DSP First 2nd Edition Solution Manual

Chapter Spectrum 3-1 Problem Solutions 32 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 3 CHAPTER 3. SPECTRUM P-3.1 DSP First 2e (a) x (t) = 11 + 14 cos(100πt − π/3) + 8 cos(350πt − π/2) (b) Since the gcd of 50 and 175 is 25, x (t) is periodic with period T0 = 1/25 = 0.04 s. (c) Negative frequencies are implicit in the cosine terms. They are needed to give a real signal when combined with their corresponding positive-frequency terms. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.2 DSP First 2e (a) A single plot labeled with complex amplitudes is sufficient. The spectrum consists of the lines {(400, 5ej π/4 ), (−400, 5e− j π/4 ), (600, 3.5e− j π/3 ), (−600, 3.5ej π/3 ), (800, 1.5), (−800, 1.5)} where the frequencies are in Hz. 5e− j π/4 5e j π/4 3.5e− j π/3 3.5ej π/3 1.5 −800 1.5 −600 −400 ✲ 0 400 600 800 f (b) The signal x (t) is periodic with fundamental frequency 200 Hz or period 1/200 = 0.005 s since the gcd of {400, 600, 800} is 200. (c) The spectrum has the added components {(500, 2.5ej π/2 ), (500, 2.5e− j π/2 )}. Now we seek the gcd of {400, 500, 600, 800} so the fundamental frequency changes to 100 Hz and the period is 0.01 s. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.3 DSP First 2e (a) x (t) = 2 cos(2t) + 4 cos(4t − π/3) + 2 cos(6t + π/4) (b) The spectrum is {(2, 1), (−2, 1), (4, 2e− j π/3 ), (−4, 2ej π/3 ), (6, ej π/4 ), (−6, e− j π/4 )} The frequencies are all in rad/s. e− j π/4 2ej π/3 1 1 −2 2 2e− j π/3 ej π/4 ✲ −6 −4 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder 0 4 6 ω May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.4 DSP First 2e (a) Determine a formula for x (t) as the real part of a sum of complex exponentials. Use Euler’s formula for the sine function obtaining . j27πt .3 e − e− j27πt sin3 (27πt) = 2j 1 . . = j27πt3 − 3ej27πt2e− j27πt + 3ej27πt e− j27πt2 − e− j27πt3 −8 j e 3 1 = 4 sin(27πt) − 4 sin(81πt) (b) What is the fundamental period for x (t)? The fundamental frequency is 27/2 so the fundamental period is 2/27. (c) Plot the spectrum for x (t). The spectrum is {(27π, − j3/8), (−27π, j3/8), (81π, j/8), (−81π, − j/8)}, where the frequencies are in rad/s. − j3/8 j3/8 − j/8 −81π j/8 −27π ©J. H. McClellan, R. W. Schafer, & M. A. Yoder ✲ 0 27π 81π ω May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.5 DSP First 2e There are seven spectral components: {(−3, 1/(1 − j3)), (−2, 1/(1 − j2)), (−1, 1/(1 − j )), (0, 1), (1, 1/(1 + j )), (2, 1/(1 + j2)), (3, 1/(1 + j3))}, where the frequencies are all in rad/s. Putting all the complex numbers in polar form gives the following plot: 1 0.707e− j0.25π 0.447e− j0.3524π 0.316e− j0.398π 0.707ej0.25π 0.447ej0.3524π 0.316ej0.398π −3 −2 −1 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder ✲ 0 1 2 3 ω May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.6 DSP First 2e (a) In this case we need to find the gcd of 36 and 84, which is 12. Thus, the fundamental frequency is ω0 = 1.2π rad/s. (b) The fundamental period is T0 = 2π/ω0 = 1/0.6 = 5/3 s. (c) The DC value is −7. (d) The ak coefficients are nonzero for k = 0, ±3, ±7. Here is the list of the nonzero Fourier series coefficients in a table. k ak −7 −3 0 3 7 3e− j π/4 4ej π/3 7ej π 4e− j π/3 3ej π/4 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.7 DSP First 2e (a) The phasor representation is z(t) = Aej2π ( fc − f∆)t + Bej2π ( fc + f∆)t (b) z(t) = ej2π fc t ( Ae− j2π f∆t + Bej2π f∆t ) = ej2π fc t ( A cos(2π f∆t) − j A sin(2π f∆t) + B cos(2π f∆t) − j B sin(2π f∆t)) = ej2π fc t [( A + B) cos(2π f∆t) − j ( A − B) sin(2π f∆t)] Therefore, the real part is x (t) = 9{z(t)} = ( A + B) cos(2π f∆t) cos(2π f ct) + ( A − B) sin(2π f∆t) sin(2π f ct) so C = A + B and D = A − B. If A = B = 1, C = 2 and D = 0, so using the trigonometric identity cos α cos β = 1 1 2 cos(α − β) + 2 cos(α + β), it follows that x (t) = 2 cos(2π f∆t) cos(2π f ct) = 2[ 1 cos(2π( f c − f∆)t) + 1 cos(2π( f c + f∆)t)] 2 2 (c) The values are A = 1 and B = −1. In this case, . j2π f∆t .. . e − e− j2π f∆t ej2π fc t − e− j2π fc t x (t) = 2 2j 2j . . = −21 ej2π ( fc + f∆)t − ej2π ( fc − f∆)t − e− j2π ( fc − f∆)t + e− j2π ( fc + f∆)t The spectrum is {(− f c − f∆, −0.5), (− f c + f∆, 0.5), ( f c − f∆, 0.5), ( f c + f∆, −0.5)}, and the plot is 0.5 0.5 −0.5 −0.5 −( f c + f ∆ ) − fc −( f c − f ∆ ) ©J. H. McClellan, R. W. Schafer, & M. A. Yoder 0 ( f c − f∆) fc ( f c + f∆) ✲ f May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.8 DSP First 2e (a) Using Euler’s relation we get x (t) = 10 + 10e j π/4 e j2π (100)t + 10e− j π/4e− j2π (100)t + 5e j2π (250)t + 5e− j2π (250)t The gcd of 100 and 250 is 50 so f0 = 50 and therefore N = 5. The nonzero Fourier coefficients are, therefore, a−5 = 5, j π/4 , a0 = 10, a2 = 10e jπ/4 , and a5 = 5. a−2 = 10e− (b) The signal is periodic because all the frequencies are multiples of 50 Hz. Therefore, the fundamental period is T0 = 1/50 = 0.02s. (c) Here is the spectrum plot of this signal versus f in Hz. 10e− j π/4 10 10ej π/4 5 −250 5 −100 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder ✲ 0 100 250 f May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.9 DSP First 2e (a) Use phasors to show that x (t) can be expressed in the form x (t) = A1 cos(ω1t + ϕ1 ) + A2 cos(ω2t + ϕ2 ) + A3 cos(ω3t + ϕ3 ) where ω1 < ω2 0. For periodicity with period T0 we require that ω0 = 2π/T0. This means that k1ω0 = ω2 − ω1 and k2ω0 = ω2 + ω1, where k1 and k2 are integers and k2 > k1. (b) Part (a) gives two equations for ω1 and ω2. If we solve them in terms of ω0 we get ω1 = (k2 − k1 )ω0/2 and ω2 = (k2 + k1 )ω0/2, so the main condition is that both ω1 and ω2 are integer multiples of ω0/2. This is the most general condition. Therefore, the relationship between ω2 and ω1 is ω2 = k2 + k1 k 2 − k1 ω1 if x (t + T0 ) = x (t). Thus, ω2 could be an integer multiple of ω1 if k2 − k1 divides into k2 + k1 with no remainder, but that is not necessary for periodicity of x (t). ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.11 DSP First 2e (a) The gcd of 40, 60, 120 is 20 so ω0 = 20π and the fundamental period is T0 = 2π/ω0 = 0.1s. The finite Fourier ∓ series has components indexed by 0, ±2, ±3, ±6 so N = 6. The coefficients are a0 = 2, a±2 = 2e jπ/5 , a±3 = 1.5e ∓ j π/2, π/3 a±6 = 2e∓ (b) The spectrum is {(−120π, 2ej π/3 ), (−60π, 1.5ej π/2 ), (−40π, 2ej π/5 ), . . . (0, 2), (40π, 2e− j π/5 ), (60π, 1.5e− j π/2 ), (120π, 2e− j π/3 )} 2e j π/3 2e j π/5 2 2e− j π/5 1.5e− j π/2 1.5ej π/2 −120π −60π 2e− j π/3 −40π ✲ 0 40π 60π 120π ω (c) Now the fundamental frequency is 10π rad/s because the gcd of 20, 40, 50, and 120 is 10. Therefore, the period is T0 = 2π/10π = 1/5 = 0.2s. The spectrum is the same as in part (b) except there are two additional components at ±50π rad/s: (−50π, 5ej π/6 ) and (50π, 5e− j π/6 ). ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.12 DSP First 2e (a) Make a table of the frequencies of the tones of the octave beginning with middle C, assuming that the A above middle C is tuned to 440 Hz. Note name Note number Frequency Note name Note number Frequency C 40 262 C# 41 277 G 47 392 F# 46 370 D 42 294 G# 48 415 Eb 43 311 A 49 440 E 44 330 Bb 50 466 F 45 349 B 51 494 F# 46 370 C 52 523 (b) The formula for the frequency f as a function of note number n is f = 440 · 2(n−49)/12 (c) The spectrum would have the form: {(−440, a∗), (−370, a∗), (−294, a∗), (294, a1 ), (370, a2 ), (440, a3 )} 3 2 1 To sound like a musical chord, the coefficients should have similar magnitudes, but the phases could be arbitrarily chosen. A chord from a real instrument would have overtones (higher harmonics) of each individual note. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.13 DSP First 2e (a) The frequency of the DC component is by definition 0. The waveform is periodic with period 25 ms so the frequency is 1/0.025 = 40 Hz. (b) The DC level is (20 − 10)/2 = 5, the amplitude of the cosine is (20 + 10)/2 = 15, and the cosine is delayed by 0.005 s, so x (t) = 5 + 15 cos(2π(40)(t − .005)) = 5 + 15 cos(80πt − 0.4π) (c) x (t) = 5 + 7.5ej (80πt−0.4π) + 7.5e− j (80πt−0.4π) = 7.5ej0.4πe− j80πt + 5 + 7.5e− j0.4πej80πt (d) Plot of the two-sided spectrum of the signal x (t). 7.5e− j0.4π 7.5ej0.4π 5 −80π ©J. H. McClellan, R. W. Schafer, & M. A. Yoder ✲ 0 80π ω May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.14 DSP First 2e (a) Using symmetry we obtain √ √ X−1 = 2 − j 2 = 2e− j π/4 j π/3 X2 = 8e ω1 = 70π ω2 = −100π (b) x (t) = 20 + 4 cos(70πt + π/4) + 16 cos(100πt + π/3) (c) The gcd of 70 and 100 is 10, so the fundamental frequency of the signal is f0 = 5 Hz and the fundamental period is T0 = 1/5 = 0.2 s. (d) Note that the −20 ≤ 4 cos(70πt + π/4)+16 cos(100πt + π/3) ≤ 20 since the individual terms satisfy −4 ≤ 4 cos(70πt + π/4) ≤ 4 and −16 ≤ 16 cos(100πt + π/3) ≤ 16. The value ±20 would be attained only if the phases of the two cosines are such that 4 cos(70π(t − t0 )) + 16 cos(100π(t − t0 )). ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.15 DSP First 2e (a) x (t) = cos2 (7πt − 0.1π) We need to express x (t) in terms of complex exponentials . .2 x (t) = 0.5ej (7πt−0.1π) + 0.5e− j (7πt−0.1π) = 0.25ej0.2πe− j14πt + 0.5 + 0.25e− j0.2πej14πt 0.5 0.25e− j0.2π 0.25ej0.2π −14π ✲ 0 ω 14π (b) y(t) = cos2 (7πt − 0.1π) cos(77πt + 0.1π) . .. . x (t) = 0.25e j0.2π e− j14πt + 0.5 + 0.25e− j0.2π e j14πt 0.5e j0.1π e j77πt + 0.5e− j0.1π e − j77πt = 0.125ej0.1πe− j91πt + 0.25e− j0.1πe− j77πt + 0.125e− j0.3πe− j63πt + 0.125ej0.3πej63πt + 0.25ej0.1πej77πt + 0.125e− j0.1πej91πt 0.25e− j0.1π 0.125ej0.1π 0.25ej0.1π 0.125ej0.3π 0.125e− j0.3π −91π −77π −63π 0.25e− j0.1π ✲ 0 63π 77π 91π ω Note: ω axis is not to scale. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.16 DSP First 2e (a) Plotting spectrum of v (t) versus f in kHz, 0.5e− j0.3π −3 0.5ej0.3π ✲ 0 f (kHz) 3 (b) The spectrum for x (t) versus f in kHz. 0.75 0.25e− j0.3π 0.75 0.25ej0.3pi −683 −680 −677 0.25e− j0.3π 0.25ej0.3π 677 683 ✲ 0 680 f (kHz) Note: f axis is not to scale. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.17 DSP First 2e (a) The gcd of 105 and 180 is 15, so the given frequencies are the 7th and 12th harmonics of f0 = 15 Hz. (b) x (t) = 22 cos(2π(105)t − 0.4π) + 14 cos(2π(180)t − 0.6π) (c) Simplify the numerical values for the complex amplitudes, i.e., phases should be in [−π, π]. x2 (t) = 22 cos(2π(105)(t − 0.05) − 0.4π) + 14 cos(2π(180)(t − 0.05) − 0.6π) = 22 cos(2π(105)t − 10.5π − 0.4π) + 14 cos(2π(180)t − 18π − 0.6π) = 22 cos(2π(105)t − 10π − 0.9π) + 14 cos(2π(180)t − 18π − 0.6π) = 22 cos(2π(105)t − 0.9π) + 14 cos(2π(180)t − 0.6π) Note that even multiples of 2π rad can be dropped from the equation. Thus, the spectrum is: {(−180, 7ej0.6π ), (−105, 11ej0.9π ), (105, 11e− j0.9π ), (180, 7e− j0.6π )} where the frequencies are in hertz. Therefore, the plot of the spectrum looks just like Fig. P-3.17 except the phase is different at frequencies ±105 Hz. (d) The effect of this operation is simply to increase all the frequencies by 105 Hz, or, in other words, to shift the spectrum of x (t) to the right by 105 Hz. Therefore, the spectrum line at −105 Hz will move to f = 0, and the new DC component is equal to the value of the spectrum originally at f = −105 Hz, i.e., 11ej0.9π . ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.18 DSP First 2e (a) The spectrum of y(t) is the spectrum of x (t) with an added DC component of size 8. 11e− j0.4π 11ej0.4π 8 7ej0.6π 7e− j0.6π ✲ −180 −105 105 0 180 f (b) The spectrum of z(t) is the same as that of x (t) with the addition of components of size 9e± j0.8π at frequencies ±40 Hz. 11ej0.4π 7ej0.6π 9e− j0.8π 9ej0.8π 11e− j0.4π 7e− j0.6π ✲ −180 −105 −40 0 105 40 180 f (c) The fundamental frequency is the gcd of 40, 105, 180, which is 5 Hz. (d) The derivative operation multiplies each spectrum component by j2π f , where f is the frequency of the complex exponential component. So we get j2π(105)11e− j0.4π − j2π(105)11ej0.4π j2π(180)7e− j0.6π − j2π(180)7ej0.6π −180 −105 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder 0 105 180 ✲ f May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.19 DSP First 2e (a) X1 = 8e− j π/3 and ω1 = −10π (b) Here is the plot of the spectrum of x (t). 8e− j π/3 8e j π/3 −10π ✲ 0 ω 10π (c) The symmetry implies that ωb = 20π and B = +4 j. Furthermore, symmetry requires that ω a = 0. To find A, ωc , and ϕ we can write y(t) as y(t) = 0.5x (t)e jϕ e jω c t + 0.5x (t)e− jϕ e − jωc t , which shows that the spectrum of y(t) will consist of the sum of scaled copies of the spectrum of x (t) shifted right (up) by ωc and left (down) by ωc . In order to have only three components we must choose ωc = 10π so that two of the shifted spectrum lines over lap at ω = 0. 4e j π/3 e jϕ + 4e− j π/3e− jϕ 4e j π/3e − jϕ 4e− j π/3 e jϕ ✲ −20π 0 20π ω Now, 4e− j π/3ejϕ = 4e− j π/2 so ϕ = −π/6. Finally, note that the DC value can be written as A = 8 cos(π/3 + ϕ) = √ √ 8 cos(π/6) = 8 3/2 = 4 3. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.20 DSP First 2e (a) The gcd of 40 and 90 is 10, so f0 = 10 Hz. (b) The fundamental period is T0 = 1/ f0 = 1/10 = 0.1 s. (c) From the plot, the DC value is 0.5. (d) With f0 = 10, the harmonics are k = 0, ±4, ±9. k −9 −4 0 4 9 ak 0.4e− j2 0.6ej1.4 0.5 0.6e− j1.4 0.4ej2 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.21 DSP First 2e (a) The instantaneous frequency is ωi (t) = dψ dt= 2αt + β, so ω1 = ωi (0) = β and ω2 = ωi (T2 ) = 2αT2 + β. (b) The instantaneous frequency versus time is ωi (t) = 80t + 27 (c) Here is the plot of the instantaneous frequency (in Hz) versus time over the range 0 ≤ t ≤ 1 sec. f (t) = i ✻ ωi (t ) 2π 107 2π ✏ ✏✏ ✏ ✏✏ 27 ✏✏ 2π ✏ ✏✏ ✏ ✏✏ ✏✏ ✏ ✏ ✏✏ ✏✏ ✏ ✏ ✏✏ ✲ 0 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder 1 t