Modulation and Demodulation: Transmitting Baseband Signals over Communication Links, Study notes of Communication

Download Modulation and Demodulation: Transmitting Baseband Signals over Communication Links and more Study notes Communication in PDF only on Docsity! MIT 6.02 DRAFT Lecture Notes Last update: April 11, 2012 Comments, questions or bug reports? Please contact {hari, verghese} at mit.edu CHAPTER 14 Modulation and Demodulation This chapter describes the essential principles behind modulation and demodulation, which we introduced briefly in Chapter 10. Recall that our goal is to transmit data over a commu- nication link, which we achieve by mapping the bit stream we wish to transmit onto analog signals because most communication links, at the lowest layer, are able to transmit ana- log signals, not binary digits. The signals that most simply and directly represent the bit stream are called the baseband signals. We discussed in Chapter 10 why it is generally un- tenable to directly transmit baseband signals over communication links. We reiterate and elaborate on those reasons in Section 14.1, and discuss the motivations for modulation of a baseband signal. In Section 14.2, we describe a basic principle used in many modulation schemes, called the heterodyne principle. This principle is at the heart of amplitude modulation (AM), the scheme we study in detail. Sections 14.3 and 14.4 describe the “inverse” process of demodulation, to recover the original baseband signal from the received version. Fi- nally, Section 14.5 provides a brief overview of more sophisticated modulation schemes. ￿ 14.1 Why Modulation? There are two principal motivating reasons for modulation. We described the first in Chap- ter 10: matching the transmission characteristics of the medium, and considerations of power and antenna size, which impact portability. The second is the desire to multiplex, or share, a communication medium among many concurrently active users. ￿ 14.1.1 Portability Mobile phones and other wireless devices send information across free space using electro- magnetic waves. To send these electromagnetic waves across long distances in free space, the frequency of the transmitted signal must be quite high compared to the frequency of the information signal. For example, the signal in a cell phone is a voice signal with a bandwidth of about 4 kHz. The typical frequency of the transmitted and received signal is several hundreds of megahertz to a few gigahertz (for example, the popular WiFi standard is in the 2.4 GHz or 5+ GHz range). 189 190 CHAPTER 14. MODULATION AND DEMODULATION 6.02 Spring 2012 Lecture 14, Slide #2 U.S. Spectrum Allocation Map http://www.wireless-technology.org/wp-content/uploads/2011/02/Wireless-Spectrum-Photo.jpg 6.02 Spring 2012 Lecture 14, Slide #3 U.S. Spectrum Allocation Map http://www.wireless-technology.org/wp-content/uploads/2011/02/Wireless-Spectrum-Photo.jpg Wi-Fi Figure 14-1: Top: Spectrum allocation in the United States (3 kHz to 300 GHz). Bot- tom: a portion of the total allocation, highlighting the 2.4 GHz ISM (Industrial, Scientific, and Medical) band, which is unlicensed spectrum that can be used for a variety of pur- poses, including 802.11b/g (WiFi), various cordless telephones, baby monitors, etc. From http://www.wireless-technology.org/wp-content/uploads/2011/02/Wireless-Spectrum-Photo.jpg One important reason why high-frequency transmission is attractive is that the size of the antenna required for efficient transmission is roughly one-quarter the wavelength of the propagating wave, as discussed in Chapter 10. Since the wavelength of the (electro- magnetic) wave is inversely proportional to the frequency, the higher the frequency, the smaller the antenna. For example, the wavelength of a 1 GHz electromagnetic wave in free space is 30 cm, whereas a 1 kHz electromagnetic wave is one million times