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6.02 Fall 2012 Lecture 16 Slide #1
6.02 Fall 2012
Lecture #16
• DTFT vs DTFS
• Modulation/Demodulation
• Frequency Division Multiplexing
(FDM)
Fast Fourier Transform (FFT) to compute
samples of the DTFT for
signals of finite duration
P−1
1
X(Ωk ) = ∑x[m]e− jΩkm
, x[n]=
m=0 P
6.02 Fall 2012 Lecture 16 Slide #2
For an x[n] that is zero outside of the interval [0,L-1], choose
P ≥ L (with P preferably a power of 2; we’ll assume that it’s at
least a multiple of 2, i.e., even):
(P/2)−1
∑ X(Ωk )ejΩkn
k=−P/2
where Ωk = k(2π/P), and k ranges from –P/2 to (P/2)–1, or over
any P successive integers. Simpler notation: X(Ωk) = Xk
Where do the Ω
6.02 Fall 2012 Lecture 16 Slide #3
Ωk live?
e.g., for P=6 (even)
– 
0
Ω0 Ω1 Ω2 Ω3
Ω3 Ω2 Ω1
exp(jΩ0)
exp(jΩ1)
exp(jΩ2)
exp(jΩ3)
= exp(jΩ3)
exp(jΩ1)
exp(jΩ2)
. 1
–1
j
–j
Fast Fourier Transform (FFT) to compute
samples of the DTFT for
signals of finite duration
P−1
1
X(Ωk ) = ∑x[m]e− jΩkm
, x[n]=
m=0 P
6.02 Fall 2012 Lecture 16 Slide #4
For an x[n] that is zero outside of the interval [0,L-1], choose
P ≥ L (with P preferably a power of 2; we’ll assume that it’s at
least a multiple of 2, i.e., even):
(P/2)−1
∑ X(Ωk )ejΩkn
k=−P/2
where Ωk = k(2π/P), and k ranges from –P/2 to (P/2)–1, or over
any P successive integers. Simpler notation: X(Ωk) = Xk
Note that X(Ω ) = X() = X(–
P/2 ) =X(Ω-P/2).
The above formulas have essentially the same structure, and are
both efficiently computed, with Plog(P) computations, by the FFT.
Some further details, assuming real x[n]
P−1
X(Ω ) = ∑x[m]e− jΩkm
k
m=0
• X(0) = sum of x[m] = real
• X() = X(–) = alternating sum of x[m] = real
• In general P-2 other complex values, but X(–Ωk) = X*(Ωk)
• So: total of P numbers to be determined, given P values of x[m]
1
x[n]=
P
6.02 Fall 2012 Lecture 16 Slide #5
(P/2)−1
∑ X(Ωk )ejΩkn
k=−P/2
• Evaluating this eqn. for n in [0,P-1] recovers the original x[n] in this
interval
• Evaluating it for n outside this interval results in periodic
replication of the values in [0,P-1], producing a periodic signal x[n]
• So this eqn. is also called a DT Fourier Series (DTFS) for the
periodic signal x[n]. Notation: Ak=X(Ωk)/P=Xk/P, Fourier coefficient.
Why the periodicity of x
6.02 Fall 2012 Lecture 16 Slide #6
x[n] is irrelevant in
many applications
h[.]
x[n] y[n]
Suppose x[n] is nonzero only over the time interval [0 , nx],
and h[n] is nonzero only over the time interval [0 , nh] .
In what time interval can the non-zero values of y[n] be
guaranteed to lie? The interval [0 , nx + nh] .
Since all the action we are interested in is confined to this
interval, choose P – 1 ≥ nx + nh .
It’s now irrelevant what happens outside [0,P–1]. So we can use
the FFT to go back and forth between samples of X(Ω), Η(Ω), Y(Ω) 
in the frequency domain and time-domain behavior in [0,P-1].
