Linearity
If $FT[x_1(t)] = X_1 (\omega)$ and $FT[x_2(t)] = X_2 (\omega)$, and if $a_1, a_2 \in \mathbb{R}$, then $FT[a_1x_1(t) + a_2x_2(t)] = a_1X_1 (\omega) + a_2 X_2 (\omega)$.
The proof of this property simply carries from the linearity of integration:
\begin{eqnarray}
FT[a_1x_1(t) + a_2x_2(t)] &=& \int_{-\infty}^{\infty} (a_1x_1(t) + a_2x_2(t)) e^{-i\omega t} dt\\
&=& \int_{-\infty}^{\infty} a_1x_1(t) e^{-i\omega t} dt + \int_{-\infty}^{\infty}a_2x_2(t) e^{-i\omega t} dt\\
&=& a_1\int_{-\infty}^{\infty} x_1(t) e^{-i\omega t} dt + a_2\int_{-\infty}^{\infty}x_2(t) e^{-i\omega t} dt\\
&=& a_1 FT[x_1(t)] + a_2 FT[x_2(t)]\\
&=& a_1X_1 (\omega) + a_2 X_2 (\omega)
\end{eqnarray}
Example: What is the FT of $(2e^{-3t}-7e^{-4t})U(t)$?
\begin{eqnarray}
FT[(2e^{-3t}-7e^{-4t})U(t)]=2FT[e^{-3t}U(t)]-7FT[e^{-4t}U(t)] = \frac{2}{3+i\omega}-\frac{7}{4+i\omega}
\end{eqnarray}
Example: What is the FT of $3 -2 \chi_{[-3,3]}(t)$?
\begin{eqnarray}
FT[3 -2 \chi_{[-3,3]}(t)]=3FT[1]-2FT[\chi_{[-3,3]}(t)] = 3\pi \delta(\omega)-4 \frac{\sin(3\omega)}{\omega}
\end{eqnarray}
Example: Compute $FT[ \cos(ft) ]$
\begin{eqnarray}
FT[ \cos(ft) ] &=& FT \left[ \frac{ e^{ift} + e^{-ift} }{2} \right]
\end{eqnarray}
By using linearity
\begin{eqnarray}
&=& \frac{1}{2}FT[ e^{ift} ] + \frac{1}{2}FT[ e^{-ift} ]
\end{eqnarray}
Using the result from the previous section, ie, $FT[ e^{ift} ]=2\pi \delta(\omega-f)$
\begin{eqnarray}
&=& \pi \delta(\omega-f)+\pi \delta(\omega+f)
\end{eqnarray}
Example: Compute $FT[ \sin(ft) ]$
\begin{eqnarray}
FT[ \sin(ft) ] &=& FT \left[ \frac{ e^{ift} – e^{-ift} }{2i} \right]
\end{eqnarray}
Time Shifting
If $FT[x(t)] = X(\omega)$ then $FT[x(t-t_0)] = e^{it_0\omega}X(\omega)$
This property can also be proven easily. Let us write the FT formula for $FT[x(t-t_0)]$
\begin{eqnarray}
FT[x(t-t_0)] &=& \int_{-\infty}^{\infty} x(t-t_0) e^{-i\omega t} dt
\end{eqnarray}
Doing the change of variables $u \rightarrow t-t_0$, $du \rightarrow dt$
\begin{eqnarray}
&=& \int_{-\infty}^{\infty} x(u) e^{-i\omega (u+t_0)} du\\
&=& \int_{-\infty}^{\infty} x(u) e^{-i\omega u0} e^{-i \omega t_0} du
\end{eqnarray}
We can take $e^{i \omega t_0}$ out of the integral, as it is not a function of $u$.
