Homework 8 Solutions

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1 EE264 Dec 3, 2004 Fall HO#27 Problem Interpolation (5 points) Homework 8 Solutions 30 points total Ω = 2π/T f(t) = sin( Ω 0 t) T f (t) DAC ˆf(t) interpolated output In this problem I ll use the notation from the lecture notes for the Fourier Transform. Preliminaries: The power of a sinusoid A sin(ω o t) or A cos(ω o t) is A 2 /2. The Fourier Transform (using Widrow s notation) of f(t) = Asin(Ω o t) is F(jω) = j2π Aδ(ω ω 2 o)+j2π Aδ(ω +ω 2 o). The power associated to a delta dirac Aδ(ω ω o ) in the frequency domain is A 2. The power 2π of a sinusoid can also be found by adding the power associated with each of its delta diracs in its Fourier Transform (Please confirm that to convince yourself). The delta function has the sifting porperty: f(ω)δ(ω ω o ) = f(ω o )δ(ω ω o ) Let H(jω) be the transfer function of the DAC. then we have that ˆF(jω) = F (jω)h(jω) The DAC Interpolator is a zero order hold with interpolating period T. Therefore its Fourier Transform is H(jω) = Te jωt/2sin(ωt/2) (ωt/2) f s (t) consists of a train of deltas obtained by sampling f(t) with a sampling period T. Therefore its Fourier Transform is F (jω) = T F(jω r2π/t) = T F(jω rω) Now, in particular we have f(t) = sin( Ω t), therefore 0 F(jω) = j2π 2 δ(ω Ω 0 ) + j2π 2 δ(ω + Ω 0 )

2 F (jω) will consist of replicas (scaled by /T ) of this pair of deltas centered at frequencies rω with r integer. That is F (jω) = r2π/t) = T F(jω j2π T 2 δ(ω Ω 0 rω) + j2π 2 δ(ω + Ω 0 rω) Try to make a picture of this, it s very useful. So we have concluded that F (jω) consists of deltas located at frequencies rω ± Ω/0 with r integer. The next step is to find ˆF(jω). ˆF(jω) = H(jω)F (jω) = H(jω) jπδ(ω Ω T 0 rω) + jπδ(ω + Ω 0 rω) ˆF(jω) = T jπh(jω)δ(ω Ω 0 rω) + jπh(jω)δ(ω + Ω 0 rω) Using the sifting property of the delta dirac function we have ˆF(jω) = jπ T H(j( Ω 0 + rω))δ(ω Ω 0 rω) + jπ T H(j( Ω 0 + rω))δ(ω + Ω 0 rω) Therefore we obtained that also ˆF(jω) consists of deltas located at frequencies rω ± Ω/0 with r integer. The power associated to one of these deltas, located at a frequency rω ± Ω/0 will be H(j(rΩ ± Ω 2T 0 )) 2 = ( H(j(rΩ ± Ω 2T 0 )) )2 For the zero order hold case we have (recall ΩT = 2π) ( 2T H(j(rΩ ± Ω 0 )) )2 = sin 2 (T(rΩ ± Ω )/2) 0 4 (T(rΩ ± Ω = sin 2 (rπ ± π ) 0 0 )/2))2 4 (rπ ± π 0 )2 The signal component in the interpolated output is the sinuoidal correponding to the deltas located at ±Ω/0. Therefore the power of the signal component in the interpolated output is: sin 2 ( π ) 0 4 ( π + sin 2 ( π ) 0 0 )2 4 ( π = 2 sin 2 ( π ) 0 0 )2 4 ( π = )2 The ripple consists of all the other undesired sinusoids correponding to the deltas located at frequencies different from ±Ω/0. The power of the ripple in the interpolated output in the vecinity of the sampling frequency correponds to the power of the sinusoid with frequencies Ω ± Ω/0 which corresponds to the power for the deltas at frequencies ±Ω ± Ω/0, that is, for the cases when r = ±. Therefore that power is sin 2 (π + π ) 0 4 (π + π + sin 2 ( π π ) 0 0 )2 4 ( π π + sin 2 ( π + π ) 0 0 )2 4 ( π + π + sin 2 (π π ) 0 0 )2 4 (π π = )2 2

