17.1.
WAVE SPEED EQUATION FOR INCOMING
SIGNALS
We saw before that we can assume that, of the frames that are in motion
relative to each other, one is motionless and the other is in motion.
The figure below was drawn
relative to the reference systems that
receive the signal (the planes and the station on the
mountain).
- The transmitter emits signals at f
0 frequency over λ
0
wavelength relative to its own reference system.
- Relative to the reference system of the plane at the top, the
transmitter moves away from itself in the direction of the red
arrow.
- Relative to the reference system of the plane at the bottom, the
transmitter approaches itself in the direction of the blue arrow.
In the figure, the speeds of the signals are shown
relative to their targets.
We have seen before that the plane at the top receives
the signal at f1 frequency and over λ1
wavelength. We also saw that, for a reference system, the incoming
signal speed
which is coming to the reference system itself is always c. Therefore,
Wave speed equation for the plane at the top is as follows: c = f1
. λ1
The plane at the bottom receives the signal at f2 frequency
and over λ2 wavelength. incoming signal speed
relative to its own reference system is c. The signal speed equation
for the plane at the bottom should be as follows: c = f2 . λ2
The station is motionless relative to the receiver. Therefore, it
receives the signal at the same f0 frequency and over the
same λ0 wavelength as the transmitter. Incoming signal
speed relative to its own reference system is c. From this, we can
write the equation: c = f0 . λ0
17.2. WAVE SPEED
EQUATION FOR OUTGOING SIGNALS
The figure below was drawn relative to the reference system of the
transmitter. We saw that the signals go to the plane moving away at c+v
speed, the plane approaching at c-v speed, and to the station on the
mountain at c speed relative to the reference system of the
transmitter. On the other hand, we also saw that, although the
transmitter was manufactured in a way that it emits signals at f
0
frequency and over λ
0 wavelength, if a signal it emits goes
to an object in motion, the signal wavelength changes at the time of
signal emission.
The change in the wavelength of a signal which is sent is obvious only
and only on the side that receives the signal. The man who stands next
to the transmitter cannot feel or measure the signals that set out from
the transmitter and go in the direction of the planes over different
wavelengths.
In the figure, the signal speeds are shown
relative to the transmitter of the signal.
We
saw that, even though the frequency of the transmitter doesn’t change,
the wavelengths of the signals going to the planes change. By using
this information in Wave Speed equation, we can find the speeds of the
signals going to the planes relative to the reference system of the
transmitter.
| Speed of the signal to the departing aircraft: |
c1 = c +v =f0 . λ1 |
| Speed of the signal to the approaching aircraft: |
c2 = c -v =f0 . λ2 |
| Speed of the signal to the station |
c = f0 . λ0 |
Therefore,
the general rule for a signal in motion is as follows: The speed of a
signal going to an object that is in motion is equal to the
multiplication of the signal frequency of the reference system that
emits the signal and the signal wavelength measured at the reference
system receiving the signal. Again, I’d like to emphasize that this
signal speed value is relative to the reference system that sends the
signal. We can briefly show this equation as follows: Outgoing Signal
Speed = Frequency of Transmitter x Wavelength at Receiver = c ± v
|
GENERAL RESULT FOR SIGNAL SPEEDS
Incoming Signal Speed = Receiver frequency x Receiver wavelength = c
Outgoing Signal Speed = Transmitter frequency x Receiver wavelength = c ± v
|
Let’s go back to the example we used on the topic “The 44th Bit” in the
part about Byte Shift. We calculated wavelength changes for signals
that go to planes that approach and move away and also signal speeds
before. Let’s calculate signal speeds with the help of wave speed
equation this time. We see the calculated signal speeds in the table
below. As can be seen, same values are obtained for signal speeds.
| Description |
Formula |
Value |
Unit |
| Transmitter factory setting |
| Frequency |
f0 |
3.18 |
GHz |
| Wavelength |
λ0 |
0.09427435786 |
m |
| Constants |
| Speed of light constant |
c |
299792458 |
m/sn |
| Aircraft speed |
v |
850 |
m/sn |
| Signal wavelengths |
| Wavelength of the signal to the departing aircraft |
λ1 |
0,09427462516 |
m |
| Wavelength of the signal to the approaching aircraft |
λ2 |
0,09427409057 |
m |
| OUTGOING signal speeds according to
transmitter (calculated using speed sum) |
| Speed of the signal to the departing aircraft |
c + v |
299793308 |
m/sn |
| Speed of the signal to the approaching aircraft |
c - v |
299791608 |
m/sn |
| OUTGOING signal speeds
according to transmitter (calculated using wave speed equation)
Electromagnetic wave speed = Transmitter
frequency x Receiver wavelength
|
| Speed of the signal to the departing aircraft |
f0
. λ1 |
299793308 |
m/sn |
| Speed of the signal to the approaching aircraft |
f0
. λ2 |
299791608 |
m/sn |
This connection between wave speed, frequency and
wavelength can be clearly seen in (c+v) (c-v) mathematics already.
(*)
Factory
setting of the signal source[1]
Wavelength
change
[2]
We obtain the following third equation from equation number 1 and
2.
Wave
Speed Equation[3]
(c±v) : Outgoing signal speed, f0 : Frequency of
transmitter, λ1 : Signal wavelength measured at the receiver
This equation gives us the speed rule that the signal that is emitted
from a transmitter is subject to. We already know that the part c±v in
the equation is the OUTGOING signal speed relative to the reference
system of the transmitter.
(*) I’m
kidding. Equation number 1 and 2 had been right in front of my eyes for
years and I couldn’t notice the equation number 3. I could have never
obtained the third equation; I know that. I feel so lucky to have
noticed it. As I said…
Physics is really difficult.