# Homework Help: Generate Polynomial from Graph

1. Oct 20, 2009

### dimava

1. The problem statement, all variables and given/known data

A linear time-invariant system has the following impulse response: (graph attached)

2. Relevant equations

a. Describe in words the effect of this system
b. What is the corresponding polynomial

3. The attempt at a solution

For part A I have "The effect of this system consists of an input of value 1/t0 which then has a “jerk” at t=0 and drops down to 0", however I'm not sure if that's what the book wants for an answer.

I understand the summation forumla for multiplying polynomials, but I have no idea how I can generate such a formula from a graph.

The chapter spends half of a page talking about what an impulse graph does and how the formula is similar to the one for multiplying polynomials. The chapter is primarily on the fast fourier transform, so this is probably somehow related.

#### Attached Files:

• ###### graph.jpg
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2. Oct 20, 2009

### dimava

A new idea popped into my head (I've been working on this for a few hours now), but again, I'm not sure if this makes sense:

= a(0)b(24-0) + a(1)b(24-1) + a(2)b(24-2) + a(3)b(24-3) + a(4)b(24-4) + a(5)b(24-5) + a(6)b(24-6) + a(7)b(24-7) + a(8)b(24-8) + a(9)b(24-9) + a(10)b(24-10) + a(11)b(24-11) + a(12)b(24-12) + a(13)b(24-13) + a(14)b(24-14) + a(15)b(24-15) + a(16)b(24-16) + a(17)b(24-17) + a(18)b(24-18) + a(19)b(24-19) + a(20)b(24-20) + a(21)b(24-21) + a(22)b(24-22) + a(23)b(24-23) + a(24)b(24-24)

For the simplicity of the following write-up we shall assign t0 = F

c0 = F/12 * 1/F = 1/12
c1 = F/12 * 1/F + 2F/12 * 1/F = 3/12 x
c2 = F/12 * 1/F + 2F/12 * 1/F + 3F/12 * 1/F = 6/12 x^2
c3 = F/12 * 1/F + 2F/12 * 1/F + 3F/12 * 1/F + 4F/12*1/F = 10/12 x^3
c4 = F/12 * 1/F + 2F/12 * 1/F + 3F/12 * 1/F + 4F/12*1/F + 5F/12*1/F = 15/12 x^4
c5 = F/12 * 1/F + 2F/12 * 1/F + 3F/12 * 1/F + 4F/12*1/F + 5F/12*1/F + 6F/12= 21/12 x^5
c6 = 28/12 x^6
c7 = 36/12 x^7
c8 = 45/12 x^8
c9 = 55/12 x^9
c10 = 66/12 x^10
c11 = 88/12 x^11
c12 = 0
c13 = 0
….
c23 = 0