3 (c) (5 points) Analogously, the ripple power in the vecinity of twice the sampling frequency, i.e., at frequencies 2Ω ± Ω/0 will be sin 2 (2π + π ) 0 4 (2π + π + sin 2 ( 2π π ) 0 0 )2 4 ( 2π π + sin 2 ( 2π + π ) 0 0 )2 4 ( 2π + π + sin 2 (2π π ) 0 0 )2 4 (2π π = )2 Problem 2 Interpolation (5 points) This problem is exactly the same as the previous problem, the only difference is that the interpolator is a First Order Hold, therefore we need to find what H(jω) is for this case and apply the same procedure as in the previous section. To find H(jω) we can observe that convolving two rectangular functions give us a triangular one, or we can just look at Fourier Transforms tables to get: H(jω) = T Hence, for the First Order Hold we have: ( sin(ωt/2) (ωt/2) ( 2T H(j(rΩ ± Ω 0 )) )2 = sin 4 (T(rΩ ± Ω )/2) 0 4 (T(rΩ ± Ω = sin 4 (rπ ± π ) 0 0 )/2))4 4 (rπ ± π 0 )4 The power of the signal component in the interpolated output is: sin 4 ( π ) 0 4 ( π + sin 4 ( π ) 0 0 )4 4 ( π = 2 sin 4 ( π ) 0 0 )4 4 ( π = )4 The power of the ripple in the interpolated output in the vecinity of the sampling frequency is sin 4 (π + π ) 0 4 (π + π + sin 4 ( π π ) 0 0 )4 4 ( π π + sin 4 ( π + π ) 0 0 )4 4 ( π + π + sin 4 (π π ) 0 0 )4 4 (π π = )4 (c) (5 points) The ripple power in the vecinity of twice the sampling frequency, i.e., at frequencies 2Ω ± Ω/0 will be sin 4 (2π + π ) 0 4 (2π + π + sin 4 ( 2π π ) 0 0 )4 4 ( 2π π + sin 4 ( 2π + π ) 0 0 )4 4 ( 2π + π + sin 4 (2π π ) 0 0 )4 4 (2π π = )4 3 ) 2

4 Problem 3 O&S 8. (20 points) x c (t) is obviously periodic since it consists of a sumation of periodic functions. For k 0, e j(2πkt/0 3) has period of 0 3 s. The period of x k c(t) is the minimum common multiple of the periods of the exponentials, therefore its period is 0 3 s. Now, x[n] is sampled with a sampling period of 0 3 s.,therefore, x[n + 6] = x 6 c( (n+6)0 3 ) = 6 x c ( n ) = x 6 c ( n0 3 ) = x[n], that is, 6 samples cover one period of x 6 c (t), which prduces the sequence to repeat every 6 samples. Therefore x[n] is periodic with period 6. Note: Since x[n] is periodic, we ll refer to it from now on as x[n]. The highest frequency component is at 2π9t/0 3 radians, the Nyquist rate is therefore 2π8t/0 3. The sampling frequency is 2π6/0 3 radians which is less than the Nyquist rate. Therefore T is not suficiently small to avoid aliasing. (c) (0 points) This problem can be done in two ways (i and ii below), the first one is applying directly the definition for the DFS coefficients and works the sumation expressions. The second one is a more insightful one, using the relation between the DTFT and the DFS. i) Using that we obtain X[k] = x c (t) = 6 k= 9 a k e j(2πkt/0 3 ) x[n]e j2πnk/6 = N 5 i= 9 x[n] = x c ( n0 3 ) = 6 a i e j(2πin/6) e j2πnk/6 = k= 9 { e j2πn(m) N N if m = ln for some integer l = 0 otherwise X[k] = 6 a 0 + a 6 + a 6 k = 0 a + a 7 + a 5 k = a 2 + a 8 + a 4 k = 2 a 3 + a 9 + a 3 + a 9 k = 3 a 4 + a 2 + a 8 k = 4 a 5 + a + a 7 k = 5 4 a k e j(2πkn/6) 5 a i i= 9 e j2πn(i k) 6

