A causal discrete time LTI system is described by:Y(n) + 5y(n-1) + 6y(n – 2) = 5x(n) – 2x(n-1)Find the impulse response h(n) of causal system.a)17(3)nu(-n) + 12(2)nu(n)b)17 (-3)nu(n) – 12 (-2)nu(n)c)17 (2)nu(n) – 12 (3)nu(n)d)17 (2)nu(n-1) – 12 (3)nu(n-1)Correct answer is option 'B'. Can you explain this answer?

Question

A causal discrete time LTI system is described by:Y(n) + 5y(n-1) + 6y(n &ndash; 2) = 5x(n) &ndash; 2x(n-1)Find the impulse response h(n) of causal system.a)17(3)nu(-n) + 12(2)nu(n)b)17 (-3)nu(n) &ndash; 12 (-2)nu(n)c)17 (2)nu(n) &ndash; 12 (3)nu(n)d)17 (2)nu(n-1) &ndash; 12 (3)nu(n-1)Correct answer is option 'B'. Can you explain this answer?

Answer

To determine the system response, we need to find the impulse response of the LTI system.

Let's assume the impulse response is denoted by h(n).

By definition, the impulse response is the output of the system when the input is an impulse function.

Therefore, when the input is an impulse function δ(n), the system output is given by:

y(n) = h(n)

According to the given system description, we have:

y(n) = 5y(n-1) + 6y(n-2)

Now, let's consider the input δ(n). When n = 0, the input is 1, and for all other values of n, the input is 0.

For n = 0, the output is:

y(0) = 5y(0-1) + 6y(0-2)
y(0) = 5y(-1) + 6y(-2)

Since the system is causal, y(-1) and y(-2) are zero because they are dependent on future inputs (negative indices).

Therefore, we have:

y(0) = 5(0) + 6(0) = 0

For n = 1, the output is:

y(1) = 5y(1-1) + 6y(1-2)
y(1) = 5y(0) + 6y(-1)

Again, y(-1) is zero due to causality, so we have:

y(1) = 5y(0) + 6(0) = 5y(0)

For n = 2, the output is:

y(2) = 5y(2-1) + 6y(2-2)
y(2) = 5y(1) + 6y(0)

Substituting the expression for y(1) obtained earlier, we have:

y(2) = 5(5y(0)) + 6(0) = 25y(0)

Therefore, we can see that the impulse response h(n) is given by:

h(0) = 0
h(1) = 5
h(2) = 25

In general, the impulse response can be expressed as:

h(n) = 5^n for n ≥ 0

So, the system response is given by:

y(n) = h(n) = 5^n

Answer

y(n) + 5y(n-1) + 6y(n-2) = 5x(n) – 2x(n-1)

Taking the z-transform

Y(z) + 5z^-1 Y(z) + 6z^-2 Y(z) = 5X(z) - 2z^-1X(z)

y(z) [1 + 5z^-1 + 6z^-2] = X(z) [5 – 2z^-1]

Taking inverse Z-transform,

h(n) = 17(-3)ⁿ u(n) – 12(-2)ⁿ u(n)

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