Question

x(n) = 2cos\left(\frac{n\pi}{4}\right) + 3sin\left(\frac{3n\pi}{4} + \frac{\pi}{8}\right)\\ h(n) = \frac{28n\left[ (n-1)\frac{\pi}{2} \right]}{(n-1)\pi}\\Find Convolution y(n) = ?\\Formula\\y(n) = \sum_{k=-N}^{N} x(k)h(n-k)

          x(n) = 2cos\left(\frac{n\pi}{4}\right) + 3sin\left(\frac{3n\pi}{4} + \frac{\pi}{8}\right)\\
h(n) = \frac{28n\left[ (n-1)\frac{\pi}{2} \right]}{(n-1)\pi}\\Find Convolution y(n) = ?\\Formula\\y(n) = \sum_{k=-N}^{N} x(k)h(n-k)

x(n) = 2cos((nπ)/(4)) + 3sin((3nπ)/(4) + (π)/(8))

h(n) = (28n[ (n-1)(π)/(2) ])/((n-1)π)
Find Convolution y(n) = ?
Formula
y(n) = ∑k=-N^N x(k)h(n-k)

Added by Steven D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

Step 1

To do this, we can use the property of exponents that states (a * b)^c = a^c * b^c. Applying this property to our expression, we have uy = w^(n-1) * y^(n-1). Now, we want to express uy in terms of v. To do this, we need to find a relationship between w, y, and Show more…

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