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Signals and Spectra
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Roadmap1. Line Spectra and Fourier Series2. Fourier Transforms and Continuous Spectra3. Time and Frequency Relations4. Convolution5. Impulses and Transforms in the Limit6. Discrete Time Signals and the Discrete Fourier Transform
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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
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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 theorem𝒆±𝒋𝜽 = 𝐜𝐨𝐬𝜽 ± 𝒋𝐬𝐢𝐧𝜽
Any sinusoid as the real part of a complex exponential𝑨𝐜𝐨𝐬ሺ𝝎𝟎𝒕+ 𝝋ሻ= 𝑨 𝑹𝒆 �𝒆𝒋(𝝎𝟎𝒕+𝝋)൧= 𝑹𝒆 [𝑨𝒆𝒋𝝋𝒆𝒋𝝎𝟎𝒕]
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Any sinusoid as the real part of a complex exponential𝑨𝐜𝐨𝐬ሺ𝝎𝟎𝒕+ 𝝋ሻ= 𝑨 𝑹𝒆 �𝒆𝒋(𝝎𝟎𝒕+𝝋)൧= 𝑹𝒆 [𝑨𝒆𝒋𝝋𝒆𝒋𝝎𝟎𝒕] This is called a phasor representation
Only three parameters completely specify a phasor: amplitude, phase angle, and rotational frequency
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A suitable frequency-domain description would be the line spectrum
Phase angles will be measured with respect to cosine waves. Hence, sine waves need to be converted to cosines via the identitysin 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- A cos ωt = A cos (ωt ± 180o)
One sided spectra
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Recalling that Re[z] = (z + z*)½If 𝒛= 𝑨𝒆𝒋𝝋𝒆𝝎𝟎𝒕
𝐴cosሺ𝜔0𝑡+ 𝜑ሻ= 𝐴2𝑒𝑗𝜑𝑒𝑗𝜔0𝑡 + 𝐴2𝑒−𝑗𝜑𝑒−𝑗𝜔0𝑡 then
Two sided spectra
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consider the signal
w(t) = 7 – 10 cos(40πt – 60o) + 4 sin 120πt
w(t) = 7 cos 2π0t + 10 cos(2π20t + 120o) + 4 cos (2π60t – 90o)
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Periodic Signals and Average Power
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∞ → lim𝑇 = ۄ����������𝑣(𝑡)ۃ������� 1𝑇 න 𝑣ሺ𝑡ሻ𝑑𝑡𝑇/2−𝑇/2
1𝑇𝑜 න = ۄ����������𝑣(𝑡)ۃ������� 𝑣ሺ𝑡ሻ𝑑𝑡𝑡1+𝑇𝑜𝑡1 = 1𝑇𝑜 න 𝑣ሺ𝑡ሻ𝑑𝑡𝑇𝑜
𝑃= ۃ�������|𝑣ሺ𝑡ሻ|21 =ۄ����������𝑇𝑜 න |𝑣ሺ𝑡ሻ|2𝑑𝑡𝑇𝑜
The average value of a power signal may be positive, negative, or zero.
The average value of any function v(t) is defined as
In case of periodic signal
The average power (normalized)
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Fourier Series
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Let v(t) be a power signal with period T0=1/f0. Its exponential Fourier series expansion is
𝑣ሺ𝑡ሻ= 𝑐𝑛𝑒𝑗2𝜋𝑛𝑓0𝑡∞𝑛=−∞ 𝑛 = 0,1,2,…
The series coefficients are related to v(t) by
𝑐𝑛 = 1𝑇0න 𝑣ሺ𝑡ሻ𝑒−𝑗2𝜋𝑛𝑓0𝑡𝑑𝑡𝑇0
the nth term of the Fourier series equation being
𝑐𝑛 = ȁI𝑐𝑛ȁI𝑒𝑗ȁrg 𝑐𝑛
𝑐𝑛𝑒𝑗2𝜋𝑛𝑓0𝑡 = ȁI𝑐𝑛ȁI𝑒𝑗ȁrg 𝑐𝑛𝑒𝑗2𝜋𝑛𝑓0𝑡
The coefficients are complex quantities in general, they can be expressed in the polar form
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|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.
