Electronic Circuits & Networks (4331101) - Winter 2022 Solution

Table of Contents

Question 1(a) [3 marks]
#

Define: 1) Branch 2) Junction 3) Mesh

Answer:

Branch: A branch is a single circuit element or a combination of elements connected between two nodes of a network.
Junction: A junction (or node) is a point in a circuit where two or more circuit elements are connected together.
Mesh: A mesh is a closed path in a network where no other closed path exists inside it.

Mnemonic: “BJM: Branches Join at junctions to Make meshes”

Question 1(b) [4 marks]
#

Write voltage division and current division rule with necessary circuit diagram

Answer:

Voltage Division Rule: In a series circuit, voltage across any component is proportional to its resistance.

graph LR
    A((\+)) --- B[R1] --- C[R2] --- D((–))
    E[V1] -.- B
    F[V2] -.- C
    G[VS] -.- A

Formula: V₁ = VS × (R₁/(R₁+R₂))
Application: Used to find individual voltage drops across series components

Current Division Rule: In a parallel circuit, current through any branch is inversely proportional to its resistance.

graph LR
    A((\+)) --- B --- C((–))
    B --- D[R1] --- C
    B --- E[R2] --- C
    F[I1] -.- D
    G[I2] -.- E
    H[IS] -.- A

Formula: I₁ = IS × (R₂/(R₁+R₂))
Key concept: Current takes path of least resistance

Mnemonic: “VoSe CuPa: Voltage divides in Series, Current divides in Parallel”

Question 1(c) [7 marks]
#

Draw Graph and Tree for a network shown in fig(1). Show link currents on a graph. Also write Tie-set schedule for a tree of network shown in fig. (1)

Answer:

Graph of the Network:

graph LR
    A((A)) --- B((B))
    A --- C((C))
    A --- D((D))
    B --- C
    B --- D
    C --- D
    A -- 1 --- B
    A -- 3 --- C
    B -- 2 --- D
    C -- 5 --- D
    B -- 6 --- C
    A -- 7 --- D
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style D fill:#f9f,stroke:#333,stroke-width:2px

Tree of the Network (shown with bold edges):

graph LR
    A((A)) --- B((B))
    A --- C((C))
    C --- D((D))
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style D fill:#f9f,stroke:#333,stroke-width:2px
    linkStyle 0 stroke-width:4px,stroke:green
    linkStyle 1 stroke-width:4px,stroke:green
    linkStyle 2 stroke-width:4px,stroke:green

Link Currents (shown on remaining branches that are not part of the tree):

Link 1: Branch 2 (BD)
Link 2: Branch 6 (BC)
Link 3: Branch 7 (AD)
Link 4: Branch 5 (CD)

Tie-set Schedule:

Link/Tree Branch	Branch 1 (AB)	Branch 3 (AC)	Branch 4 (CD)	Branch 2 (BD)	Branch 6 (BC)	Branch 7 (AD)	Branch 5 (CD)
Link 1 (BD)	1	0	0	1	0	0	0
Link 2 (BC)	1	1	0	0	1	0	0
Link 3 (AD)	0	0	1	0	0	1	0
Link 4 (CD)	0	0	1	0	0	0	1

Mnemonic: “TGLT: Trees Generate Link-current Tie-sets”

Question 1(c) OR [7 marks]
#

Draw Graph and Tree for a network shown in fig(1). Show branch voltages on tree. Also write cut-set schedule for a tree of network shown on fig.(1)

Answer:

Graph of the Network:

graph LR
    A((A)) --- B((B))
    A --- C((C))
    A --- D((D))
    B --- C
    B --- D
    C --- D
    A -- 1 --- B
    A -- 3 --- C
    B -- 2 --- D
    C -- 5 --- D
    B -- 6 --- C
    A -- 7 --- D
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style D fill:#f9f,stroke:#333,stroke-width:2px

