Incidence matrix and an electrical network
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 1
A connected graph has vertices 1, 2, 3, 4, 5 and oriented edges
𝑒1 : 1 → 3, 𝑒 2 : 1 → 4, 𝑒 3 : 3 → 4, 𝑒 4 : 3 → 5, 𝑒 5 : 4 → 5, 𝑒 6 : 5 → 2.
Let 𝐴 be the 6 × 5 incidence matrix with −1 at the tail and +1 at the head of each edge.
1
3
4
5 2
𝑒1
𝑒2
𝑒3
𝑒4
𝑒5
𝑒6
(a) Write down 𝐴. Find its rank, the dimensions of its right and left null spaces, and give a basis for the left null
space. (8 marks)
(b) Regard the same graph as an electrical network. Give conductances 2 , 1, 1, 1, 2, 1 respectively to edges
𝑒1, . . . , 𝑒6. Node 1 is held at voltage 1, node 2 is grounded, and KCL holds at nodes 3 , 4, 5. Find the three
unknown voltages, the currents in edges 𝑒3 and 𝑒5, and the effective conductance. (12 marks)
(Total: 20 marks)
5
Worked solution and marking guidance
Question 1(a). With columns ordered 1 , 2, 3, 4, 5,
A =
−1 0 1 0 0
−1 0 0 1 0
0 0 −1 1 0
0 0 −1 0 1
0 0 0 −1 1
0 1 0 0 −1
.
The graph is connected, so rank A = 5 − 1 = 4. Thus
dim N (A) = 1, dim N (AT ) = 6 − 4 = 2.
Two independent cycle vectors are
c1 = (1, −1, 1, 0, 0, 0)T , c 2 = (0, 0, 1, −1, 1, 0)T ,
and they span the left null space. 8 marks, A
Question 1(b). Let the unknown voltages be v3, v4, v5. KCL gives
2(v3 − 1) + (v3 − v4) + (v3 − v5) = 0,
(v4 − 1) + (v4 − v3) + 2(v4 − v5) = 0,
(v5 − v3) + 2(v5 − v4) + v5 = 0.
Solving,
v3 = 5
6 , v 4 = 3
4 , v 5 = 7
12 .
Hence
ie3 = v3 − v4 = 1
12 (3 → 4),
ie5 = 2(v4 − v5) = 1
3 (4 → 5),
and
Ceff = 2(1 − v3) + (1 − v4) = 1
3 + 1
4 = 7
12 .
12 marks, B
Bridge networks and random walks
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 2
All drawn conductances are implemented by the corresponding number of parallel unit edges.
(a) A two-node bridge network has conductances
𝐶+𝐴 = 1, 𝐶 +𝐵 = 2, 𝐶 𝐴− = 2, 𝐶 𝐵− = 1, 𝐶 𝐴𝐵 = 2.
Find its effective conductance. (8 marks)
+
𝐴
𝐵
−
1
2
2
1
2
(b) A second network has two parallel branches, each consisting of three unit edges in series. Find its effective
conductance. (6 marks)
+ −
3 unit edges in series
3 unit edges in series
(c) Put the two networks in parallel between common terminals + and −. A simple random walker starts at +
and chooses uniformly among incident unit edges. Find the probability of reaching − before returning to +. (6
marks)
+ −
𝐴
𝐵
bridge network
two length-3 branches
(Total: 20 marks)
6
Worked solution and marking guidance
Question 2(a). Let potentials at A, B be a, b. KCL gives
(a − 1) + 2a + 2(a − b) = 0, 2(b − 1) + b + 2(b − a) = 0.
Thus 5a − 2b = 1, −2a + 5b = 2, whence
a = 3
7 , b = 4
7 .
The current from + is
CA = (1 − a) + 2(1 − b) = 4
7 + 6
7 = 10
7 .
8 marks, C
Question 2(b). Each branch has conductance 1 /3; in parallel,
CB = 2
3 .
6 marks, A
Question 2(c). Parallel conductances add:
C = CA + CB = 10
7 + 2
3 = 44
21 .
4
MATH40007 Equal-Difficulty Replacement Pack Solutions and Marking Guide
The starting vertex has five incident unit edges (three from the first network and two from the second), so
pesc = C
5 = 44
105 .
