New answers tagged continuous-signals
1
Let $x(t)=\big[u(t-a)-u(t-b)\big]$, and $s(t)=x(t)\cos(2\pi f_0t)$. If we can assume that $f_0\gg 1/(b-a)$, then we can approximate the analytic signal of $s(t)$ (i.e., the signal with all negative frequency components removed) by
$$s_a(t)\approx x(t)e^{j2\pi f_0t}\tag{1}$$
The Hilbert transform of $s(t)$ is the imaginary part of the analytic signal, i.e., ...
0
I think I figured it out, dumb mistake in writing the system in continuous time.
$\dot{l} = r_1(u - l)$
$\dot{s} = r_2(\dot{l} - s)$
$\ \ = r_2(r_1(u - l) + r_1 s - s)$
$\ \ = r_2(r_1(u - l) + (r_1 - 1)s)$
$\ \ = r_2 r_1(u - l) + r_2(r_1 - 1)s$
So $$
A = \begin{bmatrix}
-r_1 & 0\\
-r_1r_2 & -r_1r_2\\
\end{bmatrix}
$$
which is stable for all ...
2
If $x(t)$ is a finite-energy signal with Fourier transform $X(f)$, then $x(t)\cos(2\pi f_c t)$ is also a finite-energy signal with Fourier transform $\left.\left.\frac 12 \right[X(f-f_c) + X(f+f_c)\right]$. This is just the modulation theorem of Fourier transform theory.
The energy spectral density of $x(t)$ is $S_x(f) = |X(f)|^2$ while the energy spectral ...
0
For audio signals, a periodic signal can have a single "pitch frequency" (detectable repeat rate, or 1/period), even though it might consist of multiple sinusoidal harmonic frequency elements, and even with a missing fundamental spectral element at the pitch frequency. e.g. the pitch frequency can be different from the dominant spectral frequency or ...
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" ...
0
You simply forgot the low-pass filtering step applied after modulation with a carrier that other derivations do.
1
Actually AC and DC are terms that come from descriptions of types of electrical supply currents. There is an interesting history between Tesla and Edison on this, you can look it up. AC stands for alternating current, in signal processing terms it is ideally a pure sinusoid. DC stands for direct current, ideally rock steady. The terms have been co-opted ...
1
When you have a continuous-time periodic signal, you should use CTFS (continuous-time Fourier series) to determine whether the signal is DC, AC or a mixture (AC + DC).
Given the periodic signal $x(t) \neq 0$, perform a CTFS analysis; $x(t) \longleftrightarrow a_k$ :
if $a_0 \neq 0$, but all remaning $a_k = 0$ for $k = \pm1,\pm2,...$ then it's a pure DC ...
2
The signal in question is a rectified sine-wave. This signal has a non-zero average, i.e. a DC component. It also has AC components at f, 3f, 5f, etc. where f is the frequency of the non-rectified signal.
1
The canonical definition of independence of two random variables $X$ and $Y$ is
$X$ and $Y$ are called independent random variables if
for every choice of Borel sets $B_1, B_2$, the events $\{X \in B_1\}$ and $\{Y \in B_2\}$ are independent events, that is, $$P\{X \in B_1, Y \in B_2\} = P\{X \in B_1\}P\{Y \in B_2\}
\tag{1}$$
If you don't know what ...
2
It happens that if $X$ and $Y$ are independent then so will their functions $g(X)$ and $h(Y)$ be; but not $g(X,Y)$ and $h(X,Y)$.
2
Because of the diode at the input, you get a DC offset at the output. Notice that a single diode is already a primitive rectifier, because it blocks the negative half wave.
But you usually will want a purely AC output, that's why you put a HPF with very low cutoff frequency after the LPF. It will block the DC and yield a pure AC output signal. You can see ...
1
I believe it is interesting to look at such operator properties from an algebraic point of view. Properties are for instance, for a generic operator $\bigcirc$:
commutativity: $a \bigcirc b = b \bigcirc a $
associativity: $a \bigcirc( b \bigcirc c) = (a \bigcirc b) \bigcirc c $
alternativity: $a \bigcirc( a \bigcirc b) = (a \bigcirc a) \bigcirc b $ ...
1
there's an ISO for that.
https://www.iso.org/obp/ui/#iso:std:iso:18431:-4:ed-1:v1:en
You must buy. No, I need not.
Anyway, as FAT32 says, they are just symbols. Which symbols stand for what is just convention, and as you have pointed out different "Authoritays" have slightly different conventions.
What is important is how the math works and that is ...
2
D ear N atalie, as you also know that these are just letters of English (org Greek) alphabet which by themselves have no meaning or explanations other than what you have arbitrarily imposed on them and the following is by convention what's being imposed on them in the mathematics, physics and the standard DSP literature as accepted unless otherwise stated ...
1
The term LTI system is a bit broad, so perhaps restricting ourselves to single input-single output systems makes sense. Let's just look at $s$ (Laplace) for now. The $Z$ transform follows in a straightforward way.
Also, if the system is not stable, a practical inverse filter will not recover the input. Convolution commutes so following a stable filter with ...
2
You have to be clear what you mean by "invertible". Commonly, you want the inverse system to be causal and stable, and that puts certain restrictions on the original system.
In the case of systems with rational transfer functions, you just have to look at the zeros of the transfer function, because they become the poles of the inverse system. If all zeros ...
3
In general LTI System is invertible if it has neither zeros nor poles in the Fourier Domain (Its spectrum).
The way to prove it is to calculate the Fourier Transform of its Impulse Response.
The intuition is simple, if it has no zeros in the frequency domain one could calculate its inverse (Element wise inverse) in the frequency domain.
Few remarks for the ...
0
A typical approach would be the Pan-tompkins algorithm. You can also apply a first order low pass filter or a mean filter to improve results.
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The mass-spring-damper combination is an LTI system described by the following continuous-time linear differential equation
$$ \ddot{x} = - \frac{k}{m} x - \frac{b}{m} \dot{x} + \frac{1}{m} u $$
where $u$ is the deterministic input force (N), $k$ is the spring constant (N/m) and $b$ is the damping coefficient (N.s/m).
Assuming two states as the position $...
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