This preview shows page 1. Sign up to view the full content.
Unformatted text preview: The Fourier Transform Do not distribute these notes 189 Prof. R.T. M’Closkey, UCLA Background Consider a continuoustime 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 nonzero 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
energydensity 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 steadystate 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 bandpass 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 discretetime 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 righthand 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 nonzero
t = [0:M1]*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 nonzero
t = [0:M1]*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 Energydensity 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 u2 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 u2 (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 energydensity spectrum
0 5 0
seconds 5 10 Energy density spectrum (V2/Hz2) 0.5
10 u2
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−ct
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 uE = −∞ 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 uE = ∞ −∞ 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 Timeinvariant
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 ...
View
Full
Document
This note was uploaded on 08/08/2011 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.
 Spring '06
 TSAO

Click to edit the document details