Problem Set Solution 12: Filtering

12 Filtering Solutions to Recommended Problems S12.1 (a) The impulse response is real because h(t) = H(w)ew' dw, - 21­ f h*(t) = H*(w)e ~i"' dw = H(w)e -i-t dw - H(-w)e-' do = --- 1 H(w)eij' dw = h(t) where we used the fact that H(co) = H*(o) = H(-w). The impulse response is even because H(w)e'wt dw, h(t) = -- 21 h(-t) = ao H(w)e -jwt dw - 2,7 ­ 21-o = H(-w)e' t dw Since H(-w) = H(w), H(w)ej' dw h(-t) = = h(t) The impulse response is noncausal because h(- t) = h(t) # 0. (b) x(t) = b(t - 9n), ( x(t) = aejitsl2ir k= -00 ak an = = M~-j[(2"k*)ITI 9odt x(t)e Here T = 9, s0 ak = - 9 and 2rk 5J{ei(2*kt)!TI) = T) Consequently, the Fourier transform of the filter input is as shown in Figure S12.1-1. X(W) 27r 9 47T 97 27r 9 0 2In 9 4nf 9 Figure S12.1-1 S12-1 Signals and Systems S12-2 Since Y(w) = H(w)X(w), the Fourier transform of the filter output is as shown in Figure S12.1-2. Y(W) 2ni 9 2n 0 2r 9 9 Figure S12.1-2 (c) We determine y(t) by performing an inverse Fourier transform on Y(W) as found in part (b). Using superposition, we have 21rt 9 1 2 = -+-co y(t)YO=9 +9 CO S12.2 From the filter frequency response plots we can determine that H(w) = 0.25e -j(r/8) 4 H(w) = 0.5e -(T/) at w = wi = 7, at w = W2 =2r Using superposition, we easily determine y(t) to be y(t) = 0.25 sin(irt + 1/8) + cos( 2rt ­ 71 S12.3 dvo (a) RC+v dt = v, Taking the Fourier transform of this equation, we have (RCjo + 1)Ve(w) = V,(w) We now define H, ( Ve(w) V,(w) 1 1 + jwRC We can see from this expression that vc(t) is a lowpass version of v,(t). Filtering / Solutions S12-3 The magnitude and phase of Hi(w) are given in Figure S12.3-1. OdB -20 -40 -60 0.1/r 1/r 1O/T 100/r 0.1/r 1/7- 10/k 100/r r = RC -37r/4 L r = RC Figure S12.3-1 (b) RCd(v - dt Vr) + V, - RCjwV,(w) - RCjwV,(w) - Vr =Vs, Vr(W) = 0, (jwRC)V,(w) = (1 + jwRC)Vr(w), jwRC V,(w) _ V,(w) 1 + jwRC Signals and Systems S12-4 The magnitude and phase of H 2(w) are given in Figure S12.3-2. 3 0 0 0 0 CN 1 R C 10 RC Figure S12.3-2 (c) The cutoff frequencies are we = 1/RC in both cases. (d) V(M) V,(w) 1 - H1) = jwRC 1 + jwRC = 11= H2(W) This is the same frequency response as sketched in part (b). We have trans­ formed a lowpass into a highpass filter by a feed-forward system. The cutoff frequency, as in part (c), is w, = 1/RC. S12.4 Consider 0 go !5 r. In this range, the gain of the filter IH(Q) I is go. The phase shift for the positive frequency component is +7r/2 and the shift for the negative fre­ quency component is -7r/2. Since 1 2 x[n] = cos (gon + 0) = -[e(On+0) + e -j6on+Oj, y[n] = 0 [eI1On+O+(/ 2 2 ) + e -j[Uon+O+(ir/2) (on+") - e ~j(fl0 n+') 2 y[n] = -9 0 sin (gon + 0) =j [e It is apparent from this expression that H(Q) is a discrete-time differentiator. A sim­ ilar result holds for -7r ! go s 0. Filtering / Solutions S12-5 If go is outside the range -7r :5 go go within this range. For example, r, we can express x[n] identically using a x[n] = cos (n = cos y[n] = 7 + - sin ) n + - n , + ) S12.5 (a) We see by examining yi[n] and y 2[n] that y[n] averages x[n] and thus tends to suppress changes while y 2[n] tends to suppress components that have not var­ ied from x[n - 1] to x[n]. Therefore, the yi[n] system is lowpass and y 2[n] is