ConvolutionThe convolution is a mathematical operation for combining two signals to form a third signal. In other words, the convolution is a mathematical way which is used to express the relation between the input and output characteristics of an LTI system.Mathematically, the convolution of two signals is given by, $$\mathrm{x_{1}\left ( t \right )\ast x_{2}\left ( t \right )=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right )x_{2}\left ( t-\tau \right )d\tau =\int_{-\infty }^{\infty }x_{2}\left ( \tau \right )x_{1}\left ( t-\tau \right )d\tau}$$CorrelationThe correlation is defined as the measure of similarity between two signals or functions or waveforms. The correlation is of two ... Read More

For a continuous-time function 𝑥(𝑡), the Fourier transform of 𝑥(𝑡) can be defined as, $$\mathrm{X\left ( \omega \right )=\int_{-\infty }^{\infty }x\left ( t \right )e^{-j\omega t}\: dt}$$Time Reversal Property of Fourier TransformStatement – The time reversal property of Fourier transform states that if a function 𝑥(𝑡) is reversed in time domain, then its spectrum in frequency domain is also reversed, i.e., if$$\mathrm{x\left ( t \right )\overset{FT}{\leftrightarrow}X\left ( \omega \right )}$$Then, according to the time-reversal property of Fourier transform, $$\mathrm{x\left ( -t \right )\overset{FT}{\leftrightarrow}X\left ( -\omega \right )}$$ProofForm the definition of Fourier transform, we have, $$\mathrm{F\left [ x\left ( t \right ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Unit Impulse FunctionThe unit impulse function is defined as, $$\mathrm{\delta(t)=\begin{cases}1 & for\:t=0 \0 & for\:t ≠ 0 \end{cases}}$$If it is given that$$\mathrm{x(t)=\delta(t)}$$Then, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}\delta(t)e^{-j\omega t}dt}$$As the impulse function exists only at t= 0. Thus, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}\delta(t) e^{-j\omega t}dt=\int_{−\infty}^{\infty}1\cdot e^{-j\omega t}dt=e^{-j\omega t}|_{t=0}=1}$$$$\mathrm{\therefore\:F[\delta(t)]=1\:\:or\:\:\delta(t) \overset{FT}{\leftrightarrow}1}$$That is, the Fourier transform of a unit impulse function is unity.The magnitude and phase representation of the Fourier transform of unit impulse function are as follows −$$\mathrm{Magnitude, |X(\omega)|=1;\:\:for\:all\:\omega}$$$$\mathrm{Phase, \angle X(\omega)=0;\:\:for\:all\:\omega}$$The graphical representation of the ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Two-Sided Real Exponential FunctionLet a two-sided real exponential function as, $$\mathrm{x(t)=e^{-a|t|}}$$The two-sided or double-sided real exponential function is defined as, $$\mathrm{e^{-a|t|}=\begin{cases}e^{at} & for\:t ≤ 0\e^{-at} & for\:t ≥ 0 \end{cases} =e^{at}u(-t)+e^{-at}u(t) }$$Where, the functions $u(t)$ and $u(-t)$ are the unit step function and time reversed unit step function, respectively.Now, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}e^{-a|t|}e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}[e^{at}u(-t)+e^{-at}u(t)]e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{0}e^{at}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{-\infty}^{0}e^{(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-(a-j\omega)t}dt+\int_{0}^{\infty}e^{-(a+j\omega)t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\left[\frac{e^{-(a-j\omega)t}}{-(a-j\omega)}\right]_{0}^{\infty}+\left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty}=\left[\frac{e^{-\infty}-e^{0}}{-(a-j\omega)} \right]+\left[\frac{e^{-\infty}-e^{0}}{-(a+j\omega)} \right]}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{a-j\omega}+\frac{1}{a+j\omega}=\frac{2a}{a^{2}+\omega^{2}}}$$Therefore, the