Let's solve each part step-by-step: --- ### **4(a) Why is the output only due to \( x_1[n] \)?** #### **Given:** - The input \( x[n] = 10 + x_1[n] \). - The first system is a **5-point moving averager**. - The second system is a **first difference**. #### **Step 1: System Linearity and Time-Invariance** - Both systems are **linear** and **time-invariant** (LTI). - For LTI systems: If input is the sum of two signals, output is the sum of their individual outputs. #### **Step 2: Output Due to Constant Input** - First, let’s process the constant \( 10 \) through the two systems: **First System (Moving Averager):** - For any constant \( a \), the moving average is still \( a \). - So, after the first system, the output due to \( 10 \) is still \( 10 \). **Second System (First Difference):** - The first difference is \( y[n] = x[n] - x[n-1] \). - For a constant input: \( y[n] = 10 - 10 = 0 \). #### **Step 3: Output Due to \( x_1[n] \)** - Let \( y_1[n] \) be the output due to \( x_1[n] \) alone. - Since the system is linear, overall output is the sum of the outputs due to \( 10 \) and \( x_1[n] \): \[ y[n] = \text{output due to 10} + \text{output due to } x_1[n] = 0 + y_1[n] = y_1[n] \] **Conclusion:** The constant part is eliminated by the first-difference system, so only \( x_1[n] \) affects the output. --- ### **4(b) Frequency-Response Function of the Overall System** #### **Step 1: Frequency Response of Each System** **1. 5-point Moving Averager:** - Impulse response: \( h_1[n] = \frac{1}{5},\ n=0,1,2,3,4 \) - Frequency response: \[ H_1(e^{j\omega}) = \sum_{k=0}^{4} \frac{1}{5} e^{-j\omega k} = \frac{1}{5} \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \] (This is a geometric series.) **2. 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: Cascade System Frequency Response** - The overall frequency response is the **product**: \[ H(e^{j\omega}) = H_1(e^{j\omega}) \cdot H_2(e^{j\omega}) \] Substituting: \[ H(e^{j\omega}) = \left( \frac{1}{5} \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \right) (1 - e^{-j\omega}) \] \[ = \frac{1}{5}(1 - e^{-j5\omega}) \] --- ### **4(c) Sketch Frequency Response (Magnitude & Phase)** #### **1. 5-point Moving Averager (\( H_1 \)):** - **Magnitude:** Low-pass filter (sinc shape), peaks at \( \omega=0 \), zeros at multiples of \( 2\pi/5 \). - **Phase:** Linear decrease. #### **2. First Difference (\( H_2 \)):** - \( |H_2(e^{j\omega})| = |1 - e^{-j\omega}| = 2|\sin(\omega/2)| \) - **Magnitude:** High-pass (zero at \( \omega=0 \), max at \( \omega=\pi \)) - **Phase:** \( -\omega/2 \) except for phase flip at \( \omega = 0, \pi \). #### **3. Cascade System (\( H \)):** \[ H(e^{j\omega}) = \frac{1}{5}(1 - e^{-j5\omega}) \] - **Magnitude:** \( \frac{1}{5} |1 - e^{-j5\omega}| = \frac{2}{5} |\sin(5\omega/2)| \) - **Phase:** Linear phase, jumps at frequencies where \( \sin(5\omega/2) = 0 \). --- ### **Summary Table** | System | Frequency Response (\( H(e^{j\omega}) \)) | Magnitude (at key points) | Phase | |-----------------------|---------------------------------------------------|----------------------------------------|-------------------------| | Moving Averager | \( \frac{1}{5} \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \) | \( 1 \) at \( \omega=0 \), zeros at \( \omega = \frac{2\pi}{5}k \) for \( k=1,..,4 \) | Linear | | First Difference | \( 1 - e^{-j\omega} \) | 0 at \( \omega=0 \), 2 at \( \omega=\pi \) | \( -\omega/2 \) | | Overall System | \( \frac{1}{5}(1 - e^{-j5\omega}) \) | 0 at \( \omega=0, \frac{2\pi}{5}k \), max at \( \omega=\pi/5 \) | Linear | --- ### **Sketches (qualitative):** #### **Moving Averager (Magnitude)** - Peaks at \( \omega=0 \), drops to zero at \( \omega = \frac{2\pi}{5}, \frac{4\pi}{5}, ..., 2\pi \) #### **First Difference (Magnitude)** - Zero at \( \omega=0 \), rises to max at \( \omega=\pi \) #### **Overall System (Magnitude)** - Sinc-like, but with main lobe shifted (zero at \( \omega=0 \), max at \( \omega=\pi/5 \), zeros at \( \omega = \frac{2\pi}{5}, \frac{4\pi}{5}, ..., 2\pi \)) --- ## **Final Answers:** **(a) Output due to constant input is zero, so output is only due to \( x_1[n] \).** **(b) Frequency-response function of cascade:** \[ \boxed{H(e^{j\omega}) = \frac{1}{5}(1 - e^{-j5\omega})} \] **(c) Sketches:** - See explanation above for qualitative plots of magnitude and phase for each system and overall cascade. Let me know if you want detailed plots!