Signals and Systems - Imperial College London

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Signals and Systems Lecture 8 DR TANIA STATHAKI READER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSING IMPERIAL COLLEGE LONDON

Transcript of Signals and Systems - Imperial College London

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Signals and Systems

Lecture 8

DR TANIA STATHAKIREADER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSINGIMPERIAL COLLEGE LONDON

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β€’ The forward and inverse Fourier transform are defined for aperiodic

signals as:

𝑋 πœ” = β„± π‘₯ 𝑑 = ΰΆ±βˆ’βˆž

∞

π‘₯ 𝑑 π‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘

π‘₯ 𝑑 = β„±βˆ’1 𝑋 πœ” =1

2πœ‹ΰΆ±βˆ’βˆž

∞

𝑋 πœ” π‘’π‘—πœ”π‘‘π‘‘πœ”

β€’ You can immediately observe the functional similarity with Laplace

transform.

β€’ For periodic signals we use Fourier Series.

π‘₯𝑇0 𝑑 =

𝑛=βˆ’βˆž

∞

π·π‘›π‘’π‘—π‘›πœ”0𝑑

𝐷𝑛 =1

𝑇0𝑇0/2βˆ’π‘‡0/2 π‘₯𝑇0 𝑑 π‘’βˆ’π‘—π‘›πœ”0𝑑𝑑𝑑 or

𝐷𝑛 =1

𝑇0one full periodπ‘₯𝑇0 𝑑 π‘’βˆ’π‘—π‘›πœ”0𝑑𝑑𝑑 , πœ”0 =

2πœ‹

𝑇0

Definition of Fourier transform

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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 βˆ’ 2 π‘₯ π‘₯ <1

2

β€’ Interpolation function sinc(π‘₯):

sinc(π‘₯) =sin(π‘₯)

π‘₯or sinc(π‘₯) =

sin(πœ‹π‘₯)

πœ‹π‘₯

Define three useful functions

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β€’ The sinc(π‘₯) function is an even function of π‘₯.

β€’ sinc π‘₯ = 0 when sin π‘₯ = 0, i.e.,

π‘₯ = Β±πœ‹,Β±2πœ‹,Β±3πœ‹,… except when π‘₯ = 0

where sinc 0 = 1. This can be proven by the

L’Hospital’s 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œ‹ with amplitude decreasing as 1/π‘₯.

More about the 𝐬𝐒𝐧𝐜(x) function

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β€’ Evaluation:

𝑋 πœ” = β„± π‘₯ 𝑑 = ΰΆ±βˆ’βˆž

∞

rect𝑑

πœπ‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘

β€’ Since rect𝑑

𝜏= 1 for

βˆ’πœ

2< 𝑑 <

𝜏

2and 0 otherwise we have:

𝑋 πœ” = β„± π‘₯ 𝑑 = ΰΆ±βˆ’πœ2

𝜏2π‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘ = βˆ’

1

π‘—πœ”π‘’βˆ’π‘—πœ”

𝜏2 βˆ’ π‘’π‘—πœ”

𝜏2 =

2sin(πœ”πœ2)

πœ”

= 𝜏sin(

πœ”πœ

2)

(πœ”πœ

2)= 𝜏sinc(

πœ”πœ

2) β‡’ β„± rect

𝑑

𝜏= 𝜏sinc(

πœ”πœ

2) or rect

𝑑

πœβ‡” 𝜏sinc(

πœ”πœ

2)

β€’ The bandwidth of the function rect𝑑

𝜏is approximately

2πœ‹

𝜏.

β€’ Observe that the wider(narrower) the pulse in time the narrower(wider)

the lobes of the sinc function in frequency.

Fourier transform of 𝒙 𝒕 = 𝐫𝐞𝐜𝐭(𝒕/𝝉)

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β€’ Using the sampling property of the impulse we get:

𝑋 πœ” = β„± 𝛿(𝑑) = ΰΆ±βˆ’βˆž

∞

𝛿 𝑑 π‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘ = 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.)

𝛿 𝑑 ⇔ 1

Fourier transform of the unit impulse 𝒙 𝒕 = 𝜹(𝒕)

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β€’ Using the sampling property of the impulse we get:

β„±βˆ’1 𝛿 πœ” =1

2πœ‹βˆžβˆ’βˆž

𝛿 πœ” π‘’π‘—πœ”π‘‘π‘‘πœ” =1

2πœ‹

β€’ Therefore, the spectrum of a constant signal π‘₯ 𝑑 = 1 is an impulse 2πœ‹π›Ώ πœ” .1

2πœ‹β‡” 𝛿 πœ” or 1 ⇔ 2πœ‹π›Ώ πœ”

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 𝜹(𝝎)

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β€’ Using the sampling property of the impulse we get:

β„±βˆ’1 𝛿 πœ” βˆ’ πœ”0 =1

2πœ‹βˆžβˆ’βˆž

𝛿 πœ” βˆ’ πœ”0 π‘’π‘—πœ”π‘‘π‘‘πœ” =1

2πœ‹π‘’π‘—πœ”0𝑑

β€’ The spectrum of an everlasting exponential π‘’π‘—πœ”0𝑑 is a single impulse

located at πœ” = πœ”0 .1

2πœ‹π‘’π‘—πœ”0𝑑 ⇔ 𝛿 πœ” βˆ’ πœ”0

π‘’π‘—πœ”0𝑑 ⇔ 2πœ‹π›Ώ πœ” βˆ’ πœ”0

π‘’βˆ’π‘—πœ”0𝑑 ⇔ 2πœ‹π›Ώ πœ” + πœ”0

Inverse Fourier transform of 𝜹(𝝎 βˆ’πŽπŸŽ)

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β€’ Remember the Euler’s formula:

cosπœ”0𝑑 =1

2(π‘’π‘—πœ”0𝑑 + π‘’βˆ’π‘—πœ”0𝑑)

β„± cosπœ”0𝑑 = β„±1

2(π‘’π‘—πœ”0𝑑 + π‘’βˆ’π‘—πœ”0𝑑) =

1

2β„± π‘’π‘—πœ”0𝑑 +

1

2β„± π‘’βˆ’π‘—πœ”0𝑑

β€’ Using the results from previous slides we get:

cosπœ”0𝑑 ⇔ πœ‹[𝛿 πœ” + πœ”0 + 𝛿 πœ” βˆ’ πœ”0 ]

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 πœπ¨π¬πŽπŸŽπ’•

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β€’ The Fourier series of a periodic signal π‘₯(𝑑) with period 𝑇0 is given by:

π‘₯ 𝑑 = Οƒβˆ’βˆžβˆž 𝐷𝑛 𝑒

π‘—π‘›πœ”0𝑑, πœ”0 =2πœ‹

𝑇0

β€’ By taking the Fourier transform on both sides we get:

𝑋(πœ”) = 2πœ‹

𝑛=βˆ’βˆž

∞

𝐷𝑛 𝛿 πœ” βˆ’ π‘›πœ”0

Fourier transform of any periodic signal

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β€’ Consider an impulse train

𝛿𝑇0 𝑑 = Οƒβˆ’βˆžβˆž 𝛿 𝑑 βˆ’ 𝑛𝑇0

β€’ The Fourier series of this impulse train can be shown to be:

𝛿𝑇0 𝑑 = Οƒβˆ’βˆžβˆž 𝐷𝑛 𝑒

π‘—π‘›πœ”0𝑑 where πœ”0 =2πœ‹

𝑇0and 𝐷𝑛 =

1

𝑇0

β€’ Therefore, using results from slide 8 we get:

𝑋 πœ” = β„± 𝛿𝑇0 𝑑 =1

𝑇0Οƒβˆ’βˆžβˆž β„±{π‘’π‘—π‘›πœ”0𝑑} =

1

𝑇0σ𝑛=βˆ’βˆžβˆž 2πœ‹π›Ώ πœ” βˆ’ π‘›πœ”0 , πœ”0 =

2πœ‹

𝑇0

𝑋 πœ” = πœ”0σ𝑛=βˆ’βˆžβˆž 𝛿 πœ” βˆ’ π‘›πœ”0 = πœ”0 π›Ώπœ”0

πœ”

β€’ The Fourier transform of an impulse train in time (denoted by 𝛿𝑇0 𝑑 ) is an

impulse train in frequency (denoted by π›Ώπœ”0πœ” ) .

β€’ The closer (further) the pulses in time the further (closer) in frequency.

Fourier transform of a unit impulse train

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Linearity and conjugate properties

β€’ Linearity

If π‘₯1(𝑑) ⇔ 𝑋1 πœ” and π‘₯2(𝑑) ⇔ 𝑋2 πœ” , then

π‘Ž1π‘₯1 𝑑 +π‘Ž2π‘₯2 𝑑 ⇔ π‘Ž1𝑋1 πœ” + π‘Ž2𝑋2 πœ”

β€’ Property of conjugate of a signal

If π‘₯(𝑑) ⇔ 𝑋 πœ” then π‘₯βˆ—(𝑑) ⇔ π‘‹βˆ— βˆ’πœ” .

β€’ Property of conjugate symmetry

If π‘₯ 𝑑 is real then π‘₯βˆ— 𝑑 = π‘₯(𝑑) and therefore, from the property above we

see that 𝑋 πœ” = π‘‹βˆ— βˆ’πœ” or 𝑋(βˆ’πœ”) = π‘‹βˆ— πœ” .We can write 𝑋 πœ” = 𝐴 πœ” π‘’π‘—πœ™(πœ”).o 𝐴 πœ” , πœ™(πœ”) are the amplitude and phase spectrum respectively.