Back to Modulation/Demodulation
• You have: a signal x[n] at baseband (i.e., centered around 0
frequency)
• You want: the same signal, but centered around some specific
frequency Ωc
• Modulation: convert from baseband up to Ωc , to get t[n]
• Demodulation: convert from Ωc down to baseband
6.02 Fall 2012 Lecture 16 Slide #7
Re(Xk)
Im(Xk)
+Ωm
Ωm
Re(Tk)
Im(Tk)
+Ωc
Ωc
modulation
demodulation
Signal centered at 0 Signal centered at Ωc
Modulation by Heterodyning or
Amplitude Modulation (AM)
6.02 Fall 2012 Lecture 16 Slide #8
×
x[n]
cos(Ωcn)
t[n]
Im(Tk)
+Ωc
Ωc
B/2
k
A/2
Re(T )
i.e., just replicate baseband
signal at ±Ωc, and scale
by ½.
k
Im(Xk)
+Ωm
Ωm
A
B
Re(X )
To get this nice picture, the
baseband signal needs to be
band-limited to some range of
frequencies [–Ωm,Ωm], where
Ωm ≤ Ωc
6.02 Fall 2012 Lecture 16 Slide #9
0 Hz
1000 Hz
–1000 Hz
Not great band-limiting, but maybe we can get away with it!
At the Receiver: Demodulation
• In principle, this is (as easy as) modulation again:
If the received signal is
r[n] = x[n]cos(Ωcn) = t[n],
(no distortion or noise) then simply compute
d[n] = r[n]cos(Ωcn)
= x[n]cos2(Ωcn)
= 0.5 {x[n] + x[n]cos(2Ωcn)}
If there is distortion (i.e., r[n] ≠ t[n]), then write y[n] instead of
x[n] (and hope that in the noise-free case y[.] is related to x[.] by
an approximately LTI relationship!)
• What does the spectrum of d[n], i.e., D(Ω), look like?
• What constraint on the bandwidth of x[n] is needed for perfect
recovery of x[n]?
6.02 Fall 2012 Lecture 16 Slide #10
Demodulation Frequency Diagram
6.02 Fall 2012 Lecture 16 Slide #11
Im(Tk)
+Ωc
Ωc
A/2
B/2
k
Re(T )
Im(Dk)
2Ωc
A/2
B/2
+2Ωc
Ωm
Ωm
Re(Dk)
R(Ω)=T(Ω)
D(Ω)
What we want
Note combining of signals around 0
results in doubling of amplitude
Demodulation + LPF
6.02 Fall 2012 Lecture 16 Slide #12
×
r[n]
cos(Ωc n)
d[n] LPF x[n]
Cutoff @ ±Ωc
Gain = 2
Phase Error In Demodulation
When the receiver oscillator is out of phase with the transmitter:
d[n]= r[n]⋅cos(Ωcn −ϕ) = x[n]⋅cos(Ωcn)⋅cos(Ωcn −ϕ)
6.02 Fall 2012 Lecture 16 Slide #13
y[n]= x[n].cos(ϕ)
So a phase error of φ results in amplitude scaling by cos(φ).
Note: in the extreme case where φ=/2, we are demodulating
by a sine rather than a cosine, and we get y[n]=0 .
But
cos(Ωcn).cos(Ωcn −ϕ) = 0.5{cos(ϕ)+cos(2Ωcn −ϕ)}
It follows that the demodulated output, after the LPF of
gain 2, is
Demodulation with sin(Ω
6.02 Fall 2012 Lecture 16 Slide #14
Ωcn)
+
+2Ωc
–2Ωc
D(Ω)
Im(Tk)
+Ωc
Ωc
A/2
B/2
Re(Tk)
R(Ω)
… produces
6.02 Fall 2012 Lecture 16 Slide #15
Note combining of signals around 0
results in cancellation!
Channel Delay
6.02 Fall 2012 Lecture 16 Slide #16
×
t[n] d[n]
LPF y[n]
Cutoff @ ±kin
Gain = 2
×
cos(Ωcn)
x[n] D
Time delay of D samples
d[n]= t[n − D]⋅cos(Ωcn)
= x[n − D]⋅cos[Ωc (n − D)]⋅cos(Ωcn)
Passing this through the LPF: Looks like a phase error
of ΩcD
y[n]= x[n − D]⋅cos(ΩcD)
D)
cos(Ωcn)
Very similar math to the previous “phase error” case:
If ΩcD is an odd multiple of /2, then y[n]=0 !!