Fourier transform of a two-sided real exponential function is, $$\mathrm{F[e^{-a|t|}]=X(\omega)=\frac{2a}{a^{2}+\omega^{2}}}$$Or, it can also be represented as, $$\mathrm{e^{-a|t|}\overset{FT}{\leftrightarrow}\frac{2a}{a^{2}+\omega^{2}}}$$Magnitude ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{x(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t }dt}$$Fourier Transform of Sine FunctionLet$$\mathrm{x(t)=sin\:\omega_{0} t}$$From Euler’s rule, we have, $$\mathrm{x(t)=sin\:\omega_{0} t=\left[\frac{ e^{j\omega_{0} t}- e^{-j\omega_{0} t}}{2j} \right]}$$Then, from the definition of Fourier transform, we have, $$\mathrm{F[sin\:\omega_{0} t]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}sin\:\omega_{0}\: t\: e^{-j\omega t}dt}$$$$\mathrm{ \Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}\left[ \frac{e^{j\omega_{0} t}-e^{-j\omega_{0} t}}{2j}\right] e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2j}\left[ \int_{−\infty}^{\infty}e^{j\omega_{0} t}e^{-j\omega t} dt-\int_{−\infty}^{\infty} e^{-j\omega_{0} t}e^{-j\omega t} dt\right]}$$$$\mathrm{=\frac{1}{2j}\{F[e^{j\omega_{0} t}] -F[e^{-j\omega_{0} t}]\}}$$Since, the Fourier transform of complex exponential function is given by, $$\mathrm{F[e^{j\omega_{0} t}]=2\pi\delta(\omega-\omega_{0})\:\:and\:\:F[e^{-j\omega_{0} t}]=2\pi\delta(\omega+\omega_{0})}$$$$\mathrm{ \therefore\:X(\omega)=\frac{1}{2j}[2\pi\delta(\omega-\omega_{0})-2\pi\delta(\omega+\omega_{0})]}$$$$\mathrm{\Rightarrow\:X(\omega)=-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$Therefore, the Fourier transform of the sine wave is, $$\mathrm{F[sin\:\omega_{0}\:t]=-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$Or, it can also be represented as, $$\mathrm{sin\:\omega_{0}\:t\overset{FT}{\leftrightarrow}-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$The graphical representation of the sine function with ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)= \int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of One-Sided Real Exponential FunctionA single-sided real exponential function is defined as, $$\mathrm{x(t)=e^{-a t}u(t)}$$Where, $u(t)$ is the unit step signal and is defined as, $$\mathrm{u(t)=\begin{cases}1 & for\:t≥ 0 \0 & for\:t < 0 \end{cases}}$$Then, from the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}e^{-at}u(t)e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{0}^{\infty}e^{-(a+j\omega)t} dt=\left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)} \right]_{0}^{\infty}}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{-(a+j\omega)}[e^{-\infty}-e^{0}]=\frac{0-1}{-(a+j\omega)}=\frac{1}{a+j\omega}}$$Therefore, the Fourier transform of a single-sided real exponential function is, $$\mathrm{F[e^{-at}u(t)]=\frac{1}{a+j\omega}}$$Or, it can also be represented as, $$\mathrm{e^{-at}u(t)\overset{FT}{\leftrightarrow}\frac{1}{a+j\omega}}$$Magnitude and phase representation of the Fourier transform of a single-sided real exponential function The Fourier transform of the ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as,$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Signum FunctionThe signum function is represented by $sgn(t)$ and is defined as$$\mathrm{sgn(t)=\begin{cases}1 & for\:t>0\-1 & for\:t