\begin{eqnarray}
&=& e^{-i \omega t_0}\int_{-\infty}^{\infty} x(u) e^{-i\omega u} du\\
&=& e^{-i \omega t_0} X(\omega)
\end{eqnarray}
Example: Compute $FT[ U(t-7) ]$
Example: Compute $FT[ \chi_{[a,b]}(t)]$
Note that we have computed the FT for the symmetric case, ie $FT[ \chi_{[-T,T]}(t)]$. Now we can write
\begin{eqnarray}
\chi_{[a,b]}(t) = \chi_{[ -(b-a)/2, (b-a)/2]} (t-\frac{a+b}{2})
\end{eqnarray}
Then
\begin{eqnarray}
FT(\chi_{[a,b]}(t) ) &=& e^{-i\omega \frac{a+b}{2} } FT \left[ \chi_{[ -(b-a)/2, (b-a)/2]}(t) \right]\\
&=& e^{-i\omega \frac{a+b}{2} } 2 \frac{ \sin \left( \frac{b-a}{2}\omega \right)}{\omega}
\end{eqnarray}
Differentiation
If $FT[x(t)] = X(\omega)$ then $FT[\frac{dx(t)}{dt}] = j\omega X(\omega)$
This property can be proven by starting from the inverse FT
\begin{eqnarray}
x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{i \omega t} d \omega
\end{eqnarray}
Differentiating both sides,
\begin{eqnarray}
\frac{dx(t)}{dt} &=& \frac{d}{dt} \left( \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{i \omega t} d \omega \right)\\
&=& \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) \frac{d e^{i \omega t}}{dt} d \omega\\
&=& \frac{1}{2\pi} \int_{-\infty}^{\infty} i \omega X(\omega) e^{i \omega t} d \omega
\end{eqnarray}
which proves the claim.
Integration
if $FT[ \int_{-\infty}^t x(s) ds] = \frac{1}{j\omega}X(\omega) + \pi X(0) \delta(\omega)$
\begin{eqnarray}
FT \left[ \int_{-\infty}^t x(s) ds \right] &=& \int_{-\infty}^{\infty} \left( \int_{-\infty}^t x(s) ds \right) e^{-i\omega t} dt
\end{eqnarray}
We will proceed by integrarion by parts. Remember that
\begin{eqnarray}
\int_a^b u dv = \left. uv \right|_a^b – \int_a^b v du
\end{eqnarray}
Choose
\begin{eqnarray}
u = \int_{-\infty}^t x(s) ds \qquad \qquad dv = e^{-i\omega t} dt
\end{eqnarray}
Then,
\begin{eqnarray}
du = x(t)dt \qquad \qquad \qquad \qquad v = \frac{e^{-i\omega t}}{-j\omega}
\end{eqnarray}
Therefore
\begin{eqnarray}
FT \left[ \int_{-\infty}^t x(s) ds \right] &=& \left. \left(\int_{-\infty}^t x(s) ds \right) \frac{e^{-i\omega t}}{-j\omega} \right|_{-\infty}^{\infty} -\int_{-\infty}^{\infty} \frac{e^{-i\omega t}}{-j\omega} x(t)dt
\end{eqnarray}
Note that
\begin{eqnarray}
\int_{-\infty}^{-\infty} x(s) ds = 0
\end{eqnarray}
Therefore on the first term we only need to compute $+\infty$ limit..
\begin{eqnarray}
\left. \left(\int_{-\infty}^t x(s) ds \right) \frac{e^{-i\omega t}}{-j\omega} \right|_{-\infty}^{\infty} = \left(\int_{-\infty}^{\infty} x(s) ds \right)\lim_{t \rightarrow \infty}\frac{e^{-i\omega t}}{-j\omega}=\left(\int_{-\infty}^{\infty} x(s) ds \right) \pi \delta(\omega)= \pi X(0) \delta(\omega)
\end{eqnarray}
For the rightmost equality, recall the definition of FT: $X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt$, therefore $X(0)=\int_{-\infty}^{\infty}x(t)dt$. Putting it all together
\begin{eqnarray}
FT \left[ \int_{-\infty}^t x(s) ds \right] &=& \pi X(0) \delta(\omega) + \frac{1}{j\omega} X(\omega)
\end{eqnarray}
Time-Frequency scaling
If $FT[x(t)] = X(\omega)$ then $FT[x(at)] = \frac{1}{|a|} X(\frac{\omega}{a})$.
