1. **Problem Statement:** We have a message signal $$m(t) = \begin{cases} 3\cos(50\pi t) + \sin(10\pi t), & t \leq 2 \\ 0, & t > 2 \end{cases}$$ and a carrier signal $$c(t) = \cos(500\pi t)$$ with amplitude $$A_C = 12$$ volts. We want to: - (a) Sample the message signal at 100 Hz to get 256 points. - (b) Sample the conventional AM signal $$x_{AM}(t) = A_C[1 + m(t)]c(t)$$ at 1000 Hz to get 2048 points. - (c) Compute and plot the normalized magnitude spectrum of the message signal. - (d) Compute and plot the normalized magnitude spectrum of the AM signal. 2. **Formulas and Important Rules:** - Sampling: $$m[n] = m\left(\frac{n}{f_s}\right)$$ where $$f_s$$ is the sampling frequency. - Conventional AM signal: $$x_{AM}(t) = A_C[1 + m(t)]c(t)$$. - Discrete Fourier Transform (DFT): $$M[k] = \sum_{n=0}^{N-1} m[n] e^{-j2\pi kn/N}$$. - Normalized magnitude spectrum: $$\frac{|M[k]|}{\max(|M[k]|)}$$. - Use fftshift to center zero frequency component. 3. **Step-by-step Solution:** **(a) Message Signal Sampling:** - Sampling frequency $$f_s = 100$$ Hz. - Sampling interval $$T_s = \frac{1}{f_s} = 0.01$$ seconds. - Number of points $$N = 256$$. - Time vector: $$t_n = nT_s, n=0,1,...,255$$. - For $$t_n \leq 2$$, $$m[n] = 3\cos(50\pi t_n) + \sin(10\pi t_n)$$; else $$m[n] = 0$$. **(b) Conventional AM Signal Sampling:** - Sampling frequency $$f_{s,AM} = 1000$$ Hz. - Sampling interval $$T_{s,AM} = 0.001$$ seconds. - Number of points $$N_{AM} = 2048$$. - Time vector: $$t_{AM,n} = nT_{s,AM}, n=0,1,...,2047$$. - Compute $$m(t_{AM,n})$$ similarly as above. - Compute $$x_{AM}[n] = A_C [1 + m(t_{AM,n})] \cos(500\pi t_{AM,n})$$. **(c) Message Signal Spectrum Analysis:** - Compute DFT: $$M[k] = \text{fft}(m[n])$$. - Normalize magnitude: $$|M[k]|_{norm} = \frac{|M[k]|}{\max(|M[k]|)}$$. - Use fftshift to center zero frequency. **(d) Conventional AM Signal Spectrum Analysis:** - Compute DFT: $$X_{AM}[k] = \text{fft}(x_{AM}[n])$$. - Normalize magnitude: $$|X_{AM}[k]|_{norm} = \frac{|X_{AM}[k]|}{\max(|X_{AM}[k]|)}$$. - Use fftshift to center zero frequency. 4. **MATLAB/GNU Octave Code Snippets:** ```matlab % (a) Message Signal Sampling fs = 100; N = 256; t = (0:N-1)/fs; m = zeros(1,N); idx = t <= 2; m(idx) = 3*cos(50*pi*t(idx)) + sin(10*pi*t(idx)); figure; stem(t,m); title('Message Signal m[n]'); xlabel('Time (s)'); ylabel('Amplitude'); % (b) Conventional AM Signal Sampling fs_AM = 1000; N_AM = 2048; t_AM = (0:N_AM-1)/fs_AM; m_AM = zeros(1,N_AM); idx_AM = t_AM <= 2; m_AM(idx_AM) = 3*cos(50*pi*t_AM(idx_AM)) + sin(10*pi*t_AM(idx_AM)); A_C = 12; c = cos(500*pi*t_AM); x_AM = A_C*(1 + m_AM).*c; figure; stem(t_AM,x_AM); title('Conventional AM Signal x_{AM}[n]'); xlabel('Time (s)'); ylabel('Amplitude'); % (c) Message Signal Spectrum Analysis M = fft(m); M_shift = fftshift(M); M_mag_norm = abs(M_shift)/max(abs(M_shift)); k = -N/2:N/2-1; figure; stem(k,M_mag_norm); title('Normalized Magnitude Spectrum of m[n]'); xlabel('Frequency Bin'); ylabel('Normalized Magnitude'); % (d) Conventional AM Signal Spectrum Analysis X_AM = fft(x_AM); X_AM_shift = fftshift(X_AM); X_AM_mag_norm = abs(X_AM_shift)/max(abs(X_AM_shift)); k_AM = -N_AM/2:N_AM/2-1; figure; stem(k_AM,X_AM_mag_norm); title('Normalized Magnitude Spectrum of x_{AM}[n]'); xlabel('Frequency Bin'); ylabel('Normalized Magnitude'); ``` 5. **Explanation:** - We sample the message and AM signals at specified frequencies to get discrete sequences. - The DFT reveals frequency components; fftshift centers the zero frequency. - Normalization helps visualize relative magnitudes. - The plots show time-domain samples and frequency-domain spectra. **Final answers:** - Sampled message sequence $$m[n]$$ with 256 points. - Sampled AM sequence $$x_{AM}[n]$$ with 2048 points. - Normalized magnitude spectra of both signals computed and plotted.