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=Relationship between the Time Domain and Frequency Domain=

'''Summary'''

This is a summary of the relationship 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:

<math> x(t)*h(t)=y(t)  </math>

<math> \phantom{a} \downarrow \phantom{aaa} \downarrow \phantom{aaa} \downarrow \phantom{aaa}  \textrm{Fourier transform} </math>

<math> X( \omega) H( \omega) = Y( \omega) </math>


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'''

<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

<math> \phantom{aaaa} = \int_{-\infty}^{\infty} 
\left( \frac{1}{2}e^{j2\pi440(t-7)} + \frac{1}{2}e^{-j2\pi440(t-7)}  \right)
\delta\left ( t - \tau - 7 \right ) d\tau </math>
because the result is only non-zero when τ = t - 7

<math> \phantom{aaaa} = \left( \frac{1}{2}e^{j2\pi440(t-7)} + \frac{1}{2}e^{-j2\pi440(t-7)}  \right)
\int_{-\infty}^{\infty} 
\delta\left ( t - \tau - 7 \right ) d\tau </math> 

<math> \phantom{aaaa} = \frac{1}{2}e^{j2\pi440(t-7)} + \frac{1}{2}e^{-j2\pi440(t-7)}  </math>

<math> \phantom{aaaa} = \cos \left (2 \pi 440(t - 7) \right ) </math>



'''Method 2: Using the Frequency Domain'''

<math> Y(\omega) = X(\omega) H(\omega) </math>


Fourier Transform of x(t):

Because x(t) is a periodic function, the traditional equation to calculate its Fourier transform should not be applied. We will guess the Fourier transform, and apply the equation for the inverse Fourier transform to confirm that our guess is correct.

guess: <math> X(\omega) = \pi \delta(\omega - 2\pi440) + \pi \delta(\omega + 2\pi440) </math>

Check that the guess is correct: 

<math> F^{-1}(X(\omega)) = \frac{1}{2\pi}\int_{-\infty}^{\infty} 
\pi \delta(\omega - 2\pi440) + \pi \delta(\omega + 2\pi440)
e^{j \omega t} d\omega </math>

<math> \phantom{aaaaaaaaa} = \frac{1}{2} \int_{-\infty}^{\infty} \delta(\omega - 2\pi440) e^{j \omega t} d\omega
+ \frac{1}{2} \int_{-\infty}^{\infty} \delta(\omega + 2\pi440) e^{j \omega t} d\omega </math>

<math> \phantom{aaaaaaaaa} = \frac{1}{2}e^{j2\pi 440t} + \frac{1}{2}e^{-j2\pi 440t} </math>
by the sifting property of δ

<math> \phantom{aaaaaaaaa} = x(t) </math>

Therefore X(ω) = the Fourier transform of x(t)


Fourier Transform of h(t):

h(t) is not a periodic function, therefore we can apply the equation for the Fourier transform directly. 

<math> F(h(t)) = \int_{-\infty}^{\infty}h(t)e^{-j \omega t} dt </math>
by the definition of a Fourier Transform

<math> \phantom{aaaaaa} = \int_{-\infty}^{\infty} \delta(t-7) e^{-j \omega t} dt </math>

<math> \phantom{aaaaaa} = \int_{-\infty}^{\infty} e^{-j \omega 7} dt </math>

<math> \phantom{aaaaaa} = e^{-7 j \omega} </math>


Calculate Y(ω):

<math> Y( \omega) = X( \omega) H(\omega) </math>

<math> \phantom{aaaa} = (\pi \delta(\omega - 2\pi440) + \pi \delta(\omega + 2\pi440)) e^{-7 j \omega} </math>
by applying X(ω) and H(ω) from before


Calculate y(t) from Y(ω):

<math> y(t) = F^{-1}(Y(\omega)) </math>

<math> \phantom{aaa} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} Y(\omega) e^{j \omega t} d \omega </math>

<math> \phantom{aaa} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} 
(\pi \delta(\omega - 2\pi440) + \pi \delta(\omega + 2\pi440)) e^{-7 j \omega}
e^{j \omega t} d \omega </math>

<math> \phantom{aaa} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \pi \delta(\omega - 2\pi440)e^{j \omega (t - 7)} d \omega 
+ \frac{1}{2 \pi} \int_{-\infty}^{\infty} \pi \delta(\omega + 2\pi440)e^{j \omega (t - 7)} d \omega </math> 

<math> \phantom{aaa} = \frac{1}{2} \int_{-\infty}^{\infty} e^{j 2 \pi 440 (t - 7)} d \omega 
+ \frac{1}{2} \int_{-\infty}^{\infty} e^{-j 2 \pi 440(t - 7)} d \omega </math> 
by the sifting property of δ

<math> \phantom{aaa} = \frac{1}{2}e^{j 2 \pi 440 (t - 7)} + \frac{1}{2}e^{-j 2 \pi 440 (t - 7)} </math>

<math> \phantom{aaa} = \cos (2 \pi 440 (t-7)) </math>

<math> \phantom{aaa} = y(t) </math>
found using Method 1


Thus it can be seen that you will come to the same answer if you calculate the response of the system via the time domain or the frequency domain.








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