\begin{eqnarray}
FT[x(at)] = \int_{-\infty}^{\infty}x(at)e^{-i\omega t}dt
\end{eqnarray}
do a change of variables. $u \rightarrow at$, $du \rightarrow adt$. There are two cases:
- $a>0$: $ FT[x(at)] = \int_{-\infty}^{\infty}x(u)e^{-i\omega \frac{t}{a}}\frac{dt}{a} = \frac{1}{a} X(\frac{\omega}{a})$
- $a<0$: $ FT[x(at)] = \int_{\infty}^{-\infty}x(u)e^{-i\omega \frac{u}{a}}\frac{du}{a} = -\int_{-\infty}^{\infty}x(u)e^{-i\omega \frac{u}{a}}\frac{du}{a} =-\frac{1}{a} X(\frac{\omega}{a})$
Hence the two formulas can be brought together as
\begin{eqnarray}
FT[x(at)] = \frac{1}{|a|} X(\frac{\omega}{a})
\end{eqnarray}
Duality
Note that FT and IFT formulas are very similar to each other. This gives us the duality property, which is very useful in finding the FT’s of some functions. Duality can be expressd as follows:
If $FT[ a(t) ] =b(\omega) $, then $FT[ b(t) ] =2\pi a(-\omega)$.
Example: Find $FT( \frac{\sin(Kt)}{t} )$ where $K \in \mathbb{R}$.
Trying to compute the fourier transform directly results in the integral:
\begin{eqnarray}
FT( \frac{\sin(Kt)}{t} ) = \int_{t=-\infty}^{\infty} \frac{\sin(Kt)}{t} e^{-i\omega t}dt
\end{eqnarray}
Instead of evaluating this integral, we can simply remember the FT we have calculated in the previous chapter:
\begin{eqnarray}
FT[ \chi_{(-K,K)}(t) ] = 2 \frac{\sin(\omega K)}{\omega}
\end{eqnarray}
by using duality, we can immediately conclude that
\begin{eqnarray}
FT[ 2 \frac{\sin(t K)}{t} ] = 2\pi \chi_{(-K,K)}(-\omega)
\end{eqnarray}
or
\begin{eqnarray}
FT[ \frac{\sin(t K)}{t} ] = \pi \chi_{(-K,K)}(\omega)
\end{eqnarray}
Convolution
If $FT[x(t)] = X(\omega)$ and $FT[y(t)] = Y(\omega)$ then $FT[x(t)*y(t)] = FT[\int X(s)y(t-s)ds] =X(\omega)Y(\omega)$
\begin{eqnarray}
FT[x(t)*y(t)] &=& FT\left[ \int_{-\infty}^{\infty}x(s)y(t-s)ds\right]\\
&=& \int_{t=-\infty}^{\infty} \left( \int_{s=-\infty}^{\infty}x(s)y(t-s)ds \right) e^{-i\omega t} dt
\end{eqnarray}
rearranging the integrals
\begin{eqnarray}
&=& \int_{s=-\infty}^{\infty} x(s) \left( \int_{t=-\infty}^{\infty}y(t-s) e^{-i\omega t}dt \right) ds
\end{eqnarray}
in the innermochanging variables $u \rightarrow t-s$, $du \rightarrow dt$ results in
\begin{eqnarray}
&=& \int_{s=-\infty}^{\infty} x(s) \left( \int_{t=-\infty}^{\infty}y(u) e^{-i\omega (u+s)}du \right) ds\\
&=& \left(\int_{s=-\infty}^{\infty} x(s) e^{-i\omega s}ds \right) \left( \int_{u=-\infty}^{\infty}y(u) e^{-i\omega u}du \right)\\
&=&X(\omega)Y(\omega)
\end{eqnarray}
Modulation
$ FT[x(t)y(t)] = \frac{1}{2\pi} X(\omega)*Y(\omega) $ where $*$ denotes convolution.
Proven in the same way as in the convolution property.