C2. Design of Digital Filters

Transcription

1 C2. Design of Digital Filters Objectives Unerstan what is an ieal filter vs a non-ieal filter Given signal an noise specifications in the frequency omain, how to esign a igital filter; For the same problem, how to esign a igital filter using appropriate computer software; How to implement a filter in a real problem 1. Introuction In the previous chapter we efine the concepts of a system an, in particular of Linear Time Invariant (LTI) systems. We saw that a system represents an operation we o on a signal (or a number of signals). Of particular interest is the esign an implementation of Digital Filters, with the goal of removing or attenuating a isturbance. In this section we aress a problem of esigning a igital filter from specifications in the frequency omain. The general problem is to separate signal from noise base on the corresponing frequency spectra. The following sections present two classes of igital filter esign techniques: one base on an analytical approach, which can be implemente by pencil an paper, an the other technique base on computer optimization software, available on a number of computer aie esign packages such as Matlab. The latter turns out to be very convenient since it allows the esign of a very general class of filters Finally we see an application on how to etermine the specifications for the particular problem at han an how to esign an appropriate filter. 2. Design of FIR Low Pass Filters: Analytical Approach VIDEO: Typical Filtering Problem (8:51) In the typical filtering problem we want to separate a signal from an aitive isturbance. In general within the frequency spectrum we can see frequencies in which the signal is ominant an frequencies where the noise is ominant. This is shown in the figure below.

2 Signal an Noise in the Digital Frequency Spectrum We have seen that the effectiveness of an FIR Filter (an this is vali for any Linear Time Invariant (LTI) System) is etermine by its frequency response. In the case shown above, we want to esign a filter which passes only the frequencies where the signal is ominant an reject all frequencies where the noise is ominant. In the case (like in the figure) where the signal occupies the low frequencies, we nee to esign a Low Pass Filter, which passes only the Low Frequencies. In what follows we will be concentrating on the esign an implementation of Low Pass Filters. Other classes of filters (Ban Pass, Stop Ban, High Pass an so on) can be treate in a very analogous way. Ieally we nee a filter with the frequency response shown below. Frequency Response of an Ieal Low Pass Filter. We represent only the positive frequencies since we assume the filter to have real coefficients an therefore the negative part of the spectrum is reunant. The ieal filter has only two regions: Pass Ban within the interval 0 ω < ωp, an Stop Ban within the interval ω P ω < π. The frequency response of the ieal filter is constant within each of the two regions. Clearly it has to be zero within the Stop Ban. A non Ieal Filter, on the other han, is something we can esign an implement, has a frequency response which is more complicate, an it is shown in the figure below.

3 Frequency Response of a Non Ieal Filter We can immeiately see the ifferences between the two frequency responses. In the non Ieal case the frequency response in the Stop Ban is small, but not zero; the frequency response in the Pass Ban is not exactly a constant but it has a ripple. Finally there is also a transition region between the Pass Ban an the Stop Ban. It turns out that (as we will see) the narrower the Transition region (ie the sharper the transition between Passban an Stopban) the higher the complexity of the filter. In what follows we want to see how we can etermine the impulse response of of a Finite Impulse Response (FIR) Filter to approximate the frequency response of an ieal filter. VIDEO: Ieal Filters (9:46) Let s start with the ieal frequency response of a Low Pass Filter, an call it H (ω), with for esire. The goal is to etermine the impulse response sequence h [ n], n = 0,..., N so that the frequency response approaches the ieal frequency response, as H ( ω ) H ( ω) = N n= 0 h[ n] e In orer to o so, let us look at the Discrete Time Fourier Transform expansion of the ieal frequency response H (ω) shown below. jωn

4 Ieal Frequency Response Then, by the DTFT, we can fin an infinite sequence h [ n], n =,..., + such that H { } + h [ n] = n= ( ω ) = DTFT h [ n] e This sequence is compute by the Inverse DTFT (IDTFT) as h + π 1 [ n] = IDTFT{ H ( ω) } = H ( ω) 2π π Substitute for H (ω) as in the figure above an we obtain jωn e jωn ω where we efine + ω P 1 h [ n] = e 2π ωp jωn ( ω n) sin ω= πn P ωp ω P = sinc n π π sin(π x) sinc ( x) = π x A plot of h [n] with ω P = π / 4 is shown in the figure below. Impulse response of Ieal Low Pass Filter (Pass Ban ω = π / 4 P )

