1. Signals and Spectra

2. Roadmap
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1. Line Spectra and Fourier Series
2. Fourier Transforms and Continuous Spectra
3. Time and Frequency Relations
4. Convolution
5. Impulses and Transforms in the Limit
6. Discrete Time Signals and the Discrete Fourier Transform
2

3. 11/30/2012 8:18 AM
LINE SPECTRA AND FOURIER SERIES
• Phasors and Line Spectra
• Periodic Signals and Average Power
• Fourier Series
• Convergence Conditions and Gibbs Phenomenon
• Parseval’s Power Theorem
3

4. Phasors and Line Spectra
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v(t) = A cos(ω0t + φ)
The phasor representation of a sinusoidal signal comes from Euler’s theorem
𝒆±𝒋𝜽 = 𝐜𝐨𝐬 𝜽 ± 𝒋 𝐬𝐢𝐧 𝜽
Any sinusoid as the real part of a complex exponential
4
𝒋(𝝎 𝟎 𝒕+𝝋) 𝒋𝝋 𝒋𝝎 𝟎 𝒕
𝑨 𝐜𝐨𝐬 𝝎 𝟎 𝒕 + 𝝋 = 𝑨 𝑹𝒆 𝒆 = 𝑹𝒆 [𝑨𝒆 𝒆 ]

5. Any sinusoid as the real part of a complex exponential
𝑨 𝐜𝐨𝐬 𝝎 𝟎 𝒕 + 𝝋 = 𝑨 𝑹𝒆 𝒆 𝒋(𝝎 𝟎 𝒕+𝝋) = 𝑹𝒆 [𝑨𝒆 𝒋𝝋 𝒆 𝒋𝝎 𝟎 𝒕 ]
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This is called a phasor representation
Only three parameters completely specify a phasor: amplitude, phase angle, and
rotational frequency 5

6. A suitable frequency-domain description would be the line spectrum
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One sided spectra
Phase angles will be measured with respect to cosine waves. Hence, sine waves
need to be converted to cosines via the identity
sin ωt = cos (ωt – 90o)
We regard amplitude as always being a positive quantity. When negative signs
appear, they must be absorbed in the phase using 6
- A cos ωt = A cos (ωt ± 180o)

7. Recalling that Re[z] = ½ (z + z*)
If 𝒛 = 𝑨𝒆 𝒋𝝋 𝒆 𝝎 𝟎 𝒕
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then 𝐴 𝑗𝜑 𝑗 𝜔0 𝑡
𝐴 −𝑗𝜑 −𝑗 𝜔0 𝑡
𝐴 cos 𝜔0 𝑡 + 𝜑 = 𝑒 𝑒 + 𝑒 𝑒
2 2
Two sided spectra
7

8. consider the signal
w(t) = 7 – 10 cos(40πt – 60o) + 4 sin 120πt
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w(t) = 7 cos 2π0t + 10 cos(2π20t + 120o) + 4 cos (2π60t – 90o)
8

9. 11/30/2012 8:18 AM
9

10. Periodic Signals and Average
Power
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The average value of any function v(t) is defined as
𝑇/2
1
𝑣(𝑡) = lim 𝑣 𝑡 𝑑𝑡
𝑇→∞ 𝑇 −𝑇/2
In case of periodic signal
𝑡 1 +𝑇 𝑜
1 1
𝑣(𝑡) = 𝑣 𝑡 𝑑𝑡 = 𝑣 𝑡 𝑑𝑡
𝑇𝑜 𝑡1 𝑇𝑜 𝑇𝑜
The average power (normalized)
1
𝑃 = |𝑣 𝑡 |2 = |𝑣 𝑡 |2 𝑑𝑡
𝑇𝑜 𝑇𝑜
10
The average value of a power signal may be positive, negative, or zero.

