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Exercise Questions
Q. y(n)0.4 y(n1) = x(n). find the anticausal impulse response? h(n)=0 for n≥ 0
h(n1) = 2.5 [h(n) δ (n) ]
h(1) = 2.5 [h(0) δ (0) ] = 2.5
h(2) = 2.5^{2} . …….. h(n) = 2.5^{n }valid for n ≤ 1
Q. x(n)={1,2,3} y(n)={3,4} Obtain difference equation from i/p & o/p information
y(n) + 2 y(n1) + 3 y(n2) = 3 x(n) + 4 x(n1) (Ans)
Q. x(n) = {4,4,}, y(n)= x(n) 0.5x(n1). Find the difference equation of the inverse system. Sketch the realization of each system and find the output of each system.
Solution:
The original system is y(n)=x(n)0.5 x(n1)
The inverse system is x(n)= y(n)0.5 y(n1)
y (n) = x (n) – 0.5 x(n1)
Y (z) = X (z) [10.5Z^{1}]
System
Inverse System
y (n) – 0.5 y(n1) =x(n)
Y (z) [10.5 Z^{1}] = X (z)
g (n) = 4 δ (n)  2δ(n1) + 4δ (n1)  2δ (n2) = 4δ (n) + 2δ (n1)  2δ (n2)
y (n) = 0.5 y(n1) + 4δ (n) + 2δ (n1) – 2δ (n2)
y (0) = 0.5y(1) + 4δ (0) = 4
y(1) = 4
y(2) = 0.5 y(1)  2δ (0) = 0
y(n) = {4, 4} same as i/p
Non Recursive filters  Recursive filters 
y(n) for causal system For causal i/p sequence  Present response is a function of the present and past N values of the excitation as well as the past N values of response. It gives IIR o/p but not 
y(n) = a_{k} x(nk) Present response depends only on present i/p & previous i/ps but not future i/ps. It gives FIR o/p  always. (n) – y(n1) = x(n) – x(n3) 
Q. y(n) = 1/3[x (n+1) + x (n) + x (n1)] Find the given system is stable or not?
Let x(n) = δ (n)
h(n) = 1/3 [ δ (n+1) + δ (n) + δ(n1)]
h(0) = 1/3
h(1) = 1/3
h(1) = 1/3
s=∑h(n) <∞ therefore Stable.
Q. y(n) = a y(n1) + x(n) given y(1) = 0
Let x(n) = δ (n)
h(n) = y(n) = a y(n1) + δ (n)
h(0) = a y(1) +δ (0) = 1 = y(0)
h(1) = a y(0) + δ (1) = a
h(2) = a y(1) + δ (2) = a^{2} . . . . . . . h(n) = a^{n} u(n) stable if a<1.
y(n1) =1/a [ y(n) – x(n)]
y(n) = 1/a [ y(n+1) – x(n+1)]
y(1) = 1/a [ y(0) – x(0)]=0
y(2) = 0
Q. y(n) = 1/n+1 y(n1) + x(n) for n ≥ 0
= 0 otherwise. Find whether given system is time variant or not?
Let x(n) = δ (n)
h (0) = 1 y(1) +δ (0) = 1
h(1) = ½ y(0) + δ(1) = ½
h(2) = 1/6
h(3) = 1/24
if x(n) = δ (n1)
y(n) = h(n1)
h(n1) = y(n) =1/n+1 h(n2) + δ (n1)
n=0 h(1) = y(0) = 1 x 0+0 =0
n=1 h(0) = y(1) = ½ x 0 +δ (0)= 1
n=2 h(1) = y(2) = 1/3 x 1 + 0 = 1/3
h(2) = 1/12
∴ h (n, 0) ≠ h (n,1) ∴ TV
Q. y (n) = 2n x(n) Time varying
Q. y (n) =1/3 [x (n+1) + x (n) + x (n1)] Linear
Q. y (n) = 12 x (n1) + 11 x(n2) TIV
Q. y (n) = 7 x^{2}(n1) non linear
Q. y (n) = x^{2}(n) non linear
Q. y (n) = n^{2} x (n+2) linear
Q. y (n) = x (n^{2}) linear
Q. y (n) = e^{x(n)} non linear
Q. y (n) = 2^{x(n)} x (n) non linear, TIV
(If the roots of characteristics equation are a magnitude less than unity. It is a necessary & sufficient condition) Non recursive system, or FIR filter are always stable
Q. y (n) + 2 y^{2}(n) = 2 x(n) – x(n1) non linear, TIV
Q. y (n)  2 y (n1) = 2^{x(n)} x (n) non linear, TIV
Q. y (n) + 4 y (n) y (2n) = x (n) non linear, TIV
Q. y (n+1) – y (n) = x (n+1) is causal
Q. y (n)  2 y (n2) = x (n) causal
Q. y (n)  2 y (n2) = x (n+1) non causal
Q. y (n+1) – y (n) = x (n+2) non causal
Q. y (n2) = 3 x (n2) is static or Instantaneous.
Q. y (n) = 3 x (n2) dynamic
Q. y (n+4) + y (n+3) = x (n+2) causal & dynamic
Q. y (n) = 2 x (αη )
If α=1 causal, static
α <1 causal, dynamic
α >1 non causal, dynamic
α≠ 1 TV
Q. y (n) = 2(n+1) x (n) is causal & static but TV.
Q. y (n) = x (n) TV
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