Solve Convolution Systems: Integrate Impulse Response & Input

In summary, convolution is a commutative operation that is applied to a system to create a new system. The integrals should reduce to this I think: \int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx.
  • #1
STEMucator
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Hi everyone, I'm having some trouble understanding how convolutions are applied to systems.

Suppose I'm given the impulse response of a system as ##g(t) = u(t)## and I'm also given the system input ##x(t) = u(t) - u(t-2)##.

The system output should then be given by:

$$y(t) = g(t) * x(t) = \int_0^t g(t - x) x(x) \space dx = \int_0^t u(t-x)u(x) \space dx - \int_0^t u(t-x)u(x-2) \space dx$$

The integrals should reduce to this I think:

$$\int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx$$

How do I finish this integration? I thought about setting ##v = t - x##, but I am unsure.
 
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  • #3
Svein said:
Your convolution formula is not according to the standard (https://en.wikipedia.org/wiki/Convolution). It should be [itex]y(t)=g(t)*x(t)=\int_{-\infty}^{\infty}g(z)x(t-z)dz=\int_{-\infty}^{\infty}u(z)(u(t-z)-u(t-z-2))dz [/itex].

Why would this matter? The convolution is a commutative operation: ##f*g = g*f##.

Also, in the context of systems it is usually taken for granted that ##0 \leq t < \infty## and the convolution limits are taken between ##0## and ##t##.

I believe I have figured out the next logical step in the integration:

$$\int_0^t u(t-x) \space dx - \int_2^t u(t-x) \space dx = \int_0^2 u(t-x) \space dx$$

Okay wait then the area under this function for ##t \geq 2## would simply be equal to ##2##. It would be equal to ##t## for ##0 \leq t \leq 2##.

Therefore:

$$\int_0^2 u(t-x) \space dx = t[u(t) - u(t - 2)] + 2u(t - 2) = tu(t) + [2 - t]u(t - 2)$$

I guess the answer was to think about the area under the function.
 
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  • #4
Zondrina said:
Why would this matter? The convolution is a commutative operation: fg=gff*g = g*f.
I do not question that. What I did was to use the definition of convolution on your problem.
Zondrina said:
Also, in the context of systems it is usually taken for granted that 0≤t<∞0 \leq t < \infty and the convolution limits are taken between 00 and tt.
Maybe - but you need to show that this is applicable to the problem.
 

1. What is a convolution system?

A convolution system is a mathematical model used to describe the relationship between an input signal and an output signal. It involves convolving the input signal with the system's impulse response to produce the output signal.

2. What is an impulse response?

An impulse response is the output of a system when an impulse signal (a sudden, brief change in input) is applied to it. It is often used to characterize the behavior of a system and can be used in convolution to find the system's output for any input signal.

3. How do you solve convolution systems?

To solve a convolution system, you first need to find the impulse response of the system. Then, you can use the convolution integral to integrate the impulse response with the input signal to find the output signal. This can be done analytically or numerically using a computer.

4. What is the purpose of integrating the impulse response and input signal in convolution?

The purpose of integrating the impulse response and input signal in convolution is to find the output signal of a system. This is important in many fields, including engineering, physics, and signal processing, where understanding how a system responds to different inputs is crucial.

5. What are some common applications of solving convolution systems?

Solving convolution systems has a wide range of applications, such as image and audio processing, communication systems, and control systems. It is also used in fields like physics and engineering to model and analyze the behavior of systems.

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