Lecture 9, Multirate Signal Processing, z-domain Effects

Transcription

1 Lecture 9, Multirate Signal Processing, z-domain Effects Last time we saw the effects of downsampling and upsampling. Observe that we can perfectly reconstruct the high pass signal in our example if we use ideal filters, using upsampling and ideal high pass filtering. In this way we have for the analysis and synthesis the following picture -pi pi -pi pi -pi pi Analysis -pi pi Synthesis Observe that we violate the conventional Nyquist criterium, because our high pass passes the high frequencies. But then the sampling mirrors those frequencies to the lower range, such that we can apply the traditional Nyquist sampling theorem. This method is also

2 known as bandpass Nyquist. This is an important principle for filter banks and wavelets. It says that we can perfectly reconstruct a bandpass signal in a filter bank, if we sample with twice the rate as the bandwidth of our bandpass signal (assuming ideal filters). Observe that this simple assumption only works for real valued filters if we have bandpass filters which start at frequencies / N k (integer multiples of / N ). Otherwise we could have overlap to the aliased components! Example: We have a real valued bandpass filter which starts its passband at and ends at / N (since it is real values the passband also appears at the same negative frequencies, to / N ). After multiplication with the delta train, we get one (out of N) aliasing component by shifting the negative passband by 2 / N to the positive frequencies, which results to the range of / N to 2 / N. Hence we have an overlap from / N to / N and we could not perfectly reconstruct our signal! Hence, sampling of twice the bandwidth for real valued signals only works if the bandwidths are aligned with / N k. What could we do otherwise to avoid overlapping aliasing components? We could simply increas the

3 sampling rate, for instance to twice the usual sampling rate (4 times the bandwidth), to have a "safety margin". Another possibility would be to shift the bandpass signals in frequency, such that they are now aligned with the above mentioned grid. Observe that this restriction is not needed for complex signals or filter banks. That is also why complex filter banks are used for instance in acoustic echo cancellation, because there the sampling rate can be increased by a certain fraction (less than a factor of 2) to reduce aliasing artifacts, and still have a lower complexity. overlap Summary: if our band boarders are alligned with multiples of / N then we can donwsample by N, otherwise we are on the save side by using N/2 as downsampling rate for real valued signals. For complex signals we can always downsample by N, regardles of the exact placement of the bandpass filter.

4 Compare with the standard Nyquist case: here we have a lowpass signal which we downsample and reconstruct: Lowpass Lowpass Lowpass filtered input spectrum (original) Aliasing Component Reconstruc. Original spectrum Matlab/Octave Example: Just take an audio signal and read it into Matlab/Octave, and filter it with our previous low pass filter (e.g. Kaiser Window with Beta=8), x=wavread('04_topchart.wav'); %listen to it: sound(x,32000) %Look at its spectrum: freqz(x)

5 This is the spectrum of our original. Observe its already low pass characteristic. Now we can low pass filter it. We take our Kaiser Window low pass filter design from slides Nr. 6, n=0:31; %ideal impulse response: h=sin(0.33*pi*(n-15.5))./(pi*(n-15.5)); %Kaiser window: hk=kaiser(32,8)'; %multiply ideal filter and Kaiser window: hfilt=hk.*h;

6 xlp=conv(x,hfilt); freqz(xlp) Observe that beginning at frequency 0.5 we have indeed much attenuation and hardly any signal left. Now we can down-sample it by a factor of N=2, xds=xlp(1:2:end); freqz(xds)

7 Observe that we now obtain the streched spectrum. Listen to it: sound(xds,16000) It should sound more muffled, but otherwise the same, but at now half the sampling rate! Now we can upsample again, xups=zeros(2*max(size(xds)),1); xups(1:2:end)=xds; freqz(xups);

8 Observe the shrinking and periodic continuation of the spectrum, the aliasing component at the high frequencies. Listen to the signal including aliasing, sound(xups,32000); Now low pass filter the result, xupslp=conv(xups,hfilt); freqz(xupslp)

9 Observe that now we removed the aliasing component at high frequencies. Now listen to it, sound(xupslp,32000); Observe: it should now sound the same as at the lower sampling rate, but now at the higher sampling rate of 32 khz! (Possible differences are due to not sufficiently attenuated aliasing).

