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EE235 Signals and Systems University of Washington Fall 2004 Dept. of Electrical Engineering Final Exam : Tuesday, December 14, 2004 Your Name: Cornell Theodore Samuel Peterson Student ID: 3.141592653589793238462643383 Problem # 1 (10 points) Recall the notion of duality, that if you do A in the time domain, and B happens in frequency, then if you do B in the time domain, then something similar to A will happen in frequency. Using duality, ﬁnd the Fourier transform of the following signal: 1 y(t) = a + jt Explain and justify each step very clearly, and indicate how and where duality is used. Note that: 1 y(t) = = F (t) a + jt where 1 F (w) = a + jω But we know that we have the FT pair f (t) ←→ F (ω), where f (t) = 2πe−at u(t). Therefore, we use form of duality that says that if x(t) ←→ X(w), then X(t) ←→ 2πx(−ω), and we get that Y (ω) = 2πx(−ω) = 2πeaω u(−ω) Problem # 2 (16 points) In class, we described the sampling problem where a signal x(t) is multiplied by a periodic impulse train which corresponds to convolving X(ω) by a periodic impulse train in the frequency domain. In this problem, you are given a signal x(t) that is bandlimited so that X(ω) = 0 for |ω| > ωB . You multiply x(t) by a periodic square wave p(t) with period T = 2π/ω0 where ω0 = 2ωB + ǫ, where ǫ is a tiny positive but non-zero value, just large enough to make ω0 > 2ωB . Speciﬁcally, one period of p(t) is deﬁned as: 1, |t| < T /4 p(t) = 0, T /4 < |t| < T /2 p(t) 1 -2T -T -T/2 -T/4 0 T/4 T/2 T -2T t The result you get is the signal y(t) = x(t)p(t). (2a) (6pts) Clearly and concisely explain, and show in pictures what is happening in the frequency domain. Specif- ically, with your explanation, plot a possible spectrum of X(w), the spectrum of p(t), and the resulting spectrum of y(t). 1 of 11 2 of 11 Forming y(t) = x(t)p(t) is multiplication in time, which means convolution in frequency, but we convolve with an impulse train since p(t) is periodic. To get the coefﬁcients of the impulse train in frequency, we need the Fourier series, speciﬁcally: T /2 T /4 1 T /4 1 1 T −T /4 dt = 1/2, k=0 Ck = p(t)e−jkω0 t dt = e−jkω0 t dt = T /4 T −T /2 T −T /4 1 T −T /4 e −jkω0 t dt = sin(k/2) , kπ k=0 This as you know is a sampled sinc function, and we convolve this with the spectrum X(ω). This is shown in the following ﬁgure where the spectrum X(ω) is convolved with an impulse train located at multiples of ω0 , with heights of 2π times the Fourier series coefﬁcients as shown in the central plot. X(ω) 0.5 0.4 Ck X(ω) 0.3 0.2 0.1 −3ω0 3ω0 0 −ωΒ ωΒ ω -0.1 −5ω0 −ω0 ω0 5ω0 -0.2 -6 -4 -2 0 2 4 6 k (2b) (3pts) Is it possible to perfectly reconstruct x(t) given only y(t) and knowledge of T ? If not, precisely explain why not. If so, show using both diagrams and equations how you can get x(t) back from y(t). Show this precisely in the frequency domain. You may use ideal ﬁlters. In the above resulting frequency domain ﬁgure for Y (ω), we just take an ideal low-pass ﬁlter centered at zero and that has a bandwidth equal to ωB . The gain of the ﬁlter must counteract the effect of C0 , so we take the gain of the ﬁlter to be 1/C0 = 2. (2c) (4pts) Notice that y(t) has half of x(t) cut off because half of the time p(t) is zero. Argue intuitively from a time- domain point of view why it is possible to essentially remove half of x(t), but then still be able to fully reconstruct x(t). This is similar to sampling. When we sample a band-limited time signal x(t) with an impulse train, and saw in class how the band-limiting of x(t) means that nothing interesting is going to happen in x(t) between the sample points. I.e., we don’t need to know the values of x(t) between samples since the band-limited nature of x(t) means that x(t) is everywhere smooth, so we don’t need to know those regions. Here, we are multiplying x(t) by a square wave and the same argument applies. Basically, there is nothing new or interesting taking place in x(t) during the portions that it is being multiplied by the