Fourier transform of typical signals

• Impulse

  As shown above,
  

  $${\cal F}[\delta(t)]=1$$
  
  

• Unit Step

  As shown above,
  

  $${\cal F}[u(t)]=\frac{1}{j\omega}+\pi \delta(\omega)$$
  
  

• Constant

  As shown above,
  

  $${\cal F}[x(t)]=\int_{-\infty}^\infty e^{-j\omega t} dt=2\pi\;\delta(\omega)$$
  
  
  This is a useful formula.

• Complex exponential

  The spectrum of a complex exponential can be found from the above due to the frequency shift property:
  

  $${\cal F}[e^{j\omega_0 t}]=2\pi\;\delta(\omega-\omega_0)$$
  
  

• Sinusoids

  
  

  $$x(t)=cos\;\omega_0 t=\frac{1}{2}(e^{j\omega_0 t}+e^{-j\omega_0 t})$$
  
  
  
  
  $$X(j\omega)={\cal F}[cos\;\omega_0 t]=\frac{1}{2}({\cal F}[e^{... ...ga_0 t}])=\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]$$
  
  
  Similarly, we have
  
  $$X(j\omega)={\cal F}[sin\;\omega_0 t]=\frac{\pi}{j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]$$
  
  

• Exponential decay - right-sided
  
  $$x(t)=e^{-at} u(t) \;\;\;\;\;((a>0)$$
  
  
  
  

  $${\cal F}[e^{-at}u(t)] = \int_{-\infty}^\infty x(t)e^{-j\omega t}dt=\int_0^\infty e^{-at}e^{-j\omega t}dt$$

  $$= \frac{-1}{a+j\omega} e^{-(a+j\omega)t}\vert _0^{\infty}=\frac{1}{a+j\omega}$$

  
  

• Exponential decay - left-sided Due to the time reversal property, we also have (for $a>0$):
  
  $${\cal F}[ e^{at} u(-t)]=\frac{1}{a-j\omega}$$
  
  
  or
  
  $${\cal F}[-e^{at} u(-t)]=\frac{1}{j\omega-a}$$
  
  

• Exponential decay - two-sided
  
  $$x(t)=e^{-a\vert t\vert} \;\;\;\;\;((a>0)$$
  
  
  As the two-sided exponential decay is the sum of the right and left-sided exponential decays, its spectrum of $x(t)$ is the sum of their spectra due to linearity:
  
  $${\cal F}[e^{-a\vert t\vert}] =\frac{1}{a+j\omega}+\frac{1}{a-j\omega}=\frac{2a}{a^2+\omega^2}$$
  
  

• Comb function

  The comb function is defined as
  

  $$comb(t)=\sum_{n=-\infty}^\infty \delta(t-nT)$$
  
  
  Its Fourier series coefficient is:
  
  $$COMB[k]=\frac{1}{T}\int_{-T/2}^{T/2} comb(t)e^{-j\omega_0t} d... ...1}{T}\int_{-T/2}^{T/2} \delta(t)e^{-j\omega_0t} dt=\frac{1}{T}$$
  
  
  and its spectrum is:
  
  $$COMB(j\omega)=\frac{2\pi}{T}\sum_{k=-\infty}^\infty \delta(\o... ..._0) =\frac{1}{T}\sum_{k=-\infty}^\infty \delta(f-\frac{k}{T})$$
  
  
  We see that the spectrum of an impulse train with time interval $T$ is also an impulse train with frequency interval $\omega_0=2\pi/T$. Also, according to the definition of the Fourier transform, we have
  

  $$COMB(j\omega) = \int_{-\infty}^\infty comb(t) e^{-j\omega t} dt = \int_{-\infty}^\infty [\sum_{n=-\infty}^\infty \delta(t-nT)]\; e^{-j\omega t} dt$$

  $$= \sum_{n=-\infty}^\infty \int_{-\infty}^\infty \delta(t-nT) e^{-j\omega t} dt = \sum_{n=-\infty}^\infty e^{-j\omega nT}$$

  
  
  Therefore we have this equation
  
  $$\sum_{n=-\infty}^\infty e^{-j\omega nT}= \frac{2\pi}{T}\sum_... ..._0) =\frac{1}{T}\sum_{k=-\infty}^\infty \delta(f-\frac{k}{T})$$
  
  
  which can be compared with the equation in continuous case:
  
  $$\int_{-\infty}^\infty e^{-j\omega t}dt=2\pi \delta(\omega)=\delta(f)$$
  
  

