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If an LTI system is described by the difference equation y(n)=ay(n-1)+bx(n), 0 < a < 1, then what is the parameter ‘b’ so that the maximum value of | H(ω)| is unity?
- a)a
- b)1-a
- c)1+a
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If an LTI system is described by the difference equation y(n)=ay(n-1)+...
Explanation: We know that,
Since the parameter ‘a’ is positive, the denominator of | H(ω)| becomes minimum at ω=0. So, | H(ω)| attains its maximum value at ω=0. At this frequency we have,
(|b|)/(1-a) =1 =>b=±(1-a).
(|b|)/(1-a) =1 =>b=±(1-a).
Most Upvoted Answer
If an LTI system is described by the difference equation y(n)=ay(n-1)+...
If the LTI system is described by the difference equation y(n) = ay(n-1) + bx(n), where a and b are constants, and the input signal x(n) is given by:
x(n) = {1, 2, 3, 4, ...}
We can find the output signal y(n) by substituting the values of x(n) into the difference equation and solving for y(n) recursively.
For n = 0:
y(0) = ay(-1) + bx(0)
y(0) = a * 0 + b * 1
y(0) = b
For n = 1:
y(1) = ay(0) + bx(1)
y(1) = a * b + b * 2
y(1) = b(a + 2)
For n = 2:
y(2) = ay(1) + bx(2)
y(2) = a * [b(a + 2)] + b * 3
y(2) = b(a^2 + 2a) + 3b
Continuing this pattern, we can find the output signal for any value of n.
In summary, the output signal y(n) for the given input signal x(n) and the LTI system described by the difference equation y(n) = ay(n-1) + bx(n) is:
y(n) = b(a^(n-1) + 2^(n-1)) + 3^(n-2)b + ... + n^(n-1)b
x(n) = {1, 2, 3, 4, ...}
We can find the output signal y(n) by substituting the values of x(n) into the difference equation and solving for y(n) recursively.
For n = 0:
y(0) = ay(-1) + bx(0)
y(0) = a * 0 + b * 1
y(0) = b
For n = 1:
y(1) = ay(0) + bx(1)
y(1) = a * b + b * 2
y(1) = b(a + 2)
For n = 2:
y(2) = ay(1) + bx(2)
y(2) = a * [b(a + 2)] + b * 3
y(2) = b(a^2 + 2a) + 3b
Continuing this pattern, we can find the output signal for any value of n.
In summary, the output signal y(n) for the given input signal x(n) and the LTI system described by the difference equation y(n) = ay(n-1) + bx(n) is:
y(n) = b(a^(n-1) + 2^(n-1)) + 3^(n-2)b + ... + n^(n-1)b
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