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**Unformatted text preview: **EE 261 The Fourier Transform and its Applications Fall 2009 Solutions to Problem Set Four 1. (10 points) Eva and Rajiv continue their conversation about convolution: Rajiv: You know, convolution really is a remarkable operation, the way it imparts properties of one function onto the convolution with another. Take periodicity: If f ( t ) is periodic then ( f ∗ g )( t ) is periodic with the same period as f . That was a homework problem last week Eva: There’s a problem with that statement. You want to say that if f ( t ) is a periodic function of period T then ( f ∗ g )( t ) is also periodic of period T . Rajiv: Right. Eva: What if g ( t ) is also periodic, say of period R ? Then doesn’t ( f ∗ g )( t ) have two periods, T and R ? Rajiv: I suppose so. Eva: But wouldn’t this lead right to a contradiction? I mean, for example, you can’t have a function with two periods, can you? Rajiv: I think we’ve found a fundamental contradiction in mathematics. Eva: Why don’t we look at a simple, special case first. What happens if you convolve sin 2 πt with itself? Rajiv: OK, both functions have period 1 so for the convolution you get a function that’s periodic of period 1, no problem. Eva: No, you don’t. Something goes wrong. What’s going on? With whom do you agree and why? What do think about that statement ‘If f ( t ) is periodic then f ∗ g is periodic.’ Solution: The problem is that the integral defining the convolution of two periodic functions won’t converge (maybe it will for some values, though I doubt it, but certainly not in general). Taking f ( t ) = sin 2 πt as an example, the convolution is ( f ∗ f )( t ) = integraldisplay ∞ −∞ sin2 πτ sin 2 π ( t − τ ) dτ . Let t = 1. Then because of periodicity ( f ∗ f )(1) = integraldisplay ∞ −∞ sin2 πτ sin 2 π (1 − τ ) dτ = integraldisplay ∞ −∞ sin2 πτ ( − sin 2 πτ ) dτ = − integraldisplay ∞ −∞ sin 2 2 πτ dτ , 1 and this integral is infinite. The only thing we can say in general is that if f is periodic and if f ∗ g exists (i.e. if the integral converges) then f ∗ g is periodic with the same period as f . 2. (10 points) Equivalent width: Still another reciprocal relationship The equivalent width of a signal f ( t ), with f (0) negationslash = 0, is the width of a rectangle having height f (0) and area the same as under the graph of f ( t ). Thus W f = 1 f (0) integraldisplay ∞ −∞ f ( t ) dt. This is a measure for how spread out a signal is. Show that W f W F f = 1. Thus, the equivalent widths of a signal and its Fourier transform are reciprocal. From the Internet Encyclopedia of Science: Equivalent width A measure of the strength of a spectral line. On a plot of intensity against wavelength, a spectral line appears as a curve with a shape defined by the line profile. The equivalent width is the width of a rectangle centered on a spectral line that, on a plot of intensity against wavelength, has the same area as the line.as the line....

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