Recent content by Fred Wright

For what it's worth, you can modulate a solid state laser by making its power source voltage proportional to the modulating signal voltage. No coils or weirdness involved.

In the paper you refer to the authors state in equ. 1 that the complex polarizability is given by, $$ \alpha=6\pi \epsilon_0 c^3 \frac{\frac{\Gamma}{\omega_0^2}}{(\omega_0^2-\omega^2)-i\Gamma \frac{\omega^3}{\omega_0^2}} $$ They state in equ. 2 that, $$ \left| \alpha \right|^2=\frac{6\pi...

Isn't the volume you seek equal to the cross-sectional area of the smaller ellipse multiplied by the length of the perimeter about the central axis of the bigger ellipse? Let ##a_1## and ##b_1## be the semi-major and semi-minor axis of the small ellipse and likewise ##a_2## and ##b_2## for...

I think what you are asking is that you have, $$ \int_0^{\infty} A(t)[1 + e^{i\omega t}+ \frac{e^{2i\omega t}}{2!} +\frac{e^{3i\omega t}}{3!} + ...]dt $$ and wish to take the FFT of your time series against each succeeding term in the expansion. Firstly, you must apply the scaling theorem for...

Many years ago when I was an undergraduate student there was a commercial on tv advertising a shampoo called "Prell". On the commercial a lady dropped a pearl in to a tube of Prell and watched it slowly descend. She called it "The Prell pearl test". One day I came to lab class and the...

IMO (for what it's worth) the author's explanation of zero bias tunneling anomalies is incomplete. I think a better explanation can be derived from the paper Surface magnetism of gallium arsenide nanofilms. From the abstract: "Gallium arsenide (GaAs) is the most widely used second-generation...

Observe that you can rewrite your integral, $$ I=(3-4i)\int_0^{1} [(3-4i)t + 5i]^7dt $$ $$ (3-4i)=5e^{-i\tan^{-1}(\frac{4}{3})}=5z_0 $$ Expand the integral with binomial thm, $$ I=5^8z_0\int_0^{1} [(z_0t + i]^7dt=5^8z_0\sum_{n=0}^{7}\int_0^{1}\begin{pmatrix} 7\\ n \end{pmatrix} z_0^{n}t^{n}...

For brevity of notation let ##a=\frac{2}{3}##, ##b=\frac{1}{3}##, ##C_{k,n} ## be the binomial coefficient of the k'th row and n'th column of Pascal's triangle and ##l_k## be the expected length for the k'th row of Pascal's triangle. For part a of the problem the expected length of the k'th...

I have an issue with the function ##r=r_0 e^{k|\phi |} ## representing a closed electrical circuit. As the angle increases the value of ##r## increases exponentially and doesn't close a circuit. The picture you drew appears to be a cardioid whose equation is, $$ r=2a(1-\cos(\phi)) $$

The authors describe the polar angle as $$ \hat r \cdot \hat e = \cos(\theta) $$ Therefore taking the sum, $$ \mathbf u^s (\hat r)=B_1(\cos(\theta)\hat r - \hat e)P_1^{'}(\cos(\theta)) $$ $$ + \frac{1}{3} B_2(\cos(\theta)\hat r - \hat e)P_1^{'}(\cos(\theta)) $$ with $$...

In this case you can define a bound state to be one that the interaction energy is greater than the sum of each individual particle's energy and likewise an unbound state has the sum of individual energies greater than the interaction energy.

This is just an observation and I'm probably wrong, but I find equ. (1) highly suggestive of the Ostwald-Freundich equation.

Hint: $$ \frac{dy}{dx}=\frac{y}{x} $$

Observe that for ##x \gt 1## the integrand is complex. $$ I=\int \sqrt{1-x^2}dx=i\int \sqrt{x^2-1}dx $$ If you want to go further with this; let ##x=\cosh(u)##, ##dx=\sinh(u)du##. $$ \sqrt{\cosh^2(u) -1}=\sinh(u) $$ $$ I=i\int \sinh^2(u)du $$ $$ \sinh^2(u)=\frac{1}{2}(\cosh(2u)-1) $$ $$...

I apologize for wasting your time. I didn't think I was being evasive. I posted the data rate and didn't think the frequency of the rf carried was relevant to my problem. I received some good advice so I humbly advise that you close the thread so that others don't waste their time on my...