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### hw10_solution_2

Course: EE EL7133, Spring 2012
School: NYU Poly
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Word Count: 916

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Can 11.7 you change the order of an up-sampler and a down-sampler with out change the total system? In other words, are the following two systems equivalent? System A: M L System B: L M Consider the following cases: (a) M = 2, L = 2. (b) M = 2, L = 3. (c) M = 2, L = 4. Try an example in each case with a simple test input signal. Solution: (a) M = 2, L = 2. You should nd that System A and System B are not...

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Can 11.7 you change the order of an up-sampler and a down-sampler with out change the total system? In other words, are the following two systems equivalent? System A: M L System B: L M Consider the following cases: (a) M = 2, L = 2. (b) M = 2, L = 3. (c) M = 2, L = 4. Try an example in each case with a simple test input signal. Solution: (a) M = 2, L = 2. You should nd that System A and System B are not equivalent. (b) M = 2, L = 3. You should nd that System A and System B are equivalent. (c) M = 2, L = 4. You should nd that System A and System B are not equivalent. It turns out that the order of a down-sampler and an up-sampler does not matter when L and M are mutually prime (co-prime). That is, when the greatest common divisor of L and M is 1. 232 11.8 Simplify each of the following systems. (a) 2 3 (b) 2 2 (c) 5 10 2 Solution: (a) 6 (b) 4 (c) 2 2 This has the eect of setting the odd values of the input to zero. 233 11.9 Suppose the DTFT of x(n) is X f ( ): X f ( ) = 1 | | , for | | Suppose we generate the sequences y (n) and s(n) from x(n) with the following system s( n ) x(n) H (z ) 2 H (z ) 2 4 y (n) where H f ( ) = 1, | | < /2 0, /2 | | < Sketch X f ( ), H f ( ), S f ( ) and Y f ( ). Solution: X() 1 0.75 0.5 0.25 0 0.5 0 0.5 0.5 H() 1 0.75 0.5 0.25 0 0.5 0 234 AFTER H() 1 0.75 0.5 0.25 0 0.5 0 0.5 0.5 0.5 AFTER FIRST DOWNSAMPLER 0.5 0.25 0 0.5 0 AFTER SECOND H() 0.5 0.375 0.25 0.125 0 0.5 0 235 S() 0.25 0.1875 0.125 0.0625 0 0.5 0 0.5 0.5 Y() 0.25 0.1875 0.125 0.0625 0 0.5 0 236 12.1 Scaling Suppose the signal x(n) is bounded by 1, |x(n)| < 1. The impulse response of an FIR digital lter is h(n) = {0.2, 0.5, 0.1, 0.1, 0, 0, . . . } To ensure that the output signal y (n) does not exceed 1, how should x(n) be scaled before ltering? Show your work. Solution: We can use the time-domain scaling procedure, because we are given the upper bound of the input signal in the signal domain. ||h||1 = |h(n)| = 0.2 + 0.5 + 0.1 + 0.1 = 0.9 n So x(n) can be multiplied by 019 = 1.11 and you can still be sure that the output will not exceed 1 in . absolute value. In this case, x(n) is amplied a bit, which will increase the SNR in the presence of quantization eects in the implementation of h. 251 12.2 You have developed a system using an 8 bit uniform quantizer. For this system, the SNR (ratio of signal power to quantization noise power) turns out to be 10 dB, although the target SNR is 20 dB. How many additional bits do you expect is needed in the uniform quantizer in order to meet that goal? Solution: Generally, each additional bit will increase the SNR by about 6 dB. To increase the SNR by 10 dB will therefore require two additional bits in the quantizer. 252 12.3 SNR of a quantizer. In this problem we assume the original x(n) signal is Gaussian (normally distributed). We will measure the SNR SNR = 10 log10 2 x 2 e under dierent conditions. We will use the Matlab command randn to generate random numbers according to the Gaussian bell shape distribution. To simulate the quantization, we will use the Matlab program fxquant (available on the webpage). This program allows various options for how overows are treated. We will use the saturation mode, in which overows are clipped to the maximum and minimum. y = fxquant(x,b,round,sat); The program fxquant returns y with 1 y < 1. 2 (a) Choose x = 0.01 and use 10000 samples. Measure and plot the SNR in dB as a function of the number of bits b for 1 b 12. Check to see if the plot you obtain agrees with the expression obtained in the lecture notes, 2 SNR = 10 log10 x + 20 b log10 2 10 log10 2 Rf s 12 by plotting both curves on the same graph. (Note that fxquant uses Rf s = 2.) Does the SNR improve by 6 dB per bit? 2 Repeat with x = 0.1. Explain any deviation you nd from the SNR formula. What assumptions are violated? (b) The following experiment will be done with an 8 bit wordlength. Measure and plot the SNR in 2 2 dB as a function of x in dB for 70dB x 10dB. In a certain range, the SNR increases 2 linearly with x in dB. Check to see if the plot you obtain agrees with the expression obtained in the lecture notes, 2 2 SNR = 10 log10 x 10 log10 e by plotting both curves on the same graph. Explain any deviation you nd from SNR formula. What assumptions are violated? Solution: 2 (a) With x = 0.01, we have the following plot. 253 SNR VERSES WORDLENGTH, 2 = 0.01 x 60 50 SNR 40 30 20 10 0 10 0 2 4 6 8 NUMBER OF BITS 10 12 The dashed line shows the predicted value. There is a deviation from the predicted value. This is explained as follows. Recall that the predicted value of the quantization error variance came from a model which makes several assumptions. One of those assumptions was that the quantization step size is small compared to the signal amplitude. But that assumption is not satised when 2 x = 0.01 and the quantizer has only a couple of bits. Notice that where the assumptions are satised, the increase in SNR is 6dB per bit. 254 2 With x = 0.1, we have the following plot. SNR VERSES WORDLENGTH, 2 = 0.1 x 70 60 SNR 50 40 30 20 10 0 0 2 4 6 8 NUMBER OF BITS 10 12 The gures shows again a deviation from the predicted value for high values of b. In this case it is because some overow occurs, preventing the SNR from increasing beyond a certain value. (b) With an 8 bit word-length we get the following gure. SNR VERSUS 2, (8 BITS) x 70 60 50 SNR (dB) 40 30 20 10 0 10 20 70 60 50 40 30 20 2 (dB) x 10 0 10 It deviates from the predicted value where i. the signal amplitude is small compared to the quantization step size, and where ii. the signal amplitude is large compared to Rf s (non-negligible overow problem). 255
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