fourier_transform

fourier_transform - The Fourier Transform Do not distribute...

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Unformatted text preview: The Fourier Transform Do not distribute these notes 189 Prof. R.T. M’Closkey, UCLA Background Consider a continuous-time signal u on the interval (−∞, ∞). This signal has finite energy if the limit τ lim τ →∞ −τ |u(t)|2 dt, (69) exists. We will call signals that satisfy (69) “energy signals". For example, the impulse response of any stable linear system is an energy signal. Other examples inlcude any bounded signal that is non-zero for only a finite interval of time. Note that periodic signals do not have finite energy on the interval (−∞, ∞) because (69) does not exist. The periodic signals studied in this course, however, have finite energy in one period. A power signal is one which satisfies 1 0 < lim τ →∞ 2τ τ −τ |u(t)|2 dt < ∞, (70) which is consistent with the notion that power is defined as energy per unit time. If u is periodic with period T then 1 lim τ →∞ 2τ τ −τ |u(t)|2 dt = 1 T T |u(t)|2 dt = 1 u T E. Thus, u is a power signal if it is periodic. Finally, note that energy signals have zero power. This section is concerned with energy signals and their Fourier transforms. To motivate the Fourier transform let’s return to periodic signals. Let u be a periodic signal with period T . The Fourier series coefficients are given by u(t)e T −jkω0 t T /2 dt = −T /2 u(t)e−jkω0 t dt, where ω0 = 2π/T . Imagine taking one period of u and then inserting zeros for some duration and repeating this process in order to make a periodic signal with longer period. For example, a periodic signal is shown in Fig. 1 and a modified version of the signal is shown in Fig. 2. The interesting fact is that the Fourier series coefficients for each case lie on the envelope function that we associate with the original periodic signal in Fig. 1. Note that in Fig. 2 the Do not distribute these notes 190 Prof. R.T. M’Closkey, UCLA effective period is larger so that frequency spacing of the Fourier series coefficients along the envelope function is actually smaller. This is evident when comparing the amplitude spectrum plots in Figs. 1 and 2. If we insert more zeros as shown in Figs. 3 and 4, the spacing between Fourier series coefficients is reduced even further. Finally, in the limit as T → ∞, we are left with a single pulse and its associated Fourier coefficients which have in fact “converged" to the envelope function. We are not submitting to a formal proof of this limit function but rather that this argument seems reasonable. Thus, the Fourier transform of an energy signal u, denoted u, is defined to be ˆ ∞ u(ω ) = ˆ u(t)e−jωt dt. −∞ Previously in these notes we denoted this as “χ(ω )". We already showed that the frequency response function of an asymptotically stable linear system is actually the Fourier transform of its impulse response. The Fourier transform can be applied to any energy signal, though, not just those associated with the impulse response of a stable system. An energy signal can also be computed from its Fourier transform just like a periodic signal can be represented as a Fourier series: if u is an energy signal and u its Fourier transform, then ˆ 1 u(t) = 2π = ∞ ∞ u(ω )ejωt dω ˆ −∞ u(2πf )e ˆ j 2πf t (ω in rad/s) (71) df (f in Hz) −∞ These integrals are typically evaluated by complex contour integration. The energy of u is calculated as ∞ −∞ which can be shown to be equivalent to 1 2π |u(t)|2 dt, ∞ −∞ |u(ω )|2 dω. ˆ These relations are just generalizations of what we derived for periodic signals. The term |u(ω )|2 is called the ˆ energy-density spectrum and it may