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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 #
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 #
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-
192 CHAPTER 14. MODULATION AND DEMODULATION
medium, it would be difficult for their intended receivers to extract the signals reliably because of interference. One approach to reduce this interference, known as frequency- division multiplexing , allocates different carrier frequencies to different users (or for dif- ferent uses, e.g., one might separate out the frequencies at which police radios or emer- gency responders communicate from the frequencies at which you make calls on your mobile phone). In fact, the US spectrum allocation map shown in Figure 14-1 is the result of such a frequency-division strategy. It enables users (or uses) that may end up with sim- ilar looking baseband signals (those that will interfere with each other) to be transmitted on different carrier frequencies, eliminating interference. There are two reasons why frequency-division multiplexing works:
- Any baseband signal can be broken up into a weighted sum of sinusoids using Fourier decomposition (Chapter 13). If the baseband signal is band-limited, then there is a finite maximum frequency of the corresponding sinusoids. One can take this sum and modulate it on a carrier signal of some other frequency in a simple way: by just multiplying the baseband and carrier signal (also called “mixing”). The result of modulating a band-limited baseband signal on to a carrier is a signal that is band-limited around the carrier, i.e., limited to some maximum frequency deviation from the carrier frequency.
- When transmitted over a linear, time-invariant (LTI) channel, and if noise is negli- gible, each sinusoid shows up at the receiver as a sinusoid of the same frequency, as we saw in Chapter 12. The reason is that an LTI system preserves the sinusoids. If we were to send a baseband signal composed of a sum of sinusoids over the channel, the output will be the sum of sinuoids of the same frequencies. Each receiver can then apply a suitable filter to extract the baseband signal of interest to it. This insight is useful because the noise-free behavior of real-world communication channels is often well-characterized as an LTI system.
14.2 Amplitude Modulation with the Heterodyne Principle
The heterodyne principle is the basic idea governing several different modulation schemes. The idea is simple, though the notion that it can be used to modulate signals for transmission was hardly obvious before its discovery!
Heterodyne principle: The multiplication of two sinusoidal waveforms may be written as the sum of two sinusoidal waveforms, whose frequencies are given by the sum and the difference of the frequencies of the sinusoids being multiplied.
This result may be seen from standard high-school trigonometric identities, or by (per- haps more readily) writing the sinusoids as complex exponentials and performing the mul- tiplication. For example, using trigonometry,
cos(Ω (^) s n) · cos(Ω (^) c n) =
cos(Ω (^) s + Ω (^) c )n + cos(Ω (^) s − Ω (^) c )n
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 (^2) Ω^ π 1 for now—as the sum of different sinusoids of frequencies Ω (^) s 1 = k 1 Ω 1 , Ω (^) s 2 = k 2 Ω 2 ,... and treating the carrier as a sinusoid of frequency Ω (^) c = k (^) c Ω 1. Here, Ω 1 is the fundamental frequency of the baseband signal.
6.02 Fall 2011 (^) Lecture 15, Slide #
Modulation
x[n]
cos(kc! 1 n)
t[n]
t [ n ] = Ake
jk! 1 n
k =" kx
kx
e
jkc! 1 n
e
$ " jkc! 1 n
Ake
j (^) ( k + kc )! 1 n
k =" kx
kx
# +^
Ake
j (^) ( k " kc )! 1 n
k =" kx
kx
Re(ak)
Im(ak)
kc +kc
A/
A/
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(Ω (^) c n) = cos(k (^) c Ω 1 n) 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:
- 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 [− k (^) x , − k (^) x ], then we can divide the frequency spectrum into non-overlapping ranges of size 2k (^) x 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] =
k (^) x
k =− k (^) x
A (^) k e jkΩ^1 n^. (14.2)
Notice how applying the input filter ensures that high-frequency components are zero; the frequency range of the baseband is now [− k (^) x Ω 1 , k (^) x Ω 1 ] radians/sample.
- Mixing step. Multiply x[n] (called the baseband modulating signal) by a carrier, cos(k (^) c Ω 1 n), to produce the signal ready for transmission, t[n]. Using the DTFS form,
SECTION 14.3. DEMODULATION: THE SIMPLE NO-DELAY CASE 195
Modulation
t [ n ] = Ake jk!^1 n k =" kx
kx
$
%
& &
'
(
) )
1 2
e jkc! 1 n
1 2
e
$ " jkc! 1 n % &
' ( )
=
1 2
Ake j (^) ( k + kc )! 1 n k =" kx
kx
+^
1 2
Ake j (^) ( k " kc )! 1 n k =" kx
kx
Re(ak)
Im(ak)
kc +kc
A/
A/
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(35Ω 1 n), 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,
196 CHAPTER 14. MODULATION AND DEMODULATION
6.02 Fall 2011 (^) Lecture 15, Slide #
Example: Modulation
(freq domain picture)
Band-limited x[n] cos( 35! 1 n) t[n]
Figure 14-6: Frequency-domain representation of Figure 14-4, showing how the DTFS components (real and imaginary) of the real-valued band-limited signal x[n] after input filtering to produce shaped pulses (left), the purely cosine sinusoidal carrier signal (middle), and the heterodyned (mixed) baseband and carrier at two frequency ranges whose widths are the same as the baseband signal, but that have been shifted ±kc in frequency, and scaled by 1 / 2 each (right). We can avoid interference with another signal whose baseband overlaps in frequency, by using a carrier for the other signal sufficiently far away in frequency from kc.
