AYDIN ADNAN MENDERES UNIVERSITY FACULTY OF ......EE213 –Transform Techniques With Computer...

AYDIN ADNAN MENDERES UNIVERSITY

FACULTY OF ENGINEERINGDepartment of Electrical and Electronics Engineering

EE213 – TRANSFORM TECHNIQUESWITH COMPUTER APPLICATIONS

2020-2021, Fall(ONLINE)

Week 8

Dr. Adem Ükte

Fourier Series Representation of CT Periodic Signals

EE213 – Transform Techniques With Computer Applications Dr. Adem Ükte

A real periodic signal 𝑥 𝑡 can be defined as a summation of complex exponentials:

Expone

ntialForm

Fourie

rSerie

s Pair

Analysis equation

Synthesis equation

Exponential Fourier Series Coefficients

𝑥 𝑡 =

𝑘=−∞

∞

𝐶𝑘𝑒𝑗𝑘𝜔0𝑡

𝐶𝑘 =1

𝑇න

𝑇

𝑥(𝑡)𝑒−𝑗𝑘𝜔0𝑡𝑑𝑡

The fundamental period of 𝑥 𝑡 is the minimum positive, nonzero value of 𝑇, and the value 𝜔0 = 2𝜋/𝑇 isreferred to as the fundamental frequency.

The coefficient 𝐶0 is the dc or constant component of 𝑥 𝑡 with 𝑘 = 0, and it is simply the average value of 𝑥 𝑡over one period:

𝐶0 =1

𝑇න

𝑇

𝑥(𝑡) 𝑑𝑡

𝐶𝑘 = 𝐶−𝑘∗



Since the FS coefficients 𝐶𝑘’s are complex valued, they have magnitude and phase components.

𝐶𝑘 = 𝐶𝑘 𝑒𝑗𝜃𝑘 𝐶𝑘 = 𝐶−𝑘

∗ 𝐶−𝑘 = 𝐶𝑘 𝑒−𝑗𝜃𝑘

𝐶−𝑘𝑒−𝑗𝑘𝜔0𝑡 + 𝐶𝑘𝑒

𝑗𝑘𝜔0𝑡 = 2 𝐶𝑘 cos 𝑘𝜔0𝑡 + 𝜃𝑘

𝑥 𝑡 =

𝑘=−∞

∞


𝑥 𝑡 = 𝐶0 +

𝑘=1

∞

2 𝐶𝑘 cos 𝑘𝜔0𝑡 + 𝜃𝑘Combined Trigonometric Form ofFourier Series Representation

𝑥 𝑡 = 𝐴0 +

𝑘=1

∞

𝐴𝑘 cos 𝑘𝜔0𝑡 + 𝐵𝑘 sin 𝑘𝜔0𝑡Trigonometric Form of

Fourier Series Representation

Exponential Form ofFourier Series Representation

2𝐶𝑘 = 𝐴𝑘 − 𝑗𝐵𝑘 𝐶0 = 𝐴0



𝑥 𝑡 = 𝐴0 +

𝑘=1

∞

𝐴𝑘 cos 𝑘𝜔0𝑡 + 𝐵𝑘 sin 𝑘𝜔0𝑡Trigonometric Form of

Fourier Series Representation

𝐴0 = 𝐶0 =1

𝑇න

𝑇

𝑥(𝑡) 𝑑𝑡

𝐴𝑘 =2

𝑇න

𝑇

𝑥(𝑡) cos 𝑘𝜔0𝑡 𝑑𝑡

𝐵𝑘 =2

𝑇න

𝑇

𝑥(𝑡) sin 𝑘𝜔0𝑡 𝑑𝑡

Trigonom

etric

Fourie

rSerie

s Coe

fficients

EE213 – Transform Techniques With Computer ApplicationsDr. Adem Ükte

Ex:

Consider the signal;

Find the exponential fourier series coefficients?