They are real functions.

o π‘‹βˆ— πœ” = 𝐴 πœ” π‘’βˆ’π‘—πœ™(πœ”) and π‘‹βˆ— βˆ’πœ” = 𝐴 βˆ’πœ” π‘’βˆ’π‘—πœ™(βˆ’πœ”)

o Based on the last bullet point, for a real function we have:

𝑋 πœ” = π‘‹βˆ— βˆ’πœ” β‡’ 𝐴 πœ” π‘’π‘—πœ™(πœ”) = 𝐴 βˆ’πœ” π‘’βˆ’π‘—πœ™(βˆ’πœ”) β‡’β–ͺ 𝐴 πœ” = 𝐴 βˆ’πœ” β‡’ for a real signal, the amplitude spectrum is even.β–ͺ πœ™ πœ” = βˆ’πœ™(βˆ’πœ”) β‡’ for a real signal, the phase spectrum is odd.

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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

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Duality property

β€’ If π‘₯(𝑑) ⇔ 𝑋 πœ” then 𝑋(𝑑) ⇔ 2πœ‹π‘₯ βˆ’πœ”

Proof

From the definition of the inverse Fourier transform we get:

π‘₯ 𝑑 =1

2πœ‹ΰΆ±βˆ’βˆž

∞

𝑋 πœ” π‘’π‘—πœ”π‘‘π‘‘πœ”

Therefore,

2πœ‹π‘₯ βˆ’π‘‘ = ΰΆ±βˆ’βˆž

∞

𝑋 πœ” π‘’βˆ’π‘—πœ”π‘‘π‘‘πœ”

Swapping 𝑑 with πœ” and using the definition of forward Fourier transform we

have:

𝑋(𝑑) ⇔ 2πœ‹π‘₯ βˆ’πœ”

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β€’ Consider the Fourier transform of a rectangular function

rect𝑑

πœβ‡” 𝜏sinc(

πœ”πœ

2)

⇔

𝜏sinc(πœπ‘‘

2) ⇔ 2πœ‹ rect

βˆ’πœ”

𝜏= 2πœ‹ rect

πœ”

𝜏

⇔

Duality property example

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Scaling property

β€’ If π‘₯(𝑑) ⇔ 𝑋 πœ” then for any real constant π‘Ž the following property holds.

π‘₯(π‘Žπ‘‘) ⇔1

π‘Žπ‘‹

πœ”

π‘Ž

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.

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Time-shifting property with example

β€’ If π‘₯(𝑑) ⇔ 𝑋 πœ” then the following property holds.

π‘₯(𝑑 βˆ’ 𝑑0) ⇔ 𝑋 πœ” π‘’βˆ’π‘—πœ”π‘‘0

β€’ Find the Fourier transform of the gate pulse π‘₯(𝑑) given by rectπ‘‘βˆ’

3𝜏

4

𝜏.

β€’ By using the time-shifting property we get 𝑋 Ο‰ = 𝜏sinc(πœ”πœ

2)π‘’βˆ’π‘—πœ”

3𝜏

4 .

β€’ Observe the amplitude (even) and phase (odd) of the Fourier transform.

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Frequency-shifting property

β€’ If π‘₯(𝑑) ⇔ 𝑋 πœ” then π‘₯(𝑑)π‘’π‘—πœ”0𝑑 ⇔ 𝑋 πœ” βˆ’πœ”0 . This property states that

multiplying a signal by π‘’π‘—πœ”0𝑑 shifts the spectrum of the signal by πœ”0.

β€’ In practice, frequency shifting (or amplitude modulation) is achieved by

multiplying π‘₯(𝑑) by a sinusoid. This is because:

π‘₯ 𝑑 cos πœ”0𝑑 =1

2[π‘₯(𝑑)π‘’π‘—πœ”0𝑑 + π‘₯(𝑑)π‘’βˆ’π‘—πœ”0𝑑]

π‘₯(𝑑) cos πœ”0𝑑 ⇔1

2[𝑋 πœ” βˆ’ πœ”0 + 𝑋 πœ” + πœ”0 ]

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β€’ Find and sketch the Fourier transform of the signal π‘₯ 𝑑 cos10𝑑 where

π‘₯ 𝑑 = rect𝑑

4. We know that rect

𝑑

4⇔ 4sinc(2πœ”)

π‘₯ 𝑑 cos 10𝑑 =1

2[π‘₯(𝑑)𝑒𝑗10𝑑 + π‘₯(𝑑)π‘’βˆ’π‘—10𝑑]

π‘₯(𝑑) cos 10𝑑 ⇔1

2[𝑋 πœ” βˆ’ 10 + 𝑋 πœ” + 10 ]

π‘₯ 𝑑 cos 10𝑑 ⇔ 2 {sinc 2 πœ” βˆ’ 10 + sinc 2 πœ” + 10 }

Frequency-shifting example

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Is the phase important?

