EE Homework 13 - Solutions

Transcription

1 EE Homework 3 - Solutions. (a) The Laplace transform of e t u(t) is s+. The pole of the Laplace transform is at which lies in the left half plane. Hence, the Fourier transform is simply the Laplace transform evaluated at s = jω, i.e., the Fourier transform of e t u(t) is jω+. (b) In the lecture notes on Fourier transform, it was shown that the Fourier transform of e jω 0t is 2πδ(ω ω 0 ). Since sin(t) = 2j [ejt e jt ], the Fourier transform of sin(t) is 2j [2πδ(ω ) 2πδ(ω+)] which is equal to π j [δ(ω ) δ(ω + )]. (c) The Fourier transform of x(t) = e t sin(t)u(t) can be found by either substituting sin(t) = 2j [ejt e jt ] or by using the fact that e t sin(t)u(t) is the product of e t u(t) and sin(t). Both methods are shown below: Method : Since sin(t) = 2j [ejt e jt ], we can write e t sin(t)u(t) = 2j [e(j )t e (j+)t ]. () Therefore, the Laplace transform of e t sin(t)u(t) is X(s) = [ ] 2j s (j ) = s + (j + ) s 2 + 2s + 2. (2) The poles of X(s) are +j and j which are in the left half plane. Hence, the Fourier transform of x(t) is simply the Laplace transform evaluated at s = jω. Therefore, the Fourier transform of x(t) is X f (ω) = ω 2 + 2jω + 2. (3) Method 2: Since x(t) = e t sin(t)u(t) is the product of e t u(t) and sin(t), the Fourier transform of e t sin(t)u(t) is 2π times the convolution of the Fourier transforms of e t u(t) and sin(t). Therefore, X f (ω) = 2π jω + π [δ(ω ) δ(ω + )] j = [ ] 2j j(ω ) + = j(ω + ) + ω 2 +2jω+2.(4)

2 (d) The signal x(t) = sin(t)u(t) is the product of sin(t) and u(t). Hence, the Fourier transform of sin(t)u(t) should be 2π times the convolution of the Fourier transforms of sin(t) and u(t). Recalling that the Fourier transform of u(t) is [ jω + πδ(ω)], the Fourier transform of sin(t)u(t) is X f (ω) = π [δ(ω ) δ(ω + )] [ 2π j jω + πδ(ω)] = [ ] 2j j(ω ) j(ω + ) + πδ(ω ) πδ(ω + ) = ω 2 + π [δ(ω ) δ(ω + )]. (5) 2j 2. Consider g(t) = δ(t). Then the Fourier transform of g(t) is f(ω) =. Therefore, by the duality theorem, the Fourier transform of f(t) = is 2πg( ω) = 2πδ( ω) = 2πδ(ω). Hence, the inverse Fourier transform of δ(ω) is 2π. 3. By the sampling theorem, the sampling frequency should be bigger than twice the maximum frequency in the signal to be able to reconstruct the signal from the sampled data. In this case, the signal to be sampled is sin(00πt) which has frequency 50 Hz. Therefore, to ensure that we can recover the original signal back from the sampled data, we have to sample faster than 00 Hz, i.e., 00 samples per second. To reconstruct the original signal from the sampled data, we can use a low pass filter with cutoff frequency a little over 50 Hz. Specifically, the cutoff frequency of the low pass filter should be bigger than 50 Hz to grab the original signal (which has frequency 50 Hz), but should be small enough to remove all the copies of frequency content introduced due to sampling, i.e., if we have sampled at f Hz, then the cutoff frequency of the low pass filter should be smaller than (f 50) Hz to ensure that we remove the copies of the frequency content. Hence, if we sample at 0 Hz, then the cutoff frequency of the low pass filter should be between 50 Hz and 5 Hz. 4. This problem illustrates the phenomenon of aliasing. You can understand aliasing from two viewpoints: Firstly, if we sample a signal too slowly (i.e., if the sampling frequency is less than twice the maximum frequency in the input signal), then when 2

3 we look at the sampled data, the signal appears to have a lower frequency than the actual frequency of the signal. This introduction of apparent lower frequencies is called aliasing. Secondly, in the context of reconstructing the original signal from the sampled data, the phenomenon of aliasing means that we cannot uniquely reconstruct the original signal from the sampled data unless we are told that the input signal does not have frequencies above half the sampling frequency. In other words, if we are only given the sampled data, then there are many possibilities for the original signal. However, if we are given the additional data that the highest frequency in the input signal is less than half the sampling frequency, then we can find the original signal uniquely. This corresponds to the fact that if we only have the sampled data, then there are infinitely many ways to connect the dots (Figure 2). In this problem, we are given that the signal x(t) was sampled at 0 Hz and that the sampled data is x(0.n) = sin(3n), i.e., we know that at time 0.n, the value of the signal is sin(3n). The given data is shown in Figure. The question is to find x(t). Since we are not given data on the maximum frequency in the input signal, we cannot uniquely find out what the original signal x(t) was. In fact, x(t) could have been any one of an infinite set of possibilities as shown below. Since the sampled data looks like a sinusoid, we can guess that x(t) is a sinusoid, i.e., x(t) = sin(2πf t) with f being the frequency. Since we know that the value of the signal at time t = 0.n is sin(3n), we should have 2πf(0.n) = 3n + 2k n π (6) where k n are constant integers. Therefore, f = 30 2π + 0k n n (7) Since f is a constant (i.e., does not vary with the sampling instant n), the ratio kn n should be a constant integer. Denoting kn n = c, we get f = c. (8) 2π 3

4 Hence, the original signal could have been x(t) = sin(2πft) = sin(30t + 20cπt) with c being any integer. In other words, any signal of the form sin(30t+20cπt), when sampled at 0 Hz gives the same sampled data sin(3n). Therefore, given just the sampled data, it is not possible to find out which one of the possibilities the original signal x(t) was. Specifically, x(t) could have been sin(30t) or sin(30t + 20πt) = sin(92.839t) or sin(30t + 40πt) = sin( t) or sin(30t 20πt) = sin( t) or sin(30t 40πt) = sin( t) or any other of the infinite set of posibilities sin(30t + 20cπt). This is illustrated in Figure 2. In fact, since each one of sin(30t + 20cπt) gives the same sampled data, any linear combination of these possibilities of the form c= a c sin(30t + 20cπt) with a c being constants such that c= a c = would also give the same sampled data. Furthermore, x(t) might contain an additional component x(t) such that its values at time instants 0.n are all zero, i.e., x(t) might have been of the form x(t) = x(t) + c= a c sin(30t + 20cπt) (9) where x(t) is any signal such that x(0.n) = 0 for all n. It is clear that there are an infinite number of such signals x(t) since it can take any values in the intervals between the sampling instants. All the possibilities for the original signal x(t) which might have resulted in the observed sampled data x(0.n) = sin(3n) are given by (9). To recapitulate, when we are given just the sampled data, we do not know which one of the possibilities in (9) the original signal x(t) was. If we were given the additional data that the maximum frequency in x(t) is less than half the sampling frequency, i.e., if the maximum frequency in x(t) was given to be less than 5 Hz, then we can be sure that the original signal was sin(30t) since all other possibilities in (9) have components at frequencies higher than 5 Hz. 4

5 x(t) t Figure : The given sampled data y[n] = x(0.n) = sin(3n). 5

6 x(t) t Figure 2: A few possibilities for the original input signal. Green: sin(30t), Yellow: sin(92.839t), Blue: sin( t), Black: sin( t), Cyan: sin( t). The given sampled data y[n] = x(0.n) = sin(3n) is shown by the red * s. 6