0.8 Interpolation and upsampling

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 $x_n=x\left(n\tau \right)$ of a band-limited signal $x\left(t\right)$ taken at frequency $f_e=1/\tau$ is above the Nyquist rate $2B$ . We want to compute the samples $z_k$ of $x\left(t\right)$ at a higher sampling rate $f_u=N·f_e$ . This means that the new sampling step should be $1/f_u=1/\left(f_{e}N\right)=\tau /N$ or, $z_k=x(k\tau /N)$ .

Since the original sampling rate $f_e>2B$ is above Nyquist, we can in theory reconstruct the entire signal $x\left(t\right)$ using the reconstruction formula [link] using $f_c=f_e/2$ . Once the continuous-time (finite energy) signal $x\left(t\right)$ is obtained, we only need to sample it at $t=k\tau /N=k/f_u$ . This reads as follows

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 $X_e\left(f\right)$ 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

where the new sequence $y_m$ is obtained by “upsampling” and is given as:

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”)
   $$\begin{array}{ccc}\hfill (x_0,x_1,x_2,...)& \to & \uparrow N\to (x_0,\underset{N-1}{\underbrace{0,..,0}},x_1,\underset{N-1}{\underbrace{0,..,0}},x_2,\underset{N-1}{\underbrace{0,..,0}},x_3...)=(y_0,y_1,y_2,...)\hfill \end{array}$$
2. Multiplication by $N$ and Low-pass filtering at cutoff frequency $\frac{1}{2N}$ using the ideal filter $\frac{1}{N}\text{sinc}\left(\frac{m}{N}\right)$ :
   $$\begin{array}{ccc}\hfill (y_0,y_1,y_2,...)& \to & ·N\to \cap \frac{1}{2N}\to \left\{y_m\right\}*\left\{\text{sinc}\left(\frac{m}{N}\right)\right\}=(z_0,z_1,z_2,...)\hfill \end{array}$$

Spectral picture of Interpolation by $N$

In analogy to the decimation we set $z\left(t\right)=x(t/N)$ . Then, the samples of $z\left(t\right)$ taken at the same rate $f_e$ constitute the samples of $x$ taken at rate $f_u$ . Analogously to the decimation we find quickly

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 $X_e$ .

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

The corresponding signal $y\left(t\right)$ (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 $X\left(Nf\right)$ (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 $y\left(t\right)=\sum y_m\text{sinc}\left(\frac{t-m\tau }{\tau }\right)$ . Using that $\text{sinc}\left(0\right)=1$ while $\text{sinc}\left(m\right)=0$ for all integer $m\ne 0$ we find quickly that the samples of $y\left(t\right)$ are indeed $y\left(k\tau \right)=y_k$ , 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 $N{X}_e\left(Nf\right)$ which consists of copies of $NX\left(fN\right)$ , or $Z\left(f\right)$ , at distance $f_e/N$ (as for $Y_e$ , there are $N$ copies in one period).

(iii) The digital low-pass filtering of $\left\{N{y}_m\right\}$ at cut-off frequency $1/\left(2N\right)$ removes all of the copies of $NX\left(fN\right)$ except the ones centered at 0, $f_e$ , $2f_e$ etc. and leaves only one copy per period, in other words, only copies at distance $f_e$ . What remains is exactly $Z_e\left(f\right)$ 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 $Z\left(f\right)=NX\left(fN\right)$ 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 $f_e>2B$ , 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

Decimation