5 VIDEO: Non Ieal Filters (7:51) Now notice the following. As n ±, h [ n] 0, which means that the coefficients get smaller an smaller. Then we can use only the ominant values, say h [ L],..., h [ L], with L a chosen parameter, to approximate the ieal filter as H + L n= L ( ω ) h [ n] e Now let the filter impulse response be efine as h[ n] = h [ n L] jωn Then the frequency response of the filter we obtain is the following H ( ω) = 2L n= 0 h[ n] e jωn 2L n= 0 h [ n L] e jωn L jωn jωl jωl = h [ n] e e H ( ω) e n= L This means that magnitue an phase of the FIR filter are relate to the ieal as = H ( w) H ( w) H ( w) = wl within the passban The last expression on the phase is vali within the passban where the phase of the ieal filter is zero. Finally choose the impulse response of the FIR filter as where N = 2L. ωp ω P h[ n] = h [ n L] = sinc ( n L), n = 0,..., N π π See an example. VIDEO: Example using Matlab (7.44) Example. We want esign a Low Pass Filter with Pass Ban ω P = π / 4 raians an orer N = 40. Then L = 20 an we obtain the impulse response 1 1 h [ n] = sinc ( n 20), n = 0,..., We can easily generate this sequence in Matlab as

6 N=40; L=N/2; % Filter Orer. It has to be even % time elay n=0:n; % generate the vector of time ineces h=(1/4)*sinc((n-20)/4); % Impulse response freqz(h,1) % show frequency response The plot of the frequency response is shown in the figure below. Notice two things: a. the filter is Low Pass, in the sense that it passes the low frequencies an it attenuates the high frequencies. This can be seen from the magnitue plot, showing a banwith of ω = 0. 25π as expecte; b. the phase within the passban (ie in the interval 0 < ω < 0. 25π ) is linear as ωl with L = 20. Frequency Response of the Filter in the Example Notice that the attenuation in the stopban is not more than 20B, in general not very acceptable. Since we are truncating the impulse response abruptly, we create high frequency components in the frequency response of the filter, which is responsible for the poor attenuation of the filter in the stopban. Following an approach similar to what we i with the DFT, where we were trying to lower the sielobes of the frequency spectrum, we can improve the response in the stopban by smoothing the impulse response with a winow.

7 VIDEO: Filter Design using a Winow (7:06) In this case the impulse response of the filter will be the same as before, but multiplie by one of the many winows, say the hamming winow, to obtain the impulse response wp w P h[ n] = sinc ( n L) w[ n], n = 0,..., N π π where w [n], n = 0,..., N is the winow sequence of length N +1. The expression for the hamming winow is 2π n wn [ ] = cos, n= 0,..., N N Let s continue with the example. Following the previous example, we want to esign a filter with passaban frequency ω P = π / 4, orer N = 40 an a hamming winow. Then the impulse response of the filter is 1 1 2πn h[ n] = sinc ( n 20) cos, n = 0,..., Let us see the frequency response using matlab: N=40; L=N/2; % Filter Orer. It has to be even % time elay n=0:n; % generate the vector of time ineces w=hamming(n+1) ; % vector of hamming winow values h=(1/4)*sinc((n-20)/4).*w; % Impulse response freqz(h,1) % show frequency response The plot of the frequency response is shown in the figure below. Notice the attenuation in the stopban, which is better than 50B! The phase shift of the filter within the banwith is still the same as 20ω.

8 Frequency Response using Hamming Winow. The impulse response of this filter is shown below. If you compare with the impulse response without winow, notice that the winow forces the impulse response to become small (close to zero) at the eges ( n = 0 an n = N ). This gives a smooth transition from the values outsie the interval, when it is force to be zero an the values insie the interval of efinition. Impulse Response with Hamming Winow.

9 Now let us pull everything together an summarize the steps in the esign of a Low Pass Filter from specifications in the frequency omain. There are two parameters we can easily set: the with of the transition region ω = ω S ω P an the attenuation in the stopban. These are shown in the figure below, together with the corresponing numerical values. Attenuation an Transition Region in terms of filter orer N an winow. Again the expression for the impulse response of the filter with passban ωp is wp w P h[ n] = sinc ( n L) w[ n], n = 0,..., N π π where w [ n], n 0,..., N is the winow. Although there is a very wie choice of winows, we just liste three significative ones: Rectangular winow w [ n] = 1, n = 0,..., N is just no winow; 2π n Hamming winow w [ n] = cos, n = 0,..., N N 2π n 4π n Blackman winow w [ n] = cos cos, n = 0,..., N N N VIDEO: Example of FIR Filter Design (9:16) Example. We want to esign a filter with the following characterisics: Passban 0 F 4kHz Stopban F > 5kHz with at least 40B attenuation Sampling Frequency F S = 20kHz

