New answers tagged

1

Here is an attempt at a constructive answer. The claim is that a pure sinusoid is the only periodic waveform that has the property $\forall A_1,\tau_1 \exists A_2,\tau_2$ such that $x(t) + A_1x(t+\tau_1) = A_2 x(t+\tau_2)\;\forall t$ where $x(t)$ is $T$-periodic. I am going to limit myselfto "well-behaved" functions which can be uniquely expressed ...


1

The phase is:$$\arg\left[ \frac { e^{-i 2 \pi f T }} { 1 + i 2 \pi f T }\right]=\arg(e^{-i2\pi f T} )-\arg(1+i2\pi f T) =-2\pi f T- \tan^{-1} 2\pi f T .$$


0

If you have an inequality $a \le x \le b$ and $f(x)$ is a monotonically increasing function, then the inequality $f(a) \le f(x) \le f(b)$ holds. This is not true if $f(x)$ is not monotonically increasing. The Fourier transform of any function does not result in a monotonically increasing function, in general. So, the inequality does not hold in general when ...


0

The Lerch Transcendent is appealing for it's mathematical beauty but tricky to implement numerically. Harmonic Numbers are much easier to work with, especially on computers. For faster approximations check out BLIT, BLEP, or miniBLEP. For quality I like the following method of generating bandlimited sawtooth waves with an arbitrary number of harmonics $k$ : $...


0

I was taught that we should think of the sensing matrix as an experimental tool that a practitioner can choose. Each row of the matrix corresponds to a single experiment, or measurement. Each of these experiments can be designed by the practitioner. What if, instead of using a binary matrix -- which is just observing the superposition of a few entries of $x$ ...


1

Let $x_1(t)$ and $x_2(t)$ be inputs and $y_1(t)$ and $y_2(t)$ the corresponding outputs. We have $$z(t)= \int_{t}^{t-T} (ax_1(\tau )+bx_2(\tau )) d\tau = a\int_{t}^{t-T} x_1(\tau )d\tau + b\int_{t}^{t-T} x_2(\tau )d\tau = ay_1(t) + by_2(t)$$ So the system is linear. It's also time-invariant since if $x(t)$ is the input and $y(t)$ is the corresponding output ...


0

I assume that $f(t)\,\overset{F}\longleftrightarrow\,g(\omega)$ means that the Fourier transform of $f(t)$ is $g(\omega)$: $$ g(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt $$ By the Fourier inversion theorem one has $$ f(t)\,\overset{F}\longleftrightarrow\,g(\omega)\,\Rightarrow\,g(t)\,\overset{F}\longleftrightarrow\,2\pi\,f(-\omega) $$ so since ...


1

Just break up the integrations into intervals at the behavior changes: $$f(t) = \begin{cases} \int_{-\infty}^t \Pi(\tau-1.5)d\tau - \int_{-\infty}^t \delta(\tau-3)d\tau &= 0 - 0 & t < 1\\ \\ \int_{1}^t d\tau - \int_{-\infty}^t \delta(\tau-3)d\tau &= t-1 - 0 & 1 \le t \le 2 \\ \\ \int_{-\infty}^t \Pi(\tau-1.5)d\tau - \int_{-\infty}^t \...


0

Let's start by assuming that A, B, and C are situated at the 12:00, 8:00, and 4:00 positions, respectively, and that A is traveling clockwise while B and C are traveling counterclockwise. Clearly, B and C will only intersect at the 4:00 position, and only when C has traveled $(9k-3)X$ for some positive integer. But in that time A will have traveled $(3k-1)X$,...


Top 50 recent answers are included