**Transcribing the problem statement** I need to read problem statement from the and start by transcribing the text. Zooming in, I see the message: "Let the baseband message be m(t) = 1.2 cos(2π·1000 t) + 0.8 sin(2π·2000 t), and the carrier be c(t) = A_c cos(2π f_c t), A_c = 10, f_c = 100 kHz." It’s unclear about whether f_c is truly 100 kHz or possibly something else due to the blurriness. The last part mentions computing the Hilbert transform m_h(t).**Computing the Hilbert transform** The task is to compute the Hilbert transform, m_H(t), of m(t). I need to use the Hilbert method to form the upper sideband (USB) signal and simplify φ_USB(t), which might actually be s_USB(t). We have to compute s_USB(t) = m(t) cos(2π f_c t) - m_h(t) sin(2π f_c t). In terms of SSB modulation, s_USB(t) can also be represented with the sin component included. Next, I need to sketch the amplitude spectrum of s_USB(t) with labeled frequencies and find the average transmitted power of the modulated signal. Finally, I'll identify the occupied bandwidth, keeping in mind that A_c might be the amplitude of the carrier in SSB suppressed-carrier.**Understanding the Hilbert method** The Hilbert method involves using m(t) multiplied by cos and subtracting the Hilbert-transformed term multiplied by sin. This leads to the SSB-SC (suppressed carrier) signal. I’m trying to figure out how the amplitude A_c factors in; typically, it’s 10, but with SSB-SC, the modulated signal's amplitude isn’t scaled by A_c. In the formula for s_USB(t), the amplitude of the cos and sin carriers is generally set to 1. However, if A is 10, then possibly the overall amplitude should reflect that. I'll double-check the text to clarify how it applies to the setup for s_SUSB(t).**Applying the Hilbert method** If we define a physical carrier as c(t) = A_c cos(2π f_c t), we can use the Hilbert method to produce the SSB with a suppressed carrier amplitude, A_c. Many texts refer to A_c for double sideband suppressed carrier (DSBSC) amplitude scaling. For SSB, the formula s_USB(t) = A_c [m(t) cos(2π f_c t) - m̂(t) sin(2π f_c t] helps us understand how to apply scaling. I must remember to ensure units are consistent. Next, I’ll produce an explicit expression for m_H(t), the Hilbert transform of the tone components. The Hilbert transforms tell us that H{cos(ωt)} = sin(ωt), and I need to check the signs and apply the VPT property to clarify shifts in phase.**Calculating frequency and power** The Hilbert transformations reveal that H{cos(ωt)} = sin(ωt) and H{sin(ωt)} = -cos(ωt). I need to compute m(t) = 1.2 cos(2π 1000 t) + 0.8 sin(2π 2000 t), leading to m_h(t) = 1.2 sin(2π 1000 t) - 0.8 cos(2π 2000 t). The SSB formula s_USB(t) = A_c [m(t) cos(ω_c t) - m_h(t) sin(ω_c t)] helps identify upper sideband frequencies using product-to-sum identities. For average power, I derive P_avg = A_c^2 * 1.04 and find the SSB bandwidth to be 2 kHz, determined by the highest frequency component.**Determining bandwidth and power** For SSB signals, the occupied bandwidth typically equals the highest baseband frequency, in this case, 2 kHz based on the two tones at 1 kHz and 2 kHz. However, since the actual signal has discrete lines only at these frequencies, I’d label the occupied bandwidth as 1 kHz. But for simplicity, I’ll go with 2 kHz based on the highest frequency for the exam context. Now, after calculating the Hilbert Transform, I find that the average transmitted power is 104 watts, considering the amplitude, and I’ll ensure to present the amplitude spectrum details correctly.**Calculating power and bandwidth** I’ve found that the average power \( P_{avg} \) is calculated as \( P_{avg} = (A_c^2 / 2)(1.2^2 + 0.8^2) = 1.04 A_c^2 \). Now, regarding occupied bandwidth for SSB, I’m noting the highest baseband frequency is 2 kHz. The passband goes from \( f_c \) to \( f_c + B \). Even though the actual energy occupies a width of 1 kHz between 101 kHz and 102 kHz, I will still state the occupied bandwidth as 2 kHz with the necessary disclaimers for clarity.