• Square wave

  A square wave or rectangular function of width $a$ can be considered as the difference between two unit step functions
  

  $$rect_a(t)=u(t+a)-u(t-a)=\left\{ \begin{array}{ll} 1 & -a \le t < a 0 & \mbox{else} \end{array} \right.$$
  
  
  and due to linearity, its Fourier spectrum is the difference between the two corresponding spectra:
  

  $${\cal F}[rect_a(t)] = {\cal F}[u(t+a)]-{\cal F}[u(t-a)] =\frac{1}{j\omega}e^{ja\omega}-\frac{1}{j\omega}e^{-ja\omega}$$

  $$= \frac{2}{\omega}\;\;\frac{e^{ja\omega}-e^{-ja\omega}}{2j}=\frac{2}{\omega}sin(a\omega)$$

  
  

• Sinc function

  The spectrum of an ideal low-pass filter is
  

  $$X(j\omega)=\left\{ \begin{array}{ll} 1 & \mbox{$-a \le \omega < a$} 0 & \mbox{else} \end{array} \right.$$
  
  
  and its impulse response can be found by inverse Fourier transform:
  
  $$x(t)=\frac{1}{2\pi}\int_{-\infty}^\infty X(j\omega) e^{j\omeg... ...ga =\frac{1}{2\pi jt}( e^{jat}-e^{-jat})=\frac{sin(at)}{\pi t}$$
  
  

• Triangle function
  
  $$x(t)=triangle(t)=\left\{ \begin{array}{ll} 1-\vert t\vert & \vert t\vert<1 0 & \vert t\vert\ge 1 \end{array} \right.$$
  
  
  As $x(t)$ is an even function, its Fourier transform is
  

  $${\cal F}[x(t)] = \int_{-\infty}^\infty x(t)e^{-j\omega t}dt =2\int_0^1(1-t)cos(\omega t) dt$$

  $$= 2[\int_0^1 cos(\omega t) dt-\int_0^1 t\;cos(\omega t) dt] =\frac{2}{\omega}[sin(\omega)-\int_0^1 t\;d\;sin(\omega t)]$$

  $$= \frac{2}{\omega}[sin(\omega)-(t\;sin(\omega t)\vert _0^1-\int_0^1sin(\omega t)dt)] =\frac{2}{\omega^2}cos(\omega t)\vert _1^0$$

  $$= \frac{2}{\omega^2}(1-cos(\omega)) =(\frac{2}{\omega})^2sin^2(\frac{\omega}{2}) =\frac{sin^2(\pi f)}{(\pi f)^2}=sinc^2(\omega)$$

  
  
  Alternatively, as the triangle function is the convolution of two square functions ($a=1/2$), its Fourier transform can be more conveniently obtained according to the convolution theorem as:
  
  $${\cal F}[triangle(t)]={\cal F}[rect(t)*rect(t)]=sinc(\omega)\;sinc(\omega) =(\frac{2}{\omega})^2sin^2(\frac{\omega}{2})$$
  
  

• Gaussian function

  The Fourier transform of a Gaussian or bell-shaped function $x(t)=e^{-\pi t^2}$ is
  

  $$X(j\omega) = {\cal F}[x(t)]=\int_{-\infty}^\infty [e^{-\pi t^2}]e^{-j\omega t}dt$$

  $$= \int_{-\infty}^\infty e^{-\pi(t^2+j2ft)} dt =e^{\pi(jf)^2}\int_{-\infty}^\infty e^{-\pi[t^2+j2ft+(jf)^2]} dt$$

  $$= e^{-\pi f^2}\int_{-\infty}^\infty e^{-\pi(t+jf)^2} d(t+jf)=e^{-\pi f^2}$$

  
  
  Here we have used the identity
  
  $$\int_{-\infty}^\infty e^{-\pi x^2}dx=1$$
  
  
  We see that the Fourier transform of a bell-shaped function is also a bell-shaped function:
  
  $${\cal F}[e^{-\pi t^2}]=e^{-\pi f^2}$$
  
  
  Note that the area underneath either $x(t)$ or $X(j\omega)$ is unity. Moreover, due to the property of time and frequency scaling, we have:
  
  $${\cal F}[a\;e^{-\pi (at)^2}]=e^{-\pi (f/a)^2}$$
  
  
  (Note that if $a=1/\sqrt{2\pi \sigma^2}$, then $a\;x(at)$ above is a normal distribution with variance $\sigma^2$ and mean $\mu=0$.) If we let $a \rightarrow \infty$, $x(t)$ becomes narrower and taller and approaches $\delta(t)$, and its spectrum $e^{-\pi (f/a)^2}$ becomes wider and approaches constant $1$. On the other hand, if we rewrite the above as
  
  $${\cal F}[e^{-\pi (at)^2}]=\frac{1}{a}e^{-\pi (f/a)^2}$$
  
  
  and let $a\rightarrow 0$, $x(t)$ approaches 1 and $X(j\omega)$ approaches $\delta(\omega)$.