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
𝑐ሺ0ሻ= 1𝑇𝑜 න 𝑣ሺ𝑡ሻ𝑑𝑡𝑇𝑜 ۄ����������𝑣(𝑡)ۃ������� =
3. If v(t) is a real (noncomplex) function of time, then𝑐−𝑛 = 𝑐𝑛∗ = ȁI𝑐𝑛ȁI𝑒𝑗ȁrg 𝑐𝑛 With replacing n by - n ȁI𝑐ሺ−𝑛𝑓0ሻȁI= ȁI𝑐ሺ−𝑛𝑓0ሻȁI ȁrg𝑐ሺ−𝑛𝑓0ሻ= − ȁrg𝑐ሺ−𝑛𝑓0ሻ
which means that the amplitude spectrum has even symmetry and the phase spectrum has odd symmetry.
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trigonometric Fourier Series
𝑣ሺ𝑡ሻ= 𝑐0 + |2𝑐𝑛|cos (2𝜋𝑛𝑓0𝑡+ ȁrg𝑐𝑛)∞𝑛=1
𝑣ሺ𝑡ሻ= 𝑐0 + 𝑎𝑛 cos2𝜋𝑛𝑓0𝑡+ 𝑏𝑛 sin2𝜋𝑛𝑓0𝑡∞𝑛=1
or
ȁn = Re[cn] ȁnd bn = Im[cn]
a one-sided spectrum
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
න 𝑣𝑛ሺ𝑡ሻ𝑣𝑚ሺ𝑡ሻ𝑑𝑡𝑡2𝑡1 = ቄ0 𝑛 ≠ 𝑚𝐾 𝑛 = 𝑚 𝑤𝑖𝑡ℎ 𝐾 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
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The integration for cn often involves a phasor average in the form1𝑇0න 𝑒𝑗2𝜋𝑓𝑡𝑑𝑡𝑇 2Τ−𝑇 2Τ = 1𝑗2𝜋𝑓𝑇 ൫𝑒𝑗𝜋𝑓𝑇 − 𝑒−𝑗𝜋𝑓𝑇൯= 1𝜋𝑓𝑇sin𝜋𝑓𝑇
we’ll now introduce the sinc function defined by
𝑠𝑖𝑛𝑐 𝜆 ≜ sin𝜋𝜆𝜋𝜆
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 𝑠𝑖𝑛𝑐 𝜆= ൜ 0 𝜆= 01 𝜆= ±1,±2,…
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EXAMPLE: Rectangular Pulse Train
𝑣ሺ𝑡ሻ= ൜𝐴 ȁI𝑡ȁI< 𝜏/20 ȁI𝑡ȁI> 𝜏/2
To calculate the Fourier coefficients
𝑐𝑛 = 1𝑇0න 𝑣ሺ𝑡ሻ𝑒−𝑗2𝜋𝑛𝑓0𝑡𝑑𝑡𝑇0/2𝑇0/2 = 1𝑇0න 𝐴𝑒−𝑗2𝜋𝑛𝑓0𝑡𝑑𝑡𝜏/2
𝜏/2
= 𝐴−𝑗𝜋𝑛𝑓0𝑇0 (𝑒−𝑗𝜋𝑛𝑓0𝜏 − 𝑒+𝑗𝜋𝑛𝑓0𝜏) = 𝐴𝑇0sin𝜋𝑛𝑓0𝜏𝜋𝑛𝑓0
Multiplying and dividing by t finally gives
𝑐𝑛 = 𝐴𝜏𝑇0 𝑠𝑖𝑛𝑐 𝑛𝑓0𝜏
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The amplitude spectrum obtained from |c(nf0)| = |cn| = Af0 τ|sinc nf0 τ|
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). Incidentally, τ/T0 equals the ratio of “on” time to period, frequently designated as the duty cycle in pulse electronics work
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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.
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
𝑣ሺ𝑡ሻ= 𝐴4 + ξ2𝐴𝜋 cos𝜔0𝑡+ 𝐴𝜋cos2𝜔0𝑡+ ξ2𝐴3𝜋 cos3𝜔0𝑡+ …
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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.
