Let's solve each part step-by-step: --- ### (a) Explanation: Why DC Component is Eliminated **Given:** - Input: \( x[n] = 10 + x_1[n] \) - The system is a cascade of: 1. **5-point moving averager**: \( h_1[n] = \frac{1}{5} ( \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3] + \delta[n-4] ) \) 2. **First difference**: \( h_2[n] = \delta[n] - \delta[n-1] \) #### **Linearity and Time-Invariance** Both subsystems are linear and time-invariant (LTI), so for any input \( x[n] = x_a[n] + x_b[n] \), the output is \( y[n] = y_a[n] + y_b[n] \). #### **Behavior with DC (Constant) Input** - Pass a **constant** input \( x[n] = 10 \) through the moving averager: - Output is also constant: \( y_1[n] = 10 \) - Pass this constant output through the first-difference system: - \( y_2[n] = y_1[n] - y_1[n-1] = 10 - 10 = 0 \) So, the overall system **completely eliminates the DC (constant) component**. #### **Conclusion** - The output \( y[n] \) for input \( x[n] = 10 + x_1[n] \) is **only due to \( x_1[n] \)**. - Thus, \( y[n] = y_1[n] \), where \( y_1[n] \) is the output for input \( x_1[n] \). --- ### (b) Frequency-Response Function of the Cascade System Let’s find the frequency responses of each system and then multiply them (since they are in cascade). #### **1. 5-point Moving Averager** Impulse response: \[ h_1[n] = \frac{1}{5} ( \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3] + \delta[n-4] ) \] 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 a **finite impulse response (FIR) filter**. #### **2. First Difference** Impulse response: \[ h_2[n] = \delta[n] - \delta[n-1] \] Frequency response: \[ H_2(e^{j\omega}) = 1 - e^{-j\omega} \] #### **3. Overall System** Multiply the two: \[ H(e^{j\omega}) = H_1(e^{j\omega}) \cdot H_2(e^{j\omega}) \] \[ = \frac{1}{5} (1 + e^{-j\omega} + e^{-j2\omega} + e^{-j3\omega} + e^{-j4\omega}) (1 - e^{-j\omega}) \] --- ### (c) Sketch Frequency-Response (Magnitude and Phase) #### **1. 5-point Moving Averager** \[ H_1(e^{j\omega}) = \frac{1}{5} \sum_{k=0}^4 e^{-j\omega k} \] This can be written as: \[ H_1(e^{j\omega}) = \frac{1}{5} \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} \] - **Magnitude:** Peaks at DC (\(\omega = 0\)), nulls at multiples of \(2\pi/5\). - **Phase:** Linear phase (delays). #### **2. First Difference** \[ H_2(e^{j\omega}) = 1 - e^{-j\omega} = e^{-j\omega/2} \cdot 2j \sin(\omega/2) \] - **Magnitude:** \(2|\sin(\omega/2)|\) - **Phase:** \(-\omega/2 + \frac{\pi}{2} \cdot \text{sgn}(\sin(\omega/2))\) #### **3. Overall Cascade** - **Magnitude:** Product of the above magnitudes. - **Phase:** Sum of the above phases. #### **Sketches:** 1. **Moving Averager:** Lowpass filter: high at DC, falls off with increasing \(\omega\), with zeros at \(2\pi/5, 4\pi/5, ...\). 2. **First Difference:** Highpass filter: zero at DC, peaks at \(\omega = \pi\). 3. **Cascade:** Bandpass-like; DC is suppressed (due to first difference), but high frequencies are attenuated (due to moving average). --- ## **Final Answers (Summary)** ### (a) **Explanation** - The overall system eliminates the DC component, so the output depends only on the non-DC part \(x_1[n]\). ### (b) **Overall Frequency-Response Function** \[ \boxed{ H(e^{j\omega}) = \frac{1}{5} (1 + e^{-j\omega} + e^{-j2\omega} + e^{-j3\omega} + e^{-j4\omega}) (1 - e^{-j\omega}) } \] ### (c) **Sketches** - **Moving Averager:** Lowpass - **First Difference:** Highpass - **Cascade:** Bandpass (attenuates both DC and high frequency) (You can plot these using MATLAB, Python, or by hand for \(-\pi < \omega \leq \pi\)). Would you like plotted graphs or analytic expressions for magnitude and phase?