GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING

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1 GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING ECE 06 Summer 08 Final Exam July 7, 08 NAME: Important Notes: Do not unstaple the test. Closed book and notes, except for three double-sided pages (8.5 ) of hand-written notes. Calculators are allowed, but no smartphones/wifi/etc. JUSTIFY your reasoning CLEARLY to receive partial credit. Express all angles as a fraction of p. For example, write 0.p as opposed to 8 or 0.34 radians. You must write your answer in the space provided on the exam paper itself. Only these answers will be graded. Circle your answers, or write them in the boxes provided. If more space is needed for scratch work, use the backs of the previous pages. Problem Points Score TOTAL: 00

2 PROB. Su8-Fin.. Let x ( t ) = cos(00p(t 0.00)). Let x ( t ) be the signal whose spectrum is shown below: j + j 00p 0 00p w [rad/s] Let x 3 ( t ) be the sinusoidal signal sketched below: t The sum of these three signals can be written as x ( t ) + x ( t ) + x 3 ( t ) = Acos(00pt + j), where A = and j =.

3 PROB. Su8-Fin.. Suppose the output of a linear and time-invariant discrete-time filter in response to a finite-length input of x[ n ] = f g is y[ n ] = f g, as shown below: x[ n ] x[ n ] LTI y[ n ] SYSTEM y[ n ] n 0 5 n The filter s impulse response h[ n ] has values: h[ 0 ] =, h[ ] =, h[ ] =, h[ 3 ] =, h[ 4 ] =

4 PROB. Su8-Fin.3. Consider the following system in which a continuous-time input x( t ) is sampled, producing x[ n ], then filtered, producing y[ n ], and finally converted back to continuous time, yielding an overall output y( t ): x( t ) x[ n ] LTI SYSTEM y[ n ] H( z ) IDEAL C-to-D CONVERTER IDEAL D-to-C CONVERTER y( t ) SAMPLE RATE f s The system function of the discrete-time LTI system is H( z ) = z + z. (a) Which of the plots (A... H) on the next page shows the magnitude response of this filter? (b) A constant input x( t ) = (for all t) will result in the constant output y( t ) = (for all t). (c) Specify a sample rate f s = samples/s so that a C-to-D input of x( t ) = cos(0pt) + cos(4pt) results in a zero output, y( t ) = 0 (for all t).

5 E A F B G C H D

6 PROB. Su8-Fin.4. Consider an FIR filter whose difference equation is: y[ n ] = ( + b)x[ n] + ( b)x[n ], where b is an unspecified real constant. If the output of this system in response to the sinusoidal input x[ n ] = cos(0.7pn) is y[ n ] = cos(0.7pn + q), then it must be that b = and q =.

7 PROB. Su8-Fin.5. Consider an LTI system described by the recursive difference equation: (a) The system is [ IIR ][ FIR ] (circle one). y[ n ] = x[ n ] 0.y[n ]. (b) The system transfer function is H( z ) =. (c) Sketch below the pole-zero plot for H( z ), carefully labeling the locations of all zeros and poles: Imfz g Refz g (UNIT CIRCLE) (d) If the output of this system in response to the input x[ n ] = 0d[ n ] + Bd[ n ] is y[ n ] = 0d[ n ] (or in other words, as lists of numbers, the input B 0... results in the output ), then it must be that the constant B is B =.

8 PROB. Su8-Fin.6. Shown below are eleven different outcomes that result from executing the MATLAB code: stem(abs(fft(ones(,l),n))); Match each plot with the corresponding values for the variables N and L by writing a letter (A through K) in each answer box. A B L =, N = C L =, N = 8 L = 3, N = D L = 3, N = 8 E L = 4, N = F L = 4, N = 8 G L = 5, N = L = 5, N = 8 H L = 8, N = 8 I L =, N = J L =, N = 8 K

