6.003 signal processing · 𝐴∙, in terms of f[k]? previously(lecture 2b) let x[k] represent the...

6.003 Signal Processing

Week 3, Lecture A:CTFS properties

6.003 Fall 2020

Continuous Time Fourier Series

Synthesis:

Analysis:

Today: useful properties of CTFS

Properties of CTFS: Linearity

• Consider 𝑦 𝑡 = 𝐴𝑥1 𝑡 + 𝐵𝑥2 𝑡 , where 𝑥1 𝑡 and 𝑥2 𝑡 are periodic in T. What are the CTFS coefficients 𝑌[𝑘], in terms of 𝑋1[𝑘]and 𝑋2[𝑘] ?

6.003 Fall 2020

First, 𝑦 𝑡 must also be periodic in T

Properties of CTFS: Time flip

• Consider 𝑦 𝑡 = 𝑥 −𝑡 , where 𝑥 𝑡 is periodic in T. What are the CTFS coefficients 𝑌[𝑘], in terms of 𝑋[𝑘]?

6.003 Fall 2020

First, 𝑦 𝑡 must also be periodic in T

𝑥 𝑡 =

𝑘=−∞

∞

𝑋[𝑘]𝑒𝑗2𝜋𝑘𝑡𝑇 𝑦 𝑡 = 𝑥(−𝑡) =

𝑘=−∞

∞

𝑋[𝑘]𝑒𝑗2𝜋𝑘(−𝑡)

𝑇

Let 𝑚 = −𝑘

𝑦 𝑡 = 𝑥(−𝑡) =

𝑚=−∞

∞

𝑋[−𝑚]𝑒𝑗2𝜋𝑚𝑡𝑇

𝑦 𝑡 =

𝑚=−∞

∞

𝑌[𝑚]𝑒𝑗2𝜋𝑚𝑡𝑇

Since we know

𝑌[𝑘] = 𝑋[−𝑘]

If 𝑦 𝑡 = 𝑥 −𝑡 , 𝑌[𝑘] = 𝑋[−𝑘]

Symmetric and Antisymmetric Parts

• If 𝑓 𝑡 = 𝑓𝑆 𝑡 + 𝑓𝐴(𝑡) is a real-valued signal and periodic in time with fundamental period T, what are the Fourier coefficients of 𝑓𝑆 ∙ and 𝑓𝐴 ∙ , in terms of F[k]?

6.003 Fall 2020

𝑓 𝑡 = 𝑓𝑆 𝑡 + 𝑓𝐴(𝑡)

Any arbitrary signal 𝑓 𝑡 can be written into two parts:

𝑓𝑆 𝑡 is symmetric about 𝑡 = 0, 𝑖𝑓 𝑓𝑆 𝑡 = 𝑓𝑆 −𝑡 for all t.

𝑓𝐴 𝑡 is antisymmetric about 𝑡 = 0, 𝑖𝑓 𝑓𝐴 𝑡 = −𝑓𝑆 −𝑡 for all t.

𝑓𝑆 𝑡 =𝑓 𝑡 + 𝑓(−𝑡)

2𝑓𝐴 𝑡 =

𝑓 𝑡 − 𝑓(−𝑡)

2

Real-valued periodic signal

6.003 Fall 2020

𝐹[𝑘] =1

𝑇න𝑇

𝑓(𝑡)𝑒−𝑗2𝜋𝑘𝑡𝑇 𝑑𝑡

If 𝑓 𝑡 is real valued periodic signal:

𝐹[−𝑘] =1

𝑇න𝑇

𝑓(𝑡)𝑒𝑗2𝜋𝑘𝑡𝑇 𝑑𝑡

𝐹∗[−𝑘] =1

𝑇න𝑇

𝑓(𝑡)𝑒−𝑗2𝜋𝑘𝑡𝑇 𝑑𝑡

If 𝑓 𝑡 is real valued periodic signal, 𝐹 𝑘 = 𝐹∗[−𝑘]

= 𝐹[𝑘]

Symmetric and Antisymmetric Parts

If 𝑓 𝑡 is real valued periodic signal, 𝐹 𝑘 = 𝐹∗[−𝑘]

𝑓𝑆 𝑡 =𝑓 𝑡 + 𝑓(−𝑡)

2

𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑡𝑦𝐹𝑆 𝑘 =

𝐹 𝑘 + 𝐹 −𝑘

2=𝐹 𝑘 + 𝐹∗ 𝑘

2=2𝑅𝑒(𝐹 𝑘 )

2= 𝑅𝑒(𝐹 𝑘 )

𝑓𝐴 𝑡 =𝑓 𝑡 − 𝑓(−𝑡)

2

𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑡𝑦𝐹𝐴 𝑘 =

𝐹 𝑘 − 𝐹 −𝑘

2=𝐹 𝑘 − 𝐹∗ 𝑘

2=2𝑗 ∙ 𝐼𝑚(𝐹 𝑘 )

2= 𝑗 ∙ 𝐼𝑚(𝐹 𝑘 )

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 𝑓 𝑡 = 𝑓𝑆 𝑡 + 𝑓𝐴(𝑡) is a real valued signal and periodic in time with fundamental period T, what are the Fourier coefficients of 𝑓𝑆 ∙ and 𝑓𝐴 ∙ , in terms of F[k]?

