Chapter 6: Problem Solutions

Transcription

1 Chapter 6: Problem s Multirate Digital Signal Processing: Fundamentals Sampling, Upsampling and Downsampling à Problem 6. From the definiton of downsampling, yn xn a) yn n n b) yn n 0 c) yn n un un d) yn e j0.n e) yn e j0.n un yn e j0.n un f) yn cos0.4n g) yn cos0.5 n cosn n h) yn sinn 0 i) yn cosn j) yn sinn 0

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10 4 s_chapter6[].nb à Problem 6.3 X () x[n] y[n] B In all cases Y X X for all Whene there is no aliasing, ie X 0 for then this relation simpilfies to a) B 5 Y X. Then there is no aliasing after downsampling and therefore Y is as shown below X () 5 x[n] y[n] Y () 5 b) B. Then again there is no aliasing after downsampling and therefore Y is as shown below X () x[n] y[n] Y () c) B 3 4. In this cases there is aliasing and we have to account for it. Best way to do it is proceed in two steps: sampling and then downsampling.

11 s_chapter6[].nb 5 The sampling operation yields Y X X The figure below shows both X and X. X ( ) Then downsampling yields Y Y, just a rescaling of the frequency axis. The final results is shown below. Y () Y () / / 3

12 6 s_chapter6[].nb d) B. Same reasoning as in c). This time it is easy to see that Y X X and therefore Y Y for all. for all à Problem 6.4 Recall that yn x n n x n n x n ejn a) yn n; b) yn n un c) yn d) yn e) yn which becomes ej0.05n ej0.05n ej0.05n ej0.05n un 4 ej0.05n 4 ej0.05n un yn cos0.05nun cos0.95nun f) similarly yn cos0.05n cos0.95n à Problem ej0.05n 4 ej0.05n un x[n] s[n] y[n] Using the DTFT. For upsampling S X and downsampling Substitute for S to obtain Y S S

13 s_chapter6[].nb 7 S X X S X X Therefore Y X from the periodicity of the DTFT. X X à Problem 6.6 From the diagram it is easy to verify that yn xn if n even 0 if n odd Therefore yn xn n, and Yz Xz Xz, or equivalently, Y X X. One way we can verify this is the following: call vn the output of the downsampler. Then V X X Since yn is the output of the upsampler then YV X X as we expect. à Problem 6.7 X () x [n] x M [n] y[n] 4 s[n] The effect of modulation on the frequency spectrum is as follows DTFTx M n X 0 X 0

14 8 s_chapter6[].nb where x M n xncos 0 n. After upsampling by the signal yn has DTFT Y X M X 0 X 0 a) 0 4. Then X M and Y are shown below. X M () / / 4 Y () 8 b) 0. Then X M and Y are shown below. X M () / / 4 Y () 4 c) 0. Then X M and Y are shown below.

15 s_chapter6[].nb 9 X M () Y () 3 4 In this case notice the maximum amplitude of the DTFT being "one" (rather then / as in the previous cases). This is due to the fact that X X. à Problem 6.8 In this problem we need to increase the sampling frequency from F s 8kHz to F s khz, ie by a factor. Therefore with ideal filters the scheme is as shown below. 8 3 x[n] 3 ) H ( s[n] y[n] H () 3

16 0 s_chapter6[].nb à Problem 6.9 In this case the Low Pass Filter H is a non ideal FIR filter. The whole problem is to choose the correct specifications for the filter. The stopband has to be S 3, since the purpose of this filter is to stop the frequency artifacts generated by the upsampling operation. The passband has to be decided on the basis of the bandwirdth of the signal we want to pass and the desired complexity of the filter. For a window based, recall that we need a hamming window (from the desired attenuation) with transition region 8 N. Therefore an FIR filter of length N will have a passband p à Problem N. The digital signal has frequencies at 6 3 rad and 6 3 rad. Therefore the output signal has frequencies at 3, 3 and also at 3 3, 3 3, 3 3 and 3 3. All components at 3 and 3 are going to sum with each other. à Problem 6. a) Hz 4z 3 6z 6 z 3 5z 3 z 6 z z 3 b) Hz z 5z 4 6z 6 z 3 4z z 4 z 6 c) Hz z z 5z 4 6z 6 z z 3 4z z 4 z 6 d) Hz z 3 4z 3 6z 6 z z 3 5z 3 z 6 z z 6 3z 3 z 3 e) Hz 0.5 n z n 0.5 n z n z 0.5 n z n whixh yields n0 n0 n0 Hz 0.5z z z z f) Hz z0.8 which can be written as Hz 0.8 n z n 0.8 n z n z 0.8 n z n. This becomes n0 n0 n0 Hz 0.64z z z The same result can be obtained by an alternative way:

