4/4/2012web.mit.edu/6.02/www/s2012/handouts/l15_slides.pdf · 4/4/2012 1 6.02 spring 2012 lecture...
TRANSCRIPT
4/4/2012
1
6.02 Spring 2012 Lecture 15, Slide #1
6.02 Fall 2011
Lecture #15
• More on signal spectra
• Modulation & demodulation
6.02 Spring 2012 Lecture 15, Slide #2
Again, by inspection: since the cos
and sin are at different frequencies,
we can analyze them separately.
Again, P is odd here (=11), so the
end points of the frequency scale
are at ±(π– (π/P)), not ±π.
A0 = average value = 1
A±3 = 2(1/2) = 1 [from cos term]
A-5 = –3(j/2) = –1.5j [from sin term]
A5 = –3(–j/2) = 1.5j
Ak = 0 otherwise
6.02 Spring 2012 Lecture 15, Slide #3
The DTFS is also good for
finite-duration signals!
Claim: Over any contiguous interval of length P that we may be
interested in --- say n=0,1,…,P–1 for concreteness --- an arbitrary
DT signal x[n] can be written in the form
What’s going on here? If we know we will only be interested in
the interval [0,P–1], then it doesn’t matter that our
representation above will create periodically repeating
extensions outside the interval of interest.
6.02 Spring 2012 Lecture 15, Slide #4
Spectrum of Digital Transmissions
|Ak|
Ak
n
k
Time Sequence
Frequency Spectrum P=196
4/4/2012
2
6.02 Spring 2012 Lecture 15, Slide #5
Application
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 , then use the DTFS to
represent x[n] and y[n] over this interval.
This is actually the much more common use of the DTFS!
6.02 Spring 2012 Lecture 15, Slide #6
Spot Quiz
Determine non-zero spectral components:
1. Ak=? , when P=5, k= -2 … 2
2. Ak=?, when P=10, k= -5 … 4
)5
2cos(][ nnx
For P=5, W1 = 2/5, and
so A-1 = 0.5 and A1 = 0.5 and Ak =0 for other k.
For P=10, W1 = 2/10, and
so A-2 = 0.5 and A2 = 0.5 and Ak =0 for other k.
njnjeenx 11 5.05.0][
WW
njnjnjnjeeeenx 2211 5.05.05.05.0][
22 WWWW
6.02 Spring 2012 Lecture 15, Slide #7
From Baseband to Modulated Signal, and
Back
codeword
bits in
codeword
bits out
101110101 DAC
ADC
NOISY & DISTORTING ANALOG CHANNEL
modulate
101110101 demodulate
& filter
generate
digitized
symbols
sample &
threshold
x[n]
y[n]
6.02 Spring 2012 Lecture 15, Slide #8
Using Some Piece of the Spectrum
• You have: a band-limited signal x[n] at baseband
(i.e., centered around 0 frequency).
Re(ak)
Im(ak)
+kx −kx
Re(ak)
Im(ak)
+kc −kc
Signal centered at 0 Signal centered at kc
modulation
demodulation
Spectrum of
baseband
signal
Spectrum of
transmitted
signal
• You want: the same signal, but centered around
some specific frequency kcW1.
• Modulation: convert from baseband up to kcW1.
• Demodulation: convert from kcW1 down
to baseband
4/4/2012
3
6.02 Spring 2012 Lecture 15, Slide #9
Modulation
× x[n]
cos(kcW1n)
t[n]
WW
W
njknjk
k
kk
njk
kcc
x
x
eeeAnt 111
2
1
2
1][
Re(ak)
Im(ak)
+kc −kc
A/2
A/2
For band-limited signal
Ak are nonzero only for
small range of ±kx i.e., just replicate baseband
signal at ±kc, and scale
by ½.
W
W
x
x
c
x
x
c
k
kk
nkkj
k
k
kk
nkkj
k eAeA 11
2
1
2
1
6.02 Spring 2012 Lecture 15, Slide #10
Effect of Band-limiting a Transmission
6.02 Spring 2012 Lecture 15, Slide #11
How Low Can We Go?
7 samples/bit → 14 samples/period → k=(N/14)=(196/14)=14
6.02 Spring 2012 Lecture 15, Slide #12
Example: Modulation (time) Shaped pulses!
Chosen to band-limit
the transmit signal Baseband input x[n]
Carrier signal
Transmitted signal t[n]
4/4/2012
4
6.02 Spring 2012 Lecture 15, Slide #13
Example: Modulation
(freq domain picture) Band-limited x[n] cos(35W1n) t[n]
6.02 Spring 2012 Lecture 15, Slide #14
Demodulation
× t[n]
cos(kcW1n)
z[n]
What we want
Hmm. So z[n] has what
we want at baseband,
but has signal we don’t
want at ±2kcW1
Assuming no
distortion or
noise on
channel, so
what was
transmitted
is received
6.02 Spring 2012 Lecture 15, Slide #15
Demodulation Frequency Diagram
Re(ak)
Im(ak)
+kc −kc
A/2
A/2
Re(ak)
Im(ak)
−2kc
A/2
A/2
+2kc
t[n]
z[n]
Re(ak)
Im(ak)
+kc −kc
1/2
cos(kcW1n)
Re(ak)
Im(ak)
−2kc
A/2
A/2
+2kc
+
6.02 Spring 2012 Lecture 15, Slide #16
Demodulation Frequency Diagram
Re(ak)
Im(ak)
+kc −kc
A/2
A/2
Re(ak)
Im(ak)
−2kc
A/2
A/2
+2kc
t[n]
z[n]
What we want
Note combining of signals around 0
results in doubling of amplitude
Re(ak)
Im(ak)
+kc −kc
1/2
cos(kcW1n)
4/4/2012
5
6.02 Spring 2012 Lecture 15, Slide #17
Example: Demodulation (freq)
x[n] t[n] z[n]
Only want these frequencies…
6.02 Spring 2012 Lecture 15, Slide #18
Demodulation + LPF
× t[n] z[n] LPF y[n]
Cutoff @ ±kx
Gain = 2 cos(kcW1n)
6.02 Spring 2012 Lecture 15, Slide #19
Demodulation
× y[n]
cos(kcW1n)
z[n]
W
W
W
WW
W
W
WW
WW
WW
WW
x
x
x
x
c
x
x
c
cc
x
x
c
x
x
c
cc
k
kk
njk
kcchcch
k
kk
nkkj
kcch
k
kk
nkkj
kcch
njknjkk
kk
nkkj
kcch
k
kk
nkkj
kcch
njknjk
eAkkHkkH
eAkkHeAkkH
eeeAkkHeAkkH
eenynz
1
11
1111
11
))()((4
1
)(4
1)(
4
1
2
1
2
1)(
2
1)(
2
1
2
1
2
1][][
11
2
1
2
1
11
What we want
Hmm. So z[n] has what
we want at baseband,
but has signal we don’t
want at ±2kcW1
t[n] Channel
Hch(W)