George Mason University ECE 201: Introduction to Signal Analysis Spring 2017

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1 Assigned: March 7, 017 Due Date: Week of April 10, 017 George Mason University ECE 01: Introduction to Signal Analysis Spring 017 Laboratory Project #7 Due Date Your lab report must be submitted on blackboard by 5:00 PM on the day of your lab on the week that it is due. Late lab reports will be penalized as noted in the Lab syllabus. Preparation Before going to lab you should, at a minimum, read the pre-lab to get an understanding of the operators and programs that you will be using. Ideally, you will spend some time using MATLAB to perform some of the exercises in the pre-lab either at home or using the GMU Virtual Commuting Lab (VCL) before your lab session. Get started early! Lab Report Your lab report for this lab will consist of answers and complete documentation of the questions and exercises in Section 3. The pre-lab is designed to get you ready for the problems in Section 3. Honor Code Forgeries and plagiarism are a violation of the honor code and any reasonable suspicion of an honor code violation will be reported. You are allowed to discuss lab exercises with other students, but the submitted work should be original and it should be your own work. 1 Introduction When the input to a linear time-invariant filter is a complex sinusoid x[n] = e jnˆω the output will be a complex sinusoid of exactly the same frequency, multiplied by a complex number whose value depends on the frequency, ˆω, H(e j ˆω )e jnˆω This complex-valued function is called the frequency response. When the input to an LTI system is a sinusoid, x[n] = cos(nˆω) the output will be a sinusoid of exactly the same frequency, but scaled in amplitude and shifted in phase, A cos(nˆω + φ) where A is the magnitude of the frequency response at frequency ˆω and φ is the phase. One of the goals of this lab is to study the response of FIR filters to inputs such as complex exponentials and sinusoids, and to better understand the concept of a frequency response function.

2 In this lab you will learn how to use freqz to find the frequency response of a filter, and learn how to characterize a filter by knowing how it reacts to different frequency components in the input. This lab introduces two practical filters: averaging filters and nulling filters. Averaging filters are used to smooth out noise in a signal and are commonly used in plots of stock market data to help remove the erratic daily fluctuations in stock market prices. Nulling filters can be used to remove sinusoidal interference, such as jamming signals in a radar, or harmonic interference from power lines. Pre-Lab.1 Frequency Response of FIR Filters Systems described by a difference equation of the form M b k x[n k] are linear and time-invariant and have a impulse response that is finite-in-length. The frequency response of these filters can be found by using a general complex exponential as an input and finding the response. For example, to find the frequency response of the two-point averaging filter 1 x[n] + 1 x[n 1] let x[n] = e j ˆωn and use the difference equation to find the output, which is 1 ej ˆωn + 1 { = e j ˆωn e j ˆω} (1) ej ˆω(n 1) In (1) there are two terms, the original input and a term that is a function of ˆω. This second term is the frequency response of the system, is denoted by H(e j ˆω ), and is equal to { H(e j ˆω ) = e j ˆω} () Thus, H(e j ˆω )e j ˆωn Once the frequency response is known, the effect of the filter on any complex exponential may be found by evaluating H(e j ˆω ) at the corresponding frequency. For example, if ˆω = π/ then the frequency response of the two-point averaging at ˆω = π/ is { H(e jπ/ ) = e jπ/} = 1 (1 j) = e jπ/4 and it follows that the response of the filter to the complex exponential x[n] = e jnπ/ will be e jπ/4 e jnπ/ = ej(nπ/ π/4)

3 Thus, the complex exponential is scaled by / and is shifted in phase by π/4. If, on the other hand, the input were a cosine of frequency π/, x[n] = cos(nπ/) then the output would be For a more general FIR filter of the form cos(nπ/ π/4) the frequency response is M b k x[n k] H(e j ˆω ) = M b k e j ˆωk (3) Since the frequency response is a complex-valued function of the frequency variable ˆω, it is often expressed in terms of its magnitude and phase, H(e j ˆω ) = H(e j ˆω ) e jφ(ˆω) and to plot the magnitude and phase as a function of ˆω. For the two-point averaging filter defined above, the magnitude and phase is shown in the following figure, where the magnitude is given in db, which is defined to be 0 log H(e j ˆω ). 0 Magnitude (db) Phase (degrees) Normalized Frequency ( π rad/sample) Normalized Frequency ( π rad/sample)

