17 Interpolation. Solutions to Recommended Problems

Transcription

1 17 Interpolation Solutions to Recommended Problems S17.1 It is more convenient to solve this problem in the time domain than in the frequency domain. Since x,(t) = x(t)p(t) and p(t) is an impulse train, x,(t) is a sampled version of x(t), as shown in Figure S x,(t) xp\ t Figure S x(t) = x(t) bt - n) n= -oo = Z x(n) 5(t - n) n= -oo Since y(t) = x,(t) * h(t) and x,(t) is impulsive, the convolution carried out in the time domain is as shown in Figure S y(t) Figure S \t Here we have that x(t) is sampled by a rectangular pulse train as opposed to an impulse train. S17-1

2 Signals and Systems S17-2 Since h(t) * (1/)h(t), shown in Figure S17.1-3, is wider than the sampling period, the resultant w(t) is not a triangularly sampled version of x(t). w(t) consists of the superposition of waveforms shown in Figure S \ t 0 2 Figure S w(t) --4 i 6 Figure S We note that this superposition is actually a linear interpolation between the samples of x(t). For example, Figure S convolved with Figure S equals Figure S x,(t) 2 1 Ji t Figure S17.1-5

3 Interpolation / Solutions S17-3 t t Figure S Figure S Now adding the shifted and scaled triangles yields Figure S17.1-8, which we see is the linear interpolation between samples of x,(t). 0 3 Figure S Now since (1/)h(t) * h(t) is Figure S17.1-9, we expect that w(t) is the linear interpolation of x,(t) shifted right by, shown in Figure S w (t) 2 Figure S /000 1'0-4 6 Figure S

4 Signals and Systems S17-4 S17.2 y(t) in all cases is the superposition of two signals. (a) y 1 (t) Figure S (b) Y 2 (t) _ -- Figure S (c) y 3 (t) sinin(t 1) (t -1) sin n (t - 2) 2 (t - 2) Figure S17.2-3

5 Interpolation / Solutions S17-5 S17.3 (a) 4 P(W) XP(wj) \ A'-Z 2Ir WCm ~+ Figure S WO We have 22r 27 - COM> Win, CO,> 2 wm, where w, is the sampling frequency. o ensure no aliasing we require that < 7r/om. (b) o recover x(t) from x,(t), we must interpolate. We have previously shown that by lowpass filtering, the spectrum is recovered, assuming that sampling has been performed at a sufficiently high rate. he interpolation may be done in many different ways, however, depending on the cutoff frequency we choose for the lowpass filter. For example, any of the filters HI(o), H 2 (w), and H 3 (w) in Figures S to S may be used to interpolate x,(t), whose Fourier transform is shown in Figure S17.3-2, to yield x(t).

6 Signals and Systems S17-6 Xp(o) Figure S HI(o) 2 2 Figure S H2(w) 7 7 I7 2 2 Figure S H3(w) 2 2 Figure S17.3-5

7 Interpolation / Solutions S17-7 (c) If x,(t) = ( x(n) b(t - n) n = -00 and H(w) is an ideal lowpass filter whose Fourier transform is shown in Figure S17.3-6, then and h(t) = sin irt ==ct WC sine c 7r x(t) = xp(t) * h(t) = wo -- x(n) sine (t - n) irn=-o sr H(w) -Coc 'oc Figure S S17.4 xxt X X t g (t) No H(cj) - y (t) p(t) Figure S17.4 We want to choose H(w) so that the cascade of the two filters is an ideal lowpass filter. In this example, G(x) = 2 sin A/ e ~jwa2 so that H(w) = eja/2 2 sin wa/2 H(w) = 0 otherwise IcoI < r/a,

8 Signals and Systems S17-8 S17.5 X,(Go) 1-41r X rx X 104 X(E2) 1-27r r Y(2) 1-27r 4 0 2r Yc() 0 X104 Figure S17.5 Solutions to Optional Problems S17.6 (a) We want ho[n] such that x,[n] * ho[n] = x 0 [n]. We sketch ho[n] in Figure S ho [n] 0 N-I Figure S17.6-1

9 Interpolation / Solutions S17-9 (b) xo[n] = x,[n] * ho[n] If there is no aliasing, we can recover x[n] from xo[n] by proper filtering. Since we require Ho(Q) = 1 -- e -" jqn. e jq(n- 1)/2 sin N/2 1 -u sin 0/2 H(Q) = N/HO(Q), A 0, 101 -<Iir N( eju(n -- 1)/2 sinw/2' ~ N 0, N< (c)' hl[n] = (1/N) (ho[n] * ho[-n]), so hi[n] is a triangular discrete-time pulse, as shown in Figure S h, [n]..?iiii I it.. -N 0 N Figure S (d) We require that H(Q) = N fori I < ~N' 0, for -7 < Q I < ir From part (c), [ N 2 (sin N/2)' 9/2 2 N ' H () = N 2 HO(U) 2 )1 =H( o N < I0I < r S17.7 (a) aking Fourier transforms of both sides of the LCCDE yields jy(wo) + Ye(W) = 1, 1 Ye(o) =., 1 + Jw'

10 Signals and Systems S17-10 so yc(t) = e-'u(t). By examining the sampler followed by conversion to an impulse train, we note that y[n] = yc(n) = e -" u[n] (b) Since y[n] = e -nu[n], we can take the Fourier transform to yield Y(Q) = e -" e -jfn n=o 1 - e (+jq) In order for w[n] = 6[n], we require W(R) = 1 for all Q. hus, Y(Q)H(Q) = W(Q) = 1, H(Q) = 1 - e -(+jq) Now since H(Q) = 2=_oo h[n]e -", we see by inspection that h[o] = 1, h[1] =-e h[n]= 0, n # 0, 1 S17.8 (a) Since X,(w) = X(w)P(w), we conclude that X,(w) is as shown in Figure S X,(w) K HwHs Figure S X,(w) = X(w) 6(co - no,) (b) From the convolution theorem, = Z X(nwo) 6(w - nwo) and y(t) = x(t)*p(t) 2 rn p(t) = - - Ws

11 Interpolation / Solutions S17-11 he fact that can be easily verified. herefore, {pxt)}= 6 - nw) 71= -OD 1 02,7rn\ y(t) = x(t) *- (s n =-OD 5 Kt- n, (c) We see from the sketch in Figure S that for no time-domain aliasing, 27r -- -, so W_s W, y (t) I WS 27 0 WS Figure S '// a (d) x(t) may be recovered from y(t), assuming that no time-domain aliasing has occurred, by low-time filtering y(t) from t = - to and applying a gain of os.

12 MI OpenCourseWare Resource: Signals and Systems Professor Alan V. Oppenheim he following may not correspond to a particular course on MI OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our erms of Use, visit: