Fourier SeriesIf $x(t)$ is a periodic function with period $T$, then the continuous-time exponential Fourier series of the function is defined as, $$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t}… (1)}$$Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by, $$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)e^{-jn\omega_{0} t}dt… (2)}$$Time Shifting Property of Fourier SeriesLet $x(t)$ is a periodic function with time period $T$ and with Fourier series coefficient $C_{n}$. Then, if$$\mathrm{x(t)\overset{FS}{\leftrightarrow}C_{n}}$$Then, the time shifting property of continuous-time Fourier series states that$$\mathrm{x(t-t_{0})\overset{FS}{\leftrightarrow}e^{-jn\omega_{0} t_{0}}C_{n}}$$ProofFrom the definition of continuous-time Fourier series, we get, $$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t}…(3)}$$Replacing $t$ by $(t− t_{0})$ in equation (3), we have, $$\mathrm{x(t− t_{0})=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0}(t− t_{0})}}$$$$\mathrm{\Rightarrow\:x(t− t_{0})=\sum_{n=−\infty}^{\infty}(C_{n}e^{-jn\omega_{0}t_{0}})e^{jn\omega_{0}t}… (4)}$$$$\mathrm{∵\:\sum_{n=−\infty}^{\infty}(C_{n}e^{-jn\omega_{0}t_{0}})e^{jn\omega_{0}t}=FS^{-1}[C_{n}e^{-jn\omega_{0}t_{0}}]… (5)}$$From equations (4) & ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$And the inverse Fourier transform is defined as, $$\mathrm{x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)e^{j\omega t}d \omega}$$Time Differentiation Property of Fourier TransformStatement – The time differentiation property of Fourier transform states that the differentiation of a function in time domain is equivalent to the multiplication of its Fourier transform by a factor $j\omega$ in frequency domain. Therefore, if$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$Then, according to the time differentiation property, $$\mathrm{\frac{d}{dt}x(t)\overset{FT}{\leftrightarrow}j\omega\cdot X(\omega)}$$ProofFrom the definition of inverse Fourier transform, we have, $$\mathrm{x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)e^{j\omega t} d\omega}$$Taking time differentiation on both sides, we get, $$\mathrm{\frac{d}{dt}x(t)=\frac{d}{dt}\left [ \frac{1}{2\pi} \int_{−\infty}^{\infty}X(\omega)e^{j\omega t} d\omega\right ]}$$$$\mathrm{\Rightarrow\:\frac{d}{dt}x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)\frac{d}{dt}[e^{j\omega t}]d\omega=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)j\omega ... Read More

Fourier SeriesIf $x(t)$ is a periodic function with period $T$, then the continuous-time exponential Fourier series of the function is defined as, $$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}\:e^{jn\omega_{0} t}… (1)}$$Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by, $$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)e^{-jn\omega_{0} t}dt… (2)}$$Time Differentiation Property of Fourier SeriesIf $x(t)$ is a periodic function with time period T and with Fourier series coefficient $C_{n}$. If$$\mathrm{x(t)\overset{FS}{\leftrightarrow}C_{n}}$$Then, the time differentiation property of continuous-time Fourier series states that$$\mathrm{\frac{dx(t)}{dt}\overset{FS}{\leftrightarrow}jn\omega_{0}C_{n}}$$ProofBy the definition of continuous time Fourier series, we get, $$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t}… (3)}$$By taking time differentiation on both sides of the equation (3), we have, $$\mathrm{\frac{dx(t)}{dt}=\sum_{n=−\infty}^{\infty}C_{n}\frac{d(e^{jn\omega_{0} t})}{dt}}$$$$\mathrm{\Rightarrow\:\frac{dx(t)}{dt}=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t}(jn\omega_{0})}$$$$\mathrm{\Rightarrow\:\frac{dx(t)}{dt}=\sum_{n=−\infty}^{\infty}(jn\omega_{0}C_{n})e^{jn\omega_{0} t}… (4)}$$$$\mathrm{∵\: \sum_{n=−\infty}^{\infty}(jn\omega_{0}C_{n})e^{jn\omega_{0}t}=FS^{-1}[jn\omega_{0}C_{n}]… ... Read More

Fourier SeriesConsider a periodic signal 𝑔(𝑡) be periodic with period T, then the Fourier series of the function 𝑔(𝑡) is defined as, $$\mathrm{g(t)=\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega_{0}t}\:\:\:\:....(1)}$$Where, 𝐶𝑛 is the Fourier series coefficient and is given by, $$\mathrm{C_{n}=\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}g(t)e^{-jn\omega_{0}t}dt\:\:\:\:....(2)}$$Derivation of Fourier Transform from Fourier SeriesLet 𝑥(𝑡) be a non-periodic signal and let the relation between 𝑥(𝑡) and 𝑔(𝑡) is given by, $$\mathrm{X(t)=\lim_{T\rightarrow \infty}g(t)\:\:\:\:.....(3)}$$Where, T is the time period of the periodic signal 𝑔(𝑡).By rearranging eq. (2), we get, $$\mathrm{TC_n=\int_{\frac{-T}{2}}^{\frac{T}{2}}g(t)e^{-jn\omega_{0}t}dt}$$The term 𝐶𝑛 represents the magnitude of the component of frequency nω0.Let nω0 = ω at 𝑇 → ∞. Then, we have, $$\mathrm{\omega_0=\frac{2\pi}{t}|_{T\rightarrow \infty}\rightarrow 0}$$Thus, the discrete ... Read More