6 marks, D
Three-mass spring system
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 3
Three unit masses move horizontally between two fixed walls. Unit springs connect the left wall to mass 1, mass
1 to mass 2, mass 2 to mass 3, and mass 3 to the right wall. Let 𝑋 = (𝑋1, 𝑋2, 𝑋3)𝑇.
1 2 3
all masses and springs have unit value
(a) Write down the stiffness matrix 𝐾3, and find an orthonormal set of eigenvectors and the corresponding
eigenvalues. Hence give the natural frequencies. (8 marks)
(b) A constant external force 𝑓 = (1, 0, 1)𝑇 is applied. Find the static equilibrium 𝑋𝑠 by modal expansion,
including the modal coefficients. (8 marks)
(c) The system is released from rest at 𝑋 (0) = 𝑋𝑠 after the force is removed. Write the free oscillation 𝑋 (𝑡) in
modal form. (4 marks)
(Total: 20 marks)
7
Worked solution and marking guidance
Question 3(a). The stiffness matrix is
K3 =
2 −1 0
−1 2 −1
0 −1 2
.
An orthonormal eigensystem is
Φ1 = 1
2
1√
2
1
, λ 1 = 2 −
√
2,
Φ2 = 1√
2
1
0
−1
, λ 2 = 2,
Φ3 = 1
2
1
−
√
2
1
, λ 3 = 2 +
√
2.
The natural frequencies are ωj =
p
λj. 8 marks, A
Question 3(b). Write Xs =P ajΦj. Since K3Xs = f,
aj = ΦT
j f
λj
.
Here ΦT
1 f = 1, ΦT
2 f = 0, ΦT
3 f = 1, so
a1 = 1
2 −
√
2 , a 2 = 0, a 3 = 1
2 +
√
2 .
Substitution gives
Xs = (1, 1, 1)T .
8 marks, B
Question 3(c). With zero initial velocity after removal of the force,
X(t) = a1 cos(
q
2 −
√
2 t)Φ1 + a3 cos(
q
2 +
√
2 t)Φ3 .
4 marks, C
Complex potential and boundary current
MATH40007 Equal-Difficulty Replacement Pack Questions
Question 4
The upper half-plane 𝑦 > 0 is a conductor of unit conductivity. Let
𝜙(𝑥, 𝑦) = Re ℎ(𝑧), ℎ (𝑧) = − 𝑚
2𝜋 log
𝑧 − 𝑖
𝑧 + 𝑖
, 𝑧 = 𝑥 + 𝑖𝑦,
where 𝑚 > 0.
𝑥
𝑦
conducting region 𝑦 > 0
grounded boundary 𝑦 = 0
source at 𝑧 = 𝑖
image point 𝑧 = −𝑖
𝜙 = Re ℎ(𝑧)
(a) Identify the singularity in the conductor, show that the boundary 𝑦 = 0 is grounded, and find the complex
current density 𝐽𝑥 − 𝑖𝐽 𝑦 = −ℎ′ (𝑧). (10 marks)
(b) Find the normal current density on 𝑦 = 0, locate its maximum magnitude, and calculate the total current
leaving the conductor through the boundary. Explain why the result is consistent with conservation of current.
(10 marks)
(Total: 20 marks)
8
Worked solution and marking guidance
Question 4(a). The logarithm has a singularity at z = i, which is a point current source of strength m. The image
singularity at z = −i is outside the conductor. For z = x ∈ R,
x − i
x + i
= 1,
so ϕ(x, 0) = 0. Also
Jx − iJy = −h′(z) = m
2π
1
z − i − 1
z + i
= mi
π(z2 + 1).
10 marks, A
Question 4(b). On y = 0 (z = x),
Jx − iJy = mi
π(x2 + 1),
so
Jx = 0, J y = − m
π(x2 + 1).
The outward normal is −ey, hence outward current density is −Jy, maximal at x = 0 with magnitude m/π. The
total outward current is Z ∞
−∞
m
π(x2 + 1) dx = m.
This equals the strength of the sole source inside the conductor, as required by current conservation. 10 marks, D
5
MATH40007 Equal-Difficulty Replacement Pack Solutions and Marking Guide