Tree of the Network (shown with bold edges and branch voltages):

graph LR
    A((A)) --"V₁"--> B((B))
    A --"V₃"--> C((C))
    C --"V₄"--> D((D))
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style D fill:#f9f,stroke:#333,stroke-width:2px
    linkStyle 0 stroke-width:4px,stroke:green
    linkStyle 1 stroke-width:4px,stroke:green
    linkStyle 2 stroke-width:4px,stroke:green

Cut-set Schedule:

Cut-set/Branch	Branch 1 (AB)	Branch 3 (AC)	Branch 4 (CD)	Branch 2 (BD)	Branch 6 (BC)	Branch 7 (AD)	Branch 5 (CD)
Cut-set 1 (AB)	1	0	0	-1	-1	0	0
Cut-set 2 (AC)	0	1	0	0	1	-1	0
Cut-set 3 (CD)	0	0	1	1	0	1	1

Mnemonic: “CGVS: Cut-sets Generate Voltage Sources”

Question 2(a) [3 marks]
#

Define: 1) Active and passive network 2)Unilateral and Bilateral network.

Answer:

Active Network: A network containing one or more sources of EMF (voltage/current sources) that supply energy to the circuit.
Passive Network: A network containing only passive elements like resistors, capacitors, and inductors with no energy sources.
Unilateral Network: A network in which the properties and performance change when input and output terminals are interchanged.
Bilateral Network: A network in which the properties and performance remain unchanged when input and output terminals are interchanged.

Diagram:

graph LR
    subgraph "Network Types"
    A[Active: Contains sources]
    B[Passive: No sources]
    C[Unilateral: Diodes/Transistors]
    D[Bilateral: R, L, C elements]
    end

Mnemonic: “APUB: Active Provides energy, Unilateral Blocks reversal”

Question 2(b) [4 marks]
#

Write equation for Z parameter and derive Z11, Z12, Z21, Z22 from that equation.

Answer:

Z-parameters define the relationship between port voltages and currents in a two-port network:

Equations:

V₁ = Z₁₁I₁ + Z₁₂I₂
V₂ = Z₂₁I₁ + Z₂₂I₂

Derivation:

Z₁₁ = V₁/I₁ (with I₂ = 0): Input impedance with output port open-circuited
Z₁₂ = V₁/I₂ (with I₁ = 0): Reverse transfer impedance with input port open-circuited
Z₂₁ = V₂/I₁ (with I₂ = 0): Forward transfer impedance with output port open-circuited
Z₂₂ = V₂/I₂ (with I₁ = 0): Output impedance with input port open-circuited

Mnemonic: “Z Impedance: Open circuit gives correct Parameters”

Question 2(c) [7 marks]
#

Derive equation of characteristic impedance(ZOT) for a standard T network.

Answer:

For a standard T-network:

graph LR
    A((Port-1)) --- B[Z1] --- C((Junction))
    C --- D[Z2] --- E((Port-2))
    C --- F[Z3] --- G((Ground))

Derivation Steps:

For a symmetric T-network, Z₁ = Z₂
Under matched condition, input impedance equals characteristic impedance
Z₀ₜ = Z₁ + (Z₁×Z₃)/(Z₁ + Z₃)
For balanced T-network where Z₁ = Z₂ = Z/2 and Z₃ = Z:
Z₀ₜ = Z/2 + (Z/2×Z)/(Z/2 + Z)
Z₀ₜ = Z/2 + (Z²/2)/(Z + Z/2)
Z₀ₜ = Z/2 + (Z²/2)/(3Z/2)
Z₀ₜ = Z/2 + Z²/3Z
Z₀ₜ = Z/2 + Z/3
Z₀ₜ = (3Z + 2Z)/6
Z₀ₜ = √(Z₁(Z₁ + 2Z₃))

Final Equation: Z₀ₜ = √(Z₁(Z₁ + 2Z₃))

Mnemonic: “TO Impedance: Two arms Over middle branch”

Question 2(a) OR [3 marks]
#

Define: 1)Driving point impedance 2) Transfer impedance

Answer:

Driving Point Impedance: The ratio of voltage to current at the same port/pair of terminals when all other independent sources are set to zero.
Transfer Impedance: The ratio of voltage at one port to the current at another port when all other independent sources are set to zero.

