Category: signals And Systems

Laplace Transforms (LT) Complex Fourier transform is also called as Bilateral Laplace Transform. This is used to solve differential equations. Consider an LTI system exited by a complex exponential signal of the form x(t) = Gest. Where s = any complex number = $sigma + jomega$, σ = real of s, and ω = imaginary of s The response of LTI can be obtained by the convolution of input with its impulse response i.e. $ y(t) = x(t) times h(t) = int_{-infty}^{infty}, h (tau), x (t-tau)dtau $ $= int_{-infty}^{infty}, h (tau), Ge^{s(t-tau)}dtau $ $= Ge^{st}. int_{-infty}^{infty}, h (tau), e^{(-s tau)}dtau $ $ y(t) = Ge^{st}.H(S) = x(t).H(S)$ Where H(S) = Laplace transform of $h(tau) = int_{-infty}^{infty} h (tau) e^{-stau} dtau $ Similarly, Laplace transform of $x(t) = X(S) = int_{-infty}^{infty} x(t) e^{-st} dt,…,…(1)$ Relation between Laplace and Fourier transforms Laplace transform of $x(t) = X(S) =int_{-infty}^{infty} x(t) e^{-st} dt$ Substitute s= σ + jω in above equation. $→ X(sigma+jomega) =int_{-infty}^{infty},x (t) e^{-(sigma+jomega)t} dt$ $ = int_{-infty}^{infty} [ x (t) e^{-sigma t}] e^{-jomega t} dt $ $therefore X(S) = F.T [x (t) e^{-sigma t}],…,…(2)$ $X(S) = X(omega) quadquad for,, s= jomega$ Inverse Laplace Transform You know that $X(S) = F.T [x (t) e^{-sigma t}]$ $to x (t) e^{-sigma t} = F.T^{-1} [X(S)] = F.T^{-1} [X(sigma+jomega)]$ $= {1over 2}pi int_{-infty}^{infty} X(sigma+jomega) e^{jomega t} domega$ $ x (t) = e^{sigma t} {1 over 2pi} int_{-infty}^{infty} X(sigma+jomega) e^{jomega t} domega $ $= {1 over 2pi} int_{-infty}^{infty} X(sigma+jomega) e^{(sigma+jomega)t} domega ,…,…(3)$ Here, $sigma+jomega = s$ $jdω = ds → dω = ds/j$ $ therefore x (t) = {1 over 2pi j} int_{-infty}^{infty} X(s) e^{st} ds,…,…(4) $ Equations 1 and 4 represent Laplace and Inverse Laplace Transform of a signal x(t). Conditions for Existence of Laplace Transform Dirichlet”s conditions are used to define the existence of Laplace transform. i.e. The function f(t) has finite number of maxima and minima. There must be finite number of discontinuities in the signal f(t),in the given interval of time. It must be absolutely integrable in the given interval of time. i.e. $ int_{-infty}^{infty} |,f(t)|, dt lt infty $ Initial and Final Value Theorems If the Laplace transform of an unknown function x(t) is known, then it is possible to determine the initial and the final values of that unknown signal i.e. x(t) at t=0+ and t=∞. Initial Value Theorem Statement: if x(t) and its 1st derivative is Laplace transformable, then the initial value of x(t) is given by $$ x(0^+) = lim_{s to infty} ⁡SX(S) $$ Final Value Theorem Statement: if x(t) and its 1st derivative is Laplace transformable, then the final value of x(t) is given by $$ x(infty) = lim_{s to infty} ⁡SX(S) $$ Learning working make money

Distortion Less Transmission Transmission is said to be distortion-less if the input and output have identical wave shapes. i.e., in distortion-less transmission, the input x(t) and output y(t) satisfy the condition: y (t) = Kx(t – td) Where td = delay time and k = constant. Take Fourier transform on both sides FT[ y (t)] = FT[Kx(t – td)] = K FT[x(t – td)] According to time shifting property, = KX(w)$e^{-j omega t_d}$ $ therefore Y(w) = KX(w)e^{-j omega t_d}$ Thus, distortionless transmission of a signal x(t) through a system with impulse response h(t) is achieved when $|H(omega)| = K ,, text{and} ,,,,$ (amplitude response) $ Phi (omega) = -omega t_d = -2pi f t_d ,,, $ (phase response) A physical transmission system may have amplitude and phase responses as shown below: Learning working make money