Fixing Phase Problems in the Receiver
So phase errors and channel delay both result in a scaling of the
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
6.02 Fall 2012 Lecture 16 Slide #17
×
cos(Ωcn)
LPF I[n] = x[n-D]·cos(θ)
× LPF Q[n] = x[n-D]·sin(θ)
From
channel
θ = φ - ΩcD
sin(Ωcn)
Quadrature Demodulation
If we let
w[n]= I[n]+ jQ[n]
then
w[n]
6.02 Fall 2012 Lecture 16 Slide #18
= I[n]2
+Q[n]2
=| x[n − D]| cos2
θ +sin2
θ
=| x[n − D]|
x[n-D]cos(θ)
x[n-D]sin(θ)
I
θ
jQ
Constellation diagrams:
transmitter receiver
I
Q
I
Q
x[n] = { 0, 1 }
OK for recovering x[n] if it
never goes negative, as in
on-off keying
Dealing With Phase Ambiguity
in Bipolar Modulation
In bipolar modulation (x[n]=±1), also called Binary
Phase Shift Keying (BPSK) since the modulated
carrier changes phase by /2 when x[n] switches
levels, the received constellation will be rotated
with respect to the transmitter’s constellation.
Which phase corresponds to which bit?
6.02 Fall 2012 Lecture 16 Slide #19
Q
I
Different fixes:
1. Send an agreed-on sign-definite preamble
2. Transmit differentially encoded bits, e.g., transmit a “1” by
stepping the phase by , transmit a “0” by not changing the
phase
QPSK Modulation
6.02 Fall 2012 Lecture 16 Slide #20
×
cos(Ωcn)
×
sin(Ωcn) + t[n]
I[n]
Q[n]
I
Q
(-1,1) (1,1)
(-1,-1) (1,-1)
We can use the quadrature scheme at the transmitter too:
Samples from first bit stream
Samples from second bit stream
http://en.wikipedia.org/wiki/Phase-shift_keying
6.02 Fall 2012 Lecture 16 Slide #21
Phase Shift Keying underlies many
familiar modulation schemes
The wireless LAN standard, IEEE 802.11b-1999, uses a variety of different PSKs depending on the data-
rate required. At the basic-rate of 1 Mbit/s, it uses DBPSK (differential BPSK). To provide the extended-rate
of 2 Mbit/s, DQPSK is used. In reaching 5.5 Mbit/s and the full-rate of 11 Mbit/s, QPSK is employed, but has
to be coupled with complementary code keying. The higher-speed wireless LAN standard,
IEEE 802.11g-2003 has eight data rates: 6, 9, 12, 18, 24, 36, 48 and 54 Mbit/s. The 6 and 9 Mbit/s modes
use OFDM modulation where each sub-carrier is BPSK modulated. The 12 and 18 Mbit/s modes use
OFDM with QPSK. The fastest four modes use OFDM with forms of quadrature amplitude modulation.
Because of its simplicity BPSK is appropriate for low-cost passive transmitters, and is used
in RFID standards such as ISO/IEC 14443 which has been adopted for biometric passports, credit cards
such as American Express's ExpressPay, and many other applications.
Bluetooth 2 will use (p/4)-DQPSK at its lower rate (2 Mbit/s) and 8-DPSK at its higher rate (3 Mbit/s) when
the link between the two devices is sufficiently robust. Bluetooth 1 modulates with
Gaussian minimum-shift keying, a binary scheme, so either modulation choice in version 2 will yield a higher
data-rate. A similar technology, IEEE 802.15.4 (the wireless standard used by ZigBee) also relies on PSK.
IEEE 802.15.4 allows the use of two frequency bands: 868–915 MHz using BPSK and at 2.4 GHz using
OQPSK.
Multiple Transmitters:
Frequency Division Multiplexing (FDM)
6.02 Fall 2012
×
xB[n]
×
xR[n]
×
xG[n]
cos(ΩBn)
cos(ΩRn)
cos(ΩGn)
Lecture 16 Slide #22
+
Channel “performs addition” by
superposing signals (“voltages”) from
different frequency bands.
Choose bandwidths and Ωc’s so as to
avoid overlap! Once signals combine at
a given frequency, can’t be undone…
LPF cutoff needs to be half the minimum
separation between carriers
-ΩG -ΩR -ΩB ΩB ΩR ΩG
AM Radio
AM radio stations are on 520 – 1610 kHz (‘medium wave”)
in the US, with carrier frequencies of different stations
spaced 10 kHz apart.
Physical effects very much affect operation. e.g., EM signals
at these frequencies propagate much further at night (by
“skywave” through the ionosphere) than during the day (100’s
of miles by “groundwave” diffracting around the earth’s
surface), so transmit power may have to be lowered at night!
