What is the impulse and step response of the given difference equation: #y(n)-y(n-1)+09y(n-2) = x(n)#?

1 Answer
Cesareo R.
Dec 5, 2017

See below.

Explanation:

Applying the #Z# transform

#Y(z)-z^-1Y(z)+0.9z^-2Y(z)=X(z)# or

#Y(z)(1-z^-1+0.9 z^-2) = X(z)#

and then

#Y(z) = 1/(1-z^-1+0.9 z^-2) X(z)#

now

#X(z) = 1# for unit impulse and
#X(z) = 1/(1-z^-1)# for an step function

and the response is obtained by anti-transforming

#Y_1(z) = (1/(1-z^-1+0.9 z^-2)) (1) = z^2/((z-0.5+0.806666i)(z-0.5-0.806226i))#
#Y_2(z) = (1/(1-z^-1+0.9 z^-2)) (1/(1-z^-1)) = z^3/((z-0.5+0.806666i)(z-0.5-0.806226i)(z-1))#

with

#y_n = (0.5 + 0.310087 i) e^((-0.0526803 - 1.01568 i) n) + (0.5 - 0.310087 i) e^((-0.0526803 + 1.01568 i) n)#

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as the impulsive response and

#y_n=1.11111 - (0.0555556 - 0.58572 i) e^((-0.0526803 - 1.01568 i) n) - (0.0555556 + 0.58572 i) e^((-0.0526803 + 1.01568 i) n)#

enter image source here

as the step function response.

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