Preview Extract

Chapter
Spectrum
3-1 Problem Solutions
32
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3
CHAPTER 3. SPECTRUM
P-3.1
DSP First 2e
(a) x (t) = 11 + 14 cos(100ฯt โ ฯ/3) + 8 cos(350ฯt โ ฯ/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.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
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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 ฯ/4 ), (โ400, 5eโ j ฯ/4 ), (600, 3.5eโ j ฯ/3 ), (โ600, 3.5ej ฯ/3 ), (800, 1.5), (โ800, 1.5)}
where the frequencies are in Hz.
5eโ j ฯ/4
5e j ฯ/4
3.5eโ j ฯ/3
3.5ej ฯ/3
1.5
โ800
1.5
โ600
โ400
โฒ
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 ฯ/2 ), (500, 2.5eโ j ฯ/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.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
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CHAPTER 3. SPECTRUM
P-3.3
DSP First 2e
(a) x (t) = 2 cos(2t) + 4 cos(4t โ ฯ/3) + 2 cos(6t + ฯ/4)
(b) The spectrum is {(2, 1), (โ2, 1), (4, 2eโ j ฯ/3 ), (โ4, 2ej ฯ/3 ), (6, ej ฯ/4 ), (โ6, eโ j ฯ/4 )}
The frequencies are all in rad/s.
eโ j ฯ/4
2ej ฯ/3
1
1
โ2
2
2eโ j ฯ/3
ej ฯ/4
โฒ
โ6
โ4
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
0
4
6
ฯ
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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 Eulerโs formula for the sine function obtaining
. j27ฯt
.3
e
โ eโ j27ฯt
sin3 (27ฯt) =
2j
1 .
.
=
j27ฯt3 โ 3ej27ฯt2eโ j27ฯt + 3ej27ฯt eโ j27ฯt2 โ eโ j27ฯt3
โ8 j e
3
1
=
4
sin(27ฯt) โ
4
sin(81ฯt)
(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 {(27ฯ, โ j3/8), (โ27ฯ, j3/8), (81ฯ, j/8), (โ81ฯ, โ j/8)}, where the frequencies are in rad/s.
โ j3/8
j3/8
โ j/8
โ81ฯ
j/8
โ27ฯ
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
โฒ
0
27ฯ
81ฯ
ฯ
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CHAPTER 3. SPECTRUM
P-3.5
DSP First 2e
There are seven spectral components:
{(โ3, 1/(1 โ j3)), (โ2, 1/(1 โ j2)), (โ1, 1/(1 โ 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
0.707eโ j0.25ฯ
0.447eโ j0.3524ฯ
0.316eโ j0.398ฯ
0.707ej0.25ฯ
0.447ej0.3524ฯ
0.316ej0.398ฯ
โ3
โ2
โ1
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
โฒ
0
1
2
3
ฯ
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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 ฯ0 = 1.2ฯ rad/s.
(b) The fundamental period is T0 = 2ฯ/ฯ0 = 1/0.6 = 5/3 s.
(c) The DC value is โ7.
(d) The ak coefficients are nonzero for k = 0, ยฑ3, ยฑ7.
Here is the list of the nonzero Fourier series coefficients in a table.
k
ak
โ7
โ3
0
3
7
3eโ j ฯ/4
4ej ฯ/3
7ej ฯ
4eโ j ฯ/3
3ej ฯ/4
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
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CHAPTER 3. SPECTRUM
P-3.7
DSP First 2e
(a) The phasor representation is z(t) = Aej2ฯ ( fc โ fโ)t + Bej2ฯ ( fc + fโ)t
(b)
z(t) = ej2ฯ fc t ( Aeโ j2ฯ fโt + Bej2ฯ fโt )
= ej2ฯ fc t ( A cos(2ฯ fโt) โ j A sin(2ฯ fโt) + B cos(2ฯ fโt) โ j B sin(2ฯ fโt))
= ej2ฯ fc t [( A + B) cos(2ฯ fโt) โ j ( A โ B) sin(2ฯ fโt)]
Therefore, the real part is
x (t) = 9{z(t)} = ( A + B) cos(2ฯ fโt) cos(2ฯ f ct) + ( A โ B) sin(2ฯ fโt) sin(2ฯ f ct)
so
C = A + B and
D = A โ B. If A = B = 1, C = 2 and D = 0, so using the trigonometric identity cos ฮฑ cos ฮฒ =
1
1
2 cos(ฮฑ โ ฮฒ) + 2 cos(ฮฑ + ฮฒ), it follows that
x (t) = 2 cos(2ฯ fโt) cos(2ฯ f ct) = 2[ 1 cos(2ฯ( f c โ fโ)t) + 1 cos(2ฯ( f c + fโ)t)]
2
2
(c) The values are A = 1 and B = โ1. In this case,
. j2ฯ fโt
..
