Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 1 (Q.No. 12)
12.
c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2p(1008 x 103t)] in the FM signal (in terms of the Bessel Coefficients) is
5 J4(3)
J8(3) J8(4)5 J4(6)
Answer: Option
Explanation:
As, we known FM, s(t)
Frequency deviation = 3 . (AM Bandwidth) = 6 fm
and β= 6 is given in (d) only.
ωc + n ωm = (1008 x 103)2p
2p x 106 + n. 4p x 103 = (1008 x 103).2p
n = 4
So, β = 6, n = 4 hence answer is 5j4(6).
Discussion:
8 comments Page 1 of 1.
Froilan cataluna said:
2 months ago
I am not getting this clearly. Please explain me in detail.
Adil Tanveer said:
4 years ago
I don't know from where it takes the value of wc and wm.
Apart from this nWm =n*4p, at the place of 2, he has taken the value 4. How it becomes? Please explain it.
Apart from this nWm =n*4p, at the place of 2, he has taken the value 4. How it becomes? Please explain it.
(1)
Chand said:
6 years ago
What is the value of wm wc?
(1)
Jaya said:
8 years ago
Can anyone please explain from where wc, wm are taken?
Sonali soni said:
8 years ago
From where the values of wc and wm have been taken.
Janani said:
8 years ago
I don't know how it comes. Please explain.
Ms malhotra said:
8 years ago
2p*10^6 + n*4p*10^3 = 1008*10^3*2p,
2p*10^3 (10^3+2n) = 1008 * 10^3 *2p,
{2p and 10^3 are cancel),
10^3 + 2n = 1008.
2n = 1008-1000 {10^3 = 1000}.
2n = 8.
n = 4.
2p*10^3 (10^3+2n) = 1008 * 10^3 *2p,
{2p and 10^3 are cancel),
10^3 + 2n = 1008.
2n = 1008-1000 {10^3 = 1000}.
2n = 8.
n = 4.
Varsha said:
1 decade ago
ωc + n ωm = (1008 x 103)2p (where p= pie= 22/7).
2p x 106 + n.4p x 103 = (1008 x 103).2p.
Can anybody please explain me from where did we took the value of 'ωc' & 'ωm'?
2p x 106 + n.4p x 103 = (1008 x 103).2p.
Can anybody please explain me from where did we took the value of 'ωc' & 'ωm'?
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