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Unformatted text preview: Fourier series is then
∞
1
f (t) =
ej 2πnt/T
T
n=−∞ = ∞
1 jnω0t
e
.
T n=−∞ EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 9 The Fourier transform of δT (t) is then
F [δT (t)] = ∞
1
2πδ (ω − nω0)
T n=−∞ = ω0 ∞
n=−∞ δ (ω − nω0) = ω0δω0 (ω )
since ω0 = 2π /T .
We then have the transform pair
δT (t) ⇔ ω0δω0 (ω )
The Fourier transform of an array evenly spaced δ ’s is another array of
evenly spaced δ ’s!
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 10 The delta train δT (t) is
!T (t ) 1 3T 2T T 0 T 2T 3T t and its Fourier transform ω0δω0 (t) is
!0"!0 (!) !0 −3!0 −2!0 −!0 !0 0 !0 = 2"/T 2!0 3!0 ! These are very useful functions! EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 11 Ideal Sampling
One of the most important uses of δT (t) is to represent sampling.
If f (t) is a signal, then
f (t)δT (t) = f (t) ∞
n=−∞ = ∞
n=−∞ = ∞
n=−∞ δ (t − nT ) f (t)δ (t − nT )
f (nT )δ (t − nT ) where we have used the fact that f (t)δ (t − T ) = f (T )δ (t − T ) for the last
step.
EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 12 Original Signal 3T f (t ) 2T T 0 T 2T 3T t 2T 3T t !T (t ) 1
2T 3T t × Sampling Function 3T 2T T 0 T =
Sampled Signal 3T 2T f (t )!T (t ) T 0 T EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 13 Example: Square Wave, Revisited
We can write the square wave function from earlier today as a convolution:
f (t) = rect(t) ∗ δ2(t)
which is illustrated below
rect(t )
−3 −3 −2 −2 −1 −1 0 ∗
0 = 1 2 3t 2 3t 2 3t !2(t )
1
f (t ) −3 −2 −1 0 EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 1 14 Using the convolution theorem, the Fourier transform of f (t) is
F [f (t)] = F [rect(t)] F [δ2(t)] = sinc(ω /2π )ω0δω0 (ω )
= π sinc(ω /2π )δπ (ω ) where ω0 = 2π /T = 2π /2 = π .
To get this into the form we found earlier, expand δπ (ω )
F [f (t)] = π sinc(ω /2π )
=π ∞
n=−∞ =π ∞
n=−∞ ∞
n=−∞ δ (ω − π n) sinc(ω /2π )δ (ω − π n)
sinc(n/2)δ (ω − π n) EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 15 which is the same thing we obtained before.
In general, if f1(t) is one cycle of a periodic function with period T , then
f (t) is
f (t) = f1(t) ∗ δT (t)
and the Fourier transform is F (j ω ) = ω0F1(j ω )δω0 (ω )
Recall that multiplying by δω0 (ω ) samples F1(j ω ) at multiples of ω0.
The Fourier transform of the periodic signal is the sampled Fourier transform
of one period EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 16 This is what we saw the periodic rect signal example:
F1( j!) −12!0 −8!0 −4!0 !0F1( j!)"!0 (!) 0 4!0 8!0 EE102A:Signal Processing and Linear Systems I; Win 0910, Pauly 12!0 ! 17 Sampling Theorem
What is the Fourier transform of a signal that has been sampled...
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This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.
 Fall '13
 Mukamel
 Signal Processing

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