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For tapped-delay line FIR filters, if you want to know "What is the constraint on real- and complex-valued FIR filters that guarantee linear phase behavior in the frequency domain?", see the following web page: https://www.dsprelated.com/showarticle/808.php


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Keep reading the text: To see that this filter is linear phase, note that the coefficients are symmetric about the center point of the filter that is defined by $\mu(k) = 1/2$.


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Linear interpolation is equivalent to a triangular filter kernel, scaled/sampled such that only two non-zero samples are present. A triangular analog prototype is linear phase because it is symmetric. -k


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Why is this the case? Because you start your time vector at 1 and not at zero. You should use ts = 0:(q-1). Starting at one is equivalent to a delay of -1 sample delay the transfer function of which is $e^{j 2\pi \omega*T}$ where T is your sample period. For the FFT that comes out to be $e^{j 2\pi k/100}$


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For what values of the model parameters, does it generate a real signal? That's probably not possible. Roughly speaking, your frequencies are all positive and a real signal must have a conjugate symmetric spectrum, i.e. equal amount of positive and negative frequencies. If you make the sum run from $-K$ to $+K$ and set $a_k = a_{-k}, \alpha_k = \alpha_{-k}, ...


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Are there situations in signal processing where one would like to identify e.g. a time-domain signal from only it's power spectrum? Well, your described optical algorithm doesn't "only" depend on the power spectrum, but on the power spectrum under different phase shifts – which is a way of estimating the Fourier transform of the time (or spatial) ...


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