PS-6-2007-Solutions

PS-6-2007-Solutions - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2007 Solutions to Problem Set 6 1. (20 points) Nyquist rate. The signal f ( t ) has the Fourier transform F ( s ) as shown below. 0 B1 B2 -B1 -B2 s F ( s ) The Nyquist frequency is 2 B 2 since the highest frequency in the signal is B 2 . The Sampling Theorem tells us that if we sample above the Nyquist rate, no aliasing will occur. Is it possible, however, to sample at a lower frequency in this case and not get aliasing effects? If it is possible, then explain how it can be done and specify at least one range of valid sampling frequencies below the Nyquist rate that will not result in aliasing. If it is not possible, explain why not. Solution: The idea here is to somehow exploit the gap between B 1 and B 1 . First of all, let us consider what happens when we sample at the Nyquist rate. The spectrum gets replicated at integer multiples of the sampling rate, i.e. ,2 B 2 as shown below: s B 1 B 2 2 B 2 2 B 2 B 2 B 1 Now, what happens as we sample a little bit below the Nyquist rate? Well, the spectrum still gets replicated at integer multiples of the sampling rate, but there is now overlap, i.e. , aliasing: 1
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B 1 B 2 2 B 2 B 2 B 1 s 2 B 2 However, what happens as we decrease the sampling frequency even further? If B 2 2 B 1 , then we clearly will continue to have overlap for all sampling rates all the way down to 0. A simple way to think about it is that the gap between the two spectral islands is simply not big enough to contain another two spectral islands of the same width. Therefore, in that case, the Nyquist rate is indeed the slowest sampling rate allowed for perfect reconstruction. However, if B 2 < 2 B 1 , then we can find a sampling rate lower than the Nyquist frequency and still avoid aliasing. For example, consider the case when the sampling rate is 2 B 1 : B 1 B 2 2 B 2 B 2 B 1 s 2 B 2 This can continue until the sampling frequency reaches B 2 and we are on the verge of yet another overlap: B 1 B 2 2 B 2 B 2 B 1 s 2 B 2 So, clearly one possible range for f s is B 2 <f s < 2 B 1 . Note that depending on what B 2 ,B 1 ultimately are, we may be able to exploit the gap further. 2. (20 points) Natural sampling. Suppose the signal f ( t ) is band-limited with F f ( s )=0fo r | s |≥ B . Instead of sampling with a train of δ ’s we sample f ( t )w ithatra ino fverynarrow pulses. The pulse is given by a function p ( t ), we sample at a rate T , and the sampled signal 2
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then has the form g ( t )= f ( t ) ± ² k = −∞ Tp ( t kT ) ³ (a) Is it possible to recover the original signal f from the signal g ? (b) If not, why not. If it is possible, what conditions on the parameters T and B ,andon the pulse p ( x ) make it possible. Solution: It’s generally easier to think about sampling in the frequency domain, so we’ll begin by finding the Fourier transform of g ( t ). The trick to this problem is noting that our natural sampling function is periodization of the pulse and can be expressed as a convolution of the pulse with a Shah function.
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PS-6-2007-Solutions - EE 261 The Fourier Transform and its...

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