11. Fourier Series
Let v(t) be a power signal with period T0=1/f0. Its exponential Fourier series expansion is
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∞
𝑣 𝑡 = 𝑐 𝑛 𝑒 𝑗 2𝜋𝑛𝑓0 𝑡 𝑛 = 0, 1, 2, …
𝑛=−∞
The series coefficients are related to v(t) by
1
𝑐𝑛 = 𝑣 𝑡 𝑒 −𝑗 2𝜋𝑛𝑓0 𝑡 𝑑𝑡
𝑇0 𝑇0
The coefficients are complex quantities in general, they can be expressed in the polar form
𝑐 𝑛 = 𝑐 𝑛 𝑒 𝑗 arg 𝑐𝑛
the nth term of the Fourier series equation being
11
𝑗 2𝜋𝑛𝑓0 𝑡
𝑐𝑛 𝑒 = 𝑐𝑛 𝑒 𝑗 arg 𝑐 𝑛 𝑒 𝑗 2𝜋𝑛𝑓0 𝑡

12. |c(nf0)| represents the amplitude spectrum as a function of f, and arg c(nf0) represents the
phase spectrum.
Three important spectral properties of periodic power signals are listed below.
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1. All frequencies are integer multiples or harmonics of the fundamental frequency f0=1/T0.
Thus the spectral lines have uniform spacing f0.
2. The DC component equals the average value of the signal, since setting n = 0
1
𝑐 0 = 𝑣 𝑡 𝑑𝑡 = 𝑣(𝑡)
𝑇𝑜 𝑇𝑜
3. If v(t) is a real (noncomplex) function of time, then
𝑐−𝑛 = 𝑐 ∗ = 𝑐 𝑛 𝑒 𝑗 arg
𝑛
𝑐𝑛
With replacing n by - n
𝑐 −𝑛𝑓0 = 𝑐 −𝑛𝑓0 arg 𝑐 −𝑛𝑓0 = − arg 𝑐 −𝑛𝑓0
which means that the amplitude spectrum has even symmetry and the phase spectrum has odd 12
symmetry.

13. trigonometric Fourier Series a one-sided spectrum
∞
𝑣 𝑡 = 𝑐0 + |2𝑐 𝑛 |cos⁡
(2𝜋𝑛𝑓0 𝑡 + arg 𝑐 𝑛 )
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𝑛=1
or ∞
𝑣 𝑡 = 𝑐0 + 𝑎 𝑛 cos 2𝜋𝑛𝑓0 𝑡 + 𝑏 𝑛 sin 2𝜋𝑛𝑓0 𝑡
𝑛=1
an = Re[cn] and bn = Im[cn]
These sinusoidal terms represent a set of orthogonal basis functions,
Functions vn(t) and vm(t) are orthogonal over an interval from t1 to t2 if
𝑡2
0 𝑛≠ 𝑚
𝑣 𝑛 𝑡 𝑣 𝑚 𝑡 𝑑𝑡 = 𝑤𝑖𝑡ℎ 𝐾 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑡1 𝐾 𝑛= 𝑚
13

14. The integration for cn often involves a phasor average in the form
𝑇 2
1 1 1
𝑒 𝑗 2𝜋𝑓 𝑡 𝑑𝑡 = 𝑒𝑗 𝜋𝑓𝑇
− 𝑒 −𝑗 𝜋𝑓𝑇
= sin 𝜋𝑓𝑇
𝑇0 −𝑇 2 𝑗2𝜋𝑓𝑇 𝜋𝑓𝑇
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we’ll now introduce the sinc function defined by
sin 𝜋𝜆
𝑠𝑖𝑛𝑐 𝜆 ≜
𝜋𝜆
sinc λ is an even function of λ having its peak at λ = 0 and zero crossings at all other integer
values of λ, so
0 𝜆=0
𝑠𝑖𝑛𝑐 𝜆 =
1 𝜆 = ±1, ±2, …
14

15. EXAMPLE: Rectangular Pulse Train
𝐴 𝑡 < 𝜏/2
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𝑣 𝑡 =
0 𝑡 > 𝜏/2
To calculate the Fourier coefficients
𝑇0 /2 𝜏/2
1 −𝑗 2𝜋𝑛 𝑓0 𝑡
1
𝑐𝑛 = 𝑣 𝑡 𝑒 𝑑𝑡 = 𝐴𝑒 −𝑗 2𝜋𝑛 𝑓0 𝑡
𝑑𝑡
𝑇0 𝑇0 /2 𝑇0 𝜏/2
𝐴 𝐴 sin 𝜋𝑛𝑓0 𝜏
= (𝑒 −𝑗 𝜋𝑛𝑓0 𝜏 − 𝑒 +𝑗 𝜋𝑛𝑓0 𝜏 ) =
−𝑗𝜋𝑛𝑓0 𝑇0 𝑇0 𝜋𝑛𝑓0
Multiplying and dividing by t finally gives
𝐴𝜏 15
𝑐𝑛 = 𝑠𝑖𝑛𝑐 𝑛𝑓0 𝜏
𝑇0