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}\:dt}$$Fourier Transform of Rectangular FunctionConsider a rectangular function as shown in Figure-1.It is defined as, $$\mathrm{rect\left(\frac{t}{τ}\right)=\prod\left(\frac{t}{τ}\right)=\begin{cases}1 & for\:|t|≤ \left(\frac{τ}{2}\right)\0 & otherwise\end{cases}}$$Given that$$\mathrm{x(t)=\prod\left(\frac{t}{τ}\right)}$$Hence, from the definition of Fourier transform, we have, $$\mathrm{F\left[\prod\left(\frac{t}{τ}\right) \right]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}\:dt=\int_{−\infty}^{\infty}\prod\left(\frac{t}{τ}\right)e^{-j\omega t}\:dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−(τ/2)}^{(τ/2)}1\cdot e^{-j\omega t}\:dt=\left[\frac{e^{-j\omega t}}{-j\omega} \right]_{-τ/2}^{τ/2}}$$$$\mathrm{\Rightarrow\:X(\omega)=\left[ \frac{e^{-j\omega (τ/2)}-e^{j\omega (τ/2)}}{-j\omega}\right]=\left[ \frac{e^{j\omega (τ/2)}-e^{-j\omega (τ/2)}}{j\omega }\right]}$$$$\mathrm{\Rightarrow\:X(\omega)=\left[ \frac{2τ[e^{j\omega (τ/2)}-e^{-j\omega (τ/2)}]}{j\omega\cdot (2τ) }\right]=\frac{τ}{\omega(τ/2)}\left[\frac{e^{j\omega (τ/2)}-e^{-j\omega (τ/2)}}{2j} \right]}$$$$\mathrm{\because \:\left[\frac{e^{j\omega (τ/2)}-e^{-j\omega (τ/2)}}{2j} \right]=sin\:\omega (τ/2)}$$$$\mathrm{\therefore\:X(\omega)=\frac{τ}{\omega(τ/2)}\cdot sin \omega (τ/2)=τ \left[\frac{sin\omega (τ/2)}{\omega (τ/2)}\right]}$$$$\mathrm{\because\:sinc \left(\frac{\omega τ}{2}\right)=\frac{sin\omega (τ/2)}{\omega (τ/2)}}$$$$\mathrm{\therefore\:X(\omega)=τ\cdot sinc \left(\frac{\omega τ}{2}\right)}$$Therefore, the Fourier transform of the rectangular function is$$\mathrm{F\left[\prod\left(\frac{t}{τ}\right)\right]=τ\cdot sinc \left(\frac{\omega τ}{2}\right)}$$Or, it can also be ... Read More

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as,$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)\:e^{-j\omega t}dt}$$Fourier Transform of a Triangular PulseA triangular signal is shown in Figure-1 −And it is defined as,$$\mathrm{\Delta \left(\frac{t}{τ}\right)=\begin{cases}\left( 1+\frac{2t}{τ}\right); & for\:\left(-\frac{τ}{2}\right)

The infinite series of sine and cosine terms of frequencies $0, \omega_{0}, 2\omega_{0}, 3\omega_{0}, ....k\omega_{0}$is known as trigonometric Fourier series and can written as, $$\mathrm{x(t)=a_{0}+\sum_{n=1}^{\infty}a_{n}\:cos\:n\omega_{0} t+b_{n}\:sin\:n\omega_{0} t… (1)}$$Here, the constant $a_{0}, a_{n}$ and $b_{n}$ are called trigonometric Fourier series coefficients.Evaluation of a0To evaluate the coefficient $a_{0}$, we shall integrate the equation (1) on both sides over one period, i.e., $$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt=a_{0}\int_{t_{0}}^{(t_{0}+T)}dt+\int_{t_{0}}^{(t_{0}+T)}\left(\sum_{n=1}^{\infty}a_{n}\:cos\:n\omega_{0} t+b_{n}\:sin\:n\omega_{0} t\right)dt}$$$$\mathrm{\Rightarrow\:\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt=a_{0}T+\sum_{n=1}^{\infty}a_{n}\int_{t_{0}}^{(t_{0}+T)}cos\:n\omega_{0} t\:dt+\sum_{n=1}^{\infty}b_{n}\int_{t_{0}}^{(t_{0}+T)}sin\:n\omega_{0} t\:dt… (2)}$$As we know that the net areas of sinusoids over complete periods are zero for any non-zero integer n and any time $t_{0}$. Therefore, $$\mathrm{\int_{t_{0}}^{(t_{0}+T)}cos\:n\omega_{0} t\:dt=0\:\:and\:\:\int_{t_{0}}^{(t_{0}+T)}sin\:n\omega_{0} t\:dt=0}$$Hence, from equation (2), we get, $$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt=a_{0}T}$$$$\mathrm{\therefore\:a_{0}=\frac{1}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt… (3)}$$Using equation (3), ... Read More