# 5.1 Interpolation, decimation, and rate changing by integer fractions

 Page 1 / 1

## Interpolation: by an integer factor l

Interpolation means increasing the sampling rate, or filling in in-between samples. Equivalent to sampling abandlimited analog signal $L$ times faster. For the ideal interpolator,

${X}_{1}(\omega )=\begin{cases}{X}_{0}(L\omega ) & \text{if \left|\omega \right|< \frac{\pi }{L}}\\ 0 & \text{if \frac{\pi }{L}\le \left|\omega \right|\le \pi }\end{cases}$
$y(m)=\begin{cases}{X}_{0}(\frac{m}{L}) & \text{if m=\{0, ±(L), ±(2L), \dots \}}\\ 0 & \text{otherwise}\end{cases}$
The DTFT of $y(m)$ is
$Y(\omega )=\sum_{m=-\omega }$ y m ω m n x 0 n ω L n n x n ω L n X 0 ω L
Since ${X}_{0}({\omega }^{\prime })$ is periodic with a period of $2\pi$ , ${X}_{0}(L\omega )=Y(\omega )$ is periodic with a period of $\frac{2\pi }{L}$ (see [link] ). By inserting zero samples between the samples of ${x}_{0}(n)$ , we obtain a signal with a scaled frequency response that simply replicates ${X}_{0}({\omega }^{\prime })$ $L$ times over a $2\pi$ interval!

Obviously, the desired ${x}_{1}(m)$ can be obtained simply by lowpass filtering $y(m)$ to remove the replicas.

${x}_{1}(m)=(y(m), {h}_{L}(m))$
Given ${H}_{L}(m)=\begin{cases}1 & \text{if \left|\omega \right|< \frac{\pi }{L}}\\ 0 & \text{if \frac{\pi }{L}\le \left|\omega \right|\le \pi }\end{cases}$ In practice, a finite-length lowpass filter is designed using any of the methods studied so far ( [link] ).

## Decimation: sampling rate reduction (by an integer factor m)

Let $y(m)={x}_{0}(Lm)$ ( [link] )

That is, keep only every $L$ th sample ( [link] ) In frequency (DTFT):
$Y(\omega )=\sum_{m=()}$ y m ω m m x 0 M m ω m n M m n x 0 n k δ n M k ω n M ω ω M n x 0 n k δ n M k ω n DTFT x 0 n DTFT δ n M k
Now $\mathrm{DTFT}(\sum \delta (n-Mk))=2\pi \sum_{k=0}^{M-1} X(k)\delta ({\omega }_{\prime }-\frac{2\pi k}{M})$ for $\left|\omega \right|< \pi$ as shown in homework #1, where $X(k)$ is the DFT of one period of the periodic sequence. In this case, $X(k)=1$ for $k\in \{0, 1, \dots , M-1\}$ and $\mathrm{DTFT}(\sum \delta (n-Mk))=2\pi \sum_{k=0}^{M-1} \delta ({\omega }_{\prime }-\frac{2\pi k}{M})$ .
$(\mathrm{DTFT}({x}_{0}(n)), \mathrm{DTFT}(\sum \delta (n-Mk)))=({X}_{0}({\omega }^{\prime }), 2\pi \sum_{k=0}^{M-1} \delta ({\omega }_{\prime }-\frac{2\pi k}{M}))=\frac{1}{2\pi }\int_{-\pi }^{\pi } {X}_{0}({\mu }^{\prime })2\pi \sum_{k=0}^{M-1} \delta ({\omega }_{\prime }-{\mu }_{\prime }-\frac{2\pi k}{M})\,d {\mu }^{\prime }=\sum_{k=0}^{M-1} {X}_{0}({\omega }_{\prime }-\frac{2\pi k}{M})$
so $Y(\omega )=\sum_{k=0}^{M-1} {X}_{0}(\frac{\omega }{M}-\frac{2\pi k}{M})$ i.e. , we get digital aliasing .( [link] ) Usually, we prefer not to have aliasing, so the downsampler is preceded by a lowpass filter to remove all frequencycomponents above $\left|\omega \right|< \frac{\pi }{M}$ ( [link] ).

## Rate-changing by a rational fraction l/m

This is easily accomplished by interpolating by a factor of $L$ , then decimating by a factor of $M$ ( [link] ).

The two lowpass filters can be combined into one LP filterwith the lower cutoff, $H(\omega )=\begin{cases}1 & \text{if \left|\omega \right|< \frac{\pi }{\max\{L , M\}}}\\ 0 & \text{if \frac{\pi }{\max\{L , M\}}\le \left|\omega \right|\le \pi }\end{cases}$ Obviously, the computational complexity and simplicity of implementation will depend on $\frac{L}{M}$ : $2/3$ will be easier to implement than $1061/1060$ !

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!