CHAPTER 5 Signals, Systems, and Transforms
5.8 The Fourier Transform
5.8.1 Properties of the Fourier transform
We will now state some useful properties of the Fourier transform. Proofs of these properties can be derived from first prin- ciples using the definition of the transform and can be found in standard textbooks on the subject. These properties, along with the table of Fourier transforms in Table 2 allow us to compute the transforms of most functions that arise in practice without having to compute the integral in Equation 37 explicitly.
x t( )dt
∞ –
∫
∞ <∞X jω( ) δ( )et –jωt
∞ –
∞
∫
dt= τ = 0
X( )τ =e–jωτ 0 = 1 δ( )t ↔1
x t( ) 1
2π--- X jω( )ejωt
∞ –
∞
∫
dω=
δ ω ω( – 0)
x t( ) 1 2π
--- δ ω ω( – 0)ejωt
∞ –
∞
∫
dω= ejω0t
ejω0t
---2π ↔δ ω ω( – 0) ejω0t↔2πδ ω ω( – 0)
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145
1. Linearity: If and , then for any constants a and b,
. Note that a and b can be complex numbers.
EXAMPLE 14: LINEARITYOFTHE FOURIERTRANSFORM
Compute the Fourier transform of the eternal sinusoid .
Solution: We use Equation 5 to rewrite . From the previous example and the linearity prop- erty, we get . Thus, a sinusoid in the time domain corresponds to two impulses in the frequency domain. These are at the two frequencies and and represent counter-rotating phasors with the same frequency. The imaginary components of these phasors are always equal and opposite and therefore their sum is a real sinu- soid.
2. Time-shifting: If then . That is, shifting the signal forward in time by t0 units results in multiplying the transform by .
EXAMPLE 15: FOURIERTRANSFORMOFATIME-SHIFTEDIMPULSE
Compute the Fourier transform of an impulse that occurs at time t0.
Solution: Since , we immediately get .
3. Time-scaling: If then , where a is an arbitrary constant. Intuitively, x(at) is a times
‘faster’ than x(t). This results in the transform having a magnitude that is a times smaller and a frequency scale that is a times
‘slower.’For instance, if a = 5, what happens at time 10 in the unscaled function happens at time 2 in the scaled function. In contrast, a feature present at 100 radians/second in the unscaled transform is present at 500 radians/second in the scaled transform.
EXAMPLE 16: FOURIERTRANSFORMOFASCALEDFUNCTION
Compute the Fourier transform of a rectangular pulse as in Figure 11 but with a pulse width of .
Solution: The transform of a pulse of width is . When the pulse is twice as wide, the pulse starts at time instead of at time , expanding the time-scale by a factor of 2, so that a = 1/2. The transform of the longer pulse is
. This has zero crossings when , that is, when , which is twice as fast as before.
x1( )t ↔X1( )jω x2( )t ↔X2( )jω ax1( )t +bx2( )t ↔aX1( )jω +bX2( )jω
ω0t
( )
cos ω0t
( )
cos 1
2---(ejω0t+e–jω0t)
= ω0t
( )
cos ↔π δ ω ω( ( + 0) δ ω ω+ ( – 0))
ω0 –ω0
x t( )↔X jω( ) x1(t–t0)↔e–jωt0X1( )jω e–jωt0
δ( )t ↔1 δ(t–t0)↔e–jωt0
x t( )↔X jω( ) x at( ) 1 ---Xa jω
---a
⎝ ⎠⎛ ⎞
↔
2τ τ τsinc ωτ
---2
⎝ ⎠
⎛ ⎞ –τ
τ 2--- –
2τsinc(ωτ) ωτ = kπ ω kπ
---τ
=
DRAFT - Version 2 - The Fourier Transform
4. Duality: If then and . This property allows us to use knowl-
edge of a transform to compute new transforms.
EXAMPLE 17: DUALITY
What is the Fourier transform of the constant function 1(t)?
Solution: Since , . But is symmetric about the origin, so, we have
which is an impulse centered on the origin, i.e., at zero frequency. This means that all the energy of the constant function is at a frequency of 0, which is what one would expect.
5. Differentiation: If then .
EXAMPLE 18: FOURIERTRANSFORMOFADERIVATIVE
Compute the Fourier transform of the signal .
Solution: Since , .
Table 2 presents some standard transform pairs. These are derived using both first principles as well as the properties of the Fourier transform discussed earlier.
