FEG2D3 -INW- 2016 · Ekspresi deret Fourier diberikan berikut ini
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Transcript of FEG2D3 -INW- 2016 · Ekspresi deret Fourier diberikan berikut ini
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FEG2D3
-INW-
2016
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convert time-domain signals into frequency-domain (or spectral) representations.
In 1807, Jean Baptiste Joseph Fourier submitted a paper using trigonometric series to represent “any” periodic signals
But Lagrange rejected it!!
In 1822, Fourier published a book “The Analytical Theory of Heat”
He also claimed that “any” periodic signal could be represented byFourier series.
He however obtained a representation for aperiodic signalsi.e., Fourier integral or transform
Fourier did not actually contribute to the mathematical theoryof Fourier series
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a continuous-time signal x(t) to be periodic if there is a positive nonzero value of T for which:
𝑥 𝑡 + 𝑇 = 𝑥 𝑡 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡
The fundamental period 𝑇0 of 𝑥(𝑡) is the smallest positive value of T
𝑓0 =1
𝑇0is referred to as the fundamental frequency
Two basic examples of periodic signals are the real sinusoidal signal
And the complex exponential signal𝑥 𝑡 = 𝑒𝑗𝜔𝑡
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The complex exponential Fourier series representation of a periodic signal x(t) with fundamental period T0 is given by
𝑥 𝑡 =
𝑘=−∞
∞
𝑐𝑘𝑒𝑗𝑘𝜔0𝑡 ; 𝜔0 =
2𝜋
𝑇0
where c, are known as the complex Fourier coefficients and are given by
𝑐𝑘 =1
𝑇0න
𝑇0
𝑥(𝑡)𝑒−𝑗𝑘𝜔0𝑡𝑑𝑡
Where𝑇0denotes the integral over any one period and 0 to To or -To/2 to T0/2 is
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Set k = 0, then:
𝑐0 =1
𝑇0න
𝑇0
𝑥(𝑡) 𝑑𝑡
which indicates that c0 equals the average value of x(t) over a period.
When x(t) is real, then it follows that:
𝑐−𝑘 = 𝑐𝑘∗
where the asterisk indicates the complex conjugate.
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The trigonometric Fourier series representation of a periodic signal x(t) with fundamental period T0 is given by
where ak and bk are the Fourier coefficients given by
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If a periodic signal x(t) is even, then bk = 0 and its Fourier series contains only cosine terms:
If x(t) is odd, then ak = 0 and its Fourier series contains only sine terms:
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Sinyal Sinusoid Kompleks
x(t) = 𝒆𝒋𝒌ω𝒐𝒕 = 𝒄𝒐𝒔𝒌ω𝒐𝒕+ j sin𝒌ω𝒐𝒕
Konjugate
x*(t) = 𝒆−𝒋𝒌ω𝒐𝒕= 𝒄𝒐𝒔𝒌ω𝒐𝒕- j sin𝒌ω𝒐𝒕
Rumus Euler
cos 𝒌ω𝒐𝒕 = 𝒆𝒋𝒌ω𝒐𝒕+𝒆−𝒋𝒌ω𝒐𝒕
2sin 𝒌ω𝒐𝒕 =
𝒆𝒋𝒌ω𝒐𝒕−𝒆−𝒋𝒌ω𝒐𝒕
2𝑗
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Tentukan koefisien DF untuk sinyal x(t)
Perioda x(t) adalah T=2, ω𝒐 =𝟐π
𝑻=
𝟐π
𝟐= π
x(t) = 𝑒−2𝑡, 0 ≤ 𝑡 ≤ 2 ; x(t) = 𝟏
𝑻𝟎𝑻𝒙 𝒕 𝒆−𝒋𝒌ω𝒐𝒕𝒅𝒕
x(t) = 𝟏
𝟐𝟎𝟐𝑒−2𝑡 𝒆−𝒋𝒌ω𝒐𝒕𝒅𝒕
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X[k] = 𝟏
𝟐𝟎𝟐𝑒−2𝑡 𝒆−𝒋𝒌π𝒕𝒅𝒕
X[k] = 𝟏
𝟐𝟎𝟐𝑒−(2+𝒋𝒌π)𝑡 𝒅𝒕
X[k] = −1
2(2+𝒋𝒌π)𝑒−(2+𝒋𝒌π)𝑡 |
20
X[k] = −1
4+𝒋𝒌𝟐π𝑒− 2+𝒋𝒌π 2 − 𝑒− 2+𝒋𝒌π 0
X[k] = −1
4+𝒋𝒌𝟐π𝑒− 2+𝒋𝒌π 2 − 𝑒0
X[k] = −1
4+𝒋𝒌𝟐π𝑒−4−𝟐𝒋𝒌π − 1
X[k] = −1
4+𝒋𝒌𝟐π𝑒−4𝑒−𝟐𝒋𝒌π − 1
X[k] = −1
4+𝒋𝒌𝟐π𝑒−4(1) − 1
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𝑒−𝟐𝒋𝒌π = 𝒄𝒐𝒔𝟐𝒌π − 𝒋𝒔𝒊𝒏𝒌π𝑒−𝟐𝒋𝒌π = 𝟏 − 𝟎𝑒−𝟐𝒋𝒌π = 𝟏 , 𝒖𝒏𝒕𝒖𝒌 0 ≤ 𝑘 ≤ ~
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Tentukan representasi Deret Fourier Sinyal :
x(t) = 3 cos (π
2𝑡 +
π
4) memakai metoda inspeksi.
