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What is the unit step response of the system described by the difference equation
y(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?
y(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?
- a)[1.099+1.088(0.9)n.cos(πn/3+5.2o)]u(n)
- b)[1.099+1.088(0.9)n.cos(πn/3-5.2o)]u(n)
- c)[1.099+1.088(0.9)n.cos(πn/3-5.2o)].
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
What is the unit step response of the system described by the differen...
Explanation: The system function is H(z)=1/(1-0.9z-1+0.81z-2 )
The system has two complex-conjugate poles at p1=0.9ejπ/3 and p2=0.9e -jπ/3
The z-transform of the unit step sequence is
X(z)=1/(1-z-1 )
Therefore,
Yzs(z) = 1/((1-0.9e^(jπ/3) z-1)(1-0.9e-jπ/3 z-1 )(1-z-1))
= (0.542-j0.049)/((1-0.9ejπ/3 z-1) ) + (0.542-j0.049)/((1-0.9e^(jπ/3) z-1 ) ) + 1.099/(1-z-1 )
and hence the zero state response is yzs(n)= [1.099+1.088(0.9)n.cos(πn/3-5.2o)]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)= yzs(n).
The system has two complex-conjugate poles at p1=0.9ejπ/3 and p2=0.9e -jπ/3
The z-transform of the unit step sequence is
X(z)=1/(1-z-1 )
Therefore,
Yzs(z) = 1/((1-0.9e^(jπ/3) z-1)(1-0.9e-jπ/3 z-1 )(1-z-1))
= (0.542-j0.049)/((1-0.9ejπ/3 z-1) ) + (0.542-j0.049)/((1-0.9e^(jπ/3) z-1 ) ) + 1.099/(1-z-1 )
and hence the zero state response is yzs(n)= [1.099+1.088(0.9)n.cos(πn/3-5.2o)]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)= yzs(n).
Most Upvoted Answer
What is the unit step response of the system described by the differen...
0.435n)+0.011(0.9)n.sin(0.435n)]u(n)
To find the unit step response, we need to solve for y(n) when x(n) is a unit step (u(n)). Starting with the difference equation:
y(n) = 0.9y(n-1) - 0.81y(n-2) x(n)
we substitute u(n) for x(n):
y(n) = 0.9y(n-1) - 0.81y(n-2) u(n)
Then, we can use the Z-transform to solve for Y(z), the Z-transform of y(n):
Y(z) = 0.9z^-1 Y(z) - 0.81z^-2 Y(z) U(z)
Solving for Y(z), we get:
Y(z) = U(z) / (1 - 0.9z^-1 + 0.81z^-2)
We can use partial fraction decomposition to simplify this expression:
Y(z) = [A/(1-0.9z^-1) + B/(1-0.9z^-1)] / (1-0.9z^-1+0.81z^-2)
Y(z) = [0.1/(1-0.9z^-1) - 0.1/(1-0.81z^-1)] / (1-0.9z^-1+0.81z^-2)
Using the inverse Z-transform, we can find y(n):
y(n) = 0.1(0.9)^n u(n) - 0.1(0.81)^n u(n)
So, the unit step response is:
y(n) = 0.1(0.9)^n u(n) - 0.1(0.81)^n u(n)
This can also be written as:
y(n) = [1.099(0.9)^n cos(0.435n) + 1.088(0.9)^n sin(0.435n)] u(n)
To find the unit step response, we need to solve for y(n) when x(n) is a unit step (u(n)). Starting with the difference equation:
y(n) = 0.9y(n-1) - 0.81y(n-2) x(n)
we substitute u(n) for x(n):
y(n) = 0.9y(n-1) - 0.81y(n-2) u(n)
Then, we can use the Z-transform to solve for Y(z), the Z-transform of y(n):
Y(z) = 0.9z^-1 Y(z) - 0.81z^-2 Y(z) U(z)
Solving for Y(z), we get:
Y(z) = U(z) / (1 - 0.9z^-1 + 0.81z^-2)
We can use partial fraction decomposition to simplify this expression:
Y(z) = [A/(1-0.9z^-1) + B/(1-0.9z^-1)] / (1-0.9z^-1+0.81z^-2)
Y(z) = [0.1/(1-0.9z^-1) - 0.1/(1-0.81z^-1)] / (1-0.9z^-1+0.81z^-2)
Using the inverse Z-transform, we can find y(n):
y(n) = 0.1(0.9)^n u(n) - 0.1(0.81)^n u(n)
So, the unit step response is:
y(n) = 0.1(0.9)^n u(n) - 0.1(0.81)^n u(n)
This can also be written as:
y(n) = [1.099(0.9)^n cos(0.435n) + 1.088(0.9)^n sin(0.435n)] u(n)
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What is the unit step response of the system described by the difference equationy(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?a)[1.099+1.088(0.9)n.cos(πn/3+5.2o)]u(n)b)[1.099+1.088(0.9)n.cos(πn/3-5.2o)]u(n)c)[1.099+1.088(0.9)n.cos(πn/3-5.2o)].d)None of the mentionedCorrect answer is option 'B'. Can you explain this answer?
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