In a nutshell, multirate signal processing addresses the question of; how to change the underlying sampling rate of a discrete time sequence entirely in the discrete time domain. Consider the following scenario. You have an input signal x of t, and you convert this into a discrete time sequence via a sampler at a sampling rate of T1. We will denote the discrete time sequence as x_T1 of n to express the dependence on the sampling period. Now, if you change the sampling period, you will get a different type of sequence. So here's the question. Given a sample sequence x_T1 of n, can we build a discrete time machine that produces another sequence, let's call it x_T2 of n, once we know the relationship between T1 and T2? In multirate signal processing, we will address two scenarios. The first is when T1 is a multiple of T2, and the second one is when T2 is a multiple of T1, where N and n are integers. In the first case, the second sequence will be faster than the first one, so we will increase the number of samples per second. In the second case, we will have to reduce the number of samples per second. But in reality, the actual physical values of T1 and T2 are immaterial, all that matters is their relationship. In the first case as we will see, we will simply output capital N points in the new sequence for each input sample. In the second case, we will discard capital N minus one input points, for every output sample. Let's start with the easier case, in which we increase the number of samples. This is called upsampling, and we can create an upsample sequence from any discrete time sequence, simply by building a sequence where we output one sample of the original sequence followed by capital N minus 1 new samples. What do we put in this new samples? Well, we don't know yet. So for now, let's just put zeros. The upsample sequence indicated by x_NU of n where capital N here is the upsampling factor, will be equal to a sample from the original sequence, when the index is a multiple of capital N, and zero otherwise. Graphically, we indicate the upsampling operator with a circle containing the upsampling factor, and an arrow pointing up. Let's look at an example. Assume this is the original sequence, of course you know it's an infinite sequence, so it continues outside of the plot, to build an upsample sequence by a factor of four, we take each sample in turn the original sequence, we output it, and we follow it by three zeros. The result is something like this. So you see here the indexing has changed. The values remain the same, but their indexes are now different. This is 0, 1, 2, 3, 4, 5, 6, 7, 8, and so on so forth. So all multiples of four are points that belong to the original sequence, all the others are zeros. What is the spectral representation of an upsample sequence? This is very easy to find if we use the z-transform as a first step. So we write the z-transform of the upsample sequence, and this is a standard summation that defines a z-transform, but now we know that in this sum, only one term out of n will be non-zero. So we can simplify this sum here by using a change of variable, and by eliminating old zero terms in this sum, we end up with this summation for k that goes to minus infinity to infinity, of x of k. So samples of the original sequence times z to the power of minus capital Nk. So this is the z-transform of the original sequence computed in z to the power of capital N. If we restrict the z-transform to the unit circle, to find the Fourier transform, we have that the spectrum of the upsample sequence is simply the spectrum of the original sequence, where we have contracted the frequency axis by a factor of capital N. So what used to be, say the minus pi pi interval, will become the minus Npi to Npi. We can see this graphically. Here in the top plot, we show the spectrum of the original sequence. Now, in multirate signal processing is fundamental to remember that discrete times spectra r to pi periodic. To make that more explicit, we can show the same spectrum over a larger interval. So here we have the classic two pi periodic pattern for discrete time spectrum. As I said, the interval minus pi pi, will now be contracted, so that it will include the minus two pi two pi interval. This will be mapped to the minus pi pi interval in the upsample sequence. So the result is Bactrim before the upsample sequence, will be this one. Let's change the upsampling factor to get a better feel for what upsampling does in the frequency domain. Same pattern here, original spectrum. Original spectrum showed over a larger interval, and now we're upsampling by three. So it will be this interval here, that will be mapped onto the minus pi pi interval in the upsample spectrum, and it will look like so. In general, for an upsampling factor of capital N, you will have capital N copies of the original spectrum pulled in the minus pi pi interval for the new upsample sequence. If n is odd as in this case, you will have three neat copies like so, and if n is even like in the previous example, you will have one full copy plus one-half here and one-half there. Let's look at another example with slightly different spectral shape that is not full band, and an upsampling factor of four. Again, the periodicity made explicit looks like so, and when we upsample by four, we will take the minus four pi to four pi interval, and map it to minus pi pi, to obtain this spectrum here. Again, sampling factor is even so you'll have three copies of the spectrum plus one-half