Unformatted text preview: e Response h[kJ () = t an 1 (0.172) =  0.17 rad. C onsider a n n thorder s ystem s pecified b y t he e quation S ubstituting () =  0.17 r adian in eeos () = 2.308 yields e = 2.34 a nd
yo[kJ = 2.34(0.9)k cos (~k  0.17) U nit I mpulse R esponse ( En + a n_IE n  (bnEn k:::: 0 O bserve t hat here we have used radian u nit for b oth fJ a nd (). We also could have used
t he degree unit, although i t is n ot recommended. T he i mportant c onsideration is to be
consistent a nd t o use t he s ame units for b oth fJ a nd (). •
E xercise E 9.2
Find and sketch the zeroinput response for the systems described by the following equations:
( a) y[k + 1 ] O.8y[k] = 3 f[k + I] ( b) y[k + 1] + O.8y[k] = 3 f[k + 1].
In each case the initial condition is Y[I] = 10. Verify the solutions by computing the first
three terms using the iterative method.
Answer: (a)8(O.8)k ( b)8(O.8)k.
V 1 + . .. + a lE + a o)y[k] = + b n_IE n  1 + ... + b lE + b o)f[k] ( 9.27a) or
Q [E]y[k] = P [E]f[k] ( 9.27b) T he u nit i mpulse r esponse h [k] is t he s olution o f t his e quation for t he i nput 6[k]
w ith a ll t he i nitial c onditions z ero; t hat is, b. b. E xercise E 9.3
Find the zeroinput response of a system described by the equation
y[k] + O.3y[k  1]  O.ly[k  2] = f [k] + 2 f[k  1] The initial conditions are y o[l] = 1 and Yo[2] = 33. Verify the solution by computing the
first three terms iteratively.
Answer: York] = (O.2)k + 2( O.5)k V
b. E xercise E 9.4
Find the zeroinput response of a system described by the equation
y[k] s ubject t o i nitial c onditions
h [l] = h [2] = . .. = h [n] = 0
( 9.29)
E quation ( 9.28) c an b e s olved t o d etermine h [k] i teratively o r i n a c losed form. T he
f ollowing e xample d emonstrates t he i terative s olution.
. Example 9 .4: I terative D etermination o f h [kj
Find h[kJ, t he u nit impulse response o f a s ystem described by t he e quation
y[kJ  0.6y[k  IJ  0.16y[k  2J = 5f[kJ (9.30) To determine t he u nit impulse response, we m ust l et t he i nput f[kJ = 6[kJ a nd t he
o utput y[kJ = h[kJ in t he above equation. T he r esulting equation is
(9.31) s ubject t o zero initial s tate; t hat is, h [IJ = h [2J = O.
S etting k = 0 i n t his equation yields
h[OJ  0.6(0)  0.16(0) = 5(1) =} h[OJ = 5 Next, setting k = 1 i n Eq. (9.31) a nd using h[OJ = 5, we o btain <::) C omputer E xample C 9.2
F ind a nd sketch t he z eroinput response for t he s ystem described by + 0.81)y[kJ u sing t he i nitial conditions y[IJ=2, a nd y[2J=I.
y =[l 2J; y =y';
k =2:16; k =k';
f or m =1:length(k)2
Y = 1.56*y(m+l)O.81 * y(m);
Y zi=[Yzi;Yj;
e nd
s tem(k,Yzi) 0 ( 9.28) h[kJ  0.6h[k  IJ  0.16h[k  2J = 56[kJ + 4y[k  2J = 2f[kJ T he initial conditions are yo[IJ =  2~ and yo[2J = ~. Verify the solution by computing
t he first three terms iteratively.
Answer:yo[k] = (2)kcos(~k ~) V (E2  I.56E Q [E]h[k] = P [E]6[k] = (E + 3)f[kJ h[IJ  0.6(5)  0.16(0) = 5(0) =} h[lJ = 3 • C ontinuing t his way, we c an d etermine a ny n umber o f t erms o f h [k]. U nfortunately, s uch a s olutio...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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