Fourier TransformThe Fourier transform of a continuous-time function 𝑥(𝑡) can be defined as, $$\mathrm{X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Convolution Property of Fourier TransformStatement – The convolution of two signals in time domain is equivalent to the multiplication of their spectra in frequency domain. Therefore, if$$\mathrm{x_1(t)\overset{FT}{\leftrightarrow}X_1(\omega)\:and\:x_2(t)\overset{FT}{\leftrightarrow}X_2(\omega)}$$Then, according to time convolution property of Fourier transform, $$\mathrm{x_1(t)*x_2(t)\overset{FT}{\leftrightarrow}X_1(\omega)*X_2(\omega)}$$ProofThe convolution of two continuous time signals 𝑥1(𝑡) and 𝑥2(𝑡) is defined as, $$\mathrm{x_1(t)*x_2(t)=\int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau}$$Now, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}[x_1(t)*x_2(t)]e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}[\int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau]e^{-j \omega t}dt }$$By interchanging the order of integration, we get, $$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}x_1(\tau)[\int_{-\infty}^{\infty}x_{2}(t-\tau)e^{-j \omega t}dt]d\tau }$$By replacing (𝑡 − 𝜏) = 𝑢 in the second integration, ... Read More

Fourier SeriesIf 𝑥(𝑡) is a periodic function with period T, then the continuous-time Fourier series of the function is defined as, $$\mathrm{x(t)=\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_{0}t}\:\:\:\:\:.....(1)}$$Where, 𝐶𝑛 is the exponential Fourier series coefficient, that is given by$$\mathrm{C_n=\frac{1}{T}\int_{t_0}^{t_0+T}x(t)e^{-jn\omega_0t}dt\:\:\:\:\:.....(2)}$$Convolution Property of Fourier SeriesAccording to the convolution property, the Fourier series of the convolution of two functions 𝑥1(𝑡) and 𝑥2(𝑡) in time domain is equal to the multiplication of their Fourier series coefficients in frequency domain.If 𝑥1(𝑡) and 𝑥2(𝑡) are two periodic functions with time period T and with Fourier series coefficients 𝐶𝑛 and 𝐷𝑛. Then, if$$\mathrm{x_1(t)\overset{FS}{\leftrightarrow}C_n}$$$$\mathrm{x_2(t)\overset{FS}{\leftrightarrow}D_n}$$Then, the convolution property of continuous time Fourier series states that$$\mathrm{x_1(t)*x_2(t)\overset{FS}{\leftrightarrow}TC_nD_n}$$ProofBy ... Read More

Fourier TransformFor a continuous-time function x(t), the Fourier transform of x(t) can be defined as, $$\mathrm{X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Conjugation Property of Fourier TransformStatement − The conjugation property of Fourier transform states that the conjugate of function x(t) in time domain results in conjugation of its Fourier transform in the frequency domain and ω is replaced by (−ω), i.e., if$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$Then, according to conjugation property of Fourier transform, $$\mathrm{x^*(t)\overset{FT}{\leftrightarrow}X^*(-\omega)}$$ProofFrom the definition of Fourier transform, we have$$\mathrm{X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Taking conjugate on both sides, we get$$\mathrm{X^*(\omega)=[\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt]^*}$$$$\mathrm{\Rightarrow X^*(\omega)=\int_{-\infty}^{\infty}x^*(t)e^{j\omega t}dt}$$Now, by replacing (ω) by (−ω), we obtain, $$\mathrm{X^*(-\omega)=\int_{-\infty}^{\infty}x^*(t)e^{-j\omega t}dt=F[x^*(t)]}$$$$\mathrm{\therefore F[x^*(t)]=X^*(-\omega)}$$Or, it can also be represented as, $$\mathrm{x^*(t)\overset{FT}{\leftrightarrow}X^*(-\omega)}$$Autocorrelation Property ... Read More

Exponential Fourier SeriesPeriodic signals are represented over a certain interval of time in terms of the linear combination of orthogonal functions. If these orthogonal functions are the exponential functions, then the Fourier series representation of the function is called the exponential Fourier series.The exponential Fourier series is the most widely used form of the Fourier series. In this representation, the periodic function x(t) is expressed as a weighted sum of the complex exponential functions. The complex exponential Fourier series is the convenient and compact form of the Fourier series, hence, its findsextensive application in communication theory.ExplanationLet a set of complex ... Read More

When a voltage of V volts is applied across a resistance of R Ω, then a current I flows through it. The power dissipated in the resistance is given by, $$\mathrm{P=I^2R=\frac{V^2}{R}\:\:\:\:\:\:....(1)}$$But when the voltage and current signals are not constant, then the power varies at every instant, and the equation for the instantaneous power is given by, $$\mathrm{P=i^2(t)R=\frac{V^2(t)}{R}\:\:\:\:\:\:....(2)}$$Where, 𝑖(𝑡) and 𝑣(𝑡) are the corresponding instantaneous values of current and voltage respectivelyNow, if the value of the resistance (R) is 1 Ω, then the instantaneous power can be represented as, $$\mathrm{p=i^2(t)=v^2(t)\:\:\:\:\:\:....(3)}$$Therefore, the instantaneous power of a signal x(t) can be given ... Read More

Fourier TransformFor a continuous-time function x(t), the Fourier transform of x(t) can be defined as$$\mathrm{X(\omega)= \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Duality Property of Continuous-Time Fourier TransformStatement – If a function x(t) has a Fourier transform X(ω) and we form a new function in time domain with the functional form of the Fourier transform as X(t), then it will have a Fourier transform X(ω) with the functional form of the original time function, but it is a function of frequency.Mathematically, the duality property of CTFT states that, if$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$Then, according to duality property, $$\mathrm{X(t)\overset{FT}{\leftrightarrow}2\pi x(-\omega)}$$ProofFrom the definition of inverse Fourier transform, we have$$\mathrm{x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega }$$$$\mathrm{\Rightarrow 2\pi.x(t)=\int_{-\infty}^{\infty}X(\omega)e^{j\omega ... Read More