Question 3
Fig. 10.1 shows the variation with frequency f of
the power P of a radio signal.
Fig. 10.1
(a) State the name of
(i) the type of modulation of this radio signal, [1]
(ii) the component of frequency 50 kHz, [1]
(iii) the components of frequencies 45 kHz
and 55 kHz. [1]
(b) State the bandwidth of the radio signal. [1]
(c) On the axes of Fig. 10.2, sketch a graph to show the
variation with time t of the signal
voltage of Fig. 10.1. [3]
Reference: Past Exam Paper – November 2007 Paper 4 Q10
Solution:
(a)
(i) Amplitude
(modulated)
(ii) Carrier (frequency / wave)
(iii) Sideband (frequency)
(b) Bandwidth
= 10kHz
(c)
The sketch should have a
general shape, i.e. any wave that is amplitude modulated
with a correct period for
the modulating waveform (200μs) {corresponding
to frequency 5 kHz}
and a correct period for the carrier waveform (20μs) {corresponding to frequency 50kHz 
Why is the period for the modulating waveform 200 microseconds? Also, I really appreciate your efforts!
ReplyDeletethe difference between the two sidebands 45 kHz and 55 kHz gives twice the frequency of the modulating waveform. So, frequency = 5 kHz and the corresponding period = 1 / f = 200 μs
DeleteDoesnt the graph show the period of the modulating waveform as 100 microseconds instead of 200?
ReplyDeleteno, it's actually 200.
Deletefrom 0-100, you could say that this is where the 'crest' (upper section of the sine curve) is and from 100-200, we have the second (below x-axis) section of the sine curve.
try to draw the envelope wave and you will understand what i mean
EXCUSE ME . I think there is a problem with the graph part. even the application booklet of CIE goes against your graph. IF the period is 200 ms then it must be like somehow the time for minimum to maximum then back to minimum (or max-min-max )= 200ms but ur graph shows 2 cycles(min-max-min-max-min)=200. it is surely wrong i guess as compared to what CAIE has published in application booklet
ReplyDeletethanks. that was a mistake. It has already been corrected
Delete