Signals Basic Operations There are two variable parameters in general: Amplitude Time The following operation can be performed with amplitude: Amplitude Scaling C x(t) is a amplitude scaled version of x(t) whose amplitude is scaled by a factor C. Addition Addition of two signals is nothing but addition of their corresponding amplitudes. This can be best explained by using the following example: As seen from the diagram above, -10 -3 3 Subtraction subtraction of two signals is nothing but subtraction of their corresponding amplitudes. This can be best explained by the following example: As seen from the diagram above, -10 -3 3 Multiplication Multiplication of two signals is nothing but multiplication of their corresponding amplitudes. This can be best explained by the following example: As seen from the diagram above, -10 -3 3 The following operations can be performed with time: Time Shifting x(t $pm$ t0) is time shifted version of the signal x(t). x (t + t0) $to$ negative shift x (t – t0) $to$ positive shift Time Scaling x(At) is time scaled version of the signal x(t). where A is always positive. |A| > 1 $to$ Compression of the signal |A| Note: u(at) = u(t) time scaling is not applicable for unit step function. Time Reversal x(-t) is the time reversal of the signal x(t). Learning working make money

Fourier Transforms The main drawback of Fourier series is, it is only applicable to periodic signals. There are some naturally produced signals such as nonperiodic or aperiodic, which we cannot represent using Fourier series. To overcome this shortcoming, Fourier developed a mathematical model to transform signals between time (or spatial) domain to frequency domain & vice versa, which is called ”Fourier transform”. Fourier transform has many applications in physics and engineering such as analysis of LTI systems, RADAR, astronomy, signal processing etc. Deriving Fourier transform from Fourier series Consider a periodic signal f(t) with period T. The complex Fourier series representation of f(t) is given as $$ f(t) = sum_{k=-infty}^{infty} a_k e^{jkomega_0 t} $$ $$ quad quad quad quad quad = sum_{k=-infty}^{infty} a_k e^{j {2pi over T_0} kt} … … (1) $$ Let ${1 over T_0} = Delta f$, then equation 1 becomes $f(t) = sum_{k=-infty}^{infty} a_k e^{j2pi k Delta ft} … … (2) $ but you know that $a_k = {1over T_0} int_{t_0}^{t_0+T} f(t) e^{-j komega_0 t} dt$ Substitute in equation 2. (2) $ Rightarrow f(t) = Sigma_{k=-infty}^{infty} {1 over T_0} int_{t_0}^{t_0+T} f(t) e^{-j komega_0 t} dt, e^{j2pi k Delta ft} $ Let $t_0={Tover2}$ $ = Sigma_{k=-infty}^{infty} [ int_{-Tover2}^{Tover2} f(t) e^{-j2 pi k Delta ft} dt ] , e^{j2 pi k Delta ft}.Delta f $ In the limit as $T to infty, Delta f$ approaches differential $df, k Delta f$ becomes a continuous variable $f$, and summation becomes integration $$ f(t) = lim_{T to infty} ⁡left{ Sigma_{k=-infty}^{infty} [ int_{-Tover2}^{Tover2} f(t) e^{-j2 pi k Delta ft} dt ] , e^{j2 pi k Delta ft}.Delta f right} $$ $$ = int_{-infty}^{infty} [ int_{-infty}^{infty},f(t) e^{-j2pi ft} dt] e^{j2pi ft} df $$ $$f(t) = int_{-infty}^{infty}, F[omega] e^{jomega t} d omega$$ $text{Where},F[omega] = [ int_{-infty}^{infty}, f(t) e^{-j2 pi ft} dt]$ Fourier transform of a signal $$f(t) = F[omega] = [int_{-infty}^{infty}, f(t) e^{-jomega t} dt]$$ Inverse Fourier Transform is $$f(t) = int_{-infty}^{infty},F[omega] e^{jomega t} d omega$$ Fourier Transform of Basic Functions Let us go through Fourier Transform of basic functions: FT of GATE Function $$F[omega] = AT Sa({omega T over 2})$$ FT of Impulse Function $FT [omega(t) ] = [int_{- infty}^{infty} delta (t) e^{-jomega t} dt] $ $quad quad quad quad = e^{-jomega t}, |, t = 0 $ $quad quad quad quad = e^{0} = 1 $ $quad therefore delta (omega) = 1 $ FT of Unit Step Function: $U(omega) = pi delta (omega)+1/jomega$ FT of Exponentials $ e^{-at}u(t) stackrel{mathrm{F.T}}{longleftrightarrow} 1/(a+jω)$ $ e^{-at}u(t) stackrel{mathrm{F.T}}{longleftrightarrow} 1/(a+jomega )$ $ e^{-a,|,t,|} stackrel{mathrm{F.T}}{longleftrightarrow} {2a over {a^2+ω^2}}$ $ e^{j omega_0 t} stackrel{mathrm{F.T}}{longleftrightarrow} delta (omega – omega_0)$ FT of Signum Function $ sgn(t) stackrel{mathrm{F.T}}{longleftrightarrow} {2 over j omega }$ Conditions for Existence of Fourier Transform Any function f(t) can be represented by using Fourier transform only when the function satisfies Dirichlet’s conditions. i.e. The function f(t) has finite number of maxima and minima. There must be finite number of discontinuities in the signal f(t),in the given interval of time. It must be absolutely integrable in the given interval of time i.e. $ int_{-infty}^{infty}, |, f(t) | , dt Discrete Time Fourier Transforms (DTFT) The discrete-time Fourier transform (DTFT) or the Fourier transform of a discrete–time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence $e^{jomega n}$. The DTFT sequence x[n] is given by $$ X(omega) = Sigma_{n= -infty}^{infty} x(n)e^{-j omega n} ,, …,… (1) $$ Here, X(ω) is a complex function of real frequency variable ω and it can be written as $$ X(omega) = X_{re}(omega) + jX_{img}(omega) $$ Where Xre(ω), Ximg(ω) are real and imaginary parts of X(ω) respectively. $$ X_{re}(omega) = |, X(omega) | costheta(omega) $$ $$ X_{img}(omega) = |, X(omega) | sintheta(omega) $$ $$ |X(omega) |^2 = |, X_{re} (omega) |^2+ |,X_{im} (omega) |^2 $$ And X(ω) can also be represented as $ X(omega) = |,X(omega) | e^{jtheta (ω)} $ Where $theta(omega) = arg{X(omega) } $ $|,X(omega) |, theta(omega)$ are called magnitude and phase spectrums of X(ω). Inverse Discrete-Time Fourier Transform $$ x(n) = { 1 over 2pi} int_{-pi}^{pi} X(omega) e^{j omega n} domega ,, …,… (2)$$ Convergence Condition: The infinite series in equation 1 may be converges or may not. x(n) is absolutely summable. $$ text{when},, sum_{n=-infty}^{infty} |, x(n)|, An absolutely summable sequence has always a finite energy but a finite-energy sequence is not necessarily to be absolutely summable. Learning working make money

