<< Chapter < Page | Chapter >> Page > |

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,

$${X}_{1}(\omega )=\begin{cases}{X}_{0}(L\omega ) & \text{if $\left|\omega \right|< \frac{\pi}{L}$}\\ 0 & \text{if $\frac{\pi}{L}\le \left|\omega \right|\le \pi $}\end{cases}$$

We wish to accomplish this digitally. Consider
[link] and
[link] .
$$y(m)=\begin{cases}{X}_{0}(\frac{m}{L}) & \text{if $m=\{0, \pm (L), \pm (2L), \dots \}$}\\ 0 & \text{otherwise}\end{cases}$$

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] ).
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
$Y(\omega )=\sum_{k=0}^{M-1} {X}_{0}(\frac{\omega}{M}-\frac{2\pi k}{M})$
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 $1061/1060$ !-
100% Free
*Android Mobile*Application - Complete Textbook by OpenStax
- Multiple Choices Questions (MCQ)
- Essay Questions Flash Cards
- Key-Terms Flash Cards

Source:
OpenStax, Dspa. OpenStax CNX. May 18, 2010 Download for free at http://cnx.org/content/col10599/1.5

Google Play and the Google Play logo are trademarks of Google Inc.

*Notification Switch*

Would you like to follow the *'Dspa'* conversation and receive update notifications?