Introduction to Fourier Transforms. Lecture 7 ELE 301: Signals and Systems. Fourier Series. Rect Example

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1 Introduction to Fourier ransforms Lecture 7 ELE 3: Signals and Systems Fourier transform as a limit of the Fourier series Inverse Fourier transform: he Fourier integral theorem Prof. Paul Cuff Princeton University Fall - Example: the rect and sinc functions Cosine and Sine ransforms Symmetry properties Periodic signals and δ functions Cuff (Lecture 7) ELE 3: Signals and Systems Fall - / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - / Fourier Series Suppose x(t) is not periodic. We can compute the Fourier series as if x was periodic with period by using the values of x(t) on the interval t [ /, /). Rect Example For example, assume x(t) = rect(t), and that we are computing the Fourier series over an interval, f(t) = rect(t) ak = / x(t)e jπkft dt, / x (t) = ake jπkft, k= where f = /. he two signals x and x will match on the interval [ /, /) but x(t) will be periodic. What happens if we let increase? Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 3 / / / t he fundamental period for the Fourier series in, and the fundamental frequency is f = /. he Fourier series coefficients are where sinc(t) = sin(πt) πt. ak = sinc (kf) Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 4 /

2 he Sinc Function Rect Example Continued ake a look at the Fourier series coefficients of the rect function (previous slide). We find them by simply evaluating sinc(f ) at the points f = kf. = ω =π = ω = π -8π -4π / 4π 8π ω = n ω -4-4 t -8π -4π 4π 8π ω = n ω =4 ω = π/ /4-8π -4π 4π 8π ω = n ω More densely sampled, same sinc() envelope, decreased amplitude. Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 5 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 6 / Fourier ransforms We can interpret this as the result of expanding x(t) as a Fourier series in an interval [ /, /), and then letting. he Fourier series for x(t) in the interval [ /, /): Given a continuous time signal x(t), define its Fourier transform as the function of a real f : X (f ) = x(t)e jπft dt his is similar to the expression for the Fourier series coefficients. Note: Usually X (f ) is written as X (iπf ) or X (iω). his corresponds to the Laplace transform notation which we encountered when discussing transfer functions H(s). x (t) = ake jπkft k= where ak = / x(t)e jπkft dt. / Define the truncated Fourier transform: X (f ) = x(t)e jπft dt so that ak = X (kf) = X ( k ). Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 7 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 8 /

3 he Fourier series is then x (t) = X (kf)ejπkft k= he limit of the truncated Fourier transform is Continuous-time Fourier ransform Which yields the inversion formula for the Fourier transform, the Fourier integral theorem: X (f ) = lim X (f ) he Fourier series converges to a Riemann integral: x(t) = lim = lim = x (t) k= X X (f )e jπft df. ( k ) e jπ k t X (f ) = x(t)e jπft dt, x(t) = X (f )e jπft df. Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 9 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - / Cosine and Sine ransforms Comments: here are usually technical conditions which must be satisfied for the integrals to converge forms of smoothness or Dirichlet conditions. he intuition is that Fourier transforms can be viewed as a limit of Fourier series as the period grows to infinity, and the sum becomes an integral. X (f )ejπft df is called the inverse Fourier transform of X (f ). Notice that it is identical to the Fourier transform except for the sign in the exponent of the complex exponential. If the inverse Fourier transform is integrated with respect to ω rather than f, then a scaling factor of /(π) is needed. Assume x(t) is a possibly complex signal. X (f ) = x(t)e jπft dt = x(t) (cos(πft) j sin(πft)) dt = x(t) cos(ωt)dt j x(t) sin(ωt) dt. Cuff (Lecture 7) ELE 3: Signals and Systems Fall - / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - /

4 Fourier ransform Notation Duality For convenience, we will write the Fourier transform of a signal x(t) as F [x(t)] = X (f ) Notice that the Fourier transform F and the inverse Fourier transform F are almost the same. and the inverse Fourier transform of X (f ) as F [X (f )] = x(t). Note that and at points of continuity of x(t). F [F [x(t)]] = x(t) Duality heorem: If x(t) X (f ), then X (t) x( f ). In other words, F [F [x(t)]] = x( t). Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 3 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 4 / Example of Duality Since rect(t) sinc(f ) then sinc(t) rect( f ) = rect(f ) (Notice that if the function is even then duality is very simple) f (t) F(ω) π π / / t ω π f (t) F(ω) π π Generalized Fourier ransforms: δ Functions A unit impulse δ(t) is not a signal in the usual sense (it is a generalized function or distribution). However, if we proceed using the sifting property, we get a result that makes sense: F [δ(t)] = δ(t)e jπft dt = so δ(t) his is a generalized Fourier transform. It behaves in most ways like an ordinary F. δ(t) t / / ω t ω Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 5 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 6 /

5 Shifted δ A shifted delta has the Fourier transform F [δ(t t)] = δ(t t)e jπft dt = e jπtf so we have the transform pair δ(t t) e jπtf Constant Next we would like to find the Fourier transform of a constant signal x(t) =. However, direct evaluation doesn t work: F [] = = e jπft jπf e jπft dt and this doesn t converge to any obvious value for a particular f. δ(t t) t t e jωt ω We instead use duality to guess that the answer is a δ function, which we can easily verify. F [δ(f )] = δ(f )e jπft df =. Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 7 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 8 / Sinusoidal Signals So we have the transform pair δ(f ) πδ(ω) t ω If the δ function is shifted in frequency, F [δ(f f)] = δ(f f)e jπft df = e jπft so e jπft δ(f f) his also does what we expect a constant signal in time corresponds to an impulse a zero frequency. e jωt πδ(ω ω) t ω ω Cuff (Lecture 7) ELE 3: Signals and Systems Fall - 9 / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - /

6 Cosine With Euler s relations we can find the Fourier transforms of sines and cosines [ F [cos(πft)] = F (e jπft + e jπft)] = ( [ F e jπft] + F [e jπft]) Sine Similarly, since sin(ft) = j (ejπft e jπft ) we can show that sin(ft) j (δ(f + f) δ(f f)). so = (δ(f f) + δ(f + f)). cos(πft) (δ(f f) + δ(f + f)). sin(ωt) jπδ(ω + ω) ω t ω ω cos(ωt) πδ(ω + ω) πδ(ω ω) jπδ(ω ω) t ω ω ω he Fourier transform of a sine or cosine at a frequency f only has energy exactly at ±f, which is what we would expect. Cuff (Lecture 7) ELE 3: Signals and Systems Fall - / Cuff (Lecture 7) ELE 3: Signals and Systems Fall - /