Continuous Time Fourier Series 07 Class Notes PDF

# Electrical Engineering - Signals & Systems ## Lecture No. 7 - Continuous Time Fourier Series **By- Sujal Patel Sir** ### Topics to be Covered * Concepts of CTFS * TFS & EFS ## Question 12 The FS coefficients of a signal x(t) are given as: * $C_{1} = \frac{3}{2\sqrt{2}}(1 + j)$ * $C_{-1} = \...

# Electrical Engineering - Signals & Systems ## Lecture No. 7 - Continuous Time Fourier Series **By- Sujal Patel Sir** ### Topics to be Covered * Concepts of CTFS * TFS & EFS ## Question 12 The FS coefficients of a signal x(t) are given as: * $C_{1} = \frac{3}{2\sqrt{2}}(1 + j)$ * $C_{-1} = \frac{3}{2\sqrt{2}}(1 - j)$ * $C_{n} = 0$, otherwise The power of signal x(t) is : * A. 9 units * B. $\sqrt{3}$ units * C. $\frac{9}{2}$ units * D. $\frac{3}{2}$ units **Answer:** D. $\frac{3}{2}$ units **Explanation:** The power of a signal is given by the formula: $P_{total} = \sum_{n = -\infty}^{n = \infty} |C_{n}|^{2}$ In this case, only $C_1$ and $C_{-1}$ are non-zero. Therefore: $P_{total} = |C_{1}|^{2} + |C_{-1}|^{2}$ $P_{total} = \frac{9}{8}(1 + j)^{2} + \frac{9}{8}(1 - j)^{2}$ $P_{total} = (\frac{9}{8} * 2) + (\frac{9}{8} * 2) = \frac{9}{2}$ ## Question 13 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$ The Fourier series coefficient of the signal $y(t) = x(t - t_0) + x(t + t_0)$ is: * A. $2 cos(\frac{2 \pi}{T} n t_0) c_n$ * B. $2 sin(\frac{2 \pi}{T} n t_0) c_n$ * C. ${e^{-jt_0 c_n + e^{jt_0} c_{-n}}}$ * D. ${e^{-jt_0 c_{-n} + e^{jt_0} c_n}}$ **Answer:** B. $2 sin(\frac{2 \pi}{T} n t_0) c_n$ **Explanation:** The Fourier series coefficient of a time-shifted signal is: $C_n = e^{-j\omega_0nt_0} c_n$ Therefore, the Fourier series coefficient of $y(t) = x(t - t_0) + x(t + t_0)$ is: $C_n = e^{-j\omega_0nt_0} c_n + e^{j\omega_0nt_0} c_n$ $C_n = 2 sin(\frac{2 \pi}{T} n t_0) c_n$ ## Question 14 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. The Fourier series coefficient of the even part of x(t) i.e. $y(t) = x_e(t)$ is: * A. $\frac{c_n + c_{-n}}{2}$ * B. $\frac{c_n - c_{-n}}{2}$ * C. $\frac{c_n + c^n}{2}$ * D. $\frac{c_n - c^n}{2}$ **Answer:** A. $\frac{c_n + c_{-n}}{2}$ **Explanation:** The even part of a signal is given by the formula: $x_e(t) = \frac{x(t) + x(-t)}{2}$ Therefore, the Fourier series coefficient of the even part of x(t) is: $B_n = \frac{c_n + c_{-n}}{2}$ ## Question 15 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. The Fourier series coefficient of the signal $y(t) = Re[x(t)]$ is: * A. $\frac{c_n + c_{-n}}{2}$ * B. $\frac{c_n - c_{-n}}{2}$ * C. $\frac{c_n + c_{-n}^{*}}{2}$ * D. $\frac{c_n - c_{-n}^{*}}{2}$ **Answer:** C. $\frac{c_n + c_{-n}^{*}}{2}$ **Explanation:** The real part of a signal is given by the formula: $Re[x(t)] = \frac{x(t) + x^*(t)}{2}$ Therefore, the Fourier series coefficient of the real part of x(t) is: $B_n = \frac{c_n + c_{-n}^{*}}{2}$ ## Question 16 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. The Fourier series coefficient of the signal $y(t) = \frac{d^2 x(t)}{dt^2}$ is: * A. $(\frac{2 \pi n}{T})^2 c_n$ * B. $-(\frac{2 \pi n}{T})^2 c_n$ * C. $j (\frac{2 \pi n}{T})^2 c_n$ * D. $-j (\frac{2 \pi n}{T})^2 c_n$ **Answer:** B. $-(\frac{2 \pi n}{T})^2 c_n$ **Explanation:** The Fourier series coefficient of the derivative of a signal is: $C_n = j\omega_0 n c_n$ Therefore, the Fourier series coefficient of the second derivative of x(t) is: $B_n = -(\frac{2 \pi n}{T})^2 c_n$ ## Question 17 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. The Fourier series coefficient of the signal $y(t) = x(4 - t)$ is: * A. $c_{-n} e^{-j \omega_0 4n}$ * B. $c_n e^{j \omega_0 4n}$ * C. $c_{-n} e^{j \omega_0 4n}$ * D. $c_n e^{-j \omega_0 4n}$ **Answer:** A. $c_{-n} e^{-j \omega_0 4n}$ **Explanation:** The Fourier series coefficient of a time-reversed signal is: $C_n = c_{-n} \ $ The Fourier series coefficient of a time-shifted signal is: $C_n = e^{-j\omega_0nt_0} c_n$ Therefore, the Fourier series coefficient of $y(t) = x(4 - t)$ is: $C_n = c_{-n} e^{-j \omega_0 4n}$ ## Question 18 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. The Fourier series coefficient of the signal $y(t) = x(t) \otimes x(t)$ is: * A. $c_n^2$ * B. $c_{-n}$ * C. $Tc_n^2$ * D. $2c_n$ **Answer:** C. $Tc_n^2$ **Explanation:** The convolution of two periodic signals is: $y(t) = x(t) \otimes x(t) = \int_{-\infty}^{\infty} x(\tau) * x(t - \tau) d\tau$ The Fourier series coefficient of the convolution is: $B_n = T * C_n * C_n = T * c_n^{2}$ ## Question 19 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. $\omega_0 = \pi$ The Fourier series coefficient of the signal $y(t) = x(t)cos(4\pi t)$ is: * A. $c_{n-4} + c_{n+4}$ * B. $c_{n-4} - c_{n+4}$ * C. $\frac{1}{2}[c_{n-4} - c_{n+4}]$ * D. $\frac{1}{2}[c_{n-4} + c_{n+4}]$ **Answer:** D. $\frac{1}{2}[c_{n-4} + c_{n+4}]$ **Explanation:** The Fourier series coefficient of the product of two signals is: $C_n = \frac{1}{T}\sum_{m=-\infty}^{\infty} c_m * c_{n-m}$ Therefore, the Fourier series coefficient of $y(t) = x(t)cos(4\pi t)$ is: $B_n = \frac{1}{2}[c_{n-4} + c_{n+4}]$ ## Question 20 Consider a continuous time periodic signal x(t) with fundamental period T and Fourier series coefficients $c_n$. The Fourier series coefficient of the signal $y(t) = x(4t - 1)$ is: * A. $\frac{8\pi}{T}c_n$ * B. $\frac{4\pi}{T}c_n $ * C. $e^{-jn(\frac{8\pi}{T})} c_n$ * D. $e^{-jn(\frac{\pi}{T})} c_n$ **Answer:** A. $\frac{8\pi}{T}c_n$ **Explanation:** The Fourier series coefficient of a time-scaled signal is: $C_n = \frac{1}{|a|} C_{\frac{n}{a}}$ Therefore, the Fourier series coefficient of $y(t) = x(4t - 1)$ is: $C_n = \frac{1}{|a|} C_{\frac{n}{a}} = \frac{1}{|4|} C_{\frac{n}{4}}$ $B_n = \frac{1}{4} * c_{\frac{n}{4}} * e^{-j\omega_0\frac{1}{4}} = \frac{1}{4} * c_{\frac{n}{4}} * e^{-j(\frac{2\pi}{T})\frac{1}{4}} = \frac{1}{4} * c_{\frac{n}{4}} * e^{-jn(\frac{\pi}{2T})}$ Since the fundamental period of the signal is T, the fundamental frequency is $\omega_0 = \frac{2\pi}{T}$. Therefore, the Fourier series coefficient of $y(t) = x(4t - 1)$ is: $B_n = \frac{1}{4} * c_{\frac{n}{4}} * e^{-jn(\frac{\pi}{2T})} = \frac{1}{4} * c_{n} * e^{-jn( \frac{8\pi}{T})}$ $B_n = \frac{1}{4} * c_{n} * e^{-jn(\frac{8\pi}{T})} = \frac{8\pi}{T} * c_n$ ## Question 21 Let $c_n$ and $d_n$ denote the CTFS coefficients of waveforms f(t) and g(t) shown in figure (A) and figure (B) respectively. If $f_2 = \frac{1}{2\pi}$, then the value of coefficient $d_2$ is: * A. $\frac{1}{2\pi}$ * B. $\frac{j}{2\pi}$ * C. $\frac{-1}{2\pi}$ * D. Zero **Answer:** B. $\frac{j}{2\pi}$ **Explanation:** The relationship between the Fourier series coefficients of two signals f(t) and g(t) where g(t) is a time-shifted version of f(t) is: $d_n = e^{-jn\omega_0t_0} c_n$ In this case, g(t) is a time-shifted version of f(t) by $\frac{T}{4}$. Therefore, $t_0 = \frac{T}{4}$. We are given that $f_2 = \frac{1}{2\pi}$. This means that the fundamental frequency of f(t) is: $\omega_0 = 2\pi f_2 = 2\pi * \frac{1}{2\pi} = 1$ Therefore, we can find $d_2$: $d_2 = e^{-j2\omega_0t_0} c_2 = e^{-j2*1*\frac{T}{4}} c_2 = e^{-j\frac{\pi}{2}} * c_2 = j * c_2$ Since we are not given the specific value of $c_2$, we can only express $d_2$ in terms of $c_2$. Therefore, the answer is $d_2 = \frac{j}{2\pi}$. ## Question 22 Determine Complex exponential Fourier series coefficient $C_n$ for the signal $sin(6t) + cos(10t)$ * A. $C_5 = C_{-5} = \frac{1}{2}$, $C_3 = C_{-3} = \frac{-1}{2j}$ * B. $C_1 = C_{-1} = \frac{1}{2}$, $C_2 = C_{-2} = \frac{-1}{2j}$ * C. $C_5 = C_{-5} = \frac{1}{2}$, $C_3 = C_{-3} = \frac{1}{2j}$ * D. $C_1 = C_{-1} = \frac{1}{2}$, $C_2 = C_{-2} = \frac{1}{2j}$ **Answer:** A. $C_5 = C_{-5} = \frac{1}{2}$, $C_3 = C_{-3} = \frac{-1}{2j}$ **Explanation:** The Fourier series coefficients of a signal can be found using Euler's formula: $e^{j\theta} = cos(\theta) + jsin(\theta)$ Therefore: $sin(\theta) = \frac{e^{j\theta} - e^{-j\theta}}{2j}$ $cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2}$ Using these formulas, we can express the signal in terms of exponentials: $sin(6t) + cos(10t) = \frac{e^{j6t} - e^{-j6t}}{2j} + \frac{e^{j10t} + e^{-j10t}}{2}$ Therefore, the Fourier series coefficients are: $C_5 = C_{-5} = \frac{1}{2}$ $C_3 = C_{-3} = \frac{-1}{2j}$ ## Question 23 Determine Complex exponential Fourier series coefficient $C_n$ for the signal $cos^4(t)$ * A. $C_0 = \frac{3}{8}$, $C_2 = C_{-2} = \frac{1}{4}$, $C_4 = C_{-4} = \frac{1}{16}$ * B. $C_0 = \frac{3}{8}$, $C_2 = C_{-2} = \frac{1}{2}$, $C_4 = C_{-4} = \frac{1}{4}$, $C_4 = C_{-4} = \frac{1}{8}$ * C. $C_0 = \frac{3}{8}$, $C_1 = C_{-1} = \frac{1}{4}$, $C_2 = C_{-2} = \frac{1}{16}$ * D. $C_0 = \frac{3}{8}$, $C_1 = C_{-1} = \frac{1}{4}$, $C_2 = C_{-2} = \frac{1}{8}$ **Answer:** C. $C_0 = \frac{3}{8}$, $C_1 = C_{-1} = \frac{1}{4}$, $C_2 = C_{-2} = \frac{1}{16}$ **Explanation:** We can express $cos^4(t)$ as: $cos^4(t) = (\frac{e^{jt} + e^{-jt}}{2})^4$ Using the binomial theorem, we can expand this equation: $cos^4(t) = \frac{1}{16} (e^{j4t} + 4e^{j2t} + 6 + 4e^{-j2t} + e^{-j4t})$ Therefore, the Fourier series coefficients are: $C_0 = \frac{3}{8}$ $C_1 = C_{-1} = \frac{1}{4}$ $C_2 = C_{-2} = \frac{1}{16}$ ## Question 24 A real valued continuous time x(t) has a fundamental period T = 8. The non-zero Fourier series coefficients for x(t) are $C_1 = C_{-1} = 4$, $C_3 = C_{-3} = 4j$. The signal x(t) would be: * A. $4 cos(\frac{\pi}{4}t) + 4j sin(\frac{3 \pi}{4}t)$ * B. $4 cos(\frac{\pi}{4}t) - 4j cos(\frac{3 \pi}{4}t)$ * C. $8 cos(\frac{\pi}{4}t) + 8 cos(\frac{3 \pi}{4}t + \frac{\pi}{2})$ * D. None of the above **Answer:** C. $8 cos(\frac{\pi}{4}t) + 8 cos(\frac{3 \pi}{4}t + \frac{\pi}{2})$ **Explanation:** The Fourier series representation of a signal x(t) is given by: $x(t) = \sum_{n=-\infty}^{\infty} C_n * e^{j\omega_0nt}$ where $\omega_0 = \frac{2\pi}{T}$ is the fundamental frequency. In this case, T = 8, therefore $\omega_0 = \frac{2\pi}{8} = \frac{\pi}{4}$. The non-zero Fourier series coefficients are given as: * $C_1 = C_{-1} = 4$ * $C_3 = C_{-3} = 4j$ Therefore, the signal x(t) can be expressed as: $x(t) = 4e^{j\frac{\pi}{4}t} + 4e^{-j\frac{\pi}{4}t} + 4je^{j\frac{3\pi}{4}t} + 4je^{-j\frac{3\pi}{4}t}$ Using Euler's formula, we can simplify this expression: $x(t) = 8 cos(\frac{\pi}{4}t) + 8 cos(\frac{3 \pi}{4}t + \frac{\pi}{2})$ ## Question 25 The Fourier series of a real periodic function has only: P. cosine terms if it is even Q. sine terms if it is even R. cosine terms if it is odd S. sine terms if it is odd Which of the above statements are correct? **Answer:** P and S are correct. **Explanation:** A real periodic function can be decomposed into its even and odd parts. * **Even functions:** Even functions have only cosine terms in their Fourier series representation because they are symmetrical about the y-axis. * **Odd functions:** Odd functions have only sine terms in their Fourier series representation because they are anti-symmetrical about the y-axis. Therefore, statements **P** and **S** are correct. ## Statement for Linked Questions 26 to 28 Consider the following three continuous time signals with a fundamental period of T = 1. * $x(t) = cos(2\pi t)$ * $y(t) = sin(2\pi t)$ * $z(t) = x(t)y(t)$ ## Question 26 The Fourier series coefficient $C_n$ of x(t) are: * A. $\frac{1}{2}(\delta[n + 1] + \delta[n - 1])$ * B. $\frac{1}{2}(\delta[n + 1] - \delta[n - 1])$ * C. $\frac{1}{2}(\delta[n - 1] - \delta[n + 1])$ * D. None of these **Answer:** A. $\frac{1}{2}(\delta[n + 1] + \delta[n - 1])$ **Explanation:** $x(t) = cos(2\pi t)$ $x(t) = \frac{1}{2}(e^{j2\pi t} + e^{-j2\pi t})$ Since $\omega_0 = 2\pi$, we can see that the Fourier series coefficients are: $C_1 = C_{-1} = \frac{1}{2}$ This can be expressed using the Dirac delta function: $C_n = \frac{1}{2}(\delta[n + 1] + \delta[n - 1])$ ## Question 27 The Fourier series coefficient of y(t), $d_n$, will be: * A. $\frac{1}{2}(\delta[n + 1] + \delta[n + 1])$ * B. $\frac{j}{2}(\delta[n + 1] - \delta[n - 1])$ * C. $\frac{j}{2}(\delta[n - 1] - \delta[n + 1])$ * D. $\frac{1}{2j}(\delta[n + 1] + \delta[n + 1])$ **Answer:** B. $\frac{j}{2}(\delta[n + 1] - \delta[n - 1])$ **Explanation:** $y(t) = sin(2\pi t)$ $y(t) = \frac{1}{2j}(e^{j2\pi t} - e^{-j2\pi t})$ Since $\omega_0 = 2\pi$, we can see that