Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 3 (Q.No. 39)
39.
If the amplitude of the voltage is kept constant, but its frequency is raised to 6 kHz, the new deviation will be for the modulation index is, if fm = 500 Hz, Δf = 2.25
Answer: Option
Explanation:
Δf = β x fm 
4.5 x 6 kHz 27 kHz.
Discussion:
7 comments Page 1 of 1.
Shadan said:
8 years ago
@All.
Modulation Index(mf)=Δf/fc
=2.25/500.
=4.5*10 to the power -3.
Now new frequency deviation =mf*fm.
=4.5*10to power -3 *6*10to power -3
= 27 khz.
Modulation Index(mf)=Δf/fc
=2.25/500.
=4.5*10 to the power -3.
Now new frequency deviation =mf*fm.
=4.5*10to power -3 *6*10to power -3
= 27 khz.
Old Kage said:
1 decade ago
Given here should be fm = 500 Hz and Freq Dev = 2.25 Khz, thus will get modulation index equal to 2.5.
Solving the new frequency deviation we multiply the Modulation Index to 6 Khz (new modulation freq), thus getting 27 Khz.
Right?
Solving the new frequency deviation we multiply the Modulation Index to 6 Khz (new modulation freq), thus getting 27 Khz.
Right?
Vishaldeep said:
1 decade ago
2.25/500 = kf1.
As no change in kf if frequency changes.
So = f2=kf1*6 khz.
= [2.25/500 ]*6 khz.
= 0.027 khz.
So Freq deviation must be in khz to get the right answer, correct it.
As no change in kf if frequency changes.
So = f2=kf1*6 khz.
= [2.25/500 ]*6 khz.
= 0.027 khz.
So Freq deviation must be in khz to get the right answer, correct it.
Ankit Kumar said:
5 years ago
The modulation index is, if fm is 500Hz, Δ f = 2.25kHz.
Shanil prakash said:
8 years ago
i) m = 4.5.
ii) delta f = 27kHz.
ii) delta f = 27kHz.
Bagya said:
7 years ago
The answer is 27Hz.
Monika said:
7 years ago
Thanks @Shadan.
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