May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.22 DSP First 2e (a) The general form for the chirp signal is x (t) = cos(αt2 + βt + ϕ). The instantaneous frequency of this signal is ωi (t) = 2αt + β. From this we observe that ω1 = 2π f1 = 2π(4800) = ωi (0) = β. To obtain α, we note that ω2 = 2π(800) = ωi (2) = 2α(2) + β = 4α + 9600π so α = −2000π. Therefore, the signal is x (t) = cos(−2000πt2 + 9600πt + ϕ) where ϕ is an arbitrary phase constant. (b) The instantaneous frequency is ωi = 800πt + 500π, so ω1 = ωi (0) = 500π and ω2 = ωi (3) = 800π(3) + 500π = 2900π rad/s. ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.23 DSP First 2e (a) The instantaneous frequency is ωi (t) = 2αt + β. Substituting the given parameters gives α = 4π and β = 2π, so the signal with the given parameters is x (t) = cos(4πt2 + 2πt + ϕ). frequency in Hz (b–f) The solution to this problem is given in the following figure. Note that the times at which f i (t) is equal to 4 Hz and 8 Hz are indicated with dashed lines. Careful scrutiny of the plots confirms that the waveform of the chirp signal does match the waveforms of the 4 Hz and 8 Hz constant-frequency sinusoids at the corresponding two times. Instantaneous Frequency of Chirp Signal 10 5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 4 Hz Signal 1 0.5 0 -0.5 -1 0 0.2 0.4 0.6 0.8 1 Chirp Signal 1 0 -1 0 0.2 0.4 0.6 0.8 1 8 Hz Signal 1 0.5 0 -0.5 -1 0 0.2 0.4 0.6 0.8 1 time in seconds ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.24 DSP First 2e (a) f1 (t) = 1 √ 2π (t) (c) f3 (t) = e2t /π (d) f4 (t) = − sin(2πt) frequency in Hz 1 frequency in Hz frequency in Hz 0.8 frequency in Hz (b) f2 (t) = t/π Solution to Problem 3.24 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 0.5 0 -0.5 -1 -3 -2 -1 0 1 2 3 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 150 100 50 0 1 0.5 0 -0.5 -1 time in seconds ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.25 DSP First 2e (a) Let x (t) be given by the Fourier series x (t) = x (0) = ∞ . k=−∞ ✿ ✘✘✘ 1 . ∞ (2π✘ /T0 )k (0) = ak✘ e j✘ ∞ . ak ej (2π/T0 )kt . Then it follows that k=−∞ ak . k=−∞ (b) Let f3 = − f2 = f0 and f4 = − f 1 = 3 f0 so that from the spectrum we can write x (t) = 12 cos(2π f0t + π/4) + 4 cos(6π f0t + 3π/4) √ √ √ 2 Therefore x (0) = 12 cos(π/4) + 4 cos(3π/4) = 6 − 2 2 = 4 2. Now if we add the coefficients of the Fourier series we get √ a1 + a2 + a3 + a4 = 2e− j3π/4 + 6e− j π/4 + 6ej π/4 + 2ej3π/4 = 12 cos(π/4) + 4 cos(3π/4) = 4 2 ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.26 DSP First 2e The equations corresponding to the spectra are: The matches are (a) (b) (c) (d) (e) (3) (1) (2) (5) (4) ©J. H. McClellan, R. W. Schafer, & M. A. Yoder x1 (t) = 4 cos(4πt + π) + 4 cos(6πt + π/2) x2 (t) = 2 cos(4πt + π/4) + 4 cos(6πt − 0.333π) x3 (t) = −3 + 2 cos(4πt + π/4) x4 (t) = −2 + 4 cos(4πt + π) x5 (t) = 4 cos(2πt + π) + 4 cos(4πt + π) May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM P-3.27 DSP First 2e (a) x (t) = cos(−250πt2 ) Spectrogram (2) (d) x (t) = cos(100πt) cos(400πt) Spectrogram (1) (b) x (t) = cos(100πt − π/4) + cos(400πt) Spectrogram (5) (e) x (t) = cos(200πt2 ) Spectrogram (6) (c) x (t) = cos(1000πt − 250πt2 ) Spectrogram (4) (f) x (t) = cos(30e2t ) Spectrogram (3) ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CHAPTER 3. SPECTRUM ©J. H. McClellan, R. W. Schafer, & M. A. Yoder May 20, 2016 © 2016 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher