6.003 signal processing · ðŽâ, in terms of f[k]? previously(lecture 2b) let x[k] represent the...
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![Page 1: 6.003 Signal Processing · ðŽâ, in terms of F[k]? Previously(lecture 2B) Let X[k] represent the Fourier series coefficients of the following signal: ... sin( ) cos( ) Previously(Lecture](https://reader033.vdocuments.us/reader033/viewer/2022053100/605a38fb9ccae26dd03856ee/html5/thumbnails/1.jpg)
6.003 Signal Processing
Week 3, Lecture A:CTFS properties
6.003 Fall 2020
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Continuous Time Fourier Series
Synthesis:
Analysis:
Today: useful properties of CTFS
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Properties of CTFS: Linearity
⢠Consider ðŠ ð¡ = ðŽð¥1 ð¡ + ðµð¥2 ð¡ , where ð¥1 ð¡ and ð¥2 ð¡ are periodic in T. What are the CTFS coefficients ð[ð], in terms of ð1[ð]and ð2[ð] ?
6.003 Fall 2020
First, ðŠ ð¡ must also be periodic in T
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Properties of CTFS: Time flip
⢠Consider ðŠ ð¡ = ð¥ âð¡ , where ð¥ ð¡ is periodic in T. What are the CTFS coefficients ð[ð], in terms of ð[ð]?
6.003 Fall 2020
First, ðŠ ð¡ must also be periodic in T
ð¥ ð¡ =
ð=ââ
â
ð[ð]ðð2ððð¡ð ðŠ ð¡ = ð¥(âð¡) =
ð=ââ
â
ð[ð]ðð2ðð(âð¡)
ð
Let ð = âð
ðŠ ð¡ = ð¥(âð¡) =
ð=ââ
â
ð[âð]ðð2ððð¡ð
ðŠ ð¡ =
ð=ââ
â
ð[ð]ðð2ððð¡ð
Since we know
ð[ð] = ð[âð]
If ðŠ ð¡ = ð¥ âð¡ , ð[ð] = ð[âð]
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Symmetric and Antisymmetric Parts
⢠If ð ð¡ = ðð ð¡ + ððŽ(ð¡) is a real-valued signal and periodic in time with fundamental period T, what are the Fourier coefficients of ðð â and ððŽ â , in terms of F[k]?
6.003 Fall 2020
ð ð¡ = ðð ð¡ + ððŽ(ð¡)
Any arbitrary signal ð ð¡ can be written into two parts:
ðð ð¡ is symmetric about ð¡ = 0, ðð ðð ð¡ = ðð âð¡ for all t.
ððŽ ð¡ is antisymmetric about ð¡ = 0, ðð ððŽ ð¡ = âðð âð¡ for all t.
ðð ð¡ =ð ð¡ + ð(âð¡)
2ððŽ ð¡ =
ð ð¡ â ð(âð¡)
2
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Real-valued periodic signal
6.003 Fall 2020
ð¹[ð] =1
ðනð
ð(ð¡)ðâð2ððð¡ð ðð¡
If ð ð¡ is real valued periodic signal:
ð¹[âð] =1
ðනð
ð(ð¡)ðð2ððð¡ð ðð¡
ð¹â[âð] =1
ðනð
ð(ð¡)ðâð2ððð¡ð ðð¡
If ð ð¡ is real valued periodic signal, ð¹ ð = ð¹â[âð]
= ð¹[ð]
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Symmetric and Antisymmetric Parts
If ð ð¡ is real valued periodic signal, ð¹ ð = ð¹â[âð]
ðð ð¡ =ð ð¡ + ð(âð¡)
2
ð¿ððððððð¡ðŠð¹ð ð =
ð¹ ð + ð¹ âð
2=ð¹ ð + ð¹â ð
2=2ð ð(ð¹ ð )
2= ð ð(ð¹ ð )
ððŽ ð¡ =ð ð¡ â ð(âð¡)
2
ð¿ððððððð¡ðŠð¹ðŽ ð =
ð¹ ð â ð¹ âð
2=ð¹ ð â ð¹â ð
2=2ð â ðŒð(ð¹ ð )
2= ð â ðŒð(ð¹ ð )
The real part of F[k] comes from the symmetric part of the signal, the imaginary part of F k comes from the antisymmetric part of the signal
⢠If ð ð¡ = ðð ð¡ + ððŽ(ð¡) is a real valued signal and periodic in time with fundamental period T, what are the Fourier coefficients of ðð â and ððŽ â , in terms of F[k]?
