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 ...
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 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 ...
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 ...
answered Apr 28 '15 at 21:58
robert bristow-johnson
13.6k3 gold badges23 silver badges58 bronze badges
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
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
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
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. $(2)$ ...
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 ...
4
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.
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
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
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.
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 $s(t)$.
3
Neither of these tones is an exact number of samples at 48 kHz, but you can calculate the periods as follows:
$\frac{48000}{441} = 108.84$ samples
$\frac{48000}{700} = 68.57$ samples
3
It is a matter of convention. The answer depends on the definition of the step function $u(t)$. In signal processing, it is common to define $u(0)=1/2$. Under this definition, we have that $Ev(\cos(4\pi t)u(t)) = 1/2(0.5+0.5)=1/2$, so the signal is indeed periodic.
If $u(0)$ is defined as either 0 or 1, then you'd be correct and the signal wouldn't be ...
3
If you have two periodic functions $f(t)$ and $g(t)$, where $f(t)$ has fundamental frequency $f_0$ and $g(t)$ has fundamental frequency $2f_0$, then their sum $h(t)=f(t)+g(t)$ is periodic with fundamental frequency $f_0$. Without any further knowledge about $f(t)$ or $g(t)$ there is no way to separate the two, because if the only knowledge you have about $f(...
3
With reference to FFT windowing of periodic signals, an integer number of cycles means that a signal repeats a whole number of times (3, 4, 5, etc.) within the FFT aperture, with no fractional remainder (3.1, 4.51, 5.99, etc.). For a frequency of X Hz, the FFT width in time would have to be some whole integer multiple of 1/X, which is the period of the ...
3
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 ...
3
If we take the Fourier transform of any constant signal, we get an
impulse at zero, which says that its frequency is zero and, hence, it
is non-repeating and its period is infinity.
No, this does not work for a zero signal (Fourier is flat-flat, no impulse). Plus, something over zero is traditionally undefined, and could be any number. And THIS is the ...
3
The first term is just a constant, so it's not relevant to periodicity. The sequence as written is periodic with a fundamental frequency of 1/6.
Your cited solution makes no sense to me.
3
In my world, frequency is inverse to period, i.e., a sinusoid with fundamental period $T$ has a frequency $f = \frac 1T$. Of course, angular frequencies are also common, in which case we have $\omega = 2\pi f = \frac{2\pi}{T}$, I guess this is what you were referring to.
If a signal is periodic but not purely sinusoidal, it does not have "just one" ...
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