except that is scaled by 1 /2. How would we recover this middle piece alone and ignore the left and right clusters, which are centered at frequencies that are at twice the carrier frequency in the positive and negative directions? We have already studied a technique in Chapter 12: a low-pass filter. By applying a low-pass filter whose cut-off frequency lies between kx and 2kc − kx, we can recover the original signal faithfully. We can reach the same conclusions by doing a more painstaking calculation, similar to the calculations we did for the modulation, leading to Equation (14.3). Let z[n] be the signal obtained by multiplying (mixing) the local replica of the carrier cos(kcΩ 1 n) and the received signal, r[n] = t[n], which is of course equal to x[n] cos(kcΩ 1 n). Using Equation 14.3, we can express z[n] in terms of its DTFS coefficients as follows:
198 CHAPTER 14. MODULATION AND DEMODULATION
y[n] !"
cos(kc! 1 n)
t[n] Channel H(!) z[n]
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 kΩ 1. 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(kcΩ 1 n) = y[n]
ejkcΩ^1 n^ +
e−^ jkcΩ^1 n
kx ∑ k=−kx
H((k + kc)Ω 1 )Akej(k+kc)Ω^1 n^ +
kx ∑ k=−kx
H((k − kc)Ω 1 )Akej(k−kc)Ω^1 n
ejkcΩ^1 n^ +
e−^ jkcΩ^1 n
kx ∑ k=−kx
AkejkΩ^1 n
H((k + kc)Ω 1 ) + H((k − kc)Ω 1 )
kx ∑ k=−kx
Akej(k+^2 kc)Ω^1 n
H((k + kc)Ω 1 ) + H((k − kc)Ω 1 )
kx ∑ k=−kx
Akej(k−^2 kc)Ω^1 n
H((k + kc)Ω 1 ) + H((k − kc)Ω 1 )
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 ± 2 kc 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)Ω 1 ) + H((k − kc)Ω 1 ). 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 #
Demodulation + LPF
t[n] ! z[n] LPF y[n]
Cutoff @ ±kx
Filter gain depends on H values
cos(kc! 1 n)
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
SECTION 14.4. HANDLING CHANNEL DELAY: QUADRATURE DEMODULATION 201
6.02 Spring 2012 Lecture 16, Slide #
output amplitude, where the magnitude of the scaling can’t necessarily be determined at system design time:
- Channel delay varies on mobile devices
- Phase difference between transmitter and receiver is arbitrary
One solution: quadrature demodulation
!
cos(Ωcn-!)
LPF I[n] = x[n-D]·cos(θ) Cutoff @ ±kin Gain = 2
!
sin(Ωcn-!)
LPF Q[n] = x[n-D]·sin(θ) Cutoff @ ±kin Gain = 2
From channel
θ = ΩcD - φ
tD[n]=t[n-D]
Figure 14-11: Quadrature demodulation to handle D -sample channel delay.
lead to no baseband signal—you should verify this fact by writing
z[n] = t[n] sin(k (^) c Ω 1 n) = t[n]
(^) − j 2 e jk^ c^ Ω^1 n^ +
j 2 e −^ jk^ c^ Ω^1 n^
and expanding t[n] using its DTFS. Hence, multiplying by the sin when the carrier is a cos will not always work; it will work only when the phase error is a fortunate value (≈ π/ 2). This observation leads us to a solution to this problem, called quadrature demodula- tion , depicted in Figure 14-11 for the case of channel delay but no channel distortion (so we can apply a gain of 2 on the LPFs rather than factors dependent on H(Ω)). The idea is to multiply the received signal by both cos(Ω (^) c n) (where Ω (^) c = k (^) c Ω 1 is the carrier frequency), and sin(Ω (^) c n). This method is a way of “hedging” our bet: we cannot be sure which term, cos or sin would work, but we can be sure that they will not be 0 at the same time! We can use this fact to recover the signal reliably always, as explained below. For simplicity (and convenience), suppose that x[n] ≥ 0 always (at the input). Then, using the notation from Figure 14-11, define w[n] = I[n] + jQ[n] (the I term is generally called the in-phase term and the Q term is generally called the quadrature term). Then,
| w[n]| =
(I[n])^2 + (Q[n])^2
= | x[n − D]|
(cos 2 θ + sin^2 θ ) = | x[n − D]| (14.7) = x[n − D] becausex[n] ≥ 0 (14.8)
Hence, the quadrature demodulator performs the following step, in addition to the ones for the no-delay case explained before: compute I[n] and Q[n], and calculate | w[n]| using Equation (14.8). Return this value, thresholding (to combat noise) at the mid-point between the voltage levels corresponding to a “0” and a “1”. With quadrature demodula- tion, suppose the sender sends 0 volts for a “0” and 1 volt for a “1”, the receiver would, in general, demodulate a rotated version in the complex plane, as shown in Figure 14-12. However, the magnitude will still be 1, and quadrature demodulation can successfully recover the input. Figure 14-13 summarizes the various steps of the quadrature demodulator that we de- scribed in this section.
