Signals and Systems - Imperial College London
Transcript of Signals and Systems - Imperial College London
Signals and Systems
Lecture 8
DR TANIA STATHAKIREADER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSINGIMPERIAL COLLEGE LONDON
β’ The forward and inverse Fourier transform are defined for aperiodic
signals as:
π π = β± π₯ π‘ = ΰΆ±ββ
β
π₯ π‘ πβπππ‘ππ‘
π₯ π‘ = β±β1 π π =1
2πΰΆ±ββ
β
π π ππππ‘ππ
β’ You can immediately observe the functional similarity with Laplace
transform.
β’ For periodic signals we use Fourier Series.
π₯π0 π‘ =
π=ββ
β
π·πππππ0π‘
π·π =1
π0π0/2βπ0/2 π₯π0 π‘ πβπππ0π‘ππ‘ or
π·π =1
π0one full periodπ₯π0 π‘ πβπππ0π‘ππ‘ , π0 =
2π
π0
Definition of Fourier transform
β’ A unit rectangular window (also called a unit gate) function rect π₯ :
rect π₯ =
0 π₯ >1
21
2π₯ =
1
2
1 π₯ <1
2
β’ A unit triangle function Ξ π₯ :
Ξ π₯ =0 π₯ β₯
1
2
1 β 2 π₯ π₯ <1
2
β’ Interpolation function sinc(π₯):
sinc(π₯) =sin(π₯)
π₯or sinc(π₯) =
sin(ππ₯)
ππ₯
Define three useful functions
β’ The sinc(π₯) function is an even function of π₯.
β’ sinc π₯ = 0 when sin π₯ = 0, i.e.,
π₯ = Β±π,Β±2π,Β±3π,β¦ except when π₯ = 0
where sinc 0 = 1. This can be proven by the
LβHospitalβs rule.
β’ sinc(π₯) is the product of an
oscillating signal sin(π₯) and a
monotonically decreasing function 1/π₯.
Therefore, it is a damping oscillation with period
2π with amplitude decreasing as 1/π₯.
More about the π¬π’π§π(x) function
β’ Evaluation:
π π = β± π₯ π‘ = ΰΆ±ββ
β
rectπ‘
ππβπππ‘ππ‘
β’ Since rectπ‘
π= 1 for
βπ
2< π‘ <
π
2and 0 otherwise we have:
π π = β± π₯ π‘ = ΰΆ±βπ2
π2πβπππ‘ππ‘ = β
1
πππβππ
π2 β πππ
π2 =
2sin(ππ2)
π
= πsin(
ππ
2)
(ππ
2)= πsinc(
ππ
2) β β± rect
π‘
π= πsinc(
ππ
2) or rect
π‘
πβ πsinc(
ππ
2)
β’ The bandwidth of the function rectπ‘
πis approximately
2π
π.
β’ Observe that the wider(narrower) the pulse in time the narrower(wider)
the lobes of the sinc function in frequency.
Fourier transform of π π = π«πππ(π/π)
β’ Using the sampling property of the impulse we get:
π π = β± πΏ(π‘) = ΰΆ±ββ
β
πΏ π‘ πβπππ‘ππ‘ = 1
β’ As we see the unit impulse contains all frequencies (or, alternatively, we
can say that the unit impulse contains a component at every frequency.)