larger, 300 km, which would make for an impractically huge antenna and transmitter power to transmit signals of that frequency! ￿ 14.1.2 Sharing using Frequency-Division Figure 14-1 shows the electromagnetic spectrum from 3 kHz to 300 GHz; it depicts how portions of spectrum have been allocated by the U.S. Federal Communications Commis- SECTION 14.2. AMPLITUDE MODULATION WITH THE HETERODYNE PRINCIPLE 193 We apply the heterodyne principle by treating the baseband signal —think of it as periodic with period 2π Ω1 for now—as the sum of different sinusoids of frequencies Ωs1 = k1Ω1,Ωs2 = k2Ω2, . . . and treating the carrier as a sinusoid of frequency Ωc = kcΩ1. Here, Ω1 is the fundamental frequency of the baseband signal. 6.02 Fall 2011 Lecture 15, Slide #19 Modulation !x[n] cos(kc!1n) t[n] t[n]= Ake jk!1n k="kx kx # $ % & & ' ( ) ) 1 2 e jkc!1n + 1 2 e" jkc!1n $ %& ' () = 1 2 Ake j k+kc( )!1n k="kx kx # + 1 2 Ake j k"kc( )!1n k="kx kx # Re(ak) Im(ak) +kc －kc A/2 A/2 For band-limited signal Ak are nonzero only for small range of ±k i.e., just replicate baseband signal at ±kc, and scale by !. Figure 14-3: Modulation involved “mixing”, or multiplying, the input signal x[n] with a carrier signal (cos(Ωcn) = cos(kcΩ1n) here) to produce t[n], the transmitted signal. The application of the heterodyne principle to modulation is shown schematically in Figure 14-3. Mathematically, we will find it convenient to use complex exponentials; with that notation, the process of modulation involves two important steps: 1. Shape the input to band-limit it. Take the input baseband signal and apply a low- pass filter to band-limit it. There are multiple good reasons for this input filter, but the main one is that we are interested in frequency division multiplexing and wish to make sure that there is no interference between concurrent transmissions. Hence, if we limit the discrete-time Fourier series (DTFS) coefficients to some range, call it [−kx,−kx], then we can divide the frequency spectrum into non-overlapping ranges of size 2kx to ensure that no two transmissions interfere. Without such a filter, the baseband could have arbitrarily high frequencies, making it hard to limit interfer- ence in general. Denote the result of shaping the original input by x[n]; in effect, that is the baseband signal we wish to transmit. An example of the original baseband signal and its shaped version is shown in Figure 14-4. We may express x[n] in terms of its discrete-time Fourier series (DTFS) representation as follows, using what we learned in Chapter 13: x[n] = kx ∑ k=−kx Ake jkΩ1n. (14.2) Notice how applying the input filter ensures that high-frequency components are zero; the frequency range of the baseband is now [−kxΩ1, kxΩ1] radians/sample. 2. Mixing step. Multiply x[n] (called the baseband modulating signal) by a carrier, cos(kcΩ1n), to produce the signal ready for transmission, t[n]. Using the DTFS form, 194 CHAPTER 14. MODULATION AND DEMODULATION Baseband input x[n]: shaped pulses to band-limit signal Carrier signal Transmitted signal t[n]: “mix” (multiply x[n] and carrier) Figure 14-4: The two modulation steps, input filtering (shaping) and mixing, on an example signal. we get t[n] = ￿ kx ∑ k=−kx Ake jkΩ1n ￿￿1 2 (e jkcΩ1n + e− jkcΩ1n) ￿ = 1 2 kx ∑ k=−kx Ake j(k+kc)Ω1n + 1 2 kx ∑ k=−kx Ake j(k−kc)Ω1n. (14.3) Equation (14.3) makes it apparent (see the underlined terms) that the process of mix- ing produces, for each DTFS component, two frequencies of interest: one at the sum and the other at the difference of the mixed (multiplied) frequencies, each scaled to be one-half in amplitude compared to the original. We transmit t[n] over the channel. The heterodyne mixing step may be explained math- ematically using Equation (14.3), but you will rarely need to work out the math from scratch in any given problem: all you need to know and appreciate is that the (shaped) baseband signal is simply replicated in the frequency domain at two different frequencies, ±kc, which are the nonzero DTFS coefficients of the carrier sinusoidal signal, and scaled by 1/2. We show this outcome schematically in Figure 14-5. The time-domain representation shown in Figure 14-4 is not as instructive as the frequency-domain picture to gain intuition about what modulation does and why frequency- division multiplexing avoids interference. Figure 14-6 shows the same information as Fig- ure 14-4, but in the frequency domain. The caption under that figure explains the key insights. SECTION 14.3. DEMODULATION: THE SIMPLE NO-DELAY CASE 195 Modulation t[n]= Ake jk!1n k="kx kx # $ % & & ' ( ) ) 1 2 e jkc!1n + 1 2 e" jkc!1n $ %& ' () = 1 2 Ake j k+kc( )!1n k="kx kx # + 1 2 Ake j k"kc( )!1n k="kx kx # Re(ak) Im(ak) +kc －kc A/2 A/2 For band-limited signal Ak are nonzero only for small range of ±k I.e., just replicate baseband signal at ±kc, and scale by !. Figure 14-5: Illustrating the heterodyne principle. This completes our discussion of the modulation process, at least for now (we’ll revisit it in Section 14.5), bringing us to the question of how to extract the (shaped) baseband signal at the receiver. We turn to this question next. ￿ 14.3 Demodulation: The Simple No-Delay Case Assume for simplicity that the receiver captures the transmitted signal, t[n], with no distor- tion, noise, or delay; that’s about as perfect as things can get. Let’s see how to demodulate the received signal, r[n] = t[n], to extract x[n], the shaped baseband signal. The trick is to apply the heterodyne principle once again: multiply the received signal by a local sinusoidal signal that is identical to the carrier! An elegant way to see what would happen is to start with Figure 14-6, rather than the time-domain representation. We now can pretend that we have a “baseband” signal whose frequency components are as shown in Figure 14-6, and what we’re doing now is to “mix” (i.e., multiply) that with the carrier. We can accordingly take each of the two (i.e., real and imaginary) pieces in the right-most column of Figure 14-6 and treat each in turn. The result is shown in Figure 14-7. The left column shows the frequency components of the original (shaped) baseband signal, x[n]. The middle column shows the frequency components of the modulated signal, t[n], which is the same as the right-most column of Figure 14-6. The carrier (cos(35Ω1n), so the DTFS coefficients of t[n] are centered around k = −35 and k = 35 in the middle column. Now, when we mix that with a local signal identical to the carrier, we will shift each of these two groups of coefficients by ±35 once again, to see a cluster of coefficients at −70 and 0 (from the −35 group) and at 0 and +70 (from the +35 group). Each piece will be scaled by a further factor of 1/2, so the left and right clusters on the right-most column in Figure 14-7 will be 1/4 as large as the original baseband components, while the middle cluster centered at 0, with the same spectrum as the original baseband signal, will be scaled by 1/2. What we are interested in recovering is precisely this middle portion, centered at 0, be- cause in the absence of any distortion, it is exactly the same as the original (shaped) baseband, 198 CHAPTER 14. MODULATION AND DEMODULATION !"y[n] cos(kc!1n) z[n] t[n] Channel H(!) Figure 14-8: Demodulation in the presence of channel distortion characterized by the frequency response of the channel. often the case), then one can extend the approach described above. The difference is that each of the Ak terms in Equation (14.4), as well as Figure 14-7, will be multiplied by the frequency response of the channel, H(Ω), evaluated at a frequency of kΩ1. So each DTFS coefficient will be scaled further by the value of this frequency response at the relevant frequency. Figure 14-8 shows the model of the