202 CHAPTER 14. MODULATION AND DEMODULATION
x[n-D]cos(θ)
x[n-D]sin(θ)
I
jQ
θ
Constellation diagrams:
transmitter receiver
x[n-D] = { 0, 1 }
I
Q
I
Q
Figure 14-12: Quadrature demodulation. The term “constellation diagram” refers to the values that the sender can send, in this case just 0 and 1 volts. The receiver’s steps are shown in the picture.
This concludes our discussion of the basics of demodulation. We turn next to briefly survey more sophisticated modulation/demodulation schemes.
14.5 More Sophisticated (De)Modulation Schemes
We conclude this chapter by briefly outlining three more sophisticated (de)modulation schemes.
14.5.1 Binary Phase Shift Keying (BPSK)
In BPSK, as shown in Figure 14-14, the transmitter selects one of two phases for the carrier, e.g. −π/2 for “0” and π/2 for “1”. The transmitter does the same mixing with a sinusoid as explained earlier. The receiver computes the I and Q components from its received waveform, as before. This approach “almost” works, but in the presence of channel delays or phase errors, the previous strategy to recover the input does not work because we had assumed that x[n] ≥ 0. With BPSK, x[n] is either +1 or −1, and the two levels we wish to distinguish have the same magnitude on the complex plane after quadrature demodula- tion! The solution is to think of the phase encoding as a differential, not absolute: a change in phase corresponds to a change in bit value. Assume that every message starts with a “0” bit. Then, the first phase change represents a 0 → 1 transition, the second phase change a 1 → 0 transition, and so on. One can then recover all the bits correctly in the demodulator using this idea, assuming no intermediate glithces (we will not worry about such glitches here, which do occur in practice and must be dealt with).
14.5.2 Quadrature Phase Shift Keying (QPSK)
Quadrature Phase Shift Keying is a clever idea to add a “degree of freedom” to the system (and thereby extracting higher performance). This method, shown in Figure 14-15, uses a quadrature scheme at both the transmitter and the receiver. When mapping bits to voltage values in QPSK, we would choose the values so that the amplitude of t[n] is constant.
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)
x[n] (-1,1)
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
msg[0::2] (-A,A)!
cos(Ωcn)
sin(Ωcn)
LPF
+ t[n]
I[n]
msg[1::2] (^) (-A,A) LPF^ Q[n]
I
Q (A,A)
(-A,-A) (A,-A)
(-A,A)
Odd bits Map bit into voltage value
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 #
QAM Modulation
(-3A,-A, msg[0::4] (^) A, 3A)! cos(Ωcn)
!
sin(Ωcn)
LPF
I[n]
Msg[2::4] (^) LPF^ Q[n]
I
Q
Odd pairs of bits Map bits into voltage value
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
- 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 × 103 t), 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 × 103 t) , (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
SECTION 14.5. MORE SOPHISTICATED (DE)MODULATION SCHEMES 207
DT signal w[n] = 10 −^3 y[n − M] = 10 −^3 x[n − M] cos
Ω (^) c (n − M)
for an appropriately chosen integer M.
C. Determine the value of M in Equation (14.11). D. Noting your answer from part B , determine for precisely which intervals of the frequency axis the Fourier series coefficients of the signal y[n − M] in Equation (14.11) are non-zero. You need not find the actual coefficients , only the fre- quency range over which these coefficients will be non-zero. Also state whether or not the Fourier coefficients will be real. Explain your answer. E. The demodulation step to obtain the DT signal x[n − M] from the received sig- nal w[n] now involves multiplying w[n] by a DT carrier-frequency signal, fol- lowed by appropriate low-pass filtering (with the gain of the low-pass filter in its passband being chosen to scale the signal to whatever amplitude is desired). Which one of the following six DT carrier-frequency signals would you choose to multiply the received signal by? Circle your choice and give a brief explana- tion. i. cos
Ω (^) c n
ii. cos
Ω (^) c (n − M)
iii. cos
Ω (^) c (n + M)
iv. sin
Ω (^) c n
v. sin
Ω (^) c (n − M)
vi. sin
Ω (^) c (n + M)
208 CHAPTER 14. MODULATION AND DEMODULATION
M -sample delay
x 1 [n]
x 2 [n]
y[n]
w[n]
v[n]
-1000 1000
-500 -250 250 500
Figure 14-17: System for problem 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 x 1 in the modulation/demodulation diagram. Be sure to label key features such as values and coefficient indices for peaks.
