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0

So who is right? Both, I think. On first looks version [3] and version [4] use different definitions of $A(z)$. [3] conjugates the zeros and [4] conjugates the poles. Either one will probably work but the definition of the coefficients is different. Specifically the $a$ coefficients of version [4] or conjugates of those of version [3]. So you have $$a_{n,i,...


1

Since I gave you a wrong hint first, let me try to make it up. You have the correct transfer function and the rest is (I think) just to slog through the algebra. Here is an outline how this can be approached. We can use the geometric sum property to simply things a bit $$4\cdot H(\omega) = 1 + e^{-j\omega} + e^{-j2\omega} e^{-j2\omega} = \frac{1-e^{-4j\...


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try short Short-time cross-correlation from matlab. There is the length of coherence. You need as short as length of coherence Lc<0.5 and You receive the dozen of the same pictures but not (-0.5 +0.5) on abscissa but -Lc,+Lc anв you must recieve max min bands


4

The filter will introduce phase shift to the signal that will appear as a total time delay. Not really. For an IIR filter the phase shift will be a function of frequency and it will (in general) not correspond to a single delay. It's a "frequency dependent delay" such that the maximum correlation is achieved when the signal is shifted back using ...


6

I am thinking of two numbers. They add to the number 15. Tell me what the two numbers are.


0

To ensure smooth continuity we generate $$ x_0 = \cos(\omega_0 t), \ t_0 \leq t < t_1 \\ x_1 = \cos(\omega_1 t + t_1 \cdot (\omega_0 / \omega_1)), \ t_1 \leq t < t_2 \\ $$ where $\omega_0 t$ is input phase for $x_0$. Example It cannot be perfectly smooth for any arbitrary $t_1$ per discretization limitations (but $t_1$ can be chosen to maximize ...


0

Generate the second sine wave as $$x(t) = sin(\omega_2 t + \omega_1t_0)$$ where $t_0$ is the time when the frequency changes from $\omega_1$ to $\omega_2$ .


0

So if I understand right, x is the phase function thus a signal fas the form of f(x) = sin(x) which x is the phase function and can be watever and f(x) is the amplitude funciton.


1

As per the question, the modulated signal is $x(t + \frac{\phi_{\mathrm{PM}}(t)}{2\pi f_c})$. In the limit of a weak phase deviation ($\Delta\phi\ll1$), we can expand to 1st order: $$ x(t + \frac{\Delta\phi \sin(2\pi f_{\mathrm{PM}}t)}{2\pi f_c}) \approx x(t) + \dot{x}(t) \frac{\Delta\phi}{2\pi f_c}\sin(2\pi f_{\mathrm{PM}}t)$$ The first term is simply the ...


0

IMO this not a particularly well written paper with confusing terminology and needless complications, so I recommend to look somewhere else. So is x the signal or the phase function? It is the phase function. Choosing a sine wave as a phase function is certainly confusing, but it IS a phase function. "amplitude" here just refers to the magnitude ...


2

Since the result is small angle FM modulation, for a single tone there will only be two significant sidebands each with magnitude close to $\beta/2$ (for the single sinusoidal modulation case with small $\beta$), where $\beta$ is the modulation index (peak angle). The two sidebands would be centered about the carrier spaced by the modulation rate - ...


2

the z transform changes formulas to be in terms of z, the complex sinusoid. That's a narrow, and not really mathematically correct, presentation. The z transform takes a signal in the time domain and transforms it into the z domain. It's better to think of the z domain as a frequency domain. When you work in this domain, $z$ does not represent a complex ...


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