Continuous Time Fourier Transform

Introduction

Recall: Fourier series representation of a periodic signal $\tilde{x(t)}$ with time period $'T'$ is given by:-

(1)

\begin{align} \tilde{x}(t)=\sum_{k=-\infty}^{\infty}\tilde{X}[k]e^{jkw_0t} \end{align}

(2)

\begin{align} \tilde{X}[k]=\frac{1}{T}\int_{0}^{T}\tilde{x}(t)e^{-jkw_0t}dt \end{align}

Suppose, $x(t)$ is not periodic.Is there a representation for $x(t)$ as a linear combination of complex exponentials?

The main idea is to think of $x(t)$ as the limit of $\tilde{x}(t)$ when $T \rightarrow \infty$ i.e $$\lim_{T \to \infty}\tilde{x}(t)$$.

Summary:-
1. F.S representation applies to periodic signals i.e A signal contains only frequencies which are integer multiples of a fundamental frequency.
2. F.T representation applies to Non-periodic (and periodic) signals i.e The signal may contain a continuum of frequencies $X(j\omega)$ refers to the F.T,where $\omega$ is a continuously changing variable.

So,the Analysis and Synthesis Equations respectively are given by:-

(3)

\begin{align} X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \end{align}

(4)

\begin{align} x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega \end{align}

Examples

1. $x(t)=e^{-at}u(t) \hspace{1cm} a>0$

(5)

\begin{aligned} X(jw) & =\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{-\infty}^{\infty}e^{-at}u(t)e^{-j\omega t}dt \\ \nonumber & =\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt=\int_{0}^{\infty}e^{-(a+j\omega) t}dt \\ \nonumber & =\bigg[\frac{e^{-(a+j\omega) t}}{-(a+j\omega)}\bigg]_{0}^{\infty}=\frac{0-1}{-(a+j\omega)} \\ \nonumber & =\frac{1}{a+j\omega}=\frac{a-j\omega}{a^2+\omega^2} \nonumber \end{aligned}

(6)

\begin{aligned} |X(jw)| & =\bigg|\frac{1}{a+j\omega}\bigg| \nonumber & =\frac{1}{|a+j\omega |}=\frac{1}{\sqrt{a^2+\omega^2}} \\ \nonumber \end{aligned}

$\angle X(jw)=0-tan^{-1}\big(\frac{w}{a}\big)$

2. $x(t)=e^{-a|t|} \hspace{1cm} a>0$

(7)

\begin{cases} x(t) = e^{-at} & \text{if}\ t>0 \\ \nonumber e^{at} & \text{if}\ t<0\\ \nonumber \end{cases}

(8)

\begin{aligned} X(jw) & =\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{-\infty}^{0}e^{at}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt \\ \nonumber & =\int_{-\infty}^{0}e^{(a-j\omega) t}dt+\int_{0}^{\infty}e^{-(a+j\omega t)}dt \\ \nonumber & =\bigg[\frac{e^{(a-j\omega) t}}{(a-j\omega)}\bigg]_{-\infty}^{0}+\bigg[\frac{e^{-(a+j\omega) t}}{-(a+j\omega)}\bigg]_{0}^{\infty} \\ \nonumber & =\frac{1-0}{a-j\omega}+\frac{0-1}{-(a+j\omega)}=\frac{1}{a-j\omega}+\frac{1}{a+j\omega} \\ \nonumber & =\frac{2a}{a^2+\omega^2} \\ \nonumber \end{aligned}

3. $x(t)=\delta(t)$

$X(jw) =\int_{-\infty}^{\infty}\delta(t)e^{-j\omega t}dt=e^{-j\omega t}\big|_{t=0} =1$

Let, $e^{-j\omega t}=g(t)$.
Recall, $\int_{-\infty}^{\infty}\delta(t)g(t)dt=g(0)=g(t)\big|_{t=0}$.
Therefore,

$X(j\omega)=1 \hspace{0.2cm} \text{i.e} \hspace{0.2cm} \delta(t) \longleftrightarrow 1$

Properties

1. Linearity:
Let $x(t)\longleftrightarrow X(j\omega)$ and $y(t) \longleftrightarrow Y(j\omega)$ then,

(9)

\begin{align} z(t)=ax(t)+by(t) \longleftrightarrow aX(j\omega)+bY(j\omega) \end{align}

2. Time Shifting:
Let $x(t)\longleftrightarrow X(j\omega)$ then,

(10)

\begin{align} z(t)=x(t-t_0) \longleftrightarrow e^{-j\omega t_0}X(j\omega) \end{align}

Proof:

$Z(j\omega)=\int x(t-t_0)e^{-jwt_0}dt$

Substituting, $\gamma=t-t_0$,we get

$Z(j\omega)=\int x(\gamma)e^{-jw(\gamma+t_0)}dt=e^{-jwt_0}X(j\omega)$

So,magnitude remains unchanged but phase is different.

