New answers tagged continuous-signals
2
I would like to add to the previous answers that the difference between a second-order highpass and a second-order lowpass filter is generally NOT an allpass filter. The resulting transfer function
$$H(s)=\frac{s^2-\omega_0^2}{s^2+\frac{\omega_0}{Q}s+\omega_0^2}\tag{1}$$
has two zeros at $s_0=\pm\omega_0$ and two poles which are either real-valued (for $Q\le\...
0
$\frac{s^n + b_{n-1}s^{n-1}+\cdots+b_0}{s^n + a_{n-1}s^{n-1}+\cdots+a_0} = 1 + \frac{(b_{n-1}-a_{n-1})s^{n-1}+\cdots+b_0 - a_0}{s^n + a_{n-1}s^{n-1}+\cdots+a_0}$, and the inverse Laplace transform of 1 is $\delta(t)$.
Is that enough to get on with?
3
Just adding to the answer. The sign for recombining L/R filter alternates, so it's '-' for second order '+' for 4th order and so forth.
The algebra is isn't all that pleasant so I will only go through the 2nd order. A second order L/R lowpass is simply the cascade of two first order Butterworth lowpass. The first order butterworths are
$$L = \frac{1}{1+jx}, ...
2
D'oh. This turned out to be pretty obvious, and I should have gotten it from reading more closely. I would have deleted this out of shame, but instead will share the correction in hopes it saves someone else the same headache. The controlling equations for the LR2 high and low pass transfer functions are correct, but one of the signals needs to be ...
2
The total power of an amplitude modulated signal is
$$\begin{align}\overline{s^2_{AM}(t)}&=\overline{\big(A+m(t)\big)^2\cos^2(2\pi f_ct)}\\&=\frac12\overline{\big(A+m(t)\big)^2}+\frac12\overline{\big(A+m(t)\big)^2\cos(4\pi f_ct)}\tag{1}\end{align}$$
The second term on the right-hand side of $(1)$ is zero if $m(t)$ is a lowpass signal, and if $f_c$ is ...
1
The solution to the differential equation is given by the sum of a particular solution and the solution of the homogeneous differential equation. The particular solution is a solution to the non-homogeneous equation
$$T\dot{x}(t)+x(t)=k_vu(t)\tag{1}$$
which is easily found as
$$x_p(t)=k_v,\qquad t>0\tag{2}$$
The homogeneous solution is the solution of the ...
1
Given a transfer function
$$G_v(s) = \frac{k_v}{1 + sT} \tag{1}$$
the corresponding LCCDE, with $y(t)$ being the solution, and $x(t)$ being the input, will be
$$ T ~\dot{y}(t) + y(t) = k_v ~x(t) \tag{2} $$
Your formulation replaces $x(t)$ with a unit-step $u(t)$, and $y(t)$ with $x(t)$, yielding
$$ T ~\dot{x}(t) + x(t) = k_v ~u(t) \tag{3} $$
or equivalently
$...
0
Don't make this more complicated than it really is. $x(t)$ is non-zero in the interval $t\in[0,1]$, and $x(t+\tau)$ is non-zero in the interval $t\in[-\tau,1-\tau]$. The integrand is non-zero only if the two functions overlap. There is no overlap for $1-\tau<0$ and $-\tau>1$, i.e., for $|\tau|>1$. So for $|\tau|>1$ the autocorrelation is zero.
...
1
If you're referring to energy spectral density, then you need to compute the squared magnitude of the Fourier transform of $g(t)$:
$$\big|G(\omega)\big|^2=\left|\int_{-\infty}^{\infty}g(t)e^{-j\omega t}dt\right|^2\tag{1}$$
0
We have signal $y(t) = f(t)cos(\pi t-\pi)$. You are modulating amplitude where bias is $\pi$. By using Euler's formula you can rewrite singal as:
$$
y(t) = f(t)\frac{1}{2}\left(e^{j(\pi t - \pi)} + e^{-j(\pi t - \pi)}\right)
$$
$$
y(t) = \frac{1}{2}f(t)e^{j(\pi t - \pi)} + \frac{1}{2}f(t)e^{-j(\pi t - \pi)}
$$
$$
y(t) = \frac{1}{2}f(t)e^{j\pi t}e^{-j\pi} + \...
