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Demodulation

Module by: Behnaam Aazhang

Summary: This module serves as an introduction to the problem of demodulation and detection: the decision procedure for deciding which signal of a set of signals was transmitted despite noise and attenuation.

Demodulation

Convert the continuous time received signal into a vector without loss of information (or performance).
r t = s m t+ N t r t s m t N t (1)
r t =∑n=1N s m n ψ n t+∑n=1N η n ψ n t+ N t ˜ r t n 1 N s m n ψ n t n 1 N η n ψ n t N t ˜ (2)
r t =∑n=1N s m n + η n ψ n t+ N t ˜ r t n 1 N s m n η n ψ n t N t ˜ (3)
r t =∑n=1N r n ψ n t+ N t ˜ r t n 1 N r n ψ n t N t ˜ (4)
proposition 1 
The noise projection coefficients η n η n 's are zero mean, Gaussian random variables and are mutually independent if N t N t is a white Gaussian process.
Proof
μ η n=E η n =E∫0T N t ψ n tdt μ η n η n t 0 T N t ψ n t (5)
μ η n=∫0TE N t ψ n tdt=0 μ η n t 0 T N t ψ n t 0 (6)
E η k η n ¯=E∫0T N t ψ k tdt∫0T N t ′ ¯ ψ k t ′ ¯d t ′ =∫0T∫0T N t N t ′ ¯ ψ k t ψ n t ′ dtd t ′ η k η n t 0 T N t ψ k t t ′ 0 T N t ′ ψ k t ′ t ′ 0 T t 0 T N t N t ′ ψ k t ψ n t ′ (7)
E η k η n ¯=∫0T∫0T R N t- t ′ ψ k t ψ n ¯dtd t ′ η k η n t ′ 0 T t 0 T R N t t ′ ψ k t ψ n t ′ (8)
E η k η n ¯= N 0 2∫0T∫0Tδt- t ′ ψ k t ψ n t ′ ¯dtd t ′ η k η n N 0 2 t ′ 0 T t 0 T δ t t ′ ψ k t ψ n t ′ (9)
E η k η n ¯= N 0 2∫0T ψ k t ψ n t¯dt= N 0 2 δ k n = N 0 2ifk=n0ifk≠n η k η n N 0 2 t 0 T ψ k t ψ n t N 0 2 δ k n N 0 2 k n 0 k n (10)
η k η k 's are uncorrelated and since they are Gaussian they are also independent. Therefore, η k ≈Gaussian0 N 0 2 η k Gaussian 0 N 0 2 and R η kn= N 0 2 δ k n R η k n N 0 2 δ k n
proposition 2 
The r n r n 's, the projection of the received signal r t r t onto the orthonormal bases ψ n t ψ n t 's, are independent from the residual noise process N t ˜ N t ˜ .
The residual noise N t ˜ N t ˜ is irrelevant to the decision process on r t r t .
Proof
Recall r n = s m n + η n r n s m n η n , given s m t s m t was transmitted. Therefore,
μ r n=E s m n + η n = s m n μ r n s m n η n s m n (11)
Var r n =Var η n = N 0 2 Var r n Var η n N 0 2 (12)
The correlation between r n r n and N t ˜ N t ˜
E N t ˜ r n ¯=E N t -∑k=1N η k ψ k t s m n + η n ¯ N t ˜ r n N t k 1 N η k ψ k t s m n η n (13)
E N t ˜ r n ¯=E N t -∑k=1N η k ψ k t s m n +E η k η n ¯-∑k=1NE η k η n ¯ ψ k t N t ˜ r n N t k 1 N η k ψ k t s m n η k η n k 1 N η k η n ψ k t (14)
E N t ˜ r n ¯=E N t ∫0T N t ′ ¯ ψ n t ′ ¯d t ′ -∑k=1N N 0 2 δ k n ψ k t N t ˜ r n N t t ′ 0 T N t ′ ψ n t ′ k 1 N N 0 2 δ k n ψ k t (15)
E N t ˜ r n ¯=∫0T N 0 2δt- t ′ ψ n t ′ d t ′ - N 0 2 ψ n t N t ˜ r n t ′ 0 T N 0 2 δ t t ′ ψ n t ′ N 0 2 ψ n t (16)
E N t ˜ r n ¯= N 0 2 ψ n t- N 0 2 ψ n t=0 N t ˜ r n N 0 2 ψ n t N 0 2 ψ n t 0 (17)
Since both N t ˜ N t ˜ and r n r n are Gaussian then N t ˜ N t ˜ and r n r n are also independent.
The conjecture is to ignore N t ˜ N t ˜ and extract information from r 1 r 2 … r N r 1 r 2 … r N . Knowing the vector r r we can reconstruct the relevant part of random process r t r t for 0≤t≤T 0 t T
r t = s m t+ N t =∑n=1N r n ψ n t+ N t ˜ r t s m t N t n 1 N r n ψ n t N t ˜ (18)
Figure4-17.png
Figure 1
Figure4-18.png
Figure 2
Once the received signal has been converted to a vector, the correct transmitted signal must be detected based upon observations of the input vector. Detection is covered elsewhere.

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