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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.
Under this condition, the series converges in the mean such that if
𝑣𝑁ሺ𝑡ሻ= 𝑐 𝑒𝑗2𝜋𝑛𝑓0𝑡𝑁𝑛=−𝑁
lim𝑁 → ∞ න ȁI𝑣ሺ𝑡ሻ− 𝑣𝑁ሺ𝑡ሻȁI2𝑑𝑡𝑇0 = 0
then
In other words, the mean square difference between v(t) and the partial sum vN(t) vanishes as more terms are included.
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Regardless of whether v(t) is absolutely integrable or square integrable, the series exhibits a behavior known as Gibbs phenomenon at points of discontinuity.
Gibbs ears,Height is independent of N
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Parseval’s Power Theorem
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Parseval’s theorem relates the average power P of a periodic signal to its Fourier coefficients.
𝑃= 1𝑇𝑜 න |𝑣ሺ𝑡ሻ|2𝑑𝑡𝑇𝑜 = 1𝑇𝑜 න 𝑣ሺ𝑡ሻ𝑣∗(𝑡)𝑑𝑡𝑇𝑜
𝑣∗ሺ𝑡ሻ= 𝑐𝑛𝑒𝑗2𝜋𝑛𝑓0𝑡∞𝑛=−∞ ൩
∗ = 𝑐𝑛∗𝑒−𝑗2𝜋𝑛𝑓0𝑡∞𝑛=−∞
𝑃= 1𝑇𝑜 න 𝑣ሺ𝑡ሻ 𝑐𝑛∗𝑒−𝑗2𝜋𝑛𝑓0𝑡∞𝑛=−∞ ൩𝑑𝑡𝑇𝑜
= ቈ1𝑇0න 𝑣ሺ𝑡ሻ𝑒−𝑗2𝜋𝑛𝑓0𝑡𝑑𝑡 𝑇0 𝑐𝑛∗
∞𝑛=−∞
𝑃= 𝑐𝑛𝑐𝑛∗ ∞𝑛=−∞ = |𝑐𝑛|2∞
𝑛=−∞
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DRILL PROBLEMS
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Problem 1.1Find the fundamental frequency of the periodic signal
.
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is the sum of two periodic functions: and , of periods
Hence, is periodic if and only if
𝑇 1=2𝜋2 =𝜋
𝑇 2=2𝜋3
respectively.
𝑚𝑇1=𝑛𝑇2
where and are relatively prime integers.
Solution
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𝑇1
𝑇2= 𝑛𝑚= 𝜋
2𝜋 /3=3
2
and 2 and 3 are relatively prime integers, then and .
The fundamental period of therefore is simply
𝑇 0=𝑚𝑇 1=𝑛𝑇 2=2𝜋
and its fundamental frequency is just
𝑓 0=1𝑇0
=1
2𝜋 Hz .
But
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Problem 1.2Sketch the two-sided amplitude spectrum of the signal of Problem 1.1, then calculate its average power.
1Solution
𝑣 (𝑡 )=𝑒 𝑗2 𝑡+cos3 𝑡=𝑒 𝑗 2𝑡+12 𝑒
𝑗3 𝑡+12 𝑒
− 𝑗 3𝑡
0 1𝜋
32𝜋− 3
2𝜋
12
12
1
𝑓
Amplitude
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The above exponential series expansion of is just its Fourier series expansion,
Why?
with Fourier coefficientsBecause it consists of
frequency components whose frequencies are only
integer multiples of the fundamental frequency .
𝑐2=1 𝑐3=𝑐− 3=12
for all other
Applying Parseval’s power theorem then yields
𝑃= ⟨|𝑣 (𝑡 )|2 ⟩= ∑𝑛=−∞
∞
|𝑐𝑛|2=1+( 1
2 )2
+( 12 )
2
=1. 5 .
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Problem 1.3Sketch the two-sided amplitude spectrum of the even symmetric triangular wave listed in Table T.2 in the textbook.
