🎧 New: AI-Generated Podcasts Turn your study notes into engaging audio conversations. Learn more

Continuous Time Fourier Series 02 Class Notes PDF

Loading...
Loading...
Loading...
Loading...
Loading...
Loading...
Loading...

Document Details

AwedPhiladelphia

Uploaded by AwedPhiladelphia

Sujal Patel Sir

Tags

continuous time fourier series signals and systems electrical engineering electronics and communication engineering

Summary

These notes provide a detailed explanation of Continuous Time Fourier Series. They cover topics such as polar form, exponential Fourier series, and calculations for finding coefficients and power spectrum. The content is suitable for an undergraduate-level electrical engineering course.

Full Transcript

# Electrical Engineering ## Electronics and Communication Engineering ### SIGNALS & SYSTEMS #### Lecture No. 2 ##### Continuous Time Fourier Series By- Sujal Patel Sir ### Topics to be Covered * Concepts of CTFS * TFS & EFS ### Polar Form or Cosine of Fourier Series $x(t) = a_0 + \sum_{n=1}^\in...

# Electrical Engineering ## Electronics and Communication Engineering ### SIGNALS & SYSTEMS #### Lecture No. 2 ##### Continuous Time Fourier Series By- Sujal Patel Sir ### Topics to be Covered * Concepts of CTFS * TFS & EFS ### Polar Form or Cosine of Fourier Series $x(t) = a_0 + \sum_{n=1}^\infty a_n cos(n \omega_0 t) + b_n sin(n \omega_0 t)$ # Polar form: $x(t) = d_0 + \sum_{n=1}^\infty d_n Cos(n \omega_0 t + \theta_n)$ $\implies x(t) = d_0 + \sum_{n=1}^\infty d_n cos\theta_n cos(n \omega_0 t) - d_n sin\theta_n sin(n \omega_0 t)$ $\implies a_n = d_n cos\theta_n \implies (a_n)^2 + (b_n)^2 = d_n^2cos^2\theta_n + d_n^2sin^2\theta_n = d_n^2$ $\implies -b_n = d_n sin\theta_n \implies d_n^2 = a_n^2 + (-b_n)^2 \implies d_n sin\theta_n/ d_n cos\theta_n = - b_n / a_n$ $\implies |d_n| = \sqrt{(a_n)^2 + (b_n)^2} \implies \theta_n = tan^{-1}(b_n/a_n)$ $\implies d_n = a_n - jb_n$ $\implies \theta_n = tan^{-1}(-b_n/a_n) = -tan^{-1}(b_n/a_n)$ $\implies d_n^* = a_n + jb_n$ # Magnitude Spectrum $x(t) = d_0 + \sum_{n=1}^\infty d_n cos(n \omega_0 t + \theta_n)$ $\theta_n = \angle d_n$ # Phase Spectrum ### Exponential Fourier Series $x(t) = a_0 + \sum_{n=1}^\infty a_n cos(n \omega_0 t) + b_n sin(n \omega_0 t)$ $\implies x(t) = a_0 + \sum_{n=1}^\infty a_n [\frac{e^{jn \omega_0 t} + e^{-jn \omega_0 t}}{2}] + b_n [\frac{e^{jn \omega_0 t} e^{-jn \omega_0 t}}{2j}]$ $\implies x(t) = a_0 + \sum_{n=1}^\infty (\frac{a_n - jb_n}{2}) e^{jn \omega_0 t} + (\frac{a_n + jb_n}{2})e^{-jn \omega_0 t}$ Let, $C_n = \frac{a_n - jb_n}{2}$ & $C_{-n} = \frac{a_n + jb_n}{2} \implies C_{-n} = C_n^* \implies |C_n| = |C_{-n}|$ $\implies x(t) = C_0 + \sum_{n=1}^\infty C_n e^{jn \omega_0 t} + \sum_{n = - \infty}^{-1} C_n e^{-jn \omega_0 t}$ $\implies x(t) = C_0 + \sum_{n=1}^\infty C_n e^{jn \omega_0 t} + \sum_{n= - \infty}^{-1} C_n e^{jn \omega_0 t}$ $\implies x(t) = \sum_{n=- \infty}^\infty C_n e^{jn \omega_0 t} \longrightarrow Exponential Fourier Series of x(t)$ Where, $C_n$ is EFS Coefficient. # $C_n = \frac{a_n - jb_n}{2}$ $\implies C_n = \frac{1}{2} [\frac{1}{T} \int_{T} x(t). Cosn\omega_0 t dt - \frac{j}{2} \frac{1}{T} \int_{T} x(t). Sinn\omega_0 t dt ]$ $\implies C_n = \frac{1}{T} \int_{T} x(t). (Cosn\omega_0 t - jSinn\omega_0 t). dt $ $\implies C_n = \frac{1}{T} \int_{T} x(t). e^{-jn \omega_0 t} dt$ $\implies d_n = a_n - jb_n \longrightarrow |C_n| = \sqrt{\frac{a_n^2 + b_n^2}{2}} = \frac{|d_n|}{2}$ $\implies \theta_n = tan^{-1} (-b_n/ a_n)$ $\implies C_n = \frac{a_n - jb_n}{2} = \frac{d_n}{2}$ $\implies C_{-n} = \frac{a_n + jb_n}{2} = \frac{d_n^*}{2}$ $\implies |C_{-n}| = \sqrt{\frac{a_n^2 + b_n^2}{2}} = \frac{|d_n^*|}{2} = \frac{|d_n|}{2}$. # $|C_n| = |C_{-n}| \longrightarrow Even$ $\implies \angle C_n = \theta_n = tan^{-1} (-b_n/a_n) = -tan^{-1}(b_n/a_n)$ $\implies a_n = C_{-n} + C_n$ $\implies \angle C_{-n} = \theta_{-n} = tan^{-1}(b_n / a_n)$ # $\theta_n = -\theta_{-n} \longrightarrow Odd$ $\implies jb_n = C_{-n} - C_n$ # Magnitude Spectrum # Phase Spectrum # $x(t) = a_0 + \sum_{n=1}^\infty a_n Cosn\omega_0 t + b_n Sinn\omega_0 t$ $\implies P_{total} = \frac{1}{T} \int_{T} |x(t)|^2 dt = a_0^2 + \sum_{n=1}^\infty \frac{a_n^2 + b_n^2}{2}$ # $x(t) = d_0 + \sum_{n=1}^\infty d_n Cos(n\omega_0 t + \theta_n)$ $d_n = a_n - jb_n$ $P_{total} = \frac{1}{T} \int_{T} |x(t)|^2 dt = d_0^2 + \sum_{n=1}^\infty \frac{|d_n|^2}{2}$ $|d_n|^2 = a_n^2 + b_n^2$ # $x(t) = \sum_{n= - \infty}^\infty (C_n e^{jn \omega_0 t})$ $P_{x(t)} = \frac{1}{T} \int_{T} |x(t)|^2 dt = \sum_{n= - \infty}^\infty (\frac{1}{T} \int_{T} |C_n|^2 e^{jn\omega_0 t} dt)$ $P_{x(t)} = \sum_{n = -\infty }^\infty \frac{1}{T} \int_{T} |C_n|^2 dt = \sum_{n = - \infty}^\infty |C_n|^2$ $P_{x(t)} = \frac{1}{T} \int_{T} |x(t)|^2 dt = \sum_{n= - \infty}^\infty |C_n|^2 = |C_0|^2 + \sum_{n= - \infty}^{-1} |C_n|^2 + \sum_{n=1}^\infty |C_n|^2$ $P_{x(t)} = |C_0|^2 + 2 \sum_{n=1}^\infty |C_n|^2$ # $x(t) \longleftarrow FS \longrightarrow C_n, a_0, a_n, b_n, d_n$ $P_{total} = \frac{1}{T} \int_{T} |x(t)|^2 dt = a_0^2 + \sum_{n=1}^\infty \frac{a_n^2 + b_n^2}{2} = d_0^2 + \sum_{n=1}^\infty \frac{|d_n|^2}{2}$ $ = \sum_{n = - \infty}^\infty |C_n|^2 = |C_0|^2 + 2 \sum_{n=1}^\infty |C_n|^2$ # Power up to 5th harmonic $P_{upto 5th} = a_0^2 + \sum_{n=1}^{5}\frac{a_n^2 + b_n^2}{2} = d_0^2 + \sum_{n=1}^ 5\frac{|d_n|^2}{2}$ $= \sum_{n = -5}^5 |C_n|^2 = |C_0|^2 + 2 \sum_{n=1}^5 |C_n|^2$ # Power Spectrum

Use Quizgecko on...
Browser
Browser