Let's solve the problem step by step: --- ## **(a) Why does a constant in the input not affect the output?** **Given:** - Input: \( x[n] = 10 + x_1[n] \) - First system: 5-point moving averager - Second system: First difference ### **Step 1: Examine the systems** **5-point moving average:** This system averages the last 5 inputs: \[ h_1[n] = \frac{1}{5}(\delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3] + \delta[n-4]) \] The output is: \[ y_1[n] = \frac{1}{5}\sum_{k=0}^{4} x[n-k] \] **First difference:** This system computes the difference between the current and previous value: \[ h_2[n] = \delta[n] - \delta[n-1] \] The output is: \[ y[n] = y_1[n] - y_1[n-1] \] ### **Step 2: Substitute the input** Let \( x[n] = 10 + x_1[n] \). Apply the moving average: \[ y_1[n] = \frac{1}{5} \sum_{k=0}^{4} (10 + x_1[n-k]) = \frac{1}{5}(5 \times 10) + \frac{1}{5} \sum_{k=0}^{4} x_1[n-k] = 10 + \frac{1}{5} \sum_{k=0}^{4} x_1[n-k] \] ### **Step 3: Apply the first difference** Apply the second system: \[ y[n] = y_1[n] - y_1[n-1] = \big(10 + \frac{1}{5} \sum_{k=0}^{4} x_1[n-k]\big) - \big(10 + \frac{1}{5} \sum_{k=0}^{4} x_1[n-1-k]\big) \] The constant \(10\) cancels: \[ y[n] = \frac{1}{5} \sum_{k=0}^{4} x_1[n-k] - \frac{1}{5} \sum_{k=0}^{4} x_1[n-1-k] \] So the output depends only on \(x_1[n]\). --- ## **(b) Frequency-Response Function of Overall Cascade** ### **Step 1: Find frequency response of each system** #### **5-point Moving Averager:** Impulse response: \( h_1[n] = \frac{1}{5} \) for \( n = 0,1,2,3,4 \), 0 otherwise. Frequency response: \[ H_1(e^{j\omega}) = \frac{1}{5} \sum_{n=0}^{4} e^{-j\omega n} = \frac{1}{5} \cdot \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \] Or, using the geometric sum formula, and simplifying: \[ H_1(e^{j\omega}) = \frac{1}{5} e^{-j2\omega} \cdot \frac{\sin(2.5\omega)}{\sin(0.5\omega)} \] #### **First Difference:** Impulse response: \( h_2[n] = \delta[n] - \delta[n-1] \) Frequency response: \[ H_2(e^{j\omega}) = 1 - e^{-j\omega} \] ### **Step 2: Overall frequency response** Cascaded system: \[ H(e^{j\omega}) = H_1(e^{j\omega}) \cdot H_2(e^{j\omega}) \] \[ \boxed{H(e^{j\omega}) = \frac{1}{5} \cdot \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \cdot (1 - e^{-j\omega}) = \frac{1}{5}(1 - e^{-j5\omega})} \] --- ## **(c) Sketch Frequency Responses** Let's describe the sketches: ### **1. 5-point Moving Averager (\(H_1(e^{j\omega})\)):** - **Magnitude:** Peaks at DC (\(\omega=0\)), zeros at multiples of \(2\pi/5\) (except 0), typical low-pass response, main lobe centered at 0. - **Phase:** Linear, due to the delay term \(e^{-j2\omega}\). ### **2. First Difference (\(H_2(e^{j\omega})\)):** - **Magnitude:** \( |1 - e^{-j\omega}| = 2|\sin(\omega/2)| \) Zero at \(\omega=0\) (DC), max at \(\omega=\pi\) (high-pass). - **Phase:** Jumps from 0 to \(-\pi\), as this is a first difference (differentiator). ### **3. Overall Cascade (\(H(e^{j\omega})\)):** - **Magnitude:** \( \frac{1}{5} |1 - e^{-j5\omega}| = \frac{2}{5} |\sin(2.5\omega)| \) - Zeros at \(\omega = 0, \pm \frac{2\pi}{5}, \pm \frac{4\pi}{5}, \pm \pi\) - Maximum at midpoints between zeros - **Phase:** Linear phase, with jumps at the zeros. --- ### **Summary Table** | System | Magnitude (\(\omega\)) | Phase (\(\omega\)) | |-----------------------|-------------------------------------------|-------------------------------| | 5-pt average | \( \left|\frac{1}{5} \frac{\sin(2.5\omega)}{\sin(0.5\omega)}\right| \) | Linear (due to \(e^{-j2\omega}\)) | | First difference | \( 2|\sin(\omega/2)| \) | \( -\omega/2 \) or jump at \(\omega=0\) | | Cascade (overall) | \( \frac{2}{5}|\sin(2.5\omega)| \) | Linear | --- ## **Final Answers** ### **(a)** The output depends only on \( x_1[n] \) because both systems are linear and time-invariant, and the first difference system removes any constant (DC) component from its input. ### **(b)** \[ \boxed{H(e^{j\omega}) = \frac{1}{5}(1 - e^{-j5\omega})} \] ### **(c)** - **5-pt average:** Low-pass, main lobe at DC, zeros at \(\omega = \frac{2\pi}{5}, \frac{4\pi}{5}, ...\) - **First difference:** High-pass, zero at DC, max at \(\omega = \pi\) - **Overall:** Bandpass-like, zeros at DC and multiples of \(\frac{2\pi}{5}\), main lobe at mid-frequencies. You can sketch these using the formulas above for \( -\pi < \omega \leq \pi \). If you need actual plots, just ask!