Well, the idea is that we need to increase the number of samples by a factor K and

obviously, the samples in the upsampled sequence will have to coincide with

the original values when the index of the upsampled sequence is a multiple of K.

There are several reasons why this is so, but probably the most intuitive one,

is that if we then discard the extra samples,

we should be able to obtain the original sequence again.

So, for lack of a better strategy, we can start by building a sequence where we put

the original samples, every K samples, and then we put 0s everywhere else in between.

So, for example, for K = 3, the upsample sequence,

say this is m = 0, will be equal to x[0] in m = 0.

Then we put two 0s, then we put x[1], then we put two 0s, then we put x[2], and

then we put two 0s.

We can see this in the time domain.

We start with the same sequence that we showed before.

And what we are doing, we're simply introducing 0s between each sample.

With this choice,

the Fourier transform of the upsampled sequence is rather easy to compute.

We just write out the standard DTFT formula, but

now here we remember that XU(m) will be equal to

0 every time that m is not a multiple of K.

And so with this we can simplify the sum and use only the non-zero terms.

And we get the sum from n that goes from minus infinity to plus infinity of x(n),

which is our original sequence, then multiplies e to the -j omega nK.

And so this is simply a scaling of the frequency axis by a factor of K.

Graphically we can plot the digital spectrum and

we know that now, since we're multiplying the frequency axis by a factor of K,

there will be a shrinkage of the frequency axis like this.

But we should never forget that the digital spectrum is 2 pi periodic.

So, let's plot this explicitly from -5 pi to 5 pi, if we choose K = 3.

We're mapping the interval from -3 pi to 3 pi back onto the -pi, pi interval.

When we do that, we get something that is very close to what we obtained going

through the analog domain, in the sense that this frequency here is again pi/4,

but we have extra copies that have crept in the main frequency interval.

Now we know what to do in this case is we apply some drastic low-pass filter to get

rid of them.

We choose an ideal low-pass filter with cut-off frequency pi/K.

Because this is where the original pi in the frequency spectrum would be mapped to.

And this will get rid of the extra copies and leave us with the spectrum

that is identical to what we obtain using an interpolator followed by a sampler.

So now let's look at the procedure back in the time domain.

So the first step is to insert K- 1 zeros after each sample

followed by an ideal low pass filter.

And we choose the cut-off frequency for

the filter to be pi/K, as we saw in the previous graph.

So now the resulting sequence

is simply the convolution of the upsampled sequence with zeroes.

And the impulse response of the filter,

that with this cut-off frequency will be simply sinc (n/K),

and if we work out the convolution sum, we have this summation here.

But again, we remember that, of these terms only one every K will be no zero.

So we'll replace i with mK, and we sum over m, and we get the sum for

m that goes from minus infinity to plus infinity of x(m) sinc (n/K- m),

which is exactly the same formula we got using an interpolator and a sampler.

As we mentioned before, if we have an upsampled sequence,

we can always recover the original sequence by downsampling.

Which means we keep only one sample out of K and throw away the rest.

Now this is obvious in the case of an upsample sequence where we just introduce

K- 1 zeros every sample.

But it is also true for a filtered sequence,

where we use an ideal filter, or any other filter

that fulfills the interpolation properties that we have seen in module six.

In other words, we want the impulse response to be equal to 1 for

n = 0 and to be equal to 0 for all multiples of K.

In general downsampling is a more complex operation than upsampling

just like sampling is more complicated than interpolation.

We have the pesky problem of aliasing, because we're throwing away information.

We will not develop the properties of the downsampling operator in detail,

because we will not need it in the following.

But you're encouraged to read about multi-rate signal processing in the book.