# Tag Info

12

As you say, the constant function is periodic. A signal $x(t)$ is said to be periodic with period $p$ or to have a period $p$ if there exists a $p>0$ such that $x(t+p)=x(t)$ for all real numbers $t$. Note that I said "a period" instead of "the period" -- periodic signals have an infinite number of periods, maybe even an uncountable infinite of them! ...

10

Most realistic signals are both random and periodic. For example, you can modulate a harmonic oscillator with a slow enough random signal that moves its frequency around a $\mu_{f}, \sigma_f$. This looks like: $$y= \sin \left( \frac{2 \pi \mathcal{N_s}(\mu_f, \sigma_f) n}{Fs} \right )$$ Where $\mathcal{N_s}(\cdot)$, denotes a normally distributed random ...

9

The cross pattern is typically a border effect, due to the periodicity induced by the standard implementation and hypotheses behind the Fast Fourier transform, when the image lacks periodicity from the right to the left, and the bottom to the top. In other words: if two opposite borders lacks continuity in values (when glued together), artifacts show. The ...

8

As for every $t_0\in\mathbb{R}$ and $k\in\mathbb{Z}$ $$\begin{eqnarray} &x(t_0+4k\pi) &=\cos(t_0+4k\pi)+\sin(t_0/2+2k\pi)\\ & &=\cos(t_0)+\sin(t_0/2)\\ & &=x(t_0) \end{eqnarray}$$ you answer is correct: $x(t)$ is periodic.

8

The argument does not work in continuous time. In discrete time the argument is that $$e^{j\omega n}=e^{j(\omega+2\pi)n},\qquad n\in\mathbb{Z}\tag{1}$$ This is true because by definition $n$ is an integer. In continuous time we generally have $$e^{j\omega t}\neq e^{j(\omega+2\pi)t},\qquad t\in\mathbb{R}\tag{2}$$ because $t$ is a real variable. Eq. $(2)$ only ...

8

I assume the model to be: $$x \left[ n \right] = \sin \left[ 2 \pi \frac{f}{ {f}_{s} } n + \phi \right] + w \left[ n \right]$$ Where $w \left[ n \right]$ is white noise uncorrelated with the signal itself. The obvious method here would be using DFT as the Maximum Likelihood Estimator of such case. It will probably be able to generate the best results in ...

6

The problem with your reasoning is that $\pi \ne \frac{22}{7}$; $\pi$ is an irrational number. There is no period $N$ for which $x[n] = x[n+N] \ \forall \ n \in \mathbb{Z}$. Hence, the sequence is not periodic.

6

well, there is always autocorrelation $$R_x(\tau)=\sum x[n] x[n+\tau]$$ or AMDF $$Q_x(\tau) = \sum |x[n] - x[n+\tau]|$$ or ASDF $$Q_x(\tau) = \sum (x[n] - x[n+\tau])^2$$ with the latter there is the relationship between autocorrelation and ASDF $$R_x(\tau)=R_x(0) - \frac12 Q_x(\tau)$$. a measure of periodicity might be $\frac{R_x(P)}{R_x(0)}$ where ...

6

You must have understood the notion of digital linear modulation or discrete time vs continuos time (see Chapter 2). Another reference. OFDM can be thought as FDM with sinc pulse whose delay-$T$-shifted versions form an orthonormal basis. In frequency domain, they are seperated by $1/T$ which is denoted $\Delta f$ , i.e. subcarrier spacing. You fix a $T$, ...

6

If you are talking about a given signal as "a deterministic realization of a phenomenon", it can be periodic, but not really random. However, some physical systems are prone to produce randomness and periodicity, like rotating machines, gears, cyclic engines, that produce signals similar to: Naturally rotating bodies (stars, planets) also produce random ...

5

The periodogram is simply the squared magnitude of the DFT. Since the periodogram is a rather poor estimate of the power density spectrum of a random process there are methods which use averaging of periodograms to obtain better estimates of the power spectrum. Two such methods are Welch's method and Bartlett's method.

5

To add a contrarian answer: If your time index, $t$, is an integer, then your signal is not periodic. The definition of periodic is: $x[t]$, $t\in\mathbb{Z}$ is periodic with period $P\in \mathbb{Z}$ iff $$x[t] = x[t+P]$$ So we need $$\cos(t) = \cos(t+P)$$ so for periodicity we require $$P = 2\pi k$$ with $k\in \mathbb{Z}$. Since $\pi$ is ...

5

The double-angle formulae for trigonometric identities tell you that $\cos \left(\frac{2t}{t}\right) = 1 - 2\sin^2(\frac{t}{2})$. You thus ave $x(t) =1 +\sin(\frac{t}{2}) - 2\sin^2(\frac{t}{2})$. Hence, your signal is composed of functions (as adds and multiplies) that all admit $4\pi$ as a period (yes, the constant function $x\to 1$ is $4\pi$ periodic as ...

5

I'd like to show you a more formal derivation. Note that the first formula for arbitrary (non-periodic) signals could be rewritten as $$P_x=\lim_{M\rightarrow\infty}\frac{1}{(2M+1)N}\sum_{n=-MN}^{(M+1)N-1}|x[n]|^2\tag{1}$$ for some integer $N\ge 1$. For $N=1$, Eq. $(1)$ is identical with the first formula in your question. If you choose $N>1$ you simply ...