highpass. (b) Taking the Fourier transforms yields Yi(Q) = X(Q) 1+ e2' Hi(9) = -(1 + e -ja 2 Figure S12.5-1 Y2(Q) = X(Q) 2e H 2(0) = -(1 - e ) 2 Signals and Systems S12-6 S12.6 (a) By inspection we see that the impulse response is given by NN [ h,[n] = (b) H 2 () = 1 E k= b[n - k] -N sin Q2N + 1) 1 ­ 2N + 1 2N + 1 Ssin(/2) (c) |H1 (2)l 2w 2T ­ 21r2 2N + 1 2N + 1 2' 2N+ 1 Figure S12.6-1 IH2WA I 2w 2N+1 I 2tr2 2N+1 Figure S12.6-2 Zero and one crossings are at 2,r (2N + 1) k. (d) H 2 (Q) is an approximation to a highpass filter. S12.7 (a) From the specification that H(O) = 1, we know that H(w) = (b) (h) ( H()| I1~) a = 1 (a2 + 4)'/2 = (a 2 + a Wa2)1/2) H()I 1,2 a a + jw 21T 21r+ 2, 2N +1 Filtering / Solutions S12-7 The low end specification is satisfied for a > 4, as shown in Figure S12.7-1. IH(w)I w=2 5 4 Figure S12.7-1 The high end specification is met for a !: 6, as shown in Figure S12.7-2. IH( o)l w =6 -f2­ 6 Figure S12.7-2 The range of a such that the total specification is met is 4 2! a 2 6. Solutions to Optional Problems S12.8 The easiest method for solving this problem is to recognize that passing x(t) through H(w) is equivalent to performing -2 dx(t) dt This is easily seen since x(t) = - 27r -2 dx(t) = 17- f dt 2,rJ -2 dx(t ) lt dt -. X(w)eiw' dw, -2jw X(w)e- t dw HCCQ) -2jwX(w) Signals and Systems S12-8 (a) -2 dx(t) dt ) (b) -2 ()2 dtdt 2 = -2jej' = y(t) dti o ut) d[(sin dt w2t)u(t)] -2 o2w(cos wt)u(t) 1 = XMx (C) jw(6 + jw) x(t) = 1[u(t) -2 dx(t) dt = = -2 - 6 _ - =6J jw , ". + 6 +jw 6t u(t) - e -6 1u(t) [16(t) + [6 6 e -6t6(t) 6 -2e- -u(t) Alternatively, for this part it is perhaps simpler to use the fact that Y(w) = H(w)X(w) = - 2jw JM(6 + jw) 2 6 + jw so that y(t) = -2e -6 1 u(t) 1 X(M) = 2 + jo x(t) = e -2 1u(t) (d) -2 dx(t) dt - 2[-2e 2 'u(t) + e -2'tt)]= 4e -2 1 u(t) - 26(t) S12.9 (a) H(Q) = H,(Q)e ~m" (i) H,(Q) is real and even: h4n] - H,.() From Table 5.1 of the text (page 335), we see that the even part of h4n] has a Fourier transform that is the real part of H,(Q). This result is easily verified: Sh4-nle-j ),h[n]e(G = so s(h,[n] + h,[-) - Ev~h,[n]} [H,(Q) + H*(n)], Re{H,.(Q)} Now since Re{H,.(Q)} = H,(Q), we have that Ev{hn]} = h,.[n], i.e., h,[n] is even, and therefore h,[n] = h,-n] Filtering / Solutions S12-9 (ii) From Table 5.1, -e x[n - no] -iflno so H,(Q)e -j" h,[n -l h[n] = hn , - M] (b) h4n] = h4-n] Since h[n] = hAn - MI, h[n + M] = hn], h[M - n] = h,(M - n) - M] = h,[-n], but hn] = h,-n] =+ h[M - n| = h[M + n] (c) h[M + n] = h[M - n] from part (b). Since h[n] is causal, h[M - n] = 0 for n > M. But if h[M + n] = h[M - n], then h[M+ n] = 0 for n > M, so h[n] = 0 for n > 2M Summarizing, we have h[n] = 0 for n < 0, n > 2M S12.10 (a) Hi (n) 1- ­ --4- Q i - I _ ( T Hv( N Figure S12.10-1 c 2i( r + Qc 7T Signals and Systems S12-10 (b) If the cutoff frequency Q. = ir/N, the total system is an identity system. N-1 (c) h[n] = L k=O N-1 ej(2 7rnk/N)ho[n] hk[n] = k=O [1- e j2rn (2nN)Ih[n], n = an integer multiple of N, n # an integer multiple of N, so r[n] is as shown in Figure S12.10-2. h[n] Nho[n], 0, (d) ho[n] = -, N) ho[n] = 0, n = 0, n an integer multiple of N, are the necessary and sufficient conditions. = S12.11 From the system diagram, Y(w) = X(w)[a - G(w)], H(w) = a - G(w) (a)