Finding impulse response of a system

[caramello]

Hi,

I have a question regarding on how to determine the impulse response of an LTI system which has the following conditions:
1).its input is x(n) = u(n) and output is y(n) = (1/2)ⁿu(n) ,and
2).x(-1) = -1/3; x(0) = 1; x(1) = 1/2, and x(n) = 0 for all other values of n

My approach:
I know that x(n) can be expressed in terms of the sum from k=-infinity to k=+infinity of x(k)delta(n-k), so:
x(n) = x(-1)delta(n+1) + x(0)delta(n) + x(1)delta(n-1/2) ---> up to here am I right?

then,

I also know that the u(n) in y(n) = (1/2)ⁿu(n) emphasized that h(n)= 0 for n<0

But after that I don't even know how to start in computing the impulse response h(n). I'm so lost here. can anyone help me with this? thank you soooo much.

 

[Valkarie]

</code>Given the information you provided, we can deduce that the LTI system has an impulse response of h(n) = (1/2)n*u(n). This follows from the input x(n) and output y(n) equations you provided. The input equation is x(n) = u(n), which means that the input is a unit step function or u(n). This means that for all values of n less than 0, x(n)=0, and for all values of n greater than or equal to 0, x(n)=1. The output equation is y(n) = (1/2)nu(n). This means that for all values of n less than 0, y(n)=0, and for all values of n greater than or equal to 0, y(n)=(1/2)n. Given this information, we can conclude that the impulse response is h(n) = (1/2)n*u(n). This is because the input equation is a unit step function, and the output equation is a scaled version of the unit step function. Therefore, the impulse response must be a scaled version of the unit step function as well. Hope this helps!

 

[Mene]

Hello,

To find the impulse response of a system, you can use the definition of an impulse response as the output of the system when the input is an impulse function, or delta function (δ(n)). In this case, the input (x(n)) is not an impulse function, but it can be expressed as a sum of delta functions as you have shown. Therefore, the output (y(n)) can also be expressed as a sum of impulse responses, which we can then solve for.

Using the definition of an impulse response, we have:
h(n) = y(n)/δ(n) = (1/2)nu(n)/δ(n)

Since u(n) is only non-zero for n≥0, we can rewrite this as:
h(n) = (1/2)nδ(n)

Now, we can use the properties of the delta function to simplify this further. Remember that δ(n) = 0 for n≠0 and ∑δ(n) = 1, so:
h(n) = (1/2)nδ(n) = (1/2)δ(n)

Since x(n) = δ(n), we have:
x(n) = x(-1)δ(n+1) + x(0)δ(n) + x(1)δ(n-1/2)

Plugging in the given values for x(-1), x(0), and x(1), we have:
δ(n) = (-1/3)δ(n+1) + δ(n) + (1/2)δ(n-1/2)

Now we can solve for δ(n) by equating the coefficients of δ(n) on both sides:
1 = -1/3 + 1 + 1/2
δ(n) = 5/6

Therefore, the impulse response of the system is:
h(n) = (1/2)δ(n) = (1/2)(5/6) = 5/12

I hope this helps. Keep in mind that this is just one approach to finding the impulse response and there may be other methods depending on the specific system. It's always important to carefully analyze the given conditions and use the appropriate equations and properties to solve the problem. Let me know if you have any further questions. Good luck!