system now. The modulated input, t[n], traverses the channel en route to the demodulator at the receiver. The result, z[n], may be written as follows: z[n] = y[n] cos(kcΩ1n) = y[n] ￿1 2 e jkcΩ1n + 1 2 e− jkcΩ1n ￿ = ￿1 2 kx ∑ k=−kx H((k + kc)Ω1)Akej(k+kc)Ω1n + 1 2 kx ∑ k=−kx H((k− kc)Ω1)Akej(k−kc)Ω1n ￿ · ￿1 2 e jkcΩ1n + 1 2 e− jkcΩ1n ￿ = 1 4 kx ∑ k=−kx Ake jkΩ1n ￿ H((k + kc)Ω1) + H((k− kc)Ω1) ￿ + 1 4 kx ∑ k=−kx Ake j(k+2kc)Ω1n ￿ H((k + kc)Ω1) + H((k− kc)Ω1) ￿ + 1 4 kx ∑ k=−kx Ake j(k−2kc)Ω1n ￿ H((k + kc)Ω1) + H((k− kc)Ω1) ￿ (14.5) Of these three terms in the RHS of Equation (14.5), the first term contains the baseband signal that we want to extract. We can do that as before by applying a lowpass filter to get rid of the ±2kc components. To then recover each Ak, we need to pass the output of the lowpass filter to another LTI filter that undoes the distortion by multiplying the kth Fourier coefficient by the inverse of H((k + kc)Ω1) + H((k − kc)Ω1). Doing so, however, will also amplify any noise at frequencies where the channel attenuated the input signal t[n], so a better solution is obtained by omitting the inversion at such frequencies. For this procedure to work, the channel must be relatively low-noise, and the receiver needs to know the frequency response, H(Ω), at all the frequencies of interest in Equation (14.5); i.e., in the range [−kc − kx,−kc + kx] and [kc − kx, kc + kx]. To estimate H(Ω), a com- mon approach is to send a known preamble at the beginning of each packet (or frame) SECTION 14.4. HANDLING CHANNEL DELAY: QUADRATURE DEMODULATION 199 6.02 Spring 2012 Lecture 13, Slide #15 Demodulation + LPF !t[n] z[n] LPF y[n] Cutoff @ ±kx Filter gain depends on H values cos(kc!1n) Figure 14-9: Demodulation steps: the no-delay case (top). LPF is a lowpass filter. The graphs show the time-domain representations before and after the LPF. of transmission. The receiver looks for this known preamble to synchronize the start of reception, and because the transmitted signal pattern is known, the receiver can deduce channel’s the unit sample response, h[·], from it, using an approach similar to the one out- lined in Chapter 11. One can then apply the frequency response equation from Chapter 12, Equation (2.2), to estimate H(Ω) and use it to approximately undo the distortion intro- duced by the channel. Ultimately, however, our interest is not in accurately recovering x[n], but rather the underlying bit stream. For this task, what is required is typically not an inverse filtering operation. We instead require a filtering that produces a signal whose samples, obtained at the bit rate, allow reliable decisions regarding the corresponding bits, despite the presence of noise. The optimal filter for this task is called the matched filter. We leave the discussion of the matched filter to more advanced courses in communication. ￿ 14.4 Handling Channel Delay: Quadrature Demodulation We now turn to the case of channel delays between the sender and receiver. This delay matters in demodulation because we have thus far assumed that the sender and receiver have no phase difference with respect to each other. That assumption is, of course, not true, and one needs to somehow account for the phase delays. Let us first consider the illustrative case when there is a phase error between the sender 200 CHAPTER 14. MODULATION AND DEMODULATION !" t[n] z[n] #$%" y[n] Cutoff @ ±kx Gain depends on H !" cos(kc!1n) x[n] &" tD[n] Time delay of D samples cos(kc!1n) Figure 14-10: Model of channel with a delay of D samples. and receiver. We will then show that a non-zero delay on the channel may be modeled exactly like a phase error. By “phase error”, we mean that the demodulator, instead of multiplying (heterodyning) by cos(kcΩ1n), multiplies instead by cos(kcΩ1n− ϕ), where ϕ is some constant value. Let us understand what happens to the demodulated output in this