EE1003_Chap2_Sem1_AY2011-12

EE1003_Chap2_Sem1_AY2011-12 - Chapter 2 Sampling & Pulse...

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Chapter 2 1 Sampling Suppose we wish to sample an analog signal v ( t ) at a sam- pling rate f 0 given by 1 /T 0 . That is, we wish to take snap- shots of v ( t ) every T 0 seconds. This may be achieved by multiplying v ( t ) with the rectangular pulse train ψ ( t )= X n = −∞ p ( t nT 0 ) where p ( t )= 1:0 t<T 0 / 2 0: otherwise. The resulting signal v s ( t )= v ( t ) ψ ( t )= v ( t ) X n = −∞ p ( t nT 0 ) consists of pulses (or snapshots of v ( t ) ) of duration T 0 / 2 and amplitudes corresponding to v ( t ) at the sampling instances. Note that the duration of the pulses is given by the duration of p ( t ) . 1
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To illustrate, we plot v s ( t ) against t for v ( t )=co s (2 πf m t ) with f m =1 : 0 0.5 1 1.5 2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 t V(t) 0 0.5 1 1.5 2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 t V s (t) ψ ( t ) is referred to as the sampling function . v s ( t ) is called a pulse amplitude modulated (PAM) signal, as it consists of amplitude-modified pulses. What will the spectrum of v s ( t ) look like for the above ex- ample? To answer this, first observe that the periodic signal described by (3) in Chap. 1 for m = coincides with ψ ( t ) . Applying sin( θ )= 1 2 j ( e e ) to this equation yields ψ ( t )= 1 2 + X k =0 1 j (2 k +1) π ± e j 2 π ((2 k +1) f 0 ) t e j 2 π ((2 k +1) f 0 ) t ² .
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v ( t )=cos(2 πf m t )= 1 2 e j 2 πf m t + 1 2 e j 2 πf m t . By expanding the product v ( t ) ψ ( t ) we obtain v s ( t )= 1 4 e j 2 πf m t + 1 4 e j 2 πf m t + 1 2 X k =0 1 j (2 k +1) π ± e j 2 π ((2 k +1) f 0 + f m ) t e j 2 π ( (2 k +1) f 0 + f m ) t + e j 2 π ((2 k +1) f 0 f m ) t e j 2 π ( (2 k +1) f 0 f m ) t ² . By applying (13) in Chap. 1, we obtain the spectrum of v s ( t ) : V s ( f )= 1 4 δ ( f f m )+ 1 4 δ ( f + f m ) + 1 2 X k =0 1 j (2 k +1) π [ δ ( f (2 k +1) f 0 f m ) δ ( f +(2 k +1) f 0 f m ) + δ ( f (2 k +1) f 0 + f m ) δ ( f +(2 k +1) f 0 + f m )] . From Example 3.2 in Chap. 1, the spectrum of v ( t ) is V ( f )= 1 2 δ ( f f m )+ 1 2 δ ( f + f m ) which yields the following simplication: V s ( f )= 1 2 V ( f )+ X k =0 1 j (2 k +1) π [ V ( f (2 k +1) f 0 ) V ( f +(2 k +1) f 0 )] . 3
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Thus, if f 0 > 2 f m , the spectrum of v s ( t ) contains an infi- nite number of non-overlapping , distorted copies of V ( f ) centered at frequencies 0 , ± f 0 , ± 3 f 0 , ± 5 f 0 , ± 7 f 0 , ... This distortion is known as the aperture effect . The aperture effect can be significantly reduced by using a
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This note was uploaded on 11/30/2011 for the course EEE 1001 taught by Professor Phoon during the Spring '11 term at National University of Singapore.

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EE1003_Chap2_Sem1_AY2011-12 - Chapter 2 Sampling & Pulse...

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