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Your diagnosis seems right. When I just precompute the product of the two oscillators in my head, that gives me frequency components at $\pm 10$ and at 2010, so mixing a tone with frequency 200 with that yields frequency components at 190, 210, -190, -210, 1810, 2210, -1810 and -2210, But nothing in $[-90;90]$. PS: it's better to keep the 2 with the $\pi$, ...


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I start with BPSK achieving 1 bits/sec/Hz over passband AWGN. Factoring in 1/3 rate coding this becomes 0.333 bits/sec/Hz. This is not correct. The Shannon noisy channel coding theorem states that the reliable discrete-time rate $r$ (whose unit is bits per symbol, or bits per channel-use, or bpcu) is upper-bounded $$r \lt \frac{1}{2}\log_2\left(1 + \frac{S}{...


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You've computed the bound on the capacity of an AWGN channel. But with the additional constraint that the signal must use BPSK modulation the capacity will be lower (I don't know offhand what it is but there should be a reference in the text). Edit: There's a good description of the difference between continuous- and discrete-input AWGN channels here: ...


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Pure MSK has a linear phase trajectory from symbol to symbol over 90 degrees (resulting in the minimum frequency separation where the frequencies over a symbol duration are still orthogonal) while GMSK the phase trajectory is the integration of a Gaussian pulse resulting in no abrupt transitions but still transitions from 0 to 90 degrees over a symbol ...


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Your approach confuses frequency with phase; the correct formulation is $$ \sin(2\pi \phi(t)) $$ where $\phi(t) = \int \omega(t)dt$. Related post. I derived the most general form for a linear chirp here; Python code: def lchirp(N, fmin=0, fmax=None, tmin=0, tmax=1): fmax = fmax if fmax is not None else N/2 t = np.linspace(0, 1, N) a = (fmin - ...


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The phase of your signal is $$\phi(t) = 2\pi c\dot (a + b\cdot t) \cdot t = 2\pi \cdot (a\cdot t + b\cdot t^2) $$ The frequency is the derivative of the phase with respect to time NOT phase divided by time. So we get $$\omega(t) = \frac{d \phi}{dt} = 2\pi \cdot (a + 2b\cdot t) $$ Solving for Nyquist, i.e. $\omega(t_N)= \pi$ $$t_N = \frac{0.5-a}{2b} = 300$$


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