Diagram:

graph LR
    subgraph "Impedance Types"
    A[Driving Point: V₁/I₁ or V₂/I₂]
    B[Transfer: V₂/I₁ or V₁/I₂]
    end

Mnemonic: “DTSS: Driving at Terminal Same, Transfer at Separate”

Question 2(b) OR [4 marks]
#

Explain Kirchhoff’s voltage law with example.

Answer:

Kirchhoff’s Voltage Law (KVL): The algebraic sum of all voltages around any closed loop in a circuit is zero.

Mathematically: ∑V = 0 (around a closed loop)

Circuit Example:

graph LR
    A((\+)) --"10V"--> B
    B --"R₁ = 2Ω"--> C
    C --"R₂ = 3Ω"--> D
    D --"R₃ = 5Ω"--> A
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style D fill:#f9f,stroke:#333,stroke-width:2px

If I = 1A, then:

V₁ = 1A × 2Ω = 2V
V₂ = 1A × 3Ω = 3V
V₃ = 1A × 5Ω = 5V

Applying KVL: 10V - 2V - 3V - 5V = 0 ✓

Mnemonic: “VACZ: Voltages Around Closed loop are Zero”

Question 2(c) OR [7 marks]
#

Derive equation to convert π network into T network.

Answer:

π Network to T Network Conversion:

graph TD
    subgraph "π Network"
    A1((A)) --- B1((B))
    A1 --- Y1[Ya] --- C1
    B1 --- Y2[Yb] --- C1
    A1 --- Y3[Yc] --- B1
    C1((C))
    end

    subgraph "T Network"
    A2((A)) --- Z1[Za] --- D2((D))
    B2((B)) --- Z2[Zb] --- D2
    D2 --- Z3[Zc] --- C2((C))
    end

Conversion Equations:

Za = (Ya × Yc) / Y∆
Zb = (Yb × Yc) / Y∆
Zc = (Ya × Yb) / Y∆

Where Y∆ = Ya + Yb + Yc

Derivation:

Start with Y-parameters of π-network
Express Y-parameters in terms of branch admittances
Convert to Z-parameters using matrix inversion
Express T-network impedances in terms of Z-parameters
Simplify to get the conversion formulas above

Mnemonic: “PIE to TEA: Product over sum for opposite branch”

Question 3(a) [3 marks]
#

Explain Kirchhoff’s current law with example.

Answer:

Kirchhoff’s Current Law (KCL): The algebraic sum of all currents entering and leaving a node must equal zero.

Mathematically: ∑I = 0 (at any node)

Circuit Example:

graph TD
    A[I₁ = 5A] --> B((Node))
    C[I₂ = 2A] --> B
    B --> D[I₃ = 3A]
    B --> E[I₄ = 4A]
    style B fill:#f9f,stroke:#333,stroke-width:2px

Applying KCL at node B:

Currents entering: I₁ + I₂ = 5A + 2A = 7A
Currents leaving: I₃ + I₄ = 3A + 4A = 7A
Therefore: I₁ + I₂ - I₃ - I₄ = 5 + 2 - 3 - 4 = 0 ✓

Mnemonic: “CuNoZ: Currents at Node are Zero”

Question 3(b) [4 marks]
#

Explain mesh analysis with required equations.

Answer:

Mesh Analysis: A circuit analysis technique that uses mesh currents as variables to solve a circuit with multiple loops.

Steps:

Identify all meshes (closed loops) in the circuit
Assign a mesh current to each mesh
Apply KVL to each mesh
Solve the resulting system of equations

Example Circuit:

graph LR
    A((A)) -- R₁ --- B((B))
    B -- R₃ --- C((C))
    A -- R₂ --- C
    A -- V₁ --- D
    D -- \+ --- A
    C -- V₂ --- E
    E -- \+ --- C
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px

Equations:

Mesh 1: V₁ = I₁R₁ + I₁R₂ - I₂R₂
Mesh 2: V₂ = I₂R₂ + I₂R₃ - I₁R₂

Mnemonic: “MILK: Mesh Is Loop with KVL”

Question 3(c) [7 marks]
#

State and explain Thevenin’s theorem.