6.02 Fall 2012 Lecture 16 Slide #23
MIT OpenCourseWare
http://ocw.mit.edu
6.02 Introduction to EECS II: Digital Communication Systems
Fall 2012
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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Digital communcation for engineerings.pdf

  • 1. 6.02 Fall 2012 Lecture 16 Slide #1 6.02 Fall 2012 Lecture #16 • DTFT vs DTFS • Modulation/Demodulation • Frequency Division Multiplexing (FDM)
  • 2. Fast Fourier Transform (FFT) to compute samples of the DTFT for signals of finite duration P−1 1 X(Ωk ) = ∑x[m]e− jΩkm , x[n]= m=0 P 6.02 Fall 2012 Lecture 16 Slide #2 For an x[n] that is zero outside of the interval [0,L-1], choose P ≥ L (with P preferably a power of 2; we’ll assume that it’s at least a multiple of 2, i.e., even): (P/2)−1 ∑ X(Ωk )ejΩkn k=−P/2 where Ωk = k(2π/P), and k ranges from –P/2 to (P/2)–1, or over any P successive integers. Simpler notation: X(Ωk) = Xk
  • 3. Where do the Ω 6.02 Fall 2012 Lecture 16 Slide #3 Ωk live? e.g., for P=6 (even) – 0 Ω0 Ω1 Ω2 Ω3 Ω3 Ω2 Ω1 exp(jΩ0) exp(jΩ1) exp(jΩ2) exp(jΩ3) = exp(jΩ3) exp(jΩ1) exp(jΩ2) . 1 –1 j –j
  • 4. Fast Fourier Transform (FFT) to compute samples of the DTFT for signals of finite duration P−1 1 X(Ωk ) = ∑x[m]e− jΩkm , x[n]= m=0 P 6.02 Fall 2012 Lecture 16 Slide #4 For an x[n] that is zero outside of the interval [0,L-1], choose P ≥ L (with P preferably a power of 2; we’ll assume that it’s at least a multiple of 2, i.e., even): (P/2)−1 ∑ X(Ωk )ejΩkn k=−P/2 where Ωk = k(2π/P), and k ranges from –P/2 to (P/2)–1, or over any P successive integers. Simpler notation: X(Ωk) = Xk Note that X(Ω ) = X() = X(– P/2 ) =X(Ω-P/2). The above formulas have essentially the same structure, and are both efficiently computed, with Plog(P) computations, by the FFT.
  • 5. Some further details, assuming real x[n] P−1 X(Ω ) = ∑x[m]e− jΩkm k m=0 • X(0) = sum of x[m] = real • X() = X(–) = alternating sum of x[m] = real • In general P-2 other complex values, but X(–Ωk) = X*(Ωk) • So: total of P numbers to be determined, given P values of x[m] 1 x[n]= P 6.02 Fall 2012 Lecture 16 Slide #5 (P/2)−1 ∑ X(Ωk )ejΩkn k=−P/2 • Evaluating this eqn. for n in [0,P-1] recovers the original x[n] in this interval • Evaluating it for n outside this interval results in periodic replication of the values in [0,P-1], producing a periodic signal x[n] • So this eqn. is also called a DT Fourier Series (DTFS) for the periodic signal x[n]. Notation: Ak=X(Ωk)/P=Xk/P, Fourier coefficient.
  • 6. Why the periodicity of x 6.02 Fall 2012 Lecture 16 Slide #6 x[n] is irrelevant in many applications h[.] x[n] y[n] Suppose x[n] is nonzero only over the time interval [0 , nx], and h[n] is nonzero only over the time interval [0 , nh] . In what time interval can the non-zero values of y[n] be guaranteed to lie? The interval [0 , nx + nh] . Since all the action we are interested in is confined to this interval, choose P – 1 ≥ nx + nh . It’s now irrelevant what happens outside [0,P–1]. So we can use the FFT to go back and forth between samples of X(Ω), Η(Ω), Y(Ω) in the frequency domain and time-domain behavior in [0,P-1].
