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The output of this program is shown in [link] . The slowly increasing sinusoidal “message” w ( t ) is modulated by the carrier c ( t ) at f c = 1000 Hz. The heart of the modulation is the point-by-point multiplication of the message and the carrierin the fifth line. This product v ( t ) is shown in [link] (c). The enveloping operation is accomplished byapplying a lowpass filter to the real part of 2 v ( t ) e j 2 π f c t . This recovers the original message signal, though it is offset by 1 and delayedby the linear filter.

Using AMlarge.m , plot the spectrum of the message w ( t ) , the spectrum of the carrier c ( t ) , and the spectrum of the received signal v ( t ) . What is the spectrum of the envelope? How close are your results to the theoretical predictions in [link] ?

One of the advantages of transmissions using AM with large carrier is that there is no needto know the (exact) phase or frequency of the transmitted signal. Verify this using AMlarge.m .

  1. Change the phase of the transmitted signal; for instance, let c=cos(2*pi* fc*t+phase) with phase=0.1, 0.5, pi/3, pi/2, pi , and verify that the recovered envelope remains unchanged.
  2. Change the frequency of the transmitted signal; for instance, let c=cos(2* pi*(fc+g)*t) with g=10, -10, 100, -100 , and verify that the recovered envelope remains unchanged.Can g be too large?
An undulation message (top) is modulated by a carrier (b). The composite signal is shown in (c), and the output of an envelope detector is shown in (d).
An undulation message (top) is modulated by a carrier (b). The composite signal is shown in (c), andthe output of an envelope detector is shown in (d).

Create your own message signal w ( t ) , and rerun AMlarge.m . Repeat Exercise  [link] with this new message. What differences do you see?

In AMlarge.m , verify that the original message w and the recovered envelope envv are offset by 1, except at the end points where the filter does not have enough data.Hint: the delay induced by the linear filter is approximately fl /2.

The principal advantage of transmission systems that use AM with a large carrier is that exact synchronization is not needed;the phase and frequency of the transmitter need not be known at the receiver, as was demonstratedin Exercise  [link] . This means that the receiver can be simplerthan when synchronization circuitry is required. The main disadvantage is thatadding the carrier into the signal increases the power needed for transmission but does not increase the amount ofuseful information transmitted.Here is a clear engineering tradeoff; the value of the wasted signal strength must be balanced againstthe cost of the receiver.

Amplitude modulation with suppressed carrier

It is also possible to use AM without adding the carrier.Consider the transmitted/modulated signal

v ( t ) = A c w ( t ) cos ( 2 π f c t )

diagrammed in [link] (a), in which the message w ( t ) is mixed with the cosine carrier. Direct application of the frequency shift property of Fouriertransforms [link] shows that the spectrum of the received signal is

V ( f ) = 1 2 A c W ( f + f c ) + 1 2 A c W ( f - f c ) .

As with AM with large carrier, the upconverted signal v ( t ) for AM with suppressed carrier has twice the bandwidth of the original message signal. If the original messageoccupies the frequencies between ± B Hz, then the modulated messagehas support between f c - B and f c + B , a bandwidth of 2 B . See [link] .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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