Post on 10-Apr-2022
Signals and Systems
Lecture 8
DR TANIA STATHAKIREADER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSINGIMPERIAL COLLEGE LONDON
โข The forward and inverse Fourier transform are defined for aperiodic
signals as:
๐ ๐ = โฑ ๐ฅ ๐ก = เถฑโโ
โ
๐ฅ ๐ก ๐โ๐๐๐ก๐๐ก
๐ฅ ๐ก = โฑโ1 ๐ ๐ =1
2๐เถฑโโ
โ
๐ ๐ ๐๐๐๐ก๐๐
โข You can immediately observe the functional similarity with Laplace
transform.
โข For periodic signals we use Fourier Series.
๐ฅ๐0 ๐ก =
๐=โโ
โ
๐ท๐๐๐๐๐0๐ก
๐ท๐ =1
๐0๐0/2โ๐0/2 ๐ฅ๐0 ๐ก ๐โ๐๐๐0๐ก๐๐ก or
๐ท๐ =1
๐0one full period๐ฅ๐0 ๐ก ๐โ๐๐๐0๐ก๐๐ก , ๐0 =
2๐
๐0
Definition of Fourier transform
โข A unit rectangular window (also called a unit gate) function rect ๐ฅ :
rect ๐ฅ =
0 ๐ฅ >1
21
2๐ฅ =
1
2
1 ๐ฅ <1
2
โข A unit triangle function ฮ ๐ฅ :
ฮ ๐ฅ =0 ๐ฅ โฅ
1
2
1 โ 2 ๐ฅ ๐ฅ <1
2
โข Interpolation function sinc(๐ฅ):
sinc(๐ฅ) =sin(๐ฅ)
๐ฅor sinc(๐ฅ) =
sin(๐๐ฅ)
๐๐ฅ
Define three useful functions
โข The sinc(๐ฅ) function is an even function of ๐ฅ.
โข sinc ๐ฅ = 0 when sin ๐ฅ = 0, i.e.,
๐ฅ = ยฑ๐,ยฑ2๐,ยฑ3๐,โฆ except when ๐ฅ = 0
where sinc 0 = 1. This can be proven by the
LโHospitalโs rule.
โข sinc(๐ฅ) is the product of an
oscillating signal sin(๐ฅ) and a
monotonically decreasing function 1/๐ฅ.
Therefore, it is a damping oscillation with period
2๐ with amplitude decreasing as 1/๐ฅ.
More about the ๐ฌ๐ข๐ง๐(x) function
โข Evaluation:
๐ ๐ = โฑ ๐ฅ ๐ก = เถฑโโ
โ
rect๐ก
๐๐โ๐๐๐ก๐๐ก
โข Since rect๐ก
๐= 1 for
โ๐
2< ๐ก <
๐
2and 0 otherwise we have:
๐ ๐ = โฑ ๐ฅ ๐ก = เถฑโ๐2
๐2๐โ๐๐๐ก๐๐ก = โ
1
๐๐๐โ๐๐
๐2 โ ๐๐๐
๐2 =
2sin(๐๐2)
๐
= ๐sin(
๐๐
2)
(๐๐
2)= ๐sinc(
๐๐
2) โ โฑ rect
๐ก
๐= ๐sinc(
๐๐
2) or rect
๐ก
๐โ ๐sinc(
๐๐
2)
โข The bandwidth of the function rect๐ก
๐is approximately
2๐
๐.
โข Observe that the wider(narrower) the pulse in time the narrower(wider)
the lobes of the sinc function in frequency.
Fourier transform of ๐ ๐ = ๐ซ๐๐๐ญ(๐/๐)
โข Using the sampling property of the impulse we get:
๐ ๐ = โฑ ๐ฟ(๐ก) = เถฑโโ
โ
๐ฟ ๐ก ๐โ๐๐๐ก๐๐ก = 1
โข As we see the unit impulse contains all frequencies (or, alternatively, we
can say that the unit impulse contains a component at every frequency.)