  • 7. Back to Modulation/Demodulation • You have: a signal x[n] at baseband (i.e., centered around 0 frequency) • You want: the same signal, but centered around some specific frequency Ωc • Modulation: convert from baseband up to Ωc , to get t[n] • Demodulation: convert from Ωc down to baseband 6.02 Fall 2012 Lecture 16 Slide #7 Re(Xk) Im(Xk) +Ωm Ωm Re(Tk) Im(Tk) +Ωc Ωc modulation demodulation Signal centered at 0 Signal centered at Ωc
  • 8. Modulation by Heterodyning or Amplitude Modulation (AM) 6.02 Fall 2012 Lecture 16 Slide #8 × x[n] cos(Ωcn) t[n] Im(Tk) +Ωc Ωc B/2 k A/2 Re(T ) i.e., just replicate baseband signal at ±Ωc, and scale by ½. k Im(Xk) +Ωm Ωm A B Re(X ) To get this nice picture, the baseband signal needs to be band-limited to some range of frequencies [–Ωm,Ωm], where Ωm ≤ Ωc
  • 9. 6.02 Fall 2012 Lecture 16 Slide #9 0 Hz 1000 Hz –1000 Hz Not great band-limiting, but maybe we can get away with it!
  • 10. At the Receiver: Demodulation • In principle, this is (as easy as) modulation again: If the received signal is r[n] = x[n]cos(Ωcn) = t[n], (no distortion or noise) then simply compute d[n] = r[n]cos(Ωcn) = x[n]cos2(Ωcn) = 0.5 {x[n] + x[n]cos(2Ωcn)} If there is distortion (i.e., r[n] ≠ t[n]), then write y[n] instead of x[n] (and hope that in the noise-free case y[.] is related to x[.] by an approximately LTI relationship!) • What does the spectrum of d[n], i.e., D(Ω), look like? • What constraint on the bandwidth of x[n] is needed for perfect recovery of x[n]? 6.02 Fall 2012 Lecture 16 Slide #10
  • 11. Demodulation Frequency Diagram 6.02 Fall 2012 Lecture 16 Slide #11 Im(Tk) +Ωc Ωc A/2 B/2 k Re(T ) Im(Dk) 2Ωc A/2 B/2 +2Ωc Ωm Ωm Re(Dk) R(Ω)=T(Ω) D(Ω) What we want Note combining of signals around 0 results in doubling of amplitude
  • 12. Demodulation + LPF 6.02 Fall 2012 Lecture 16 Slide #12 × r[n] cos(Ωc n) d[n] LPF x[n] Cutoff @ ±Ωc Gain = 2
  • 13. Phase Error In Demodulation When the receiver oscillator is out of phase with the transmitter: d[n]= r[n]⋅cos(Ωcn −ϕ) = x[n]⋅cos(Ωcn)⋅cos(Ωcn −ϕ) 6.02 Fall 2012 Lecture 16 Slide #13 y[n]= x[n].cos(ϕ) So a phase error of φ results in amplitude scaling by cos(φ). Note: in the extreme case where φ=/2, we are demodulating by a sine rather than a cosine, and we get y[n]=0 . But cos(Ωcn).cos(Ωcn −ϕ) = 0.5{cos(ϕ)+cos(2Ωcn −ϕ)} It follows that the demodulated output, after the LPF of gain 2, is
  • 14. Demodulation with sin(Ω 6.02 Fall 2012 Lecture 16 Slide #14 Ωcn) + +2Ωc –2Ωc D(Ω) Im(Tk) +Ωc Ωc A/2 B/2 Re(Tk) R(Ω)
  • 15. … produces 6.02 Fall 2012 Lecture 16 Slide #15 Note combining of signals around 0 results in cancellation!
  • 16. Channel Delay 6.02 Fall 2012 Lecture 16 Slide #16 × t[n] d[n] LPF y[n] Cutoff @ ±kin Gain = 2 × cos(Ωcn) x[n] D Time delay of D samples d[n]= t[n − D]⋅cos(Ωcn) = x[n − D]⋅cos[Ωc (n − D)]⋅cos(Ωcn) Passing this through the LPF: Looks like a phase error of ΩcD y[n]= x[n − D]⋅cos(ΩcD) D) cos(Ωcn) Very similar math to the previous “phase error” case: If ΩcD is an odd multiple of /2, then y[n]=0 !!
