Lesson 18: Intro to Modulation, Exams of Communication

Download Lesson 18: Intro to Modulation and more Exams Communication in PDF only on Docsity! Lesson 18: Intro to Modulation Objectives: (a) Define the term baseband signal and describe some potential limitations associated with transmitting baseband signals directly. (b) Discuss the role of modulation in signal transmission and the methods of modulating a sinusoidal carrier. (c) Using a plot of an AM signal, determine Vmax, Vmin, Vm, Vc and m. (d) Create a frequency plot for and determine the sidebands and bandwidth of an AM signal where the information signal consists of one or more distinct sinusoids. (e) Determine the condition for overmodulation in an AM signal, and the consequence of overmodulation. Connection to Cyber Security In Lesson 17, we defined a communications system and learned that the wireless section of EC312 will focus on communications in which the communication channel is free space. We also learned that in order to send any information through a communications system it must be in the form of a signal (which is the name given to the function that conveys our information), and if our communication channel is free space, it means we’re dealing with signals carried in the Electromagnetic Spectrum (EM). Finally, we learned that signals can be represented as a function of either time or frequency. Wireless channels have different vulnerabilities than we saw in the host section of the course, because of the frequency of transmission. It is possible that a wireless network can be attacked like we saw in the networks section of the course, but to see how such a cyber-attack can be carried out on a wireless network or a wireless communication in general, you must first understand how information is carried through the wireless channel. 1. Baseband Signals If you’re sitting in your EC312 classroom in the basement of Rickover Hall and you speak to the student next to you, will that person be able to hear you? Will you be heard across the room? How about at the end of that loooong Rickover passageway? Does anyone in Bancroft have a chance of hearing what you’re grumbling about in your EC312 classroom? Why not? Your voice doesn’t travel as far as you might like it to—your voice creates pressure waves in the air, and the strength of these waves attenuates over distance. The louder you yell the farther you’ll be heard, but this of course has its limits. As an alternative to walking around screaming all day, you might consider speaking at a more normal volume into a microphone. A microphone is a device that transforms sound pressure waves into electrical signals. You could then send the electrical output of the microphone to an antenna. Then your voice frequencies would travel as electromagnetic waves (“radio waves”), and as long as you provided enough power to the antenna, you could presumably greatly extend the geographic range of your EC312 musings. In this chapter, we’ll see that the latter approach is on the right track, but in order to be practical, it requires a bit more finesse. In this example our voice signal, which you’ll recall is comprised of frequencies roughly in the range between 300 Hz and 3 kHz, is what we call a baseband signal. Baseband signals are information signals at their original frequencies, typically low frequencies. To transmit a baseband signal directly as is, we use baseband transmission…as you’ll see in shortly, communication systems typically will upshift the frequency spectrum of baseband signals to a higher range of frequencies to allow transmission through the atmosphere. In general, before signals can be transmitted effectively, they must first be converted to a form that is compatible with the communication medium. One facet of this conversion is transducing the signal from its natural physical form into an electrical signal. For example, • Microphones convert acoustic pressure waves (sound) into electrical signals. • Video cameras convert light patterns into electrical signals. • Computer keyboards convert physical input (typing) into an electrical signals. But, as alluded to above, even after converting your voice signal to, say, a voltage signal using a microphone, attempting to transmit it over the air as a baseband signal is impractical. Why? Let’s look at an example that will point us in the right direction. Practice Problem 18.1 Physics dictates that antenna length is intrinsically tied to the wavelength of the signal it is transmitting or receiving. To transmit a signal through the atmosphere with an antenna efficiently, the length of the antenna must be at least a tenth of a wavelength long. What is the approximate length of the antenna required to transmit the sound of a tuning fork (which creates musical note A = 440 Hz)? Note: this sound must be transduced into an electrical signal first before it is transmitted. Solution: Wait! To transmit that lousy tuning fork signal my antenna needs to be at least 68 km? That’s over 42 miles! We’d need an antenna that extends into the upper parts of the atmosphere for that. Clearly, that’s not going to work. Well, I know that if I want to listen to the Navy game on the radio (because for some reason I avoided the mandatory fun), I can tune in to AM radio station 1430 WNAV. Recall that from Chapter 17, when referring to a commercial AM radio station, such as 1430 WNAV, that the 1430 refers to the center of its transmission frequency in kHz. So what size antenna does WNAV use? Practice Problem 18.2 (a) What is the wavelength of an AM radio station whose transmission frequency is 1430 kHz? Solution: (b) What is the approximate antenna length if the station uses an antenna that is half the wavelength long? Solution: 105 meters? Okay, that’s still big – it’s about 115 yards - almost an entire football field… but at least you don’t need an antenna that reaches into outer space now. In reality, WNAV’s antenna, pictured at right, is 117 meters. We’re close! The purpose of those two examples was to demonstrate that we need to somehow get our baseband information to a higher frequency (shorter wavelength) in order to be able to transmit it across our channel. Higher frequencies give us reasonable antenna sizes plus some added benefits: first, signals will attenuate less quickly if the higher frequencies are well-chosen. Second, multiple people - whose voices all occupy the same baseband frequencies - can communicate without interfering if each transmits on a different higher frequency range. We can shift baseband information to higher frequencies for transmission using a process called modulation. A zoomed-in plot of the resulting modulated AM signal is as follows, showing the graphical relation between Vm and Vc: The envelope of the modulating signal (which is drawn onto the AM signal below in a dashed red line) varies above and below the unmodulated carrier amplitude, Vc. It is the envelope that carries the information signal; the receiver must separate the envelope from the received AM signal to recover the information that was transmitted. In this case, the envelope is in the shape of a sine wave, which is the same as the information signal. The values of Vm and Vc are related by the modulation index (m). 3.1 Modulation Index The relationship between the information signal amplitude, Vm , and the unmodulated carrier amplitude, Vc, is expressed as a ratio called the modulation index (m), defined as: Sometimes m is expressed as a percentage: percent modulation = m x 100%. The following figure shows the AM signal at three different values of percent modulation: 20%, 50% and 90%. Overall, the greater the value of m, the closer the envelope gets to the horizontal (time) axis. We can also mathematically determine the modulation index m from the maximum and minimum values of the envelope of vAM(t) as follows, where Vmax is the maximum value of the envelope and Vmin is the minimum value: In order for the AM signal to convey the original signal accurately and prevent distortion, the information signal amplitude (Vm) must be less than the unmodulated carrier signal amplitude (Vc). Here again, the unmodulated carrier refers to the AM signal if the information signal amplitude is equal to 0 (Vm = 0), in which case, ( ) cos(2 )m m mv t V f tπ= . The maximum usable modulation index is m = 1.0, corresponding to 100% modulation, when Vm is equal to Vc. When Vm is greater than Vc (that is, m > 1), overmodulation occurs. Overmodulation, depicted below, results in distortion of the AM signal’s envelope, and since the envelope holds the information, the recovered information signal is also distorted. Practice Problem 18.3 If a carrier signal vc(t) = 9 cos(2π5000t) Volts is modulated by a cosine wave vm(t) =7.5 cos(2π440t) V, what is the percentage modulation of the resulting AM signal? Solution: Now that we have a basic understanding of how Amplitude Modulation works in the time domain, let’s look at AM in the frequency domain. 4. AM in Frequency Domain Recall the equation for the amplitude modulated waveform if the information signal is a single sine wave is given by: . ( )( ) ( )( ) cos 2 cos 2 .AM c m m cv t V V f t f tπ π= + We already know the frequency domain representations of the modulating signal (vm(t)) and the carrier signal (vc(t)), but how does the amplitude modulated signal look in the frequency domain? m C Vm V = max min max min max min max min 2 2 m c m c V VV V VV V V Vm V V V − = + = − = = + To answer this question, recall the trig identity for the product of two sine waves: 1 1 2 2cos cos cos( ) cos( )A B A B A B= + + − Applying this trig identity for product of two sine waves to the AM signal results in: ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( )2 2 ( ) cos 2 cos 2 cos 2 cos 2 cos 2 cos 2 cos 2 cos 2m m AM c m m c c c m c m V V c c c m c m v t V V f t f t V f t V f t f t V f t f f t f f t π π π π π π π π = + = + = + + + − This means that when a single sine wave information signal is used to modulate the carrier in AM, the resulting AM signal contains three sinusoids (labeled above): one at the carrier frequency (1), one fm Hz above the carrier frequency (2), and one fm Hz below the carrier frequency (3). For the tuning fork example, we have: fc = 5 kHz, fc − fm = 4.560 kHz and fc + fm = 5.440 kHz. The trig identity puts the amplitudes at frequencies fc − fm and fc + fm at one half that of Vm. This means the resulting frequency domain plot for this tuning fork example looks like the following (the labels follow the ones above): The process of modulating a carrier creates an upper and a lower sideband that is apparent in the frequency plot. The lower sideband (or LSB) is that portion of the transmitted signal that has frequency content less than the carrier frequency, and the upper sideband (or USB) has frequency content greater than the carrier frequency. For the tuning fork example, the USB is the 5440 Hz cosine, and the LSB is the 4560 Hz cosine. On a frequency plot of an AM signal, the lower sideband is a mirror image of the upper sideband centered about the carrier frequency. What is the AM signal’s bandwidth? Since bandwith is the highest transmitted frequency minus the lowest frequency transmitted, it is (fc + fm) − (fc − fm) = 2 fm = 880 Hz. This is twice the bandwith of the information signal we started with…if we didn’t modulate the information signal, the transmission bandwidth would have only been fm (440 Hz in this case)1. This means that by transmitting with AM, we have doubled the required bandwith to transmit the signal. Why is this a concern? Bandwidth is the #2 limiting factor in communications systems, and can be expensive to use… so we’re going to want to send as much information as possible while occupying the minimum amount of bandwith possible. (The #1 limiting factor in communications systems is noise, but that’s a subject for a future lesson.) Let’s look at a slightly more complex example – suppose the information signal was comprised of two cosine waves: 1 The bandwidth of a baseband signal is considered to be its maximum frequency content. In this case, if the message is a single sinusoid at a frequency of fm Hz, we say its bandwidth is fm Hz. 1 2 3 1 2 3