# Am modulation the dsb sc systems weve been looking at

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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 09-10, 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 09-10, 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 09-10, 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 09-10, 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 09-10, 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 09-10, 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 09-10, 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 09-10, 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 09-10, 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.

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