A binary bit '1' is assigned a power level of (or energy ), while a binary bit '0' is assigned zero power (or no energy).
Simulator for Binary ASK Constellation Diagram
Noisy Modulated Signal (ASK)
Original Modulated Signal (ASK)
Energy per bit (Eb) (Tb = bit duration):
We know that all periodic signals are power signals. Now we’ll find the
energy of ASK for the transmission of binary ‘1’.
Eb = ∫0Tb(Ac.cos(2П.fc.t))2
dt
= ∫0Tb(Ac)2.cos2(2П.fc.t)
dt
Using the identity cos2x = (1 + cos(2x))/2:
= ∫0Tb((Ac)2/2)(1 +
cos(4П.fc.t)) dt
= ((Ac)2/2) ∫0Tb(1) dt
+ ((Ac)2/2) ∫0Tbcos(4П.fc.t)
dt
= ((Ac)2/2) * Tb + 0 (The integral of
cos(4П.fc.t) over a full period is zero, assuming
Tb is an integer multiple of 1/(2fc))
Eb = (Ac2/2).Tb (where
Tb is the bit duration)
** where Ac is the amplitude of the carrier signal and
fc is the carrier frequency in Hz.
To save transmitter energy, Eb should be small.
** for transmission of binary ‘0’
Eb = ∫0Tb(S2(t))2dt = 0
** Constellation Diagram
First, we define the orthonormal basis function for this system:
φ1(t) = √(2/Tb) cos(2Пfct) for 0
≤ t ≤ Tb.
The energy of this basis function is 1.
Now, we can represent our signaling waveforms using this basis function:
For binary '1': S1(t) = Ac cos(2Пfct) =
[Ac * √(Tb/2)] * φ1(t)
The coordinate for S1(t) in the constellation diagram is
g11 = Ac * √(Tb/2).
The energy of S1(t) is Eb = g112 =
(Ac2 * Tb)/2.
Therefore, g11 = √(Eb).
For binary '0': S2(t) = 0. The coordinate for S2(t)
is g21 = 0.
So, in the constellation diagram:
1 => point at √(Eb) along the φ1 axis
0 => point at 0 (the origin)