Energy of Sampling Signal [sa(t)]

$latex \displaystyle E = \int_{-\infty}^{\infty} |x^2(t)| dt &s=2 &bg=ffffff$ By the help of Parsavel's relationship $latex \displaystyle E = \int_{-\infty}^{\infty} |x^2(t)| dt = \frac{1}{2 \pi} \int_{-\infty}^{\infty} |X^2(\omega)| d \omega &s=2 &bg=ffffff$ $latex \displaystyle (A) rect(\frac{t}{\tau}) \rightleftharpoons (A \tau) sa(\frac{\omega \tau}{2}) &s=2 &bg=ffffff$ $latex \displaystyle sa(\frac{t \tau}{2}) \rightleftharpoons (\frac{2 \pi}{\tau}) rect(\frac{\omega}{\tau}) &s=2 &bg=ffffff$ $latex \displaystyle sa(t) … Continue reading Energy of Sampling Signal [sa(t)]

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Fourier Transform of Periodic Signals

Exponential fourier series representation of x(t) $latex \displaystyle x(t) = \sum_{n=-\infty}^{\infty} C_n e^{jn \omega_0 t} &s=2 &bg=ffffff$ $latex \displaystyle 1 \rightleftharpoons 2 \pi \delta(\omega) &s=2 &bg=ffffff$ $latex \displaystyle C_n \rightleftharpoons 2 \pi C_n \delta(\omega) &s=2 &bg=ffffff$ $latex \displaystyle y(t) = \sum_{n=-\infty}^{\infty} C_n \rightleftharpoons \sum_{n=-\infty}^{\infty} 2 \pi C_n \delta(\omega) &s=2 &bg=ffffff$ $latex y(t) \displaystyle e^{jn \omega_0 … Continue reading Fourier Transform of Periodic Signals

Fourier Transform of a Rectangular Pulse and Sampling Function

$latex \displaystyle x(t) = A rect(\frac{t}{\tau}) &s=2 &bg=ffffff$ Where $latex \displaystyle \tau &s=2 &bg=ffffff$ is the duration of rectangular pulse. FIRST METHOD $latex \displaystyle X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j \omega t} dt&s=2 &bg=ffffff$ $latex \displaystyle X(\omega) = \int_{-\frac{\tau}{2}}^{\frac{\tau}{2}} A e^{-j \omega t} dt &s=2 &bg=ffffff$ $latex \displaystyle X(\omega) = \frac{-2A}{\omega} [\frac{e^{-\frac{j \omega \tau}{2}} - e^{\frac{j … Continue reading Fourier Transform of a Rectangular Pulse and Sampling Function

Basics of Fourier Transform

$latex \Rightarrow &s=2 &bg=ffffff$ Fourier transform - transforms the signal from time domain to frequency domain. $latex \Rightarrow &s=2 &bg=ffffff$ Fourier transform is used for the analysis of Energy and power signals. $latex \Rightarrow &s=2 &bg=ffffff$ Stable and Marginally stable system can be analysed by the use of Fourier transform. $latex x(t) \rightleftharpoons X(\omega) &s=2 … Continue reading Basics of Fourier Transform

Special Fourier Transforms related to sin(ωt)

Bilateral Fourier transform of $latex \displaystyle sin(\omega_0 t) &s=2 &bg=ffffff$ Fourier Transform of $latex \displaystyle sin(\omega_0 t) &s=2 &bg=ffffff$ $latex \displaystyle 1 \rightleftharpoons 2 \pi \delta(\omega) &s=2 &bg=ffffff$ $latex \displaystyle e^{-j \omega_0 t} \rightleftharpoons 2 \pi \delta(\omega + \omega_0) &s=2 &bg=ffffff$ $latex \displaystyle e^{j \omega_0 t} \rightleftharpoons 2 \pi \delta(\omega - \omega_0) &s=2 &bg=ffffff$ $latex … Continue reading Special Fourier Transforms related to sin(ωt)

Fourier Transform of Some basic Signals

Fourier Transform of $latex \displaystyle \delta(t)&s=2 &bg=ffffff$ $latex \displaystyle X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt &s=2 &bg=ffffff$ $latex \displaystyle X(\omega) = \int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt &s=2 &bg=ffffff$ $latex \displaystyle \delta(t) x(t) = x(0) \delta(t) &s=2 &bg=ffffff$ $latex \displaystyle X(\omega) = \int_{-\infty}^{\infty} \delta(t) dt &s=2 &bg=ffffff$ $latex \displaystyle \int_{-\infty}^{\infty} \delta(t) = 1 &s=2 &bg=ffffff$ … Continue reading Fourier Transform of Some basic Signals