zero part of p(t). In fact, multiplying x(t) by this square wave retains even more of the temporal portion of x(t) than does multiplying x(t) by a time impulse train, so it is not surprising that we have full recoverability here as well. (2d) (3pts) Would this be a good sample scheme for say using in CD player? Fully explain why or why not? This would not be a good scheme to use in a CD player. The point of sampling is to be able to store a continuous time signal on a computer (and store it say in an array). By sampling with an impulse train, we take snapshots of the signal x(t). By multiplying x(t) by a square wave, as we do here, however, we are not taking snapshots since x(t) can vary within the portion of p(t) that is not zero. This doesn’t help us store x(t) on a computer since we are really interested in only one value per unit time (sample) rather than lots of values for each time unit. Problem # 3 (14 points) You are given the following impulse response: 1/R, −R/2 < t < R/2 h(t) = 0, else which can be used in a system y(t) = h(t) ∗ x(t) with input x(t) and output y(t). 3 of 11 (3a) (2pts) Is this system LTI? Clearly and concisely explain why or why not. The system is expressed by the convolution of the input with an impulse response, so it is clearly LTI since all LTI systems can be characterized in this way. (3b) (6pts) What kind of ﬁlter is this? In particular, is it more like a low-pass, a high-pass, or a band-pass ﬁlter? Concisely, clearly, and convincingly argue from both the time domain and frequency domain point of view. Answer 1: time domain argument: h(t) is a moving average window, and when it is convolved with x(t), it take the average value of x(t) over a window of width R. In other words, you can look at the convolution integral as: t+R/2 ∞ 1 y(t) = h(t − τ )x(τ )dτ = x(τ )dτ −∞ R t−R/2 which is the average value of x(t) in a window of width R centered at t. Such an average will tend to smooth out any high-frequency variation, but any slow variation (especially that which varies much slower than the window width) will not be effected by this ﬁlter. Answer 2: frequency domain argument: The Fourier transform of h(t) is a sinc function in frequency, and a sinc function in frequency is more like a low-pass ﬁlter than any of the others. We can see this by looking at the magnitude of the sync function, which has most of the energy near zero frequency (ω = 0), and has side lobes that decrease as ω increases. (3c) (6pts) The impulse response is clearly not causal and so is not realizable. How would you make it causal (and thus realizable)? Deﬁne h1 (t) as your modiﬁed causal (and realizable) version of h(t). How would you compare the Fourier transform of h(t) with your realizable modiﬁcation h1 (t)? All we need to do is to delay h(t) by R/2. Lets do this by convolving with a delta, i.e., h1 (t) = h(t) ∗ δ(t − R/2). Now h1 (t) is causal (and therefore realizable). This delay in time corresponds to just multiplying by a complex exponential in the frequency domain, i.e.: h1 (t) ←→ H1 (ω) = H(ω)e−jωR/2 So the magnitude has not changed (|H1 (ω)| = |H(ω)|) and the phase has only changed by a linear difference −ωR/2. Problem # 4 (14 points) Consider a set of real signals indicated by x(t), and each with a Laplace transform which is a ratio of polynomials in s, and all having the same pole/zero plot as follows: s-plane j -3 -2 1 -j (4a) (7pts) How many possible regions of convergence could this plot correspond to? Clearly and neatly sketch all of these regions of convergence. 