EE 261 The Fourier Transform and its Applications Fall 2007 Problem Set Eight Solutions

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1 EE 6 he Fourier ransform and its Applications Fall 7 Problem Set Eight Solutions. points) A rue Story: Professor Osgood and a graduate student were working on a discrete form of the sampling theorem. his included looking at the DF of the discrete rect function {, n N f[n] 4, N + n < N 4, N 4 < n N he grad student, ever eager, said Let me work this out. A short time later the student came back saying I took a particular value of N and I plotted the DF using MALAB their FF routine). Here are plots of the real part and the imaginary part. 8 6 real part of Fk) k imaginary part of Fk) k a) Produce these figures.

2 Professor Osgood said, hat can t be correct. b) Is Professor Osgood right to object? If so, what is the basis of his objection, and produce the correct plot. If not, explain why the student is correct. Solution he student executed the following MALAB command. f [ ]; F fftshiftfftfftshiftf))); DF in MALAB is done assuming DC at the first component of the array. herefore the command fftshift is used to alter the center of the input array. here is a fine detail here that is worth noting. When the array size is N, fftshift shifts the ceiln/)+ component to be the first component of the array. herefore, to obtain the desired result, the rect function should be centered around the ceiln/)+ component. he following command will give F without any imaginary component. f [ ]; F fftshiftfftfftshiftf))); Here s the correct plot:

3 8 6 real part of Fk) k imaginary part of Fk) k. points) Linearity and time-invariance. State whether the following systems are linear or non-linear, time-invariant or time-variant, and why. Assume that vt) is the input and wt) is the output for all systems. No credit will be given for answers without explanations or proofs! a) wt) vt)cosωt) b) wt) sinvt)) c) wt) vτ)e πitτ dτ d) wt) d dt vt) e) wt) cosωt + vt)) Solution: In each case, we need to use the definition of linearity and time-invariance to prove or disprove the claim. A system is linear if {α v t) + α v t)} α {v t)} + α {v t)}. Given that wt) {vt)}, a system is time-invariant if wt τ) {vt τ)}. a) Linearity: α v t) + α v t))cosωt) α v t)cosωt) + α v t)cosωt), so the system is clearly linear. ime-invariance: vt τ) cosωt) wt τ), so the system is time-variant. b) Linearity: sinα v t)+α v t)) α sinv t))+α sinv t)), so the system is non-linear. ime-invariance: sinvt τ)) wt τ), so the system is time-invariant. 3

4 c) Linearity: α v τ)+α v τ))e πitτ dτ α v τ)e πitτ dτ+α v τ)e πitτ dτ, so the system is linear. ime-invariance: vτ α)e πitτ dτ wt α), so the system is time-variant. d d) Linearity: dt α d v t) + α v t)) α dt v d t) + α dt v t), so the system is linear. d ime-invariance: dtvt τ) wt τ), so the system is time-invariant. e) Linearity: cosωt + α v t) + α v t)) α cosωt + v t)) + α cosωt + v t)), so the system is non-linear. ime-invariance: cosωt + vt τ)) wt τ), so the system is time-variant points) Zero-order hold. he music on your CD has been sampled at the rate 44. khz. his sampling rate comes from the Sampling heorem together with experimental observations that your ear cannot respond to sounds with frequencies above about khz. he precise value 44. khz comes from the technical specs of the earlier audio tape machines that were used when CDs were first geting started.) A problem with reconstructing the original music from samples is that interpolation based on the sinc function is not physically realizable for one thing, the sinc function is not timelimited. Cheap CD players use what is known as zero-order hold. his means that the value of a given sample is held until the next sample is read, at which point that sample value is held, and so on. Suppose the input is represented by a train of δ-functions, spaced /44. msec apart with strengths determined by the sampled values of the music, and the output looks like a staircase function. he system for carrying out zero-order hold then looks like the diagram, below. he scales on the axes are the same for both the input and the output.) zero-order hold a) Is this a linear system? Is it time invariant for shifts of integer multiples of the sampling period? b) Find the impulse response for this system. c) Find the transfer function. Solution: 4

5 a) Say the music is represented by a signal ut). We sample ut) with time spacing and so the input can be written as vt) un)δt n). he output, wt), takes the value un) and holds it for seconds, then takes the value un + )) and holds it for seconds, and so on. We can express this as follows. Start with the rect function of width, that is Π t), and center it on the interval from n to n + ), that is Π t n + ) ). his shifted rect looks like n n + ) n + ) t hen the output is wt) Lvt) un)π t n + ) ). From this expression we can easily see that the system is linear, for Lv t) + v t)) u n) + u n))π t n + ) ) Lv t) + Lv t) u n)π t n + ) ) + u n)π t n + ) ) Similarly, Lαvt)) αlvt). Next, suppose we shift the time by an integer multiple of, say m. hen the input is vt m) un)δt m n) k un)δt n + m)) now let k m + n to write the sum as) uk m))δt k) 5

6 If wt) Lvt) then the output resulting from the shift is Lvt m)) uk m))π t k + ) ) k now, undoing what we did before, let n k m to write the sum as) un)π t n + m + ) ) wt m) un)π t n + ) ) m hus the system is time invariant for shifts by integer multiples of. b) In order to determine the impulse response, we input the delta function, δt), and have to determine the output. he input is concentrated at t with strength and is thereafter. So, starting at t, the output holds the value for seconds and is then for t. he output is also for t <. In other words, the impulse response is {, t < ht), otherwise. c) We can now calculate the transfer function from the obtained impulse response. his is Hs) Fhs) e πist πis e πist dt ] e πis πis πis sinπs e πs e πis sincs). ht)e πist dt 4. points) Let wt) Lvt) be an LI system with impulse response ht); thus wt) h v)t). a) If ht) is bandlimited show that wt) Lvt) is bandlimited regardless of whether vt) is bandlimited. b) Suppose vt), ht), and wt) Lvt) are all bandlimited with spectrum contained in / < s < /. According to the sampling theorem we can write vt) vk) sinct k) ht) hk) sinct k) wt) wk) sinct k) k k k Express the sample values wk) in terms of the sample values of vt) and ht). You will need to consider the convolution of shifted sincs.) 6

