Found 189 Articles for Signals and Systems

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

What is Fourier Series?In the domain of engineering, most of the phenomena are periodic in nature such as the alternating current and voltage. These periodic functions could be analysed by resolving into their constituent components by a process called the Fourier series.Therefore, the Fourier series can be defined as under −“The representation of periodic signals over a certain interval of time in terms of linear combination of orthogonal functions (i.e., sine and cosine functions) is known as Fourier series.”The Fourier series is applicable only to the periodic signals i.e. the signals which repeat itself periodically over an interval from $(-\infty\:to\:\infty)$and ... Read More

The cosine form of Fourier series is the alternate form of the trigonometric Fourier series. The cosine form Fourier series is also known as polar form Fourier series or harmonic form Fourier series.The trigonometric Fourier series of a function x(t) contains sine and cosine terms of the same frequency. That is, $$\mathrm{x(t)=a_{0}+\sum_{n=1}^{\infty}a_{n}\:cos\:n\omega_{0} t+b_{n}\:sin\:n\omega_{0} t… (1)}$$Where, $$\mathrm{a_{0}=\frac{1}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt}$$$$\mathrm{a_{n}=\frac{2}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:cos\:n\omega_{0} t\:dt}$$$$\mathrm{b_{n}=\frac{2}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:sin\:n\omega_{0} t\:dt}$$In equation (1), by multiplying the numerator and denominator of the sine and cosine terms with ($\sqrt{a_{n}^{2}+b_{n}^{2}}$), we get, $$\mathrm{x(t)=a_{0}+\sum_{n=1}^{\infty}\left( \sqrt{a_{n}^{2}+b_{n}^{2}}\right)\left( \frac{a_{n}}{\sqrt{a_{n}^{2}+b_{n}^{2}}}cos\:n\omega_{0} t+\frac{b_{n}}{\sqrt{a_{n}^{2}+b_{n}^{2}}}sin\:n\omega_{0} t\right)… (2)}$$Putting the values in the equation (2) as, $$\mathrm{a_{0}=A_{0}}$$$$\mathrm{\sqrt{a_{n}^{2}+b_{n}^{2}}=A_{n}… (3)}$$$$\mathrm{\frac{a_{n}}{\sqrt{a_{n}^{2}+b_{n}^{2}}}=cos\:\theta_{n}\:\:and\:\:\frac{b_{n}}{\sqrt{a_{n}^{2}+b_{n}^{2}}}=-sin\:\theta_{n}}$$We obtain, $$\mathrm{x(t)=A_{0}+\sum_{n=1}^{\infty}A_{n}(cos\:\theta_{n}\:cos\:n\omega_{0} t-sin\:\theta_{n}\:sin\:n\omega_{0} t)}$$$$\mathrm{\Rightarrow\:x(t)=A_{0}+\sum_{n=1}^{\infty}A_{n}\:cos(n\omega_{0} t+\theta_{n})… (4)}$$Where, $$\mathrm{\theta_{n}=-tan^{-1} \left(\frac{b_{n}}{a_{n}}\right)… (5)}$$The ... Read More

Quarter Wave SymmetryA periodic function $x(t)$ which has either odd symmetry or even symmetry along with the half wave symmetry is said to have quarter wave symmetry.Mathematically, a periodic function $x(t)$ is said to have quarter wave symmetry, if it satisfies the following condition −$$\mathrm{x(t)=x(-t)\:or\:x(t)=-x(-t)\:and\:x(t)=-x\left (t ± \frac{T}{2}\right )}$$Some examples of periodic functions having quarter wave symmetry are shown in Figure-1.The Fourier series coefficients for the function having quarter wave symmetry are evaluated as follows −Case I – When n is odd$$\mathrm{x(t)=-x(-t)\:and\:x(t)=-x\left (t ± \frac{T}{2}\right )}$$For this case, $$\mathrm{a_{0}=0\:\:and\:\:a_{n}=0}$$And, $$\mathrm{b_{n}=\frac{8}{T} \int_{0}^{T/4}x(t)\:sin\:n\omega_{0}\:t\:dt}$$Case II – When n is even$$\mathrm{x(t)=x(-t)\:and\:x(t)=-x\left (t ± \frac{T}{2}\right ... Read More