ELG317 Introdution to Communiation Systems Conventional AM
Disadvantages o DSB-SC The reeiver must generate a replia o the arrier in order to demodulate a DSB-SC signal. Any phase and/or requeny error will result in a distorted estimate o the message signal. It is diiult to generate a peret replia o the transmitted arrier. A simple modiiation to the tehnique results in a less eiient transmission but simpliies the detetion proess greatly. Conventional AM uses nonoherent demodulation. Detetion is possible even with requeny and phase errors.
Conventional AM Consider a message signal m(t) where M()=0 or =0, s AM ( t) A [1 k a m( t)]os t where A is the arrier amplitude, k a is the amplitude sensitivity, and is the arrier requeny. Also, >> B m where B m is the bandwidth o m(t).
Modulation index Let us assume that m p m(t) m p. For onventional AM, k a m(t) < 1, or -1 < k a m(t) < 1. Thereore, 0< k a m p < 1, or 0 < k a < 1/m p. The modulation index is m a = k a m p. For onventional AM, 0 < m a < 1. Thereore or onventional AM, [1 + k a m(t)] > 0.
Example 1 We wish to transmit m(t) = ost + os(1.4)t using onventional AM. The arrier is (t) = os10t. Find the value o k a so that the modulation index is 0.. Solution We an show that -3 m(t) 3, thereore m p = 3. thereore or m a = 0., k a = 1/6. The resulting AM signal is s AM (t) = [1+(1/6)m(t)]os10t
3 m(t) 1 0-1 - -3 -. - -1. -1-0. 0 0. 1 1.. 10 s AM (t) 0 - -10 -. - -1. -1-0. 0 0. 1 1..
The envelope o an AM signal The instantaneous amplitude o s AM (t) is A [1+k a m(t)]. This instantaneous amplitude is alled the signal s envelope. The message signal m(t) an be extrated diretly rom the envelope o s AM (t).
Overmodulation I m a > 1, we say that s AM (t) is overmodulated. An overmodulated signal annot be deteted using the nonoherent method that is used or onventional AM. Take our previous example with k a = 0.8. In this ase, m a = k a m p = 0.8 3 =.4. Thereore, sometimes k a m(t) < -1 leading to A [1+k a m(t)] < 0.
4 1+0.8m(t) 3 1 0-1 - -. - -1. -1-0. 0 0. 1 1.. [1+0.8m(t)]os0t 0 10 0-10 -0 -. - -1. -1-0. 0 0. 1 1..
Spetrum o Conventional AM signals We an express the AM signal as : Its Fourier transorm is S AM () = F{s AM (t)} whih is given by: t m t A k t A t s a AM )os ( os ) ( ) ( ) ( ) ( ) ( ) ( a a AM M A k M A k A A S
Spetrum o AM signal Assuming that m(t) has no DC omponent, then M(- ) has no spetral omponent at = and M(+ ) has no spetral omponent at = -. M() m o -B m B m S AM () A / A / (A k a /)m o - -B m - -+B m -B m +B m
Example Find the spetrum o s AM (t) = [1+(1/6)m(t)]os10t where m(t) = ost + os(1.4)t SOLUTION The signal m(t) has spetrum M() = ½(-1) + ½(+1) + (-1.4) + (+1.4). Thereore the spetrum o s AM (t) is: S AM ( ) ( 10) ( 10) M ( 10) M ( 10) 1 1 ( 10) ( 10) ( 11) ( 9) ( 11.4) 4 4 1 ( 8.6) ( 11) ( 9) ( 11.4) ( 8.6) 1 4 4 1 1
S AM ()
Power o a onventional AM signal Let s assume that m(t) is a power signal. I m(t) has no DC omponent, then we an ind that: where A / is the power o the arrier and A k a P m / is the power o the omponent that arries the message. The eiieny o a modulation sheme is the ratio o the power dediated to the transmission o message to the total power o the transmission. Thereore the eiieny o onventional AM is: m a s P k A A P m a m a m a m a P k P k P k A A P k A 1
Nonoherent detetion We know that the envelope o s AM (t) is A [1+k a m(t)]. An envelope detetor is a iruit that outputs the envelope o a bandpass signal. A hal-wave retiier with a apaitor ilter is a simple envelope detetor. I >> B m, the output o a halwave retiier ressembles a sampled version o the envelope. s AM (t) Halwave retiier x(t)
m(t) 3 1 0-1 - -3 -. - -1. -1-0. 0 0. 1 1.. 10 s AM (t) 0 - -10 -. - -1. -1-0. 0 0. 1 1..
8 7 x(t) 6 4 3 1 0-1 -. - -1. -1-0. 0 0. 1 1..
Simple envelope detetor Applying a low pass ilter at the output o the halwave retiier should give us a signal that ressembles A [1+k a m(t)]. Plaing a apaitor at the output o the retiier provides a simple solution. s AM (t) + - + A [1+k a m(t)] + v r (t) - The power o v r (t) is inversely proportional to.
Output o envelope detetor ompared to atual envelope. 8 7 6 4 3 1 0-1 -. - -1. -1-0. 0 0. 1 1..
Obtaining m(t) rom the envelope I we neglet v r (t), the output o the envelope detetor is A + A k a m(t). Passing this through a devie that bloks DC omponents, suh as a transormer, we have A k a m(t) at the output as long as m(t) has no DC omponent. s AM (t) déteteur d enveloppe A k a m(t) Envelope detetor