𝑥(𝑡) = sin 𝜔0𝑡

Solution:

𝑥(𝑡) = sin 𝜔0𝑡 =1

2𝑗𝑒𝑗𝜔0𝑡 −

1

2𝑗𝑒−𝑗𝜔0𝑡

In our case, we have only 𝐶1and 𝐶−1

𝑥 𝑡 =

𝑘=−∞

∞

𝐶𝑘𝑒𝑗𝑘𝜔0𝑡 = ⋯+ 𝐶−2 𝑒

𝑗(−2)𝜔0𝑡 +𝐶−1 𝑒𝑗(−1)𝜔0𝑡 +𝐶0 𝑒

𝑗(0)𝜔0𝑡 +𝐶1 𝑒𝑗(1)𝜔0𝑡 +⋯

𝑥(𝑡) = sin 𝜔0𝑡 =1

2𝑗𝑒𝑗𝜔0𝑡 −

1

2𝑗𝑒−𝑗𝜔0𝑡

𝑘 = 1 𝑘 = −1

𝐶1 =1

2𝑗𝐶−1 = −

1

2𝑗

𝐶𝑘 = 0, 𝑘 ≠ ±1


Ex:

Consider the signal;


𝑥 𝑡 = 1 + sin 𝜔0𝑡 + 2 cos 𝜔0𝑡 + cos 2𝜔0𝑡 +𝜋

4

Solution:

𝑥 𝑡 = 1 + sin 𝜔0𝑡 + 2 cos 𝜔0𝑡 + cos 2𝜔0𝑡 +𝜋

4

𝑘 = 0

= 1 +1

2𝑗𝑒𝑗𝜔0𝑡 − 𝑒−𝑗𝜔0𝑡 + 𝑒𝑗𝜔0𝑡 + 𝑒−𝑗𝜔0𝑡 +

1

2𝑒𝑗 2𝜔0𝑡+

𝜋4 + 𝑒

−𝑗 2𝜔0𝑡+𝜋4

= 1 + 1 +1

2𝑗𝑒𝑗𝜔0𝑡 + 1 −

1

2𝑗𝑒−𝑗𝜔0𝑡 +

1

2𝑒𝑗

𝜋4𝑒𝑗2𝜔0𝑡 +

1

2𝑒−𝑗

𝜋4𝑒−𝑗2𝜔0𝑡

𝑘 = 1 𝑘 = −1 𝑘 = 2 𝑘 = −2


𝐶𝑘 =

1 = 1𝑒𝑗0° , 𝑘 = 0

1 +1

2𝑗= 1.12𝑒−𝑗26.57° , 𝑘 = 1

1 −1

2𝑗= 1.12𝑒𝑗26.57° , 𝑘 = −1

1

2𝑒𝑗

𝜋4 =

1

2 21 + 𝑗 = 0.5𝑒𝑗45° , 𝑘 = 2

1

2𝑒−𝑗

𝜋4 =

1

2 21 − 𝑗 = 0.5𝑒−𝑗45° , 𝑘 = −2

0 , otherwise


Ex:


𝑥 𝑡 =

𝑘=−∞

∞



✓ These equations play the same role for DT periodic signals as the role of CTFS pair equations in CT case.

✓ As in CT, the DTFS coefficients 𝐶𝑘 are often referred to as the spectral coefficients of 𝑥 𝑛 .

✓ These coefficients specify a decomposition of 𝑥[𝑛] into a sum of N harmonically related complexexponentials.

✓ Unlike CTFS coefficients, DTFS coefficients are periodic in 𝑘 domain, with the period of 𝑁.

𝑥[𝑛] =

𝑘=

𝐶𝑘𝑒𝑗𝑘𝜔0𝑛 =

𝑘=

𝐶𝑘𝑒𝑗𝑘 ൗ2𝜋 𝑁 𝑛

𝐶𝑘 =1

𝑁

𝑛=

𝑥[𝑛]𝑒−𝑗𝑘𝜔0𝑛 =1

𝑁

𝑛=

𝑥[𝑛]𝑒−𝑗𝑘 ൗ2𝜋

𝑁 𝑛Analysis equation

Synthesis equation


Ex:

Consider the signal

This signal is periodic with period N, and we can expand x[n] directly in terms of complex exponentials by using Euler’s formula;

Collecting terms, we find that

𝑪𝟏 𝑪−𝟏 𝑪𝟐 𝑪−𝟐

𝑪𝟎


𝐶𝑘 = 0 for other values of k in the interval of summation (that is oneperiod, N) in the synthesis equation.