4 of 11 Since there are three vertical sets of poles in the S-plane, then there are 4 possible regions of convergence. These are: ROC1) Re(s) < −3 ROC2) −3 < Re(s) < −2 ROC3) −2 < Re(s) < 1 ROC4) 1 < Re(s) (4b) (7pts) How many possible time signals could correspond to this pole/zero plot? Clearly and neatly write down the equation for each of these time signals, giving a very brief explanation as to why each one is valid. Indicate if it is a left, right, or two-sided signal. Note: you do NOT need to explicitly compute any constant multipliers in these time signals, meaning it is ﬁne to write the signals as a sum of amplitude scaled time signals, where the amplitude scalings are non-computed constants. Hint: you do not need to compute the inverse Laplace transform to solve this problem. Since there are 4 ROCs, there are 4 signals, each one corresponding to a different ROC. Basically, for each of the poles given in the ﬁgure, there is a signal that takes the form: xi (t) = βi e−αi t u(βi t) where βi = ±1, and αi is an appropriate value. Because x(t) is real, if αi is real, it stands alone, and if it is complex, then there is another αj that is complex conjugate of αi . βi determines if the signal is right sided (βi = 1) or left sided (βi = −1), and αi determines the vertical position of the pole (the horizontal position is at s = −α). Since i αi 1 α1 = −3 there are 4 poles, we have 4 such signals. Furthermore, we know that the signals are such that: 2 α2 = −2 + j . 3 α3 = −2 − j 4 α4 = 1 In other words, each x(t) signal corresponding to each of the ROCs above has the form of a sum of these basic signals: 4 xj (t) = Ai βi e−αi t u(βi t) i=1 using the α values given in the table and for some constants Ai (which you do not need to compute in this problem). The only thing you need to worry about is which xi (t) is right sided (βi = 1) and which one is left sided (βi = −1) to get each ROC. We have the following results: ROC sidedness βi ROC1) left-sided β1 = −1, β2 = −1, β3 = −1, β4 = −1 ROC2) two-sided β1 = +1, β2 = −1, β3 = −1, β4 = −1 ROC3) two-sided β1 = +1, β2 = +1, β3 = +1, β4 = −1 ROC4) right-sided β1 = +1, β2 = +1, β3 = +1, β4 = +1 Problem # 5 (12 points) Two signals x1 (t) and x2 (t) are multiplied, yielding a new signal y(t) = x1 (t)x2 (t). The spectra of the two signals are as shown below: 5 of 11 The resulting signal y(t) is then sampled using an impulse train of period T (frequency ωs ). Determine minimum sampling frequency ωs such that the signal y(t) may be reconstructed from its samples by using an ideal low-pass ﬁlter. Also determine the speciﬁcations (cut-off frequency and magnitude) of the ﬁlter. Recall that multiplication of signals in the time domain is the same as convolution in the frequency domain. Thus, ∞ Y (ω) = −∞ X1 (τ )X2 (ω − τ )dτ . As the signals are bandlimited, the convolution will lie between −(ωA + ωB ) and (ωA + ωB ). Thus from Nyquist Theorem in order to avoid aliasing we have ωs ≥ 2(ωA + ωB ) and hence the minimum sampling frequency is ωs = 2(ωA + ωB ). The signal may be recovered by passing the sampled signal 1 through an ideal low-pass ﬁlter with cut-off frequency 2(ωA + ωB ) and magnitude ωs (to undo the amplitude effect of the sampling). Problem # 6 (13 points) (6a) (4pts) Compute the Fourier Transform of f (t) = 4u(t) − 2u(t − 1) − 2u(t − 2) Based on the result (i.e.,F (ω)) can you verify the claim “a signal that is time-limited has inﬁnite bandwidth”. 1 2 2e−jωt dt= 4−2e −jω −2e−j2ω F (ω) = 0 4e−jωt dt + 1 jω It can be seen that F (ω) has inﬁnite bandwidth. (6b) (4pts) Compute the Inverse Fourier Transform of F (ω) = cos(ω) u(t + π ) − u(t − π ) . Based on the result 2 2 can you verify the claim “a signal that is band-limited is inﬁnite in time”. Note: A signal is said to be inﬁnite in time if there does not exist a to such that, 0, for all t > to and to < ∞; f (t) = 0, for all t < to and to > −∞; π f (t) = 2π − π cos(ω)ejωt dω = π(1−t2 ) cos( πt ) It can be seen that f (t) is inﬁnite in time as it is a periodic signal 1 2 1 2 2 multiplied by another signal that is also not ﬁnite. (6c) (5pts) Can a signal be inﬁnite in both time and frequency domains? Clearly explain why this is possible/not possible by giving an example. No credit will be given for a simple “yes/no”. Once again note that a signal is said to be inﬁnite in time if there does not exist a to such that, 0, for all t > to and to < ∞; f (t) = 0, for all t < to and to > −∞; A signal is inﬁnite in the frequency domain, if it has inﬁnite bandwidth. 