Discrete convolution evalutation.

In summary, the conversation is about a homework question that involves computing a convolution, specifically y[n] = x[n] * h[n]. The attempt at a solution involves sketching the graphs and breaking down the problem into two intervals, n <= 6 and n > 6. The first interval is solved using summation and a calculator, but there is confusion about the second interval and its upper limits. The conversation ends with a clarification about the unit step function and the given answer for n <= 6.
  • #1
perplexabot
Gold Member
329
5
Hello all. I have a homework question but since I have no idea how to go about solving it I have started with an exercise problem from the book (with the solution and vague steps provided). Here is my attempt of the exersize question.

Homework Statement



Compute the convolution y[n] = x[n] * h[n]
where:
x[n] = u[n - 4] (-1/2)^n
h[n] = u[2 - n] 4^n


Homework Equations



y[n] = summation of x[k]h[n - k] from k = -inf to k = inf


The Attempt at a Solution


Sketching the graphs, I was able to realize that there are two intervals. n <= 6 and n> 6. I continue on and imagine "sweeping" or shifting the h[n] function over x[n] for the n <= 6 region:
summation of x[n]h[n - k] from k = -inf to k = inf
=> summation of ( (-1/2)^k ) * ( 4^(n-k) ) from k = -inf to k = 6 (where "*" means multiply)
=> solving the summations using a calculator yields ( (.5^n)/3 ) - ( (2^n)(4/3) )

I could continue on to solve for n > 6 region but there is no point since my convolution is wrong. Can anyone lead me to the right path? I will be working on it in the mean time. Thank you.

Note: I have attached an image of the solution for convenience. I do not understand it.
 

Attachments

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  • #2
What's the sequence u?
 
  • #3
Sorry, I should have mentioned that u[n] is a unit step function.
 
  • #4
I don't understand why the second summation upper limits go to 3 and (n - 1) for both equations. I would have thought 4 and n. Anyone?
 
Last edited:
  • #5
perplexabot said:
Sorry, I should have mentioned that u[n] is a unit step function.
Ok. u[0] = 0.5?
 
  • #6
no, u[0] = 1.
 
  • #7
In that case I get the given answer for n <= 6.
Where we diverge is in the summation range for k. I have k > 3.
 

FAQ: Discrete convolution evalutation.

1. What is discrete convolution evaluation?

Discrete convolution evaluation is a mathematical operation used to combine two discrete functions to create a new function. It is commonly used in signal processing and image processing to analyze and manipulate data.

2. How is discrete convolution different from continuous convolution?

The main difference between discrete convolution and continuous convolution is that discrete convolution is performed on discrete functions, while continuous convolution is performed on continuous functions. This means that discrete convolution is used for digital data, while continuous convolution is used for analog data.

3. What are the steps involved in discrete convolution evaluation?

The steps involved in discrete convolution evaluation are as follows:

  • 1. Flip one of the two discrete functions.
  • 2. Shift the flipped function by one sample at a time.
  • 3. Multiply the two functions at each shifted position.
  • 4. Sum up all the products to obtain the final output.

4. What are some applications of discrete convolution evaluation?

Discrete convolution evaluation has various applications in fields such as signal processing, image processing, and engineering. It is used for tasks such as filtering, smoothing, and feature extraction.

5. Is there a difference between convolution and correlation?

Yes, there is a difference between convolution and correlation. While both operations involve multiplying two functions and summing the results, the main difference is that convolution flips one of the functions before performing the operation, while correlation does not. This results in different outputs for the two operations.

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