Problem 3. (6%) In our discussion on synchronised variants of AM, we have always been assum- ing perfect synchronisation

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Problem 3. (6%) In our discussion on synchronised variants of AM, we have always been assum- ing perfect synchronisation

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Problem 3 6 In Our Discussion On Synchronised Variants Of Am We Have Always Been Assum Ing Perfect Synchronisation 1
Problem 3 6 In Our Discussion On Synchronised Variants Of Am We Have Always Been Assum Ing Perfect Synchronisation 1 (104.9 KiB) Viewed 33 times
Problem 3. (6%) In our discussion on synchronised variants of AM, we have always been assum- ing perfect synchronisation is guaranteed. This problem considers the opposite. Consider a generic synchronous detector: s(t) u(t) m(t) Prod. Mod L.P.F. 2(t) Figure: Synchronous Detector. wherein the input signal s(t) is assumed to be a DSB-SC modulated transmission signal. Suppose the local oscillator at the modulator produces a carrier signal c(t) = A cos(29ft), while the one at the demodulator produces the same frequency, but with some phase difference, i.e. (t) = Ā_COS(27fc4+ ) The objective of this problem is to investigate a phase detector. We mention that the phase difference can be mitigated by sending some prior header. Assume the header is a constant direct current signal m(t) = 1, which implies s(t)=1.c(t) = A.cos(2xft). We claim the system: a(t) Prod. Mod J ( )dt u c(t) L.O. c(t) arctan(5) 90 phase u b(t) Prod. Mod đ ( dt Figure: Phase Detector. is capable of detecting the phase difference . Let us verify this by the following steps. (a) Evaluate the signal a(t) into a form such that it is a sinusoidal signal adding a constant, i.e. a(t) = 41 +0.008(21f*++ **) where C1, C2,8* ©* are some real constants. Then, do the same on b(t). (b) Compute u = = a(t)dt for T = 4/fe and v = "(t)dt for T = = 4/8 (c) Verify that the system output = 0.
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