6 of 11 Yes, it is possible. Any periodic signal is inﬁnite in both time and frequency domains. Recall that the Fourier transform of any periodic signal f (t) is given by F (ω) = 2πCk δ(ω − kωs ) (0.1) all k where, Ck are the Fourier series coefﬁcients of f (t). Thus, any periodic signal is inﬁnite in both time and frequency domains. Problem # 7 (10 points) You are given a system with input x(t) and output y(t), and with the input/output relationship expressed by the following: y(t) ∗ z(t) = x(t) In other words, the input and output relationship is expressed by saying that when you convolve the output y(t) with some signal z(t), then you get the input x(t). (7a) (5pts) Assuming that this system is invertible, is it LTI? Clearly explain why or why not? The system is expressing the input/output relationship in terms of convolution it is just that the input and output have switched places. Since we are given that the system is invertible, we can look at this in the frequency domain to get some help. Speciﬁcally, the above implies that Y (ω)Z(ω) = X(ω) or Y (ω) = X(ω)/Z(ω) and we can deﬁne H(ω) = 1/Z(ω), and if we take the inverse FT, this means that there is some h(t) such that h(t) ∗ z(t) = δ(t). Therefore, if we convolve both sides of the equation with this h(t), we will get: (y(t) ∗ z(t)) ∗ h(t) = x(t) ∗ h(t) or y(t) = x(t) ∗ h(t) which is the standard convolution form for an LTI system. Therefore, the system is LTI. (7b) (5pts) Regardless of your answer in part (a), come up with an expression in both the time and frequency domains for the relationship of the inverse system. Part (a) already gave the expression in both the time and frequency domain. Problem # 8 (11 points) Consider the following parallel system with input x(t), sent to two LTI systems in parallel h1 (t), and h2 (t), and the results are combined by addition to produce output y(t). h(t) 1 x(t) + y ( t) h(t) 2 You are given that x(t) = y(t), and that x(t) is a non bandlimited signal (i.e., there is no frequency value ω above which x(t) does not have energy). 7 of 11 You are also given that H1 (ω) is is a low-pass ideal ﬁlter with unity gain, and cutoff frequency ωc , meaning: 1, |ω| < ωc H1 (ω) = 0, |ω| > ωc (8a) (5pts) Give an example of an x(t) that satisﬁes the properties given above. One simple example is for x(t) = δ(t) which is an impulse and has inﬁnite spectral extent. (8b) (6pts) Come up with an expression for either h2 (t) a time-domain answer or for H2 (ω), a frequency domain answer. Clearly and concisely explain why this expression is valid. The picture above depicts the following LTI system: y(t) = x(t) ∗ (h1 (t) + h2 (t)) = x(t) ∗ h(t) where h(t) = h1 (t) + h2 (t) is the resulting impulse response of the combined system. Regardless of what you choose for x(t), one simple way of making sure that y(t) = x(t) is to choose h2 (t) = δ(t) − h1 (t), so that h2 (t) is an ideal high-pass ﬁlter, with Fourier transform: 1, |ω| > ωc H1 (ω) = 0, |ω| < ωc Problem # 9 Extra credit. (p = 0.5/X(0) + 5 points, where X(s) = ∞ −∞ e e u(t)dt, −st −αt and α is the answer to this question.) The answer to this question is α, where α = −2 ∗ (p − 5), where p is the point worth of this problem, and where X(s) is as given in the deﬁnition of the point worth of this problem. Your task is to ﬁnd α. To get the point worth of the problem p = −α/2 + 5, you need to compute α, but to compute α you need to compute α = −2 ∗ (p − 5) = −1/X(0). Therefore, we get that: ∞ α = −1/X(0) = −1/ e−αt u(t)dt 0 This says that if α > 0 so that the integral converges to something ﬁnite and positive, then this implies that α < 0, so we can’t have that α > 0. If, on the other hand, α < 0 so that the integral is inﬁnite, then α = 0, so we can’t have that α < 0. If α = 0, then the integral is again inﬁnite but −1/∞ = 0, so α = 0 is a logically consistent solution, so the point worth of this problem is p = 0/2 + 5 = 5 points. To get the 5 extra credit points, you need to correctly compute how many extra credit points you will get if you get the problem correct.