ω𝒐 = π
2, T=
2π
ω𝒐=
2ππ
2
= 𝟒 (Perioda fundamental/Perioda Dasar)
x(t) = 3 cos (π
2𝑡 +
π
4) = 3
𝑒𝑗(π2𝑡+
π4)+𝑒
−𝑗(π2𝑡+
π4)
2
x(t) = 3
2𝑒𝑗
π
4𝑒𝑗π
2𝑡 +
3
2𝑒−𝑗
π
4𝑒−𝑗π
2𝑡
Maka X[k] =
3
2𝑒−𝑗
π
4 , 𝒖𝒏𝒕𝒖𝒌 𝒌 = −𝟏
3
2𝑒𝑗
π
4 , 𝒖𝒏𝒕𝒖𝒌 𝒌 = 𝟏
𝟎 , 𝒏𝒊𝒍𝒂𝒊 𝒍𝒂𝒊𝒏
x(t) = σ𝑘=−~~ X[k] 𝒆𝒋𝒌ω𝒐𝒕
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Tentukan deret Fourier dari bentuk gelombang periodik berikut ini
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Ekspresi deret Fourier diberikan berikut ini
Tujuan kita adalah mendapatkan koefisien deret a0, an, dan bn
Dari gambar diperoleh
T=2
ω0=2π/T= π
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Dengan mensubtitusikan koefisien diperoleh:
Dapat diringkas sebagai:
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Jika sinyal periodik tersebut diketahui simetrik genap
maka diperoleh
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Jika sinyal periodik tersebut diketahui simetrik ganjil
maka diperoleh
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Jika sinyal periodik tersebut diketahui simetrik ganjil setengah gelombang maka diperoleh
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Let x( t ) be a nonperiodic signal of finite duration, that is,
𝑥 𝑡 = 0 ; 𝑡 > 𝑇1
Such a signal is shown below . Let xT0(t) be a periodic signal formed by repeating x(r) with fundamental period T,
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If we let 𝑇0 → ∞
The complex exponential Fourier series of 𝑥𝑇0(𝑡) is given by
where
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Since xT0 t = x(t) for t < T0/2 and also since x( t) = 0 outside this interval
Let us define 𝑋(𝜔) as
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the complex Fourier coefficients ck can be expressed as
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The function 𝑋 𝜔 is called the Fourier transform of 𝑥 𝑡
While the invers Fourier Transform is
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1. Linearity
2. Time Shifting
the effect of a shift in the time domain is simply to add a linear term −ωt0, to the original phase spectrum θ(ω). This is known as a linear phase shift of the Fourier transform X ω
3. Frequency ShiftingThe multiplication of x(t) by a complex exponential signal 𝑒𝑗𝜔0𝑡is sometimes called complex modulation. Thus, Equation below shows that complex modulation in the time domain corresponds to a shift of X ω in the frequency domain.
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4. Time Scaling
where 𝑎 is a real constant.
This equation indicates that scaling the time variable t by the factor 𝑎
causes an inverse scaling of the frequency variable ω by 1
𝑎
as well as an amplitude scaling of X𝜔
𝑎by
1
𝑎(. Thus, the scaling
property implies that time compression of a signal (a > 1) results in its spectral expansion and that time expansion of the signal (a < 1) results in its spectral compression.
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5. Time Reversal
time reversal of x(t) produces a like reversal of the frequency axis for 𝑋(𝜔). Equation below is readily obtained by setting 𝑎 = −1
6. Duality or symmetry
7. Differentiation in the Time Domain
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8. Differentiation in the Frequency Domain
9. Integration in the time domain
10. Convolution
11. Multiplication
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The output 𝑦(𝑡) of a continuous-time LTI system equals the convolution of the input 𝑥(𝑡) with the impulse response ℎ(𝑡)
𝑦 𝑡 = 𝑥 𝑡 ∗ ℎ(𝑡)
Applying the convolution property, we obtain:𝑌 𝜔 = 𝑋 𝜔 𝐻 𝜔
Relationships between inputs and outputs in an LTI system shown by:
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Where 𝑌 𝜔 , 𝑋(𝜔) and 𝐻(𝜔) are the Fourier transforms of 𝑦 𝑡 , 𝑥(𝑡) and ℎ(𝑡), respectively we have
𝐻 𝜔 =𝑌(𝜔)
𝑋(𝜔)
The function 𝐻(𝜔) is called the frequency response of the system.