Answer:

Thevenin’s Theorem: Any linear network with voltage and current sources can be replaced by an equivalent circuit consisting of a voltage source (VTH) in series with a resistance (RTH).

graph TD
    subgraph "Original Network"
    A((A)) --- B[Complex Network] --- C((B))
    end
    subgraph "Thevenin Equivalent"
    D((A)) --- E[VTH] --- F((\+))
    F --- G[RTH] --- H((B))
    end

Steps to Find Thevenin Equivalent:

Remove the load from the terminals of interest
Calculate the open-circuit voltage (VOC) across these terminals (= VTH)
Calculate the resistance looking back into the circuit with all sources replaced by their internal resistances (= RTH)
The Thevenin equivalent consists of VTH in series with RTH

Example Application:

Original complex circuit with load RL
Remove RL and find VOC = VTH
Deactivate sources and find RTH
Reconnect RL to simplified Thevenin equivalent

Mnemonic: “TORV: Thevenin’s Open-circuit Resistance and Voltage”

Question 3(a) OR [3 marks]
#

State and explain reciprocity theorem.

Answer:

Reciprocity Theorem: In a linear, bilateral network, if a voltage source in one branch produces a current in another branch, then the same voltage source, if placed in the second branch, will produce the same current in the first branch.

graph LR
    subgraph "Original Circuit"
    direction LR
    A((A)) --- B[V] --- C((B))
    C --- D[Network] --- E((C))
    E --- F[Ammeter] --- A
    end

    subgraph "Reciprocal Circuit"
    direction LR
    G((A)) --- H[Ammeter] --- I((B))
    I --- J[Network] --- K((C))
    K --- L[V] --- G
    end

Mathematically: If a voltage V₁ in branch 1 produces current I₂ in branch 2, then voltage V₁ in branch 2 will produce current I₂ in branch 1.

Limitations: Applies only to networks with:

Linear elements
Bilateral elements (no diodes, transistors)
Single independent source

Mnemonic: “RESWAP: REciprocity SWAPs Position with identical results”

Question 3(b) OR [4 marks]
#

Explain nodal analysis with required equations.

Answer:

Nodal Analysis: A circuit analysis technique that uses node voltages as variables to solve a circuit.

Steps:

Choose a reference node (ground)
Assign voltage variables to remaining nodes
Apply KCL at each non-reference node
Solve the resulting system of equations

Example Circuit:

graph LR
    A((Node 1)) -- G₁ --- B((Ground))
    C((Node 2)) -- G₂ --- B
    A -- G₃ --- C
    A -- I₁ --> B
    C -- I₂ --> B
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#f9f,stroke:#333,stroke-width:2px

Equations:

Node 1: I₁ = V₁G₁ + (V₁-V₂)G₃
Node 2: I₂ = V₂G₂ + (V₂-V₁)G₃

Mnemonic: “NKCV: Nodal uses KCL with Voltage variables”

Question 3(c) OR [7 marks]
#

State and prove maximum power transfer theorem.

Answer:

Maximum Power Transfer Theorem: A load connected to a source will extract maximum power when its resistance equals the internal resistance of the source.

graph LR
    A((\+)) --- B[VS] --- C((X))
    C --- D[RS] --- E((Y))
    E --- F[RL] --- G((Z))
    G --- A
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style E fill:#f9f,stroke:#333,stroke-width:2px

Proof:

Current in the circuit: I = VS/(RS + RL)
Power delivered to load: P = I²RL = (VS²RL)/(RS + RL)²
For maximum power, dP/dRL = 0
Solving: (VS²(RS + RL)² - VS²RL·2(RS + RL))/(RS + RL)⁴ = 0
Simplifying: (RS + RL)² = 2RL(RS + RL)
Further simplifying: RS + RL = 2RL
Therefore: RS = RL

Maximum Power: Pmax = VS²/(4RS)

Mnemonic: “MaRLRS: Maximum power when load Resistance equals Source Resistance”

Question 4(a) [3 marks]
#

Why series resonance circuit act as voltage amplifier and parallel resonance circuit act as current amplifier?

