In digital signal processing, upsampling, expansion, and interpolation are terms associated with the process of resampling in a multi-rate digital signal processing system. Upsampling can be synonymous with expansion, or it can describe an entire process of expansion and filtering (interpolation). When upsampling is performed on a sequence of samples of a signal or other continuous function, it produces an approximation of the sequence that would have been obtained by sampling the signal at a higher rate (or density, as in the case of a photograph). For example, if compact disc audio at 44,100 samples/second is upsampled by a factor of 5/4, the resulting sample-rate is 55,125.
Rate increase by an integer factor
L
xL[n],
x[n],
L-1
xL[n]=x[n]\uparrow.
In this application, the filter is called an interpolation filter, and its design is discussed below. When the interpolation filter is an FIR type, its efficiency can be improved, because the zeros contribute nothing to its dot product calculations. It is an easy matter to omit them from both the data stream and the calculations. The calculation performed by a multirate interpolating FIR filter for each output sample is a dot product:
where the
h
K
k
h[j+kL]
y[m]=
infty | |
\sum | |
r=-infty |
xL[m-r] ⋅ h[r]
xL[m-r]
m-r
L.
m-r=l\lfloor\tfrac{m}{L}r\rfloorL-kL
k,
\begin{align} y[m]&=
infty | |
\sum | |
k=-infty |
xL\left[l\lfloor\tfrac{m}{L}r\rfloorL-kL\right] ⋅ hl[\overbrace{m-l\lfloor\tfrac{m}{L}r\rfloorL+kL}rr]\\ &=
infty | |
\sum | |
k=-infty |
x\left[l\lfloor\tfrac{m}{L}r\rfloor-k\right] ⋅ h\left[m-l\lfloor\tfrac{m}{L}r\rfloorL+kL\right] \stackrel{m \triangleq j+nL}{\longrightarrow} y[j+nL]=
K | |
\sum | |
k=0 |
x[n-k] ⋅ h[j+kL], j=0,1,\ldots,L-1 \end{align}
In the case
L=2,
h
L
L
L
x
L
For completeness, we now mention that a possible, but unlikely, implementation of each phase is to replace the coefficients of the other phases with zeros in a copy of the
h
xL[n]
L
L-1
L
y
L-1
L-1
Let
X(f)
x(t),
T,
x[n]
x[n]
X(f):
When
T
f
L
T/L
L:
which is also the desired result of interpolation. An example of both these distributions is depicted in the first and third graphs of Fig 2.
When the additional samples are inserted zeros, they decrease the sample-interval to
T/L.
\sumn=0,{}x(nT/L) e-i \stackrel{m \triangleq n/L}{\longrightarrow}\summ=0,{}x(mT) e-i,
which is equivalent to regardless of the value of
L.
L/T
L=3,
x[n]
\tfrac{0.5}{T},
Let L/M denote the upsampling factor, where L > M.
Upsampling requires a lowpass filter after increasing the data rate, and downsampling requires a lowpass filter before decimation. Therefore, both operations can be accomplished by a single filter with the lower of the two cutoff frequencies. For the L > M case, the interpolation filter cutoff,
\tfrac{0.5}{L}