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Demodulation is about recovering the original signal--Crystal Radio ExampleSignal Flow in Crystal Radio--Circuit Level IssuesAbout “Peak Detection” and WaveformsCondition for Optimum RCAbout the Equation for “optimum”…And the answer is…What we DON’T want--Clipping1Demodulation is about recovering the original signal--Crystal Radio ExampleTuningCircuitDemodulationCircuitFilter(Mechanical)FMAMAntenna = Long WireA simple Diode!Basically a “tapped”Inductor (L) and variableCapacitor (C)(envelop ofAM Signal)We’ll not spend a lot of time on the AM “crystal radio”, although I love it dearly as a COOL, ultra-minimal piece of electronics--Imagine, you get radio FREE with no batteries required.But…The things we will look at and actually do a bit in lab is to consider the “peak detector” (I.e. the means for demodulating the AM signal)From a block diagram point of view, the circuit has a tuning component (frequency selective filter) attached to the antenna (basically a wire for the basic X-tal radio).The demodulation consists of a diode (called the “crystal” from the good old days of “Empire of the Air”…movie we’ll watch) and an R-C filter to get rid of the carrier frequency.In the Radio Shack version there is no “C” needed; your ear bones can’t respond to the carrier so they act as “the filter”.The following slide gives a more electronics-oriented view of the circuit…AM Demodulation (peak detect.)2Signal Flow in Crystal Radio--Circuit Level Issues+V-Vtimetimemusicground=0V+V (only)Wire=AntennaBWfoFilter:•fo set by LC•BW set by RLCfrequency“KX” “KY” “KZ”“tuning”So, here’s the incoming (modulated) signal and the parallel L-C (so-called “tank” circuit) that is hopefully selective enough (having a high enough “Q”--a term that you’ll soon come to know and love) that “tunes” the radio to the desired frequency.Selective enough means that you receive “KX” and don’t also get KY and KZ (for AM you definitely won’t get KZSU:)The diode rectified signal looks as shown; basically we keep thepositive side of the signals (referenced to GROUND)[Comment about X-tal Radios; To get a good signal, you do indeed need a solid ground…an interesting challenge unto itself]Back to the detection…Now, our challenge is to keep the envelop and “get rid of” the carrier…basically to filter it out.Per the NEW EE101A diodes are used to create “power supplies” (a lab experience now in progress:). Here we are using the incoming AM signal to create a power supply (I.e. no battery needed) where the “ripple” is the information (music etc.) that we want to hear.3idνdtodνdtGenerally..we..want :odνdt≥idνdtAbout “Peak Detection” and WaveformsSo, let’s look at a cycle of the “music” that rides on top of the much higher frequency carrier.By analogy to the power supply example we will use an R-C filter to decay at a rate that hopefully follows the modulated signal but doesn’t decay too fast and therefore follow the carrier.This plot shows us that the modulated signal has a slope and theresult of the R-C filter will also have a slope.Generally speaking, we want the slope of the filtering to be steeper than the envelope.If it is NOT steeper, we’re not following the modulating signal (the very last slide in this set corresponds to “Diagonal Clipping”-- the consequence of going too slowly).If it’s TOO STEEP, we’re not filtering out the carrier.OK…Let’s try and put that in a more formal (mathematical) form4Condition for Optimum RCReqVo(t)RC∆ν“refresh” needed fromnext phase pulse, via diode1mω= RC ⋅112m        − 1cωTmωTIncoming SignalWhat happens “peak to peak” (after diode)Here we define the circuit to be considered (and used in lab!) a bit more formally.We have the diode that gets us half-wave rectification.Going from one carrier peak to the next, we have an RC fall-off as shown.At the end of the following few pages we will determine an “optimum” C value (in terms of R, m and ωm)This figure simply is showing graphically both the circuit RC inrelationship to the carrier period and also how that compares to the period of the modulating signal.The BOXED equation tells us the final result in terms of how theRC and modulation index should ultimately relate to the modulating frequency…Now let’s take a very quick stroll through the derivation of the real inequality that is involved.5About the Equation for “optimum”…vi= Vi(1 + mcosωmt)vo= Ve−tTRCVTRCe−tTRC≥ Vimωmsinωmtequating_ vi= vo_(at _ some _ t)1+ mcosωmtTRC≥ mωmsinωmtGenerally..we ..want:odνdt≥idνdtGiven that we want…This is THE constraint equation…now let’s make it useful.Assume that the incoming (envelop) waveform looks as shown above (first equation)The R-C circuit will have a response that looks like that shown in the second equation.Taking the derivative of both equations with respect to time andapplying the desired inequality (per the previous slide), the third equation is obtained.Also, at some point in time the top two equations can be equatedand that result, combined with the third equation, gives the fourth equation--an inequality that relates:•RC time constant TRC•Modulation index “m”•Modulation frequency ωmUnfortunately, how to work with this equation is NOT so easy andanother page of “equation hacking” is needed.[We won’t spend much time on the “hacking”…but we need to get to the final result!]6And the answer is…after_some_"trig"_manipulations...1− m2()12m≥ωmTRCwhere _ TRC= RC _ then :C ≤1− m2()12m1ωmR=1m2−1ωmRThe KEY equation for C (in terms of:m,ωm&R)There are some trigonometric identities that allow us to simplify the inequality from the previous page to the one shown here.The RC time constant is defined as shown.The [bracked/boxed] equation tells us how small to make C (in the RC) in terms of ωm, m and R.If C is larger, then we start to “loose” information in the envelop due to diagonal clipping.If C is too small (I.e. why not make it ZERO?) then we certainlyfollow the envelop but we are NOT getting rid of the carrier…Basically, if C is too small we’ve kept too much and we haven’t really “demodulated” the signal.7What we DON’T want--Clippingenvelope“DiagonalClipping” tt’=0Ve(t)Considering “the dark side”… what we DON’T want…If the slopes in the above inequality


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Stanford EE 133 - AM Demodulation

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