Answer:

Series Resonance as Voltage Amplifier:

At resonance, series circuit impedance is minimum (just R)
Voltage across L or C can be much larger than source voltage
Voltage magnification factor = Q = XL/R = 1/R√(L/C)
Voltage across L or C = Q × Source voltage

Parallel Resonance as Current Amplifier:

At resonance, parallel circuit impedance is maximum
Current in L or C can be much larger than source current
Current magnification factor = Q = R/XL = R√(C/L)
Current through L or C = Q × Source current

Table:

Circuit Type	Impedance at Resonance	Amplification
Series	Minimum (R only)	Voltage (VL or VC = Q×VS)
Parallel	Maximum (R²/r)	Current (IL or IC = Q×IS)

Mnemonic: “SeVoPa: Series Voltage, Parallel current amplification”

Question 4(b) [4 marks]
#

Derive equation of Q of coil.

Answer:

Q-factor of a Coil:

graph LR
    A((A)) --- B[R] --- C((B))
    C --- D[L] --- A
    style C fill:#f9f,stroke:#333,stroke-width:2px

Derivation:

Q-factor is defined as: Q = Energy stored / Energy dissipated per cycle
Energy stored in inductor = (1/2)LI²
Power dissipated in resistor = I²R
Energy dissipated per cycle = Power × Time period = I²R × (1/f)
Therefore: Q = ((1/2)LI²) / (I²R × (1/f))
Simplifying: Q = 2π × (1/2)LI² × f / (I²R)
Q = 2πf × L / R = ωL / R

Final Equation: Q = ωL / R = 2πfL / R = XL / R

Mnemonic: “QualityEDR: Quality equals Energy stored Divided by energy lost per Radian”

Question 4(c) [7 marks]
#

Derive equation of series resonance frequency for series R-L-C circuit.

Answer:

Series R-L-C Circuit:

graph LR
    A((Input)) --- B[R] --- C[L] --- D[C] --- E((Output))
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style E fill:#f9f,stroke:#333,stroke-width:2px

Derivation:

Impedance of series RLC circuit: Z = R + j(XL - XC)
Where: XL = ωL and XC = 1/ωC
At resonance, XL = XC (inductive and capacitive reactances are equal)
Therefore: ωL = 1/ωC
Solving for ω: ω² = 1/LC
Resonant frequency: ω₀ = 1/√(LC)
In terms of frequency f: f₀ = 1/(2π√(LC))

Characteristics at Resonance:

Impedance is minimum (purely resistive: Z = R)
Current is maximum (I = V/R)
Power factor is unity (circuit appears resistive)
Voltages across L and C are equal and opposite

Mnemonic: “RES: Reactances Equal at Series resonance”

Question 4(a) OR [3 marks]
#

What is coupled circuits? Define self-inductance and mutual inductance.

Answer:

Coupled Circuits: Two or more circuits that are magnetically linked such that energy can be transferred between them through their mutual magnetic field.

graph TD
    subgraph "Primary"
    A((A)) --- B[L1] --- C((B))
    end

    subgraph "Secondary"
    D((C)) --- E[L2] --- F((D))
    end
    
    G[M] -.-> B
    G -.-> E

Self-inductance (L): The property of a circuit whereby a change in current produces a self-induced EMF in the same circuit. L = Φ/I (ratio of magnetic flux to the current producing it)

Mutual inductance (M): The property of a circuit whereby a change in current in one circuit induces an EMF in another circuit. M = Φ₂₁/I₁ (ratio of flux in circuit 2 due to current in circuit 1)

Mnemonic: “SiMu: Self in Mine, Mutual in Yours”

Question 4(b) OR [4 marks]
#

Derive equation for co-efficient of coupling (K).