4 . MATLAB Function for Frequency Response MATLAB has a built-in function called freqz for finding the frequency response of a discrete-time LTI system. The following MATLAB statements show how to use freqz to compute and plot both the magnitude (absolute value) and the phase of the frequency response of a two-point averaging system as a function of ˆω in the range π ˆω π: >> bb = [0.5, 0.5]; % filter Coefficients >> ww = -pi:(pi/100):pi; % omega hat >> H = freqz(bb, 1, ww); % aa=1 for FIR filter >> subplot(,1,1); >> plot(ww, abs(h)) >> subplot(,1,); >> plot(ww, angle(h)) >> xlabel( Radian Frequency ) For FIR filters, the second argument of freqz must be equal to 1. Also, the frequency vector ww should cover an interval of length π for ˆω, and its spacing should be fine enough to give a smooth curve for H(e j ˆω ). 1 Note: We will always use capital H for the frequency response..3 Periodicity of the Frequency Response The frequency responses of a discrete-time filter is always periodic with a period of π. Explain why this is the case by stating a definition of the frequency response and then considering two input sinusoids whose frequencies are ˆω and ˆω + π. x 1 [n] = e j ˆωn ; versus x [n] = e j(ˆω+π)n Note: The implication of periodicity is that a plot of H(e j ˆω ) only needs to extend over the interval π ˆω π..4 The MATLAB FIND Function Often signal processing functions are performed in order to extract information that can be used to make a decision. The decision process inevitably requires logical tests, which might be done with if-then constructs in MATLAB. However, MATLAB permits vectorization of such tests, and the find function is one way to do a lot of tests at once. In the following example, find extracts all the numbers that round to 3 for a vector of values that lie between 1.4 and 5 in increments of 0.33: >> xx = 1.4:0.33:5, jkl = find(round(xx)==3), xx(jkl) The argument of the find function can be any logical expression. Note that find returns a list of indices where the logical condition is true. See help on relop for information on relational operators. (a) Consider the frequency response of a seven-point averaging filter, >> ww = -pi:(pi/500):pi; >> HH = freqz( 1/7*ones(1,7), 1, ww ); 1 If the output of the freqz function is not assigned, then plots are generated automatically; however, the magnitude is given in decibels which is a logarithmic scale. For linear magnitude plots a separate call to plot is necessary.

5 Use the find command to determine the indices where HH is zero, and then use those indices to display the list of frequencies where HH is zero. Since there might be round-off error in calculating HH, the logical test should probably be a test for those indices where the magnitude (absolute value in MATLAB) of HH is less than some small number, e.g., (b) Plot the magnitude of the frequency response, >> plot(ww,abs(hh)) and verify that your results from the find function correspond to the frequencies where the magnitude of the frequency response is zero. 3 Lab Exercises In these exercises you will be finding the frequency response of two different types of filter: an L-point averager, and three-point nulling filters. 3.1 Frequency Response of Averaging Filters We have seen a number of examples involving a filter that averages input samples over certain intervals. These filters are called running average filters or averagers and they have the following form for the L-point averager: 1 L 1 x[n k] (4) L (a) Use Euler s formula and complex number manipulations to show that the frequency response of a 5-point running average filter is given by: H(e j ˆω ) = 1 + cos(ˆω) + cos(ˆω) 5 e jˆω (5) (b) Use the geometric series N 1 α k = 1 αn 1 α to show that the frequency response of the 5-point running average filter may also be expressed in the form H(e j ˆω jˆω sin(5ˆω/) ) = e sin(ˆω/) (c) Implement (5) directly in MATLAB. Specifically, create a vector that includes 400 samples between π and π for ˆω, and plot the function H(e j ˆω ) at these frequencies. Since the frequency response is a complex-valued quantity, use abs and angle to find the magnitude and phase of the frequency response for plotting. Plotting the real and imaginary parts of H(e j ˆω ) is not very informative. (d) In this part, use freqz to find H(e j ˆω ) from the filter coefficients, and plot its magnitude and phase versus ˆω. Write the appropriate MATLAB code to plot both the magnitude and phase of H(e j ˆω ). Follow the example in Section.. The filter coefficient vector for the 5-point averager is defined by:

