Let's solve the problem **step by step** in English. --- ### **Given:** - \( T_1 = 10\,\text{ms} = 0.01\,\text{s} \) - \( T_2 = 5\,\text{ms} = 0.005\,\text{s} \) - \( x_c(t) = \cos(150\pi t) - \frac{1}{2} \cos(25\pi t + 30^\circ) \) You are to **find** \( y_r(t) \). --- ## **Step 1: Sampling** - The first block is C/D (Continuous to Discrete) conversion with sampling period \( T_1 \). So, the discrete-time signal after sampling is: \[ x[n] = x_c(nT_1) \] Calculate \( x[n] \): \[ x[n] = \cos(150\pi nT_1) - \frac{1}{2} \cos(25\pi nT_1 + 30^\circ) \] Plug in \( T_1 = 0.01 \): \[ x[n] = \cos(150\pi \cdot n \cdot 0.01) - \frac{1}{2} \cos(25\pi \cdot n \cdot 0.01 + 30^\circ) \] \[ x[n] = \cos(1.5\pi n) - \frac{1}{2} \cos(0.25\pi n + 30^\circ) \] --- ## **Step 2: Downsampling by 3** - Downsampling means keeping every 3rd sample: \[ x_1[m] = x[3m] \] So, \[ x_1[m] = \cos(1.5\pi \cdot 3m) - \frac{1}{2} \cos(0.25\pi \cdot 3m + 30^\circ) \] \[ x_1[m] = \cos(4.5\pi m) - \frac{1}{2} \cos(0.75\pi m + 30^\circ) \] --- ## **Step 3: Upsampling by 2** - Upsampling by 2 means inserting a zero between every sample: \[ x_2[k] = \begin{cases} x_1[k/2], & \text{if } k \text{ even} \\ 0, & \text{if } k \text{ odd} \end{cases} \] For even \( k = 2l \): \[ x_2[2l] = x_1[l] = \cos(4.5\pi l) - \frac{1}{2} \cos(0.75\pi l + 30^\circ) \] For odd \( k \), \( x_2[k] = 0 \). --- ## **Step 4: D/C (Discrete to Continuous) Conversion** - The new sampling period is \( T_2 = 0.005\,s \). - The output is \( y_r(t) \), which is a continuous-time signal reconstructed from \( x_2[k] \): \[ y_r(t) = x_2[k], \quad t = kT_2 \] So: - For \( t = kT_2 \) where \( k \) is even (\( k = 2l \)): \[ y_r(t) = \cos(4.5\pi l) - \frac{1}{2} \cos(0.75\pi l + 30^\circ) \] But \( l = \frac{t}{2T_2} \), so: \[ y_r(t) = \cos\left(4.5\pi \cdot \frac{t}{2T_2}\right) - \frac{1}{2} \cos\left(0.75\pi \cdot \frac{t}{2T_2} + 30^\circ\right) \] \[ = \cos\left(\frac{4.5\pi}{2T_2} t\right) - \frac{1}{2} \cos\left(\frac{0.75\pi}{2T_2} t + 30^\circ\right) \] - For odd \( k \), \( y_r(t) = 0 \). Plug in \( T_2 = 0.005 \): \[ \frac{4.5\pi}{2T_2} = \frac{4.5\pi}{0.01} = 450\pi \] \[ \frac{0.75\pi}{2T_2} = \frac{0.75\pi}{0.01} = 75\pi \] So the final answer for even \( k \) (or \( t = 2l T_2 \)): \[ \boxed{ y_r(t) = \cos(450\pi t) - \frac{1}{2} \cos(75\pi t + 30^\circ) } \] For all other times, the output is zero (if you only reconstruct at sample points). --- ### **Summary** **The output is:** \[ \boxed{ y_r(t) = \cos(450\pi t) - \frac{1}{2} \cos(75\pi t + 30^\circ) } \] for \( t = 0, 0.01, 0.02, \ldots \), i.e., at every \( 2T_2 = 0.01\,s \) interval (if ideal D/C is used, this is the reconstructed signal). **All steps and calculations are shown above.**