๐ฟ ๐ก โ 1
Fourier transform of the unit impulse ๐ ๐ = ๐น(๐)
โข Using the sampling property of the impulse we get:
โฑโ1 ๐ฟ ๐ =1
2๐โโโ
๐ฟ ๐ ๐๐๐๐ก๐๐ =1
2๐
โข Therefore, the spectrum of a constant signal ๐ฅ ๐ก = 1 is an impulse 2๐๐ฟ ๐ .1
2๐โ ๐ฟ ๐ or 1 โ 2๐๐ฟ ๐
โข By looking at current and previous slide, observe the relationship: wide
(narrow) in time, narrow (wide) in frequency.
o Extreme case is a constant everlasting function in one domain and a
Dirac in the other domain.
Inverse Fourier transform of ๐น(๐)
โข Using the sampling property of the impulse we get:
โฑโ1 ๐ฟ ๐ โ ๐0 =1
2๐โโโ
๐ฟ ๐ โ ๐0 ๐๐๐๐ก๐๐ =1
2๐๐๐๐0๐ก
โข The spectrum of an everlasting exponential ๐๐๐0๐ก is a single impulse
located at ๐ = ๐0 .1
2๐๐๐๐0๐ก โ ๐ฟ ๐ โ ๐0
๐๐๐0๐ก โ 2๐๐ฟ ๐ โ ๐0
๐โ๐๐0๐ก โ 2๐๐ฟ ๐ + ๐0
Inverse Fourier transform of ๐น(๐ โ๐๐)
โข Remember the Eulerโs formula:
cos๐0๐ก =1
2(๐๐๐0๐ก + ๐โ๐๐0๐ก)
โฑ cos๐0๐ก = โฑ1
2(๐๐๐0๐ก + ๐โ๐๐0๐ก) =
1
2โฑ ๐๐๐0๐ก +
1
2โฑ ๐โ๐๐0๐ก
โข Using the results from previous slides we get:
cos๐0๐ก โ ๐[๐ฟ ๐ + ๐0 + ๐ฟ ๐ โ ๐0 ]
โข The spectrum of a cosine signal has two impulses placed symmetrically
at the frequency of the cosine and its negative.
Fourier transform of an everlasting sinusoid ๐๐จ๐ฌ๐๐๐
โข The Fourier series of a periodic signal ๐ฅ(๐ก) with period ๐0 is given by:
๐ฅ ๐ก = ฯโโโ ๐ท๐ ๐
๐๐๐0๐ก, ๐0 =2๐
๐0
โข By taking the Fourier transform on both sides we get:
๐(๐) = 2๐
๐=โโ
โ
๐ท๐ ๐ฟ ๐ โ ๐๐0
Fourier transform of any periodic signal
โข Consider an impulse train
๐ฟ๐0 ๐ก = ฯโโโ ๐ฟ ๐ก โ ๐๐0
โข The Fourier series of this impulse train can be shown to be:
๐ฟ๐0 ๐ก = ฯโโโ ๐ท๐ ๐
๐๐๐0๐ก where ๐0 =2๐
๐0and ๐ท๐ =
1
๐0
โข Therefore, using results from slide 8 we get:
๐ ๐ = โฑ ๐ฟ๐0 ๐ก =1
๐0ฯโโโ โฑ{๐๐๐๐0๐ก} =
1
๐0ฯ๐=โโโ 2๐๐ฟ ๐ โ ๐๐0 , ๐0 =
2๐
๐0
๐ ๐ = ๐0ฯ๐=โโโ ๐ฟ ๐ โ ๐๐0 = ๐0 ๐ฟ๐0
๐
โข The Fourier transform of an impulse train in time (denoted by ๐ฟ๐0 ๐ก ) is an
impulse train in frequency (denoted by ๐ฟ๐0๐ ) .
โข The closer (further) the pulses in time the further (closer) in frequency.
Fourier transform of a unit impulse train
Linearity and conjugate properties
โข Linearity
If ๐ฅ1(๐ก) โ ๐1 ๐ and ๐ฅ2(๐ก) โ ๐2 ๐ , then
๐1๐ฅ1 ๐ก +๐2๐ฅ2 ๐ก โ ๐1๐1 ๐ + ๐2๐2 ๐
โข Property of conjugate of a signal
If ๐ฅ(๐ก) โ ๐ ๐ then ๐ฅโ(๐ก) โ ๐โ โ๐ .