Let 𝐻 𝜔 = 𝐻(𝜔) 𝑒𝑗𝜃𝐻(𝜔)
Then |𝐻 𝜔 | is called the magnitude response of the system
And 𝜃𝐻(𝜔) is called the phase response of the system𝜃𝐻 𝜔 = tan−1𝜔
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A system described by: 𝑦′ 𝑡 + 2𝑦 𝑡 = 𝑥 𝑡 + 𝑥′(𝑡)
Find the impulse response ℎ(𝑡) of the system
Answer𝑗𝜔𝑌 𝜔 + 2𝑌 𝜔 = 𝑋 𝜔 + 𝑗𝜔𝑋(𝜔)
Or 𝑗𝜔 + 2 𝑌 𝜔 = 1 + 𝑗𝜔 𝑋 𝜔
𝐻 𝜔 =𝑌(𝜔)
𝑋(𝜔)=1 + 𝑗𝜔
2 + 𝑗𝜔=2 + 𝑗𝜔 − 1
2 + 𝑗𝜔= 1 −
1
2 + 𝑗𝜔
Inverse Fourier Transform of 𝐻(𝜔), the impulse response ℎ(𝑡) is
ℎ 𝑡 = 𝛿 𝑡 − 𝑒−2𝑡𝑢(𝑡)
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𝑑𝑦(𝑡)
𝑑𝑥+ 2𝑦 𝑡 = 𝑥(𝑡)
Find the output 𝑦(𝑡) if 𝑥 𝑡 = 𝑒−1𝑢(𝑡)
Answer:𝑗𝜔𝑌 𝜔 + 2𝑌 𝜔 = 𝑋(𝜔)
𝐻 𝜔 =𝑌(𝜔)
𝑋(𝜔=
1
2 + 𝑗𝜔
𝑒−1𝑢 𝑡 ↔1
1+𝑗𝜔
𝑌(𝜔)= X(𝜔) H 𝜔 =1
(1+𝑗𝜔)(2+𝑗𝜔)=
1
1+𝑗𝜔−
1
2+𝑗𝜔↔ 𝑦 𝑡 = 𝑒−𝑡 − 𝑒−2𝑡 𝑢(𝑡)
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𝑦 𝑛 = 𝑥 𝑛 ∗ ℎ 𝑛 ↔ 𝑌 Ω = 𝑋 Ω 𝐻 Ω
𝐻 Ω =𝑌(Ω)
𝑋(Ω)→ 𝐻 Ω = |𝐻(Ω)|𝑒−𝑗𝜃𝐻(Ω)
The relationship is figured by:
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As in the continuous-time case,
the function 𝐻(Ω) is called the
frequency response of the
system. |H Ω | the magnitude
response and 𝜃𝐻(Ω) is the phase
response
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A causal discrete time LTI system is described by:
𝑦 𝑛 −3
4𝑦 𝑛 − 1 +
1
8𝑦[𝑛 − 2] = 𝑥[𝑛]
a. Determine the frequency response 𝐻 Ω
b. Find the impulse response ℎ 𝑛
𝑌 Ω −3
4𝑒−𝑗Ω𝑌 Ω +
1
8𝑒−𝑗2Ω𝑌 Ω = 𝑋(Ω)
1 −3
4𝑒−𝑗Ω +
1
8𝑒−𝑗2Ω 𝑌 Ω = 𝑋(Ω)
𝐻 Ω =1
1 −12𝑒−𝑗Ω 1 −
14𝑒−𝑗Ω
↔2
1 −12𝑒−𝑗Ω
−1
1 −14𝑒−𝑗Ω
ℎ 𝑛 = 21
2
𝑛
−1
4
𝑛
𝑢[𝑛]
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𝑦 𝑛 −1
2𝑦 𝑛 − 1 = 𝑥 𝑛 +
1
2𝑥[𝑛 − 1]
Determine 𝐻 Ω , ℎ 𝑛
𝑌 Ω −1
2𝑒−𝑗Ω𝑌 Ω = 𝑋 Ω +
1
2𝑒−𝑗Ω𝑋 Ω
𝐻 Ω =1+
1
2𝑒−𝑗Ω
1−1
2𝑒−𝑗Ω
=1
1−1
2𝑒−𝑗Ω
+1
2
𝑒−𝑗Ω
1−1
2𝑒−𝑗Ω
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