πΏ π‘ β 1
Fourier transform of the unit impulse π π = πΉ(π)
β’ Using the sampling property of the impulse we get:
β±β1 πΏ π =1
2πβββ
πΏ π ππππ‘ππ =1
2π
β’ Therefore, the spectrum of a constant signal π₯ π‘ = 1 is an impulse 2ππΏ π .1
2πβ πΏ π or 1 β 2ππΏ π
β’ 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 πΉ(π)
β’ Using the sampling property of the impulse we get:
β±β1 πΏ π β π0 =1
2πβββ
πΏ π β π0 ππππ‘ππ =1
2ππππ0π‘
β’ The spectrum of an everlasting exponential πππ0π‘ is a single impulse
located at π = π0 .1
2ππππ0π‘ β πΏ π β π0
πππ0π‘ β 2ππΏ π β π0
πβππ0π‘ β 2ππΏ π + π0
Inverse Fourier transform of πΉ(π βππ)
β’ Remember the Eulerβs formula:
cosπ0π‘ =1
2(πππ0π‘ + πβππ0π‘)
β± cosπ0π‘ = β±1
2(πππ0π‘ + πβππ0π‘) =
1
2β± πππ0π‘ +
1
2β± πβππ0π‘
β’ Using the results from previous slides we get:
cosπ0π‘ β π[πΏ π + π0 + πΏ π β π0 ]
β’ 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 ππ¨π¬πππ
β’ The Fourier series of a periodic signal π₯(π‘) with period π0 is given by:
π₯ π‘ = Οβββ π·π π
πππ0π‘, π0 =2π
π0
β’ By taking the Fourier transform on both sides we get:
π(π) = 2π
π=ββ
β
π·π πΏ π β ππ0
Fourier transform of any periodic signal
β’ Consider an impulse train
πΏπ0 π‘ = Οβββ πΏ π‘ β ππ0
β’ The Fourier series of this impulse train can be shown to be:
πΏπ0 π‘ = Οβββ π·π π
πππ0π‘ where π0 =2π
π0and π·π =
1
π0
β’ Therefore, using results from slide 8 we get:
π π = β± πΏπ0 π‘ =1
π0Οβββ β±{ππππ0π‘} =
1
π0Οπ=βββ 2ππΏ π β ππ0 , π0 =
2π
π0
π π = π0Οπ=βββ πΏ π β ππ0 = π0 πΏπ0
π
β’ The Fourier transform of an impulse train in time (denoted by πΏπ0 π‘ ) is an
impulse train in frequency (denoted by πΏπ0π ) .
β’ The closer (further) the pulses in time the further (closer) in frequency.
Fourier transform of a unit impulse train
Linearity and conjugate properties
β’ Linearity
If π₯1(π‘) β π1 π and π₯2(π‘) β π2 π , then
π1π₯1 π‘ +π2π₯2 π‘ β π1π1 π + π2π2 π
β’ Property of conjugate of a signal
If π₯(π‘) β π π then π₯β(π‘) β πβ βπ .
β’ Property of conjugate symmetry
If π₯ π‘ is real then π₯β π‘ = π₯(π‘) and therefore, from the property above we
see that π π = πβ βπ or π(βπ) = πβ π .We can write π π = π΄ π πππ(π).o π΄ π , π(π) are the amplitude and phase spectrum respectively.
They are real functions.
o πβ π = π΄ π πβππ(π) and πβ βπ = π΄ βπ πβππ(βπ)
o Based on the last bullet point, for a real function we have:
π π = πβ βπ β π΄ π πππ(π) = π΄ βπ πβππ(βπ) ββͺ π΄ π = π΄ βπ β for a real signal, the amplitude spectrum is even.βͺ π π = βπ(βπ) β for a real signal, the phase spectrum is odd.
Time-frequency duality of Fourier transform
β’ There is a near symmetry between the forward and inverse Fourier
transforms.
β’ The same observation was valid for Laplace transform.
Forward FT
Inverse FT
Duality property
β’ If π₯(π‘) β π π then π(π‘) β 2ππ₯ βπ
Proof
From the definition of the inverse Fourier transform we get:
π₯ π‘ =1
2πΰΆ±ββ
β
π π ππππ‘ππ
Therefore,
2ππ₯ βπ‘ = ΰΆ±ββ
β
π π πβπππ‘ππ
Swapping π‘ with π and using the definition of forward Fourier transform we
have:
π(π‘) β 2ππ₯ βπ
β’ Consider the Fourier transform of a rectangular function
rectπ‘
πβ πsinc(
ππ
2)
β
πsinc(ππ‘
2) β 2π rect
βπ
π= 2π rect
π
π
β
Duality property example
Scaling property
β’ If π₯(π‘) β π π then for any real constant π the following property holds.
π₯(ππ‘) β1
ππ
π
π
β’ 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
β’ If π₯(π‘) β π π then the following property holds.
π₯(π‘ β π‘0) β π π πβπππ‘0
β’ Find the Fourier transform of the gate pulse π₯(π‘) given by rectπ‘β
3π
4
π.
