Let us show that. To start, let
Then we can take the Fourier Transform of y(t) and plug in the convolution integral for y(t) (notice how we've marked the integrals with dt and dτ to keep track of them):
Now, let's switch the order of the two integrals (this is valid in all but the most pathological cases):
We've just shown that the Fourier Transform of the convolution of two functions is simply the product of the Fourier Transforms of the functions. This means that for linear, time-invariant systems, where the input/output relationship is described by a convolution, you can avoid convolution by using Fourier Transforms. This is a very powerful result.
Now, write x1 (t) as an inverse Fourier Transform.
Example 1 Find the inverse Fourier Transform of
Here is a plot of this function:
Example 2 Find the Fourier Transform of x(t) = sinc2(t) (Hint: use the Multiplication Property).
Example 3 Find the Fourier Transform of y(t) = sinc2(t) * sinc(t). Use the Convolution Property (and the results of Examples 1 and 2) to solve this Example.
This tells us that modulation (such as multiplication in time by a complex exponential, cosine wave, or sine wave) corresponds to a frequency shift in the frequency domain.