Finished ECOMMS homework 3
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@ -90,6 +90,4 @@ Frequency: 1kHz
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### 3c
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If the RF waveform appears across a 50$\Omega$ load, determine the average power and the PEP.
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The average power of the signal is:
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$$\langle s^2(t) \rangle = \frac{1}{T} \int\limits_{-T/2}^{T/2} 500\cos(\omega_c t + 20\cos(\omega_1 t)) dt$$
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The average power and the PEP are the same: $\frac{A_c^2}{2} \times \frac{1}{50} = 2.5$kW.
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import numpy as np
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import scipy as sp
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import matplotlib.pyplot as plt
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f_c = 100E+6
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f_m = 1E3
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T_c = 1/f_c
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f_s = 1E9
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T_s = 1/f_s
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A_c = 500
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D_p = 100
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D_f = 1E+6
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omega_m = 2*np.pi*f_m
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omega_c = 2*np.pi*f_c
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# Time samples to 10 seconds at a samplig frequency of 8kHz
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t = np.arange(0,10*T_c,T_s)
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m_p = 0.2*np.cos(omega_m * t)
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m_f = -2E-5*omega_m*np.sin(omega_m * t)
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M_f = 2E-5*np.cos(omega_m*t)
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theta = D_f * M_f
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# The phase modulated signal
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s_p = A_c*np.cos(omega_c*t + D_p*m_p)
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s_f = A_c*np.cos(omega_c*t + theta)
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c = A_c*np.cos(omega_c *t)
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plt.plot(t, s_p, label="Phase Modulated Signal")
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plt.plot(t, s_p-c)
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#plt.plot(t, s_f, label="Frequency Modulated Signal")
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plt.show()
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