5 ii) First let s find the relation between the DTFT of x[n], X(e jω ), and its DSF coefficients. For reasons I ll describe further, the DTFT of a periodic sequence x[n] with period N is always of the form: X(e jω ) = k= b k δ(ω k2π/n) Note that since X(e jω ) is always periodic with period 2π, the sequence b k has to be periodic with period N. We can now express x[n] in terms of the b k s using the Inverse DTFT: x[n] = 2π ɛ 2π 0 ɛ = 2π = N 2π ɛ 0 ɛ X(e jω )e jωn dω = 2π ɛ 2π 0 ɛ N 2π b ke j2πkn N k= b k δ(ω k2π/n)e jωn dω = We can also express x[n] in term of its DFS coefficients: N b k δ(ω k2π/n)e jωn dω 2π ɛ b k δ(ω k2π/n)e jωn dω 2π 0 ɛ x[n] = N N X[k]e j2πn(m) N Comparing both expressions for x[n] we can easily observe that X[k] = N 2π b k Therefore we can obtain the DFS coefficients of x[n] by finding its DTFT X(e jω ). To do that, recall the result from example 2.24 from O&S (page 54). It states that Hence, x[n] = i= 9 x[n] = e jωon = X(e jω ) = a k e j(2πin/6) = X(e jω ) = X(e jω ) = i= 9 i= 9 2πδ(ω ω o + 2πr) 2πa k δ(ω 2πi/6 + 2πr) 2πa k δ(ω 2πi/6 + 2πr) 5

6 This last equation can be interpreted as the superposition of replicas of 9 i= 9 2πa k δ(ω 2πi/6) that are centered at frequencies 2πr. So we can rewrite X(e jω ) as: X(e jω ) = 2π [(a 0 + a 6 + a 6 )δ(ω + 2πr) + (a + a 7 + a 5 )δ(ω π/6 + 2πr)+ +(a 2 + a 8 + a 4 )δ(ω 2π/6 + 2πr) + (a 3 + a 9 + a 3 + a 9 )δ(ω 2π/6 + 3πr) + +(a 4 + a 2 + a 8 )δ(ω 4π/6 + 2πr) + (a 5 + a + a 7 )δ(ω 5π/6 + 2πr)] Therefore, in the range 0 ω < 2π we have: X(e jω ) = (a 0 + a 6 + a 6 )δ(ω + 2πr) + (a + a 7 + a 5 )δ(ω π/6) + +(a 2 + a 8 + a 4 )δ(ω 2π/6) + (a 3 + a 9 + a 3 + a 9 )δ(ω 2π/6) + +(a 4 + a 2 + a 8 )δ(ω 4π/6) + (a 5 + a + a 7 )δ(ω 5π/6) From this expression is easy to identify b k for k = 0,..., 5, and therefore we obtain: X[k] = 6 a 0 + a 6 + a 6 k = 0 a + a 7 + a 5 k = a 2 + a 8 + a 4 k = 2 a 3 + a 9 + a 3 + a 9 k = 3 a 4 + a 2 + a 8 k = 4 a 5 + a + a 7 k = 5 Problem 4 O&S 8.3 (20 points) In order for X[k] to be real, x[n] should be even, that is, x[ n] = x[n]. The only sequence for which a time origin can be chosen so that happens is x 2 [n]. In order for X[k] to be imaginary (except for k an integer multiple of N), x[n] should be odd, that is, x[ n] = x[n]. For none of the sequences a time origin can be chosen so that happens. Notice that the problem does not require X[k] to be imginary for k multiple of N, so this relaxes the condition on the sequences. So we need to verify that despite of this relaxation on the condition, none of the sequences satisfies the overall requirement. This is left to be done as an exercise to you. (c) (0 points) All the sequences in these problem are periodic with period N = 8. Therefore their DFS are: X[k] = x[n]e j2πnk/8 = x[n]e j2πnm/4 = x[n](j) nm for k = 2m with m = ±, ±2, ±3 6