.
e
โ eโ j2ฯ fโt ej2ฯ fc t โ eโ j2ฯ fc t
x (t) = 2
2j
2j
.
.
= โ21 ej2ฯ ( fc + fโ)t โ ej2ฯ ( fc โ fโ)t โ eโ j2ฯ ( fc โ fโ)t + eโ j2ฯ ( fc + fโ)t
The spectrum is {(โ f c โ fโ, โ0.5), (โ f c + fโ, 0.5), ( f c โ fโ, 0.5), ( f c + fโ, โ0.5)}, and the plot is
0.5
0.5
โ0.5
โ0.5
โ( f c + f โ )
โ fc
โ( f c โ f โ )
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
0
( f c โ fโ)
fc
( f c + fโ)
โฒ
f
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CHAPTER 3. SPECTRUM
P-3.8
DSP First 2e
(a) Using Eulerโs relation we get
x (t) = 10 + 10e j ฯ/4 e j2ฯ (100)t + 10eโ j ฯ/4eโ j2ฯ (100)t + 5e j2ฯ (250)t + 5eโ j2ฯ (250)t
The gcd of 100 and 250 is 50 so f0 = 50 and therefore N = 5. The nonzero Fourier coefficients are, therefore, aโ5 = 5,
j ฯ/4
, a0 = 10, a2 = 10e jฯ/4 , and a5 = 5.
aโ2 = 10eโ
(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.
10eโ j ฯ/4
10
10ej ฯ/4
5
โ250
5
โ100
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
โฒ
0
100
250
f
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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(ฯ1t + ฯ1 ) + A2 cos(ฯ2t + ฯ2 ) + A3 cos(ฯ3t + ฯ3 )
where ฯ1 < ฯ2 0. For periodicity with period T0 we require that ฯ0 = 2ฯ/T0. This means
that k1ฯ0 = ฯ2 โ ฯ1 and k2ฯ0 = ฯ2 + ฯ1, where k1 and k2 are integers and k2 > k1.
(b) Part (a) gives two equations for ฯ1 and ฯ2. If we solve them in terms of ฯ0 we get ฯ1 = (k2 โ k1 )ฯ0/2 and
ฯ2 = (k2 + k1 )ฯ0/2, so the main condition is that both ฯ1 and ฯ2 are integer multiples of ฯ0/2. This is the most
general condition.
Therefore, the relationship between ฯ2 and ฯ1 is
ฯ2 =
k2 + k1
k 2 โ k1
ฯ1
if x (t + T0 ) = x (t). Thus, ฯ2 could be an integer multiple of ฯ1 if k2 โ k1 divides into k2 + k1 with no remainder, but
that is not necessary for periodicity of x (t).
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
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CHAPTER 3. SPECTRUM
P-3.11
DSP First 2e
(a) The gcd of 40, 60, 120 is 20 so ฯ0 = 20ฯ and the fundamental period is T0 = 2ฯ/ฯ0 = 0.1s.
The finite Fourier
โ
series has components
indexed by 0, ยฑ2, ยฑ3, ยฑ6 so N = 6. The coefficients are a0 = 2, aยฑ2 = 2e jฯ/5 , aยฑ3 = 1.5e โ j ฯ/2,
ฯ/3
aยฑ6 = 2eโ
(b) The spectrum is
{(โ120ฯ, 2ej ฯ/3 ), (โ60ฯ, 1.5ej ฯ/2 ), (โ40ฯ, 2ej ฯ/5 ), . . .
