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Give step-by-step solution with explanation and final answer: 9. (5 points) Analytically determine the spectrum X(jw) of the continuous-time signal x(t) = 2 + 5cos(200pi*t - pi/6) Show and clearly label your work.

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Give step-by-step solution with explanation and final answer: 9. (5 points) Analytically determine the spectrum X(jw) of the continuous-time signal x(t) = 2 + 5cos(200pi*t - pi/6) Show and clearly label your work.

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Asked by: Aman
Created at: 2025-12-09 00:56:14

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# Step-by-Step Solution Given signal: \[ x(t) = 2 + 5\cos(200\pi t - \frac{\pi}{6}) \] We are to find its spectrum \( X(j\omega) \), i.e., its **Fourier Transform**. --- ## 1. **Break Down the Signal** The signal consists of: - A DC component: \( 2 \) - A cosine term: \( 5\cos(200\pi t - \frac{\pi}{6}) \) --- ## 2. **Fourier Transform of Each Component** ### **A. DC Term** The Fourier Transform of a constant \( A \) is: \[ \mathcal{F}\{A\} = 2\pi A \delta(\omega) \] So, \[ \mathcal{F}\{2\} = 4\pi \delta(\omega) \] --- ### **B. Cosine Term** Recall: \[ \cos(\omega_ t + \phi) = \frac{1}{2} \left[ e^{j(\omega_ t + \phi)} + e^{-j(\omega_ t + \phi)} \right] \] The Fourier Transform of \( \cos(\omega_ t + \phi) \) is: \[ \pi \left[ e^{j\phi} \delta(\omega - \omega_) + e^{-j\phi} \delta(\omega + \omega_) \right] \] Here, - Amplitude = 5 - \( \omega_ = 200\pi \) - \( \phi = -\frac{\pi}{6} \) So, \[ \mathcal{F}\{5\cos(200\pi t - \frac{\pi}{6})\} = 5\pi \left[ e^{-j\frac{\pi}{6}} \delta(\omega - 200\pi) + e^{j\frac{\pi}{6}} \delta(\omega + 200\pi) \right] \] --- ## 3. **Combine Results** Add the two parts: \[ X(j\omega) = 4\pi \delta(\omega) + 5\pi \left[ e^{-j\frac{\pi}{6}} \delta(\omega - 200\pi) + e^{j\frac{\pi}{6}} \delta(\omega + 200\pi) \right] \] --- ## 4. **Final Answer (Labeled and Boxed)** \[ \boxed{ X(j\omega) = 4\pi \delta(\omega) + 5\pi e^{-j\frac{\pi}{6}} \delta(\omega - 200\pi) + 5\pi e^{j\frac{\pi}{6}} \delta(\omega + 200\pi) } \] --- ## 5. **Interpretation** - **At \( \omega = \):** DC component with magnitude \( 4\pi \) - **At \( \omega = +200\pi \):** Spectral line with magnitude \( 5\pi \) and phase \( -\frac{\pi}{6} \) - **At \( \omega = -200\pi \):** Spectral line with magnitude \( 5\pi \) and phase \( +\frac{\pi}{6} \) --- **Summary Table:** | Frequency (\(\omega\)) | Magnitude | Phase | |-----------------------|-----------|------------------| | | \(4\pi\) | | | \(200\pi\) | \(5\pi\) | \(-\frac{\pi}{6}\) | | \(-200\pi\) | \(5\pi\) | \(+\frac{\pi}{6}\) | --- **No further computation is necessary. This is the complete analytical spectrum.**

Question:

Final answer not in table summary

This is AI generated Response.
Asked by: Aman
Created at: 2025-12-09 00:59:13

Final Answer:

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# Step-by-Step Solution Given signal: \[ x(t) = 2 + 5\cos(200\pi t - \frac{\pi}{6}) \] We are to find its spectrum \( X(j\omega) \), i.e., its **Fourier Transform**. --- ## 1. **Break Down the Signal** The signal consists of: - A DC component: \( 2 \) - A cosine term: \( 5\cos(200\pi t - \frac{\pi}{6}) \) --- ## 2. **Fourier Transform of Each Component** ### **A. DC Term** The Fourier Transform of a constant \( A \) is: \[ \mathcal{F}\{A\} = 2\pi A \delta(\omega) \] For our DC component: \[ \mathcal{F}\{2\} = 4\pi \delta(\omega) \] ### **B. Cosine Term** Using the identity: \[ \cos(\omega_0 t + \phi) = \frac{1}{2} \left[ e^{j(\omega_0 t + \phi)} + e^{-j(\omega_0 t + \phi)} \right] \] The Fourier Transform of \( \cos(\omega_0 t + \phi) \) is: \[ \mathcal{F}\{\cos(\omega_0 t + \phi)\} = \pi \left[ e^{j\phi} \delta(\omega - \omega_0) + e^{-j\phi} \delta(\omega + \omega_0) \right] \] For our cosine component: - Amplitude = 5 - \( \omega_0 = 200\pi \) - \( \phi = -\frac{\pi}{6} \) Thus, \[ \mathcal{F}\{5\cos(200\pi t - \frac{\pi}{6})\} = 5\pi \left[ e^{-j\frac{\pi}{6}} \delta(\omega - 200\pi) + e^{j\frac{\pi}{6}} \delta(\omega + 200\pi) \right] \] --- ## 3. **Combine Results** Adding the two Fourier Transform results: \[ X(j\omega) = 4\pi \delta(\omega) + 5\pi \left[ e^{-j\frac{\pi}{6}} \delta(\omega - 200\pi) + e^{j\frac{\pi}{6}} \delta(\omega + 200\pi) \right] \] --- ## 4. **Final Answer** The complete analytical spectrum is: \[ \boxed{ X(j\omega) = 4\pi \delta(\omega) + 5\pi e^{-j\frac{\pi}{6}} \delta(\omega - 200\pi) + 5\pi e^{j\frac{\pi}{6}} \delta(\omega + 200\pi) } \] --- ## 5. **Interpretation** - **At \( \omega = 0 \):** DC component with magnitude \( 4\pi \) - **At \( \omega = +200\pi \):** Spectral line with magnitude \( 5\pi \) and phase \( -\frac{\pi}{6} \) - **At \( \omega = -200\pi \):** Spectral line with magnitude \( 5\pi \) and phase \( +\frac{\pi}{6} \) --- **Summary Table:** | Frequency (\(\omega\)) | Magnitude | Phase | |-----------------------|-----------|------------------| | \(0\) | \(4\pi\) | \(0\) | | \(200\pi\) | \(5\pi\) | \(-\frac{\pi}{6}\) | | \(-200\pi\) | \(5\pi\) | \(+\frac{\pi}{6}\) | --- This completes the analytical determination of the spectrum \( X(j\omega) \).

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