17 s_chapter6[].nb Hz z z0.8 z0.8 z0.8 z amd you can verify that the two answers are the same. z 0.64 z 0.8z z 0.64 g) You can verify that the general exprezzion for the polyphase terms is H k z hnm kz n n for k 0,..., M. Applying this formula we obtain à Problem 6. H k z sin 5 3nk n 5 3nk z n for k 0,, a) From the transfer function of the filter Hz z z z 3 z 4 z 5 z 6 and M 4 we obtain the decomposition with Hz H 0 z 4 z H z 4 z H z 4 z 3 H 3 z 4 The block diagram of the system is shown below: H 0 z z H z z H z z H 3 z x [n] y[n] 4 H 0( z ) z z z 4 H ( z ) 4 H ( z ) 4 H 3 ( z) b) Using the same filters, the realization is as follows:

18 s_chapter6[].nb x[n] H 0 ( z) 4 y[n] z H ( z ) 4 z H ( z ) 4 z H 3 ( z) 4 c) When M the polyphase decomposition becomes H 0 z z z z 3 and Hz z z Therefore the system becomes as shown below: x[n] H ( z 0 ) y[n] z H ( z) Now notice that the cascade upsampler - downsampler (both by ) is just an identity. Also the cascede upsampler - time delay - downsampler as shown gives an output of zero, no matter what the input is (easy to verify). This is shown below: z 0 Therefore the overall system looks like this one:

19 s_chapter6[].nb 3 x [n] y[n] H 0( z ) Applications of MultiRate à Problem 6.3 a) In digital frequency, the passband is P radians, and the stopband is S radians. As a consequence the transition region is S P For a 60 db attenuation we can use (say) a Kaiser window with parameters N 4, and therefore the order is 4, 6. This yields a total number of 4, 66, multiplications and additions per second b) Since we want to reject all frequencies above 30Hz, we can downsample from the orginal sampling frequency (6kHz) down to 60Hz, ie by a factor D 6, As seen in class, this can be done in three stages by factoring D 00 as (say) as shown below F Fx 6kHz F 600Hz F3 0Hz F y x[n] H 3 H 0 H 5 Now the specifications of the filters become as follows:

20 4 s_chapter6[].nb H : F p 5Hz, F S Hz, H : F p 5Hz, F S Hz, H 3 : F p 5Hz, F S Hz, From the transition bands we can determine the orders of the filters. using the Kaiser window again we have the orders N,N,N 3 of the threee filters to be. respectively, 608 N N N Finally the total number of operations per second becomes: ops sec , which is reduction of a factor of more than 90. Also notice that we are not even attempting to save even more in computation using the polyphase decomposition! à Problem 6.4 Q) Recall the Butterworth filter frequency response: H c N Then for the reconstruction filter, the passband is F p khz, 000, that is to say p 44, 000 rad sec. The stopband has to be at half the sampling frequency, ie F S kHz, that si to say S 44, 00 rad sec. From the passband ripple we determine the factor e from and therefore 0.45 We determine the order from the requirement p N H S N Solving for N gives a very large number, and the analog filter cannot be implemented. Furthermore there will be a distortion in phase from the reconstruction filter itself. Q) Now the filter has still the same expression, but with N 4, since we are restricted to a 4 pole filer. Therefore the filter is given by H 44, Since we want the attenuation to be 40dB in the stopband, we can solve for the stopband, as

21 s_chapter6[].nb 5 which yields , S 44, Therefore the stopband of this filter is at F S 5.470kHz 3.4kHz. This has to coincide with half the sampling frequency, and therefore the new sampling frequency is F s 3.4kHz 6.5kHz. In other words we have to upsample at least by 6.5/44.=5. times before the Digital to Analog conversion.