Am modulation the dsb sc systems weve been looking at

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online