Analogue QAM Receiver Signal Analysis Laboratory

Quadrature Amplitude Modulation with 38 MHz carrier and 1 kHz & 2 kHz baseband signals

CARRIER FREQUENCY
38,000 kHz (38 MHz)
BASEBAND SIGNALS
1 kHz and 2 kHz
MODULATION
Quadrature Amplitude (QAM)
LAB DURATION
Approx. 90 minutes

Pre-Lab Section

Objective

This laboratory exercise aims to analyze a receiver for a Quadrature Amplitude Modulation (QAM) signal. You will examine the time and frequency domain representations of a QAM signal with a carrier frequency of 38,000 kHz, modulated by two baseband signals of 1 kHz and 2 kHz.

Background Theory

Quadrature Amplitude Modulation (QAM) is a modulation scheme that conveys data by changing both the amplitude and phase of a carrier wave. It is widely used in modern communication systems like Wi-Fi, digital cable television, and cellular networks.

For this lab, we consider a QAM signal with two independent baseband signals:

  • In-phase component (I): 1 kHz sine wave
  • Quadrature component (Q): 2 kHz sine wave

The QAM signal can be expressed mathematically as:

s(t) = I(t)cos(2pi f_c t) + Q(t)sin(2pi f_c t)

Where:

  • f_c = 38,000  kHz} = 38  MHz  (carrier frequency)
  • I(t) = A_I cos(2pi f_I t) with f_I = 1  kHz
  • Q(t) = A_Q cos(2pi f_Q t) with f_Q = 2 kHz

At the receiver, the signal is demodulated by multiplying with in-phase and quadrature carriers followed by low-pass filtering:

$$ I_{rec}(t) = \text{LPF}\{s(t) \cdot 2\cos(2\pi f_c t)\} $$ $$ Q_{rec}(t) = \text{LPF}\{s(t) \cdot (-2\sin(2\pi f_c t))\} $$

Receiver Block Diagram

QAM Receiver Structure:

QAM Signal → [Bandpass Filter] → [Down-conversion]
→ [Mix with cos(2πf₀t)] → [Low-pass Filter] → I-channel output
→ [Mix with -sin(2πf₀t)] → [Low-pass Filter] → Q-channel output

Pre-Lab Questions

  1. What is the main advantage of QAM over simple amplitude modulation (AM)?
  2. Why do we need both in-phase and quadrature components in QAM?
  3. What will be the frequency components present in the transmitted QAM signal spectrum?
  4. Explain the purpose of the low-pass filters in the QAM receiver.

Signal Analysis Simulation

Use the controls below to adjust signal parameters and observe the effects on the time and frequency domain representations. The simulation shows the transmitted QAM signal and the recovered I and Q signals at the receiver.

38000
1.0
1.0
0.10

Time Domain Signals

Frequency Domain Signals

Interpretation of Graphs

Time Domain (QAM): Shows the modulated signal with high-frequency carrier oscillations. Time Domain (Recovered): Shows the demodulated I (1 kHz) and Q (2 kHz) signals.

Frequency Domain (QAM): Shows spectral components at f_c ± f_I and f_c ± f_Q. Frequency Domain (Recovered): Shows the baseband signals at 1 kHz and 2 kHz after demodulation.

Post-Lab Quiz

Test your understanding of QAM receiver analysis by answering the following questions. Select your answer and check it to see the explanation.

1. In a QAM receiver, what is the purpose of multiplying the received signal with both cosine and sine waveforms at the carrier frequency?
To separate the in-phase and quadrature components
To increase the signal amplitude
To reduce noise in the signal
To shift the signal to a higher frequency
Correct Answer: A
Multiplying with cosine extracts the I-component, while multiplying with sine extracts the Q-component. This is because cos² and sin² terms yield baseband signals after low-pass filtering, while cross terms produce high-frequency components that are filtered out.
2. For a QAM signal with carrier at 38 MHz and baseband signals at 1 kHz and 2 kHz, what frequency components would you expect to see in the transmitted spectrum?
1 kHz, 2 kHz, and 38 MHz only
37.999 MHz, 38.001 MHz, 37.998 MHz, and 38.002 MHz
38 MHz ± 1 kHz and 38 MHz ± 2 kHz
1 kHz, 2 kHz, 38 MHz, and 76 MHz
Correct Answer: C
The transmitted QAM signal contains sidebands at f_c ± f_I and f_c ± f_Q. So for f_c = 38 MHz, f_I = 1 kHz, and f_Q = 2 kHz, we expect components at 38 MHz ± 1 kHz and 38 MHz ± 2 kHz.
3. What would happen if the receiver's local oscillator frequency is slightly offset from the transmitter's carrier frequency (e.g., 37.999 MHz instead of 38.000 MHz)?
The recovered baseband signals would have a frequency offset of 1 kHz
The signal would be completely lost
The signal-to-noise ratio would improve
Only the Q-component would be affected
Correct Answer: A
A frequency offset causes the recovered baseband signals to be shifted by the offset amount (1 kHz in this case). This is because mixing with the wrong frequency doesn't completely demodulate the signal to baseband, leaving a residual frequency component.
4. Why is low-pass filtering necessary after the mixing stage in a QAM receiver?
To amplify the signal
To remove high-frequency components generated during mixing
To convert the signal to digital form
To match the impedance of the antenna
Correct Answer: B
The mixing process produces sum and difference frequencies. The low-pass filter removes the high-frequency sum components (around 2f_c) while passing the difference frequencies (baseband signals).
5. In the context of this lab, what would be the effect of increasing the amplitude of the Q-channel signal relative to the I-channel signal?
The 2 kHz component would become more dominant in the recovered signal
The carrier frequency would shift
The 1 kHz component would disappear
The signal would change from QAM to pure AM
Correct Answer: A
In QAM, the I and Q channels are independent. Increasing the amplitude of the Q-channel signal (2 kHz) makes this component more dominant in the overall modulated signal and in the recovered output, while the I-channel (1 kHz) remains unaffected.
Score: 0/5