1Solution
The periodic signal waveform is given by
𝑣 (𝑡 )=1− 4|𝑡|𝑇 0
for|𝑡|<𝑇 0
2, where
𝑡
𝑣 (𝑡 )
1
−1
𝑇 0−𝑇 0
𝑇 0/2−𝑇 0/2
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with complex Fourier series coefficients
𝑐𝑛=1𝑇 0
∫𝑇 0
❑
𝑣 (𝑡 )𝑒− 𝑗 2𝜋𝑛 𝑓 0 𝑡 𝑑𝑡= 1𝑇0
∫−𝑇 0/2
𝑇 0 /2
(1−4|𝑡|𝑇 0 )𝑒− 𝑗2𝜋𝑛 𝑓 0𝑡 𝑑𝑡
¿2𝑇 0
∫0
𝑇0 /2
(1− 4 𝑡𝑇0 )cos2𝜋𝑛 𝑓 0 𝑡 𝑑𝑡 by even symmetry
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¿ { 0 𝑛even(2/𝜋𝑛 )2 𝑛odd How?
Amplitude
1/𝜋 21/𝜋 2
1/ (3𝜋 )21/ (3𝜋 )2𝑓𝑓
0−𝑓 00
2 𝑓0
3 𝑓0
4 𝑓0−2 𝑓
0
−3 𝑓0
−4 𝑓0
Ponder!A time shift of will
bring the signal to odd symmetry. How will
that affect its Fourier coefficients.
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Problem 1.4Calculate the average power of the periodic signal of Problem 1.3.
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Problem 1.5Given two periodic signals and of the same fundamental period and with Fourier coefficients and , respectively, show that their product signal is also periodic of the same fundamental period and with Fourier coefficients given by.
Problem 1.6How many harmonic terms are required in the Fourier sereies of a periodic square wave with 50 percent duty cycle and amplitude to represent 99 percent of its power?
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FOURIER TRANSFORMS AND CONTINUOUS SPECTRA
• Fourier Transforms• Symmetric and Causal Signals• Rayleigh’s Energy Theorem• Duality Theorem• Transform Calculations
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Fourier Transforms
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average v(t) or |v(t)|2 over all time you’ll find that these averages equal zero.
Non-periodic signals …
normalized signal energy
𝐸 ≜ න ȁI𝑣ሺ𝑡ሻȁI2 𝑑𝑡∞−∞
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.
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To introduce the Fourier transform, we’ll start with the Fourier series representation of a periodic power signal
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
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The Fourier transform of v(t)
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
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Three major properties of V( f ) are listed below
1. The Fourier transform is a complex function, so |V(f)| is the amplitude spectrum 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
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EXAMPLE: Rectangular Pulse
and
so V(0) = Aτ, which clearly equals the pulse’s area.
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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 phenomenon, called reciprocal spreading,
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Symmetric and Causal Signals
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Using 𝑒−𝑗2𝜋𝑓𝑡 = cos𝜔𝑡− 𝑗sin𝜔𝑡
even part of V(f)
odd part of V(f)
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When v(t) has time symmetry,
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
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Consider the case of a causal signal, defined by the property that
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.
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EXAMPLE: Causal Exponential Pulse
which is a complex function in unrationalized form. Multiplying numerator anddenominator by b - j2πf yields the rationalized expression
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Conversion to polar form then gives the amplitude and phase spectrum
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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 thetotal energy.
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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
the total pulse energy is E ≈ A2τ
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Rayleigh’s theorem is actually a special case of the more general integral relationship
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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
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EXAMPLE: Sinc Pulse
z(t) = A sinc 2Wt
We’ll obtain Z(f) by applying duality to the transform pair
Re-writing
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Transform Calculations• Fouriertransform tables may be very helpful in the calculation of
the Fourier transform, especially when utilized in connection with the duality theorem.
• Similarly, Laplacetransform tables may also prove useful in the calculation of the Fourier transform when the signal in question is causalas was pointed out earlier.
• In addition, either approach could be utilized when applicable for the calculation of the Fourier series as follows.• Let be a periodic signal and let , a nonperiodic signal consisting of
one period of .• If it is possible to calculate , then the Fourier coefficients of are
simply given by (prove it!)50
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Fourier transform table.
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Fourier transform table. (continued)
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DRILL PROBLEMS
53
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54
Problem 2.1Sketch the waveform and the amplitude spectrum of the signal then determine the frequency bandwidth for which 99 percent of its energy is contained within.