โข Property of conjugate symmetry
If ๐ฅ ๐ก is real then ๐ฅโ ๐ก = ๐ฅ(๐ก) and therefore, from the property above we
see that ๐ ๐ = ๐โ โ๐ or ๐(โ๐) = ๐โ ๐ .We can write ๐ ๐ = ๐ด ๐ ๐๐๐(๐).o ๐ด ๐ , ๐(๐) are the amplitude and phase spectrum respectively.
They are real functions.
o ๐โ ๐ = ๐ด ๐ ๐โ๐๐(๐) and ๐โ โ๐ = ๐ด โ๐ ๐โ๐๐(โ๐)
o Based on the last bullet point, for a real function we have:
๐ ๐ = ๐โ โ๐ โ ๐ด ๐ ๐๐๐(๐) = ๐ด โ๐ ๐โ๐๐(โ๐) โโช ๐ด ๐ = ๐ด โ๐ โ for a real signal, the amplitude spectrum is even.โช ๐ ๐ = โ๐(โ๐) โ for a real signal, the phase spectrum is odd.
Time-frequency duality of Fourier transform
โข There is a near symmetry between the forward and inverse Fourier
transforms.
โข The same observation was valid for Laplace transform.
Forward FT
Inverse FT
Duality property
โข If ๐ฅ(๐ก) โ ๐ ๐ then ๐(๐ก) โ 2๐๐ฅ โ๐
Proof
From the definition of the inverse Fourier transform we get:
๐ฅ ๐ก =1
2๐เถฑโโ
โ
๐ ๐ ๐๐๐๐ก๐๐
Therefore,
2๐๐ฅ โ๐ก = เถฑโโ
โ
๐ ๐ ๐โ๐๐๐ก๐๐
Swapping ๐ก with ๐ and using the definition of forward Fourier transform we
have:
๐(๐ก) โ 2๐๐ฅ โ๐
โข Consider the Fourier transform of a rectangular function
rect๐ก
๐โ ๐sinc(
๐๐
2)
โ
๐sinc(๐๐ก
2) โ 2๐ rect
โ๐
๐= 2๐ rect
๐
๐
โ
Duality property example
Scaling property
โข If ๐ฅ(๐ก) โ ๐ ๐ then for any real constant ๐ the following property holds.
๐ฅ(๐๐ก) โ1
๐๐
๐
๐
โข That is, compression of a signal in time results in spectral expansion and
vice versa. As mentioned, the extreme case is the Dirac function and an
everlasting constant function.
Time-shifting property with example
โข If ๐ฅ(๐ก) โ ๐ ๐ then the following property holds.
๐ฅ(๐ก โ ๐ก0) โ ๐ ๐ ๐โ๐๐๐ก0
โข Find the Fourier transform of the gate pulse ๐ฅ(๐ก) given by rect๐กโ
3๐
4
๐.
โข By using the time-shifting property we get ๐ ฯ = ๐sinc(๐๐
2)๐โ๐๐
3๐
4 .
โข Observe the amplitude (even) and phase (odd) of the Fourier transform.
Frequency-shifting property
โข If ๐ฅ(๐ก) โ ๐ ๐ then ๐ฅ(๐ก)๐๐๐0๐ก โ ๐ ๐ โ๐0 . This property states that
multiplying a signal by ๐๐๐0๐ก shifts the spectrum of the signal by ๐0.
โข In practice, frequency shifting (or amplitude modulation) is achieved by
multiplying ๐ฅ(๐ก) by a sinusoid. This is because:
๐ฅ ๐ก cos ๐0๐ก =1
2[๐ฅ(๐ก)๐๐๐0๐ก + ๐ฅ(๐ก)๐โ๐๐0๐ก]
๐ฅ(๐ก) cos ๐0๐ก โ1
2[๐ ๐ โ ๐0 + ๐ ๐ + ๐0 ]
โข Find and sketch the Fourier transform of the signal ๐ฅ ๐ก cos10๐ก where
๐ฅ ๐ก = rect๐ก
4. We know that rect
๐ก
4โ 4sinc(2๐)
๐ฅ ๐ก cos 10๐ก =1
2[๐ฅ(๐ก)๐๐10๐ก + ๐ฅ(๐ก)๐โ๐10๐ก]
๐ฅ(๐ก) cos 10๐ก โ1
2[๐ ๐ โ 10 + ๐ ๐ + 10 ]
๐ฅ ๐ก cos 10๐ก โ 2 {sinc 2 ๐ โ 10 + sinc 2 ๐ + 10 }
Frequency-shifting example
Is the phase important?