Example:

Start with $x(t)=1$ when $-T_0<t<T_0$

$x(t)\longleftrightarrow 2T_0sinc(\frac{\omega T_0}{\pi})=\frac{2}{\omega}sinc(\omega T_0)$

So,

(11)

\begin{align} Z(j\omega)=e^{-j\omega T_1} 2 T_0 sinc(\frac{\omega T_0}{\pi}) \end{align}

3. Frequency Shifting (Modulation Property):

Let $x(t)\longleftrightarrow X(j\omega)$ then,

(12)

\begin{align} e^{j\gamma t_0}x(t) \longleftrightarrow X(j(\omega-\gamma)) \end{align}

4. Time and Frequency Scaling:

Let $x(t)\longleftrightarrow X(j\omega)$ then,

(13)

\begin{align} x(at) \longleftrightarrow \frac{1}{|a|}X(j\frac{\omega}{a}) \hspace{0.4cm} \textit{a is a real constant} \end{align}

Special case, $a=-1$

$x(-t) \longleftrightarrow X(-j\omega)$

5. Conjugation and Symmetry:

Let $x(t)\longleftrightarrow X(j\omega)$ then,

(14)

\begin{align} x^*(t) \longleftrightarrow X^*(-j\omega) \end{align}

Proof:

$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega$

$x^*(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(j\omega)e^{-j\omega t}d\omega$

Substitute, $\gamma=-\omega$ and $d\gamma=-d\omega$

(15)

\begin{aligned} x^*(t) &= \frac{-1}{2\pi}\int_{\infty}^{-\infty}X^*(-j\gamma)e^{j\gamma t}d\gamma \\ \nonumber &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(-j\gamma)e^{j\gamma t}d\gamma \\ \nonumber \end{aligned}

• If $x(t)$ is real, i.e $x(t)=x^*(t)$,then we can observe a conjugate symmetry here-

$X(j\omega)=X^*(-j\omega)$

(16)

\begin{align} X(-j\omega)=X^*(j\omega) \end{align}

Example:

$x(t)=e^{-at}u(t)\longleftrightarrow X(j\omega)=\frac{1}{a+j\omega}$

Then,

$X^*(-j\omega)= \frac{1}{a-j\omega}$ is same as $X(-j\omega)=\frac{1}{a-j\omega}$

• If $x(t)$ is strictly imaginary, i.e $x(t)=-x^*(t)$ then we can observe a conjugate anti-symmetry here-

$X(j\omega)=-X^*(-j\omega)$

(17)

\begin{align} -X(j\omega)=X^*(-j\omega) \end{align}

• If $x(t)$ has some symmetry then conjugate symmetry will be-

[[x(t)=x^*(-t)$]]

(18)

\begin{equation} x(-t)=x^*(t) \end{equation}

6. Convolution-Multiplication Duality

--------------—Put proof here-------------------——

Examples:

1. Given $x(t)=\frac{1}{\pi t}sin\pi t$ and $h(t)=\frac{1}{\pi t}sin 2 \pi t$. Find $y(t)=x(t)*h(t)$.

Solution:
We will find $Y(j\omega)=X(j\omega)H(j\omega)$ and then find $y(t)$.

$Y(j\omega)=X(j\omega)H(j\omega)=X(j\omega) \Rightarrow y(t)=x(t)$

2. Find $x(t)$ where $X(j\omega)=\frac{4}{\omega^2}sin^2\omega$.

Solution:

$X(j\omega)=(\frac{2}{\omega}sin\omega)(\frac{2}{\omega}sin\omega)=Y(j\omega)Y(j\omega)$

3. Multiplication in time (Windowing):

Suppose we have $x(t)\longleftrightarrow X(j\omega)$. If $X(j\omega)=0$ for $|\omega|> \omega_0$. We say that $x(t)$ is bandlimited if:-

$Z(j\omega)=\frac{1}{2\pi}X(j\omega)*Y(j\omega)$

Note: Any time limited signal cannot be bandlimited.

