What Is the Effect of a Linear Time-Invariant System's Impulse Response?

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The discussion focuses on analyzing the impulse response of a linear time-invariant system, noting that the system's effect includes an initial "jerk" at t=0 followed by a drop to zero. The user expresses uncertainty about how to derive a polynomial from the impulse response graph and seeks clarity on the relationship between impulse responses and polynomial multiplication. They attempt to formulate a summation for polynomial coefficients based on the impulse response, illustrating their thought process with specific calculations for coefficients c0 through c11. The conversation highlights the complexities of relating impulse responses to polynomial expressions in the context of the fast Fourier transform. Understanding these relationships is crucial for solving problems related to linear time-invariant systems.
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Homework Statement



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

Homework Equations



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


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.

thanks in advance!
 

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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
 
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