case. Working out the algebra, we can write z[n] = t[n] cos(kcΩ1n−ϕ) = x[n] cos(kcΩ1n) cos(kcΩ1n−ϕ) (14.6) But noting that cos(kcΩ1n) cos(kcΩ1n−ϕ) = 1 2 ￿ cos(2kcΩ1n−ϕ) + cosϕ ￿ , it follows that the demodulated output, after the LPF step with the suitable gains, is y[n] = x[n] cosϕ. Hence, a phase error of ϕ radians results in the demodulated amplitude being scaled by cosϕ. This scaling is problematic: if we were unlucky enough to have the error close to π/2, then we would see almost no output at all! And if x[n] could take on both positive and negative values, then cosϕ going negative would cause further confusion. A channel delay between sender and receiver manifests itself as a phase error using the demodulation strategy we presented in Section 14.3. To see why, consider Figure 14-10, where we have inserted a delay of D samples between sender and receiver. The algebra is very similar to the phase error case: with a sample delay of D samples, we find that y[n] = x[n− D] cos(kcΩ1n) = cos(kcΩ1(n− D) + kcΩ1D). The cos factor in effect looks like it has a phase error of kcΩ1D, so the output is attenuated by cos(kcΩ1D). So how do we combat phase errors? One approach is to observe that in situations where cosϕ is 0, sinϕ is close to 1. So, in those cases, multiplying (heterodyning) at the demodulator by sin(kcΩ1n) = cos(π 2 − kcΩ1n) corrects for the phase difference. Notice, however, that if the phase error were non-existent, then multiplying by sin(kcΩ1n) would SECTION 14.5. MORE SOPHISTICATED (DE)MODULATION SCHEMES 203 6.02 Spring 2012 Lecture 16, Slide #12 × cos(!cn-") LPF I[n] = x[n-D]·cos(#) Cutoff @ ±kin × sin(!cn-") LPF Q[n] = x[n-D]·sin(#) Cutoff @ ±kin # = !cD - " | |2 | |2 + sqrt() y[n]=|x[n-D]| Q[n]2 I[n]2 y[n] = sqrt(I[n]2+Q[n]2) t[n] × cos(!cn) x[n] D tD[n] Transmitter Channel Receiver Decimate & slice Received bitsy[n] Quadrature demodulator I[n] Q[n] Delay Bits to samples Transmit bits Full system view tD[n] = t[n-D] Alternative representation × LPF Cutoff @ ±kin | | |y[n]|=|x[n-D]| tD[n] Receiver Decimate & slice Received bits Quadrature demodulator )( !"# nj ce )(][][ !"#"= Dj ceDnxny Figure 14-13: Quadrature demodulation: overall system view. The “alternative representation” shown implements the quadrature demodulator using a single complex exponential multiplication, which is a more compact representation and description. Moreover, because the constellation now involves four symbols, we map two bits to each symbol. So 00 might map to (A, A), 01 to (−A, A), 11 to (−A,−A), and 10 to (A,−A) (the amplitude is therefore √ 2A). There is some flexibility in this mapping, but it is not completely arbitrary; for example, we were careful here to not map 11 to (A,−A) and 00 to (A, A). The reason is that any noise is more likely to cause (A, A) to be confused fo (A,−A), compared to (−A,−A), so we would like a symbol error to corrupt as few bits as possible. ￿ 14.5.3 Quadrature Amplitude Modulation (QAM) QAM may be viewed as a generalization of QPSK (in fact, QPSK is sometimes called QAM- 4). One picks additional points in the constellation, varying both the amplitude and the phase. In QAM-16 (Figure 14-16), we map four bits per symbol. Denser QAM constella- tions are also possible; practical systems today use QAM-4 (QPSK), QAM-16, and QAM- 64. Quadrature demodulation with the adjustment for phase is the demodulation scheme used at the receiver with QAM. For a given transmitter power, the signal levels corresponding to different bits at the input get squeezed closer together in amplitude as one goes to constellations with more points. The resilience to noise reduces because of this reduced separation, but sophisti- cated coding and signal processing techniques may be brought to bear to deal with the effects of noise to achieve higher communication bit rates. In many real-world commu- nication systems, the physical layer provides