be plotted versus ω just like the Fourier series examples we studied earlier. Do not distribute these notes 191 Prof. R.T. M’Closkey, UCLA Period T=2 seconds Spectrum; Period T=2 seconds 1 10 1 0.8 0 10 0.6 0.4 10 1 0.2 0 10 8 6 4 2 0 sec 2 4 6 8 10 10 2 5 4 3 2 1 0 Hz 1 2 3 4 5 Figure 1: Periodic signal with its associated amplitude spectrum plot. The period is T = 2 seconds so the frequency spacing of the Fourier series points is 0.5Hz. Do not distribute these notes 192 Prof. R.T. M’Closkey, UCLA Period T=4 seconds Spectrum; Period T=4 seconds 1 10 1 0.8 0 10 0.6 0.4 10 1 0.2 0 10 8 6 4 2 0 sec 2 4 6 8 10 10 2 5 4 3 2 1 0 Hz 1 2 3 4 5 Figure 2: This new periodic signal is generated from the one in Fig. 1 by inserting an interval of zeros between the “fundamental" period. In this case, the new period is T = 4 so the frequency spacing is 0.25Hz. The envelope function remains unchanged, though, so the spectrum points are “packed" more closely on the envelope function. Do not distribute these notes 193 Prof. R.T. M’Closkey, UCLA Period T=6 seconds Spectrum; Period T=6 seconds 1 10 1 0.8 0 10 0.6 0.4 10 1 0.2 0 10 8 6 4 2 0 sec 2 4 6 8 10 10 2 5 4 3 2 1 0 Hz 1 2 3 4 5 Figure 3: This periodic signal is again just and extension of the one in Fig. 2 by increasing the interval for which the signal is zero. In this case, T = 6 seconds so the that frequency spacing is now 1/6Hz. Do not distribute these notes 194 Prof. R.T. M’Closkey, UCLA Period T=8 seconds Spectrum; Period T=8 seconds 1 10 1 0.8 0 10 0.6 0.4 10 1 0.2 0 10 8 6 4 2 0 sec 2 4 6 8 10 10 2 5 4 3 2 1 0 Hz 1 2 3 4 5 Figure 4: As in Fig. 1 through 3, the period is further increased to T = 8 seconds by the expansion of the interval over with the signal is zero. The frequency spacing is now 0.125Hz in the spectrum plot. Do not distribute these notes 195 Prof. R.T. M’Closkey, UCLA Period T Spectrum; Period T 1 10 1 0.8 0 10 0.6 0.4 10 1 0.2 0 10 8 6 4 2 0 sec 2 4 6 8 10 10 2 5 4 3 2 1 0 Hz 1 2 3 4 5 Figure 5: In the limit as the period grows to be infinite, the frequency spacing reduces to zero and the spectrum converges to the envelope function. Furthermore, the spectrum of the signal now becomes a continuum. Do not distribute these notes 196 Prof. R.T. M’Closkey, UCLA Fourier Transform Properties Convolution Property The most important property of Fourier transforms is related to its effect on convolution. If u is an energy signal, then its Fourier transform, denoted u or F (u), is ˆ ∞ u(ω ) = ˆ u(t)e−jωt dt. −∞ Suppose u is the input to a stable linear system with impulse response h. The system output, denoted y , is y (t) = ∞ −∞ h(t − τ )u(τ )dτ. The Fourier transform of y is = = ∞ −∞ ∞ −∞ ∞ −∞ ∞ −∞ ∞ −∞ y (ω ) = ˆ ∞ −∞ h(t − τ )u(τ )dτ e−jωt dt h(t − τ )e−jω(t−τ ) u(τ )e−jωτ dτ dt ˆu = h(ω )ˆ(ω ), (72) h(t − τ )e−jω(t−τ ) dt u(τ )e−jωτ dτ ω ∈ (−∞, ∞). ˆ where h represents the system’s frequency response function. Thus, the Fourier transform of the output of a stable system driven by an energy signal is simply the product of systems’ frequency response function and Fourier transform of the input. Do not distribute these notes 197 Prof. R.T. M’Closkey, UCLA You can use (72) and the inverse formula (71) to solve the following kinds of problems: • If u and h are given, then 1 y (t) = 2π ∞ 1 y (ω )dω = ˆ 2π −∞ i.e. find the output given the input and impulse response. • If u and y are given, then 1 h(t) = 2π ∞ ˆu h(ω )ˆ(ω )ejωt dω, (73) −∞ ∞ 1 ˆ h(ω )dω = 2π −∞ ∞ y (ω ) jωt ˆ e dω, ˆ −∞ u(ω ) (74) i.e. find the impulse response given the input and output (typically called system identification) • If y and h are given, then 1 u(t) = 2π ∞ 1 u(ω )dω = ˆ 2π −∞ ∞ y (ω ) jωt ˆ e