Let's solve the problem step by step: ### **Given:** - \( x(t) \) is a carrier modulated signal: \( x(t) = \cos(10t) \cdot s(t) \) - The general Fourier transform pair for a time-shifted and modulated signal is provided. Let's **find the Fourier transform** of \( x(t) \). --- ## **Step 1: Recall Fourier Transform Properties** - **Fourier Transform of \( \cos(\omega_0 t) \):** \[ \mathcal{F}\{\cos(\omega_0 t)\} = \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] \] - **Modulation Theorem:** If \( x(t) \longleftrightarrow X(\omega) \), then \( x(t) \cos(\omega_0 t) \longleftrightarrow \frac{1}{2}[X(\omega - \omega_0) + X(\omega + \omega_0)] \) --- ## **Step 2: Given Signal** Suppose \( s(t) \) is the baseband signal with Fourier transform \( S(\omega) \). - The modulated signal: \( x(t) = s(t) \cos(10t) \) --- ## **Step 3: Apply Modulation Theorem** - The Fourier transform of \( x(t) \): \[ X(\omega) = \frac{1}{2} [S(\omega - 10) + S(\omega + 10)] \] --- ## **Step 4: Use the Provided Example** The form given is: \[ X(\omega) = a\pi \left\{ \text{sinc}(b\pi(\omega + c)) + \text{sinc}(b\pi(\omega - c)) \right\} e^{jd\pi\omega} \] This matches the modulation and shifting property. --- ## **Step 5: Substitute Values** If the carrier is \( \cos(10t) \), then \( c = 10 \). If the baseband \( S(\omega) \) is \( a\pi\,\text{sinc}(b\pi\omega) e^{jd\pi\omega} \): So, \[ X(\omega) = a\pi \left\{ \text{sinc}(b\pi(\omega + 10)) + \text{sinc}(b\pi(\omega - 10)) \right\} e^{jd\pi\omega} \] --- ### **Summary Table** | Step | Operation | Result | |-------------|-----------------------------------------------|----------------------------------------------------------------| | 1 | Baseband signal \( s(t) \) FT: \( S(\omega) \)| \( S(\omega) \) | | 2 | Modulate with \( \cos(10t) \) | \( \frac{1}{2}[S(\omega-10) + S(\omega+10)] \) | | 3 | Substitute \( S(\omega) \) | \( a\pi \left\{ \text{sinc}(b\pi(\omega + 10)) + \text{sinc}(b\pi(\omega - 10)) \right\} e^{jd\pi\omega} \) | --- ### **Final Answer** \[ \boxed{ X(\omega) = a\pi \left\{ \text{sinc}(b\pi(\omega + 10)) + \text{sinc}(b\pi(\omega - 10)) \right\} e^{jd\pi\omega} } \] --- **Explanation:** The spectrum of the baseband is shifted to \( \pm 10 \) due to the cosine carrier, as shown by the modulation theorem. The form matches the provided solution. If you have a specific \( s(t) \) (like a rectangular pulse), substitute its Fourier transform for \( S(\omega) \) into the above result.

Certainly! Here's the solution in 2 clear steps: **Step 1:** Since \( x(t) = s(t) \cos(10t) \), and the Fourier transform of \( \cos(10t) \) is \( \pi[\delta(\omega - 10) + \delta(\omega + 10)] \), by the modulation theorem, the Fourier transform of \( x(t) \) is the baseband spectrum \( S(\omega) \) shifted to \( \pm 10 \) and scaled by \( \frac{1}{2} \). Therefore, \[ X(\omega) = \frac{1}{2} [S(\omega - 10) + S(\omega + 10)] \] **Step 2:** Given the form of \( S(\omega) \) as involving sinc functions, the shifted spectra become: \[ X(\omega) = a\pi \left\{ \text{sinc}(b\pi(\omega + 10)) + \text{sinc}(b\pi(\omega - 10)) \right\} e^{jd\pi\omega} \] This matches the provided expression, representing the spectrum of the modulated signal.