Comments on delta/step function
▪ In chapter 1, we had the relationship of unit step and unit impulse
▪𝛿 𝑛 = 𝑢 𝑛 −𝑢 𝑛−1
(1.65)
𝑢[𝑛]=σ+∞
𝑘=0 𝛿[𝑛 − 𝑘]
(1.67)
∞,
𝑖𝑓 𝑡 = 0
the Dirac delta 𝛿 t = ቊ
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
∞
and ‫׬‬−∞ 𝑥(𝜏)𝛿(t - 𝜏)d 𝜏 = 𝑥 𝑡 . 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑖𝑠 Eq, we may derive (2.6) from (2.2).
the FT of Dirac delta is white noise; or inverse FT constant is the delta:
1 ∞ 𝑗𝜔𝑡
𝛿(𝑡)= ‫׬‬−∞ 𝑒
d𝜔, often used in mathematical physics [1].
2𝜋
1,
𝑖𝑓 𝑛 = 0,
(1.63)
Unit impulse 𝛿[𝑛] = ቊ
0,
𝑖𝑓 𝑛 ≠ 0
1, 𝑖𝑓 𝑡 > 0,
1, 𝑖𝑓 𝑛 ≥ 0, (Fig 1.18)
(1.70),
▪ In contrast,u 𝑡 = ቊ
u[n] = ቊ
(Fig 1.29)
0, 𝑖𝑓 𝑡 < 0.
0, 𝑖𝑓 𝑛 < 0.
[1]. G. Pan and L. Zhang, “Closed Form Solution to the Incident Power of Gaussian-Like Beam for Scattering Problems,”
IEEE TRANSACTIONS ON ANTENNAS AND PROPAGATION, VOL. 67, NO. 2, FEBRUARY 2019.
1
Properties of Dirac delta function
• Dirac is a distribution (function)
properties of 𝛿 𝑥
• Show that 𝑡𝛿 𝑡 = 0.
EE 3316
Lecture 3
2
Chapter 1 outline
❑Periodic function
(1.11)
▪
𝑥 𝑡 = 𝑥 𝑡 + 𝑇 , 𝑇-period
▪
𝑥[𝑛] = 𝑥[𝑛 + 𝑁], 𝑁-period
▪ Complex exponential and sinusoidal
𝑒
𝑗𝜔0 𝑡
=𝑒
𝑗𝜔0 (𝑡+𝑇)
(1.12)
(1.22)
,
2𝜋
𝑇=
𝜔0
Period
(1.25)
(1.26)
▪ Sinusoidal 𝑥 𝑡 = 𝐴𝑐𝑜𝑠 𝜔0 𝑡 + 𝜙
▪ Euler relation 𝑒 𝑗𝜔0𝑡 = 𝑐𝑜𝑠𝜔0 𝑡 + 𝑗𝑠𝑖𝑛𝜔0 𝑡
❑Even- and odd- functions
𝑥 −𝑡 = 𝑥 𝑡 , even, symmetric about y-axis
(1.14)
(1.16)
𝑥 −𝑡 = −𝑥 𝑡 , odd, antisymmetric
▪ Any signal can be decomposed as even- and odd1
(1.18)
▪ 𝑥𝑒 𝑡 : = [𝑥 𝑡 + 𝑥 −𝑡 ], verify that 𝑥𝑒 is even,
2
1
2
▪ 𝑥𝑜 𝑡 : = [𝑥 𝑡 − 𝑥 −𝑡 ], verify that 𝑥𝑜 is odd.
EE 3316
𝑥 𝑡 = 𝑥𝑒 𝑡 + 𝑥𝑜 𝑡
Lecture 3
(1.19)
3
❑Discrete complex exponentials
▪ 𝑒 𝑗(𝜔0 +2π)𝑛 = 𝑒 𝑗𝜔0 𝑛 ,
0≤ 𝜔0 < 2𝜋, or -𝜋 ≤ 𝜔0 < 𝜋
▪ Fundamental period N,
2𝜋
N=𝑚
𝜔0
(1.51)
Fig 1.42 cascade- and
parallel- interconnection
(1.58)
❑Interconnects of systems
▪ Cascade,
▪ Parallel,
▪ Series-parallel,
▪ Feedback.
EE 3316
Fig 1.43 feedback
interconnection
Lecture 3
4
Chapter 2 LTI: Convolution sum and convolution integral
Section 2.1. 1 discrete signal in terms of impulses
(2.2)
▪ output
(2.6)
EE 3316
Lecture 3
5
convolution sum
❑ Discrete vs continuous convolution
𝑥[𝑛]
▪ Discrete: convolution sum
A signal x(t) is represented by series of
discrete samples. Mathematically, in the
convolution sum, each term is a scalar
product of x[k] and shifted
unit “area” delta 𝛿[𝑛 − 𝑘].
For
n =0, 1, 2, …
n =-1, -2, …
𝛿[𝑛 − 𝑘] right shift
𝛿[𝑛 − 𝑘] left shift
As showing on next slide,
EE 3316
Lecture 3
6
Discrete convolution summation
▪ Signal x(t) passing through a system of ℎ 𝑛 = 𝛼 𝑛 u(n), and let 𝛼=0.9,
x[0]=1, x[1]=.8, x[2]=.6, x[3]=.7,…
𝑥[𝑘] h[𝑛 − 𝑘]
𝑥[𝑛]
x[-1]=.3, x[-2]=.5, x[-3]=.2, …
1.0
0.9
h[-1]=0,h[0]=1, h[1]=.9,
h[2]=.81, h[3]=.729, …
0.8
▪ Find the output, y[n].
