The phase ZH(e) can be described as: ∠H(e) angle of vector connecting i th zero to e of poles angle of vec ng j th pole to e) Again for our specific case, using the vectors u and u2 defined above we have ∠H(e-)=-(∠v1+∠v2), The phase starts off at 0 when w=0 and decreases to when w=3 symmetric poles, the phase keeps decreasing to -2T at w = T. The phase plot is bele ∠H(e) Problem 2(O&W 10.34) ym]=yn-1]+yn-2]+r{n-1] Taking the z-transform of this equation Y(z)=z-1Y()+x-2Y(2)+x-1Xx(x2) H(z) X 1+√5 1-√5 H(z) has a zero at z=0 and poles at z1 and Since the system is causal, the ROC of H(a)will be outside the circle containing its outermost pole:[=>|a1 The pzmap and ROC are depicted belowThe phase ∠H(e jω ) can be described as: ∠H(e jω ) = # X of zeros i=1 ( angle of vector connecting i th zero to e jω ) − # X of poles j=1 ( angle of vector connecting j th pole to e jω ). Again for our specific case, using the vectors v1 and v2 defined above we have ∠H(e jω ) = −(∠v1 + ∠v2), The phase starts off at 0 when ω = 0 and decreases to −π when ω = π 2 . Because of the symmetric poles, the phase keeps decreasing to −2π at ω = π. The phase plot is shown below: ∠H(e jω ) ω π 2 π −π −2π Problem 2 (O&W 10.34) (a) y[n] = y[n − 1] + y[n − 2] + x[n − 1] Taking the z-transform of this equation: Y (z) = z −1Y (z) + z −2Y (z) + z −1X(z) H(z) = Y (z) X(z) = z −1 1 − z −1 − z −2 = z z 2 − z − 1 = z z − 1 + √ 5 2 ! z − 1 − √ 5 2 ! H(z) has a zero at z = 0 and poles at z1 = 1+√ 5 2 and z2 = 1− √ 5 2 . Since the system is causal, the ROC of H(z) will be outside the circle containing its outermost pole: |z| > |z1|. The pzmap and ROC are depicted below: 3