WBL7 N=n/a
10.3 Priority Three ‘Retaining and Developing the Modern, Professionalised Workforce’
92 Example 22
A DSBSC transmitter radiates 1 kW when the modulation percentage is 60%. How much of carrier power (in kW) is required if we want to transmit the same message by an AM transmitter?
Solution
Givenñ~2~ = 1KW and m = 0.6.
Carrier power, x = ñ~2~FïG = 1 F7.. G = 5.56 è.
We require 5.56 kW to transmit the carrier component along with the existing 1 kW for the sidebands when m=0.6.
93 wanted sideband. Therefore still all the three versions of AM, namely, AM, DSBSC and SSB coexist in the analog communication field.
3.2.1. Frequency Spectrum of the SSB Wave
One way of viewing SSB is DSBSC followed by bandpass filtering, as illustrated in Figure2.6.
The mathematical treatment here follows this assumption. The situation of instantaneous value of SSB wave is same as in DSB, illustrated in Figure 2.5, which shows how the DSBSC modulated voltage is made to vary with modulating voltage changes. From Figure 2.5. It is possible to write an equation for the amplitude of the DSBSC modulated voltage.
Figure 2.6. Block diagram representation of SSB generation by band pass filtering
_
ñ~2~=
4Ðcos(
x−
4) a −
4Ðcos(
x+
4) a (2.21)
Now, for generating the SSB, the DSBSC is passed through the bandpass filter. Depending on the cut-off frequencies, either LSB or USB comes out of the bandpass filter if the cut-off frequencies are
(f
c-f
m) and f
c,
then LSB is chosen for transmission and instantaneous voltage of SSB signal is given by_
~~2=
4Ðcos(
x−
4) a (2.22)
Alternatively, if the cut-off frequencies are f. and if. + “1", the instantaneous voltage of the USB chosen for transmission is given by
_
~~2=
4Ðcos(
x+
4) a (2.23)
It has thus been shown that the equation of SSB wave contains one term, that is, either LSB or USB. The bandwidth required for SSB is the frequency of the modulating signal. That is,
~~2= (f
c-f
m) - f
c= f
c- (f
c-f
m) = f
m(2.24)
94 The frequency spectrum of SSB wave is shown in Figure. 2.7. Using the equations of SSB as illustrated, SSB, consists of one discrete frequency either at
f
c-f
mor f
c+ f
m.
Figure. 2.7. Frequency spectrum of the SSB wave. Spectrum for (a) SSB = USB, and (b) SSB = LSB.
3.2.2. Time Domain Representation of the SSB Wave
Figure 2.8 shows the time domain representation of SSB wave for one cycle of message signal.
The modulated wave will have only one sine wave. The only wave to distinguish is to compare with carrier signal. Its frequency will be either lower or more than carrier frequency by an amount of modulating signal frequency. The envelope of SSB does not contain message and hence a simple envelope detector circuit is not useful for recovering the message. This is the price we pay by suppressing the carrier and one of the sidebands. Of course, here also, there are ways to overcome this problem to recover message.
Figure 2.8. Time domain representation of the SSB wave.
95 3.2.3. Power Computations in the SSB Wave
Form the discussions above, It has been shown that the carrier component and one sideband are suppressed in the SSB wave. The modulated wave contains energy only due to one sideband component. Since amplitude of the sideband depends on the modulation index Ï
Ðthe total power in the modulated wave will depend on the modulation index also. The total power in the SSB modulated wave will be
~~2=
g=
Ýg(2.25)
Where all the voltages are rms values and R is the resistance in which the power is dissipated.
w~2=
Ü~2=
g= é
ÏbÐ√ê ÷ $ =
4"Ð (2.26)Substituting Equ. (2.25) in (2.26), we have
~~2=
4"Ð (2.27) ~~2=
4"(2.28)
Equ. (2.28) relates the total power in the SSB modulated wave to the unmodulated carrier power.
It is interesting to know from Equ.(2.28) that the maximum power in the SSB wave is
~~2=
ùØ
"when m = 1. Thus, we need only maximum of 25% of unmodulated carrier power for the transmission of SSB wave. This is correct also, because, in case of SSB wave, one-sixth of the total power is utilized by the sideband and this constitutes 25% of unmodulated carrier power.
Example 23.
A 400 W carrier is amplitude modulated to a depth of 100%. Calculate the total power in case of SSB technique. How much power saving (in W) is achieved for SSB compared to AM and DSBSC of Example 20? If the depth of modulation is changed to 75%, then how much power (in W) is required for transmitting the SSB wave? Compare the powers required for SSB in both the cases and comment on the reason for change in the power levels.
Solution
Case 1 Given, Pc = 400 W and m = 1.
Total power in SSB, PSSB= xF4"G = 400F8G = 100 è. Power saving (in W) compared to AM = PAM- PSSB = 500 W.
Power saving (in W) compared to DSBSC = PDSBSC - PSSB = 100 W.
Thus, we require only 100 W in case of SSB which is one-sixth of total AM power.
96 Case 2 Given, Pc = 400 W and m = 0.75
Total power in SSB,
P
SSB=
xF
4"G = 400 F
(7.>r)"G = 56.25 è
The power required in this case is lower than m = 1 case. This infers that the total power in SSB also depends on the depth of modulation. It will be maximum, that is, one-sixth of total AM power when m = 1 and less for m<1.
Example 24
A SSB transmitter radiates 0.5 kW when the modulation percentage is 60%. How much of carrier power (in kW) is required if we want to transmit the same message by an AM transmitter?
Solution
Given, PSSB = 0.5 kW and m = 0.6.
Carrier power,x = ~~2F4"G = 0.5 F7.." G = 5.56 è..
We require 5.56 kW to transmit the carrier component along with the existing 0.5 kW for one sideband and 0.5 kW more for another sideband when m = 0.6. In total 6.56 kW is required by the AM transmitter.
Example 25
Calculate the percentage power saving when the carrier and one of the sidebands are suppressed in an AM wave modulated to a depth of (a) 100 percent, and (b) 50 percent.
Solution
(a) kï = xF1 +4G = xF1 +4G = 1.5x
~~2 = xF4"G = xF8"G = 0.25x
Û_¤¥§ =8.r7.r8.r =8.r8.r = 0.833 = 83.3%
(b) kï = F1 +7.rG = 1.125x
~~2 = xF7.r"G = 0.0625x
Û_¤¥§ =8.8r7.7.r8.8r =8.7.r8.8r = 0.944 = 94.4%