- #1
kosmocomet
- 11
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Homework Statement
Homework Equations
- y(t)=x(t)*h(t)=∫x(λ)⋅h(t-λ)⋅dλ
The Attempt at a Solution
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Is what I have the correct interpretation or or am I wrong?
Thanks
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dRic2 said:This is what I would do. I don't think there is the need to bring up Laplace or other transforms.
Take the integral of the convolution:
$$\int_{-\infty}^{+\infty} x(t-\tau)h(\tau)d\tau$$
Let's see what happens if ##\tau < 0##, well, of course, ##h(\tau < 0) = 0## by definition. So the lower limit of the integral becomes 0. Now what about the upper limit?
Looking again at at ##h(\tau)## if ##\tau > 1## again ##h(\tau > 1 ) = 0## by definition. So here we would be tempted to set the upper limit to ##1##. But wait a sec: what about the other function? If ##\tau > t## then ##x((t-\tau) < 0) = 0## and the integral is zero. From this we conclude that we must break the integral in two parts and in the end you will have a case defined function like this:
$$(x * h)(t) = \begin{cases}
\text{if } t < 1 : \int_0^{t} x(t-\tau)h(\tau)d\tau \\
\text{if } t > 1: \int_0^{1} x(t-\tau)h(\tau)d\tau
\end{cases}
$$
Yes, you need an expression for the decreasing line. Use point-slope or some method to get a valid expression. You will only integrate it in the limits which are shown.kosmocomet said:Thanks for the help. I think I understand the set up. The issue is when evaluating the integral, I get this:
Using the t>1 case, the integral becomes ∫(t-λ)*(1)⋅dλ , where x(t-λ)=(t-λ) and h(λ)=1. This is because the x graph is only valid when λ is between 1 and 0. Since t is between [1,2], then it has to use the decreasing line from the original graph which has the points (t-1,1) and (t,0). Use slope intercept to find equation. Is this correct?
I looked into that & did not find it so "straightforward". Integrating all those complex jt ejwt terms gave me a headache!Dr Transport said:If you use Fourier transforms the problem is fairly straightforward.
Using graphical techniqe you should be able to come up with the four integrals of post 10. I hope they taught you that technique.dvscrobe said:Did we ever get the answer to this? What is y(t)?
A convolution problem is a mathematical operation that combines two functions to produce a third function. It is commonly used in signal processing and image processing to modify or filter signals.
Triangular and rectangular pulses are two types of waveforms commonly used in signal processing. Triangular pulses have a triangular shape, while rectangular pulses have a rectangular shape. Both types of pulses are commonly used in communication systems and digital signal processing.
In convolution problems, triangular and rectangular pulses are used as one of the input functions. The other input function is typically a signal or a filter. The output of the convolution operation is a modified version of the input signal or filter.
The main difference between triangular and rectangular pulses in convolution problems is the shape of the pulse. Triangular pulses have a smooth and continuous shape, while rectangular pulses have a sharp and sudden change in amplitude. This difference can affect the output of the convolution operation.
Convolution problems with triangular and rectangular pulses have various applications in signal processing, such as filtering, noise reduction, and image enhancement. They are also used in communication systems to improve the quality of transmitted signals and in digital signal processing to extract information from signals.