7 With this observation is easy to verify that X [k] = X 3 [k] = 0 for k = ±2, ±4, ±6, disregarding where the time origin is. Problem 5 O&S 8.7 (0 points) 5 X [k] = X(z) z=e j2πk/4 = x[n]e j2πnk/4 for k = 0,, 2, 3 X [k] is regarded as a 4-point sequence when its inverse DFT is taken to obtain a 4-point sequence x [n]. So we have x [n] = 3 [k]e 4 X j2πnk/4 = 4 3 As in problem part a) version i), using the fact that N 5 i=0 x[i]e j2πki/4 e j2πnk/4 = 4 { e j2πn(m) N N if m = ln for some integer l = 0 otherwise 5 3 i=0 x[i]e j2πk(n i) 4 we can conclude that x [n] = x[0] + x[4] = 2 n = 0 x[] + x[5] = 2 n = x[2] = k = 2 x[3] = k = 3 The sketch of x [n] is trivial. Problem 6 O&S #7.(5 points) (a)(5 points) From tables or by finding the poles of H c (s) and doing partial fraction expansion we can find that s + a H c (s) = = h (s + a) 2 + b 2 c (t) = 2 e( a+jb)t + 2 e( a jb)t = e at cos(bt) h [n] = h c (nt) = 2 e( a+jb)nt + 2 e( a jb)nt = H (z) = 2 e ( a jb)t z + 2 e ( a+jb)t z After some manipulations it simplifies to H (z) = e at cos(bt)z 2e at cos(bt)z + e 2aT z 2 7

8 Also, you can do this problem by doing the Partial fraction expasion of H c (s) and then applying equation (7.2) from O&S, just remove the T d factor in the numerators since for that formula they defined h[n] = T d h c (nt d ), and in our case we have h [n] = h c (nt) only. Anyway, please take a look at the deduction procedure for equation (7.2) and you ll realize that is equivalent to what we did. (b)(5 points) The continuous-time step response of the system is S c (s) = s H c(s) = s + a s((s + a) 2 + b 2 ) = a a 2 +b 2 s Applying the same procedure as in part a) we obtain 2(a+jb) s + (a + jb)t 2(a jb) s + (a jb)t Finally (c)(5 points) S 2 (z) = a a 2 +b 2 z 2(a+jb) e (a+jb)t z 2(a jb) e (a jb)t z z H 2(z) = S 2 (z) = H 2 (z) = ( z )S 2 (z) You can verify that and H 2 (z) H (z) = S (z) = Problem 7 O&S #7.2 (20 points) h 2 [n] h [n] z H (z) S 2 (z) Whe we use impulse invarinace method, the relation between the continuous-time frequency Ω and the discrete-time frequency ω is ω = ΩT. Therefore, the specifications for the discrete-time system imply the following specifications for the continuous-time system: H(jΩ), 0 Ω 0.2π T d H(jΩ) , 0.3π T d Ω π T d Notice the similarity with equations (7.4a) and (7.4b) from example 7.2 from O&S. 8

9 For the sketch, please look at the Answers to Basic Problems for chapter 7 (O&S page 842). Notice that we can use the same procedure indicated in example 7.2 from O&S just replacing π by π/t d in equations (7.4) to (7.7). Therefore we can conclude that N = 6 and ΩT d = (c) (0 points) Please look at the Answers to Basic Problems for chapter 7 (O&S page 842), there it is explained why the poles of the discrete-time filter remain unchanged after using T d, but what about the residues corresponding to each pole? Well, it happens that scaling the poles of the continuous time filter by T d requires to scale the filter by /T d in order to keep the same gain of the filter and since the impulse invariance method defines h[n] = T d h c (nt d ) (Notice the factor T d ), the factor T d gets also canceled on the residues. Another way to look at this is in the frequency domain. Sampling h c (t) with T d scales the replicas of H c (jω) by /T d, but the factor T d from h[n] = T d h c (nt d ) cancels that out, producing the same gain in the discrete-time filter. Problem 8 O&S #7.9 (5 points) ω c = Ω c T d = (2π000)(.0002) = 0.4πrad Problem 9 O&S #7.0 (5 points) ω c = 2 tan ( Ω ct d 2 ) = 2 tan ( 2π2000(0.0004) ) = 2tan (0.8π) = rad 2 Problem 0 O&S #7.2 (5 points) Ω c = 2 T d tan( ω c 2 ) = 2000 tan(π 4 ) = 2000rad/s 9