# ch03 - Chapter 3 Basic Modulation Techniques 3.1 Problems...

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Unformatted text preview: Chapter 3 Basic Modulation Techniques 3.1 Problems Problem 3.1 The demodulated output, in general, is y D ( t ) = Lp { x c ( t ) 2cos[ ω c t + θ ( t )] } where Lp {&} denotes the lowpass portion of the argument. With x c ( t ) = A c m ( t ) cos [ ω c t + φ ] the demodulated output becomes y D ( t ) = Lp { 2 A c m ( t ) cos [ ω c t + φ ] cos [ ω c t + θ ( t )] } Performing the indicated multiplication and taking the lowpass portion yields y D ( t ) = A c m ( t ) cos [ θ ( t ) − φ ] If θ ( t ) = θ (a constant), the demodulated output becomes y D ( t ) = A c m ( t ) cos [ θ − φ ] Letting A c = 1 gives the error ε ( t ) = m ( t ) [1 − cos ( θ − φ )] The mean-square error is- ε 2 ( t ) f = D m 2 ( t ) [1 − cos ( θ − φ )] 2 E 1 2 CHAPTER 3. BASIC MODULATION TECHNIQUES where h•i denotes the time-average value. Since the term [1 − cos ( θ − φ )] is a constant, we have- ε 2 ( t ) f =- m 2 ( t ) f [1 − cos ( θ − φ )] 2 Note that for θ = φ , the demodulation carrier is phase coherent with the original modu- lation carrier, and the error is zero. For θ ( t ) = ω t we have the demodulated output y D ( t ) = A c m ( t ) cos ( ω t − φ ) Letting A c = 1 , for convenience, gives the error ε ( t ) = m ( t ) [1 − cos ( ω t − φ )] giving the mean-square error- ε 2 ( t ) f = D m 2 ( t ) [1 − cos ( ω t − φ )] 2 E In many cases, the average of a product is the product of the averages. (We will say more about this in Chapters 4 and 5). For this case- ε 2 ( t ) f =- m 2 ( t ) f D [1 − cos ( ω t − φ )] 2 E Note that 1 − cos ( ω t − φ ) is periodic. Taking the average over an integer number of periods yields D [1 − cos ( ω t − φ )] 2 E =- 1 − 2 cos ( ω t − φ ) + cos 2 ( ω t − φ ) f = 1 + 1 2 = 3 2 Thus- ε 2 ( t ) f = 3 2- m 2 ( t ) f Problem 3.2 Multiplying the AM signal x c ( t ) = A c [1 + am n ( t )] cos ω c t by x c ( t ) = A c [1 + am n ( t )] cos ω c t and lowpass & ltering to remove the double frequency (2 ω c ) term yields y D ( t ) = A c [1 + am n ( t )] cos θ ( t ) 3.1. PROBLEMS 3 ( ) C x t ( ) D y t C R Figure 3.1: For negligible demodulation phase error, θ ( t ) ≈ , this becomes y D ( t ) = A c + A c am n ( t ) The dc component can be removed resulting in A c am n ( t ) , which is a signal proportional to the message, m ( t ) . This process is not generally used in AM since the reason for using AM is to avoid the necessity for coherent demodulation. Problem 3.3 A full-wave recti & er takes the form shown in Figure 3.1. The waveforms are shown in Figure 3.2, with the half-wave recti & er on top and the full-wave recti & er on the bottom. The message signal is the envelopes. Decreasing exponentials can be drawn from the peaks of the waveform as depicted in Figure 3.3(b) in the text. It is clear that the full-wave recti & ed x c ( t ) de & nes the message better than the half-wave recti & ed x c ( t ) since the carrier frequency is e f ectively doubled....
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ch03 - Chapter 3 Basic Modulation Techniques 3.1 Problems...

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