(0, 2), (40ฯ, 2eโ j ฯ/5 ), (60ฯ, 1.5eโ j ฯ/2 ), (120ฯ, 2eโ j ฯ/3 )}
2e j ฯ/3
2e j ฯ/5
2
2eโ j ฯ/5
1.5eโ j ฯ/2
1.5ej ฯ/2
โ120ฯ
โ60ฯ
2eโ j ฯ/3
โ40ฯ
โฒ
0
40ฯ
60ฯ
120ฯ
ฯ
(c) Now the fundamental frequency is 10ฯ rad/s because the gcd of 20, 40, 50, and 120 is 10.
Therefore, the period is
T0 = 2ฯ/10ฯ = 1/5 = 0.2s.
The spectrum is the same as in part (b) except there are two additional components at
ยฑ50ฯ rad/s: (โ50ฯ, 5ej ฯ/6 ) and (50ฯ, 5eโ j ฯ/6 ).
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
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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 ยท 2(nโ49)/12
(c) The spectrum would have the form:
{(โ440, aโ), (โ370, aโ), (โ294, aโ), (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.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
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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 โ 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(2ฯ(40)(t โ .005)) = 5 + 15 cos(80ฯt โ 0.4ฯ)
(c) x (t) = 5 + 7.5ej (80ฯtโ0.4ฯ) + 7.5eโ j (80ฯtโ0.4ฯ) = 7.5ej0.4ฯeโ j80ฯt + 5 + 7.5eโ j0.4ฯej80ฯt
(d) Plot of the two-sided spectrum of the signal x (t).
7.5eโ j0.4ฯ
7.5ej0.4ฯ
5
โ80ฯ
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
โฒ
0
80ฯ
ฯ
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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
โ
โ
Xโ1 =
2 โ j 2 = 2eโ
j ฯ/4
j ฯ/3
X2 = 8e
ฯ1 = 70ฯ
ฯ2 = โ100ฯ
(b) x (t) = 20 + 4 cos(70ฯt + ฯ/4) + 16 cos(100ฯt + ฯ/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 โ20 โค 4 cos(70ฯt + ฯ/4)+16 cos(100ฯt + ฯ/3) โค 20 since the individual terms satisfy โ4 โค 4 cos(70ฯt +
ฯ/4) โค 4 and โ16 โค 16 cos(100ฯt + ฯ/3) โค 16. The value ยฑ20 would be attained only if the phases of the two cosines
are such that 4 cos(70ฯ(t โ t0 )) + 16 cos(100ฯ(t โ t0 )).
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
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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 (7ฯt โ 0.1ฯ)
We need to express x (t) in terms of complex exponentials
.
.2
x (t) = 0.5ej (7ฯtโ0.1ฯ) + 0.5eโ j (7ฯtโ0.1ฯ) = 0.25ej0.2ฯeโ j14ฯt + 0.5 + 0.25eโ j0.2ฯej14ฯt
0.5
0.25eโ j0.2ฯ
0.25ej0.2ฯ
โ14ฯ
โฒ
0
ฯ
14ฯ
(b) y(t) = cos2 (7ฯt โ 0.1ฯ) cos(77ฯt + 0.1ฯ)
.
..
.
x (t) = 0.25e j0.2ฯ eโ j14ฯt + 0.5 + 0.25eโ j0.2ฯ e j14ฯt 0.5e j0.1ฯ e j77ฯt + 0.5eโ j0.1ฯ e โ j77ฯt
= 0.125ej0.1ฯeโ j91ฯt + 0.25eโ j0.1ฯeโ j77ฯt + 0.125eโ j0.3ฯeโ j63ฯt
+ 0.125ej0.3ฯej63ฯt + 0.25ej0.1ฯej77ฯt + 0.125eโ j0.1ฯej91ฯt
0.25eโ j0.1ฯ
0.125ej0.1ฯ
0.25ej0.1ฯ
0.125ej0.3ฯ
0.125eโ j0.3ฯ
โ91ฯ โ77ฯ โ63ฯ
0.25eโ j0.1ฯ
โฒ
0
63ฯ
77ฯ
91ฯ
ฯ
Note: ฯ axis is
not to scale.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
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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,
0.5eโ j0.3ฯ
โ3
0.5ej0.3ฯ
โฒ
0
f (kHz)
3
(b) The spectrum for x (t) versus f in kHz.