  • 17. Fixing Phase Problems in the Receiver So phase errors and channel delay both result in a scaling of the 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 6.02 Fall 2012 Lecture 16 Slide #17 × cos(Ωcn) LPF I[n] = x[n-D]·cos(θ) × LPF Q[n] = x[n-D]·sin(θ) From channel θ = φ - ΩcD sin(Ωcn)
  • 18. Quadrature Demodulation If we let w[n]= I[n]+ jQ[n] then w[n] 6.02 Fall 2012 Lecture 16 Slide #18 = I[n]2 +Q[n]2 =| x[n − D]| cos2 θ +sin2 θ =| x[n − D]| x[n-D]cos(θ) x[n-D]sin(θ) I θ jQ Constellation diagrams: transmitter receiver I Q I Q x[n] = { 0, 1 } OK for recovering x[n] if it never goes negative, as in on-off keying
  • 19. Dealing With Phase Ambiguity in Bipolar Modulation In bipolar modulation (x[n]=±1), also called Binary Phase Shift Keying (BPSK) since the modulated carrier changes phase by /2 when x[n] switches levels, the received constellation will be rotated with respect to the transmitter’s constellation. Which phase corresponds to which bit? 6.02 Fall 2012 Lecture 16 Slide #19 Q I Different fixes: 1. Send an agreed-on sign-definite preamble 2. Transmit differentially encoded bits, e.g., transmit a “1” by stepping the phase by , transmit a “0” by not changing the phase
  • 20. QPSK Modulation 6.02 Fall 2012 Lecture 16 Slide #20 × cos(Ωcn) × sin(Ωcn) + t[n] I[n] Q[n] I Q (-1,1) (1,1) (-1,-1) (1,-1) We can use the quadrature scheme at the transmitter too: Samples from first bit stream Samples from second bit stream
  • 21. http://en.wikipedia.org/wiki/Phase-shift_keying 6.02 Fall 2012 Lecture 16 Slide #21 Phase Shift Keying underlies many familiar modulation schemes The wireless LAN standard, IEEE 802.11b-1999, uses a variety of different PSKs depending on the data- rate required. At the basic-rate of 1 Mbit/s, it uses DBPSK (differential BPSK). To provide the extended-rate of 2 Mbit/s, DQPSK is used. In reaching 5.5 Mbit/s and the full-rate of 11 Mbit/s, QPSK is employed, but has to be coupled with complementary code keying. The higher-speed wireless LAN standard, IEEE 802.11g-2003 has eight data rates: 6, 9, 12, 18, 24, 36, 48 and 54 Mbit/s. The 6 and 9 Mbit/s modes use OFDM modulation where each sub-carrier is BPSK modulated. The 12 and 18 Mbit/s modes use OFDM with QPSK. The fastest four modes use OFDM with forms of quadrature amplitude modulation. Because of its simplicity BPSK is appropriate for low-cost passive transmitters, and is used in RFID standards such as ISO/IEC 14443 which has been adopted for biometric passports, credit cards such as American Express's ExpressPay, and many other applications. Bluetooth 2 will use (p/4)-DQPSK at its lower rate (2 Mbit/s) and 8-DPSK at its higher rate (3 Mbit/s) when the link between the two devices is sufficiently robust. Bluetooth 1 modulates with Gaussian minimum-shift keying, a binary scheme, so either modulation choice in version 2 will yield a higher data-rate. A similar technology, IEEE 802.15.4 (the wireless standard used by ZigBee) also relies on PSK. IEEE 802.15.4 allows the use of two frequency bands: 868–915 MHz using BPSK and at 2.4 GHz using OQPSK.
  • 22. Multiple Transmitters: Frequency Division Multiplexing (FDM) 6.02 Fall 2012 × xB[n] × xR[n] × xG[n] cos(ΩBn) cos(ΩRn) cos(ΩGn) Lecture 16 Slide #22 + Channel “performs addition” by superposing signals (“voltages”) from different frequency bands. Choose bandwidths and Ωc’s so as to avoid overlap! Once signals combine at a given frequency, can’t be undone… LPF cutoff needs to be half the minimum separation between carriers -ΩG -ΩR -ΩB ΩB ΩR ΩG
  • 23. AM Radio AM radio stations are on 520 – 1610 kHz (‘medium wave”) in the US, with carrier frequencies of different stations spaced 10 kHz apart. Physical effects very much affect operation. e.g., EM signals at these frequencies propagate much further at night (by “skywave” through the ionosphere) than during the day (100’s of miles by “groundwave” diffracting around the earth’s surface), so transmit power may have to be lowered at night! 6.02 Fall 2012 Lecture 16 Slide #23
  • 24. MIT OpenCourseWare http://ocw.mit.edu 6.02 Introduction to EECS II: Digital Communication Systems Fall 2012 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.