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Unformatted text preview: ynchronization.
This can be solved with more complex receivers
(synchronous or quadrature), but this costs money.
Another alternative is to use a modulation scheme that makes demodulation
easy. This is what AM does.
If m(t) is the message, we add a constant before modulating
xAM (t) = [A + m(t)] cos(ωct)
where A is chosen so that A + m(t) is always positive. EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 15 The reason for doing this can be appreciated from the following plots:
A + m(t )
m(t ) t t
[A + m(t )] m(t ) [A + m(t )] cos(!0t ) m(t ) cos(!ct ) t
−m(t ) t
−[A + m(t )] The signal we want is the envelope of xAM (t).
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 16 Simple receivers for these signals are envelope detectors. One possible
receiver ﬁrst rectiﬁes the signal, and then lowpass ﬁlters. The system looks
like:
m(t ) +
A × Propagation [A + m(t )] cos(!ct ) 2 H (j ω ) A + m(t )
ω · cos(!ct )
Receiver Transmitter The constant A is suppressed using a DC block (capacitive coupling). EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 17 One way of understanding what this does is to assume the lowpass ﬁlter is
a simple RC circuit [A + m(t )] cos(!ct ) +
− +
C [A + m(t )] R
− If we look at what the AM signal and its rectiﬁed version,
[A + m(t )] [A + m(t )] cos(!0t ) t [A + m(t )] [A + m(t )] cos(!ct ) t −[A + m(t )] we see that if the time constant RC is short compared to the variation in
m(t) and long compared to ωc, the output will track the envelope.
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 18 This makes it look like we can recover something like [A + m(t)], but it
may not be perfect. A better perspective is to look at the spectrum of
[A + m(t)] cos(ωct),
F [(A + m(t)) cos(ωct)] = 1
[2π Aδ (ω ) + M (j ω )] ∗ F [ cos(ωct)]
2π But what is F [ cos(ωct)]? It has a Fourier series:
cos(ωct) = ∞
jnω0 t Dne = n=−∞ ∞
Dnej 2nωct n=−∞ since the period of  cos(ωct) is half of the period of cos(ωct), so ω0 = 2ωc.
Interestingly, the actual values of Dn don’t really matter as long as D0 = 0
(is it?).
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 19 The Fourier transform is then F [ cos(ωct)] = F Dnej 2nωct n=−∞ ∞
= ∞
n=−∞ Dn2πδ (ω − 2nωc) These are an array of δ ’s separated by 2ωc.
cos(ωc t)  cos(ωc t) T0 T0
2 ω0 = 2ωc t πδ (ω + ωc )
−ωc 2π D0 πδ (ω − ωc )
ωc t 2π D−2 ω EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly −4ωc 2π D1 2π D−1 −2ωc 0 2ωc 2π D2
4ωc ω 20 The output spectrum is then
F [(A + m(t)) cos(ωct)] =
= 1
[2π Aδ (ω ) + M (j ω )] ∗ F [ cos(ωct)]
2π
∞
1
[2π Aδ (ω ) + M (j ω )] ∗
Dn2πδ (ω − 2nωc)
2π
n=−∞
∞
= [2π Aδ (ω ) + M (j ω )] ∗ (Dnδ (ω − 2nωc)) n=−∞ = ∞
n=−∞ Dn[2π Aδ (ω − 2nωc) + M (j (ω − 2nωc))] EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 21 We have replicas of the spectrum of [A + m(t)] at multiples of 2ωc,
Ideal Lowpass
D−1 [2π Aδ (ω + 2ωc ) + M (j ω + 2ωc )] −2!c −!c D0 [2π Aδ (ω ) + M (j ω )]
D1 [2π Aδ (ω − 2ωc ) + M (j ω − 2ωc )] !c 2!c ! A lowpass ﬁlter perfectly extracts [2π Aδ (ω ) + M (j ω )]. We can perfectly
recover [A + m(t)]!
Does this still work if the signal is phase shifted? EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 22 Conclusion • Modulation and demodulation are important components of many
systems
– Communications
– Radar, sonar, ultrasound imaging, MRI
– Optics
• Many diﬀerent modulation methods
• Many diﬀerent approaches to extracting the signal EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 23...
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This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.
 Fall '13
 Mukamel
 Signal Processing

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