Green matrix and a parameter-dependent circuit
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 1
(a) Consider the path graph 0 − 1 − 2 − 3 − 4, with nodes 0 and 4 grounded. Its reduced Laplacian on nodes
1, 2, 3 is
𝐾3 = ©
«
2 −1 0
−1 2 −1
0 −1 2
ª®
¬
.
Find the matrix 𝑀 such that 𝑥 = 𝑀 𝑓 for any vector of current divergences 𝑓 at nodes 1, 2, 3. Briefly explain
why 𝑀 is symmetric.
0 1 2 3 4
ground ground
(b) In the parameter-dependent circuit below, node 1 is at voltage 1 and node 2 is grounded. The conductances
are
𝐶13 = 1, 𝐶 32 = 𝑐, 𝐶 14 = 𝑐, 𝐶 42 = 1, 𝐶 34 = 1,
where 𝑐 > 0. Find the voltages at nodes 3 and 4, the current in edge 3–4 including its direction, and the effective
conductance as a function of 𝑐. Check the limits 𝑐 → 0 and 𝑐 → ∞.
1
3
4
2
1 𝑐
𝑐 1
1
(Total: 20 marks)
13
Worked solution and marking guidance
Question 1(a). Direct inversion gives
M = K −1
3 =
3/4 1 /2 1 /4
1/2 1 1 /2
1/4 1 /2 3 /4
.
Equivalently, its columns are the voltage solutions for unit current injected at nodes 1, 2 and 3. It is symmetric
because K3 is symmetric and invertible; physically this is the reciprocity identity xTbf =bxT f. 8 marks, A
Question 1(b). Let the internal voltages be v3, v4. KCL gives
(v3 − 1) + cv3 + (v3 − v4) = 0,
c(v4 − 1) + v4 + (v4 − v3) = 0.
Solving,
v3 = 2
c + 3 , v 4 = c + 1
c + 3 .
The current from 4 to 3 is
v4 − v3 = c − 1
c + 3.
Thus it flows 4 → 3 for c > 1, 3 → 4 for 0 < c < 1, and vanishes at c = 1. The effective conductance is
Ceff = (1 − v3) + c(1 − v4) = 3c + 1
c + 3 .
The limits are Ceff → 1/3 as c → 0 and Ceff → 3 as c → ∞. 12 marks, B
Bridge networks, parallel paths and random walks
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 2
(a) A bridge network has conductances 𝐶+𝐴 = 1, 𝐶+𝐵 = 1, 𝐶𝐴− = 1, 𝐶𝐵− = 2, 𝐶𝐴𝐵 = 2. Find its effective
conductance.
+
𝐴
𝐵
−
1
1
1
2
2
(b) A second network consists of two parallel unit-edge paths, one of length 2 and one of length 3. Find its
effective conductance.
+ −
length 2
length 3
(c) Put the networks in (a) and (b) in series with one extra unit edge between them. A simple random walker
starts at the first + terminal. Find the probability of reaching the final − terminal before returning to the starting
terminal.
(Total: 20 marks)
14
Worked solution and marking guidance
Question 2(a). Let internal potentials be a, b. KCL gives
(a − 1) + a + 2(a − b) = 0, (b − 1) + 2b + 2(b − a) = 0.
Hence 4a − 2b = 1, −2a + 5b = 1, so
a = 7
16 , b = 3
8 .
Therefore
CA = (1 − a) + (1 − b) = 9
16 + 10
16 = 19
16 .
8 marks, C
Question 2(b). The two branch conductances are 1 /2 and 1 /3, so
CB = 5
6 .
6 marks, A
Question 2(c). The series resistance is
R = 16
19 + 1 + 6
5 = 289
95 ,
so C = 95/289. The starting vertex has degree 2, hence
pesc = 95
578 .
6 marks, D
Forced diffusion on a three-node path
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 3
Let
𝐾3 = ©
«
2 −1 0
−1 2 −1
0 −1 2
ª®
¬
.
A three-node diffusion system satisfies
ˆ𝑥′ (𝑡) = −𝐾3 ˆ𝑥(𝑡) + ©
«
0
1
0
ª®
¬
, ˆ𝑥(0) = 0.
1 2 3
source at node 2
(a) Find the steady state and an orthonormal eigensystem of 𝐾3.
(b) Find an explicit modal expression for all three components of ˆ𝑥(𝑡).
(c) State the symmetry, initial values and limiting values that a correct sketch must show.
(Total: 20 marks)
15
Worked solution and marking guidance
Question 3(a). The steady state solves K3xs = (0, 1, 0)T , giving
xs = (1/2, 1, 1/2)T .
8
MATH40007 Equal-Difficulty Replacement Pack Solutions and Marking Guide
Use
Φ1 = 1
2(1,
√
2, 1)T , λ 1 = 2 −
√
2,
Φ2 = 1√
2(1, 0, −1)T , λ 2 = 2,
Φ3 = 1
2(1, −
√
2, 1)T , λ 3 = 2 +
√
2.
8 marks, A
Question 3(b). Sinceex(0) = −xs, the modal coefficients are
b1 = −1 +
√
2
2 , b 2 = 0, b 3 =
√
2 − 1
2 .
Thus
x1(t) = x3(t) = 1
2 − 1 +
√
2
4 e−(2−
√
2)t +
√
2 − 1
4 e−(2+
√
2)t,
x2(t) = 1 − 2 +
√
2
4 e−(2−
√
2)t − 2 −
√
2
4 e−(2+
√
2)t.
8 marks, B
Question 3(c). The solution is reflection-symmetric: x1 = x3. All three values begin at zero and tend respectively
to 1/2, 1, 1/2. A correct sketch shows smooth exponential approach with the centre node above the two outer nodes
for t > 0. 4 marks, C
Right-half-plane point source and grounded wall
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 4
The right half-plane x> 0 is a conductor of unit conductivity. Let a> 0 and
ϕ(x,y ) = Reh(z), h (z) =−m
2πlog
(z−a
z +a
)
, z =x +iy.
(a) Identify the current source in the conductor, show that the wall x = 0 is grounded, and find
Jx−iJy =−h′(z). (10 marks)
(b) Find the current density on the wall, its maximum magnitude, and the total current leaving through
the wall. Interpret the answer physically. (10 marks)
(Total: 20 marks)
16
MATH40007 Equal-Difficulty Replacement Pack Questions
16
Worked solution and marking guidance
Question 4(a). The singularity at z = a is a point source of strength m inside the conductor; z = −a lies outside.
On the wall z = iy,
iy − a
iy + a
= 1,
so ϕ = 0. Moreover
Jx − iJy = −h′(z) = m
2π
1
z − a − 1
z + a
= ma
π(z2 − a2) .
10 marks, A
Question 4(b). On z = iy,
Jx = − ma
π(a2 + y2) , J y = 0.
The current is normal to the grounded wall and points outward. Its maximum magnitude is at y = 0:
|Jx(0)| = m
πa .
The total outward current is Z ∞
−∞
−Jx dy = ma
π
Z ∞
−∞
dy
a2 + y2 = m.
Thus every unit of current emitted by the interior source exits through the grounded wall. 10 marks, D
9
MATH40007 Equal-Difficulty Replacement Pack Solutions and Marking Guide