10 Then first specify the filter in the igital frequency omain ω P = 2π 4 / 20 = 2π / 5ra ω S = 2π 5/ 20 = π / 2ra Then the specifications in the igital frequency omain are shown in the figure below. Filter specifications in igital frequency From the attenuation we see that we nee to use a hamming winow. This yiels a transition region 8 π ( ω ω ) = π S P N 10 This yiels N = 80. Finally the filter impulse response 2 2 2π n hn [ ] = sinc ( n 40) cos, n= 0,..., FIR Low Pass Filter Design: Computer Base VIDEO: Computer Base Filter Design (10:00) Whenever possible, we try to use a computer base technique to esign a filter. In most cases there are a few parameters which specify the filter. These are shown in the figure below an efine as Pass Ban Frequency ω P ra Stopban Frequency ω S ra Pass Ban Ripple 20log10 ( A / B) B 20log10 A/ C B Stop Ban Attenuation ( )

11 Relevant Filter Parameters As we can easily see, these parameters attempt to quantify how close to ieal is the frequency response of a filter. One of the most effective esigns of igital filters is the equiripple filter, which is shown to be the one with the smallest maximum eviation from ieal. It requires a computerize algorithm which, in recent versions of Matlab, goes uner the name of firpm, with pm staning for Parks an MacClellan, the esigners. The same algorithm in oler versions was calle remez, with very similar parametrization. The first step in the esign is to specify the esire filter (say a Low Pass Filter, but it can be any kin of filter) by a set of parameters. Referring to the figure below, the filter is specifie by pairs of points of the frequency response. Specification of Ieal Low Pass Filter parameters Referring to the figure, given the four points, the ieal filter we want to approximate is etermine by linear interpolation of each pair. Before going into more etails, notice that the number of points has to be even, since the segments have to be isjoint. The reason is a bit like give an take. If you want to have a real goo approximation in specific regions of interest (in the passban 0 ωpass an the stopban ωstop π ) you nee to leave it loose in regions of less interest (the transition region ωpass ω STOP ). If we constrain it too much, by specifying what we want even within the transition region, we en up with an unsatisfactory esign. Again following the figure above, we see that the filter is specifie by two vectors, one specifying the amplitues

12 A=[1,1,0,0] an one specifying the frequencies efine as f = ω / π as f=[0,f1,f2,1] where f1 = ω / π an f2 = ω / π. PASS STOP Then we nee to specify the filter orer N, which can be one on the basis of an empirical formula which relates it to the esire attenuation of the filter, as N Attenuation (B) ~ 11 ω ω / π STOP PASS Finally, using these parameters, we use the function firpm to etermine the impulse response of the filter: h=firpm(n,f,a); The vector h yiels the impulse response of the filter as [ h[ 0], h[1] h[ N] ] h = Let s see an example. VIDEO: Low Pass Filter Design Example (5:44) Example.We want to esign a igital Low Pass Filter with the following specifications: Passban frequencies: Stopban frequencies: Attenuation in Stopban: Sampling Frequency: up to 3.0 khz above 3.5 khz 60 B 15.0 khz The first step is the specification of the filter in terms of igital frequencies ω = 2πF / : F S 3,000 2π ωpass = 2π = 15, ,500 7π ωstop = 2π = 15, Then we etermine an estimate of the orer of the filter

13 60 N = = ( ) 15 5 Finally we etermine the impulse response of the filter as h=firpm(82, [0, 2/5, 7/15, 1], [1, 1, 0, 0]); Now we nee to check whether we got the esire frequency response. We can see it immeiately by the comman freqz(h) which shows the frequency response as in the figure below. Frequency Response (Magnitue an Phase) of filter in the example, with N=82 As we can see from the magnitue plot, the attenuation in the stopban is 50B s, not exactly what we esire (60 B). The reason for this is the empirical formula for the filter length N is a bit conservative, so we nee to use a value larger than N=82. By some simple trial an error we can use N=95, so that the call for firpm becomes h=firpm(95, [0, 2/5, 7/15, 1], [1, 1, 0, 0]); Again using freqz(h) we plot the new frequency response as in the figure below.

