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=Moving between the Time Domain and Frequency Domain= | =Moving between the Time Domain and Frequency Domain= | ||
| − | + | '''Summary''' | |
This is a summary of how to move between the Time Domain and Frequency Domain, and an example of how one can solve for the output of a system via either the time domain or frequency domain. You will come to the same answer. | This is a summary of how to move between the Time Domain and Frequency Domain, and an example of how one can solve for the output of a system via either the time domain or frequency domain. You will come to the same answer. | ||
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Thus, given the signal x(t) and the unit impulse response h(t), one can either directly calculate the response y(t) via convolution in the time domain, or one can apply the Fourier transform to x(t) and y(t) to move into the Fourier domain. There, the Fourier transforms, X(ω) and H(ω) respectively, can be multiplied together to obtain Y(ω), and Y(ω) can be inverse Fourier transformed to find y(t). Though this process requires more steps, its computations are easier. Here is a CT example of how each of these two methods can be used to find y(t). | Thus, given the signal x(t) and the unit impulse response h(t), one can either directly calculate the response y(t) via convolution in the time domain, or one can apply the Fourier transform to x(t) and y(t) to move into the Fourier domain. There, the Fourier transforms, X(ω) and H(ω) respectively, can be multiplied together to obtain Y(ω), and Y(ω) can be inverse Fourier transformed to find y(t). Though this process requires more steps, its computations are easier. Here is a CT example of how each of these two methods can be used to find y(t). | ||
| − | Method 1: Exclusively Time Domain | + | |
| + | |||
| + | Given: a signal x(t) = cos(2π440t), and the unit impulse response h(t) = δ(t-7) | ||
| + | Find: the system response y(t) | ||
| + | |||
| + | '''Method 1: Exclusively Time Domain''' | ||
| + | |||
| + | <math> y(t) = x(t) * h(t) </math> | ||
| + | |||
| + | <math> \phantom{aaaa} = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau </math> | ||
| + | by the definition of convolution | ||
| + | |||
| + | <math> \phantom{aaaa} = \int_{-\infty}^{\infty} | ||
| + | \left( \frac{1}{2}e^{j2\pi440\tau} + \frac{1}{2}e^{-j2\pi440\tau} \right) | ||
| + | \delta\left ( t - \tau - 7 \right ) d\tau </math> | ||
| + | applying the Fourier series representation of cosine | ||
| + | |||
| + | test | ||
Revision as of 02:51, 2 December 2018
Moving between the Time Domain and Frequency Domain
Summary
This is a summary of how to move between the Time Domain and Frequency Domain, and an example of how one can solve for the output of a system via either the time domain or frequency domain. You will come to the same answer.
The relationship is summarized as follows:
$ x(t)*h(t)=y(t) $
$ \phantom{a} \downarrow \phantom{aaa} \downarrow \phantom{aaa} \downarrow \phantom{aaa} \textrm{Fourier transform} $
$ X(w) H(w) = Y(w) $
Thus, given the signal x(t) and the unit impulse response h(t), one can either directly calculate the response y(t) via convolution in the time domain, or one can apply the Fourier transform to x(t) and y(t) to move into the Fourier domain. There, the Fourier transforms, X(ω) and H(ω) respectively, can be multiplied together to obtain Y(ω), and Y(ω) can be inverse Fourier transformed to find y(t). Though this process requires more steps, its computations are easier. Here is a CT example of how each of these two methods can be used to find y(t).
Given: a signal x(t) = cos(2π440t), and the unit impulse response h(t) = δ(t-7) Find: the system response y(t)
Method 1: Exclusively Time Domain
$ y(t) = x(t) * h(t) $
$ \phantom{aaaa} = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau $ by the definition of convolution
$ \phantom{aaaa} = \int_{-\infty}^{\infty} \left( \frac{1}{2}e^{j2\pi440\tau} + \frac{1}{2}e^{-j2\pi440\tau} \right) \delta\left ( t - \tau - 7 \right ) d\tau $ applying the Fourier series representation of cosine
test
