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

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SUMMARY

The discussion centers on the impulse response of a linear time-invariant (LTI) system, specifically analyzing its effects and deriving a corresponding polynomial representation. The impulse response indicates that the system produces a "jerk" at time t=0, followed by a decay to zero. The participant attempts to formulate a polynomial based on the impulse response graph and references the fast Fourier transform (FFT) as a related concept. The polynomial coefficients are calculated using a summation formula, demonstrating the relationship between the impulse response and polynomial multiplication.

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  • Understanding of linear time-invariant (LTI) systems
  • Familiarity with impulse response concepts
  • Knowledge of polynomial multiplication
  • Basic principles of the fast Fourier transform (FFT)
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  • Learn how to derive polynomials from impulse response graphs
  • Explore the fast Fourier transform (FFT) and its applications in signal processing
  • Practice calculating impulse responses for various LTI 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 formula 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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