156L#17 19 March 2015. Further remarks on FT's also on linear and time-shift-invariantsystems, convolutions, causality, Greens functions, and Fourier Transforms.Further Examples of Fourier Transforms and how plots of € ˜ x (ω) are meaningful. Heaviside step function Η(t) = 1 ( t > 0), = 0 (t <0 )FT = 1/iω compare with its Laplace transform = 1/s Notice how H(t) is constant (for t > 0 ), and how its FT has its chief amplitude at ω = 0.Delta function δ(t-to)F.T. = exp(-iωto) compare Laplace transform = exp(-sto)Notice how δ(t) is concentrated in time, and ts FT is spread out over all frequency. Ramp function f(t) = t Η (t)F.T. = -1/ω2compare LT = 1/s2BTW – why do we plot the absolute value of the FTs ? |%x(ω) | rather than %x(ω) ?Answer: It is hard to know WHAT to plot, Real part? Imaginary part? Both? Phase?It turns out that the magnitude has most of the information we might want. We notice thatthe phase is affected by when x(t) acts, the magnitude is affected by the shape of x(t).Consider the FT of a shifted function x(t- Δ) The FT of a shifted function of time isjust a phase factor exp(-iωΔ) times the FT of the unshifted function. [ Becausex(t − Δ)exp(−iωt )dt∫ = (using the change of variables τ=t-Δ; dτ = dt)exp(−iωΔ) x(τ)exp(−iωτ)dτ∫ = exp(−iωΔ)%x(ω) ] The factor exp(−iωΔ)has unitmagnitude, thus time shifts do not affect the magnitude |%x(ω) | of the FT.157A square pulse x(t) = 1 for -T < t < T, and x(t) = 0 otherwise. T = ∫ exp(- iωt ) dt = (2/ω) sin ωT -TThis function of frequency is peaked at ω=zero , thus having a strong constant DC component(with magnitude = 2T = area under the curve in the time domain ) and dies away as ω gets to beof order 2π/T or larger. In a plot one can see that the distribution in frequency has a "width"(depending on exactly how one defines width) of order 2π/T, inversely proportional to the"duration", T, that x had in the time domain. (depending on exactly how one defines duration)This character is also seen in the other plots.The same behavior is seen for a Gaussian x(t)= exp(-t2/T2)which may be said to have a time-domain width of T. In thefrequency domain, its FT is √π T exp(-ω2T2/4) which is also aGaussian and has "width" 4/T, again inversely proportional to theduration in the time domain.For a function x(t) which is infinitely narrow in the timedomain, e.g., x(t) = δ(t-to), we find its FT %x(ω)= exp(-iωto), isinfinitely broad in the frequency domain. ( | its abs magnitude | isconstant.)For an impulse response in an undamped system,x(t) = (1/mωn) sin ωn t for t > 0;x(t) = 0 for t < 0we find its FT is our old friend %G(ω) = (1/m)/[ ωn 2 - ω2 ] ( see plot)The FT has ∞ amplitude at the natural frequency, (but non zero amplitude elsewhere.)158Some more examples of FT'sConsider a single cycle of a sine wave x(t) = sin(πt/T) for -T<t<T, and x=0 otherwise.The waveform has an apparent "frequency" of π/T corresponding to the period of the single cyclesine wave.Here is its plot vs time ( I take T = π.)It has a FT which is easily calculated to be = -2i sin ωT (π/T) / [ (π/T)2 - ω2 ]which can be seen, upon plotting vs ω, to have very broad peaks at ω = ±π/T. It also, though, hassignificant amplitude at other frequencies. Here is its -imaginary part:-----------------Consider a 10 cycle sine wave x = sin(10 πt/T) for -T<t<T, and x=0 otherwise. Suchan x(t) might be termed a "tone burst". It has a nominal frequency of 10 π /T and a FT of = -2i sin ωT (10π/T) / [ (10π/T)2 - ω2 ]Here are the plots of x(t) and | %x(ω)| ( we take T = 10 π)159Its FT is fairly strongly peaked at the apparent frequency ω = ±10π/T; but it has non zeroamplitude at other frequencies, concentrated near ω = 10π/T.-----------------Consider an ∞ cycle sine wave x(t) = sin αt for all t. In this case its FT is (π/i)[ δ(ω+α) − δ(ω−α) ] and has amplitude (absolute value) only at theapparent frequency α. It appears that the FT requires a large amount of time duration of a sinewave before it will unambiguously single out a particular ω.-------------------Similarly, the damped impulse response G(t)has a |%G(ω) | which has been plotted several times in previous lectures. |%G(ω) | peaks with afinite value at the nominal frequency, but has non-zero amplitude at other frequencies too. (Theplots are for the case Q ~ 20. Notice it takes about 20 cycles for G(t) to decay to 4%, also noticethe peak of G(ω) is about 20 times greater than its low frequency limit; also notice that the widthof the peak (between half-power points) is about one twentieth of the natural frequency. )The above many cases in which I plot the function of time and the corresponding function offrequency all illustrate the "uncertainty principle" - that we cannot simultaneously have a definiteprecise frequency and a definite unambiguous time of occurrence. This is a mathematical analogof the Heisenberg uncertainty principle - that one cannot simultaneously and precisely specifythe time of an event's occurrence and its energy. The ten-cycle sine wave had a time ofoccurrence ambiguous to within an interval of length T. It had an imprecision of frequency oforder 1/T. The product of the imprecisions is of order unity. Depending on how one quantifies"uncertainty", one can conclude with mathematical statements that the product of theuncertainties cannot be less than some quantity of order 1, regardless of the shape of theenvelope of the tone burst.160This tells us that the human notion of frequency is not necessarily the same as that of the Fouriertransform's mathematical machinery. Humans might label the frequency of the above ten-cycletone burst to be exactly ω = 1. The FT sort of agrees – as its peak value occurs near ω = ±1.But it also does not agree – as it has amplitude at other frequencies also.Using the FT to "solve" the ODE.After obtaining %x(ω) by multiplying %F(ω) and %G(ω) we might wish to get x(t). Oneway to do this is to look it up in a table ( this is often what one does with Laplace transformsx (s).) Another way is to perform the inverse FT: ( numerically or analytically) x(t ) =12π%x(ω)exp(iωt )dω−∞∞∫=12π%G(ω)%F(ω)exp(iωt )dω−∞∞∫A third approach is merely to plot | %x(ω)| and get a qualitative understanding of x(t). This