2
Solution
The signal waveform may be sketched according to the following table after noting its even symmetry, as shown below.
2/𝑎
1/𝑎2/5𝑎1/5𝑎
0 𝑎2𝑎3𝑎𝑡
𝑣 (𝑡 )
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2𝜋
2𝜋𝑒−1
2𝜋𝑒−22𝜋𝑒−3
0 12𝜋 𝑎
𝑓
𝑉 ( 𝑓 )
1𝜋𝑎
32𝜋 𝑎
55
To calculate the Fourier transform of , we first not the analogy between its mathematical form and that of the Fourier transform of the signal , namely, in the Fourier transform table.
Rewriting
with .
Then by duality it follows that
2
and
Ponder!How would you
elaborate on reciprocal spreading
for this signal?
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2By Rayleigh’s energy theorem
Because of the form of the first integral is not easy to calculate.
For a specific frequency bandwidth, , on the other hand, and because of the symmetry property of the signal spectrum, the second integral may be modified to yield
with a maximum .
Hence, a 99% energy bandwidth, , corresponds to or
and
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TIME AND FREQUENCY RELATIONS
• Superposition• Time Delay and Scale Change• Frequency Translation and Modulation• Differentiation and Integration
57
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Superposition
58
If a1 and a2 are constants and
then
generally
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Time Delay and Scale Change
59
Replacing t by t - td in v(t) produces the time-delayed signal v(t - td).
The amplitude spectrum remains unchanged in either case, since
proof
λ = t - td
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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
negative value of α yields time reversal as well as expansion or compression.
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61
proof, for case α < 0α = - |α | λ = - |α|t
t = λ/αdt = -dλ/|α|
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62
EXAMPLE: Superposition and Time Delay 3
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we have : θ1 = - πftd θ2 = - πf (td + T) θ1 – θ2 = πfTθ1 – θ2 = πft0
where t0 = td + T/2
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64
with t0 = 0T = τ
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Frequency Translation and Modulation
65
Frequency translation or complex modulation
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66
1. The significant components are concentrated around the frequency fc.
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:
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67
EXAMPLE: RF Pulse 3
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Differentiation and Integration
68
Differentiation theorem
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69
Integration theorem
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.
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70
EXAMPLE: Triangular Pulse
Has zero net area,and integration produces a triangular pulse shape
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71
Applying the integration theorem to Zb (f) 3
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The triangular function 3
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DRILL PROBLEMS
73
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74
Problem 3.1Sketch the amplitude and phase spectra for the modulated signals depicted below, where the unmodulated carrier is .
Solution
0
1
−𝜋 𝜋 𝑡
1
𝜋 3𝜋 𝑡0
(a) (b)
(a) The modulated signal may be expressed as
whose Fourier transform is simplyHow?
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75
(b) The signal here could be interpreted in one of two ways:
Method 1: Modulation followed by a time delay
The modulated signal in this case is simply a delayed version of that in (a) and it
may be experessed as
whose Fourier transform is just
obtained by applying the modulation theorem followed by the time delay
property of the Fourier transform.
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76
Method 2: Time delay followed by modulation
In this specific instance could equally well be expressed as
whose Fourier transform is
which simplifies to
as before, after noting that .
Ponder!Is it always possible to interchange the
order of the mathematical
operations as with this problem?
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77
The modulating signal
0
1
−𝜋 𝜋 𝑡
Λ ( 𝑡𝜋 )
1𝜋
2𝜋
3𝜋− 3
𝜋− 2𝜋− 1
𝜋 0
𝜋𝜋 sinc 2𝜋 𝑓
𝑓
Signal waveform
Its spectrum
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78
Modulated signal (a)
5𝜋
7𝜋
8𝜋0
𝜋
𝑓6𝜋
3𝜋
4𝜋
2𝜋− 8
𝜋− 7𝜋− 6
𝜋− 5𝜋− 4
𝜋− 3𝜋− 2
𝜋
𝑉 ( 𝑓 )
0
1
−𝜋 𝜋 𝑡
𝑣 (𝑡 )
Signal waveform
Its spectrum
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79
Modulated signal (b)
1
𝜋 3𝜋 𝑡0
𝑤 (𝑡 )
5𝜋
7𝜋
8𝜋0
𝜋
𝑓6𝜋
3𝜋
4𝜋
2𝜋− 8
𝜋− 7𝜋− 6
𝜋− 5𝜋− 4
𝜋− 3𝜋− 2
𝜋
|𝑊 ( 𝑓 )|arg𝑊 ( 𝑓 )=− 4𝜋 2 𝑓
Note: The phase spectrum in this sketch is not to scale.
