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1. A pulse train with N half sine pulse as N —1 sin(irt) f (t)= Eg(t—nT), g(t) = n=0 0 Find the Fourier transform of f(t) 0 2. a) In a spread spectrum system, what is the number of chips per bit when...

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1. A pulse train with N half sine pulse as
N —1 sin(irt) f (t)= Eg(t—nT), g(t) = n=0 0
Find the Fourier transform of f(t)
0 2. a) In a spread spectrum system, what is the number of chips per bit when the coding gain is 13.424dB? b) If the spreading code is C.[ XXXXXXXXXX XXXXXXXXXX] and the received 11-chip sequence for a bit interval is r XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX] what is the value of this binary bit (1 or —1)? Write down your procedure.
3. Let an and bn, for n.1,2,...,N, be two independent real random sequences. The value of an N N (be) is 1 or —1 of equal probability. What are the mean and variance of Ea,E bm ? n=1 m=1
4. Consider the downlink of a CDMA system, and focus on a Cell with K mobile users and a single base station. Suppose that BPSK modulation is used, and the chip pulse shape is the standard sinc pulse with zero crossings every 7', seconds. The signal at the output of the despreader of the Mth mobile user's receiver is
00
Y(z) = NTHliTcd h(r)+ETc\IK. Eq h(r — + ri(r) i=1 Where q(r) is a white gaussian noise process with power spectral density 7', AIN 0 , h(t) is the common multipath channel experienced by all signals transmitted from the base to the mth mobile unit, and d is the data of the Mth user, which takes the values ±1.
Let us assume that this system uses Walsh codes to separate the users, in addition to a common scrambling code an (the same for all users in the cell) to suppress interference from other base stations. In that case, the complex gaussian amplitudes .71 have the form
Answered Same Day Dec 23, 2021

Solution

David answered on Dec 23 2021
125 Votes
Answer
Fourier transform of g(t) will be,
G(ω) = ∫ ( )



= ∫ ( )


∫ ( )

∫ ( )


From the function limits we will left only with
= ∫ ( )



= ∫ ( )



G(ω) = ∑ ( )
Answer
(a) Coding Gain is 13.424dB
And relation between chip per bit and coding gain is given by
N = Coding gain2 / 4
No of chips per bit will be ,
N = 13.4242 / 4 = 45.04 Users per bit
(b)
The given sequence is ,
C = {1,1,1,-1,-1,-1,1,-1,-1,1,-1}
And the received sequence is ,
= {0.2,0.8,-0.1,0.2,-0.5,-0.9,-0.4,-0.1,-0.9,0.8,-0.3}
For the given sequence we can write,
= 21 + 21 + 21 + 2-1 + 2-1 +2-1 +21 +2-1 +2-1 +2-1 + 21
= 13
So the binary bit will be,
= 1101
Answer
Since anand bn are independent that means,
E(an) = 0
E(bn) = 0
So , mean of ∑

and ∑

so,
E {∑



} but since the an and bn are independent so,
E {∑



} = 0
Now Variance,
Var{∑



} =
K(∑



)(∑



) (∑



) (∑



)
Now since,
K(∑

...
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