multiple possible constellations and choice of codes; for any given set of channel conditions (e.g., the noise variance, if the channel is well-described using the AWGN model), there is some combination of constellation, 204 CHAPTER 14. MODULATION AND DEMODULATION 6.02 Spring 2012 BPSK I Q In binary phase-shift keying (BPSK), the message bit selects one of two phases for the carrier, e.g., π/2 for 0 and –π/2 for 1. ! sin(Ωcn) (-1,1) x[n] ! cos(Ωcn) ! sin(Ωcn) LPF ∠ LPF phase[n] I[n] Q[n] Figure 14-14: Binary Phase Shift Keying (BPSK). coding scheme, and code rate, which maximizes the rate at which bits can be received and decoded reliably. Higher-layer “bit rate selection” protocols use information about the channel quality (signal-to-noise ratio, packet loss rate, or bit error rate) to make this decision. ￿ Acknowledgments Thanks to Mike Perrot, Charlie Sodini, Vladimir Stojanovic, and Jacob White for lectures that helped shape the topics discussed in this chapter, and to Patricia Saylor for bug fixes. SECTION 14.5. MORE SOPHISTICATED (DE)MODULATION SCHEMES 205 6.02 Spring 2012 QPSK Modulation (-A,A) msg[0::2] ! cos(Ωcn) ! sin(Ωcn) LPF + t[n] I[n] Q[n] (-A,A) msg[1::2] LPF I Q (A,A) (A,-A) (-A,-A) (-A,A) Map bit into voltage value Odd bits Even bits Still need band limiting at transmitter When mapping bits to voltage values, we should choose the values so that the maximum amplitude of t[n] is 1. For QPSK (also referred to as QAM-4) that would mean 1 2 , 1 2 ! " # $ % &= (.707,.707) We can use the quadrature scheme at the transmitter too: Figure 14-15: Quadrature Phase Shift Keying (QPSK). 6.02 Spring 2012 Lecture 16, Slide #16 QAM Modulation (-3A,-A, A, 3A) msg[0::4] ! cos(Ωcn) ! sin(Ωcn) LPF + t[n] I[n] Q[n] Msg[2::4] LPF I Q Map bits into voltage value Odd pairs of bits Even pairs of bits Still need band-limiting at transmitter Use more levels in each arm (e.g. 4 levels per arm – 16QAM): (-3A,-A, A, 3A) Symbol/bits mapping table 00 ! -3A 01 ! -A 11 ! A 10 ! 3A Gray Code (noise movement into another constellation point only causes single bit errors) 00 01 10 11 00 01 10 11 Figure 14-16: Quadrature Amplitude Modulation (QAM). ￿ Problems and Questions Please solve the problems at http://mit.edu/6.02/www/s2012/handouts/tutprobs/modulation.html 1. The Boston sports radio station WEEI AM (“amplitude modulation”) broadcasts on a carrier frequency of 850 kHz, so its continuous-time (CT) carrier signal can be taken to be cos(2π × 850× 103t), where t is measured in seconds. Denote the CT audio signal that’s modulated onto this carrier by x(t), so that the CT signal transmitted by the radio station is y(t) = x(t) cos(2π× 850× 103t) , (14.9) as indicated schematically on the left side of the figure below. We use the symbols y[n] and x[n] to denote the discrete-time (DT) signals that would 208 CHAPTER 14. MODULATION AND DEMODULATION M-sample delay x1[n] x2[n] y[n] w[n] v[n] -1000 1000 -500 -250 250 500 Figure 14-17: System for problem 2. 2. All parts of this question pertain to the following modulation-demodulation system shown in Figure 14-17, where all signals are periodic with period P = 10000. Please also assume that the sample rate associated with this system is 10000 samples per second, so that k is both a coefficient index and a frequency. In the diagram, the modulation frequency, km, is 500. (a) Suppose the DFT coefficients for the signal y[n] in the modula- tion/demodulation diagram are as plotted in Figure 14-17. Assuming that M = 0 for the M-sample delay (no delay), plot the coefficients for the signals w and v in the modulation/demodulation diagram. Be sure to label key features such as values and coefficient indices for peaks. (b) Assuming the coefficients for the signal y[n] are the same as in part (a), please plot the DTFS coefficients for the signal x1 in the modulation/demodulation diagram. Be sure to label key features such as values and coefficient indices for peaks. SECTION 14.5. MORE SOPHISTICATED (DE)MODULATION SCHEMES 209 (c) If the M-sample delay in the modulation/demodulation diagram has the right number of samples of delay, then it will be possible to nearly perfectly recover x2[n] by low-pass filtering w[n]. Determine the smallest positive number of samples of delay that are needed and the cut-off frequency for the low-pass filter. Explain your answer, using pictures if appropriate. 