Phase from (b), Amp. from (a) Phase from (a), Amp. from (b)

(a) (b)

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Convolution properties

β€’ Time and frequency convolution.

If π‘₯1(𝑑) ⇔ 𝑋1 πœ” and π‘₯2(𝑑) ⇔ 𝑋2 πœ” , then

β–ͺ π‘₯1 𝑑 βˆ— π‘₯2 𝑑 ⇔ 𝑋1 πœ” 𝑋2 πœ”

β–ͺ π‘₯1 𝑑 π‘₯2 𝑑 ⇔1

2πœ‹π‘‹1 πœ” βˆ— 𝑋2 πœ”

β€’ Let 𝐻(πœ”) be the Fourier transform of the unit impulse response β„Ž(𝑑), i.e.,

β„Ž(𝑑) ⇔ 𝐻 πœ”

β€’ Applying the time-convolution property to 𝑦 𝑑 = π‘₯(𝑑) βˆ— β„Ž(𝑑) we get:

π‘Œ πœ” = 𝑋 πœ” 𝐻 πœ”

β€’ Therefore, the Fourier Transform of the system’s impulse response is the system’s Frequency Response.

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Frequency convolution example

β€’ Find the spectrum of of the signal π‘₯ 𝑑 cos10𝑑 where π‘₯ 𝑑 = rect𝑑

4.

β€’ We know that rect𝑑

4⇔ 4sinc(2πœ”).

Γ— βˆ—1/21/2

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Time differentiation property

β€’ If π‘₯(𝑑) ⇔ 𝑋 πœ” then the following properties hold:

β–ͺ Time differentiation property.𝑑π‘₯(𝑑)

𝑑𝑑⇔ π‘—πœ”π‘‹(πœ”)

β–ͺ Time integration property.

βˆžβˆ’π‘‘

π‘₯ 𝜏 π‘‘πœ ⇔𝑋(πœ”)

π‘—πœ”+ πœ‹π‘‹(0)𝛿(πœ”)

β€’ Compare with the time differentiation property in the Laplace domain.

π‘₯(𝑑) ⇔ 𝑋 𝑠𝑑π‘₯ 𝑑

𝑑𝑑⇔ 𝑠𝑋 𝑠 βˆ’ π‘₯(0βˆ’)

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Appendix: Proof of the time convolution property

β€’ By definition we have:

β„± π‘₯1 𝑑 βˆ— π‘₯2 𝑑 = ࢱ𝑑=βˆ’βˆž

∞

[ࢱ𝜏=βˆ’βˆž

∞

π‘₯1 𝜏 π‘₯2 𝑑 βˆ’ 𝜏 π‘‘πœ]π‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘

= ࢱ𝜏=βˆ’βˆž

∞

[ࢱ𝑑=βˆ’βˆž

∞

π‘₯1 𝜏 π‘₯2 𝑑 βˆ’ 𝜏 π‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘]π‘‘πœ

= ࢱ𝜏=βˆ’βˆž

∞

π‘₯1 𝜏 [ࢱ𝑑=βˆ’βˆž

∞

π‘₯2 𝑑 βˆ’ 𝜏 π‘’βˆ’π‘—πœ”π‘‘π‘‘π‘‘]π‘‘πœ

= ࢱ𝜏=βˆ’βˆž

∞

π‘₯1 𝜏 π‘’βˆ’π‘—πœ”πœ[ࢱ𝑑=βˆ’βˆž

∞

π‘₯2 𝑑 βˆ’ 𝜏 π‘’βˆ’π‘—πœ” π‘‘βˆ’πœ 𝑑(𝑑 βˆ’ 𝜏)]π‘‘πœ

= ࢱ𝜏=βˆ’βˆž

∞

π‘₯1 𝜏 π‘’βˆ’π‘—πœ”πœ[ࢱ𝑑=βˆ’βˆž

∞

π‘₯2 𝑣 π‘’βˆ’π‘—πœ”π‘£π‘‘π‘£]π‘‘πœ

= βˆžβˆ’=𝜏∞

π‘₯1 𝜏 π‘’βˆ’π‘—πœ”πœπ‘‹1 πœ” π‘‘πœ = 𝑋1 πœ” βˆžβˆ’=𝜏∞

π‘₯1 𝜏 π‘’βˆ’π‘—πœ”πœπ‘‘πœ = 𝑋1 πœ” 𝑋2 πœ”

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Fourier transform table 1

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Fourier transform table 2

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Fourier transform table 3

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Summary of Fourier transform operations 1

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Summary of Fourier transform operations 2