# Section 2.4.  SPECTRAL INVERSION IN BANDPASS SAMPLING

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### 2.4. SPECTRAL INVERSION IN BANDPASS SAMPLING

Some of the permissible f s values from Eq. (2-10) will, although avoiding aliasing problems, provide a sampled baseband spectrum (located near zero Hz) that is inverted from the original positive and negative spectral shapes , that is, the positive baseband will have the inverted shape of the negative half from the original spectrum. This spectral inversion happens whenever m , in Eq. (2-10), is an odd integer, as illustrated in Figures 2-9(b) and 2-9(e). When the original positive spectral bandpass components are symmetrical about the f c frequency, spectral inversion presents no problem and any nonaliasing value for f s from Eq. (2-10) may be chosen . However, if spectral inversion is something to be avoided, for example, when single sideband signals are being processed , the minimum applicable sample rate to avoid spectral inversion is defined by Eq. (2-10) with the restriction that m is the largest even integer such that f s 2 B is satisfied. Using our definition of optimum sampling rate, the expression that provides the optimum noninverting sampling rates and avoids spectral replications butting up against each other, except at zero Hz, is

where m even = 2, 4, 6, etc. For our bandpass signal example, Eq. (2-14) and m = 2 provide an optimum noninverting sample rate of = 17.5 MHz, as shown in Figure 2-9(c). In this case, notice that the spectrum translated toward zero Hz has the same orientation as the original spectrum centered at 20 MHz.

Then again, if spectral inversion is unimportant for your application, we can determine the absolute minimum sampling rate without having to choose various values for m in Eq. (2-10) and creating a table like we did for Table 2-1. Considering Figure 2-16, the question is "How many replications of the positive and negative images of bandwidth B can we squeeze into the frequency range of 2 f c + B without overlap?" That number of replications is

Equation 2-15

##### Figure 2-16. Frequency span of a continuous bandpass signal.

To avoid overlap, we have to make sure that the number of replications is an integer less than or equal to R in Eq. (2-15). So, we can define the integral number of replications to be R int where

Equation 2-16

With R int replications in the frequency span of 2 f c + B , then, the spectral repetition period, or minimum sample rate is

Equation 2-17

In our bandpass signal example, finding first requires the appropriate value for R int in Eq. (2-16) as

so R int = 4. Then, from Eq. (2-17), = (40+5)/4 = 11.25 MHz, which is the sample rate illustrated in Figures 2-9(e) and 2-12. So, we can use Eq. (2-17) and avoid using various values for m in Eq. (2-10) and having to create a table like Table 2-1. (Be careful though. Eq. (2-17) places our sampling rate at the boundary between a white and shaded area of Figure 2-12, and we have to consider the guard band strategy discussed above.) To recap the bandpass signal example, sampling at 11.25 MHz, from Eq. (2-17), avoids aliasing and inverts the spectrum, while sampling at 17.5 MHz, from Eq. (2-14), avoids aliasing with no spectral inversion.

Now here's some good news. With a little additional digital processing, we can sample at 11.25 MHz, with its spectral inversion and easily reinvert the spectrum back to its original orientation. The discrete spectrum of any digital signal can be inverted by multiplying the signal's discrete-time samples by a sequence of alternating plus ones and minus ones (1, –1, 1, –1, etc.), indicated in the literature by the succinct expression (–1) n . This scheme allows bandpass sampling at the lower rate of Eq. (2-17) while correcting for spectral inversion, thus avoiding the necessity of using the higher sample rates from Eq. (2-14). Although multiplying time samples by (–1) n is explored in detail in Section 13.1, all we need to remember at this point is the simple rule that multiplication of real signal samples by (–1) n is equivalent to multiplying by a cosine whose frequency is f s /2. In the frequency domain, this multiplication flips the positive frequency band of interest, from zero to + f s /2 Hz, about f s /4 Hz, and flips the negative frequency band of interest, from – f s /2 to zero Hz, about – f s /4 Hz as shown in Figure 2-17. The (–1) n sequence is not only used for inverting the spectra of bandpass sampled sequences; it can be used to invert the spectra of low-pass sampled signals. Be aware, however, that, in the low-pass sampling case, any DC (zero Hz) component in the original continuous signal will be translated to both + f s /2 and – f s /2 after multiplication by (–1) n . In the literature of DSP, occasionally you'll see the (–1) n sequence represented by the equivalent expressions cos( p n ) and e j p n .

##### Figure 2-17. Spectral inversion through multiplication by (–1) n : (a) original spectrum of a time-domain sequence; (b) new spectrum of the product of original time sequence and the (–1) n sequence.

We conclude this topic by consolidating in Table 2-2 what we need to know about bandpass sampling.

##### Table 2-2. Bandpass Sampling Relationships

Requirement

Sample rate expression

Conditions

Acceptable ranges of f s for bandpass sampling: Eq. (2-10)

m = any positive integer so that f s 2 B .

Sample rate in the middle of the acceptable sample rate bands: Eq. (2-12)

m = any positive integer so that 2 B .

Sample rate forcing signal to reside at one fourth the sample rate: Eq. (2-13)

m odd = any positive odd integer so that . (Spectral inversion occurs when m odd = 3, 7, 11, etc.)

Optimum sample rate to avoid spectral inversion: Eq. (2-14)

m even = any even positive integer so that .

where

Absolute minimum f s to avoid aliasing: Eq. (2-17)

.

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