Fourier Xform Properties

Contents

Introduction

In this page several properties of the Fourier Transform are introduced. Many are presented with proofs, but a few are simply stated (proofs are easily available through internet searches). Applications are not discussed here, that is done on the next page.

Linearity

$$\eqalign{
y(t) &= \alpha \cdot {x_1}(t) + \beta \cdot {x_2}(t) \cr
Y(\omega ) &= \int_{ - \infty }^{ + \infty } {\left( {\alpha \cdot {x_1}(t) + \beta \cdot {x_2}(t)} \right){e^{ - j\omega t}}dt} \cr
&= \alpha \int_{ - \infty }^{ + \infty } {{x_1}(t){e^{ - j\omega t}}dt} + \beta \int_{ - \infty }^{ + \infty } {{x_2}(t){e^{ - j\omega t}}dt} \cr
&= \alpha \cdot {X_1}(\omega ) + \beta \cdot {X_2}(\omega ) \cr
\alpha \cdot {x_1}(t) + \beta \cdot {x_2}(t)&\buildrel F \over
\longleftrightarrow \alpha \cdot {X_1}(\omega ) + \beta \cdot {X_2}(\omega ) \cr} $$

(the letter F over the double arrow denotes a Fourier Transform pair)

The Fourier Transform is linear. The Fourier Transform of a sum of functions, is the sum of the Fourier Transforms of the functions. Also, if you multiply a function by a constant, the Fourier Transform is multiplied by the same constant.

Time Scaling

$$\eqalign{
y(t) &= x\left( {{t \over a}} \right) \cr
Y(\omega ) &= \int_{ - \infty }^{ + \infty } {x\left( {{t \over a}} \right){e^{ - j\omega t}}dt} \quad \quad {\rm{let}} \ u = {t \over a}, \ t = ua, \ dt = a \cdot du \cr
&= \int_{ - \infty }^{ + \infty } {x\left( u \right){e^{ - j\omega au}}a \cdot du} = a\int_{ - \infty }^{ + \infty } {x\left( u \right){e^{ - j\left( {\omega a} \right)u}}du} = aX(\omega a) \cr
x\left( {{t \over a}} \right)&\buildrel F \over
\longleftrightarrow aX(\omega a) \cr} $$

If a function is expanded in time by a quantity a, the Fourier Transform is compressed in frequency by the same amount. This is a well known phenomenon — if you play back an audio signal expanded by a a factor of a=2 (so it plays back twice as slow; y(30)=x(30/2)=x(15)), the frequencies are compressed by a factor of 2 (so all frequencies are twice as low; Y(1000)=X(1000·2)=X(2000)).

Time Shift

$$\eqalign{
y(t) &= x\left( {t - a} \right) \cr
Y(\omega ) &= \int_{ - \infty }^{ + \infty } {x\left( {t - a} \right){e^{ - j\omega t}}dt} \quad \quad {\rm{let }}u = t - a,{\rm{ }}t = u + a,{\rm{ }}dt = du \cr
&= \int_{ - \infty }^{ + \infty } {x\left( u \right){e^{ - j\omega \left( {u + a} \right)}}adu} = {e^{ - j\omega a}}\int_{ - \infty }^{ + \infty } {x\left( u \right){e^{ - j\omega u}}du} = {e^{ - j\omega a}}X(\omega ) \cr
x\left( {t - a} \right)&\buildrel F \over
\longleftrightarrow {e^{ - j\omega a}}X(\omega ) \cr} $$

If a function is delayed in time, this corresponds to multiplication by a complex exponential. The complex exponential has a magnitude of 1, so this is equivalent to a phase shift of -ω·a radians.

Time Reversal

$$x( - t)\buildrel F \over \longleftrightarrow X( - \omega )\quad \quad \quad \quad {\rm{without}}\;{\rm{proof}}$$

Convolution

Note: this proof uses the time shift property derived previously.