0.72
▪ Solution
0.3
0.27
y[n]= x[n]*h[n]
0.5
=σ∞
𝑥
𝑘
ℎ
𝑛
−
𝑘
0.45
𝑘=0
= 𝑥 0 ℎ[𝑛]+x[1]h[n-1]+…
Summing up all rows, we got output series
EE 3316
Lecture 3
7
2.2 Convolution integral
Fig. 2.12
▪ MIT explained graphically that the Dirac delta function 𝛿(𝜏)
moves/duplicates a smooth integrand, x(𝜏), from
origin to the parameter
t location.
▪ This is continuous
▪ the Dirac delta
∞,
▪𝛿 t = ቊ
0,
EE 3316
𝑖𝑓 𝑡 = 0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(2.27)
Lecture 3
8
Comparison of continuous vs discrete
Signal x(t), output y(t)
(2.39)
(2.27)
(2.33)
(2.34)
EE 3316
Lecture 3
9
working example 1
• Discrete: y n = σ𝑘 𝑥 𝑘 ℎ 𝑛 − 𝑘 , 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑘 = 𝑢 𝑘 , ℎ 𝑛 = 𝛼 𝑛 𝑢 𝑛 .
Solution. h[n-k]=𝛼 𝑛−𝑘 𝑢 𝑛 − 𝑘 ; n=0 →h[-k]= 𝛼 −𝑘 , k=0, -1, -2, ⋯
n=1→h[1-k]= 𝛼 1−𝑘 , k=1, 0, -1, -2, ⋯
𝛼
y[0]=x[0]h[0]=1
y[1]=x[1]h[0]+x[0]h[1]=1+𝛼
y[2]=x[2]h[0]+x[1]h[1]+x[0]h[2]=1+𝛼 + 𝛼 2
y[3]=x[3]h[0]+x[2]h[1]+x[1]h[2]+x[0]h[3]=1+𝛼 + 𝛼 2 + 𝛼 3
For n<0, no overlap of x[.] and h[.]. Therefore, y[n] =0.
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Lecture 3
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Discrete: convolution sum
• By mathematic induction, or by intuition, we write the Eq of y[n]:
𝑛+1
𝑛 1−𝛼
+⋯ + 𝛼 =
,
1−𝛼
• y[n]= 1+𝛼 + 𝛼
n=0, 1, 2, ⋯
• We may restrict 𝛼<1, otherwise, the system will violate stability.
• For n<0, y[n] = 0.
Appendix: summation of geometric series
Let S = 1 + 𝛼 + 𝛼 2 + ⋯ + 𝛼 𝑛
Multiplying both sides by 𝛼, one has
𝛼S=
𝛼 + 𝛼 2 + ⋯ + 𝛼 𝑛 + 𝛼 𝑛+1 ,
Subtracting lower- from upper- Eq,
(1 − 𝛼 )S = 1 −
𝛼 + 𝛼 2 + ⋯ + 𝛼 𝑛 + 𝛼 𝑛+1 ,
2
1−𝛼𝑛+1
→ S=
1−𝛼
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Lecture 3
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Continuous: convolution integral
❑Many approaches to compute convolution analytically
▪ Laplace transform: in the transform domain, h(t)*x(t) ⇔ H(s)X(s)
▪ Using Laplace Transform tables in many books
▪ Partial fraction method
▪ Residue technique
▪ Time-domain: the integrand may be different in different integration regens
EE 3316
Lecture 3
12
Continuous convolution integration
Example 2.6 h(t)=u(t), 𝑥(𝑡)=𝑒 −𝑎𝑡 𝑢(𝑡), 𝑎>0.
Solution.
∞
𝑦(𝑡)= 𝑥 𝑡 ∗ ℎ 𝑡 = ‫=𝜏׬‬−∞ ℎ 𝜏 𝑥 𝑡 − 𝜏 𝑑𝜏
∞
∞
−𝑎 𝑡−𝜏
= ‫=𝜏׬‬−∞ 𝑢 𝜏 𝑒
𝑢(𝑡 − 𝜏)𝑑𝜏 = ‫=𝜏׬‬0 𝑒 −𝑎 𝑡−𝜏 𝑢(𝑡 − 𝜏)𝑑𝜏
first Heaviside u(τ) changes lower integral limit from 𝜏 = −∞ to τ =0
∞
𝑡
−𝑎
𝑡−𝜏
= ‫=𝜏׬‬0 𝑒
𝑢(𝑡 − 𝜏)𝑑𝜏 = ‫=𝜏׬‬0 𝑒 −𝑎 𝑡−𝜏 𝑑𝜏
2nd Heaviside u(t-τ) changes upper integral limit from 𝜏 = ∞ to τ = t
𝑡
𝑡
1 𝑎τ 𝜏 = 𝑡
−𝑎
𝑡−𝜏
−𝑎𝑡
𝑎τ
−𝑎𝑡
= ‫=𝜏׬‬0 𝑒
𝑑𝜏 = 𝑒
‫=𝜏׬‬0 𝑒 𝑑𝜏= 𝑒 [ 𝑎 𝑒 ]
τ=0
1
= 𝑎 [1 − 𝑒 −𝑎𝑡 ], t>0.
EE 33116
Lecture 3
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Convolution for continuous signals
❑ Textbook solution of Example 2.6 seems simpler
❑ Our approach utilizes the Heaviside step function extensively
EE 3316
Lecture 3
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