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Interpolation and upsampling interpolation

Let us now look at increasing the sample rate. Since it is less obvious how to achieve this, let us first consult theory.

Given are the samples x n = x ( n τ ) of a band-limited signal x ( t ) taken at frequency f e = 1 / τ is above the Nyquist rate 2 B . We want to compute the samples z k of x ( t ) at a higher sampling rate f u = N · f e . This means that the new sampling step should be 1 / f u = 1 / ( f e N ) = τ / N or, z k = x ( k τ / N ) .

Since the original sampling rate f e > 2 B is above Nyquist, we can in theory reconstruct the entire signal x ( t ) using the reconstruction formula [link] using f c = f e / 2 . Once the continuous-time (finite energy) signal x ( t ) is obtained, we only need to sample it at t = k τ / N = k / f u . This reads as follows

z k = z ( k τ ) = x k τ N = ( 28 ) n = - x n sinc k τ N - n τ τ = n = - x n sinc k - n N N

This formula allows indeed to compute z k from x n , at least in principle. However, a closer look at theory is required to understand the effect when usingonly finite many samples. The reconstruction formula [link] is best understood in the frequency domain: it amounts to removing the spectral copies of X e ( f ) via filtering with cut-off frequency f c = f e / 2 . To make this filtering step visible we need to write [link] in form of a convolution. To this end, we write

z k = m = - y m sinc k - m N = { y m } * { sinc m N } = { N · y m } * { 1 N sinc ( m N ) }

where the new sequence y m is obtained by “upsampling” and is given as:

y m = x n if m / N = n is integer 0 if m / N is not integer

The convolution [link] allows for more convenient data processing via digital filtering and for a simple spectral interpretation.

Interpolation by N or resampling at N times larger rate consists of the following steps:

  1. Up-sampling (in French: “interpolation”)
    ( x 0 , x 1 , x 2 , . . . ) N ( x 0 , 0 , . . , 0 N - 1 , x 1 , 0 , . . , 0 N - 1 , x 2 , 0 , . . , 0 N - 1 , x 3 . . . ) = ( y 0 , y 1 , y 2 , . . . )
  2. Multiplication by N and Low-pass filtering at cutoff frequency 1 2 N using the ideal filter 1 N sinc ( m N ) :
    ( y 0 , y 1 , y 2 , . . . ) · N 1 2 N { y m } * { sinc ( m N ) } = ( z 0 , z 1 , z 2 , . . . )
digital filter upsampled by n=3 in black; sinc/N for comparison IDFTI squared of filter up sampled by n=3; idfti squared of sinc/n for comparison
The result of upsampling the filter from [link] by a factor N = 3 on the left, and its power spectrum | DFT | 2 on the right (black and red). Again, the ideal sinc filter (divided by N to adjust for the zero-samples) in blue for comparison. Note that the filter's cutoff frequency has decreased by N (contraction of the spectrum by N ) and that its | DFT | 2 values have not changed (see text). The power computed from either side is 0.001'297 and has decreased by a factor N from the original power of 0.003'829: only upsampling of a signal with 2 B < f e decreases power by the upsampling factor.

Spectral picture of Interpolation by N

In analogy to the decimation we set z ( t ) = x ( t / N ) . Then, the samples of z ( t ) taken at the same rate f e constitute the samples of x taken at rate f u . Analogously to the decimation we find quickly

z ( t ) = x ( t / N ) Z ( f ) = N X ( f N )

which indicates how the spectrum at rate f u = N f e is obtained: the spectrum at rate f e gets contracted in the frequency axis by N and expanded in amplitude by N .

Note that the spectral copies of Z e are at distance f e just like those of X e .

interpolated digital filter in black; sinc for comparison IDFTI squared of interpolated filter; idfti squared of sinc for comparison
The result of interpolating (upsampling and filtering) the filter from [link] by a factor N = 3 on the left, and its power spectrum | DFT | 2 on the right (black and red). Again, the ideal sinc filter (no need to divide by N since the zero-samples are now corrected through the filtering step) in blue for comparison. Note that the fundamental period contains now only one spectral copy, as it should, as a result from the filtering. The spectrum is contracted by N in frequency and expanded in amplitude by N 2 . The power computed from either side is 0.003'891 and almost identical to the original power of 0.003'829: interpolation of a signal with 2 B < f e does not change power.

We may break the procedure down into the individual steps:

(i) Upsampling (introducing the zero-samples) leaves the Fourier transform, and thus the spectrum almost intact, leading only to a rescaling of the frequencies(contraction of X ): Indeed, the Fourier transform Y e of the samples y m becomes

Y e ( f ) = m τ y m e - j 2 π f m τ = n τ x n e - j 2 π f n N τ = X e ( N f ) .