dω, ˆ −∞ h(ω ) (75) i.e. find the input given the output and impulse response (typically called deconvolution). Comparison with Fourier series. The relation (72) is analogous to case the we derived for periodic inputs. Recall that if u is periodic with period T and its Fourier series is given by 1 uF S (t) = T ∞ ck ejkω0 t , u(t)e−jkω0 t dt, where ck = T k =−∞ then the steady-state output, y , of a stable linear system with transfer function H (s) is also expressed as a Fourier series 1 yF S (t) = T ∞ H (jkω0 )ck ejkω0 t , k =−∞ ck ˜ where ck are the Fourier series coefficients of the output. The relationship between the input and output Fourier ˜ series coefficients is ck = H (jkω0 )ck , ˜ which bears strong resemblance to (72). Do not distribute these notes k = 0, ±1, ±2, . . . , 198 Prof. R.T. M’Closkey, UCLA Example. Consider the linear system given by the following linear differential equation y (4) + 15.11y (3) + 272.0¨ + 1192y + 6234y = 114.1¨. y ˙ u (76) The ODE represents a Butterworth band-pass filter. Butterworth filters are a common choice in many filter applications. Let us furthermore assume that the input and output units are both “volts". The transfer function for the filter is 114.1s2 H (s) = 4 , s + 15.11s3 + 272.0s2 + 1192s + 6234 with unit V/V, and the frequency response function is 114.1(jω )2 , H (jω ) = (jω )4 + 15.11(jω )3 + 272.0(jω )2 + 1192jω + 6234 (77) with unit V/V. The Bode plots of this filter are shown on the next slide. Furthermore, suppose the input to this filter is an “isolated" sawtooth. The magnitude and phase of the Fourier transform of this signal is shown along with the frequency response of the filter. The Fourier transform of the filter output, denoted y , can then be found using (72). In other words, the Fourier transform of the filter output is ˆ the product of the frequency response function and the Fourier transform of the input: the magnitude plots are multiplied and the phases from these figures (see next slide) are added. The challenge now becomes that of “inverting" the Fourier transform y in order to find y . There are several options: ˆ 1. It may be possible to “look up" y in a table of Fourier transform pairs. In other words, a table of “common" time ˆ functions and their corresponding Fourier transforms can be consulted. This probably the easiest approach if you can find your transformed function in a table. Although such tables can be made, it is more common to use the Laplace transform (which we will study) for this purpose. 2. Complex contour integration and the calculus of residues can be used to invert the Fourier transform. This approach requires knowledge of complex variable theory. 3. Numerical approximation of y can be carried out in which the inverse transform 1 u(t) = 2π Do not distribute these notes ∞ u(ω )ejωt dω, ˆ −∞ 199 Prof. R.T. M’Closkey, UCLA is numerically approximated. This is a good approach for complicated functions, although the disadvantage is that the solution is given in the form of a graph (not a formula). 1 10 input spectrum Butterworth freq. resp. output spectrum magnitudes Butterworth Filter y magnitude (mixed units) u } y (t) = (h ∗ u)(t) ˆ y (ω ) = h(ω )ˆ(ω ) ˆ u ˆ h 0 10 10 frequency response function u ˆ 1 y ˆ 10 1.5 2 5 0 Hz 200 input spectrum Butterworth freq. resp. output spectrum phases u 0.5 100 phase (degrees) filter input and output (V) 150 1 5 y 0 50 0 50 100 0.5 10 5 0 seconds 5 150 10 200 5 Do not distribute these notes 200 0 Hz 5 Prof. R.T. M’Closkey, UCLA ˆ Let’s try the numerical approximation of the inverse Fourier transform. We already have a formula for h (see (77)). The Fourier transform of u is easily computed to