2
It is easy to see that using the trigonometric identity in Equation $(1)$ below
$$
\cos(\theta_1 \pm \theta_2) = \cos(\theta_1)\cos(\theta_2) \mp \sin(\theta_1)\sin(\theta_2) \tag{1}
$$
We have
\begin{align}
f(t)\cos\big(\pi(t - 1)\big) &= f(t)\overbrace{\cos(\pi t - \pi)}^{\text{use Equation}\ (1)}\\
&\equiv -f(t)\cos(\pi t)\tag{2}
\end{align}
Which ...
1
Using Duality Property, we have $X(j\omega) = \delta(\omega-\omega_{0})$. By using this and rewriting our function using $1-2k = -\omega_{0}$, we get:
$$\begin{align}
x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega-\omega_{0})e^{j\omega t}d\omega \\
&= \frac{1}{2\pi}e^{j \omega_0 t}
\end{align}$$
$$
\sum_{k=-\infty}^{\infty}\delta(\omega-\...
0
All of these squared energy terms come from the basic physics of simple harmonic motion (SHM), where there is a restoring force proportional to displacement. This can be mechanical, electric field, magnetic field, etc etc. When f(x) = -kx, energy = 0.5 k x^2 + c, since force is by definition the derivative of energy.
1
You made a small mistake in the final step. The second term in parentheses should be $2((-1)^k-1)$ because you have two terms $e^{-j\pi k}$. So your final result should be
$$X[k]=\frac{3}{2\pi jk}\big(1-(-1)^k\big)\tag{1}$$
With $(-1)^k=e^{-j\pi k}$, Eq. $(1)$ can be rewritten as
$$\begin{align}X[k]&=\frac{3}{k\pi}\frac{1-e^{-jk\pi}}{2j}\\&=\frac{3e^{...
1
The formula for the energy of a signal over an interval isn't a derivation. It's a definition. You do that integration and you call the result the signal's energy.
The reason that in the mathematical world we make that definition is because (A) it makes the math handy, and (B) for a preponderance (but by no means all) of the means of signal transmission in ...
0
From a mathematical point of view, a finite series is convergent, as answered on Can a sequence be called convergent/divergent if it has finite number of terms?.
If it is a subsequence of an infinite series, I know of no way to tell for sure from the subsequence that the whole sequence convergence. There are many counter-examples and reasons: processes ...
0
Yes and no.
To confidently determine convergence, you should always test for meeting a formal convergence criteria.
As a hack, you can numerically calculate the series until it "stops moving" or your running out of time/memory or start getting NaNs, infinites or over/underflows.
1
No, it's not that simple. In general, sampling will introduce aliasing. The correct derivation is as follows. Let $T$ be the period of the continuous-time signal $x(t)$, and let $T_s$ be the sampling period satisfying $T=NT_s$ with integer $N$. It follows that the sampled sequence $x_d[n]=x(nT_s)$ is also periodic with period $N$.
We have
$$x(t)=\sum_{k=-\...
2
You are right that the argument in the proof is not correct, or at least misleading. The fact that $e^{-st}$ becomes zero as $t\to\infty$ is true for any $s$ with $\textrm{Re}\{s\}>0$. However, the integral
$$\int_0^{\infty}f(\tau)d\tau$$
might not exist, so we can't just compute the limit by claiming that the first term becomes zero.
We have to compute
$$...
0
$H(jω)$, which is DTFT of the transfer function $h(n)$, is basically a special case of the Z transform $H(z)$, where $z=e^jω$ i.e. $|z|=1$.
In this special case, the unit circle ($|z|=1$) should be included in the region of convergence (ROC) of $H(z)$. In other words, the system (which is LTI in your example) should be BIBO stable.
The system in your example ...
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