3. Figure 14-18 shows a standard modulation/demodulation scheme where N = 100. 6.02 Spring 2011 - 2 of 10 - Quiz 3 Problem 1. Modulation/Demodulation (31 points) Consider the following standard modulation/demodulation scheme where N=100. (A) (5 points) The figure below shows a plot of the input, x[n]. Please draw the approximate time-domain waveform for y[n], the signal that is the input to the low-pass filter in the demodulator. Don’t bother drawing dots for each sample, just use a line plot to indicate the important timing characteristics of the waveform. Draw line plot for y[n] (B) (5 Points) Building on the scheme shown in the previous question, suppose there are multiple modulators and demodulators all connected to a single shared channel, with each modulator given a different modulation frequency. If the low-pass filter in each modulator is eliminated, briefly describe what the effect will be on signal z[n], the output of a demodulator tuned to the frequency of a particular transmitter. Brief description Figure 14-18: System for problem 3. (a) Figure 14-19 shows a plot of the input, x[n]. Please draw the approximate time- domain waveform for y[n], the signal that is the input to the lo -pass fil er in the demodulator. Don’t bother drawing dots for each sample, just use a line plot to indicate the important timing characteristics of the waveform. 6.02 Spring 2011 - 2 of 10 - Quiz 3 Problem 1. Modulation/Demodulation (31 points) Consider the following standard modulation/demodulation scheme where N=100. (A) (5 points) The figure below shows a plot of the input, x[n]. Please draw the approximate time-domain waveform for y[n], the signal that is the input to the low-pass filter in the demodulator. Don’t bother drawing dots for each sample, ju t use a line plot to indicate he important timing characteristics of the waveform. Draw line plot for y[n] (B) (5 Points) Building on the scheme shown in the previous question, suppose there are multiple modulators and demodulators all connected to a single shared channel, with each modulator given a different modulation frequency. If the low-pass filter in each modulator is eliminated, briefly describe what the effect will be on signal z[n], the output of a demodulator tuned to the frequency of a particular transmitter. Brief description Figure 14-19: Plot for problem 3(a). (b) Building on the scheme shown in Part (a), suppose there are multiple modu- lators and demodulators all connected to a single shared channel, with each modulator given a different modulation frequency. If the low-pass filter in each 210 CHAPTER 14. MODULATION AND DEMODULATION modulator is eliminated, briefly describe what the effect will be on signal z[n], the output of a demodulator tuned to the frequency of a particular transmitter. 4. The plot on the left of Figure 14-20 shows ak, the DTFS coefficients of the signal at the output of a transmitter with N = 36. If the channel introduces a 3-sample delay, please plot the Fourier series coefficients of the signal entering the receiver. 6.02 Spring 2011 - 3 of 10 - Quiz 3 (C) (5 Points) The plot on the left below shows ak, the Fourier seri s coefficients of the signal at the output of a transmitter where N=36. If the channel introduces a 3-sample delay, please plot the Fourier series coefficients of the signal entering the receiver. Plot Fourier series coefficients Figure 14-20: System for problem 4.