Phase from (b), Amp. from (a) Phase from (a), Amp. from (b)
(a) (b)
Convolution properties
โข Time and frequency convolution.
If ๐ฅ1(๐ก) โ ๐1 ๐ and ๐ฅ2(๐ก) โ ๐2 ๐ , then
โช ๐ฅ1 ๐ก โ ๐ฅ2 ๐ก โ ๐1 ๐ ๐2 ๐
โช ๐ฅ1 ๐ก ๐ฅ2 ๐ก โ1
2๐๐1 ๐ โ ๐2 ๐
โข Let ๐ป(๐) be the Fourier transform of the unit impulse response โ(๐ก), i.e.,
โ(๐ก) โ ๐ป ๐
โข Applying the time-convolution property to ๐ฆ ๐ก = ๐ฅ(๐ก) โ โ(๐ก) we get:
๐ ๐ = ๐ ๐ ๐ป ๐
โข Therefore, the Fourier Transform of the systemโs impulse response is the systemโs Frequency Response.
Frequency convolution example
โข Find the spectrum of of the signal ๐ฅ ๐ก cos10๐ก where ๐ฅ ๐ก = rect๐ก
4.
โข We know that rect๐ก
4โ 4sinc(2๐).
ร โ1/21/2
Time differentiation property
โข If ๐ฅ(๐ก) โ ๐ ๐ then the following properties hold:
โช Time differentiation property.๐๐ฅ(๐ก)
๐๐กโ ๐๐๐(๐)
โช Time integration property.
โโ๐ก
๐ฅ ๐ ๐๐ โ๐(๐)
๐๐+ ๐๐(0)๐ฟ(๐)
โข Compare with the time differentiation property in the Laplace domain.
๐ฅ(๐ก) โ ๐ ๐ ๐๐ฅ ๐ก
๐๐กโ ๐ ๐ ๐ โ ๐ฅ(0โ)
Appendix: Proof of the time convolution property
โข By definition we have:
โฑ ๐ฅ1 ๐ก โ ๐ฅ2 ๐ก = เถฑ๐ก=โโ
โ
[เถฑ๐=โโ
โ
๐ฅ1 ๐ ๐ฅ2 ๐ก โ ๐ ๐๐]๐โ๐๐๐ก๐๐ก
= เถฑ๐=โโ
โ
[เถฑ๐ก=โโ
โ
๐ฅ1 ๐ ๐ฅ2 ๐ก โ ๐ ๐โ๐๐๐ก๐๐ก]๐๐
= เถฑ๐=โโ
โ
๐ฅ1 ๐ [เถฑ๐ก=โโ
โ
๐ฅ2 ๐ก โ ๐ ๐โ๐๐๐ก๐๐ก]๐๐
= เถฑ๐=โโ
โ
๐ฅ1 ๐ ๐โ๐๐๐[เถฑ๐ก=โโ
โ
๐ฅ2 ๐ก โ ๐ ๐โ๐๐ ๐กโ๐ ๐(๐ก โ ๐)]๐๐
= เถฑ๐=โโ
โ
๐ฅ1 ๐ ๐โ๐๐๐[เถฑ๐ก=โโ
โ
๐ฅ2 ๐ฃ ๐โ๐๐๐ฃ๐๐ฃ]๐๐
= โโ=๐โ
๐ฅ1 ๐ ๐โ๐๐๐๐1 ๐ ๐๐ = ๐1 ๐ โโ=๐โ
๐ฅ1 ๐ ๐โ๐๐๐๐๐ = ๐1 ๐ ๐2 ๐
Fourier transform table 1
Fourier transform table 2
Fourier transform table 3
Summary of Fourier transform operations 1
Summary of Fourier transform operations 2