Let's break down the problem step by step as requested. --- # **Step 1: System Overview** - **System Transfer Function:** \[ (s) = \frac100s}{s-10)(s^2 + s + 4)} \] - **Input:** Periodic signal \( x(t) \) (not explicitly given in the image; likely a square or periodic signal as shown in the plot). - **Bandpass Filter:** - Ideal bandpass: Passes frequencies \( 4.3 \leq \omega \leq 5.1 \) rad/s - Gain: dB (unity), Phase: ° - **Output:** \( Y(s) \) --- ## **Step 2: Frequency Response (Bode Plot) for \( H(s) \)** ### **a. Find Frequency Response \( H(j\omega) \)** Substitute \( s = j\omega \): \[ H(j\omega) = \frac{100j\omega}{(j\omega-10)[(j\omega)^2 + j\omega + 4]} \] #### **Magnitude and Phase:** - **Magnitude:** \( |H(j\omega)| \) - **Phase:** \( \angle H(j\omega) \) #### **Mark Bandpass Range:** - The range \( 4.3 \leq \omega \leq 5.1 \) is relevant. - On the Bode plot, shade/mark this region. --- ## **Step 3: Fourier Series of Input** The input appears to be a periodic square wave (from the plot). **Assume:** - Period \( T \), fundamental frequency \( \omega_ \). - The nonzero Fourier coefficient will be at the harmonic \( k \) for which \( k\omega_ \) falls in \( [4.3, 5.1] \) rad/s. **General Trigonometric Fourier Series:** \[ x(t) = a_ + \sum_{k=1}^\infty [a_k \cos(k\omega_ t) + b_k \sin(k\omega_ t)] \] - **Calculate \( k \) such that \( k\omega_ \) is within the ideal bandpass filter.** --- ## **Step 4: Output of the Bandpass Filter** - Only the frequency component(s) within \( [4.3, 5.1] \) rad/s pass through. - Output will be the filtered Fourier component(s). ### **a. Output in Cosine Form:** \[ y(t) = K \cos(\omega_ t + \Phi) \] ### **b. Output in Sine Form:** \[ y(t) = K \sin(\omega_ t + \Phi) \] Where \( K \) and \( \Phi \) are determined from the Fourier coefficients. --- ## **Step 5: Example Calculation (Assume Details)** Let’s assume the period \( T = 2\pi \) (so \( \omega_ = 1 \) rad/s). - Harmonics: \( \omega_k = k \omega_ \) - The bandpass filter passes \( 4.3 \leq \omega_k \leq 5.1 \implies k = 5 \) - So, only the 5th harmonic passes. ### **Fourier Coefficient for Square Wave:** For a square wave of amplitude \( A \) and 50% duty cycle: \[ b_k = \frac{4A}{k\pi},\ \text{for odd } k \] \[ a_k = \] So, for \( k=5 \): \[ b_5 = \frac{4A}{5\pi} \] ### **Output Signal:** \[ y(t) = b_5 \sin(5t) \] If you want it in cosine form: \[ y(t) = b_5 \cos(5t - 90^\circ) \] --- # **Summary Table** | Step | Result | |------|--------| | **Transfer Function** | \( H(s) = \frac{100s}{(s-10)(s^2+s+4)} \) | | **Bandpass Range** | \( 4.3 \leq \omega \leq 5.1 \) rad/s | | **Fourier Harmonic Passing** | \( k = 5 \), \( \omega_ = 1 \Rightarrow \omega_5 = 5 \) rad/s | | **Fourier Coefficient** | \( b_5 = \frac{4A}{5\pi},\ a_5= \) | | **Filtered Output** | \( y(t) = \frac{4A}{5\pi} \sin(5t) \) or \( y(t) = \frac{4A}{5\pi} \cos(5t - 90^\circ) \) | --- # **Final Answer** ### **a. Mark the Bode plot in the range \([4.3, 5.1]\) rad/s.** ### **b. Fourier Coefficient (for square wave):** - Only the 5th harmonic passes: \( b_5 = \frac{4A}{5\pi} \) ### **c. Output of the Ideal Bandpass Filter:** \[ y(t) = \frac{4A}{5\pi} \sin(5t) \] or \[ y(t) = \frac{4A}{5\pi} \cos(5t - 90^\circ) \] --- **If your input amplitude \( A = 1 \):** \[ \boxed{y(t) = \frac{4}{5\pi}\sin(5t)} \] --- ### **Let me know if you want the Matlab code or further details for Bode plot or other steps!**

Sure! Here’s a concise explanation of each step, limited to 30 words each: --- # **Step 1: System Overview** Identify the system's transfer function and input signal. The bandpass filter allows specific frequency components to pass while attenuating others, critical for subsequent calculations. --- # **Step 2: Frequency Response (Bode Plot) for \( H(s) \)** Substitute \( s = j\omega \) to derive \( H(j\omega) \). Calculate magnitude and phase for each frequency, highlighting the ideal bandpass range on the plot. --- # **Step 3: Fourier Series of Input** Assume the input signal is periodic. Use Fourier series to express it, identifying harmonics within the bandpass filter's frequency range to ensure proper filtering later. --- # **Step 4: Output of the Bandpass Filter** Only frequencies within the filter's range contribute to the output signal. Use the Fourier coefficients to express the output as a cosine or sine function. --- # **Step 5: Example Calculation** Assuming a 2π period, determine the 5th harmonic's contribution. Calculate the Fourier coefficient, yielding the output signal, which represents the filtered response of the system. --- # **Final Answer** Summarize results, including the output signal in sine or cosine form. Specify amplitude and phase shifts to provide a complete description of the filtered output. --- Feel free to ask for any additional details or clarifications!