Question

Question 1: a) For any linear phase filter, prove that if zo is a zero, then so must zobe. Hint: Using the properties of the z-transform, write h[n] = Eh[N - n) in the z-domain, and substitute 2 = 20. b) For any Type III or Type IV filter, prove that z = 1 is a zero. c) For any Type II filter, prove that z = -1 is a zero. d) In light of the above, find the zeros of the filter hinl= 10<n<3 10 otherwise Looking at the zeros, argue whether this is a lowpass, highpass, bandpass, or bandstop filter. Hint: Since all coefficients of H(2) are real, its complex roots must come in conjugate pairs. Combine this fact with a and c. e) Repeat part d for h[n] = { 0 otherwise h[n] = { (-1)" 0<n<3

Answer 1

ANSWERS- Given that (a) for any linear phase filter, that if Zo is a zero then so must to be Here we are using the properties of Zuransform, h (n) = th[n 1] in the 7-doman. H(z)= IZ [n(n=n)).-0 Heae Z[fcn)] means Z-Transform of f(n). h (n+w) <> EN HCZ) = P(Z). h (-n+N) 6PC71) from Eruation 0 => HCZ) = + Z5NH(24) =0 HW-0 If tó is zero is also fexo (b) for any type o or Type 1 falleg Z=4 is a zero - TV filter Type 11 - Aillene-ben) = -h(n-1) (n+ even) ; n(n)= (n-1) (N-odd).

H4+) = -7-WH(71) ht)= -1) NH4) HCH) = -#(1) 14(1=0 Z=1 is zero Similarly, irrespective Ņ is Odd (ov) Even. HCI)=0 Zul Zeve for Type D N © Foy any type of filter 77 is a zero h(n)= h (N-1) Hegre N is odd, longer NH is Even HLZ) = IN H(74) for Z=-) - H(-1)=(-5NH(-) since Ñ 18 odd => 464) = -H61) HC1=0 72-) is Zeyo Of Type - filley (d) In light of the above, the zeros of the filter . Si osns 3 hen) 20Otherwise . 10

hen) = $1,113 H(E)=1+3/+24+73. DIR=H(Z)= (23+72+z+1) 73 Zeros => N(E)=0 oto-1 -> (2+1 Z3+z²+2+1=0 (ZH) (Z²+1)=0 ZEDS> 72-, Zoti (e) h(n) = $1,-1,1,-13 HCZ) =1+74+7-2+7-3 H(2)- z3 z²+Z1 Z3 H(E)=0->) 73–7+7-1=0 (20 (2²+1)=0 Zeros => z=1, ti

(©) Repeat part de for hen}={ C Osna 20 otherwise Zero's at 70-ti HCZ)= 1+z+22+23 23 HCZ) =0 et Z=- Z-u ==1= woo Z= > WET 7==) =)W=472 ; w=714=bZ-patita Z=1 =) (H(2)]=4 @ W = 0 WT1 => -->> Hz)=0 clipele w=112 Z=j=) |H(z)l=0 unit andle The ciw) It can be of @ Band pass of parties. H(Z)=1-3+72-73 Zeros -> Zrliti 1 Hetfa------- .. the filter can be High pass on Band pass.