Help with this convolution sum

[atrus_ovis]

Homework Statement

Calculate the discrete time convolution y[n] = x[n] * h[n] , where
x[n]=(1/2)^-nu[-n-1] is the input
h[n]=u[n] is the impulse response

Homework Equations

Discrete time Convolution formula, where x[n] is the input and h[n] the impulse response

Convolution x[n]*h[n] =$$\Sigma$$x[k]h[n-k] ,with k from $$-\infty$$ to $$\infty$$
u[n] is the unit step function, u[n] = 1, n>=0 , u[n] = 0 otherwise

The Attempt at a Solution

x[n]*h[n] = $$\Sigma$$x[k]h[n-k] = $$\Sigma$$ 2^ku[-k-1]u[n-k]

now for k>n, u[n-k]=0 , conv. sum is zero

for -k-1<0 =>k>-1 , u[-k-1] = 0 , conv sum is zero

so the sum boundaries, in which the sum is not zero, are -1, n.
conv. sum != 0, : k < n , k < -1

case 1: n $$\geq$$ -1
conv => $$\Sigma$$ 2^k , sums from$$-\infty$$ to -1

case 2: n$$\leq$$ -1
conv => $$\Sigma$$ 2^k , sums from $$-\infty$$ to nIs the above correct?
Also, how can i simplify the sums to an expression?

 

[KataKoniK]

Your approach is correct, but there are a few things that can be improved:

1. The unit step function u[n] is not needed in the calculation, since it is equal to 1 for all values of n in this problem.

2. The convolution formula should have a limit of -\infty to \infty, not just from k = -\infty to \infty.

3. The sum boundaries are incorrect. Since the impulse response h[n] is a unit step function, it is only non-zero for n >= 0. Therefore, the convolution sum should only be non-zero for k < 0, since the input x[n] is only non-zero for n < 0. This means that the only non-zero terms in the sum will be when k < 0 and n - k < 0, which can be simplified to -n < k < 0.

4. To simplify the sums, you can use the formula for the sum of the first n natural numbers: \Sigma n = n(n+1)/2. Using this, you can simplify the sums to:

Convolution sum = 2\left(\Sigma_{k = -n}^{-1} k + \Sigma_{k = -\infty}^{-n-1} k \right)

= 2\left(\frac{(-n)(-n+1)}{2} + \frac{(-n-1)(-n-1+1)}{2} \right)

= 2\left(\frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2} \right)

= n^2 + 3n + 2

Therefore, the final expression for the convolution is:

y[n] = x[n] * h[n] = (1/2)u[-n-1] * u[n] = (n^2 + 3n + 2)u[n]