PROBLEM SET 5. Reminder: Quiz 1will be on March 6, during the regular class hour. Details to follow. z = e jω h[n] H(e jω ) H(z) DTFT.

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1 PROBLEM SET 5 Issued: 2/4/9 Due: 2/22/9 Reading: During the past week we continued our discussion of the impact of pole/zero locations on frequency response, focusing on allpass systems, minimum and maximum-phase systems, zero-phase systems, and linear-phase systems. This material is discussed in OSYP Secs as well as the lecture notes on frequency response from pole/zero locations. We also began our discussion of continuous--time sampling, following material in OSYP Secs. 4.0 through 4.3. We will continue with sampling in discrete time, focussing on change in sampling rate in the week to come, following material in OSYP Secs. 4.6 and 4.7. Reminder: Quiz will be on March 6, during the regular class hour. Details to follow. Unit Sample Response DTFT/IDTFT z e jω h[n] H(e jω ) H(z) Iteration with x[n] δ[n] or formal solution Inspection DTFT Inspection ZT/IZT Magnitude and phase analysis Inspection Z-transform + ROC Difference Equation + Initial Conditions Pole/Zero Locations + Gain Factor Note: The diagram above summarizes the five complementary representations that we have learned to characterize discrete-time signals and systems. As you know, the unit sample response is used to characterize time-domain qualities, the DTFT is the basic frequency representation, the z-transform provides greater insight into stability, causality, and an easier approach to inverse transforms, the difference equations are used to actually implement digital filters, and the pole/zero locations provide valuable design insight. The first problem is a drill that addresses these relationships between the various representations. Feel free to use MATLAB to facilitate your solutions to the problem (tell us how you use it), unless stated otherwise.

2 8-49 Problem Set 5-2- Spring 209 Problem 5.: We know the following facts about a stable LSI system under consideration: The system has three poles located at z j 2 and z The system has one zero located at He j 0 that represents the transfer function of the system including all multiplica- (a) Write the z-transform tive constants. Hz z (b) Write the difference equation characterizing the system, including the initial conditions. (c) Determine hn, the unit sample response of the system. (Again, feel free to use MATLAB liberally here.) (d) Is the system causal? Is it linear phase? Why or why not? Problem 5.2: Problem 5.45 from OSYP, parts (a) through (i) only

3 8-49 Problem Set 5-3- Spring 209 Problem 5.3: x c (t) x s (t) Convert to Discrete Time x[n] x c (nt ) LSI system y[n] y s (t) Convert to Continuous Time n y[n]δ(t nt 2 ) Continuous- Time LPF y c (t) s(t) δ(t nt ) n h[n] H LP (j) Ideal C/D Conversion, Sample time /T Ideal D/C Conversion, Sample time /T 2 The figure above is a block diagram of a generic system that converts a continuous-time signal x c t into its discrete-time representation xn by uniform sampling, passes hn through a discrete-time filter with sample response hn and reconstructs the continuous-time signal y c t by lowpass filtering. Note that the sample times T and T 2 associated with the sampling and reconstruction could be different. x c (t) X c (j) s(t) t 2π/T W W S(j) 2π/T 4π/T x s (t) t 2π/T X s (j) 2π/T 4π/T x[n] t 2π/T W W X(e jω ) 2π T W 2π/T 4π/T n 4π WT WT 2π WT 2π 4π ω The figure above summarizes the major steps of the sampling process. Note that the functions in the time domain are arbitrary, and that the triangular frequency representation is also arbitrary and clearly is not the shape of the actual CTFTs/DTFTs depicted. (Triangles are just easy to draw.) Also, the sampling period in the figure above is indicated by the generic T rather than T. Finally, note that the function x s t is a continuous-time function that consists of a sequence of delta functions, while the function xn is a discretetime sequence with the amplitudes of the discrete-time samples of xn equal to the areas of the impulses in x c t, and both of these are equal to the original sampled values of x c t. As we discussed in class, the sampling process is summarized by the equations:

4 8-49 Problem Set 5-4- Spring 209 x s t x c st t xn x c nt x s t x c nt t nt n xn x c nt x c lt n l and in the frequency domain, Sj l k T T k Sj X s j Xe j k T T k Xj k T T k X j k T T T k and As you know, the maximum input frequency W must be less than half the sampling rate, aliasing distortion. T to avoid The reconstruction of the continuous-time output begins with the discrete-time filter output converted into a continuous-time train of delta functions y s t separated by T 2 seconds, and with areas equal to the amplitudes of the samples of yn. The filter H LP j Hj is typically an ideal lowpass filter with frequency response T T 2 0 otherwise Finally, assume that the discrete-time LSI system depicted above is also an ideal lowpass filter with frequency response

5 8-49 Problem Set 5-5- Spring 209 with He j periodic with period 2 In working this problem, assume that the input signal depicted below: He j x c t is bandlimited to 0 khz, with the spectrum X c (j) 2π0 4 2π0 4 Sketch and dimension in the frequency domain the functions X s j, Xe j, Ye j and Y c j for each of the following values of T and T 2 : (a) T T (b) T T (c) T and T Problem 5.4: In this problem we reconsider the system of Problem 5.3, but with a periodic input. Now assume that the discrete-time filter is allpass with He j for all. Let x c t cos28000t (i.e. a cosine with frequency 8 khz). Repeat Problem 5.3 (a), (b), and (c), but with this cosine function as input. Also write out the time functions yn and y c t that you obtain in each case. Note: It is important to remember that the frequency axis is scaled as signals are converted from xn x s t and back from yn to y s t. This in effect causes the delta functions to expand or contract in fre- to

6 8-49 Problem Set 5-6- Spring 209 quency, which affects their areas as discussed in Problem 4.5 (b) and (c). You need to be mindful of this in order to obtain the correct answers for this problem. Problem 5.5: Problem 4.3 in OSYP