Previously(lecture 2B)

Let X[k] represent the Fourier series coefficients of the following signal:

T=1, 𝜔0 =2𝜋

𝑇= 2𝜋

𝑋[𝑘] = ቐ

0, 𝑘 𝑖𝑠 𝑒𝑣𝑒𝑛1

𝑗𝜋𝑘, 𝑘 𝑖𝑠 𝑜𝑑𝑑

𝑥 𝑡 is antisymmetric around t=0, thus its Fourier Series coefficients are purely imaginary

Symmetric and Antisymmetric Parts in Trig form CTFS

𝑓𝑆 𝑡 =𝑓 𝑡 + 𝑓(−𝑡)

2𝑓𝐴 𝑡 =

𝑓 𝑡 − 𝑓(−𝑡)

2

The symmetric part shows up in the 𝑐𝑘 coefficients, and the antisymmetric part shows up in the 𝑑𝑘 coefficients.

• 𝑐𝑘’s (cosines) alone can only represent symmetric functions.

• 𝑑𝑘 ’s (sines) alone can only represent antisymmetric functions

sin(𝑥) cos(𝑥)

Previously(Lecture 2A)• Find the Fourier series coefficients for the following square wave:

If without 𝑐0 =1

2“DC” part, 𝑓 𝑡 is antisymmetric around t=0, thus only having non-zero 𝑑𝑘 ’s

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 𝑐𝑘 terms if k is odd. It has no effect if k is even.

𝑔 𝑡 =

𝑔 𝑡 = σ𝑘=0∞ 𝑐𝑘

′ cos(𝑘𝜔0𝑡) + σ𝑘=1∞ 𝑑𝑘

′ sin(𝑘𝜔0𝑡)

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 𝑑𝑘 terms if and only if k is odd.

𝑔 𝑡 =

𝑔 𝑡 = σ𝑘=0∞ 𝑐𝑘

′ cos(𝑘𝜔0𝑡) + σ𝑘=1∞ 𝑑𝑘

′ sin(𝑘𝜔0𝑡)

Quarter period shift?

Shifting by T /4 becomes more complicated.

ℎ 𝑡 =

ℎ 𝑡 = σ𝑘=0∞ 𝑐𝑘

′′cos(𝑘𝜔0𝑡) + σ𝑘=1∞ 𝑑𝑘

′′sin(𝑘𝜔0𝑡)

Quarter period shift?

6.003 Fall 2020

ℎ 𝑡 =

ℎ 𝑡 = σ𝑘=0∞ 𝑐𝑘

′′cos(𝑘𝜔0𝑡) + σ𝑘=1∞ 𝑑𝑘

′′sin(𝑘𝜔0𝑡)

Summary of shift result

6.003 Fall 2020

Other shifts will be even more complicated!

4ℎ 𝑡

Properties of CTFS: Time Shift• Consider 𝑦 𝑡 = 𝑥 𝑡 − 𝑡0 , where 𝑥 is periodic in 𝑇. What are the

CTFS coefficients 𝑌[𝑘], in terms of 𝑋[𝑘]?

𝑙𝑒𝑡 𝑢 = 𝑡 − 𝑡0, then 𝑡 = 𝑢 + 𝑡0,d𝑡 = 𝑑𝑢

Each coefficient 𝑌 𝑘 in the series for 𝑦(𝑡) is a constant 𝑒−𝑗𝑘𝜔0𝜏 times the corresponding coefficient 𝑋[𝑘] in the series for 𝑥(𝑡).

Properties of CTFS: Time Derivative

• Consider 𝑦 𝑡 =𝑑

𝑑𝑡𝑥 𝑡 , where 𝑥 𝑡 and 𝑦 𝑡 are periodic in 𝑇. What

are the CTFS coefficients 𝑌[𝑘], in terms of 𝑋[𝑘]?

6.003 Fall 2020

Start with the synthesis equation:

Then, from the definition of y(·), we have:

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 𝑓 𝑡 is real valued periodic signal, 𝐹 𝑘 = 𝐹∗[−𝑘]

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 𝑦 𝑡 = 𝑥 −𝑡 , 𝑌[𝑘] = 𝑋[−𝑘]

Check yourself!

Let 𝑌[𝑘] represent the Fourier series coefficients of the following signal:

6.003 Fall 2020

Check yourself!What is the relationship between the two following signals?

6.003 Fall 2020

𝑋[𝑘] = ቐ

0, 𝑘 𝑖𝑠 𝑒𝑣𝑒𝑛1

𝑗𝜋𝑘, 𝑘 𝑖𝑠 𝑜𝑑𝑑

Check yourself!The triangle waveform is the integral of the square wave.

𝑥 𝑡 =𝑑

𝑑𝑡𝑦 𝑡

𝑋 𝑘 = 𝑗2𝜋𝑘

𝑇𝑌 𝑘 , 𝑇 = 1

𝑋[𝑘] = ቐ

0, 𝑘 𝑖𝑠 𝑒𝑣𝑒𝑛1

𝑗𝜋𝑘, 𝑘 𝑖𝑠 𝑜𝑑𝑑

𝑌 𝑘 =1

𝑗2𝜋𝑘𝑋 𝑘

𝑦 𝑡 is symmetric around t=0, thus its Fourier Series coefficients are purely real

= ቐ0, 𝑘 𝑖𝑠 𝑒𝑣𝑒𝑛

−1

2𝜋2𝑘2, 𝑘 𝑖𝑠 𝑜𝑑𝑑

Check yourself!

Let 𝑌[𝑘] represent the Fourier series coefficients of the following signal:

6.003 Fall 2020

𝑌[𝑘] = ቐ0, 𝑘 𝑖𝑠 𝑒𝑣𝑒𝑛

−1

2𝜋2𝑘2, 𝑘 𝑖𝑠 𝑜𝑑𝑑

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