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

[dimava]

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!

 

[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

 

[Mene]

I would approach this problem by first understanding the concept of impulse response and its relationship to polynomials. The impulse response of a system is the output when an impulse (a very short and high-amplitude signal) is applied as an input. In other words, it represents how the system responds to a sudden change in input.

Based on the given graph, we can see that the impulse response starts at 0, then has a sudden increase at t=0, and then gradually decreases back to 0. This indicates that the system has a "jerk" or sudden change at t=0, but then settles back to its original state.

To find the corresponding polynomial, we can use the fact that the impulse response is related to the system's transfer function by the Laplace transform. The transfer function is the ratio of the output to the input in the Laplace domain. In this case, the input is an impulse, which has a Laplace transform of 1. Therefore, the transfer function is simply the Laplace transform of the impulse response.

Using the Laplace transform, we can find the transfer function as:

H(s) = L{h(t)} = L{1/t0} = 1/s

where L{} represents the Laplace transform and s is the Laplace variable.

Now, to find the corresponding polynomial, we need to inverse Laplace transform the transfer function. This can be done using partial fraction decomposition, which involves finding the roots of the denominator of the transfer function. In this case, the denominator is simply s, which has a root of 0.

Therefore, the inverse Laplace transform of H(s) is:

h(t) = L^-1{H(s)} = L^-1{1/s} = 1

This means that the corresponding polynomial is simply 1, which represents a constant value. In other words, the system's transfer function is just a constant value, indicating that the system does not change the input signal in any way.

In conclusion, the effect of this system is a "jerk" or sudden change at t=0, followed by a gradual decrease back to its original state. The corresponding polynomial is 1, representing a constant value. This indicates that the system does not affect the input signal, and therefore, has no impact on the output.