I can not for the life of me show by hand, or understand for the matter, of the following statement.

If x(t) consists of the sum of several sinusoidal functions and has the Fourier transform x(t) => X(w) then find the Transform of the following function:

$$ E[1 + m \cdot x(t))] \cdot cos(w_{c} \cdot t) $$

The solution shows: $$ F_{am}(w) = \frac{E}{2} \cdot [\delta (w-w_{c}) + mX(w-w_{c}) + \delta (w+w_{c}) + mX(w-w_{c})] $$ When I consider x(t) = cos(wt) I can see where the E/2 comes from (on some terms but not all). What I don't know is how to derive what the book came up with as shown above.

What I can do is the simple Algebra/Trig Identity as such: $$ E[1 + m \cdot cos(wt))] \cdot cos(w_{c} \cdot t) = Ecos(w_{c}t) + \frac{mE}{2} \cdot [(cos((w_{c} + w)t) + cos((w_{c} - w)t))] $$

The book is Jones, Optical Fiber Communication Systems; page 13. Equation 2.1.2

Any help in the right direction will be greatly appreciated.

Thanks in advance for everyones help!

  • 2
    $\begingroup$ You might want to use \cdot to indicate multiplication, and save $*$ for convolution. BTW, the linearity of the Fourier transform and the convolution theorem of the Fourier transform will help you out here. The tricky part is the Fourier transform of $\cos()$ which shows up as a pair of $\delta()$ functions. $\endgroup$ – Andy Walls Mar 6 '18 at 0:33
  • $\begingroup$ Will read into that now (and do problems) $\endgroup$ – James Hayek Mar 6 '18 at 0:46

The Fourier transform is all about the complex exponentials. You'll be in a much better position if you can express this problem in that language. So ask yourself:

  • How can I express $\cos \omega_c t$ as a combination of complex exponentials?
  • If I multiply a function by (for example) $e^{i\omega_c t}$, what happens to its Fourier transform?
  • 1
    $\begingroup$ cos(w_{c}t) => Re{e^{jw_{c}t} = 1/2(e^{iw_{c}t + e^{-iw_{c}t) I will try this with the convolution integral and see what I get. Thanks! $\endgroup$ – James Hayek Mar 6 '18 at 0:49

$$F\{E \left[1+m\cdot x(t)\right]\cos(\omega_c t)\} = E\left[ F\{\cos(\omega_c t)\} + m \cdot F\{x(t)\}*F\{\cos(\omega_c t)\}\right]$$

$$ = E\left[\dfrac{1}{2}\left(\delta\left(\omega + \omega_c\right) +\delta\left(\omega -\omega_c\right) \right)+ mX(\omega)*\dfrac{1}{2}\left(\delta\left(\omega + \omega_c\right) +\delta\left(\omega -\omega_c\right)\right)\right]$$

and the next step is your answer.


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