β’ By using the time-shifting property we get π Ο = πsinc(ππ
2)πβππ
3π
4 .
β’ Observe the amplitude (even) and phase (odd) of the Fourier transform.
Frequency-shifting property
β’ If π₯(π‘) β π π then π₯(π‘)πππ0π‘ β π π βπ0 . This property states that
multiplying a signal by πππ0π‘ shifts the spectrum of the signal by π0.
β’ In practice, frequency shifting (or amplitude modulation) is achieved by
multiplying π₯(π‘) by a sinusoid. This is because:
π₯ π‘ cos π0π‘ =1
2[π₯(π‘)πππ0π‘ + π₯(π‘)πβππ0π‘]
π₯(π‘) cos π0π‘ β1
2[π π β π0 + π π + π0 ]
β’ Find and sketch the Fourier transform of the signal π₯ π‘ cos10π‘ where
π₯ π‘ = rectπ‘
4. We know that rect
π‘
4β 4sinc(2π)
π₯ π‘ cos 10π‘ =1
2[π₯(π‘)ππ10π‘ + π₯(π‘)πβπ10π‘]
π₯(π‘) cos 10π‘ β1
2[π π β 10 + π π + 10 ]
π₯ π‘ cos 10π‘ β 2 {sinc 2 π β 10 + sinc 2 π + 10 }
Frequency-shifting example
Is the phase important?
Phase from (b), Amp. from (a) Phase from (a), Amp. from (b)
(a) (b)
Convolution properties
β’ Time and frequency convolution.
If π₯1(π‘) β π1 π and π₯2(π‘) β π2 π , then
βͺ π₯1 π‘ β π₯2 π‘ β π1 π π2 π
βͺ π₯1 π‘ π₯2 π‘ β1
2ππ1 π β π2 π
β’ Let π»(π) be the Fourier transform of the unit impulse response β(π‘), i.e.,
β(π‘) β π» π
β’ Applying the time-convolution property to π¦ π‘ = π₯(π‘) β β(π‘) we get:
π π = π π π» π
β’ Therefore, the Fourier Transform of the systemβs impulse response is the systemβs Frequency Response.
Frequency convolution example
β’ Find the spectrum of of the signal π₯ π‘ cos10π‘ where π₯ π‘ = rectπ‘
4.
β’ We know that rectπ‘
4β 4sinc(2π).
Γ β1/21/2
Time differentiation property
β’ If π₯(π‘) β π π then the following properties hold:
βͺ Time differentiation property.ππ₯(π‘)
ππ‘β πππ(π)
βͺ Time integration property.
ββπ‘
π₯ π ππ βπ(π)
ππ+ ππ(0)πΏ(π)
β’ Compare with the time differentiation property in the Laplace domain.
π₯(π‘) β π π ππ₯ π‘
ππ‘β π π π β π₯(0β)
Appendix: Proof of the time convolution property
β’ By definition we have:
β± π₯1 π‘ β π₯2 π‘ = ΰΆ±π‘=ββ
β
[ΰΆ±π=ββ
β
π₯1 π π₯2 π‘ β π ππ]πβπππ‘ππ‘
= ΰΆ±π=ββ
β
[ΰΆ±π‘=ββ
β
π₯1 π π₯2 π‘ β π πβπππ‘ππ‘]ππ
= ΰΆ±π=ββ
β
π₯1 π [ΰΆ±π‘=ββ
β
π₯2 π‘ β π πβπππ‘ππ‘]ππ
= ΰΆ±π=ββ
β
π₯1 π πβπππ[ΰΆ±π‘=ββ
β
π₯2 π‘ β π πβππ π‘βπ π(π‘ β π)]ππ
= ΰΆ±π=ββ
β
π₯1 π πβπππ[ΰΆ±π‘=ββ
β
π₯2 π£ πβπππ£ππ£]ππ
= ββ=πβ
π₯1 π πβππππ1 π ππ = π1 π ββ=πβ
π₯1 π πβπππππ = π1 π π2 π
Fourier transform table 1
Fourier transform table 2
Fourier transform table 3
Summary of Fourier transform operations 1
Summary of Fourier transform operations 2