be u(ω ) = ˆ 1 2 1 −j 2ω 2 e − 1 + j e−j 2ω , ω2 ω ω = 0; u(0) = 1. ˆ Thus, 1 114.1(jω )2 y (ω ) = ˆ (jω )4 + 15.11(jω )3 + 272.0(jω )2 + 1192jω + 6234 2 y (0) = 0 ˆ 2 1 −j 2ω e − 1 + j e−j 2ω , ω2 ω ω=0 For any given fixed value of t we can approximate y (t) as 1 y (t) = 2π 1 ≈ 2π 1 ≈ 2π ∞ −∞ Ω −Ω y (ω )ejωt dω ˆ y (ω )ejωt dω, ˆ Ω > 0, represents a finite range of integration (78) y (k ∆ω )ejk∆ω t ∆ω ˆ k where ∆ω is the frequency “step size" (in rad/s) we are using to approximate the integral. The index k ranges over those integers for which k ∆ω ∈ [−Ω, Ω]. This summation is reminiscent of the “synthesis" equation associated with the discrete-time Fourier series (DTFS), the only difference being the time “t" in (78) can be any time we choose. On the other hand, suppose we sum over M ∈ Z frequencies in (78) with a frequency “step size" of ∆ω rad/s. Note 1 2π ∆ω looks like the fundamental frequency parameter in the DTFS (expressed in hertz) which corresponds to a fundamental period of T = 2π/∆ω seconds. If we’re content approximating y at M equally spaced points in time with ts = T /M being the time interval between each point, then the approximation in (78) can be written as y (pts ) ≈ Do not distribute these notes 1 T 1 2 M −1 y (k ∆ω )ejk∆ω pts ∆ω , ˆ k =− 1 M +1 2 201 p∈Z (79) Prof. R.T. M’Closkey, UCLA where M is assumed even for now. The right-hand side of (79) can be evaluated using the inverse fast Fourier transform algorithm (Matlab’s ifft command) in order to compute y at times pts , p ∈ Z. The following code does this (compare it to a time simulation –you will see no difference) %%%% numerical approximation of inverse Fourier transform clear close all M = 2^14; %%% number of points Delta_om = 1/20; %%% fundamental frequency in Hz; T = 20 seconds ts = 1/Delta_om/M; %%% time step between estimated values of y k1 = [1:round(M/2)]; om1 = k1*Delta_om*2*pi; %%% frequencies in rad/s y1 = 114.1*(j*om1).^2./( (j*om1).^4 + 15.11*(j*om1).^3 + 272.0*(j*om1).^2 + 1192*j*om1 + 6234); y1 = y1.*(0.5*(1./om1.^2.*(exp(-j*2*om1) - 1) + j*2./om1.*exp(-j*2*om1))); yf = [0 y1 fliplr(conj(y1([1:1:max(k1)-1])))]/ts; %%% place into Matlab form y = real(ifft(yf)); %%% the exact solution is real, however, small non-zero t = [0:M-1]*ts; %%% imaginary parts are created from numerical errors figure(1) plot(t,y); grid on xlabel(’seconds’) Do not distribute these notes 202 Prof. R.T. M’Closkey, UCLA Example. Consider the convolution example with the impulse response and input (left and right plots, respectively) Impulse response Input to system 1.5 0.5 1 0.4 0.3 0.5 0.2 0 u output 0.1 0 0.1 0.5 0.2 0.3 1 0.4 0.5 1.5 1 0.5 0 0.5 1 1.5 seconds 2 2.5 3 3.5 4 1 0.5 0 0.5 1 1.5 seconds 2 2.5 3 3.5 4 The Fourier transform of the impulse response (in other words, the frequency response function) is 1 −j 1.5ω 1 e−jω − 1 + e − e−jω , −jω jω 1 ˆ h(0) = . 2 ˆ h(ω ) = ω=0 The Fourier transform of the input is 1 e−j 2ω − 1 , −j 2ω u(0) = 1. ˆ u(ω ) = ˆ Do not distribute these notes 203 ω=0 Prof. R.T. M’Closkey, UCLA ˆu The Fourier transform of the output, denoted y , is y (ω ) = h(ω )ˆ(ω ), ˆ ˆ y (ω ) = ˆ 1 1 −j 1.5ω e−jω − 1 + e − e−jω −jω jω 1 y (0) = . ˆ 2 1 e−j 2ω − 1 −j 2ω , ω=0 (80) With enough effort (80) can be analytically inverted. On the other hand, the numerical inversion yields a very fast solution, albeit at points in time. The following code produces the figure shown on the next page: %%%% numerical approximation of inverse Fourier transform clear close all M = 2^14; %%% number of points Delta_om = 1/10; %%% fundamental frequency in Hz; T = 20 seconds ts = 1/Delta_om/M; %%% time step between estimated values of y k1 = [1:round(M/2)]; om1 = k1*Delta_om*2*pi; %%% frequencies in rad/s y1 = 1./(-j*om1).