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Just think of the value of $t$ where $x(0)$ appears. For $x(t+t_0)$ it is at $t=-t_0$, which corresponds to a left shift if $t_0>0$. However, for $x(-t+t_0)$ the value $x(0)$ occurs at $t=t_0$, which is to the right of its original position if $t_0>0$. You can think of deriving $x(-t+t_0)$ from $x(t)$ in two different ways: invert the time axis: $x(-t)... 1 If$X(f)=\mathcal{F}\{x(t)\}$you have, in analogy with Eq.$(7)$, $$x(f)\Longleftrightarrow X(-t)\tag{1}$$ where we assume that functions with the independent variable$f$are Fourier transforms of the corresponding functions with independent variable$t$. Applying$(1)$to $$x(t-b)\Longleftrightarrow e^{-2\pi jbf} X(f)\tag{2}$$ gives $$x(f-b)\... 2 I think the confusion comes from the fact that the command hilbert in Scipy (and also in Matlab/Octave) does not just compute the Hilbert transform, but its output is the analytic signal. So if x(t) is the (real-valued) input to such a function, its (complex-valued) output is$$y(t)=x(t)+j\mathcal{H}\{x(t)\}\tag{1}$$Clearly, if you want to obtain x(t) ... 1 The autocorrelation of x(t) is$$r_x(t)=x(t)\star x(-t)\tag{1}$$where \star denotes convolution. Taking the Fourier transform of (1) gives$$S_x(\omega)=X(\omega)X^*(\omega)=|X(\omega)|^2\tag{2}$S_x(\omega)$is the energy density of$x(t)$, and according to$(2)$it equals the squared magnitude of the Fourier transform of$x(t)$. So if$x(t)$is ... 4 Let me suppose that by Nyquist, you mean that there is some frequency$F_s$such that there is a existence theorem that says: if one samples any continuous$s(t)$at$F_s$(or above),$F_s$denoting an extremal frequency or a extremal bandwidth, to get a discrete signal$s[k]$, then one can recover$s(t)\$ by some algorithm. Case 0: neither oversampling ...