$$\eqalign{
y(t) &= {x_1}(t)*{x_2}(t) = \int_{ - \infty }^{ + \infty } {{x_1}(\lambda ){x_2}(t - \lambda )d\lambda } \cr
Y(\omega ) &= \int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {{x_1}(\lambda ){x_2}(t - \lambda )d\lambda } } \right){e^{ - j\omega t}}dt} \quad \quad {\rm{switch}}\;{\rm{order}}\;{\rm{of}}\;{\rm{integration}} \cr
&= \int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {{x_1}(\lambda ){x_2}(t - \lambda )} {e^{ - j\omega t}}dt} d\lambda \quad \quad {\rm{bring}}\;{\rm{terms}}\;{\rm{without}}\;t\;{\rm{outside}}\;{\rm{inner}}\;{\rm{integral}} \cr
&= \int_{ - \infty }^{ + \infty } {{x_1}(\lambda )\left( {\int_{ - \infty }^{ + \infty } {{x_2}(t - \lambda )} {e^{ - j\omega t}}dt} \right)} d\lambda \quad \quad {\rm{apply}}\;{\rm{time}}\;{\rm{shift}}\;{\rm{property}}\;(see\;previous\;proof) \cr
&= \int_{ - \infty }^{ + \infty } {{x_1}(\lambda ){e^{ - j\omega \lambda }}{X_2}} (\omega )d\lambda \quad \quad {\rm{bring}}\;{\rm{terms}}\;{\rm{without}}\;\lambda \;{\rm{outside}}\;{\rm{integral}} \cr
&= {X_2}(\omega )\int_{ - \infty }^{ + \infty } {{x_1}(\lambda ){e^{ - j\omega \lambda }}} d\lambda = {X_2}(\omega ){X_1}(\omega ) \cr
{x_1}(t)*{x_2}(t)&\buildrel F \over
\longleftrightarrow {X_1}(\omega ){X_2}(\omega ) \cr} $$

Convolution in the time domain is equivalent to multiplication in the frequency domain.

Multiplication

$${x_1}(t){x_2}(t)\buildrel F \over
\longleftrightarrow {1 \over {2\pi }}{X_1}(\omega )*{X_2}(\omega )$$

The proof closely follows that for convolution and is not given here.

Differentiation

$$\eqalign{
y(t) &= {{dx(t)} \over {dt}} \cr
&= {d \over {dt}}\left( {{1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {X(\omega ){e^{j\omega t}}d\omega } } \right)\quad \quad \quad {\rm{synthesis}}\;{\rm{equation}}\;{\rm{for}}\;x(t) \cr
&= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {X(\omega ){d \over {dt}}{e^{j\omega t}}d\omega } \quad \quad \quad {\rm{bring}}\;{\rm{derivative}}\;{\rm{inside}}\;{\rm{integral}} \cr
&= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {X(\omega )\left( {j\omega } \right){e^{j\omega t}}d\omega } = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {Y(\omega ){e^{j\omega t}}d\omega } \quad \quad \quad {\rm{synthesis}}\;{\rm{equation}}\;{\rm{for}}\;y(t) \cr
Y(\omega ) &= \left( {j\omega } \right)X(\omega ) \cr
{{dx(t)} \over {dt}}&\buildrel F \over
\longleftrightarrow \left( {j\omega } \right)X(\omega ) \cr} $$

Repeated differentiation simply increases the power of the (jω) term.

$${{{d^n}x(t)} \over {d{t^n}}}\buildrel F \over \longleftrightarrow {(j\omega )^n}X(\omega )$$

Integration

$$\int_{ - \infty }^t {x(\lambda )d\lambda } \buildrel F \over \longleftrightarrow {{X(\omega )} \over {j\omega }} + \pi X(0)\delta (\omega )$$

This property is stated without proof; if interested, check elsewhere.

Time Multiplication

$${t^n}x(t)\buildrel F \over \longleftrightarrow {j^n}{d \over {d{\omega ^n}}}X(\omega )$$

This property is stated without proof. Note similarity to differentiation property — the proof is similar.

Parseval's Theorem

$$Energy = \int_{ - \infty }^{ + \infty } {{{\left| {x(t)} \right|}^2}dt} = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {{{\left| {X(\omega )} \right|}^2}d\omega } $$

This property is stated without proof; if interested, check elsewhere. The importance of this property is that the total energy in the time domain signal, x(t) (i.e., the left integral) can be easily calculated from the frequency domain signal, X(ω) (i.e., the right integral).

Duality

$$\eqalign{
x(t) &= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {X(\omega ){e^{j\omega t}}d\omega } \quad \quad \quad {\rm{let}}\;u = - t \cr
x( - u) &= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {X(\omega ){e^{ - j\omega u}}d\omega } \quad \quad \quad {\rm{exchange}}\;u\;{\rm{and}}\;\omega {\rm{,}}\;{\rm{bring}}\;2\pi \;{\rm{to}}\;{\rm{other}}\;{\rm{side}} \cr
2\pi x( - \omega ) &= \int_{ - \infty }^{ + \infty } {X(u){e^{ - j\omega u}}du} \cr
X(t)&\buildrel F \over
\longleftrightarrow 2\pi x( - \omega ) \cr} $$

The duality property is quite useful, but the notation can be tricky. To put it succinctly: given two functions x(t) and X(ω) that form a Fourier Transform pair,

$$x(t)\buildrel F \over \longleftrightarrow X(\omega )$$

then we can immediately assert another Fourier Transform pair between X(-t) and 2πx(ω) are also a pair.

$$X(t)\buildrel F \over \longleftrightarrow 2\pi x( - \omega )$$

An example is given on the next page.

