Problem

Wireless Communications Exercise I Q1: Why do we use the modulation technique? Q2: What is relationship between information frequency, \( f_{m} \), and carrier signal frequency \( f_{c} \) ? Q3: Write the mathematical formula for signals of the following: Carrier signal \( \mathrm{c}(\mathrm{t}) \) Amplitude Modulation \( \mathrm{M}_{\mathrm{AM}}(\mathrm{t}) \) Amplitude Modulation \( \mathrm{M}_{\mathrm{FM}}(\mathrm{t}) \) Q4: Sketch the modulated signals of the following digital modulation schemes for digitized information signal \( \mathrm{m}_{\mathrm{d}}(\mathrm{t}) \), and sketch the constellation diagram for each one? \[ \mathrm{m}_{\mathrm{d}}(\mathrm{t})=(1,0,0,1,1,0,0) \] Q5: explain the QAM modulation? Q6: Explain in detail the OFDMA ? Q7: Does \( x_{1}(t)=\cos (2 \pi f t) \), and \( x_{2}(t)=\sin (2 \pi f t) \) are orthogonal signals?

Solution

Step 1 :A1: Modulation techniques are used to represent information and transmit it over wireless channels with improved performance and robustness.

Step 2 :A2: The relationship between information frequency \(f_{m}\) and carrier signal frequency \(f_{c}\) is typically given by: \(f_c >> f_m\).

Step 3 :A3: Carrier signal: \(c(t) = A_c\cos(2\pi f_ct)\); AM: \(M_{AM}(t) = (A_c + m(t))\cos(2\pi f_ct)\); FM: \(M_{FM}(t) = A_c\cos(2\pi f_ct + k_fm\int m(\tau)d\tau)\)

Step 4 :A4: Sketch not possible with text; see resources such as lecture notes or textbooks for example diagrams.

Step 5 :A5: QAM (Quadrature Amplitude Modulation) combines two amplitude-modulated signals, with carriers in quadrature (90-degree phase shift), to transmit multiple bits per symbol.

Step 6 :A6: OFDMA (Orthogonal Frequency-Division Multiple Access) is a technique that allocates subsets of OFDM subcarriers to multiple users simultaneously, enabling efficient and flexible use of the available frequency spectrum.

Step 7 :A7: \(x_1(t)\) and \(x_2(t)\) are orthogonal signals if their integral product over a period is zero: \(\int_{0}^{T} x_1(t)x_2(t)dt = \int_{0}^{T} cos(2\pi ft)sin(2\pi ft)dt = 0\)

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