Given: - The device has input-output relation: \( e_0 = a_1 e_{in} + a_2 e_{in}^2 \) - The input to the device is the sum of a message signal \( x(t) \) and a carrier signal at 5 kHz: \[ x(t) = 0.3\cos(2\pi \cdot 150t) + 0.4\cos(2\pi \cdot 150t) \] *(But this looks like both terms are at 150 Hz – usually the message would have different frequencies. Let's proceed as is, but note it.)* - The carrier: \( c(t) = A_c \cos(2\pi f_c t) \), \( f_c = 5 \) kHz. Let the input to the device be: \[ e_{in} = x(t) + c(t) \] Let's proceed step by step: --- ### 1. **Square-Law Device Output** \[ e_0 = a_1 e_{in} + a_2 e_{in}^2 \] \[ e_{in} = x(t) + A_c \cos(2\pi f_c t) \] So, \[ e_{in}^2 = [x(t) + A_c \cos(2\pi f_c t)]^2 = x^2(t) + 2x(t)A_c \cos(2\pi f_c t) + A_c^2 \cos^2(2\pi f_c t) \] \[ e_0 = a_1 [x(t) + A_c \cos(2\pi f_c t)] + a_2 \left[ x^2(t) + 2x(t)A_c \cos(2\pi f_c t) + A_c^2 \cos^2(2\pi f_c t) \right] \] --- ### 2. **Terms of Interest for AM Generation** For AM generation, the required output is: \[ [A_c + x(t)] \cos(2\pi f_c t) \] That is, we need a component at \( f_c \) (5 kHz) whose amplitude is modulated by \( x(t) \). From the above, the cross term \( 2x(t)A_c \cos(2\pi f_c t) \) (from \( e_{in}^2 \)) is key. Let us expand the terms that result in frequency components around \( f_c \): #### a. \( a_1 A_c \cos(2\pi f_c t) \) — carrier at \( f_c \) #### b. \( a_2 \cdot 2x(t) A_c \cos(2\pi f_c t) \) — contains product of message and carrier, which produces sidebands. #### c. Other terms: - \( a_1 x(t) \), \( a_2 x^2(t) \): baseband (low-frequency) - \( a_2 A_c^2 \cos^2(2\pi f_c t) \): \[ \cos^2(2\pi f_c t) = \frac{1}{2} + \frac{1}{2}\cos(4\pi f_c t) \] — DC and 2\( f_c \). So, the sum of terms near \( f_c \) is: \[ a_1 A_c \cos(2\pi f_c t) + 2a_2 A_c x(t) \cos(2\pi f_c t) \] \[ = \left[ a_1 A_c + 2a_2 A_c x(t) \right] \cos(2\pi f_c t) \] \[ = A_c' [1 + k x(t)] \cos(2\pi f_c t) \] where \( A_c' = a_1 A_c \), \( k = \frac{2 a_2}{a_1} \) (assuming \( a_1 \neq 0 \)), i.e., a standard AM signal. --- ### 3. **Band Pass Filter (BPF) Requirements** The BPF should **select only the AM signal** centered at the carrier frequency, and pass the carrier and its sidebands (from message). It should suppress the baseband, DC, and higher harmonics. #### **Carrier frequency:** \[ f_c = 5\,\text{kHz} \] #### **Message components:** - Both terms in \( x(t) \) are at 150 Hz (as per your question, though usually they would differ). So, message bandwidth: - The highest frequency in \( x(t) \) is 150 Hz. But, both terms are at 150 Hz, so the message is at 150 Hz. #### **AM Spectrum:** - Lower sideband: \( f_c - f_m = 5000 - 150 = 4850\,\text{Hz} \) - Upper sideband: \( f_c + f_m = 5000 + 150 = 5150\,\text{Hz} \) **Bandwidth required to pass both sidebands:** \[ BW = 2 f_m = 2 \times 150 = 300\,\text{Hz} \] --- ## **Summary** ### **Central frequency of BPF:** \[ \boxed{5\,\text{kHz}} \] ### **Bandwidth of BPF:** \[ \boxed{300\,\text{Hz}} \] **So, the BPF should be centered at \( 5\,\text{kHz} \) and have a bandwidth of \( 300\,\text{Hz} \) to pass the AM signal with both sidebands.** --- #### **If the message had more/different frequencies, the bandwidth would be \( 2 f_{m,\max} \) where \( f_{m,\max} \) is the highest frequency in \( x(t) \).**

The central frequency of the band pass filter (BPF) should be equal to the carrier frequency to select the AM signal, which is **5 kHz**. The bandwidth should cover the upper and lower sidebands, which are centered at the carrier frequency and separated by the maximum message frequency. Given the message frequency components at 150 Hz, the total bandwidth needed is: \[ \text{Bandwidth} = 2 \times 150\,\text{Hz} = 300\,\text{Hz} \] **Final answer:** **Central frequency of BPF:** **5 kHz** **Bandwidth of BPF:** **300 Hz**

Step 1: Understand the device relation and input - The device output: \( e_0 = a_1 e_{in} + a_2 e_{in}^2 \) - Input signal: sum of message \( x(t) \) and carrier \( c(t) \) Given: \[ x(t) = 0.3 \cos(2\pi \cdot 150 t) + 0.4 \cos(2\pi \cdot 150 t) \] (Note: Both are at 150 Hz; typically, messages differ, but using what's provided.) Carrier: \[ c(t) = A_c \cos(2\pi \cdot 5000 t) \] Total input: \[ e_{in} = x(t) + c(t) \] --- Step 2: Expand the output \( e_0 \) \[ e_0 = a_1 e_{in} + a_2 e_{in}^2 \] Calculate \( e_{in}^2 \): \[ e_{in}^2 = (x + c)^2 = x^2 + 2 x c + c^2 \] --- Step 3: Identify frequency components in \( e_0 \) - \( a_1 e_{in} \): \[ a_1 x(t) + a_1 c(t) \] - \( a_2 e_{in}^2 \): \[ a_2 x^2 + 2 a_2 x c + a_2 c^2 \] The components relevant for AM are: - The carrier: \( a_1 c(t) \) at 5 kHz - The modulation sidebands: \( 2 a_2 x c \), which produce signals at \( f_c \pm f_m \) --- Step 4: Focus on the AM signal components The term: \[ 2 a_2 x(t) c(t) \] produces: \[ 2 a_2 \times \left( 0.3 \cos(2\pi 150 t) + 0.4 \cos(2\pi 150 t) \right) \times A_c \cos(2\pi 5000 t) \] which simplifies to (assuming sum of the two identical terms): \[ \text{Amplitude} \propto \cos(2\pi 150 t) \times \cos(2\pi 5000 t) \] Using the product-to-sum identity: \[ \cos A \cos B = \frac{1}{2} [ \cos(A+B) + \cos(A-B) ] \] we get components at: \[ f_c + f_m = 5000 + 150 = 5150\, \text{Hz} \] and \[ f_c - f_m = 5000 - 150 = 4850\, \text{Hz} \] --- Step 5: Determine the BPF center frequency To extract the AM signal, the BPF should be centered at the carrier frequency: \[ f_c = 5000\, \text{Hz} \] --- Step 6: Determine the bandwidth The sidebands are at 4850 Hz and 5150 Hz, which are 150 Hz away from the carrier: \[ \text{Bandwidth} = \text{upper sideband} - \text{lower sideband} = 5150\, \text{Hz} - 4850\, \text{Hz} = 300\, \text{Hz} \] This bandwidth covers the message frequency range. --- ### **Final answer:** - **Central frequency of BPF:** **5 kHz** - **Bandwidth of BPF:** **300 Hz**