Aside: An alternate statement of the duality principle

It is sometimes easier to use an alternate statement of the duality principle. If the functions x(t) and X(ω) form a Fourier Transform pair

$$x(t)\buildrel F \over \longleftrightarrow X(\omega )$$

then another pair is

$${1 \over {2\pi }}X( - t)\buildrel F \over \longleftrightarrow x(\omega )$$

Symmetry Properties

Consider only the case when x(t) is real (the complex case is not much more difficult, but does not pertain to the signals being considered). Represent x(t) as the sum of an even function and an odd function (recall that any function can be represented as the sum of an even part and an odd part).

$$x(t) = {x_o}(t) + {x_e}(t)$$

Express the Fourier Transform of x(t), substitute the above expression and use Euler's identity for the complex exponential.

$$\eqalign{
X(\omega ) &= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt} = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {\left( {{x_o}(t) + {x_e}(t)} \right){e^{ - j\omega t}}dt} \cr
&= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {\left( {{x_o}(t) + {x_e}(t)} \right)\left( {\cos (\omega t) - j\sin (\omega t)} \right)dt} \cr
&= {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {\left( {\left( {{x_o}(t)\cos (\omega t) + {x_e}(t)\cos (\omega t)} \right) - j\left( {{x_o}(t)\sin (\omega t) + {x_e}(t)\sin (\omega t)} \right)} \right)dt} \cr} $$

Recall that the product of two odd functions or two even functions is an even function, and the product of an odd and an even function is odd. Recall, also, that the integral of an odd function from -a to +a is zero. Since xo(t)cos(ωt) and xe(t)sin(ωt) are odd, their integrals go to zero so the previous result simplifies to

$$X(\omega ) = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {\left( {{x_e}(t)\cos (\omega t) - j{x_o}(t)\sin (\omega t)} \right)dt} $$

The even part of the function gives us a transform that is real and even (as we can see by changing the sign on ω in the integral).

$${X_e}(\omega ) = {1 \over {2\pi }}\int_{ - \infty }^{ + \infty } {{x_e}(t)\cos (\omega t)dt} = {X_e}( - \omega )$$

The odd part of the function gives us a transform that is imaginary and odd (as we can see by changing the sign on ω in the integral).

$${X_o}(\omega ) = - {j \over {2\pi }}\int_{ - \infty }^{ + \infty } {{x_o}(t)\sin (\omega t)dt} = - {X_o}( - \omega )$$

x(t) is real

If x(t) is real, then X(ω) is complex, but the real part is even and the imaginary part is odd. In other words

$$\displaylines{
X(\omega ) = X^*( - \omega ) \cr
{\mathop{\rm Re}\nolimits} \left( {X(\omega )} \right) = {\mathop{\rm Re}\nolimits} \left( {X( - \omega )} \right) \cr
{\mathop{\rm Im}\nolimits} \left( {X(\omega )} \right) = - {\mathop{\rm Im}\nolimits} \left( {X( - \omega )} \right) \cr} $$

where the asterix denotes the complex conjugate. In terms of magnitude and phase,

$$\displaylines{
\left| {X(\omega )} \right| = \left| {X( - \omega )} \right| \cr
\angle X(\omega ) = - \angle X( - \omega ) \cr} $$

x(t) is real and even

If x(t) is real, then X(ω) is real and even. In other words

$$\displaylines{
X(\omega ) = X( - \omega ) \cr
{\mathop{\rm Im}\nolimits} \left( {X(\omega )} \right) = 0 \cr} $$

x(t) is real and odd

If x(t) is real, then X(ω) is imaginary and odd. In other words

$$\displaylines{
X(\omega ) = {X^*}( - \omega ) \cr
{\mathop{\rm Re}\nolimits} \left( {X(\omega )} \right) = 0 \cr} $$

Moving on

The next page uses these properties to find the Fourier Transform of functions without needing to resort to the complex integration involved in the analysis equation.


References

© Copyright 2005 to 2015 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College