6 >> bb = 1/5*ones(1,5); Note: The function freqz(bb,1,ww) evaluates the frequency response for all frequencies in the vector ww. It uses the summation in (3), not the formula in (5). The filter coefficients are defined in the assignment to vector bb. How do your results compare with part (c)? 3. Nulling Filters for Frequency Rejection Nulling filters are filters that completely eliminate some frequency component. These can be useful to remove jamming signals or to eliminate unwanted interference, such as 60 Hz harmonic hum from power lines. The simplest possible nulling filter has three coefficients and has the form x[n] cos(ˆω 1 )x[n 1] + x[n ] (6) where ˆω 1 is the nulling frequency, i.e., H(e j ˆω ) will be equal to zero at ˆω = ˆω 1. For example, a filter designed to completely eliminate signals of the form A cos(0.5πn + φ) would have the following coefficients b 0 = 1, b 1 = cos(0.5π) = 0, b = 1 In order to remove more than one frequency, we may form a cascade of two or more nullling filters. Consider, for example the following two filters, w[n] = x[n] cos(ˆω 1 )x[n 1] + x[n ] w[n] cos(ˆω )w[n 1] + w[n ] (FIR Filter-1) (FIR Filter-) where the first filter removes sinusoids with a frequency ˆω 1 and the second removes sinusoids with a frequency ˆω. The cascade of these two filters is illustrated in the following figure. x[n] w[n] y[n] FIR FIR Filter #1 Filter # Figure 1: Cascade of two FIR nulling filters. (a) Show that the filter defined in Eq. (6) removes any complex exponential having a frequency ±ω 1, and therefore will remove any cosines of the form x[n] = A cos(nˆω 1 + φ) (b) Design two nulling filters of the form given in Eq. (6), where the first eliminates cosines with a frequency of ˆω 1 = 0.π, and the second eliminates cosines with a frequency of ˆω = 0.85π. (c) Plot the magnitude of the frequency response of the two filters in a two-panel plot using the subplot command to verify that the two filters that you designed will eliminate the desired frequencies. (d) If the input to the cascade of these two nulling filters is x[n] = 0 cos(0.πn) + 0 cos(0.85πn π/4) + 5 cos(πn π/3) (7) what is the output of the first filter, w[n]?

7 Hint: Recall that when a cosine is the input to a LTI system, the output is a cosine of the same frequency but scaled in amplitude and shifted in phase, x[n] = cos(nˆω 0 ) A cos(nˆω 0 + φ) where A is the magnitude of H(e j ˆω ) evaluated at ˆω = ˆω 0 and φ is the phase (angle) of H(e j ˆω ) evaluated at ˆω = ˆω 0. Use linearity to find the response to the sum of three cosines. (e) What is the output of the second filter, i.e., the signal y[n] coming out of the cascade of FIR Filter #1 and FIR Filter #. (f) Use MATLAB to create the signal x[n] given in Eq. (7) that is 150 samples long over the range 0 n 149. (g) Use filter or conv to process x[n] with a system that is a cascade of the two filters you designed in part (a), i.e., first filter x[n] with h 1 [n] and then filter with h [n]. Thus, the output will be y[n] where (x[n] h 1 [n]) h [n] (h) Make a plot of the output signal, and compare your result to what you found mathematically in part (e). Are the two results the same? If not, where are they different, and can you give a reason for why they are different?

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