Signals Sampling Theorem Statement: A continuous time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to the twice the highest frequency component of message signal. i. e. $$ f_s geq 2 f_m. $$ Proof: Consider a continuous time signal x(t). The spectrum of x(t) is a band limited to fm Hz i.e. the spectrum of x(t) is zero for |ω|>ωm. Sampling of input signal x(t) can be obtained by multiplying x(t) with an impulse train δ(t) of period Ts. The output of multiplier is a discrete signal called sampled signal which is represented with y(t) in the following diagrams: Here, you can observe that the sampled signal takes the period of impulse. The process of sampling can be explained by the following mathematical expression: $ text{Sampled signal}, y(t) = x(t) . delta(t) ,,…,…(1) $ The trigonometric Fourier series representation of $delta$(t) is given by $ delta(t)= a_0 + Sigma_{n=1}^{infty}(a_n cos⁡ nomega_s t + b_n sin⁡ nomega_s t ),,…,…(2) $ Where $ a_0 = {1over T_s} int_{-T over 2}^{ T over 2} delta (t)dt = {1over T_s} delta(0) = {1over T_s} $ $ a_n = {2 over T_s} int_{-T over 2}^{T over 2} delta (t) cos nomega_s, dt = { 2 over T_2} delta (0) cos n omega_s 0 = {2 over T}$ $b_n = {2 over T_s} int_{-T over 2}^{T over 2} delta(t) sin⁡ nomega_s t, dt = {2 over T_s} delta(0) sin⁡ nomega_s 0 = 0 $ Substitute above values in equation 2. $therefore, delta(t)= {1 over T_s} + Sigma_{n=1}^{infty} ( { 2 over T_s} cos ⁡ nomega_s t+0)$ Substitute δ(t) in equation 1. $to y(t) = x(t) . delta(t) $ $ = x(t) [{1 over T_s} + Sigma_{n=1}^{infty}({2 over T_s} cos nomega_s t) ] $ $ = {1 over T_s} [x(t) + 2 Sigma_{n=1}^{infty} (cos nomega_s t) x(t) ] $ $ y(t) = {1 over T_s} [x(t) + 2cos omega_s t.x(t) + 2 cos 2omega_st.x(t) + 2 cos 3omega_s t.x(t) ,…, …,] $ Take Fourier transform on both sides. $Y(omega) = {1 over T_s} [X(omega)+X(omega-omega_s )+X(omega+omega_s )+X(omega-2omega_s )+X(omega+2omega_s )+ ,…] $ $therefore,, Y(omega) = {1over T_s} Sigma_{n=-infty}^{infty} X(omega – nomega_s )quadquad where ,,n= 0,pm1,pm2,… $ To reconstruct x(t), you must recover input signal spectrum X(ω) from sampled signal spectrum Y(ω), which is possible when there is no overlapping between the cycles of Y(ω). Possibility of sampled frequency spectrum with different conditions is given by the following diagrams: Aliasing Effect The overlapped region in case of under sampling represents aliasing effect, which can be removed by considering fs >2fm By using anti aliasing filters. Learning working make money