Found 189 Articles for Signals and Systems

Trigonometric Fourier SeriesA periodic function can be represented over a certain interval of time in terms of the linear combination of orthogonal functions. If these orthogonal functions are the trigonometric functions, then it is known as trigonometric Fourier series.Mathematically, the standard trigonometric Fourier series expansion of a periodic signal is, $$\mathrm{x(t)=a_{0}+ \sum_{n=1}^{\infty}a_{n}\:cos\:\omega_{0}nt+b_{n}\:sin\:\omega_{0}nt\:\:… (1)}$$Exponential Fourier SeriesA periodic function can be 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 it is known as exponential Fourier series.Mathematically, the standard exponential Fourier series expansion for a periodic ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ is defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Inverse Fourier TransformThe inverse Fourier transform of a continuous-time function is defined as, $$\mathrm{x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)e^{j\omega t}d\omega}$$Properties of Fourier TransformThe continuous-time Fourier transform (CTFT) has a number of important properties. These properties are useful for driving Fourier transform pairs and also for deducing general frequency domain relationships. These properties also help to find the effect of various time domain operations on the frequency domain. Some of the important properties of continuous time Fourier transform are given in the table as −Property of CTFTTime Domain x(t)Frequency Domain X(ω)Linearity Property$ax_{1}(t)+bx_{2}(t)$$aX_{1}(\omega)+bX_{2}(\omega)$Time Shifting ... 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, $$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)e^{-jn\omega_{0} t}dt\:\:… (2)}$$Modulation or Multiplication PropertyLet $x_{1}(t)$ and $x_{2}(t)$ two periodic signals with time period $T$ and with Fourier series coefficient $C_{n}$ and $D_{n}$. If$$\mathrm{x_{1}(t)\overset{FS}{\leftrightarrow}C_{n}}$$$$\mathrm{x_{2}(t)\overset{FS}{\leftrightarrow}D_{n}}$$Then, the modulation or multiplication property of continuous time Fourier series states, that$$\mathrm{x_{1}(t)\cdot x_{2}(t)\overset{FS}{\leftrightarrow}\sum_{k=−\infty}^{\infty}C_{k}\:D_{n-k}}$$Proof From the definition of continuous time Fourier series, we get, $$\mathrm{FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[x_{1}(t)\cdot x_{2}(t)]e^{-jn\omega_{0} t}dt}$$$$\mathrm{\Rightarrow\:FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\left (\sum_{k=−\infty}^{\infty} C_{k} e^{jk\omega_{0} t}\right )e^{-jn\omega_{0} t}dt}$$$$\mathrm{\Rightarrow\:FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\left (\sum_{k=−\infty}^{\infty} C_{k} e^{-j(n-k)\omega_{0} t}\right )e^{-jn\omega_{0} ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Modulation Property of Fourier TransformStatement – The modulation property of continuous-time Fourier transform states that if a continuous-time function $x(t)$ is multiplied by $cos \:\omega_{0} t$, then its frequency spectrum gets translated up and down in frequency by $\omega_{0}$. Therefore, if$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$Then, according to the modulation property of CTFT, $$\mathrm{x(t)\:cos\:\omega_{0}t\overset{FT}{\leftrightarrow}\frac{1}{2}[X(\omega-\omega_{0})+X(\omega+\omega_{0})]}$$ProofUsing Euler’s formula, we get, $$\mathrm{cos\:\omega_{0}t=\left [\frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2} \right ]}$$Therefore, $$\mathrm{x(t)\:cos\:\omega_{0}t=x(t)\left [ \frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2}\right ]}$$Now, from the definition of Fourier transform, we have, $$\mathrm{F[x(t)]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega_{0} t} \:dt}$$$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} t]=\int_{−\infty}^{\infty}x(t)\:cos\:\omega_{0} t\:e^{-j\omega_{0} t}dt}$$$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} t]=\int_{−\infty}^{\infty}x(t)\left [ \frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2}\right ]e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} ... Read More

