B Pulses, waves, train waves....

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1. Apr 11, 2016

Cozma Alex

What are the difference between them?

What is that one in the photo?

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2. Apr 11, 2016

Jon B

It is a damped sine-wave pulse that appears to have been gated. A sequence of them would be a pulse train. The wave is about shape.

3. Apr 12, 2016

blue_leaf77

In terms of daily life, waves are something which are waving and are not necessarily periodic, e.g. (real) water waves, sound waves, and EM waves. Mathematically, waves are the solutions of the wave equation.
Pulses are usually used to describe a propagating disturbance but localized in space at a given instant of time. A train of pulses (rather than a train of waves) are actually a superposition of many waves, therefore it also satisfies the wave equation, and with regard to the above paragraph it can also be called a wave.

4. Apr 13, 2016

Jon B

So non-sinousoidal tone pulses are okay like the distorted waveform in the picture. My favorite pulses are always made of a superposition of waves of various frequencies and amplitudes as a Fourier series.

5. Apr 13, 2016

blue_leaf77

Yes, that's actually what I was talking about in the last sentence of my previous comment.
I don't know what you want to say there.

6. Apr 14, 2016

Jon B

I was thinking about a comment about sinusoidal wave shape not being a pulse.

7. Apr 14, 2016

blue_leaf77

As I have suggested before, that the term "pulse" is most commonly used to describe a propagating disturbance which is localized in space in a given instant of time. This is what I believe as what most people, in particular scientists or engineers working in a field relevant to this term, have in their mind. So, a pulse does not seem to have been well-defined to refer to a certain distinct physical phenomena, but it's a more convenient term to use to distinguish it from the continuous type of waves.
Following this, a continuous stationary sinusoidal function wouldn't normally be called a pulse.

8. Apr 14, 2016

Jon B

What is it when a continuous sinusoidal wave never crosses zero reference so that there is a train of pulses that doesn't produce upper harmonics like a square pulse does?

9. Apr 14, 2016

blue_leaf77

There is no sinusoidal wave which does not cross the zero reference.

10. Apr 14, 2016

Jon B

Right. It is undulating dc that looks identical to a sine wave. Like a mass in motion on a spring. The anchor is the zero reference and the sinus motion is above zero. Electronic magic.

11. Apr 14, 2016

vanhees71

Of course, a true signal is never a pure plain wave. You can have very many "wave forms", like "wave packets" or "wave trains". An example for the letter is a sinusoidal wave, switched on at a finite time $t=0$. We put the phase velocity of the wave $c=1$. Then the right-moving wave train is given by
$$f(t,x)=\Theta(t) \sin[\omega_0 (t-x)].$$
It can be written as a Fourier integral
$$f(t,x)=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} A(\omega) \exp[-\mathrm{i} \omega (t-x)].$$
We get the spectral function by the inverse Fourier transform
$$A(\omega)=\int_{\mathbb{R}} \mathrm{d} t f(t,0) \exp(\mathrm{i} \omega t) = \int_\mathbb{R} \mathrm{d} t \Theta(t) \sin(\omega_0 t) \exp(\mathrm{i} \omega t).$$
Decomposing the sine into exponentials gives
$$A(\omega)=\int_0^{\infty} \mathrm{d} t \frac{1}{2 \mathrm{i}} \{\exp[\mathrm{i}(\omega+\omega_0) t]-\exp[-\mathrm{i}(\omega+\omega_0) t] \}.$$
Now this integral is not well defined. So we regularize it by adding a damping factor to the integrand
$$A(\omega)=\int_0^{\infty} \mathrm{d} t \frac{1}{2 \mathrm{i}} \{\exp[\mathrm{i}(\omega+\omega_0) t]-\exp[\mathrm{i}(\omega-\omega_0) t] \} \exp(-\eta t), \quad \eta>0.$$
Then we get
$$A(\omega)=-\frac{\mathrm{i}}{2} \left [-\frac{1}{\mathrm{i}(\omega+\omega_0+\mathrm{i} \eta)}+\frac{1}{\mathrm{i}(\omega-\omega_0+\mathrm{i} \eta)} \right ]=\frac{\omega_0}{\omega_0^2-(\omega+\mathrm{i} 0^+)^2}.$$
As you see, that's a pretty broad spectrum around the frequency $\omega_0$.

12. Apr 14, 2016

Jon B

I was using the road for the reference instead of the axle.

13. Apr 14, 2016

blue_leaf77

Could you possibly be thinking about something like $y(t) = \cos(\omega_0 t)+1$? It might look like a sinusoidal function but it's actually not. Try to calculate the Fourier series coefficients for $-\pi/\omega_0 < x < \pi/\omega_0$, the coefficients for sine terms vanish but those for the cosine do not. In particular, the series will contain other harmonics as the sinusoid component $\omega_0$ due to the constant term $1$.

14. Apr 14, 2016

Jon B

Sounds good to me.