Signal waveform
Its spectrum
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CONVOLUTION
• Convolution Integral• Convolution Theorems
80
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Convolution Integral
81
The convolution of two functions of the same variable, say v(t) and w(t), is defined by
take the functions
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82
v(λ) has the same shape as v(t)
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.
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83
w(t - λ)
t - T t
As v(t) * w(t) is evaluated for -∞ < t < + ∞, w(t - λ) slides from left to rightwith respect to y(λ)
when t < 0
functions don’t overlap
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84
w(t - λ)
t - T t
when 0 < t < T
functions overlap for 0 < λ < t
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85
w(t - λ)
t - T t
when t > T
functions overlap for t - T < λ < t
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86
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Convolution Theorems
87
proof
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88
EXAMPLE: Trapezoidal Pulse
If τ1 > τ2, the problem breaks up into three cases: no overlap, partial overlap, and full overlap.
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89
case 1: no overlap
or
|t| > (τ1 + τ2)/2
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90
case 2: partial overlap
The region where there is partial overlap
and
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91
case 2: total overlap 4
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92
The spectrum becomes (A1τ sinc fτ) (A2τ sinc fτ) = Aτ sinc2fτ
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93
EXAMPLE: Ideal Lowpass Filter
Rectangular function v(t) = AΠ (t/τ) whose transform, V(f) = Aτ sinc fτ,
We can lowpass filter this signal at f = 1/τ by multiplying V(f) by the rectangular function
The output function is
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Review Questions
94
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Problems to Ponder
95
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IMPULSES AND TRANSFORMS IN THE LIMIT
• Properties of the Unit Impulse• Impulses in Frequency• Step and Signum Functions• Impulses in Time
96
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Properties of the Unit Impulse
97
The unit impulse or Dirac delta function δ(t)
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98
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99
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Impulses in Frequency
100
Knowing
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101
Direct application of the frequency-translation and modulation theorems yields 5
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102
then its Fourier transform is
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103
EXAMPLE: Impulses and Continuous Spectra 5
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104
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Step and Signum Functions
105
The unit step function
the signum function (also called the sign function)
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106
The signum function is a limited case of the energy signal z(t), where v(t) = e-btu(t) and
so that z(t) → sgn t if b → 0
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107
The step and signum functions are related by
Hence,
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108
An impulsive DC term appears in the integration theorem when the signalbeing integrated has nonzero net area.
since u(t - λ) = 0 for λ > 0
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Impulses in Time
109
let τ → 0
An impulsive signal with “zero” duration has infinite spectral width, whereas a constant signal with infinite duration has “zero” spectral width.
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110
show that the inverse transform does, indeed, equal v(t).
But the bracketed integral equals δ(t - λ)
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111
We relate the unit impulse to the unit step by means of the integral
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
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112
EXAMPLE: Raised Cosine Pulse
We’ll use the differentiation method to find the spectrum V(f ) and the high-frequency rolloff
has no discontinuities
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113
d2v(t)/dt2 is discontinuous at t = ±τ so
the first term of d3v(t)/dt3 can be written as w(t) = -(π/τ)2dv(t)/dt
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114
5
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Review Questions
115
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Problems to Ponder
116
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DISCRETE TIME SIGNALS AND THE DISCRETE FOURIER TRANSFORM
• Foundation• Convolution using the DFT
117
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Foundation
118
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
where x(k) is a discrete-time signal
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119
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
the discrete frequency
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120
Inverse discrete Fourier transform (IDFT) 6
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Convolution using the DFT
121
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Review Questions
122
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Problems to Ponder
123
6