7. Differentiation and Integration Properties

• Differentiation:

$\frac{d}{dt}x(t) \longleftrightarrow j\omega X(j\omega)$

(19)

\begin{align} -jtx(t) \longleftrightarrow \frac{d}{d\omega} X(j\omega) \end{align}

Proof:

$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega$

$\frac{d}{dt}x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}j \omega X(j\omega)e^{j\omega t}d\omega$

$\frac{d}{dt}x(t)\longleftrightarrow j \omega X(j\omega)$

Example:
Find the D.T of the given signal $y(t)$ using the F.T of $x(t)$.

Solution:
The F.T of $x(t)$ is

$X(j\omega)=\frac{4sin^2\omega}{\omega^2}$

Notice that,

$y(t)=\frac{d}{dt}x(t) \Rightarrow Y(j\omega)=\frac{j\omega 4 sin^2\omega}{\omega^2}$

Can we check another way?
Start with

$X(j\omega)=(e^{j\omega}-e^{-j\omega})\frac{2sin\omega}{\omega}=\frac{j\omega 4 sin^2\omega}{\omega^2}$

We will see yet another way to compute this F.T using the integration property.

• Integration:

It is also called Total area under the curve property.

$X(j\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}x(t)e^{-j\omega t}d\omega$

(20)

\begin{align} X(j0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}x(t)dt \end{align}

$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega$

(21)

\begin{align} x(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)d\omega \end{align}

Example:
What is $I=\int_{-\infty}^{\infty}\frac{1}{\pi t}sin\omega t dt$ ?

Solution: $I=X(j0)$ where $X(j\omega)$ is given as:-

FT of Periodic Signals

So far we have studied:-
1. Fourier series for Periodic signals
2. Fourier Transform of Non-Periodic Signals

By defining the FT of Periodic Signals,we can use only the FT to analyze all the signals under a common framework.We begin by recalling,

$\delta(t) \longleftrightarrow 1$

$1 \longleftrightarrow 2\pi \delta(\omega)$

$e^{j\omega_0 t} \longleftrightarrow 2\pi \delta(\omega-\omega_0)$

$\delta(t-T) \longleftrightarrow e^{-j\omega T}$

Say $x(t)$ is periodic and let $X[K]$ be the F.S coefficients.Then

$x(t)=\sum_{k=-\infty}^{\infty}X[k]e^{jkw_0t}$

Therefore,

(22)

\begin{align} X(j\omega)=2\pi \sum_{k=-\infty}^{\infty}X[k]\delta(\omega-k\omega_0) \end{align}

Examples:

1.
$x(t)=cos\omega_0t=\frac{1}{2}e^{j\omega_0t}+\frac{1}{2}e^{-j\omega_0t}$
$X[1]=\frac{1}{2}$,$X[-1]=\frac{1}{2}$
$X(j\omega)=\pi \delta(\omega-\omega_0) + \pi \delta(\omega+\omega_0)$

2.
$x(t)=sin\omega_0t=\frac{1}{2j}e^{j\omega_0t}-\frac{1}{2j}e^{-j\omega_0t}$
$X(j\omega)=\pi \delta(\omega-\omega_0) + \pi \delta(\omega+\omega_0)$

3.
$x(t)=\sum_{n=-\infty}^{\infty}\delta(t-nT_s)$

Time period=$T_s$, $\omega_0=\frac{2\pi}{T_s}$
$X[k]=\frac{1}{T_s} \int_{\frac{-T_s}{2}}^{\frac{T_s}{2}}\delta(t)e^{-j\omega t}=\frac{1}{T_s}$
So,
$X(j\omega)=\frac{2\pi}{T_s}\sum_{k=-\infty}^{\infty}\delta(\omega-k\omega_0)=\frac{2\pi}{T_s}\sum_{k=-\infty}^{\infty}\delta(\omega-\frac{2\pi}{T_s}\omega_0)=$

• Parseval's Relation:

(23)

\begin{align} \frac{1}{T}\int_{-\infty}^{\infty}|x(t)|^2dt=\bigg[\frac{1}{2\pi}\int_{-\infty}^{\infty}|X(j\omega)|^2d\omega \bigg]\frac{1}{T} \end{align}

Suppose $x(t)$ is periodic then,

$X(j\omega)=2\pi \sum_{k}\delta(\omega-k\omega_0)X[k]$

Finally,

(24)

\begin{align} \frac{1}{T}\int_{-\infty}^{\infty}|x(t)|^2dt=\sum_{k=-\infty}^{\infty}|X[k]|^2 \end{align}