0.75
0.25eโ j0.3ฯ
0.75
0.25ej0.3pi
โ683 โ680 โ677
0.25eโ j0.3ฯ
0.25ej0.3ฯ
677
683
โฒ
0
680
f (kHz)
Note: f axis is not to scale.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
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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(2ฯ(105)t โ 0.4ฯ) + 14 cos(2ฯ(180)t โ 0.6ฯ)
(c) Simplify the numerical values for the complex amplitudes, i.e., phases should be in [โฯ, ฯ].
x2 (t) = 22 cos(2ฯ(105)(t โ 0.05) โ 0.4ฯ) + 14 cos(2ฯ(180)(t โ 0.05) โ 0.6ฯ)
= 22 cos(2ฯ(105)t โ 10.5ฯ โ 0.4ฯ) + 14 cos(2ฯ(180)t โ 18ฯ โ 0.6ฯ)
= 22 cos(2ฯ(105)t โ 10ฯ โ 0.9ฯ) + 14 cos(2ฯ(180)t โ 18ฯ โ 0.6ฯ)
= 22 cos(2ฯ(105)t โ 0.9ฯ) + 14 cos(2ฯ(180)t โ 0.6ฯ)
Note that even multiples of 2ฯ rad can be dropped from the equation. Thus, the spectrum is:
{(โ180, 7ej0.6ฯ ), (โ105, 11ej0.9ฯ ), (105, 11eโ j0.9ฯ ), (180, 7eโ j0.6ฯ )} 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 ยฑ105 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 โ105 Hz will move to f = 0, and the new DC component is equal to the value of the
spectrum originally at f = โ105 Hz, i.e., 11ej0.9ฯ .
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
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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.
11eโ j0.4ฯ
11ej0.4ฯ
8
7ej0.6ฯ
7eโ j0.6ฯ
โฒ
โ180
โ105
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 9eยฑ j0.8ฯ at frequencies ยฑ40
Hz.
11ej0.4ฯ
7ej0.6ฯ
9eโ j0.8ฯ
9ej0.8ฯ
11eโ j0.4ฯ
7eโ j0.6ฯ
โฒ
โ180
โ105
โ40
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 j2ฯ f , where f is the frequency of the complex
exponential component. So we get
j2ฯ(105)11eโ j0.4ฯ
โ j2ฯ(105)11ej0.4ฯ
j2ฯ(180)7eโ j0.6ฯ
โ j2ฯ(180)7ej0.6ฯ
โ180
โ105
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
0
105
180
โฒ
f
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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 = 8eโ j ฯ/3
and
ฯ1 = โ10ฯ
(b) Here is the plot of the spectrum of x (t).
8eโ j ฯ/3
8e j ฯ/3
โ10ฯ
โฒ
0
ฯ
10ฯ
(c) The symmetry implies that ฯb = 20ฯ and B = +4 j. Furthermore, symmetry requires that ฯ a = 0. To find A, ฯc ,
and ฯ we can write y(t) as y(t) = 0.5x (t)e jฯ e jฯ c t + 0.5x (t)eโ jฯ e โ jฯc 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 ฯc and left (down) by ฯc . In order to have only three components we must choose ฯc = 10ฯ so that two of the shifted spectrum lines over lap at ฯ = 0.
4e j ฯ/3 e jฯ + 4eโ j ฯ/3eโ jฯ
4e j ฯ/3e โ jฯ
4eโ j ฯ/3 e jฯ
โฒ
โ20ฯ
0
20ฯ
ฯ
Now, 4eโ j ฯ/3ejฯ = 4eโ j ฯ/2 so ฯ = โฯ/6. Finally, note that the DC value can be written as A = 8 cos(ฯ/3 + ฯ) =
โ
โ
8 cos(ฯ/6) = 8 3/2 = 4 3.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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, ยฑ4, ยฑ9.
k
โ9
โ4
0
4
9
ak
0.4eโ j2
0.6ej1.4
0.5
0.6eโ j1.4
0.4ej2
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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 ฯi (t) =
dฯ
dt= 2ฮฑt + ฮฒ, so ฯ1 = ฯi (0) = ฮฒ and ฯ2 = ฯi (T2 ) = 2ฮฑT2 + ฮฒ.