9 PROB. Su8-Fin.7. (The different parts are unrelated.) (a) There are an infinite number of values for the delay parameter t 0 below such that: cos(00pt) + cos(00p(t t 0 )) + cos(00p(t t 0 )) = 0, for all time t. Name any two, both positive: t 0 =, or t 0 =. (b) Simplify P k= sin( 0.9pk). sin( 0.75p( n k) ) =. pk p( n k) a function of n

10 (c) The signal x( t ) = cos(88pt)cos(8pt)) + cos(40pt) is periodic with fundamental frequency f 0 = Hz. (d) If fx[ 0 ],... X[ 5]g are the DFT coefficients that result from taking a 5-point DFT of the length- signal x[ n ] = [ ] for nf0,... g, then find numerical values for the following two DFT coefficients: X[ 0 ] =, X[ 56] =.

11 Table of DTFT Pairs Time-Domain: xœn Frequency-Domain: X.e j O! / ıœn ıœn n 0 e j O!n 0 uœn uœn L sin. O! b n/ n a n uœn.jaj < / sin. L O!/ j O!.L /= sin. e O!/ ( j O!j O! b u. O! C O! b / u. O! O! b / D 0 O! b < j O!j ae j O! Table of DTFT Properties Property Name Time-Domain: xœn Frequency-Domain: X.e j O! / Periodic in O! X.e j. O!C/ / D X.e j O! / Linearity ax Œn C bx Œn ax.e j O! / C bx.e j O! / Conjugate Symmetry xœn is real X.e j O! / D X.e j O! / Conjugation x Œn X.e j O! / Time-Reversal xœ n X.e j O! / Delay (d=integer) xœn d e j O!d X.e j O! / Frequency Shift xœn e j O! 0n X.e j. O! O! 0/ / Modulation xœn cos. O! 0 n/ X.ej. O! O! 0/ / C X.ej. O!C O! 0/ / Convolution xœn hœn X.e j O! /H.e j O! / X Z Parseval s Theorem jxœn j jx.e j O! /j d O! nd Date: 8-Apr-03

12 Table of Pairs for N -point DFT Time-Domain: xœn ; n D 0;;;:::;N Frequency-Domain: XŒk ; k D 0;;;:::;N If xœn is finite length, i.e., xœn 0 only when n Œ0;N XŒk D X.e j O! / and the DTFT of xœn is X.e j O! ˇ / O!Dk=N (frequency sampling the DTFT) ıœn ıœn n 0 e j.k=n /n 0 e j.n=n /k 0 NıŒk k 0, when k 0 Œ0;N uœn uœn L, when L N sin. L.k=N // j.k=n /.L /= sin. e.k=n // sin. L. n=n // sin.. n=n // ej.n=n /.L /= N.uŒk uœk L /, when L N Table of z-transform Pairs Signal Name Time-Domain: xœn z-domain: X.z/ Impulse ıœn Shifted impulse ıœn n 0 z n 0 Right-sided exponential Decaying cosine a n uœn r n cos. O! 0 n/uœn az ; jaj < r cos. O! 0 /z r cos. O! 0 /z C r z Decaying sinusoid Ar n cos. O! 0 n C '/uœn A cos.'/ r cos. O! 0 '/z r cos. O! 0 /z C r z Table of z-transform Properties Property Name Time-Domain xœn z-domain X.z/ Linearity ax Œn C bx Œn ax.z/ C bx.z/ Delay (d=integer) xœn d z d X.z/ Convolution xœn hœn X.z/H.z/ Date: 8-April-03

13 GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING ECE 06 Summer 08 Final Exam July 7, 08 NAME: ANSWER KEY Important Notes: Do not unstaple the test. Closed book and notes, except for three double-sided pages (8.5 ) of hand-written notes. Calculators are allowed, but no smartphones/wifi/etc. JUSTIFY your reasoning CLEARLY to receive partial credit. Express all angles as a fraction of p. For example, write 0.p as opposed to 8 or 0.34 radians. You must write your answer in the space provided on the exam paper itself. Only these answers will be graded. Circle your answers, or write them in the boxes provided. If more space is needed for scratch work, use the backs of the previous pages. Problem Points Score TOTAL: 00