Again, the Fourier coefficients are periodic with period N, so, for example,

𝐶𝑁 = 1, 𝐶3𝑁−1 =3

2+

1

2𝑗, and 𝐶2−𝑁 =

1

2𝑗

Thus the Fourier series coefficients for this example are

The real and imaginary parts, and the magnitude and phase of these

coefficients for N = 10, are depicted in below figures:


Practice 10:The exponential Fourier series coefficients of the below periodic square wave signal was calculated in previous

examples.

𝑥 𝑡 =

𝑘=−∞

∞


𝐶𝑘 =

2𝑇1𝑇

, 𝑘 = 0

sin 𝑘𝜔0𝑇1𝑘𝜋

, 𝑘 ≠ 0𝜔0 =

2𝜋

𝑇

In MATLAB, using the parameters 𝑇 = 2 and 𝑇1 = 0.5, generate the original rectangular signal 𝑥 𝑡 and the

Fourier series representation of 𝑥 𝑡 (use 𝑥 𝑡 = σ𝑘=−∞∞ 𝐶𝑘𝑒

𝑗𝑘𝜔0𝑡) for a time vector of t=-3:1/1000:3. Let

the final value of 𝑘 in the summation be 10, 100 and 1000 respectively and then plot 𝑥 𝑡 for each case. Observe

the effect of increasing the upper limit (i.e. increasing the number of harmonics) and comment on it.


Practice 11:The trigonometric Fourier series coefficients of the below periodic triangular wave signal can be calculated as:

𝑥 𝑡 = 𝐴0 +

𝑘=1

∞

𝐴𝑘 cos 𝑘𝜔0𝑡 + 𝐵𝑘 sin 𝑘𝜔0𝑡 𝜔0 =2𝜋

𝑇

In MATLAB, using the parameter 𝑇 = 2, generate the original triangular signal 𝑥 𝑡 and the Fourier series

representation of 𝑥 𝑡 (use𝑥 𝑡 = 𝐴0 + σ𝑘=1∞ 𝐴𝑘 cos 𝑘𝜔0𝑡 + 𝐵𝑘 sin 𝑘𝜔0𝑡 ) for a time vector of

t=-3:1/1000:3. Let the final value of 𝑘 in the summation be 3, 10 and 100 respectively and then plot

𝑥 𝑡 for each case.

𝐴0 =1

𝑇න

𝑇

𝑥(𝑡) 𝑑𝑡 =1

𝑇න

−𝑇2

0

−2

𝑇𝑡𝑑𝑡 + න

0

𝑇22

𝑇𝑡𝑑𝑡 =

1

2

𝐴𝑘 =2

𝑇න

𝑇

𝑥(𝑡) cos 𝑘𝜔0𝑡 𝑑𝑡 =2

𝑇න

−𝑇2

0

−2

𝑇𝑡 cos 𝑘𝜔0𝑡 𝑑𝑡 + න

0

𝑇22

𝑇𝑡 cos 𝑘𝜔0𝑡 𝑑𝑡 = ቐ

−4

𝜋𝑘 2, for 𝑘: odd

0 , for 𝑘: even

𝐵𝑘 =2

𝑇න

𝑇

𝑥(𝑡) sin 𝑘𝜔0𝑡 𝑑𝑡 =2

𝑇න

−𝑇2

0

−2

𝑇𝑡 sin 𝑘𝜔0𝑡 𝑑𝑡 + න

0

𝑇22

𝑇𝑡 sin 𝑘𝜔0𝑡 𝑑𝑡 = 0

END OF WEEK 8


AYDIN ADNAN MENDERES UNIVERSITY FACULTY OF ......EE213 –Transform Techniques With Computer...

Documents

Transcript of AYDIN ADNAN MENDERES UNIVERSITY FACULTY OF ......EE213 –Transform Techniques With Computer...