An AM modulator has two inputs. The first input is a sine wave with a frequency of 500Hz and amplitude of 2V RMS. The second input is a carrier signal with RMS voltage of 4 V and frequency 1.5 MHz. Determine the equation for the modulated signal.

Here are the solutions to the questions from the exam paper:

1a.i. Determine the equation for the modulated signal.

Step 1: Identify parameters for the modulating signal.
The modulating signal has a frequency $f_m = 500$ Hz and an RMS amplitude $V_{m,rms} = 2$ V.
The peak amplitude is $V_m = V_{m,rms} \times \sqrt{2} = 2\sqrt{2}$ V.
The equation for the modulating signal is $v_m(t) = 2\sqrt{2} \sin(2\pi \times 500 t)$.

Step 2: Identify parameters for the carrier signal.
The carrier signal has a frequency $f_c = 1.5$ MHz $= 1.5 \times 10^6$ Hz and an RMS voltage $V_{c,rms} = 4$ V.
The peak amplitude is $V_c = V_{c,rms} \times \sqrt{2} = 4\sqrt{2}$ V.
The equation for the carrier signal is $v_c(t) = 4\sqrt{2} \sin(2\pi \times 1.5 \times 10^6 t)$.

Step 3: Calculate the modulation index ($m_a$).
The modulation index for AM is the ratio of the peak modulating voltage to the peak carrier voltage:

$$m_a = \frac{V_m}{V_c} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2} = 0.5$$

Step 4: Write the equation for the AM modulated signal.
The general equation for an AM modulated signal is $v_{AM}(t) = V_c [1 + m_a \sin(2\pi f_m t)] \sin(2\pi f_c t)$.
Substitute the values:

$$v_{AM}(t) = 4\sqrt{2} [1 + 0.5 \sin(2\pi \times 500 t)] \sin(2\pi \times 1.5 \times 10^6 t)$$

The equation for the modulated signal is $\boxed{v_{AM}(t) = 4\sqrt{2} [1 + 0.5 \sin(1000\pi t)] \sin(3 \times 10^6 \pi t) \text{ V}}$.

1a.ii. Define the following terms as used in radio receiver:

• Sensitivity: The ability of a radio receiver to detect and amplify very weak input signals. A highly sensitive receiver can pick up distant or low-power transmissions.
• Selectivity: The ability of a radio receiver to differentiate between a desired radio signal and other unwanted signals on adjacent frequencies. A highly selective receiver can tune into one station without interference from others.
• Fidelity: The accuracy with which a radio receiver reproduces the original modulating signal (audio or data) without distortion or loss of quality. High fidelity means the output closely matches the input.

1b. Illustrate two (2) advantages and two (2) disadvantages of using diode detector in AM broadcast receiver.

• Advantages of Diode Detector:

  1. Simplicity: Diode detectors have a very simple circuit design, making them easy and inexpensive to implement.
  2. No External Power: For basic detection, a diode detector does not require an external power supply, as it rectifies the incoming RF signal directly.

• Disadvantages of Diode Detector:

  1. Low Sensitivity: Diode detectors are generally less sensitive than other types of detectors, meaning they are not ideal for detecting very weak signals.
  2. Distortion: They can introduce distortion into the demodulated audio signal, especially at higher modulation indices, leading to lower fidelity.

2a. Sketch the equivalent circuit diagram of a high frequency practical transmission line and define the primary and secondary constants.

• Equivalent Circuit Diagram: A high-frequency practical transmission line can be represented by a series of infinitesimal sections, each containing:

  • Series Resistance ($R$)
  • Series Inductance ($L$)
  • Shunt Conductance ($G$)
  • Shunt Capacitance ($C$) (Imagine a small segment of a transmission line. It has resistance and inductance along its length, and capacitance and conductance between the two conductors.)

• Primary Constants: These are the physical electrical properties of the transmission line per unit length.

  • Resistance (R): The series resistance of the conductors per unit length, measured in Ohms per meter ($\Omega$/m).
  • Inductance (L): The series inductance of the conductors per unit length, measured in Henrys per meter (H/m).
  • Conductance (G): The shunt conductance between the conductors per unit length, representing leakage current through the dielectric, measured in Siemens per meter (S/m).
  • Capacitance (C): The shunt capacitance between the conductors per unit length, measured in Farads per meter (F/m).

• Secondary Constants: These are derived from the primary constants and describe the overall behavior of the transmission line.

  • Characteristic Impedance ($Z_0$): The impedance seen looking into an infinitely long transmission line, or a finite line terminated with its characteristic impedance. It is measured in Ohms ($\Omega$).
  • Propagation Constant ($\gamma$): A complex number that describes how the voltage and current waves propagate along the line. It determines both the attenuation and phase shift per unit length. It is measured in per meter (m$^{-1}$).

2b. Calculate the characteristics impedance, propagation constant, attenuation constant, and phase constant.

Step 1: Convert given primary constants to per meter and calculate angular frequency.
Given:
$R = 22 \, \Omega/\text{Km} = 0.022 \, \Omega/\text{m}$
$L = 5 \, \text{mH/Km} = 5 \times 10^{-3} \, \text{H/Km} = 5 \times 10^{-6} \, \text{H/m}$
$C = 0.04 \, \text{mF/Km} = 0.04 \times 10^{-3} \, \text{F/Km} = 0.04 \times 10^{-6} \, \text{F/m}$
$G = 0 \, \text{S/Km}$
Frequency $f = 1 \, \text{KHz} = 1000 \, \text{Hz}$
Angular frequency $\omega = 2\pi f = 2\pi \times 1000 = 2000\pi \, \text{rad/s}$

Step 2: Calculate the series impedance ($Z$) and shunt admittance ($Y$).
Series impedance $Z = R + j\omega L$:

$$Z = 0.022 + j(2000\pi)(5 \times 10^{-6})$$ $$Z = 0.022 + j(0.01\pi) \, \Omega/\text{m}$$ $$Z \approx 0.022 + j0.031416 \, \Omega/\text{m}$$

Shunt admittance $Y = G + j\omega C$:

$$Y = 0 + j(2000\pi)(0.04 \times 10^{-6})$$ $$Y = j(80\pi \times 10^{-6}) \, \text{S/m}$$ $$Y \approx j0.0002513 \, \text{S/m}$$

Step 3: Calculate the characteristic impedance ($Z_0$).
$Z_0 = \sqrt{\frac{Z}{Y}}$
First, convert $Z$ and $Y$ to polar form:
$|Z| = \sqrt{0.022^2 + (0.01\pi)^2} \approx \sqrt{0.000484 + 0.00098696} \approx \sqrt{0.00147096} \approx 0.03835 \, \Omega/\text{m}$
$\theta_Z = \arctan\left(\frac{0.01\pi}{0.022}\right) \approx \arctan(1.4289) \approx 54.99^\circ$
So, $Z \approx 0.03835 \angle 54.99^\circ \, \Omega/\text{m}$
$Y \approx 0.0002513 \angle 90^\circ \, \text{S/m}$
Now, calculate $\frac{Z}{Y}$:

$$\frac{Z}{Y} = \frac{0.03835 \angle 54.99^\circ}{0.0002513 \angle 90^\circ} = \left(\frac{0.03835}{0.0002513}\right) \angle (54.99^\circ - 90^\circ) \approx 152.60 \angle -35.01^\circ$$

Finally, calculate $Z_0$:

$$Z_0 = \sqrt{152.60 \angle -35.01^\circ} = \sqrt{152.60} \angle \frac{-35.01^\circ}{2}$$ $$Z_0 \approx 12.35 \angle -17.51^\circ \, \Omega$$