The corresponding signal y ( t ) (with samples y m at sampling rate f e ) is not of interest. If you want to know about it any way, see Comment 7 . Note, however, that the fundamental period of Y e amounts to f e and contains N copies of X ( N f ) (see [link] ).

Comment 7 For clarity: the Fourier transform of y is found by removing the spectral copies of Y e outside [ - f e / 2 , f e / 2 ] . These copies are caused by sampling. The Fourier transform of y consists of N contracted copies of X at distance f e / N of each other which are caused by upsampling. Using the reconstruction formula [link] with the samples y m , sample rate f e and correct pass-band f e / 2 yields the signal y ( t ) = y m sinc t - m τ τ . Using that sinc ( 0 ) = 1 while sinc ( m ) = 0 for all integer m 0 we find quickly that the samples of y ( t ) are indeed y ( k τ ) = y k , i.e., ( . . . , x 0 , 0 . . 0 , x 1 , 0 . . . 0 , x 2 , . . . ) .

(ii) Multiplication with N restores the average value of the samples. The Fourier transform is now N X e ( N f ) which consists of copies of N X ( f N ) , or Z ( f ) , at distance f e / N (as for Y e , there are N copies in one period).

(iii) The digital low-pass filtering of { N y m } at cut-off frequency 1 / ( 2 N ) removes all of the copies of N X ( f N ) except the ones centered at 0, f e , 2 f e etc. and leaves only one copy per period, in other words, only copies at distance f e . What remains is exactly Z e ( f ) as we have pointed out earlier. (see [link] ).

Power and Interpolation

Similarly as with the decimation the power of a periodic signal does not change under interpolation.In fact, we only need to use Z ( f ) = N X ( f N ) and replace 1 / M by N in the computation done with the decimation. We conclude that the power of discrete samples does not change under interpolation provided that f e > 2 B , at least approximatively.

To verify this, let us move through the 3 steps above. Step (i), upsampling, reduces power by a factor N since the sum of squares of the samples is the same (the zeros added don't contribute), but there are now N times more samples. Power is an average.

Step (ii) obviously multiplies power with N 2 . Step (iii), the low-pass filtering, removes N - 1 spectral copies and leaves only 1, thus divides power by N . All steps together leave the power as it is.

Interpolation

x sub e F
N
signal is correctly sampled ( B / f e < 1 / 2 )
Y sub e F = X sub e NF
1 2 N
inserting N - 1 zeroes contracts the Fourier transform by a factor N
X sub u NF
· N
digital low-pass filtering with cut-off frequency at 1 / ( 2 N ) leaves only one of the contracted copies per period 1, i.e., it leaves X u ( N f )
z sub e F = N X sub U NF
If B < f e 2
multiplying with N leads to N X u ( N f ) = Z e ( f ) where z ( t ) = x ( t / N ) . Sampling z at rate f e provides the desired samples at rate f u of x .
y sub e F = X sub e NF

If B < f e / 2 , then a transition band [ B N f e , 1 N - B N f e ] is feasible
Interpolation illustrated in the Spectral Domain with N = 2 .
Horizontal arrows indicate spectral copies. To keep the sample rate at the same value f e one changes from X u , resp. Y d to Z e (see d).

Decimation

x sub e F
1 2 M
1 / M marks the center of the first spectral copy after decimation
Y sub e F
M
digital low-pass filtering with (ideal) cut-off frequency at 1 / ( 2 M ) ensures no aliasing after decimation; the new utilized bandwidth is B * = f e / ( 2 M ) ; high frequency information might be lost, resulting in a new signal y .
X sub d F
adjusting sampling rate
After decimation only the samples y k M remain, corresponding to a sampling of y at f d = f e / M , with Fourier transform Y d ( f ) .
z sub e F = 1 over M times Y sub d f over m
If B * < f e 2 M The samples z k = y k M correspond to z ( t ) = y ( M t ) sampled at f e , and Z e ( f ) = ( 1 / M ) · Y d ( f / M ) (see text).
y sub d F transition band
Choosing the new utilized bandwidth to be B * < f e 2 M , then a transition band [ B * f e , 1 M - B * f e ] is feasible
Decimation illustrated in the Spectral Domain with M = 2 .
Horizontal arrows indicate spectral copies. To keep the sample rate at the same value f e one changes from Y d to Z e (see d).

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Source:  OpenStax, Sampling rate conversion. OpenStax CNX. Sep 05, 2013 Download for free at http://legacy.cnx.org/content/col11529/1.2
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