16. The amplitude spectrum obtained from |c(nf0)| = |cn| = Af0 τ|sinc nf0 τ|
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for the case of τ/T0 = τf0 =1/4
We construct this plot by drawing the continuous function Af0 τ|sinc nfτ| as a dashed
curve, which becomes the envelope of the lines.
The spectral lines at 4f0, 8f0, and so on, are ―missing‖ since they fall precisely at
multiples of 1/τ where the envelope equals zero.
The dc component has amplitude c(0) = Aτ/T0 which should be recognized as the
average value of v(t). 16
Incidentally, τ/T0 equals the ratio of ―on‖ time to period, frequently designated as
the duty cycle in pulse electronics work

17. The phase spectrum is obtained by observing that cn is always real but sometimes
negative.
Hence, arg c(nf0) takes on the values 0o and 180o, depending on the sign of sinc nf0 τ.
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Both +180o and -180o were used here to bring out the odd symmetry of the phase.
Having decomposed the pulse train into its frequency components, let’s build it back
up again. For that purpose, we’ll write out the trigonometric series, still taking
τ/T0 = f0 τ = 1/4, so c0=A/4 and |2cn|=(2A/4) |sinc n/4| = (2A/πn) |sin πn/4|. Thus
𝐴 2𝐴 𝐴 2𝐴
𝑣 𝑡 = + cos 𝜔0 𝑡 + cos 2𝜔0 𝑡 + cos 3𝜔0 𝑡 + …
4 𝜋 𝜋 3𝜋 17

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19. 11/30/2012 8:18 AM
19

20. Convergence Conditions and Gibbs
Phenomenon
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The Dirichlet conditions for Fourier series expansion are as follows:
If a periodic function v(t) has
•a finite number of maxima, minima, and discontinuities per period,
•and if v(t) is absolutely integrable, so that v(t) has a finite area per period,
then the Fourier series exists and converges uniformly wherever v(t) is continuous.
These conditions are sufficient but not strictly necessary.
20

21. An alternative condition is that v(t) be square integrable, so that |v(t)|2 has finite
area per period—equivalent to a power signal.
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Under this condition, the series converges in the mean such that if
𝑁
𝑣𝑁 𝑡 = 𝑐 𝑒 𝑗 2𝜋𝑛 𝑓0 𝑡
𝑛=−𝑁
then
2
lim 𝑣 𝑡 − 𝑣𝑁 𝑡 𝑑𝑡 = 0
𝑁→∞ 𝑇0
In other words, the mean square difference between v(t) and the partial sum vN(t)
vanishes as more terms are included. 21

22. Regardless of whether v(t) is absolutely integrable or square integrable, the series
exhibits a behavior known as Gibbs phenomenon at points of discontinuity.
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Gibbs ears,
Height is independent
of N
22

23. Parseval’s Power Theorem
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Parseval’s theorem relates the average power P of a periodic signal to its Fourier
coefficients.
1 1
𝑃= |𝑣 𝑡 |2 𝑑𝑡 = 𝑣 𝑡 𝑣 ∗ (𝑡)𝑑𝑡
𝑇𝑜 𝑇𝑜 𝑇𝑜 𝑇𝑜
∞ ∗ ∞
𝑣∗ 𝑡 = 𝑐 𝑛 𝑒 𝑗 2𝜋𝑛 𝑓0 𝑡
= 𝑐 ∗ 𝑒 −𝑗 2𝜋𝑛
𝑛
𝑓0 𝑡
𝑛=−∞ 𝑛=−∞
∞
1
𝑃= 𝑣 𝑡 𝑐 ∗ 𝑒 −𝑗 2𝜋𝑛
𝑛
𝑓0 𝑡
𝑑𝑡
𝑇𝑜 𝑇𝑜 𝑛 =−∞
∞
1
= 𝑣 𝑡 𝑒 −𝑗 2𝜋𝑛 𝑓0 𝑡
𝑑𝑡 𝑐∗
𝑛
𝑇0 𝑇0
𝑛 =−∞
∞ ∞ 23
𝑃= 𝑐 𝑛 𝑐∗ =
𝑛 |𝑐 𝑛 |2
𝑛=−∞ 𝑛=−∞

