Wave packets & Fourier analysis: minimum pulse duration w/ given frequency

[WJSwanson]

Homework Statement

I: A telephone line can transmit a range of frequencies \Delta f = 2500 Hz. Roughly what is the duration of the shortest pulse that can be sent over this line?

II: A space probe sends a picture containing 500 by 500 elements, each containing a brightness scale with 256 possible levels. This scale requires eight binary digits. Suppose that the transmitter uses a bandwidth of 1000 Hz. (For the faint signals from distant space probes the bandwidth must be kept small to reduce the effects of electronic noise that is present at all frequencies.) Roughly how long is needed to send one picture? Note that the center-to-center separation, a, of adjacent pulses must be at least the total width of anyone pulse.

Homework Equations

I:
\Delta\omega = 2\pi\Delta f
\Delta t \geq \frac{1}{2\Delta\omega}

II:
N (number of pulses) = 500 * 500 * 8 = 2.00 * 10^{6} pulses
a \geq 2\Delta t = \frac{2}{2\Delta\omega} = \frac{1}{2\pi\Delta f}

The Attempt at a Solution

I:
\Delta\omega = 2\pi\Delta f = 6.28 * 2.50 * 10^{3} Hz = 1.57 * 10^{4} Hz

\Delta t \geq \frac{1}{2\Delta\omega} = \frac{1}{3.14 * 10^{4}} s^{-1}

\Delta t \geq 3.18 * 10^{-5} s

II:
I'm not quite sure how to approach this. I know that there will have to be 2.00 * 10⁶ pulses per picture and that the individual pulses have to be spaced by a = 2\Delta t. I also would imagine that I would want to use the relation

\Delta x\Delta k \geq 1/2

by deriving

\Delta k = \frac{2\pi}{\Delta\lambda} = \frac{2\pi\Delta f}{c} = \frac{2\pi}{c\Delta t}

or else perhaps use

\Delta\omega\Delta t = \Delta t * \Delta(\frac{1}{2\pi t}) \geq 1/2

but I really don't know where to go from there, aside from multiplying my minimum time interval by 2 million.

 

[WJSwanson]

Okay, I ended up figuring out part II.

Since we know that
\Delta\omega = 2\pi\Delta f = 6.28 * 10^{3} s^{-1}

we infer that

\Delta t \geq \frac{1}{2\Delta\omega}<br /> <br /> and since the probe sends the message at a rate of 2\Delta t = \frac{t}{2.0 * 10^{6}} we know that<br /> <br /> \Delta t = \frac{t}{4.0 * 10^{6}}<br /> <br /> and since<br /> <br /> \Delta t\Delta\omega = \Delta\omega * \frac{t}{4.0 * 10^{6}} \geq \frac{1}{2}<br /> <br /> we can solve for t_min:<br /> <br /> t_{min} \geq \frac{4.0 * 10^{6}}{2\Delta\omega} \Rightarrow t_{min} \geq \frac{2.0 * 10^{6}}{6.28 * 10^{3} s^{-1}}<br /> <br /> thus,<br /> <br /> t_{min} \geq 3.18 * 10^{2} s<br /> <br /> Or alternatively,<br /> <br /> t_{min} \geq 5.3 min