14 PROB. Su8-Fin.. Let x ( t ) = cos(00p(t 0.00)). ) X = e j0.p Let x ( t ) be the signal whose spectrum is shown below: j + j ) X = e j0.5p 00p 0 00p w [rad/s] Let x 3 ( t ) be the sinusoidal signal sketched below: ) x 3 ( t ) =.5cos(00p(t t 0 ))... ) X 3 =.5e j00p t 0 =.5e j0.5p t 0.5 t 0 = The sum of these three signals can be written as x ( t ) + x ( t ) + x 3 ( t ) = Acos(00pt + j), where A = and j =. Phasor addition rule: p (= 0.05 rads =.89 ) Ae jj = X + X + X 3 = e j0.p + e j0.5p +.5e j0.5p = 4.68e j0.06p

15 PROB. Su8-Fin.. Suppose the output of a linear and time-invariant discrete-time filter in response to a finite-length input of x[ n ] = f g is y[ n ] = f g, as shown below: x[ n ] x[ n ] LTI y[ n ] SYSTEM y[ n ] n 0 5 n The filter s impulse response h[ n ] has values: h[ 0 ] = 0, h[ ] =, h[ ] =, h[ 3 ] =, h[ 4 ] = y[ n ] = S k h[k]x[n k] But y[ 0 ] = 0 ) 0 = h[ 0 ] x[ 0 ] = h[ 0 ] ) h[ 0 ] = 0 0 y[ ] = ) = h[ 0 ] x[ ] + h[ ] x[ 0 ] = h[ ] ) h[ ] = y[ ] = ) = h[ 0 ] x[] + h[ ] x[ ] + h[ ] x[ 0 ] = + h[ ] ) h[ ] = y[ 3 ] = 3 = h[ ] x[ ] + h[ ] x[ ] + h[ 3 ] x[ 0 ] = + h[ 3 ] ) h[ 3 ] = y[ 4 ] = 3 = h[ ] x[ 3 ] + h[ ] x[ ] + h[ 3 ] x[ ] + h[ 4 ] x[ 0 ] ) h[ 4 ] = 0 0.5

16 PROB. Su8-Fin.3. Consider the following system in which a continuous-time input x( t ) is sampled, producing x[ n ], then filtered, producing y[ n ], and finally converted back to continuous time, yielding an overall output y( t ): x( t ) x[ n ] LTI SYSTEM y[ n ] H( z ) IDEAL C-to-D CONVERTER IDEAL D-to-C CONVERTER y( t ) SAMPLE RATE f s The system function of the discrete-time LTI system is H( z ) = z + z. (a) Which of the plots (A... H) on the next page shows the magnitude response of this filter? Recognize this as a nulling filter, where cos( ^w ) = ) look for nulls at ^w = ±p/3 (b) A constant input x( t ) = (for all t) will result in the constant output y( t ) = (for all t). C The DC gain is b 0 + b + b = + = (c) Specify a sample rate f s = samples/s 6 so that a C-to-D input of x( t ) = cos(0pt) + cos(4pt) results in a zero output, y( t ) = 0 (for all t). There are an infinite number of solutions, the largest being f s = 6 Hz: 0p 5p p ) ^w = = = p 3 3 f s ) ^w = 4p = 7p = p -- + p 3 3 f s both reduce to ^w = ±p/3 ) both nulled

17 E A F B NULL at p/3 G C H D

18 PROB. Su8-Fin.4. Consider an FIR filter whose difference equation is: y[ n ] = ( + b)x[ n] + ( b)x[n ], where b is an unspecified real constant. If the output of this system in response to the sinusoidal input x[ n ] = cos(0.7pn) is y[ n ] = cos(0.7pn + q), p then it must be that b = and q = (alternative answer) 0.49p The frequency response must have magnitude at 0.7p ) = jh(e j0.7p )j = j( + b) + ( b)e j0.7p j = ( + b) + ( b) + ( + b)( b)cos(0.7p) = + b + ( b )cos(0.7p) ) b = cos( 0.7p) = ) b = ±0.35 cos( 0.7p) With b = +0.35, the frequency response simplifies to H(e j0.7p ) = ( + b) + ( b)e j0.7p = e j0.7p = e j0.4p Alternative answer: With b = 0.35, the frequency response simplifies to H(e j0.7p ) = ( + b) + ( b)e j0.7p = e j0.7p = e j0.4876p