No. x(t) X(jω) Notes
1 1
2 1
3
4 Note that the Fourier transform of a cosine
results in impulses in both the positive and nega- tive frequency axis
5 See note above
6
7 , and this is an exponential defined only
for positive time values
8 This is a sinusoid defined only for positive time
values x t( )↔X jω( ) X t( )↔2πx(–jω) X jt( )↔2πx(–ω)
δ( )t ↔1 1( )t ↔2πδ(–jω) δ( ). 1( )t ↔2πδ( )jω
x t( )↔X jω( ) dxn( )t dtn
---↔( )jω nX jω( )
jω0ejω0t
jω0ejω0t dejω0t ---dt
= jω0ejω0t↔2π( )δ ω ωjω ( – 0)
δ( )t
2πδ ω( ) ejω0t 2πδ ω ω( – 0)
ω0
cos t π δ ω ω( ( + 0) δ ω ω+ ( – 0))
ω0
sin t jπ δ ω ω( ( – 0) δ ω ω– ( + 0))
u t( ) πδ ω( ) 1
jω--- +
e–atu t( ) 1
a+jω
--- a>0
ω0t ( )u t( )
cos π
2---(δ ω ω( + 0) δ ω ω+ ( – 0)) jω
ω02–ω2 ---
+
DRAFT - Version 2 -Properties of the Fourier transform
147
TABLE 2. Some standard Fourier transforms
6. Convolution property: If , , and then
. That is, convolution in the time domain corresponds to multiplication in the transform domain. For a LTI system with an impulse response of h(t), we know from Equation 28 that an input of x(t) leads to an output of
. Therefore, . We can recover y(t) from this by taking the inverse transform.
The symmetric convolution property also holds: multiplication in the time domain corresponds to convolution in the trans-
form domain. That is, if , , and then
.
EXAMPLE 19: SOLVINGASYSTEMUSINGCONVOLUTION
Consider a system such that the Fourier transform H(j ) of its transfer function h(t) is . What is its response to the input ?
Solution: Because , from Row 7 of Table 2, . Therefore, . Using
the method of partial fractions (i.e., writing the expression as and solving for a and b)7, we find that . To find the inverse transform, recall that the Fourier transform is linear, so we only need to find the inverse transform of each term in isolation. From Row 7 of Table 2 again, we get .
It is clear that the Fourier transform greatly simplifies the analysis of LTI systems: instead of having to deal with the complex convolution operator, we merely need to deal with multiplying two transforms then taking the inverse. In general, we can always find the inverse of the product from first principles by using Equation 40. However, because the Fourier transform may not always exist, system analysis is typically performed using the Laplace transform.
9 This is a sinusoid defined only for positive time
values
10 This is a pulse symmetric about the origin with
width
7. More generally, for , we can write where
. Also see Section 8.12 on page 255.
No. x(t) X(jω) Notes
ω0t
( )
sin u t( ) π
2j---(δ ω ω( – 0) δ ω ω– ( + 0)) ω0
ω02–ω2 ---
+
rect t τ--
⎝ ⎠⎛ ⎞ τsinc ωτ
---2
⎝ ⎠
⎛ ⎞
τ
x1( )t ↔X1( )jω x2( )t ↔X2( )jω y t( )↔Y jω( ) y t( ) = x1( )t ⊗x2( )t Y jw( )=X1( )jw X2( )jw
y t( ) = x t( )⊗h t( ) Y jw( )= X jw( )H jw( )
x1( )t ↔X1( )jω x2( )t ↔X2( )jω y t( )↔Y jω( ) y t( ) = x1( )xt 2( )t Y jw( ) 1
2π---X
1( )jw ⊗X
2( )jw
=
ω 1
jω+2 --- e–tu t( )
x t( ) = e–tu t( ) X jω( ) 1 1+jω ---
= Y j( )ω 1
1+jω
( )(2+jω) ---
= a
1+jω --- b
2+jω --- +
F x( ) P x( ) x–λ1
( )(x–λ2)…(x–λr) ---
= F x( ) k1
x–λ1
( )
--- … kr x–λr
( )
---
+ +
= kr [(x–λr)F x( )]
x=λr
= Y jω( ) 1
1+jω --- 1
2+jω --- –
=
y t( ) = (e–t–e–2t)u t( )
DRAFT - Version 2 - The Laplace Transform
This concludes our discussion of the Fourier transform. We will now turn our attention to the more general Laplace trans- form, which shares all the properties of the Fourier transform, and can be applied to wider range of functions.