14 50 Magnitue (B) Normalize Frequency ( π ra/sample) 0 Phase (egrees) Normalize Frequency ( π ra/sample) Frequency Response of filter in the same example, with N=95 Looking at the magnitue plot, we se that the attenuation is about 60B as we require from the esign specifications. The impulse of the filter can be plotte by stem(h) an it is shown in the figure below. Impulse Response of filter in the example

15 4. Example of Application VIDEO: Example of Application (9:38) In a number of applications, such as in Raar, Sonar an Communications, we transmit information through a pulse or a sequence of pulses. At the receiver the signal is attenuate by the propagation within the meium in between (say air, space or water) an corrupte by isturbances an noise. The first step at the receiver will be to try to eliminate some of the isturbances before processing. For example suppose we transmit a pulse mae of four perios of a sinusoi as x ( t) = Asin(2π F0t), 0 t T1 = 4T where the frequency F 0 an length in time T 1 are given by F0 = 500Hz 1 T1 = 4 sec = 8 msec 500 The signal is plotte in the figure below. 0 Transmitte Pulse x (t) in the example For example for a communication signal we sen a binary message mae of zeros an ones with (say) the following assignment 1 x( t) 0 x( t) In this case a transmitte sequence woul look yiel the signal shown in the figure below.

16 Transmitte Signal in the Example Suppose we have a situation in which a consierable amount of aitive noise is affecting the signal with a Signal to Noise Ratio of SNR = 0B. In this situation both signal an noise have the same power. Let the receive noisy signal be as shown in the figure below. Receive Signal in the Example (SNR=0B) At the receiver, we sample the signal with a sampling frequency of, say, F S = 20kHz Then we attenuate the effect of the noise by esigning a igital filter. In orer to esign the filter we nee to ecie on the passban an the stopban of the filter. Since the signal is the repetition of the basic pulse x (t) shown above, we nee to look at its frequency spectrum. In this case we generate the sequence x [n] as follows

17 % Genenerate the sample pulse F0=500; % Hz Fs=20000; % Hz T0=1/F0; % perio of sinusoi Ts=1/Fs; % sampling perio w0=2*pi*f0/fs; N0=4*T0/Ts; % ata length as 4 perios n=0:n0-1; x=2*sin(w0*n); The length of the sequence is N0=160 samples. Now we want to take the DFT of the generate sequence x. In orer to have a more accurate frequency spectrum, we zero pa it up to a length of (say) 1024 samples an we take the FFT. This is one all in one comman as X=fft(x, 1024); % take 1024 frequency samples Now we plot it. Since the signal is real we nee only half the frequencies, since the secon half of the values of the DFT are just the complex conjugate of the first half of the values: k=0:511; F=k*Fs/1024; % vector of frequencies plot(f, 20*log10(abs(X(1:512)))),axis([0, max(f), -50,50]) The axis comman sets the minimum an maxim values of the horizaontal an vertical axis of the plot, for better presentation. The effect is the magnitue plot shown below. Frequency Spectrum of the pulse an its estimate banwith From this figure here we see that the signal has most of its energy at the lower frequencies. Therefore a filter has to pass only the frequencies where the signal is

18 ominant an reject the other frequencies where the noise is ominant. As a rule of thumb to etermine the banwith of the signal, let s keep the frequencies which correspon to magnitues within 20B from the maximum, as shown in the figure. This yiels a banwith of about 1,000 Hz. Then we esign the filter using firpm as follows wp=2*pi*1000/fs; % passban igital freq. in ra. ws=2*pi*1500/fs; % stopban igital freq. in ra. fp=wp/pi; fs=ws/pi; A=40; % attenuation in B N=ceil(A/(11*(fs-fp))); % estimate of filter orer N=ceil(1.1*N); % a 10 percent h=firpm(n, [0, fp, fs, 1], [1,1,0,0]); freqz(h) This yiels the frequency response shown below. Frequency Response of Filter The vector h compute by firpm is the impulse response of the filter. Finally we filter the ata using the convolution comman as we i in the previous chapter. Therefore if we call y r [n] the row receive sequence, we compute the filtere sequence y f [n] as follows y f [ n] = h[0] yr[ n] + h[1] yr[ n 1] h[ N] yr[ n N] In matlab this is reaily obtaine by the conv comman, as yf=conv(h,yr)

19 This yiels the sequence shown in the figure below, compare with the receive sequence. Top: receive noisy sequence Bottom: sequence after the filter Comparing the filtere sequence with the original transmitte sequence we see that a goo amount of noise has been eliminate by the filter.