13
(a) Product modulator
(b) respective schematic
diagram
= multiplier+ adder
14
SQUARE-LAW MODULATOR (FOR AM)
Square-law modulators are based on
nonlinear elements:
(a) functional block diagram, (b) circuit
realization
15
BALANCED MODULATOR (FOR DSB)
*By using balanced configuration non-idealities on
square-law characteristics can be compensated
resulting a high degree of carrier suppression:
*Note that if the modulating signal has a DCcomponent, it is not cancelled out and will
appear at the carrier frequency of the
modulator output
16
17
THE ENVELOPE DETECTOR
Important motivation for using AM is the possibility to use the envelope detector that
has a simple structure (also cheap) needs no synchronization (e.g. no auxiliary, un-modulated
carrier input in receiver no threshold effect SNR can be very small and
receiver still works)
18
19
3.1 Single Sinusoid AM
• An AM signal of the form
𝜋
𝑋𝑐 𝑡 = 𝐴𝑐 1 + acos 2𝜋𝑓𝑚 𝑡 +
cos(2𝜋𝑓𝑐 𝑡)
3
contains a total power of 1000 W
• The modulation index is 0.8
• Find the power contained in the carrier and the
sidebands, also
• find the efficiency
• The total power is
2
2
𝐴
𝐴
1000 = 𝑥𝑐 2 (𝑡) =
+
𝑚𝑛 2 (𝑡)
2
2
• It should be clear that in this problem
2
𝑚𝑛 𝑡 = cos 2𝜋𝑓𝑚 𝑡 , 𝑠𝑜 𝑚 𝑛 = 1/2
1
1
33
2
2
1000 = 𝐴𝑐
+ 0.64 =
𝐴𝑐
2 2
50
•
33
Thus we see that 𝐴𝑐 =1000. =1515.15
50
2
1 2
1515
• 𝑃𝑐𝑎𝑟𝑟𝑖𝑒𝑟 = 𝐴 𝑐 =
= 757.6𝑊
2
2
• 𝑃𝑠𝑖𝑑𝑒𝑏𝑎𝑛𝑑𝑠 = 1000 − 𝑃𝑐 = 242.4𝑊
• The efficiency is:242.4
𝐸𝑓𝑓 =
= 0.242 = 24.2%
1000
• The magnitude and phase spectra can be plotted by
first expanding out 𝑥𝑐 (𝑡)
𝑥𝑐 𝑡 = 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡 +
𝜋
𝑎𝐴𝑐 cos 2𝜋𝑓𝑚 𝑡 + cos(2𝜋𝑓𝑐 𝑡)
3
𝑎𝐴𝑐
𝜋
= 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡 +
cos 2𝜋 𝑓𝑐 + 𝑓𝑚 𝑡 +
2
3
𝐴𝑐
𝜋
+ 𝑎 cos[2𝜋 𝑓𝑐 − 𝑓𝑚 𝑡 − ]
2
3
Amplitude and
phase spectra
for one tone AM
3.2 Pulse Train with DC Offset
• Find 𝑚𝑛 (𝑡) and the efficiency E
• From the definition of 𝑚𝑛 𝑡 : −
𝑚(𝑡)
𝑚(𝑡)
𝑚𝑛 𝑡 =
=
= 𝑚(𝑡)
min 𝑚(𝑡)
−1
• The efficiency is :2
2
𝑎 𝑚𝑛
𝐸=
1 + 𝑎2 𝑚𝑛 2
2
• 𝑇𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝑚 𝑛 (𝑡) 𝑤𝑒 𝑓𝑜𝑟𝑚 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒
1
2
𝑚 𝑛 (𝑡) =
𝑇𝑚
• Thus :-
𝑇𝑚ൗ
3
0
2
2
𝑇𝑚
𝑑𝑡 + 𝑚𝑇ൗ −1 2 𝑑𝑡
3
1 𝑇𝑚
2𝑇𝑚
=
.4 +
.1
𝑇𝑚 3
3
4 2 6
= + = =2
3 3 3
2
2𝑎
𝐸=
1 + 2𝑎2
• The best AM efficiency we can achieve with this
waveform is when a =1
2
𝐸𝑓𝑓 ห 𝑎=1 = = 0.67 𝑜𝑟 67%
3
• Suppose that the message signal is –m(t) as given
here.
• 𝑁𝑜𝑤 min 𝑚 𝑡 = −2 𝑎𝑛𝑑
𝑚 𝑡
1
2
1
1
2
2
2
𝑚𝑛 𝑡 =
𝑎𝑛𝑑 𝑚𝑛 = (−1) + . ( ) =
2
3
3 2
2
• The efficiency in this case is :1 2
2
( )𝑎
𝑎
2
𝐸𝑓𝑓 =
=
1 2 2 + 𝑎2
1 + ( )𝑎
2
• Now when a = 1 we have𝐸𝑓𝑓 = 1=3 or just 33.3%
• Note that for 50% duty cycle squarewave the
efficiency maximum is just 50%
3.3 Multiple Sinusoids
• Suppose that m.t/ is a sum of multiple sinusoids
(multi-tone AM)
𝑀
𝑚 𝑡 =
𝐴𝑘 cos(2𝜋𝑓𝑘 𝑡 + 𝜙𝑘 )
𝑘=1
• To find mn(t)we need to find min m(t)
𝑀
σ
• A lower bound on min m(t) is − 𝑘=1 𝐴𝑘
0