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Previously(lecture 2B)
Let X[k] represent the Fourier series coefficients of the following signal:
T=1, ð0 =2ð
ð= 2ð
ð[ð] = á
0, ð ðð ðð£ðð1
ððð, ð ðð ððð
ð¥ ð¡ is antisymmetric around t=0, thus its Fourier Series coefficients are purely imaginary
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Symmetric and Antisymmetric Parts in Trig form CTFS
ðð ð¡ =ð ð¡ + ð(âð¡)
2ððŽ ð¡ =
ð ð¡ â ð(âð¡)
2
The symmetric part shows up in the ðð coefficients, and the antisymmetric part shows up in the ðð coefficients.
⢠ððâs (cosines) alone can only represent symmetric functions.
⢠ðð âs (sines) alone can only represent antisymmetric functions
sin(ð¥) cos(ð¥)
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Previously(Lecture 2A)⢠Find the Fourier series coefficients for the following square wave:
If without ð0 =1
2âDCâ part, ð ð¡ is antisymmetric around t=0, thus only having non-zero ðð âs
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Time shifts in CTFS
6.003 Fall 2020
The time shifting in f(t) is carried to the underlying basis functions of it Fourier expansion.
Half-period shift inverts ðð terms if k is odd. It has no effect if k is even.
ð ð¡ =
ð ð¡ = Ïð=0â ðð
â² cos(ðð0ð¡) + Ïð=1â ðð
â² sin(ðð0ð¡)
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Half period shift
6.003 Fall 2020
The time shifting in f(t) is carried to the underlying basis functions of it Fourier expansion.
Half-period shift inverts ðð terms if and only if k is odd.
ð ð¡ =
ð ð¡ = Ïð=0â ðð
â² cos(ðð0ð¡) + Ïð=1â ðð
â² sin(ðð0ð¡)
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Quarter period shift?
Shifting by T /4 becomes more complicated.
â ð¡ =
â ð¡ = Ïð=0â ðð
â²â²cos(ðð0ð¡) + Ïð=1â ðð
â²â²sin(ðð0ð¡)
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Quarter period shift?
6.003 Fall 2020
â ð¡ =
â ð¡ = Ïð=0â ðð
â²â²cos(ðð0ð¡) + Ïð=1â ðð
â²â²sin(ðð0ð¡)
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Summary of shift result
6.003 Fall 2020
Other shifts will be even more complicated!
4â ð¡
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Properties of CTFS: Time Shift⢠Consider ðŠ ð¡ = ð¥ ð¡ â ð¡0 , where ð¥ is periodic in ð. What are the
CTFS coefficients ð[ð], in terms of ð[ð]?
ððð¡ ð¢ = ð¡ â ð¡0, then ð¡ = ð¢ + ð¡0,dð¡ = ðð¢
Each coefficient ð ð in the series for ðŠ(ð¡) is a constant ðâððð0ð times the corresponding coefficient ð[ð] in the series for ð¥(ð¡).
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Properties of CTFS: Time Derivative
⢠Consider ðŠ ð¡ =ð
ðð¡ð¥ ð¡ , where ð¥ ð¡ and ðŠ ð¡ are periodic in ð. What
are the CTFS coefficients ð[ð], in terms of ð[ð]?
6.003 Fall 2020
Start with the synthesis equation:
Then, from the definition of y(·), we have:
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Properties of CTFS: brief summary
⢠Certain operations in the time domain have predictable effects in the frequency domain.
These can help us find the Fourier coefficients of a new signal without explicitly integrating!
If ð ð¡ is real valued periodic signal, ð¹ ð = ð¹â[âð]
The real part of F[k] comes from the symmetric part of the signal, the imaginary part of F k comes from the antisymmetric part of the signal
If ðŠ ð¡ = ð¥ âð¡ , ð[ð] = ð[âð]
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Check yourself!
Let ð[ð] represent the Fourier series coefficients of the following signal:
6.003 Fall 2020
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Check yourself!What is the relationship between the two following signals?
6.003 Fall 2020
ð[ð] = á
0, ð ðð ðð£ðð1
ððð, ð ðð ððð
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Check yourself!The triangle waveform is the integral of the square wave.
ð¥ ð¡ =ð
ðð¡ðŠ ð¡
ð ð = ð2ðð
ðð ð , ð = 1
ð[ð] = á
0, ð ðð ðð£ðð1
ððð, ð ðð ððð
ð ð =1
ð2ððð ð
ðŠ ð¡ is symmetric around t=0, thus its Fourier Series coefficients are purely real
= á0, ð ðð ðð£ðð
â1
2ð2ð2, ð ðð ððð
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Check yourself!
Let ð[ð] represent the Fourier series coefficients of the following signal:
6.003 Fall 2020
ð[ð] = á0, ð ðð ðð£ðð
â1
2ð2ð2, ð ðð ððð
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Summary
⢠Various properties of the CTFS, making it easier to find the Fourier coefficients of a new signal
⢠The Complex Exponential Form is much easier in math, will focus on this in the later part of the course
6.003 Fall 2020