EEE (Spring - 2019)

(1)

a) What do you mean by frequency response circuit? Why is frequency response so important for telecommunication engineering?

Answer:

1. Frequency Response Circuit:
   A frequency response circuit is a system that shows how the output of a circuit changes with different input frequencies. It helps to understand how well the circuit can handle various frequencies.

2. Importance in Telecommunication Engineering:

• It helps in designing circuits that can efficiently transmit signals over different frequencies.
• Ensures that the circuit can filter unwanted noise and maintain signal quality.
• Crucial for designing systems like radio, TV, and mobile communication to ensure clear and reliable communication.

c) What is a transfer function and how many types of transfer functions are there?

Answer:

1. Transfer Function:
   A transfer function is a mathematical expression that relates the output of a system to its input, showing how the system behaves for different frequencies.

2. Types of Transfer Functions:
   There are mainly two types:

   • Voltage Transfer Function: Relates output voltage to input voltage.
   • Current Transfer Function: Relates output current to input current.

(2)

(a) What do you mean by a balanced poly-phase system? How many combinations are possible between a three-phase source and a three-phase load connection?

Answer:

1. Balanced Poly-phase System:
   A balanced poly-phase system consists of multiple phases (typically three) where the voltages or currents in all phases are equal in magnitude and have a phase difference of 120 degrees.

2. Combinations of Source and Load Connections:
   There are two types of connections:

   • Star (Y) connection
   • Delta (Δ) connection
     These can be combined in six possible ways:
   • Y-connected source and Y-connected load
   • Y-connected source and Δ-connected load
   • Δ-connected source and Y-connected load
   • Δ-connected source and Δ-connected load

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(b) One line voltage of a balanced Y-connected source is VA=180∠20°V. If the source is connected to a Δ-connected load of 20∠40°, find the phase and line currents with a phasor diagram. Assume the abc sequence.

Answer:
Given:

• Line voltage of the Y-connected source: VA = 180∠20° V
• Load impedance (Δ-connected): ZL = 20∠40° Ω

1. Phase Voltage (Vph):
   Since it's a Y-connected source, phase voltage is the same as line voltage:
   Vph = 180∠20° V

2. Phase Current (Iph):
   Using Ohm's Law:
   Iph = Vph / ZL = (180∠20°) / (20∠40°)
   Iph = 9∠(-20°) A

3. Line Current (IL):
   For a Δ-connected load, the line current is √3 times the phase current:
   IL = √3 × Iph = √3 × 9∠(-20°)
   IL = 15.59∠(-20°) A

4. Phasor Diagram:
   Draw a phasor diagram showing the phase voltages and currents for both the source and the load, noting the 120° phase difference between each phase.

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(c) Draw the diagram for a 120/240 household power system.

Answer:
A 120/240 V household power system typically uses a split-phase connection, where the voltage is split into two 120 V lines and a neutral.

1. Diagram Explanation:
   • The system uses a single-phase transformer with a center-tapped secondary winding.
   • The center tap is grounded and serves as the neutral.
   • The two 120 V lines are 180 degrees out of phase with each other.
   • The total voltage between the two lines is 240 V, and each line-to-neutral voltage is 120 V.

Diagram:

• Two 120 V lines (L1 and L2) with a neutral in the middle.
• The voltage between L1 and neutral is 120 V, and between L1 and L2 is 240 V.

(3)

a) Derive the following equation for a 3-0 balanced star system: V = √3 * V1 and P = √3 * V1 * I * cos(θ).

Answer:

1. Voltage in a Balanced Star System (V = √3 * V1):
   In a balanced three-phase star system, the line voltage (V) is related to the phase voltage (V1) by the equation:
   $V = \sqrt{3} \times V_1$
   This is because in a star connection, the phase voltages are 120° apart, and the line voltage is the vector sum of two phase voltages.

2. Power in a Balanced Star System (P = √3 * V1 * I * cos(θ)):
   The total power (P) in a three-phase system is given by:
   $P = \sqrt{3} \times V_L \times I_L \times \cos(\theta)$
   Where:

   • $V_L$ is the line voltage
   • $I_L$ is the line current
   • $\theta$ is the phase angle between voltage and current

   Using the relation $V_L = \sqrt{3} \times V_1$, we substitute this into the equation for power:
   $P = \sqrt{3} \times V_1 \times I \times \cos(\theta)$
   This equation shows the total power in a balanced three-phase system.

(4)

a) What do you mean by a balanced poly-phase system? How many combinations are possible between a three-phase source and a three-phase load connection?

Answer:

1. Balanced Poly-phase System:
   A balanced poly-phase system consists of multiple phases (usually three) where the voltages or currents in all phases are equal in magnitude and separated by 120 degrees in phase.

2. Combinations between Source and Load Connections:
   There are six possible combinations between a three-phase source and a three-phase load:

   • Y-connected source and Y-connected load
   • Y-connected source and Δ-connected load
   • Δ-connected source and Y-connected load
   • Δ-connected source and Δ-connected load

(5)

(a) What do you mean by conductively coupled and electromagnetically coupled? Give examples for both of them.

Answer:

1. Conductively Coupled:
   This refers to a connection where two circuits are connected through a physical conductor, allowing direct flow of current.
   Example: Two resistors connected in series.

2. Electromagnetically Coupled:
   This refers to a connection where two circuits are linked through a magnetic field, without direct physical contact.
   Example: A transformer, where the primary coil induces current in the secondary coil through a magnetic field.

(6)

(a) Discuss an A.C. filter application in audio systems with a diagram.

Answer:
In audio systems, A.C. filters help improve sound quality by removing unwanted noise. These filters allow only certain frequencies to pass and block others.

Example in Audio System:
A low-pass filter in audio systems removes high-pitched noise (like hiss) and allows only the bass (low frequencies) to reach the speakers.

Diagram:

• A simple low-pass filter has a resistor (R) and capacitor (C) connected in series.
• The output is taken across the capacitor, allowing low frequencies to pass and blocking higher ones.

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(b) Explain the following filters with proper figures:

i) Low-pass Filter:
A low-pass filter allows low-frequency signals to pass and blocks high-frequency signals (like noise).

Diagram:

• R and C are connected in series.
• The output is taken across the capacitor, allowing lower frequencies to pass.

ii) High-Pass Filter:
A high-pass filter allows high-frequency signals to pass and blocks low-frequency signals (like hums).

Diagram:

• R and C are connected differently than in the low-pass filter.
• The output is taken across the resistor, allowing higher frequencies to pass.

iii) Pass Band:
The pass band is the range of frequencies that pass through the filter without being blocked.

Diagram:

• In the frequency graph, the pass band is the flat area where the signal remains strong.

iv) Stop Band:
The stop band is the range of frequencies that the filter blocks or reduces.

Diagram:

• In the frequency graph, the stop band is where the signal drops sharply, indicating blocked frequencies.