*(exp(-j*om1) - 1) + 1./(j*om1).*(exp(-j*1.5*om1) - exp(-j*om1)); y1 = y1.*(1./(-j*2*om1).*(exp(-j*2*om1) - 1)); yf = [0.5 y1 fliplr(conj(y1([1:1:max(k1)-1])))]/ts; %%% place into Matlab form y = real(ifft(yf)); %%% the exact solution is real, however, small non-zero t = [0:M-1]*ts; %%% imaginary parts are created from numerical errors figure(1) plot(t,y); grid on; set(gca,’FontSize’,16) xlabel(’seconds’,’FontSize’,16) axis([0 4 -0.5 1]) Do not distribute these notes 204 Prof. R.T. M’Closkey, UCLA There are M = 214 points describing the approximate solution. 1 0.5 0 0.5 0 0.5 1 1.5 2 2.5 seconds 3 3.5 4 The fft/ifft algorithm is especially efficient when the number of points is a power of 2. Take a course in signal processing to learn more about this. Do not distribute these notes 205 Prof. R.T. M’Closkey, UCLA Energy-density Spectrum If u is an energy signal, then its Fourier transform, denoted u or F (u), is ˆ ∞ u(ω ) = ˆ u(t)e−jωt dt. −∞ The spectrum of u is another name for its Fourier transform. Note that the magnitude and phase of a signal’s spectrum may be plotted versus frequency. Recall the notion of energy in the signal, ∞ −∞ |u(t)|2 dt, which can also be expressed in terms of its spectrum (analogous to the situation we derived for periodic signals), ∞ 1 |u(t)| dt = 2π −∞ ∞ 2 −∞ |u(ω )|2 dω. ˆ If we change from the frequency unit of “rad/s" to “Hz" via ω = 2πf , then the energy can also be expressed ∞ −∞ ∞ 2 |u(t)| dt = −∞ |u(2πf )|2 df. ˆ (81) The point of this change of frequency unit is to show that if we plot |u|2 versus f (expressed in Hertz), then the ˆ area under the plot is actually the total signal energy. For example, let’s consider once again the signal shown on the next slide. The square of this signal is also shown and the signal energy is simply the shaded area, i.e. signal energy = ∞ −∞ |u(t)|2 dt = shaded area in time domain plot. Note that the square of the signal has units V2 so the area has units V2 s or, equivalently, V2 /Hz. On the other hand, the spectral density is also plotted and we have the fact signal energy = ∞ −∞ Do not distribute these notes |u(2πf )|2 df = shaded area in frequency domain plot. ˆ 206 Prof. R.T. M’Closkey, UCLA This is the reason behind naming |u(ω )|2 the energy spectral density, i.e. integrating the energy density over ˆ frequency yields energy. 1 square of signal |u|2 (V2) 0.8 signal with finite energy u2 0.6 area = 0.66V 2 /Hz 0.4 1.5 0.2 signal (V) 1 u 0 1 0.5 0 0.5 0.5 1 1.5 seconds 2 2.5 3 energy-density spectrum 0 5 0 seconds 5 10 Energy density spectrum (V2/Hz2) 0.5 10 |u|2 ˆ 1 0.8 area = 0.66V 2 /Hz 0.6 0.4 0.2 0 2 Do not distribute these notes 1.5 207 1 0.5 0 Hz 0.5 1 1.5 2 Prof. R.T. M’Closkey, UCLA Lastly, if we are interested in computing the signal energy in a certain frequency band, say [−ω2 , −ω1 ] and [ω1 , ω2 ], then we simply compute signal energy in given frequency band = −ω1 −ω2 ω2 =2 ω1 Do not distribute these notes 208 2 |u(2πf )| df + ˆ ω2 ω1 |u(2πf )|2 df ˆ |u(2πf )|2 df. ˆ Prof. R.T. M’Closkey, UCLA Fourier Transform Analysis of the Unit Impulse Function The unit impulse enjoys a special place in the analysis of linear systems. You’ve seen that response of a system to a unit impulse function can be used to characterize the response to any input via convolution. It is informative to study the frequency domain properties of the unit impulse, thus, consider the four realization of the unit impulse and their corresponding Fourier transforms (c > 0 in all cases), δ1 (t) = ce−ct µ(t) −→ δ2 (t) = c2 te−ct µ(t) −→ 1 δ3 (t) = c3 t2 e−ct µ(t) −→ 2 1 −→ δ4 (t) = ce−c|t| 2 c jω + c c2 ˆ2 (ω ) = δ (jω + c)2 c3 ˆ δ3 (ω ) = (jω + c)3 2 ˆ4 (ω ) = c δ ω 2 + c2 ˆ δ1 (ω ) = The interesting fact is that for any fixed ω , ˆ lim δk (ω ) = 1, c→∞ k = 1, 2, 3, 4 Thus, as the impulse becomes narrow and taller, its Fourier transform converges to 1 over any desired frequency band. This demonstrates why the unit impulse is of fundamental importance in the analysis of linear systems: a unit impulse when applied to the system represents a “broadband" input that equally excites the system at all frequencies. A few examples are given on the next slides for c ∈ {100, 1000, 10000}. Do not distribute these notes 209 Prof. R.T. M’Closkey, UCLA 1 90 80 10 δ1 0 10 F 60 50 40 30 magnitude 70 c = 102 20 10 10 1 ˆ |δ1 | 1 100 c = 104 1 c = 103 c = 102 2 10 0 3 10 0.1 0.05 0 seconds 0.05 10 1000 0.1 3 90 80 10 δ2 0 10 F 60 50 40 magnitude 70 10 0 rad/s 500 1000 500 1000 2 ˆ |δ2 | 1 100 500 1 30 c = 102 20 10 2 10 0 3 10 0.1 0.05 0 seconds 0.05 10 1000 0.1 3 80 10 δ3 70 60 Do not distribute these notes 50 0 10 F 210 nitude 90 ˆ |δ3 | 1 100 500 0 rad/s 3 Prof. R.T. M’Closkey, UCLA 10 1 0 3 10 0.1 0.05 0 seconds 0.05 10 1000 0.1 3 80 δ3 0 10 70 F 60 50 40 c = 102 30 magnitude 90 10 20 10 10 0 rad/s 500 1000 500 1000 500 1000 3 ˆ |δ3 | 1 100 500 1 2 10 0 3 10 0.1 0.05 0 seconds 0.05 10 1000 0.1 4 90 80 10 δ4 0 10 F 60 50 40 magnitude 70 10 0 rad/s 4 ˆ |δ4 | 1 100 500 1 30 20 10 c = 10 2 10 2 0 10 0.1 3 0.05 Do not distribute these notes 0 seconds 0.05 10 1000 0.1 211 500 0 rad/s Prof. R.T. M’Closkey, UCLA Fourier Potpourri Fourier Series Fourier Transform ...the signal is periodic... used when... Analysis equation u(t + T ) = u(t), t ∈ (−∞, ∞) T is the fundamental period 2π ω0 = is the fundamental frequency in rad/s T ck = ￿ u(t)e −jkω0 t T Energy Relationships T |u(t)|2 dt = ∞ ￿ k=−∞ there is a graphical interpretation associated with the sum Do not distribute these notes −∞ |u(t)|2 dt < ∞ u(t)e−j ωt dt, ω ∈ (−∞, ∞) ￿∞ 1 u(t) = u(ω )ej ωt dω , t ∈ (−∞, ∞) ˆ 2π −∞ ￿∞ = u(2π f )ej 2πf t df ˆ k=−∞ ￿ ∞ ∞ fact: frequency is a continuous variable ∞ 1￿ uF S (t) = ck ejkω0 t T ￿u￿E = ￿ −∞ fact: frequency is a discrete variable Synthesis equation u(t) is an energy signal → u(ω ) = ˆ dt, k = 0, ±1, ±2, . . . ￿ −∞ |ck |2 f0 , f0 = 1 Hz T ￿u￿E = ￿ ∞ −∞ |u(t)| dt = 2 ￿ ∞ −∞ |u(2π f )|2 df, f in Hz ˆ there is a graphical interpretation associated with the integral fundamental frequency in hertz 212 Prof. R.T. M’Closkey, UCLA Power Tools for Linear Systems u Time-invariant Linear System y particular solution Complex exponential input y (t) = H (s)est + yh (t) u(t) = est , s ∈ C, t ∈ (−∞, ∞) homogeneous solution reqʼd if IC is specified Sinusoidal input u(t) = cos(ω t + θ), t ∈ (−∞, ∞) particular solution y (t) = |H (j ω )| cos(ω t + θ + ∠H (j ω )) + yh (t) homogeneous solution reqʼd if IC is specified particular solution Periodic input u(t + T ) = u(t), FS coefficients of input ∞ 1￿ y (t) = H (jk ω0 )ck ejkω0 t + yh (t) T −∞ t ∈ (−∞, ∞) (use Fourier Series representation) homogeneous solution reqʼd if IC is specified Energy input u is energy signal defined for t ∈ (−∞, ∞) (assume the system is asymptotically stable so that the Fourier transform of the impulse response exists) Do not distribute these notes time domain solution y (t) = ￿ ∞ −∞ freq. domain solution h(t − τ )u(τ )dτ ˆ y (ω ) = h(ω )ˆ(ω ) ˆ u * the IC is specified to be zero at t=- ∞ so no other IC can be specified (we will revisit this issue in a future lecture) 213 Prof. R.T. M’Closkey, UCLA ...
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This note was uploaded on 08/08/2011 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.

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