# Fourier transform of a sum

I have a function : \begin{equation} C(t)=\left(1.42*\exp^{-1.192t}- 12.44*\exp^{-1.192t} +11.02 \right) u(t) \end{equation} where u(t) is a unit-step function What is its fourier transform? a step by step breakdown would greatly be appreciated: notably how would you manage the fact that it is a summation, would the linearity of a fourier property apply? How would you compute the power and the phase of the sum?

would the linearity of a fourier property apply?

Of course, why not?

How would you compute the power and the phase of the sum?

Computing each term in complex numbers, summing and converting to power/phase representation

a step by step breakdown would greatly be appreciated

\begin{equation} \begin{split} C(t)&=\left(1.42 \exp^{-1.192t}- 12.44 \exp^{-1.192t} +11.02 \right) u(t) \\ & = 1.42 \exp^{-1.192t} u(t) - 12.44 \exp^{-1.192t} u(t) +11.02 u(t) \end{split} \end{equation}

\begin{equation} \mathcal{F}(C(t)) = 1.42 \mathcal{F}\left( \exp^{-1.192t} u(t)\right) - 12.44 \mathcal{F}\left(\exp^{-1.192t} u(t)\right) + 11.02\mathcal{F}\left( u(t)\right) \end{equation}

Can you go from here? And why aren't you summing your exponential terms?

• Yes I can take it from here on.....what do you mean but why not sum my exponentials they are not products here or do you mean something else? – Wazaa Jul 20 '15 at 18:34
• I just saw what you meant by summing , my mistake the second exponential exponent is -0,136*t...unfortunately I don't have the privileges to edit my own question per stackexchange Policy can somebody do it? – Wazaa Jul 20 '15 at 18:42
• So here is what I found as F(C(t)), could you please confirm if it checks out : F(C(t))= 1.42/(1.192+jw) - 12.44/(0.136+jw) +11.02 F(u(t)) – Wazaa Jul 20 '15 at 19:03

Fourier transform is a linear filter wolfram. Linearity means:

F[ax+by] = aF[x] + bF[y]

for any constants a and b, and for any signal/variable/function x and y. So of course, the linearity applies to your signal C(t).

For a signal of form

exp(-at)u(t)

its Fourier transform is readily available See Page 3.