(b) The instantaneous frequency versus time is ฯi (t) = 80t + 27
(c) Here is the plot of the instantaneous frequency (in Hz) versus time over the range 0 โค t โค 1 sec.
f (t) =
i
โป
ฯi (t )
2ฯ
107
2ฯ
โ
โโ
โ
โโ
27 โโ
2ฯ
โ
โโ
โ
โโ
โโ
โ
โ
โโ
โโ
โ
โ
โโ
โฒ
0
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
1
t
May 20, 2016
ยฉ 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(ฮฑt2 + ฮฒt + ฯ). The instantaneous frequency of this signal is
ฯi (t) = 2ฮฑt + ฮฒ. From this we observe that ฯ1 = 2ฯ f1 = 2ฯ(4800) = ฯi (0) = ฮฒ. To obtain ฮฑ, we note that ฯ2 =
2ฯ(800) = ฯi (2) = 2ฮฑ(2) + ฮฒ = 4ฮฑ + 9600ฯ so ฮฑ = โ2000ฯ. Therefore, the signal is
x (t) = cos(โ2000ฯt2 + 9600ฯt + ฯ)
where ฯ is an arbitrary phase constant.
(b) The instantaneous frequency is ฯi = 800ฯt + 500ฯ, so ฯ1 = ฯi (0) = 500ฯ and ฯ2 = ฯi (3) = 800ฯ(3) + 500ฯ =
2900ฯ rad/s.
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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 ฯi (t) = 2ฮฑt + ฮฒ. Substituting the given parameters gives ฮฑ = 4ฯ and ฮฒ = 2ฯ, so the
signal with the given parameters is x (t) = cos(4ฯt2 + 2ฯt + ฯ).
frequency in Hz
(bโf) 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
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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
โ
2ฯ (t)
(c) f3 (t) = e2t /ฯ
(d) f4 (t) = โ sin(2ฯt)
frequency in Hz
1
frequency in Hz
frequency in Hz
0.8
frequency in Hz
(b) f2 (t) = t/ฯ
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
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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) =
โ
.
k=โโ
โฟ
โโโ
1 .
โ
(2ฯโ
/T0 )k (0) =
akโ
e jโ
โ
.
ak ej (2ฯ/T0 )kt . Then it follows that
k=โโ
ak .
k=โโ
(b) Let f3 = โ f2 = f0 and f4 = โ f 1 = 3 f0 so that from the spectrum we can write
x (t) = 12 cos(2ฯ f0t + ฯ/4) + 4 cos(6ฯ f0t + 3ฯ/4)
โ
โ
โ 2
Therefore x (0) = 12 cos(ฯ/4) + 4 cos(3ฯ/4) = 6 โ 2 2 = 4 2. Now if we add the coefficients of the Fourier series
we get
โ
a1 + a2 + a3 + a4 = 2eโ j3ฯ/4 + 6eโ j ฯ/4 + 6ej ฯ/4 + 2ej3ฯ/4 = 12 cos(ฯ/4) + 4 cos(3ฯ/4) = 4 2
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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)
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
x1 (t) = 4 cos(4ฯt + ฯ) + 4 cos(6ฯt + ฯ/2)
x2 (t) = 2 cos(4ฯt + ฯ/4) + 4 cos(6ฯt โ 0.333ฯ)
x3 (t) = โ3 + 2 cos(4ฯt + ฯ/4)
x4 (t) = โ2 + 4 cos(4ฯt + ฯ)
x5 (t) = 4 cos(2ฯt + ฯ) + 4 cos(4ฯt + ฯ)
May 20, 2016
ยฉ 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(โ250ฯt2 )
Spectrogram (2)
(d) x (t) = cos(100ฯt) cos(400ฯt)
Spectrogram (1)
(b) x (t) = cos(100ฯt โ ฯ/4) + cos(400ฯt)
Spectrogram (5)
(e) x (t) = cos(200ฯt2 )
Spectrogram (6)
(c) x (t) = cos(1000ฯt โ 250ฯt2 )
Spectrogram (4)
(f) x (t) = cos(30e2t )
Spectrogram (3)
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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
ยฉJ. H. McClellan, R. W. Schafer, & M. A. Yoder
May 20, 2016
ยฉ 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

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