Let's break down the problem step by step: ## **(a) Why y[n] = y₁[n] when x[n] = 10 + x₁[n]?** ### **Step 1: Understanding the Systems** - **System 1:** 5-point moving averager (LTI system) - Impulse response: \( h_1[n] = \frac{1}{5} (u[n] - u[n-5]) \) - **System 2:** First difference (LTI system) - Impulse response: \( h_2[n] = \delta[n] - \delta[n-1] \) ### **Step 2: Input Decomposition** - Input: \( x[n] = 10 + x_1[n] \) - 10 is a constant (DC component) - \( x_1[n] \) is any signal ### **Step 3: System Linearity** - Both systems are **linear and time-invariant** (LTI). - Due to linearity: \( y[n] = T(10) + T(x_1[n]) \), where \( T \) is the overall system. ### **Step 4: DC Response Through Both Systems** - The **moving averager** passes a constant 10 as output: \( y_1[n] = 10 \) - The **first difference** of a constant is zero: \( 10 - 10 = 0 \) - Therefore, **overall output due to 10 is zero**. ### **Step 5: Output Form** - The output is **solely due to \( x_1[n] \)**, i.e., \( y[n] = y_1[n] \). **Summary:** The output is only due to the non-constant part, because the first-difference system removes any constant (DC) component. --- ## **(b) Frequency-Response Function of the Overall Cascade System** ### **Step 1: Frequency Response of Each System** #### **System 1: 5-point Moving Averager** - Impulse response: \( h_1[n] = \frac{1}{5} \) for \( n=0,1,2,3,4 \) - Frequency response: \[ H_1(e^{j\omega}) = \frac{1}{5} (1 + e^{-j\omega} + e^{-j2\omega} + e^{-j3\omega} + e^{-j4\omega}) \] - This is equivalent to: \[ H_1(e^{j\omega}) = \frac{1}{5} \cdot \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \] (Geometric sum formula.) #### **System 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: Overall (Cascade) Frequency Response** - The overall response is the **product**: \[ H_\text{total}(e^{j\omega}) = H_1(e^{j\omega}) H_2(e^{j\omega}) \] \[ H_\text{total}(e^{j\omega}) = \left[\frac{1}{5} \cdot \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}}\right] \cdot (1 - e^{-j\omega}) \] Simplifies to: \[ \boxed{ H_\text{total}(e^{j\omega}) = \frac{1}{5} \left(1 - e^{-j5\omega}\right) } \] --- ## **(c) Sketches of Frequency Response (Magnitude and Phase)** ### **Step 1: Individual Systems** #### **1. 5-point Moving Averager:** - Magnitude: Peaks at \( \omega = 0 \) (DC), zeros at \( \omega = 2\pi/5, 4\pi/5, ... \) - Phase: Linear decrease with \( \omega \) #### **2. First Difference:** - Magnitude: \( |1 - e^{-j\omega}| = 2|\sin(\omega/2)| \), zero at DC, max at \( \omega = \pi \) - Phase: \( \angle (1 - e^{-j\omega}) = \arctan\left(\frac{-\sin\omega}{1 - \cos\omega}\right) \) or simply \( -\omega/2 \) (for \( 0 < \omega < 2\pi \)) ### **Step 2: Overall System** - Magnitude: \( \frac{1}{5} |1 - e^{-j5\omega}| = \frac{2}{5} |\sin(5\omega/2)| \) - Zeros at \( \omega = 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \ldots, 2\pi \) - Phase: \( \angle (1 - e^{-j5\omega}) = -5\omega/2 \) (linear with \(\omega\)) --- ### **Summary Table** | System | Magnitude Response | Phase Response | |-------------|--------------------------------------------|--------------------| | Averager | \( \left|H_1\right| = \frac{1}{5} \frac{|\sin(2.5\omega)|}{|\sin(0.5\omega)|} \) | Decreases with \(\omega\) | | Difference | \( 2 |\sin(\omega/2)| \) | \( -\omega/2 \) | | Cascade | \( \frac{2}{5} |\sin(2.5\omega)| \) | \( -2.5\omega \) | --- ## **Final Answers** ### **(a)** The output is only due to \( x_1[n] \) (non-constant part) because the cascade includes a first difference system, which eliminates constants (DC). ### **(b)** \[ \boxed{ H_\text{total}(e^{j\omega}) = \frac{1}{5} \left(1 - e^{-j5\omega}\right) } \] ### **(c)** - **Averager:** Magnitude: peaks at DC, zeros periodically; Phase: linear. - **Difference:** Magnitude: zero at DC, max at \(\pi\); Phase: linear. - **Cascade:** Magnitude: \( \frac{2}{5} |\sin(2.5\omega)| \); Phase: \( -2.5\omega \). If you need **actual sketches**, let me know!