5

A periodic continuous-time signal satisfies $x(t)=x(t+T_0)$ for all $t$. The period $T_0$ doesn't need to be a rational number. A periodic discrete-time signal satisfies $x[n]=x[n+N]$ for all integers $n$. The period $N$ is an integer. If you sample a periodic continuous-time signal, you don't necessarily get a periodic sequence. E.g., sampling the periodic ...

5

The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind: $$J_n(x)=\frac{1}{\pi}\int_0^{\pi}\cos(nt-x\sin t)dt,\qquad n\in\mathbb{Z}\tag{1}$$ Using the cycloid parameterization $$y(t)=1-\cos t,\qquad x(t)=t-\sin t\tag{2}$$ which results in a period of $2\pi$ and a maximum value of $2$, the Fourier series of $y(t)$ ...

4

As you have correctly observed, $2N/W$ must be an integer, because the window can only have an integer number of samples. Furthermore, regardless of the upper summation limit, $$Y_k=\sum_{m=0}^Ke^{-j\frac{2\pi mk}{2N}}$$ is always $2N$-periodic because $$Y_{k+2N}=\sum_{m=0}^Ke^{-j\frac{2\pi m(k+2N)}{2N}}=\sum_{m=0}^Ke^{-j\frac{2\pi mk}{2N}e^{-j2\pi m}}=\... 4 The basic trick is to bound the series above and below. Let us do it on one side, for positive indices. For any N> 0, you can write N=kN_0+r_N, with 0\le r_N< N_0. Then if a_n (here a_n = |x_n|^2) is positive, \sum_{n=0}^{N-1} a_n is increasing. Now kN_0 \le N< (k+1)N_0, hence you have:$$ \sum_{n=0}^{kN_0-1} a_n \le \sum_{n=0}^{...

4

If the data is cyclic by its nature the best thing would work using its spectrum. You can easily build a system which checks sub set of data to verify periodic and the once you establish your groups checking the affinity of new data is easy - add it to each series. It should belong to the one creates less spread in the frequency (Namely, it follows the ...

4

You should use the synthesis equation of an impulse train with period $T$ (which is easy to derive): $$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$ That is: the Fourier coefficients for all terms is a constant ($\frac{1}{T}$). Now assume that there are two impulses with different ...

4

I suggest using Spectral Flatness, aka Wiener Entropy. It is defined as a ratio of geometric and arithmetic mean of the magnitude spectra $X(k)$: $$\Xi=\dfrac{\sqrt[K]{\prod_{k=0}^{K} X(k)}}{\frac{1}{K}\sum_{k=0}^{K}X(k)}$$ For signals which have flat spectra, its value tends towards $1$, whereas for tonal signals it is close to $0$. In your particular ...

4

When you are in doubt, use the limiting approach as an aid in your deductions: For example, you can consider a constant signal $x_C(t)=1$ as the limit of a periodic sine wave $x_p(t)= \cos(\omega_0 t)$ when the frequency goes to zero; i.e: $$x_C(t) = 1 = \lim_{\omega_0 \to 0} \cos(\omega_0 t)$$ And you now that the Fourier transform of a sine wave is a ...

4

If your top equation is really $$x(t) = 2\cos\left(\frac 45 \pi t\right)\sin^2\left(\frac{16}{3} t\right)\tag{1}$$ You gonna have a hard time getting the fundamental period/frequency as the there isn't an exact integer relating the two periods/frequencies. You can make approximations of the $\pi$ multiplier/divisor but the errors accumulates and this ...

4

No, in a conventional sense of a "periodic signal" phrase, but, if you permit me to delve into a math subtlety, differentiation can turn an aperiodic waveform to a periodic one: $$\frac{\mathrm{d}}{\mathrm{d}t}(a\cdot t + b\cdot \cos(\omega\cdot t)) = a-b\cdot\omega\cdot\sin(\omega\cdot t)$$ Notwithstanding a dubious usefulness of this excursion.

4

I haven't managed to completely evaluate the integral, but I've made some progress and perhaps someone can pick up where I leave off. The integral you gave is $$c_n = \int_0^{2\pi}\left(\frac{1-\cos\theta}{\pi}\right)^2 e^{-jn(\theta-\sin\theta)}d\theta.$$ The first thing I did is perform a Taylor expansion on that exponential term; $$c_n = \frac{1}{\pi^2}... 4 HINT:$$\sin\left(\frac{12\pi n}{5}\right)=\sin\left(\frac{2\pi n}{5}+\frac{10\pi n}{5}\right)=\sin\left(\frac{2\pi n}{5}+2\pi n\right)$$Taking into account that \sin(x) is 2\pi-periodic should make the result obvious. 4 The pitch period of a perfectly periodic function, x(t), is the smallest positive value P>0 such that$$ x(t+P) = x(t) \qquad \forall t \in \mathbb{R} $$Now, simply because a function is periodic with period P, then it is also periodic with periods 2P or 3P or 4P or any integer multiple of P, but we don't pick 2P or 3P or 4P for the ... 3 The periodicity of a signal holds if we can show x(n)=x(n+N), otherwise, the signal is nonperiodic. Simply start with$$ \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})\cos(\frac{N}{6}) - \sin(\frac{n}{6})\sin(\frac{N}{6}) \end{align} $$In order for x(n)=x(n+N) to hold, \cos(\frac{N}{6})=1 and \sin(\frac{N}{6}... 3 The average power of a signal s(t) is given by$$\overline{s^2(t)}=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt\tag{1}$$If s(t) is periodic, (1) is equivalent to the average power in one period:$$\overline{s^2(t)}=\frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt\tag{2} where $T$ now denotes the period of $s(t)$.

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