Chapter 3 - Modulation

Question 1

14. Sketch an AMDSB signal in the time and frequency domains. Label the main components of the signal.

Answer 1

Question 2

15. Contrast and compare AMDSB and DSBSC in the frequency domain.

Answer 2

1. AMDSB:
   
   • The AMDSB signal includes both the carrier frequency and two sidebands (upper and lower) containing the same information.
   • In the frequency domain, the carrier appears as a strong central peak, with sidebands symmetrically located on either side.
   • The bandwidth required for AMDSB is twice the frequency of the highest modulating signal.
2. DSBSC:
   
   • DSBSC transmits only the two sidebands without the carrier.
   • In the frequency domain, the carrier peak is absent, resulting in only the sidebands being visible.
   • The bandwidth required for DSBSC is the same as for AMDSB, but without the carrier, it is more power-efficient.

While both occupy the same bandwidth, DSBSC is more efficient as it eliminates the carrier, allowing more power to be dedicated to transmitting the information-carrying sidebands.

Question 3

16. Compare the bandwidth of AMDSB and AMISB if both sidebands are the same bandwidth.

Question 4

17. Why do we use a pilot carrier in AM?

Answer 4

1. Provide a frequency reference: Ensures accurate demodulation.
2. Enable synchronization: Keeps the receiver in sync with the transmitted signal.
3. Improve demodulation accuracy: Reduces errors in signal reconstruction.

Question 5

18. With the aid of a diagram (WTAD) demonstrate how a carrier signal is baseband modulated in an FM system. What effect occurs when an FM signal falls below a certain level?

Question 6

19. What functions accurately measure the bandwidth of an FM signal?

Answer 6

Spectrum Analyser
Carsons Rule
Fast Fourier Transform

Question 7

20. If an FM system has a modulating baseband with a period of 250 ms and the maximum and minimum frequency outputs are 100 MHz and 99.85 MHz respectively. What is the Carson Bandwidth?

Answer 7

Carsons BW = 2(∆f + fm)
∆f = (fmax - fmin) / 2
Fm = 1 / Tm

∆f = (100Mhz - 99.85Mhz) / 2 = 0.075 = 75kHz
fm = 1 / 0.25s = 4 Hz

BW = 2(75000 Hz + 4 Hz) = 158kHz