and p = N, the end atoms. All the equations
can be simplified by first dividing by B. For example, for p =j we get
Cj-1 - xc, + c,+1 = 0
(for p = j)
(3.6)
where
x = (E – a)/B.
(3.7)
Let us try the solution c, = e where i = (-1)2 and 0 is to be determined
(do not confuse i in the exponential with the label of atom i). At this stage 0
may be real, imaginary, or complex. Inserting this guess into eqn (3.6) we get
elU- 1)0 - x ev + eU+1]® = 0
or
x = (e" + e-10) = 2 cos 0.
(3.8)
If we replace e by -0 we get the same solution for x. Therefore we could
have c, = A e + Be-, where A and B are arbitrary constants and still
eqn (3.8) would be satisfied. How do we determine A and B? Well, do not
forget the end atoms. For p =1 we have c, = xc. Therefore,
A e210 + Be-2io = 2 cos 8(A eº + Be-)
(3.9)
and this equation is satisfied by setting A = - B. Thus the equation c2 = xc,
pins down A and B to the extent that c, = A(e – e-iv) where A is still an
arbitrary constant and 0 satisfies eqn (3.8). We may rewrite c, as D sin je
where D = 2iA, i.e. yet another arbitrary constant. The equation for the other
end atom, cN -1 = xCN, is used now to pin down the allowed values of 0
D sin(N – 1)0 = xD sin(N6)
= 2D cos 8 sin(NO)
or
sin(N – 1)0 = 2 cos 0 sin(N8)
= sin(N8) cos 0 - cos(NO) sin 0
(3.10)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d44ab11-7e31-4242-94cf-7d0dfa22638f%2Fabcc2559-1802-4905-ad81-0ff1721959eb%2Fk41uaxq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:40 From the finite to the infinite
matrix clements. We have
ac, + Bcz = Ec,
(for p = 1)
Bc, + ac2 + Bcz = Ec2
(for p= 2)
Bc2 + ac, + Bc, = Ec,
(for p = 3)
Bcj-1 + ac, + Bc,+1 = Ec,
(for p = j)
(3.5)
Bex-2 + aCN-1 + BCN = EcN-1
(for p = N - 1)
BCN-1+ acN = EcN
(for p = N).
There are N coupled linear equations here, and we have to solve them! But,
it is not as bad as it looks. First of all we note that the equations have all
the same form except for p = 1 and p = N, the end atoms. All the equations
can be simplified by first dividing by B. For example, for p =j we get
Cj-1 - xc, + c,+1 = 0
(for p = j)
(3.6)
where
x = (E – a)/B.
(3.7)
Let us try the solution c, = e where i = (-1)2 and 0 is to be determined
(do not confuse i in the exponential with the label of atom i). At this stage 0
may be real, imaginary, or complex. Inserting this guess into eqn (3.6) we get
elU- 1)0 - x ev + eU+1]® = 0
or
x = (e" + e-10) = 2 cos 0.
(3.8)
If we replace e by -0 we get the same solution for x. Therefore we could
have c, = A e + Be-, where A and B are arbitrary constants and still
eqn (3.8) would be satisfied. How do we determine A and B? Well, do not
forget the end atoms. For p =1 we have c, = xc. Therefore,
A e210 + Be-2io = 2 cos 8(A eº + Be-)
(3.9)
and this equation is satisfied by setting A = - B. Thus the equation c2 = xc,
pins down A and B to the extent that c, = A(e – e-iv) where A is still an
arbitrary constant and 0 satisfies eqn (3.8). We may rewrite c, as D sin je
where D = 2iA, i.e. yet another arbitrary constant. The equation for the other
end atom, cN -1 = xCN, is used now to pin down the allowed values of 0
D sin(N – 1)0 = xD sin(N6)
= 2D cos 8 sin(NO)
or
sin(N – 1)0 = 2 cos 0 sin(N8)
= sin(N8) cos 0 - cos(NO) sin 0
(3.10)
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