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Unformatted text preview: THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Spring Problem Set 2 1. When Is a Carrier a Good Carrier?: Given an information signal m ( t ) you want to gener- ate r ( t ) = Am ( t ) cos ( 2 π f c t ) (1) However, life (your professor) is unkind and you’re only allowed to use cos 3 ( 2 π f c t ) as your carrier signal. That is, the first stage of your transmitter block diagram with m ( t ) as the input is going to be a multiplication by cos 3 ( 2 π f c t ) (instead of the usual sane cos ( 2 π f c t ) ). (a) Assume you can build any linear time invariant (LTI) filter you’d like. Can you fil- ter the output of the multiplier to obtain the desired signal? If so, what is the filter characteristic? SOLUTION: In order to obtain r ( t ) = A c m ( t ) cos ( 2 π f c t ) ,we need to design a filter that give us the desired signal, therfore using the math identities such as: cos 2 ( x ) = 1 2 ( 1 + cos ( 2 x )) cos ( α ) cos ( β ) = 1 2 ( cos ( α + β ) + cos ( α- β )) cos 3 ( x ) = cos ( x ) cos 2 ( x ) = cos ( x )( 1 2 + 1 2 cos ( 2 x )) = 3 4 cos ( x ) + 1 4 cos ( 3 x ) we can design a LPF filter so that we get rid off of the third harmonic components e.g. 1 4 cos ( 3 ( 2 π f c )) thus that we are only left with 3 4 cos ( 2 π f c t ) . (b) Suppose cos 2 ( 2 π f c t ) is the carrier. Repeat the previous part. SOLUTION: cos 2 ( 2 π f c t ) does not have a component at frequency f c . So, nope! (c) Suppose we generalize the carrier to be cos n ( 2 π f c t ) for n > 2. When can you generate the desired r ( t ) using LTI filters. SOLUTION: Use the odd harmonics of the signal and we’re always guaranteed a component at frequency f c . 2. Simple Envelope Detection: Consider the following AM signal s ( t ) = A c [ 1 + λ cos ( 2 π f m t )] cos ( 2 π f c t ) (2) The modulation factor is λ = 1 and f c À f m . The AM signal is applied to an ideal envelope detector producing the output v ( t ) . 1 (a) Why did we stipulate f c À f m ? SOLUTION: An envelope detector works by allowing the incoming signal to quickly drag the envelope detector output higher, but then resisting negative swings via the diode (the output capacitor has to discharge). This means that an envelope detector tracks the peaks of the incoming signal and when the carrier frequency is high, the rec- tified signal looks like a “picket fence” representation of the low frequency information signal. If the carrier frequency is on the order of the frequencies in the information signal, there are no “peaks” to detect. (b) Find v ( t ) and V ( f ) . SOLUTION: Just for reference we compute S ( f ) = 1 4 A c 2 [ δ ( f + f c ) + δ ( f- f c )] + λ [ δ ( f + f c + f m ) + δ ( f + f c- f m )] + λ [ δ ( f- f c + f m ) + δ ( f- f c- f m )] Now to the main show. A simple envelope detector chops off the negative half-cycles of the carrier. For cosine modulation and the information signal not going negative,of the carrier....
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This homework help was uploaded on 04/03/2008 for the course ECE ECE332 taught by Professor Rose during the Spring '08 term at Rutgers.
- Spring '08