24. DRILL PROBLEMS
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24

25. 11/30/2012 8:18 AM
Solution
respectively.
25

26. 11/30/2012 8:18 AM
But
and its fundamental frequency is just
26

27. 11/30/2012 8:18 AM
Solution
Amplitude
27

28. 11/30/2012 8:18 AM
Why?
with Fourier coefficients
Applying Parseval’s power theorem then yields
28

29. Problem 1.3
Sketch the two-sided amplitude spectrum of the even symmetric
triangular wave listed in Table T.2 in the textbook.
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Solution
The periodic signal waveform is given by
29

30. with complex Fourier series coefficients
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by even symmetry
How?
Amplitude
30

31. Problem 1.4
Calculate the average power of the periodic signal of Problem 1.3.
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31

32. 11/30/2012 8:18 AM
FOURIER TRANSFORMS AND CONTINUOUS
SPECTRA
• Fourier Transforms
• Symmetric and Causal Signals
• Rayleigh’s Energy Theorem
• Duality Theorem
• Transform Calculations
32

33. Fourier Transforms
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Non-periodic signals …
average v(t) or |v(t)|2 over all time you’ll find that these averages equal zero.
normalized signal energy
∞
2
𝐸 ≜ 𝑣 𝑡 𝑑𝑡
−∞
33
When the integral exists and yields 0 < E < ∞, the signal v(t) is said to have well-
defined energy and is called a nonperiodic energy signal.

34. To introduce the Fourier transform, we’ll start with the Fourier series representation
of a periodic power signal
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Let the frequency spacing approach zero and the index n approach infinity such
that the product nf0 approaches a continuous frequency variable f
Fourier transform 34

35. The Fourier transform of v(t)
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The time function v(t) is recovered from V( f ) by the inverse Fourier transform
a nonperiodic signal will have a continuous spectrum rather than a line spectrum.
In the periodic case we return to the time domain by summing discrete-frequency
phasors, while in the nonperiodic case we integrate a continuous frequency
function
35

36. Three major properties of V( f ) are listed below
1. The Fourier transform is a complex function, so |V(f)| is the amplitude spectrum
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of v(t) and arg V(f) is the phase spectrum.
2. The value of V(f) at f = 0 equals the net area of v(t), since
which compares with the periodic case where c(0) equals the average value of v(t)
3. If v(t) is real, then
36

37. EXAMPLE: Rectangular Pulse
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and
so V(0) = Aτ, which clearly equals the pulse’s area.
37

38. 11/30/2012 8:18 AM
We may take 1/τ as a measure of the spectral ―width.‖
If the pulse duration is reduced (small τ), the frequency width is increased,
whereas increasing the duration reduces the spectral width
Thus, short pulses have broad spectra, and long pulses have narrow spectra. This 38
phenomenon, called reciprocal spreading,

39. Symmetric and Causal Signals
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Using 𝑒 −𝑗 2𝜋𝑓𝑡 = cos 𝜔𝑡 − 𝑗 sin 𝜔𝑡
even part of V(f)
odd part of V(f)
39

40. When v(t) has time symmetry,
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where w(t) stands for either v(t) cos ωt or v(t) sin ωt
If v(t) has even symmetry then v(t) cos ωt is even whereas v(t) sin ωt is odd, then
Vo(f) = 0 and
if v(t) has odd symmetry, then Ve(f) = 0 and
40

41. Consider the case of a causal signal, defined by the property that
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One-sided Laplace transform is a function of the complex variable s = σ + jω defined
by
you can get V( f ) from the Laplace transform by letting s = jω = j2πf.
41

42. EXAMPLE: Causal Exponential Pulse
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which is a complex function in unrationalized form. Multiplying numerator and
denominator by b - j2πf yields the rationalized expression
42

43. 11/30/2012 8:18 AM
Conversion to polar form then gives the amplitude and phase spectrum
43

44. 11/30/2012 8:18 AM
44

45. Rayleigh’s Energy Theorem
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Rayleigh’s energy theorem states that the energy E of a signal v(t) is related to
the spectrum V(f) by
Integrating the square of the amplitude spectrum over all frequency yields the
total energy.
45

46. The energy spectral density of a rectangular pulse, whose spectral width was claimed
to be |f|=1/τ. The energy in that band is the shaded area in the figure, namely
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the total pulse energy is E ≈ A2τ
46