Fourier TransformFor a continuous-time function $x(t)$, the Fourier transform can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Linearity Property of Fourier TransformStatement − The linearity property of Fourier transform states that the Fourier transform of a weighted sum of two signals is equal to the weighted sum of their individual Fourier transforms.Therefore, if$$\mathrm{x_{1}(t)\overset{FT}{\leftrightarrow}X_{1}(\omega)\:\:and\:\:x_{2}\overset{FT}{\leftrightarrow}X_{2}(\omega)}$$Then, according to the linearity property of Fourier transform, $$\mathrm{ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)}$$Where, a and b are constants.ProofFrom the definition of Fourier transform, we have, $$\mathrm{F[x(t)]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=F[ax_{1}(t)+bx_{2}(t)]=\int_{−\infty}^{\infty}[ax_{1}(t)+bx_{2}(t)]e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}ax_{1}(t)e^{-j \omega t} dt+\int_{−\infty}^{\infty}bx_{2}(t)e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=a\int_{−\infty}^{\infty}x_{1}(t)e^{-j \omega t} dt+b\int_{−\infty}^{\infty}x_{2}(t)e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=aX_{1}(\omega)+bX_{2}(\omega)}$$$$\mathrm{\therefore\:F[ax_{1}(t)+bx_{2}(t)]=aX_{1}(\omega)+bX_{2}(\omega)}$$Or, it can also be written as, $$\mathrm{ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)}$$Frequency Shifting Property of Fourier TransformStatement – Frequency ... 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, $$\mathrm{x(t)=\displaystyle\sum\limits_{n=−\infty}^\infty\:C_{n}\:e^{jn\omega_{0}t}\:\:\:… (1)}$$Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by, $$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}X(t)e^{-jn\omega_{0}t}\:dt\:\:… (2)}$$Linearity Property of Continuous Time Fourier SeriesConsider two periodic signals $x_{1}(t)$ and $x_{2}(t)$ which are periodic with time period T and with Fourier series coefficients $C_{n}$ and $D_{n}$ respectively. If$$\mathrm{x_{1}(t)\overset{FS}{\leftrightarrow}C_{n}}$$$$\mathrm{x_{2}(t)\overset{FS}{\leftrightarrow}D_{n}}$$Then, the linearity property of continuous-time Fourier series states that$$\mathrm{Ax_{1}(t)+Bx_{2}(t)\overset{FS}{\leftrightarrow}AC_{n}+BD_{n}}$$Proof By the definition of Fourier series of a periodic function, we get, $$\mathrm{FS[Ax_{1}(t)+Bx_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[Ax_{1}(t)+Bx_{2}(t)]e^{-jn\omega_{0}t}\:dt}$$$$\mathrm{\Rightarrow\:FS[Ax_{1}(t)+Bx_{2}(t)]=A\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}\:x_{1}(t)\:e^{-jn\omega_{0}t}\:dt\right )+B\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}\:x_{2}(t)\:e^{-jn\omega_{0}t}\:dt \right )\:\:… (3)}$$On comparing equations (2) & (3), ... Read More

What is GIBBS Phenomenon?The GIBBS phenomenon was discovered by Henry Wilbraham in 1848 and then rediscovered by J. Willard Gibbs in 1899.For a periodic signal with discontinuities, if the signal is reconstructed by adding the Fourier series, then overshoots appear around the edges. These overshoots decay outwards in a damped oscillatory manner away from the edges. This is known as GIBBS phenomenon and is shown in the figure below.The amount of the overshoots at the discontinuities is proportional to the height of discontinuity and according to Gibbs, it is found to be around 9% of the height of discontinuity irrespective ... Read More

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

Fourier TransformFor a continuous-time function $x(t)$, the Fourier transform is defined as, $$\mathrm{X(\omega)=\int_{−\infty }^{\infty}x(t)e^{−j\omega t}\:dt}$$Fourier Transform of Unit Step FunctionThe unit step function is defined as, $$\mathrm{u(t)=\begin{cases}1 & for\:t≥ 0\0 & for\:t< 0\end{cases}}$$Because the unit step function is not absolutely integrable, thus its Fourier transform cannot be found directly.In order to find the Fourier transform of the unit step function, express the unit step function in terms of signum function as$$\mathrm{u(t)=\frac{1}{2}+\frac{1}{2}sgn(t)=\frac{1}{2}[1+sgn(t)]}$$Given that$$\mathrm{x(t)=u(t)=\frac{1}{2}[1+sgn(t)]}$$Now, from the definition of the Fourier transform, we have, $$\mathrm{F[u(t)]=X(\omega)=\int_{−\infty }^{\infty}x(t)e^{-j\omega t} dt=\int_{−\infty }^{\infty} u(t)e^{-j\omega t} dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty }^{\infty} \frac{1}{2}[1+sgn(t)]e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2}\left [ \int_{−\infty }^{\infty} 1 \cdot e^{-j\omega t} dt ... Read More