Answer:

Coefficient of Coupling (k):

graph LR
    subgraph "Coupled Coils"
    A((A)) --- B[L1] --- C((B))
    D((C)) --- E[L2] --- F((D))
    G[M] -.-> B
    G -.-> E
    end

Derivation:

The mutual inductance (M) between two coils depends on:
- Self-inductances of the coils (L₁ and L₂)
- Physical arrangement (proximity and orientation)
Maximum possible mutual inductance: Mₘₐₓ = √(L₁L₂)
Coefficient of coupling is defined as: k = M/Mₘₐₓ
Therefore: k = M/√(L₁L₂)

Characteristics:

k ranges from 0 (no coupling) to 1 (perfect coupling)
k depends on geometry, orientation, and medium
Typical transformers: k = 0.95 to 0.99
Air-core coils: k = 0.01 to 0.5

Mnemonic: “KMutual: K Measures Mutual linkage proportion”

Question 4(c) OR [7 marks]
#

A series RLC circuit has R=30Ω, L=0.5H, and C=5µF. Calculate (i) series resonance frequency (2) Q Factor (3)BW

Answer:

Given:

Resistance, R = 30Ω
Inductance, L = 0.5H
Capacitance, C = 5µF = 5×10⁻⁶F

Calculations:

(i) Series Resonance Frequency:

f₀ = 1/(2π√(LC))
f₀ = 1/(2π√(0.5 × 5×10⁻⁶))
f₀ = 1/(2π√(2.5×10⁻⁶))
f₀ = 1/(2π × 1.58×10⁻³)
f₀ = 1/(9.9×10⁻³)
f₀ = 100.76 Hz
f₀ ≈ 100 Hz

(ii) Q Factor:

Q = (1/R)√(L/C)
Q = (1/30)√(0.5/(5×10⁻⁶))
Q = (1/30)√(100,000)
Q = (1/30) × 316.23
Q = 10.54

(iii) Bandwidth (BW):

BW = f₀/Q
BW = 100.76/10.54
BW = 9.56 Hz

Table:

Parameter	Formula	Value
Resonant Frequency (f₀)	1/(2π√(LC))	100 Hz
Quality Factor (Q)	(1/R)√(L/C)	10.54
Bandwidth (BW)	f₀/Q	9.56 Hz

Mnemonic: “RQB: Resonance Quality determines Bandwidth”

Question 5(a) [3 marks]
#

Classify various types of attenuators.

Answer:

Attenuators: Network of resistors designed to reduce (attenuate) signal level without distortion.

Types of Attenuators:

graph TD
    A[Attenuators] --> B[Fixed Attenuators]
    A --> C[Variable Attenuators]
    B --> D[T-type]
    B --> E[π-type]
    B --> F[Bridged-T]
    B --> G[Lattice]
    C --> H[Step Attenuators]
    C --> I[Continuously Variable]

Based on configuration:

T-type: Three resistor T-shaped configuration
π-type: Three resistor π-shaped configuration
Bridged-T: T-type with a resistor bridging across
Lattice: Balanced configuration with four resistors

Based on symmetry:

Symmetrical: Equal input and output impedance
Asymmetrical: Different input and output impedance

Mnemonic: “ATP Fixed: Attenuator Types include Pad, Tee, Lattice”

Question 5(b) [4 marks]
#

Derive relation between attenuator and neper.

Answer:

Relationship between Attenuation and Neper:

Attenuation (α): Ratio of input voltage (or current) to output voltage (or current), expressed in different units.
Neper (Np): Natural logarithmic unit of ratios, used mainly in transmission line theory.

Derivation:

For a voltage ratio V₁/V₂:
- Attenuation in Nepers = ln(V₁/V₂)
- Attenuation in Decibels = 20log₁₀(V₁/V₂)
For a power ratio P₁/P₂:
- Attenuation in Nepers = (1/2)ln(P₁/P₂)
- Attenuation in Decibels = 10log₁₀(P₁/P₂)
Relationship between dB and Neper:
- 1 Neper = 8.686 dB
- 1 dB = 0.115 Neper

Table:

Unit	Voltage Ratio	Power Ratio
Neper (Np)	ln(V₁/V₂)	(1/2)ln(P₁/P₂)
Decibel (dB)	20log₁₀(V₁/V₂)	10log₁₀(P₁/P₂)

Mnemonic: “NED: Neper Equals Decibel divided by 8.686”

Question 5(c) [7 marks]
#

Derive equations of R1 and R2 for symmetrical T attenuator.

