ECE 301 Fall 2011 Division 1 Homework 10 Solutions. { 1, for 0.5 t 0.5 x(t) = 0, for 0.5 < t 1

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1 ECE 3 Fall Division Homework Solutions Problem. Reconstruction of a continuous-time signal from its samples. Let x be a periodic continuous-time signal with period, such that {, for.5 t.5 x(t) =, for.5 < t Suppose we use the ideal impulse-train sampling model considered in class, to sample and reconstruct this signal. First, in order to avoid aliasing, we filter the signal using an ideal CT low-pass filter H with passband gain and cut-off frequency ω s /: { H, if ω ω s =, otherwise. We then sample the filtered signal at the rate of ω s /(π) samples per second, using an ideal impulsetrain sampler with sampling period T = π/ω s (i.e. multiplication by a periodic train of ideal continuous-time unit impulses with period T). From the continuous-time impulse train obtained through sampling, we reconstruct a signal y(t) using another ideal analog low-pass filter, H: { T, if ω ω s H =, otherwise. Find the reconstructed signal y(t) for the following values of ω s /(π). (i).5 samples per second (i.e., one sample is taken every two seconds). (ii).5 samples per second. (iii).5 samples per second. (iv).5 samples per second. For Parts (i), (ii), (iii), give expressions for y(t) without using MATLAB. For Part (iv), use MATLAB to plot y(t). Comment on your results. x(t) t Figure : Signal x(t) of Problem.

2 p(t) x(t) x (t) x p (t) H H y(t) Figure : Block diagram of the system in Problem. Solution. The plot of signal x(t) is shown in Fig., and the system diagram is shown in Fig.. Let us first consider the effect of the last two stages, sampling and low-pass filter H. Suppose X, X p, and Y are the CTFT of x, x p, and y, respectively. As derived in class, we have: X p = Y = HX p = T X (j(ω ω s )) { TXp (ω) if ω ωs if ω > ωs () Because of the first low-pass filter H, we have: X = for ω > ω s. Thus, there is no aliasing in X p. So () can be rewritten as follows: { T Y = T X if ω ωs if ω > ωs = X. () This means that y(t) = x (t). The last two stages together did nothing to the signal and can be omitted. Therefore, we only need to consider the effect of the first filter, to decide which frequency components of x it keeps and which ones it discards. To do that, let us compute the Fourier series expansion of x: x(t) = where T = is the fundamental period of x. a k = = = = πk j x(t)e t dt x(t)e jkπt dt = { jkπ e jkπt.5 { a k e j πk T t, (3).5.5 k.5 k = kπ sin(kπ/) k k = e jkπt dt The same result can be obtained by using the result of Example from the notes on CT Fourier transform. This is because x is the same as signal s from that example with A =, t =, and T =.

3 Figure 3: Matlab plot for Problem (iv). Notice that a k = a k, to get the following from Eq. (3): x(t) = + (a k e jkπt +a k e jkπt) k= = + a k (e jkπt +e jkπt ) k= = + a k cos(kπt) k= = + k= kπ sin(kπ/)cos(kπt) = + π cos(πt) 3π cos(3πt)+ cos(5πt)+... (4) 5π After the low-pass filter H, only those terms with angular frequency no greater than ω s / in (4) are preserved in x (t). Now let us see the results for different ω s. (i) If ωs = π, only the first term in (4) gets through H : x (t) =. (ii) If ωs =.5π, only the first terms in (4) get through H : x (t) = + π cos(πt). (iii) If ωs =.5π, only the first terms in (4) get through H : x (t) = + π cos(πt).

4 (iv) If ωs =.5π, only the first 6 terms in (4) get through H : x (t) = + π cos(πt) 3π cos(3πt)+ 5π cos(5πt) 7π cos(7πt)+ 9π cos(9πt). A plot of y(t) is shown in Figure 3. Problem. Sampling theorem. Consider sampling the signal below (which is defined for all time, < t < ). x c (t) = sin(5πt) πt cos(5πt) (a) x c (t) is sampled with sampling period T = / seconds, to produce the discrete-time signal x d [n] = x c (nt). Plot themagnitude ofthe DTFTofx d, X d (e jω ), over the interval π < Ω < π. Show as much detail as possible. (b) What is the Nyquist sampling rate for this signal? Solution. Multiplication in the time domain corresponds to convolution in the frequency domain. Breaking up x c (t) into x (t) = sin(5πt) πt and x (t) = cos(5πt), we have: { if ω < 5π, X = otherwise, X c = π X X = X = πδ(ω +5π)+πδ(ω 5π), if ω < 5π, / if 5π ω < 75π, otherwise, see the top three plots of Fig. 4. To get X d (e jω ) from X c, we can use the following formulas from the notes and the text: X p = ( X c (j ω πk )) (5) T T ( X d (e jω ) = X p j Ω ) = ( ) j(ω πk) X c. (6) T T T Note that the highest frequency in X c is 75π, which means that the Nyquist sampling rate for this signal is 75 samples per second. Since we are sampling at f s = Hz > 75 Hz, there will be no aliasing, i.e., the shifted replicas of X c /T in the formula (5) will not overlap. Thus, in order to get X p we simply put replicas of X c /T at ω =,±ω s,±ω s,... Then, to get X d (e jω ) from X p according to the formula (6), we rescale the frequency axis by a factor T = /, as shown in the bottom plot of Fig. 4. Note that X d (e jω ) =X d (e jω ), since X d (e jω ) is real (which is due to the fact that x d [n] = x c (nt) is even symmetric).

5 X.5 4π 5π 5π 4π X.5 4π 5π 5π 4π X c.5 4π 75π 5π 5π 75π 4π X d (e jω ) π 3π/8 π/8 π/8 3π/8 π Figure 4: Spectra for Problem.