19 PROB. Su8-Fin.5. Consider an LTI system described by the recursive difference equation: (a) The system is [ IIR ][ FIR ] (circle one). y[ n ] = x[ n ] 0.y[n ] z (b) The system transfer function is H( z ) =. (c) Sketch below the pole-zero plot for H( z ), carefully labeling the locations of all zeros and poles: Imfz g H( z) = z : z + 0. Refz g (UNIT CIRCLE) (d) If the output of this system in response to the input x[ n ] = 0d[ n ] + Bd[ n ] is y[ n ] = 0d[ n ] (or in other words, as lists of numbers, the input B 0... results in the output ), then it must be that the constant B is B =. Y( z ) = X( z )H( z ) = (0 + Bz ) = z B z z æ ö ç ç ç è ø B But Y( z ) = 0 ) = 0. ) B = 0

20 PROB. Su8-Fin.6. Shown below are eleven different outcomes that result from executing the MATLAB code: stem(abs(fft(ones(,l),n))); Match each plot with the corresponding values for the variables N and L by writing a letter (A through K) in each answer box. A B L =, N = B C L =, N = 8 L = 3, N = E A D L = 3, N = 8 K E L = 4, N = F F L = 4, N = 8 H G L = 5, N = G H L = 5, N = 8 L = 8, N = 8 C J I L =, N = D J L =, N = 8 I K

21 PROB. Su8-Fin.7. (The different parts are unrelated.) (a) There are an infinite number of values for the delay parameter t 0 below such that: cos(00pt) + cos(00p(t t 0 )) + cos(00p(t t 0 )) = 0, for all time t. Name any two, both positive: t 0 =, or t 0 =. Phasor addition rule, where q = 00pt 0 : 0 = + e jq + e jq ) add to zero when either e jq e jq In order for these phasors to add to zero, they must form a peace sign, with equally spaced angles p + 3m q = mp ) t 0 = f , , , ,...g p + 3k 5 8 or q = kp ) t 0 = f , , , ,...g (b) Simplify P k= sin( 0.75pn) pn sin( 0.9pk). sin( 0.75p( n k) ) =. pk p( n k) a function of n Recognize as the convolution of h[ n ] = sin( 0.9pn) with x[ n ] = sin( 0.75pn) : pn pn y[ n ] = x[ n ] * h[ n ] ) Y( e jw^ ) = X( e jw^ )H( e jw^ ) = X( e jw^ ) (narrow)(wide) (multiplying a narrow rectangle by a wide rectangle yields a narrow rectangle) ) y[ n ] = x[ n ].

22 (c) The signal x( t ) = cos(88pt)cos(8pt)) + cos(40pt) is periodic with fundamental frequency f 0 = Hz. 8 Multiplying the 44-Hz sinusoid by the 4-Hz sinusoid leads to the sum of two sinusoids, one with frequency 40 Hz and the other with frequency 48 Hz, so overall x( t ) has contributations at three frequencies: 40 Hz = (8)(5) 48 Hz = (8)(6) 0 Hz = (8)(5) ) greatest common factor is 8 (d) If fx[ 0 ],... X[ 5]g are the DFT coefficients that result from taking a 5-point DFT of the length- signal x[ n ] = [ ] for nf0,... g, then find numerical values for the following two DFT coefficients: X[ k ] = P nx[n]e jkpn/n 30 X[ 0 ] =, X[ 56] =. ) X[ 0 ] = P nx[n] = = 30 ) X[56] = P nx[n]e jpn = =