Answer:

Symmetrical T Attenuator:

graph LR
    A((Input)) --- B[R1] --- C((Junction))
    C --- D[R1] --- E((Output))
    C --- F[R2] --- G((Ground))
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style E fill:#f9f,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px

Derivation:

For a symmetrical T-attenuator with characteristic impedance Z₀:
- Input and output impedance must both equal Z₀
- Attenuation ratio N = V₁/V₂ = I₂/I₁
From circuit analysis:
- Z₀ = R₁ + (R₂(R₁))/(R₂+R₁)
- N = (R₁ + R₂ + R₁)/R₂ = (2R₁+R₂)/R₂
Solving for R₁ and R₂:
- R₁ = Z₀(N-1)/(N+1)
- R₂ = 2Z₀N/(N²-1)
For attenuation in dB (α):
- N = 10^(α/20)
- R₁ = Z₀·tanh(α/2)
- R₂ = Z₀/sinh(α)

Final Equations:

R₁ = Z₀(N-1)/(N+1)
R₂ = 2Z₀N/(N²-1)

Mnemonic: “TSR: T-attenuator Symmetry Requires equal R1 values”

Question 5(a) OR [3 marks]
#

Draw circuit diagram of symmetrical Bridge T and symmetrical Lattice attenuator.

Answer:

Symmetrical Bridge-T Attenuator:

Symmetrical Lattice Attenuator:

Characteristics:

Bridge-T: Combines features of T and π attenuators, suitable for high-frequency applications
Lattice: Balanced configuration with excellent phase and frequency response, commonly used in balanced lines

Mnemonic: “BL-BA: Bridge Ladder, Balanced Attenuators”

Question 5(b) OR [4 marks]
#

Write classification of filter based on frequency with their frequency responses showing pass band and stop band.

Answer:

Classification of Filters Based on Frequency:

graph TD
    A[Passive Filters] --> B[Low Pass Filter]
    A --> C[High Pass Filter]
    A --> D[Band Pass Filter]
    A --> E[Band Stop Filter]
    A --> F[All Pass Filter]

Frequency Responses:

Low Pass Filter: Passes frequencies below cutoff, attenuates above

Gain |
   1 |****
     |    ****
     |        ****
   0 |------------****----
     |
     +----------------------
        0     fc        f →

High Pass Filter: Passes frequencies above cutoff, attenuates below

Gain |
   1 |            ****
     |        ****
     |    ****
   0 |****-----------------
     |
     +----------------------
        0     fc        f →

Band Pass Filter: Passes frequencies within a specific band

Gain |
   1 |        ****
     |    ****    ****
     |   *          *
   0 |***-------------***--
     |
     +----------------------
        0   f1   f2     f →

Band Stop Filter: Rejects frequencies within a specific band

Gain |
  1 |***             ***
     |   *           *
     |    ***     ***
  0 |        *****
     |
     +----------------------
        0   f1   f2     f →

Mnemonic: “LHBBA: Low High Band-pass Band-stop All-pass”

Question 5(c) OR [7 marks]
#

Draw the circuit for T-section and π-section constant-K low pass filter and Derive equation of cut-off frequency.

Answer:

T-section Constant-K Low Pass Filter:

π-section Constant-K Low Pass Filter:

Derivation of Cutoff Frequency:

For a constant-K filter:
- Z₁ × Z₂ = R₀² (characteristic impedance squared)
- Z₁ = jωL (series impedance)
- Z₂ = 1/jωC (shunt impedance)
Therefore:
- R₀² = Z₁ × Z₂ = jωL × 1/jωC = L/C
- R₀ = √(L/C)
Pass band condition:
- -1 < Z₁/4Z₂ < 0
- -1 < jωL/(4 × 1/jωC) < 0
- -1 < -ω²LC/4 < 0
At cutoff frequency:
- ω²LC/4 = 1
- ωc² = 4/LC
- ωc = 2/√(LC)
- fc = ωc/2π = 1/π√(LC)