Direct Downconversion of Multiband RF Signals Using Bandpass Sampling


Direct Downconversion of Multiband RF Signals Using Bandpass Sampling

Fig.2.The dual-band signal shown in Fig.1(b)(assuming M =2)after

bandpass sampling.The 8possible replica orders are shown in (a)to (h).The parameters L i and H i (i =±1,±2)are defined in (a),(d),(f),and (g).

GSM signals is conducted to demonstrate the usage of the proposed method.

II.B ANDPASS S AMPLING D UAL -BAND RF S IGNALS Consider the problem of sampling a dual-band RF signal whose spectrum is like that shown in Fig.1(b)with M =2.To be immune from aliasing,the sampling frequency f s needs to be chosen without causing spectral overlapping in the sampled signal spectrum.This leads to the 8possible replica orders shown in Fig.2,where we use shaded trapezoids to denote the spectrum of the original signal,and solid and dashed trapezoids to denote the replicas for positive-frequency and negative-frequency spectra,respectively.

For a given replica order,the sampling frequency must satisfy two types of constraints:one is referred to as the neighbor constraint and the other is referred to as the boundary constraint in this paper.Note that,as shown in Figs.2(a)and 2(b),the replicas ‘1’and ’2’are neighbors in the first half of a segment in Case 1and in the second half of a segment in Case 2.Since the positions of replica ‘i ’and ‘−i ’are symmetric with respective to the midpoint of a segment,one can also see from Figs.2(a)and 2(b)that the replicas ‘-2’and ‘-1’are neighbors in the second half of a segment in Case 1and in the





G .2.



Case Shorthand Constraint 1 1,2 1 H 1≤ L 2f s ≤f L 2−f H 1n 2−n 1 -2,-1 22 1,-2 3 H 1≤ L −2f s ≥f H 1+f H 2n 1+n 2+1 2,-1 43

-1,2 5 H −1≤ L 2f s ≤f L 1+f L 2n 1+n 2+1 -2,1 64

-1,-2 7 H −1≤ L −2

f s ≥

f H 2−f L 1n 2−n 1

2,1 8

first half of a segment in Case 2.If those cases having the same

neighboring replicas in either the first or the second half of a segment are considered as in one group,we may categorize the 8cases in Fig.2into 4groups,as shown in Table I,where the neighboring relation between two replicas j and k is denoted by j,k .The two cases in each group have the same neighboring replicas and thus share the same neighbor constraint.However,they have different boundary constraints because the neighboring replicas show up in different halves of a segment.

For simplicity,we define L i and H i (i =±1,±2)as the relative lowest and highest frequencies of replica ‘i ’in a segment,as shown in Figs.2(a),2(d),2(f),and 2(g).By comparing Figs.2(a),2(d),2(f),and 2(g)to Fig.1(b),one can easily see that

H i =f H i −n i f s ,i =1,2,

(1) H −i =(n i +1)f s −f L i ,i =1,2,

(2) L i =f L i −n i f s ,i =1,2,

(3) L −i

=(n i +1)f s −f H i ,i =1,2.


In deriving (2)and (4),we have used the following relations

H −i =f s − L i ,

i =1,2,(5) L −i


f s − H i ,i =1,2,


which are due to the fact that the relative positions of the replicas ‘i ’and ‘−i ’in each segment are symmetric with respect to the midpoint of the segment.

Let’s consider the neighbor constraint first.From Table I we see that the neighboring relation in Group 1is 1,2 (or equivalently, −2,−1 ).This relation may be written in a shorthand representation as H 1≤ L 2(see Fig.2(a)),

which can lead to f s ≤f L

2−f H 1

n 2−n 1with the aid of (1)and (3).

Similarly,one can easily find the shorthand representations of the neighbor relations and their corresponding constraints on f s for the other three groups.The result is summarized in Table I.The boundary constraints for each case can be obtained by observing Fig.2.Taking Case 1as an example,the neighboring replicas 1,2 must be completely inside the first half of a segment.We therefore see from Fig.2(a)that the shorthand representations of the boundary constraints for spectra ‘1’and ‘2’are L 1≥0and H 2≤f s /2,respectively,

which can result in f s ≤f L 1

n 1and f s ≥f H 2n 2+1/2

by using (1)and (3).Following the same procedure,it is straightforward to derive the boundary constraints for all other cases.The result