show that u(n) = summation k=-inf to n delta(k) =summation k=0 to infinity delta(n-k)

The given equation is a mathematical representation in the field of signal processing. Here, u(n) represents the unit step function, and delta(k) represents the Dirac delta function. The Dirac delta function is a mathematical construct which is used extensively in signal processing and system analysis.

The unit step function, u(n), is defined as:

u(n) = 0 for n < 0 u(n) = 1 for n >= 0

The Dirac delta function, delta(k), is defined as:

delta(k) = 0 for k ≠ 0 delta(k) = 1 for k = 0

Now, let's prove the given equation:

u(n) = Σ (from k=-∞ to n) delta(k)

This summation will add up all the values of delta(k) from k=-∞ to n. Since delta(k) is only 1 at k=0 and 0 otherwise, the summation will be 1 for all n >= 0 (since we include k=0 in the summation) and 0 for n < 0. This is exactly the definition of the unit step function, u(n).

Now, let's prove the second part of the equation:

u(n) = Σ (from k=0 to ∞) delta(n-k)

This summation will add up all the values of delta(n-k) from k=0 to ∞. Since delta(n-k) is only 1 at n=k and 0 otherwise, the summation will be 1 for all n >= 0 (since we include n=k in the summation) and 0 for n < 0. This is again exactly the definition of the unit step function, u(n).

Therefore, we have shown that:

u(n) = Σ (from k=-∞ to n) delta(k) = Σ (from k=0 to ∞) delta(n-k)