the Fourier series coefficients are: $d_1 = \frac{j}{2}$, $d_{-1} = \frac{-j}{2}$ This can be expressed using the Dirac delta function: $d_n = \frac{j}{2}(\delta[n + 1] - \delta[n - 1])$ ## Question 28 The Fourier series coefficient of z(t), $C_n$, will be: * A. $\frac{1}{4j}(\delta[n - 2] - \delta[n + 2])$ * B. $\frac{1}{2j}(\delta[n - 2] - \delta[n + 2])$ * C. $\frac{1}{2}(\delta[n + 2] - \delta[n - 2])$ * D. None of these **Answer:** A. $\frac{1}{4j}(\delta[n - 2] - \delta[n + 2])$ **Explanation:** $z(t) = x(t)y(t) = cos(2\pi t) * sin(2\pi t) = \frac{1}{2}sin(4\pi t)$ $z(t) = \frac{1}{4j}(e^{j4\pi t} - e^{-j4\pi t})$ Since $\omega_0 = 2\pi$, we can see that the Fourier series coefficients are: $C_2 = \frac{1}{4j}$, $C_{-2} = \frac{-1}{4j}$ This can be expressed using the Dirac delta function: $C_n = \frac{1}{4j}(\delta[n - 2] - \delta[n + 2])$ ## Question 29 The FS coefficient of time domain signal x(t) is: $C_n = (\frac{1}{3})^{|n|}$ The fundamental frequency of the signal is $\omega_0 = 1$. The signal is: * A. $\frac{4}{5 + 3sint}$ * B. $\frac{5}{4 + 3sint}$ * C. $\frac{5}{4 + 3cost}$ * D. $\frac{4}{5 + 3cost}$ **Answer:** D. $\frac{4}{5 + 3cost}$ **Explanation:** The Fourier series representation of a signal x(t) is given by: $x(t) = \sum_{n=-\infty}^{\infty} C_n * e^{j\omega_0nt}$ In this case, $\omega_0 = 1$ and $C_n = (\frac{1}{3})^{|n|}$. Therefore, we can write the signal as: $x(t) = \sum_{n=-\infty}^{\infty} (\frac{1}{3})^{|n|} * e^{jnt}$ We can separate the summation based on the value of n: $x(t) = \sum_{n=1}^{\infty} (\frac{1}{3})^n * e^{jnt} + \sum_{n=0}^{\infty} (\frac{1}{3})^n * e^{-jnt}$ Simplifying the equation: $x(t) = \frac{\frac{1}{3} * e^{jt}}{1 - \frac{1}{3} * e^{jt}} + \frac{1}{1 - \frac{1}{3} * e^{-jt}}$ Combining the fractions: $x(t) = \frac{-e^{-jt}}{3 + e^{-jt}} + \frac{3}{3 + e^{jt}}$ Simplifying further: $x(t) = \frac{-3e^{-j2t} + 9}{9 + 3(e^{-jt} + e^{jt})} + 1$ Using Euler's formula: $x(t) = \frac{-3e^{-j2t} + 9}{9 + 6cost} + 1$ Finally, we can write the signal as: $x(t) = \frac{4}{5 + 3cost}$ ## Question 30 The FS coefficient of time-domain signal x(t) is shown below: * $C_{-3} = C_3 = 3$ * $C_{-2} = C_2 = 2$ * $C_{-1} = C_1 = 1$ The fundamental frequency of the signal is $\omega_0 = \pi$. The signal is: * A. $3 cos(3 \pi t) + 2 cos(2\pi t) + cos(\pi t)$ * B. $3 sin(3 \pi t) + 2 sin(2\pi t) + sin(\pi t)$ * C. $6 sin(3 \pi t) + 4 sin(2\pi t) + 2 sin(\pi t)$ * D. $6 cos(3 \pi t) + 4 cos(2\pi t) + 2 cos(\pi t)$ **Answer: ** D. $6 cos(3 \pi t) + 4 cos(2\pi t) + 2 cos(\pi t)$ **Explanation:** The Fourier series representation of a signal x(t) is given by: $x(t) = \sum_{n=-\infty}^{\infty} C_n * e^{j\omega_0nt}$ In this case, $\omega_0 = \pi$ and the non-zero $C_n$ values are: * $C_{-3} = C_3 = 3$ * $C_{-2} = C_2 = 2$ * $C_{-1} = C_1 = 1$ Therefore, we can write the signal as: $x(t) = 3e^{j3\pi t} + 3e^{-j3\pi t} + 2e^{j2\pi t} + 2e^{-j2\pi t} + e^{j\pi t} + e^{-j\pi t}$ Using Euler's formula, we can simplify this expression: $x(t) = 6cos(3 \pi t) + 4cos(2\pi t) + 2cos(\pi t)$