10 Effects in the z-domain The z-transform is a more general transform than the Fourier transform, and we will use it to obtain perfect reconstruction in filter banks and wavelets. Hence we will now look at the effects of sampling and some more tools in the z- domain. Since we usually deal with causal systems in practice, we use the 1-sided z-transform, defined as X z = n=0 x n z n First observe that we get our usual frequency response if we evaluate the z-tranform along the unit circle in the z-domain, z=e j What is now the effect of multiplying our signal with the delta impulse train in the z- domain? To see this we simply apply the z- transform, X d z = = 1 N k=0 n=0 N 1 n=0 x n N n z n x n e j 2 N k z n N 1 = 1 N k=0 X e j 2 N k z or, in short, N 1 X e j 2 N k z X d z = 1 N k=0

11 This is very similar to the Fourier transform formulation, just that we now don't have frequency shifts for the aliasing terms, but a multiplication of their exponential functions to z. The next effect is the removal or re-insertion of the zeros from or into the signal. Let's again use our definition y m =x d mn, meaning the y(m) is the signal without the zeros. Then the z-transform becomes, Y Z = y m z m = = m=0 m=0 x d mn z m = Replacing the sum index m (the lower sampling rate) by the higher sampling rate n=mn, and observing that the sequence x d n contains the zeros, with x d n =0 for n mn, this results in = n=0 x d n z n/ N = X d z 1/ N Observe the 1/N in the exponent of z! In short, we get Y Z = X d z 1/ N This exponent 1/N of z now correponds to the stretching of our frequency scale in the Fourier spectrum.

12 Another very useful new tool is the so-called modulation. This is the multiplication of our signal with a periodic function, for instance an exponential function. It can be written as x M n :=x n e j n M. Observe that the modulation function here has a periodicity of 2 / M. Its z-transform hence becomes j n X M z = x n e M n z n=0 X M z = X e j M z here again we have this rotation of z by an exponential factor. Observe that this has the effect of simply shifting the frequency response by M in the Fourier spectrum, which can be seen by simply replacing z by e j to obtain the frequency response. But instead of multiple frequency shifts and adds in the downsampled case, here we only have a single shift. An intersting case is obtained if we set M =. Then the modulating function is just a sequence of +1,-1,+1,..., (as can be seen by using e j n as a signal, with n=0,1,2,...) and multiplying this with our signal results in a frequency shift of. If we have a real valued low pass signal, for instance, this will then be shifted by, and we will obtain a high pass. In this way we can turn the impulse response of

13 a low pass into a high pass, and vice versa. We see that this is a simple method to obtain a high pass filter from a low pass filter. We simply design a low pass, the take its impulse response, and invert the sign of every second sample, and the result is a high pass! In a similar way we can also obtain a band pass from a low pass, we just need the general modulating function e j M n and multiply it with the impulse response of our low pass filter. What was around frequency zero now after modulation appears around frequency M, and we have a bandpass with center frequeny M. This is a useful "trick", because there are many design methods for low pass filters, but not so many for the other types, and because we can obtain filters with the same properties at multiple frequencys (we simply "copy" a filter to different frequencies). Observe that in general we will obtain a complex valued

14 impulse response using this approach. The resulting spectrum becomes one-sided. If we want to obtain real valued filters and impulse responses, we have to make the spectrum symmetric around zero again. We can do this by simple take the real parts of the complex impulse response (in that case that symmetric extension has the same sign as the original spectrum components) or we take the imaginary part (in that case we obtain the opposite sign for the symmetric extension). In summary: with modulation we can shift a low pass spectrum to any desired frequency to obtain a desired filter, and using the real or imaginary part we can then obtain a real valued filter. Another important tool is the reversal of the ordering of a finite length signal sequence, with length L (meaning x(n) is non-zero only for n=0,..., L 1 ), x r n :=x L 1 n. Its z-transform is X r z = n=0 x L 1 n z n we can now reverse the order of the summation (of course without affecting the result) by starting at the highest index, going to the lowest, replacing the index n by the expression L 1 n' (index substitution),

15 or, in short, X r z = n' =0 L 1 n' x n' z =z L 1 X z 1 X r z =z L 1 X z 1 So what we obtain is the inverse of z in the z- transform (which signifies the time reversal), and a factor of z L 1, which is simply a delay of L-1 samples! Important here is the inverse of z. What difference does this make in our Fourier spectrum, replacing z by e j? We obtain X instead of X. For real valued signals this only makes a difference for the signs (phases) of our frequency responses, because of the spectral symmetries for real valued signals. The magnitudes are identical. This can still be of importance, for instance in filter banks with aliasing cancellation. Here the different signs also change the sign of the aliasing components, and that can make the difference between aliasing components cancelling between different bands or adding up! An example is the 2 band Haar filter bank, where the high pass impulse responese on one side has to be time reversed to obtain perfect reconstruction. For complex valued signals, the negative and

16 positive frequencies can be completely different, and hence time-reversal would make a bigger difference.