ECE 328 Undergraduate Electrical Engineering - Communication Systems Lab
Carrier Frequency: 38,000 kHz | Signal Frequencies: 1 kHz and 2 kHz
Analyze a Quadrature Amplitude Modulation (QAM) transmitter with a carrier frequency of 38,000 kHz, modulated by two baseband signals of 1 kHz and 2 kHz. Understand the time-domain and frequency-domain characteristics of the QAM signal.
Quadrature Amplitude Modulation (QAM) is a modulation scheme that conveys data by changing both the amplitude and phase of a carrier wave. In this lab, we'll examine a simple QAM transmitter where two different frequency signals modulate two carriers in quadrature (90° phase difference).
QAM signal formula: s(t) = m₁(t)cos(2πf₀t) + m₂(t)sin(2πf₀t)
Where: f₀ = 38,000 kHz, m₁(t) = cos(2πf₁t), m₂(t) = cos(2πf₂t), f₁ = 1 kHz, f₂ = 2 kHz
Time Domain Analysis: The QAM signal appears as a complex waveform resulting from the sum of two amplitude-modulated carriers in quadrature. The envelope varies according to the combination of the two baseband signals.
Frequency Domain Analysis: The spectrum shows sidebands around the carrier frequency (38,000 kHz) at offsets of ±1 kHz and ±2 kHz. Since we're using real signals (cosines), we see symmetric sidebands on both sides of the carrier.
Test your understanding of QAM transmitter analysis with the following questions.
Which of the following best describes the QAM signal in this experiment?
Correct Answer: C
Explanation: QAM uses two carriers at the same frequency but with a 90° phase difference (in quadrature). Each carrier is modulated by a different baseband signal. In this lab, the carrier frequency is 38,000 kHz, and the two baseband signals are 1 kHz and 2 kHz.
What frequency components would you expect to see in the spectrum of the QAM signal?
Correct Answer: B
Explanation: When a carrier is amplitude modulated by a sinusoidal signal, it produces sidebands at f_c ± f_m. With two modulation signals (1 kHz and 2 kHz), we get sidebands at 38,000 kHz ± 1 kHz (37,999 kHz and 38,001 kHz) and 38,000 kHz ± 2 kHz (37,998 kHz and 38,002 kHz), plus the carrier at 38,000 kHz.
What is the total bandwidth occupied by the QAM signal in this experiment?
Correct Answer: D
Explanation: The bandwidth is determined by the highest frequency component. With modulation frequencies of 1 kHz and 2 kHz, the highest sideband is at f_c ± 2 kHz. Therefore, the total bandwidth is 2 × 2 kHz = 4 kHz.
What is the primary advantage of using QAM compared to standard AM modulation?
Correct Answer: C
Explanation: The main advantage of QAM is spectral efficiency - it can transmit two independent signals (in-phase and quadrature) within the same bandwidth that would be required for a single AM signal. This effectively doubles the data rate without increasing bandwidth.
What is critical for successfully demodulating a QAM signal at the receiver?
Correct Answer: B
Explanation: For coherent demodulation of QAM signals, the receiver must generate local carriers that are exactly in phase with the transmitted carriers. Any phase error causes crosstalk between the in-phase and quadrature channels. A phase-locked loop (PLL) is typically used to recover the exact carrier phase and frequency.