Loading in 5 sec....

The Mathematics of FM RadioPowerPoint Presentation

The Mathematics of FM Radio

- 53 Views
- Uploaded on
- Presentation posted in: General

The Mathematics of FM Radio

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

6 . 1 . 00

Randy Franks

Jon Nakashima

Matt Snider

AM = Amplitude Modulation

FM = Frequency Modulation

FM waves require line-of-sight between transmitter and receiver.

AM waves use line-of-sight, but can also reflect off the ionosphere and then to a receiver.

battery

transistor

inductor

resistor

capacitor

An arrow across a component indicates its level is variable.

Schematic of a basic transmitter

Our FM Receiver :

a Schematic

- like a battery

- stores charge

- when the device needs extra power, say for a particularly loud sound, the capacitor discharges

“dielectric”- a material placed between the plates of the capacitor to increase capacitance

“electrolytic” – type of capacitor which has strict polarity (charge may only flow in one direction through this type of capacitor)

A parallel-plate

capacitor

A variable capacitor

A cylindrical capacitor

(really, it’s just a stack of parallel plates)

(cross-section from above)

Basic Strategy: use equations for q and V separately , solve for the situation, divide and simplify results

electrostatic charge

voltage

d = separation of plates

A = area of Gaussian surface

Again, E and ds are in the same direction. Also the integral of ds from one plate to the other is equal to the total separation.

E and dA are parallel, and dA is a constant, in this case. So….

So….

(cross section)

a = inside radius

b = outside radius

r = radius of Gaussian surface

From the last derivation:

A = area of Gaussian surface, which is now cylindrical. A = 2πrl, So….

Solving the above for E will prove useful in the determination of V, So….

First, replace ds with dr, since the Gaussian surface is circular. Second, substitute for E (using result of last page).

a = inside radius

b = outside radius

r = radius of Gaussian surface

Third, change the integral limits to a and b (from the inside radius to the outside radius. After all that….

Finally, it’s time to divide the two results.

A lot of Algebra later:

Pull out constants, and….

Evaluating the integral leaves:

Well, almost….

Matt will now perform the ritual dance necessary to achieve decent radio reception on this campus.