Question

Question Three: The system function of a discrete-time system is 2 1-e-0.22-1 1 1-e-0.42-1 a) Assume that this discrete-time filter was designed by the impulse invariance method with Td = 2, i.e. h[n] = 2h.(2n), where he(t) is a real-valued impulse response of a continuous-time filter. Find the system function H.(s) of a continuous-time filter he(t) that could have been used for this design. (8 marks) b) Assume that H(2) was obtained by the bilinear transformation method with Td = 2. Find the system function He(s) of a continuous-time filter that could have been used for this design. (8 marks) c) Suppose that H (2), H2(2), and H(2) are transformed versions of continuous-time systems Ha(s), He(s), and He(s), respectively. Which one of the two methods, i.e. impulse invariance, bilinear transformation, both, or neither, will guarantee that H() = H (2) H2(2) whenever He(s) = He(s) H2(s)? Justify your answer. (9 marks) (Total: 25 marks)

Answer 1

soyle Home Theatre Using Impulse invarience method (1.2. Method) Here To = te h (n] = Td h (nId) = 2.4 (27) 22. he (2n) hc ln Td) = sampled signal of continuous lime - signal fransfer function helt) impulse invarience Transformation in case of method is given by Z= est (Stp) this can be unclesslood by following enample- Let Hts) = 1 in s-domain infilt) = et ult) in lime domain .. sampled hlt) = h (OTA) = è Pinta) un kinah (nd) = & Phd un 2. Transform - H (2) = { hindizen ns-co e é Plata), zon noo | H2) = 60ww.z-) Hence $t Pi izland (1-e

for the Ml2)= given H1 (2) - question Hz (2) l-e 0.2 27 I-e0427 H112) - PiTd = -0.2 € Tdi A1 = 2 8 Pi = 0-2 = 0.22 0.1 A1 = 2 2 = 1 :. 4,9 = 1540 H₂ (2) - P2 Td = -0.4 3) Py = 0.4 = 0.2 of Td'A2 = 1 :: My essa ja he is 0.2) Td H(s) = H, Cs) - H₂ Ls) 2 - 1 Istoly (sto2) a 2 (5+0.2)-(3+0.1) (Sto. (sto-2) Me [5) = 15102] s +0.3 (stol (sto.2)

Here in above equation Hc(s) only,(no bar on s, which i have written by mistake, so please ignore this mistake)

(6) Using In this Bilinear method Transformation Transformation Method (BLT) used is - SI = (27 212-1)= ST (2+1) 22-2 – ST. Z + ST 1 2-ST 212-5T) = (ST+2) 2= ( STAD for geven H2) = 1 o 21) - Theron 2) pufting 2- (ST+2) Andere T= Td = 2 > 2= 2 1.2- 2-ST HCS) - is a video -0.4 + S HS = 2(5+1) - (sty (5+1) - €0" (1-5) (5+1) - € 0.4 (1-5)

H(S) H (5= 2 (SU (st) s(lte ol) + (1-éo) sliteout (1-60 2(5+1) - (St) 11.95 +0.095 (1•675 +0.33) (©) Given - Hals) Heals) Hi (2) in 8-domain H2 (2) l in s-domain H12) " his) Let us consider one example where He(s) = (sten Heals) = (sters Hel (3) He2(3) = heister using partial fraction expansion - Mels) Hel) - He69 = ei testen = A ز را رها رمان B- (step Isabel He(s) = H (3) Hez (8) = 1 ) - (Stp) + ( P-pe) (StP2)

- Now In coexe of Implutse. Invarvience method (z-set) for Hejcs) = (ster , Hcz(s) = 18tes H1 12) = 1 The Pita 27) LH412). H212) = 1-e Peld zt) (lepi H(2) Again from egn. De He(s)= Hel(s) Hez (S) = (P1-P2) PP2) 18+Pi) (stP2) in H (2) = Td [in e Pita, zl) + (Pirty Td I . To 11-e P2Td z =) - (PI-P2) H (2) = - 11-é Pitd. 21) (1 - 6 P2 Td. 2-1) Hence from 40, we can guarantee H12) = H, (2). H2(2) when He(s) = He(s). Her(s)

Here, from above equation (1) and (2) (as both are not equal) so, we can say that it can not be guaranteed H(Z)=H1(Z)H2(2) for given assumption,which I have done mistake in above conclusion
Now In case of Biliner Transformation (5= 3(174) . Hells)= ste , Merløs stra) (: 3 24) - zeleno) :: HLZ) = z domain representation of Hells) stado ng istinis stalas iste) la le oth les (2+1) similarly, H2(2)= 2 1 = (2+1) . 7 Bild+Pi} 24)+P,12+1) H2 (2)= (2+1) (2-1) +P2/2+1) H, (2). Hz (2) = (2+1 3² {(2-1) +Pi(2)} { (2-1) +Py (z+u} How Hel (5) H(S) - 1steu ister substituting sales de tous How He (s) to

He (S) = Heics) He als) & He(s)= I- (stpij (stP2) . . H (2) = 2-domain representation of HCCS) → H (2) = Hels) + 8 = 124 ) = (SAP) (SHP) 5 = 21 ) 7 H (2) - H eu)*} {(21) +Pay} (2+02 22./2+1)+(2-03 {{2+1) +2/2+1)] = from & a quarantee we can say that there is of H (2) = H (2) H₃CZ) whenever Hels)= Hals). He 2(3) Bilinear Transformaliañ so, is case of Nole H(Z) = H, (2) H₂ (2) whenever HCCS) = Hel(s) Mcq (s) in Bilinear Transformation method, but it can not be guaranteed in case of Impulse. Invariance method. Se

This problem is solved.