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(Solved): The Nyquist sampling rate of a signal is the minimum rate s = 2/Ts for which the signal may be s ...



The Nyquist sampling rate of a signal is the minimum rate ωs = 2π/Ts for which the signal may be sampled and no aliasing occurs from its samples. The Nyquist rate is equal to 2 times the largest non-zero frequency in the Fourier transform. Determine the Nyquist rate ωs (in angular frequency) for the following signals. Hint: If you get stuck, check Euler’s Formula or trig identities! (a) x(t) = 600 cos(500πt + 400) + 800 cos(700πt + 600) (b) x(t) = 10 + 10 cos(20t + π/30) sin(15t + 29π/30) cos(5t − π) (c) x(t) = P∞ m=0 cos(5mt)



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