Let's carefully solve this problem step by step, correcting any previous mistakes. --- ## Given Data: - Sampling periods: - \( T_1 = 10\,\text{ms} = 0.01\,\text{s} \) - \( T_2 = 5\,\text{ms} = 0.005\,\text{s} \) - Input continuous-time signal: \[ x_c(t) = \cos(150\pi t) - \frac{1}{2} \cos(25\pi t + 30^\circ) \] --- ## Goal: Find the reconstructed output \( y_r(t) \) after the sequence of operations: - Sampling at \( T_1 \), - Downsampling by 3, - Upsampling by 2, - Converting back to continuous-time at \( T_2 \). --- ## **Step 1: Sampling \( x_c(t) \) with \( T_1 \)** Sampled discrete-time signal: \[ x[n] = x_c(nT_1) \] Calculating: \[ x[n] = \cos(150\pi n T_1) - \frac{1}{2} \cos(25\pi n T_1 + 30^\circ) \] Substitute \( T_1 = 0.01\,s \): \[ x[n] = \cos(150\pi n \times 0.01) - \frac{1}{2} \cos(25\pi n \times 0.01 + 30^\circ) \] Simplify: \[ x[n] = \cos(1.5\pi n) - \frac{1}{2} \cos(0.25\pi n + 30^\circ) \] --- ## **Step 2: Downsampling by 3** Downsampling by 3: \[ x_1[m] = x[3m] \] Evaluate: \[ x_1[m] = \cos(1.5\pi \times 3m) - \frac{1}{2} \cos(0.25\pi \times 3m + 30^\circ) \] Calculate: \[ x_1[m] = \cos(4.5\pi m) - \frac{1}{2} \cos(0.75\pi m + 30^\circ) \] --- ## **Step 3: Upsampling by 2** Upsampling by 2: \[ x_2[k] = \begin{cases} x_1[k/2], & \text{if } k \text{ even} \\ 0, & \text{if } k \text{ odd} \end{cases} \] Expressed as: - For even \(k=2m\): \[ x_2[2m] = x_1[m] = \cos(4.5\pi m) - \frac{1}{2} \cos(0.75\pi m + 30^\circ) \] - For odd \(k\), \(x_2[k] = 0\). --- ## **Step 4: Reconstruction at \( T_2 \)** The final step involves converting the discrete-time sequence \( x_2[k] \) into continuous-time signal \( y_r(t) \) with sampling period \( T_2 = 0.005\,s \). The reconstructed \( y_r(t) \) is a **zero-order hold** or ideal reconstruction of \( x_2[k] \): \[ t = k T_2 \] Since \( x_2[k] \) is non-zero only at even \(k=2m\): \[ t = 2m \times T_2 = 2m \times 0.005 = 0.01 m \] Define: \[ m = \frac{t}{0.01} \] Express the signal: \[ y_r(t) = \begin{cases} \cos(4.5 \pi m) - \frac{1}{2} \cos(0.75 \pi m + 30^\circ), & \text{if } t = 0.01 m \\ 0, & \text{if } t \text{ is not a multiple of } 0.01 \end{cases} \] Replace \(m = t/0.01\): \[ \boxed{ \textbf{At } t = 0.01 m: } \] \[ y_r(t) = \cos\left(4.5 \pi \times \frac{t}{0.01}\right) - \frac{1}{2} \cos\left(0.75 \pi \times \frac{t}{0.01} + 30^\circ\right) \] Simplify: \[ y_r(t) = \cos(450 \pi t) - \frac{1}{2} \cos(75 \pi t + 30^\circ) \] --- ## **Final Answer:** \[ \boxed{ \boxed{ y_r(t) = \cos(450 \pi t) - \frac{1}{2} \cos(75 \pi t + 30^\circ) } } \] This expression describes the reconstructed continuous-time output \( y_r(t) \). --- ## **Summary:** - The original signal's two frequency components are scaled by the sampling and processing steps. - The final reconstructed signal contains two cosines with frequencies: - \( f_1 = \frac{450 \pi}{2\pi} = 225\, \text{Hz} \), - \( f_2 = \frac{75 \pi}{2\pi} = 37.5\, \text{Hz} \). - The phase shift of the second component is preserved. --- **Note:** This derivation assumes ideal sampling and reconstruction. Actual systems may introduce additional effects like aliasing or filtering, but based on the given problem, this is the correct mathematical derivation.