47. Rayleigh’s theorem is actually a special case of the more general integral
relationship
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47

48. Duality Theorem
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The theorem states that if v(t) and V( f ) constitute a known transform pair, and if
there exists a time function z(t) related to the function V( f ) by
then
48

49. EXAMPLE: Sinc Pulse
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z(t) = A sinc 2Wt
We’ll obtain Z(f) by applying duality to the transform pair
Re-writing
49

50. Transform Calculations
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•
50

51. Fourier transform table.
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51

52. Fourier transform table. (continued)
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52

53. DRILL PROBLEMS
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53

54. 11/30/2012 8:18 AM
Solution
The signal waveform may be sketched according to the following table after
noting its even symmetry, as shown below.
54

55. 11/30/2012 8:18 AM
Ponder!
How would you
elaborate on
reciprocal spreading
for this signal?
and
55

56. 11/30/2012 8:18 AM
56

57. 11/30/2012 8:18 AM
TIME AND FREQUENCY RELATIONS
• Superposition
• Time Delay and Scale Change
• Frequency Translation and Modulation
• Differentiation and Integration
57

58. Superposition
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If a1 and a2 are constants and
then
generally
58

59. Time Delay and Scale Change
Replacing t by t - td in v(t) produces the time-delayed signal v(t - td).
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The amplitude spectrum remains unchanged in either case, since
proof
λ = t - td
59

60. Scale change, produces a horizontally scaled image of v(t) by replacing t with αt.
The scale signal v(αt) will be expanded if |α| < 1 or compressed if |α| > 1; a
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negative value of α yields time reversal as well as expansion or compression.
60

61. proof, for case α < 0
α = - |α |
λ = - |α|t
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t = λ/α
dt = -dλ/|α|
61

62. EXAMPLE: Superposition and Time Delay
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62

63. 11/30/2012 8:18 AM
we have : θ1 = - πftd
θ2 = - πf (td + T)
θ1 – θ2 = πfT
θ1 – θ2 = πft0
where t0 = td + T/2
63

64. with
T=τ
t0 = 0
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64

65. Frequency Translation and
Modulation
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Frequency translation or complex modulation
65

66. 1. The significant components are concentrated around the frequency fc.
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2. Though V(f) was bandlimited in W, V (f - fc) has a spectral width of 2W.
Translation has therefore doubled spectral width. Stated another way, the
negative-frequency portion of V(f) now appears at positive frequencies.
3. V(f - fc) is not hermitian but does have symmetry with respect to translated
origin at f = fc.
These considerations are the basis of carrier modulation, we have the following
modulation theorem:
66

67. EXAMPLE: RF Pulse
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67

68. Differentiation and Integration
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Differentiation theorem
68

69. Integration theorem
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Differentiation enhances the high-frequency components of a signal, since
|j 2πfV(f )| > |V(f)| for |f | > 1/2π.
Conversely, integration suppresses the high-frequency components.
69

70. EXAMPLE: Triangular Pulse
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Has zero net area,
and integration produces a triangular pulse shape
70

71. Applying the integration theorem to Zb (f)
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71

72. The triangular function
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72

73. DRILL PROBLEMS
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73

74. 11/30/2012 8:18 AM
(a) (b)
Solution
74
How?

75. 11/30/2012 8:18 AM
75

76. 11/30/2012 8:18 AM
Ponder!
Is it always possible
to interchange the
order of the
mathematical
operations as with
this problem?
76

77. The modulating signal
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Signal waveform
Its spectrum
77

78. Modulated signal (a)
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Signal waveform
Its spectrum
78

79. Modulated signal (b)
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Signal waveform
Its spectrum
79
Note: The phase spectrum in this sketch is not to scale.

80. 11/30/2012 8:18 AM
CONVOLUTION
• Convolution Integral
• Convolution Theorems
80

81. Convolution Integral
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The convolution of two functions of the same variable, say v(t) and w(t), is
defined by
take the functions
81

82. v(λ) has the same shape as v(t)
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But obtaining the picture of w(t - λ) as a function of requires two steps:
First, we reverse w(t) in time and replace t with λ to get w(λ);
Second, we shift w(λ) to the right by t units to get w[-(λ-t)] = w(t-λ) for a given
value of t.
82

83. As v(t) * w(t) is evaluated for -∞ < t < + ∞, w(t - λ) slides from left to right
with respect to y(λ)
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when t < 0
w(t - λ)
t-T t
functions don’t overlap
83