here and one-half here for a total of four copies brought in the minus pi pi interval. Okay. So we have created an upsampling machine, that fulfills the basic requirement of outputting capital N samples for every input sample. But the result, we don't really like for a couple of reasons. In the time domain, the signal doesn't look natural. You have one non-zero sampled then followed by a run of zeros samples. This is not a signal you would find in nature. Similarly, in the frequency domain, we have this extra copies of the spectrum that are brought in the band. We don't like that. That's extra spurious content that we want to eliminate. Now, the cool thing is that the two problems are exactly one and the same. To understand why that is so, let's look again at the upsampling problem. Let's take the same signal as before, but now we're looking at it over a larger interval. So we can see that the signal is smooth. Now, we upsample it, say by a factor of four, and let's focus on the smaller region of this smooth signal. So we start with this. We upsample, and we introduced this zeros. Now, the signal is no longer smooth. Because it jumps from non-zero to zero, every four samples. How can we make it smoother? I'll suppose that instead of inserting zeros, what we do is repeat the previous non-zero samples three times. The resulting signal looks like so. So we repeat each sample four time, and we got a signal that, although still a little bit blocky, certainly looks better than jumping back to zero every four samples. How can we obtain this? Well, it's a simple case of zero-order hold interpolation done in the discrete time domain. We take our signal, we do the standard upsampling where we insert zeros, and then we convolve this with the impulse response of the discrete time zero-order hold, that looks exactly like so. Four ones, from zero to three. Of course for a generic upsampling factor of capital N, we will use a zero-order interpolator, whose impulse response is equal to one for n that goes from zero to capital N minus one. In the frequency domain, the response of this filter is well-known, we've studied before, and for capital N equal to four is shown in this picture here. So you'll have a low pass characteristic that as you can already foresee, will help us eliminate the copies of the spectrum that are brought in the minus pi pi interval. We will see that in a second. But first, let's consider what happens if we want an even smoother upsample signal. As per usual, the next step in smoothness, is moving from the zero-order hold interpolator, to the first-order interpolator. Which as you remember, corresponds to connecting the dots in the output signal. In discrete time, this is exactly the same thing. So we will get a signal is like this, although it appears very very smooth. If you look closely, you will see that the intermediate samples in the upsample sequence, are simply a linear interpolation between the two neighboring samples. We know how to obtain this. It's simply the convolution of the standard upsample sequence where we insert zeros, with a first-order interpolator, whose impulse response in discrete time is the familiar triangular shape. So in this case, for n equal to four, it will be a triangle that goes from minus three to plus three, like so. We also remember that to obtain this triangular characteristic, all we need to do is to convolve two zero-order hold impulse responses of half the length. This helps us find the frequency response of the first-order interpolator in discrete time, which in magnitude, is the square magnitude of a zero-order hold. So what you have, is the same general shape of the zero-order hold, but with a steeper low-pass characteristic, due to the fact that the frequency response is raised to the second power. Let's look at what happens in the frequency domain. Remember, this is the spectrum of the original sequence shown with explicit periodicity in the second panel, and this is the spectrum of the standard upsample sequence where we introduce zeros. Now, when we convolve with the discrete time interpolator, we are multiplying the spectrum of the upsample sequence with the frequency response of the interpolator. In the case of the zero-order hold, we have a bland low-pass characteristics, that will make the signals smoother in the time domain, and will almost remove the copies of the spectrum in the frequency domain. This is probably not entirely acceptable, so we move on to the first-order interpolator in most cases, and we will have a higher rejection of the high frequency component. It is exactly the same problem as interpolation in the continuous time domain except that here we're working entirely in the discrete time domain. In the limit, if we would like to have a perfect rejection of the extra copies of the spectrum that are brought into minus pi pi interval, we should use an ideal low-pass with a cutoff frequency of pi over capital N. This means that the ideal digital interpolator is composed of two elements, an upsampler by capital N, where we insert zeros in the original sequence, and an ideal low-pass filter with an impulse response that is a sinc. Of course we cannot achieve this in practice, but what we can do, is use as good as an approximation as possible to the ideal low-pass with cutoff frequency, pi over n. This is our digital interpolator that will increase the underlying sampling frequency of a discrete time sequence entirely in discrete time domain. This is a discrete time filter.