7 Solution: a) ake the Fourier transform of w h v to obtain Fw Fh)Fv). From here we see that the spectrum of w can extend no farther than the spectrum of h, and hence if h is bandlimited so too is w whether v is or not. b) We work in the time domain and consider the convolution m wm)sinct m) wt) h v)t) k k, hk)sinct k) ) hk)vn)sinct k) sinct n) vn)sinct n) We had to use different indices on the sums on the right to combine them in this way. You ll see in a moment why we used still another index for the sum on the left. Now let s work with the convolution of the shifted sincs. aking the Fourier transform we have Fsinct k) sinct n)) e πiks Πs)e πins Πs) e πik+n) Π s) e πik+n) Πs), since Π Π. And now taking the inverse Fourier transform we get hus m sinct k) sinct n) sinct k + n)). wm)sinct m) k, hk)vn)sinct k + n)) ake a particular m and let t m. hen the left hand side is just wm ) so we have wm ) k, hk)vn)sincm k + n)). Next, sincm k + n)) will be zero except when k + n m, in which case it s. So the double sum on the right hand side reduces to a sum over indices k and n for which k + n m. Or, writing n m k, we can express the result as wm ) k hk)vm k) ) 7

8 Since this holds for all m let s just write this as wm) k hk)vm k). Yep. he values of the output wt) at the sample points are obtained by taking the convolutions of the infinite) sequences of the sample points of the impulse response and the input. 5. points) Let a > and define the continuous) running average of a signal ft) to be Lft) a t+a t a fx)dx a) Show that L is an LI system and find its impulse response and transfer function. b) What happens to the system as a? Justify your answer based on properties of the impulse response. c) Find the impulse response and the transfer function for the cascaded system M LLf). Simplify your answer as much as possible, i.e., don t simply express the impulse response as a convolution. d) hree signals are plotted approximately) below. One is an input signal ft). One is Lft) and one is Mft). Which is which, and why? i) ii) iii) Solution: a) First let s show that L is linear: Lα f t) + α f t)) a α a t+a t a t+a t a α f x) + α f x))dx f x)dx + α a α Lf t)) + α Lf t)) t+a t a f x)dx 8

9 Hence, L is linear. Now let s show that L is time-invariant: Lτ t ft)) a a t+a t a t t )+a t t ) a τ t Lft)) fx t )dx fy)dy where τ t is the shift-operator, i.e., τ t ft) ft t ). Hence, we have shown that L is indeed time-invariant. If we define h a t) a Π t a ), then clearly Lft)) h at) ft). Hence, the impulse response of the system is given by ht), and the transfer function is simply Hs) sincas). b) If we let a, then the impulse response function becomes: ht) lim a a Π t a ) δt) In the frequency domain, the transfer function becomes: which makes perfect sense. Hs) lim a sincas) c) Note that Mt) LLft))) h a t) h a t) ft)) h a t) h a t)) ft). Hence, the impulse response of M is given by: h a t) h a t) a Π t ) a ) a Π t ) a ) a t a ) and its transfer function is given by Hs) sinc as). d) From part b), we know that the operator Lft)) atenuates high frequencies smoothing operator). It s reasonable to think that the third plot corresponds to ft), the second plot to Lft)) and the first plot to Mft)) LLft)))). Let s show this more formally. Let ft) ut) the rect function). From part b), we know that Lft)) ) t a ut) Π a ) t τ Π dτ a a,t a + t a, a < t a,t > a which is piecewise linear on t. Similarly, it can be shown that Mft)) is piecewise quadratic on t. Hence, our guess was correct. 9

10 6. 5 points) Upsampling and Downsampling discrete signals: We are given a discrete-time signal x n, which was obtained by sampling a continuous-time xt) at frequency much higher than the Nyquist rate. Let x n xn s ). Sometimes for an application, we need the samples of the signal xt) at a different sampling rate, say α s, where α >. If α >, it is downsampling else it is upsampling. One way to obtain the sample points at a different sampling frequency, is to reconstruct the signal xt) which can be done as the original samples were obtained by sampling above the Nyquist rate) from the samples x n and then resample it at the new frequency. However, due to non-idealities of the D/A and the A/D converters, this approach is not desirable. In this question, we want to obtain the new sample points directly by using x n. a) Find the expression for w n xn M s )) in terms of x n, where M is an integer. b) Find the expression for y n x n s M )) in terms of xn, where M is an integer. If the resulting expression looks complicated, it will motivate the use of linear interpolation or nearest-neighbor interpolation as used in HW-5. c) Find the expression for z n xn α s )) in terms of x n, where α > is a rational number α P Q, where P and Q are integers. Give some insight into how would you accomplish a change in sampling frequency for any α > real number). Solution: a) w n x nm b) y n k sin π )) n km M x k π ) n km M c) For any rational P and Q, z n can be obtained by a combination of upsampling and downsampling as shown in parts a) and b). Depending upon the values of P and Q, we can decide if we first downsample or upsample. For any real number α, we can choose P and Q such that to approximate α closely.