84. when 0 < t < T
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w(t - λ)
t-T t
functions overlap for 0 < λ < t
84

85. when t > T
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w(t - λ)
t-T t
functions overlap for t - T < λ < t
85

86. 11/30/2012 8:18 AM
86

87. Convolution Theorems
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proof
87

88. EXAMPLE: Trapezoidal Pulse
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If τ1 > τ2, the problem breaks up into three cases: no overlap, partial overlap, and
full overlap.
88

89. case 1: no overlap
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or
|t| > (τ1 + τ2)/2
89

90. case 2: partial overlap
The region where there is partial overlap
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and
90

91. case 2: total overlap
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91

92. 11/30/2012 8:18 AM
The spectrum becomes (A1τ sinc fτ) (A2τ sinc fτ) = Aτ sinc2fτ
92

93. EXAMPLE: Ideal Lowpass Filter
Rectangular function v(t) = AΠ (t/τ) whose transform, V(f) = Aτ sinc fτ,
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We can lowpass filter this signal at f = 1/τ by multiplying V(f) by the rectangular
function
The output function is
93

94. Review Questions
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94

95. Problems to Ponder
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95

96. 11/30/2012 8:18 AM
IMPULSES AND TRANSFORMS IN THE LIMIT
• Properties of the Unit Impulse
• Impulses in Frequency
• Step and Signum Functions
• Impulses in Time
96

97. Properties of the Unit Impulse
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The unit impulse or Dirac delta function δ(t)
97

98. 11/30/2012 8:18 AM
98

99. 11/30/2012 8:18 AM
99

100. Impulses in Frequency
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Knowing
100

101. Direct application of the frequency-translation and modulation theorems yields
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101

102. 11/30/2012 8:18 AM
then its Fourier transform is
102

103. EXAMPLE: Impulses and Continuous Spectra
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103

104. 11/30/2012 8:18 AM
104

105. Step and Signum Functions
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The unit step function
the signum function (also called the sign function)
105

106. The signum function is a limited case of the energy signal z(t), where
v(t) = e-btu(t) and
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so that z(t) → sgn t if b → 0
106

107. The step and signum functions are related by
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Hence,
107

108. An impulsive DC term appears in the integration theorem when the signal
being integrated has nonzero net area.
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since u(t - λ) = 0 for λ > 0
108

109. Impulses in Time
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let τ → 0
An impulsive signal with ―zero‖ duration has infinite spectral width,
whereas a constant signal with infinite duration has ―zero‖ spectral width.
109

110. 11/30/2012 8:18 AM
show that the inverse transform does, indeed, equal v(t).
But the bracketed integral equals δ(t - λ)
110

111. We relate the unit impulse to the unit step by means of the integral
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Repeatedly differentiate the signal in question until one or more stepwise
discontinuities first appear. The next derivative, say the nth, then includes an
impulse Ak δ(t - tk) for each discontinuity of height Ak at t = tk, so
where w(t) is a nonimpulsive function 111

112. EXAMPLE: Raised Cosine Pulse
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We’ll use the differentiation method to find the spectrum V(f ) and the high-
frequency rolloff
112
has no discontinuities

113. d2v(t)/dt2 is discontinuous at t = τ so
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the first term of d3v(t)/dt3 can be written as w(t) = -(π/τ)2dv(t)/dt
113

114. 11/30/2012 8:18 AM
114

115. Review Questions
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115

116. Problems to Ponder
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116

117. 11/30/2012 8:18 AM
DISCRETE TIME SIGNALS AND THE
DISCRETE FOURIER TRANSFORM
• Foundation
• Convolution using the DFT
117

118. Foundation
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If we sample a signal at a rate at least twice its bandwidth, then it can be completely
represented by its samples
sampled at rate fs = 1/Ts
118
where x(k) is a discrete-time signal

119. 11/30/2012 8:18 AM
Function X(n) is the Discrete Fourier transform (DFT)
The DFT spectrum repeats itself
every N samples or every fs Hz
The DFT is computed only for
positive N
Note the interval from
n→(n+1) represents fs /N
Hz
119
the discrete frequency

120. Inverse discrete Fourier transform (IDFT)
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120

121. Convolution using the DFT
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121

122. Review Questions
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122

123. Problems to Ponder
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123