SAMPLING THEOREM
SAMPLING THEOREM: STATEMENT [1/3]
1. Statement of Sampling Theorem
• Given: Continuous-time signal x(t).
2. Derivation of Sampling Theorem
• That’s: Bandlimited to B Hertz.
3. Ideal Reconstruction from Samples
• Means: Maximum frequency is B Hertz.
4. Determining Signal Bandwidths
• Means: X(ω) = F{x(t)}=0 for |ω| ≥ 2πB.
5. Finite Pulse Width Sampling
RADIAN
• Means: Bandwidth=B Hertz=2πB SECOND
.
6. Undersampling and Aliasing
SAMPLING THEOREM: STATEMENT [2/3]
SAMPLING THEOREM: STATEMENT [3/3]
• Given: Knowledge of only the samples {x(nT )} of x(t).
• Then: x(t) can be reconstructed from its samples {x(nT )}
SAMPLE
.
• Means: x(t) sampled every T seconds, at rate S= T1 SECOND
• If: Sampling rate S =
Signal x(t)
✬✩
Pulse Train p(t)
✲
×
t
✻✻✻✻✻✻✻✻✻
✫✪
✲
=
t
❄
❄ ❄
❄
✲
• Note: Co-discovered by Claude Shannon (UM Class of 1938)
t
SAMPLING THEOREM: EXAMPLE #1
SAMPLE
1
x(t)=cos(2π1000t) sampled at S=8000 SECOND
. T = S1 = 8000
sec.
1000
π
Set t=nT =n/8000: x(nT )=cos(2π 8000 n)=cos( 4 n).
Can reconstruct x(t) from its samples {x(nT )}: (How? See below.)
n
... 0 1 2
3
4 ...
x(nT ) . . . 1.0 .71 0.0 –.71 –1.0 . . .
• Note: Digital Signal Processing is possible because of this.
SAMPLING THEOREM: EXAMPLE #2
Do we need S > 2B or only S ≥ 2B?
SAMPLE
1
x(t)=sin(2π500t) sampled at S=1000 SECOND
. T = S1 = 1000
sec.
500
Set t=nT =n/1000: x(nT )=sin(2π 1000 n)=sin(πn)=0!
Can’t reconstruct x(t) from its samples {x(nT )}!
1
0.5
1
0
0.5
−0.5
0
−1
0
−0.5
−1
0
0.5
1
1.5
2
2.5
sec
3
3.5
4
4.5
5
−3
x 10
> 2B=2(bandwidth).
• Where: S > 2B Here 2B is the Nyquist sampling rate.
Sampled Signal x(t)p(t)
✻✻✻
✻
✻
1 SAMPLE
T SECOND
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
0.01
SAMPLING THEOREM: PROOF [1/3]
P∞
• DEF: p(t)=
n=−∞ δ(t–nT )=train
SAMPLING THEOREM: PROOF [2/3]
• Fourier series expansion: p(t)= pk ej2πkt/T = T1
(series) of impulses.
P
• DEF: x(t)p(t)=sampled signal=train of impulses weighted by {x(nT )}.
• Fourier coefficient formula: pk = T1
• Note: x(t)p(t)=x(t) δ(t–nT )= x(nT )δ(t–nT ).
• Then: F{x(t)p(t)}= T1
• So: Only need samples {x(nT )} to create x(t)p(t).
• S= T1 → F{x(t)p(t)}= T1
P
Signal x(t)
✬✩
P
Pulse Train p(t)
×
✲ t
✻✻✻✻✻✻✻✻✻
✫✪
Sampled Signal x(t)p(t)
=
✲ t
✻
✻ ✻
✻
✻
❄
❄❄❄
✲
F{x(t)}
✟✟❍❍
❍❍
✟✟
✲
t
–B
B
P
ej2πkt/T .
R T /2
−j2πkt/T
dt= T1 .
−T /2 δ(t)e
F{x(t)ej2πkt/T }= T1
P
P
X(2π(f – Tk ))= T1
P
X(ω–2π Tk ).
X(2π(f –kS)).
P
F{x(t)p(t)}=Spectrum of Sampled Signal
f
✟✟❍❍
❍❍
✟✟
–B–S
✟✟❍❍
❍❍
✟✟
B–S –B
✟✟❍❍
❍❍
✟✟
✲
B S–B
SAMPLING THEOREM: PROOF [3/3]
SAMPLING THEOREM: EXAMPLE
• Can: Reconstruct x(t) from x(t)p(t) if (S–B) > B → S > 2B .
• Given: Continuous-time x(t) is bandlimited to 4 kHz.
• By: Low-pass filtering x(t)p(t). Cutoff frequency=B Hertz.
SAMPLE
• Sample: 10 “kHz”=10000 SECOND
> 2(4 kHz).
• Formula: x(t)= 2B
2B T
X
|
x(nT )δ(t
− nT )} ∗
{z
SAMPLED SIGNAL x(t)p(t)
sin(2πBt)
.
πt
|
{z
}
• F{x(t)p(t)}=Spectrum of sampled signal. Repeats in f !
• Reconstruct: Low-pass filter with cutoff=4 kHz.
LPF h(t)
F{x(t)}
2πB(t−nT )
.
• Interpolation Formula: x(t)= x(nT )(2BT ) sin2πB(t−nT
)
P
✟❍
❍❍
✟✟
✟
❍ ✲
–4
SAMPLING AND RECONSTRUCTION: SUMMARY
p(t)
p(t)
↓
↓
O
O
x(t) → → x(t)p(t)
→ {x(nT )}} → → x(t)p(t) → LOWPASS
FILTER → x(t)
|
{z
|
SAMPLING
✬✩
×
✲
t
✫✪
✟✟❍❍
❍❍ ✲
✟✟
–B
B
f
{z
✟✟❍❍
❍❍
✟✟
–B–S
}
RECONSTRUCTION
✻✻✻✻✻✻✻✻✻
=
✲
t
✻
✻ ✻
✻
✻
✟✟❍❍
❍❍
✟✟
B–S –B
✲
t
x(t)p(t)
❄
❄ ❄
❄
✟✟❍❍
❍❍✲
✟✟
B S–B
f
S+B
F{x(t)p(t)}
f
S+B
4
F{x(t)p(t)}=Spectrum of Sampled Signal
f
✟❍
❍❍
✟✟
✟
❍
–14
–6
✟❍
❍❍
✟✟
✟
❍
–4
✟❍
❍❍
✟✟
✟
❍✲
4 6
14
f kHz
DETERMINING SIGNAL BANDWIDTH: EXAMPLES
1. x(t)= 3t+7
sin(2π5t): Rewrite: x(t)=3 sin(2π5t)+7π sin(2π5t)
.
t
πt
2. x(t)=sin(2π3t) sin(2π2t): Rewrite: x(t)= 12 [cos(2π1t)–cos(2π5t)].
1
[rect( ω6 ) ∗ rect( ω4 )].
3. x(t)= sin(2π3t)t2sin(2π2t) : Then X(ω)= 2π
Convolve length=6 with length=4→length=10.
SAMPLE
For all 3: Bandlimited to ω = 10π or f =5 Hertz. Need 10 SECOND
.
SAMPLING WITH FINITE-WIDTH PULSES [1/2]
SAMPLING WITH FINITE-WIDTH PULSES [2/2]
• Problem: The ideal impulse train p(t)= δ(t–nT ) doesn’t exist!
• Fourier series expansion: pq (t)= pq,k ej2πkt/T . Pulse width: ∆.
• Sol’n: Use pq (t)= q(t–nT ) where q(t)=short pulse (does exist).
• Fourier coefficient formula: pq,k = T1
• Then: pq (t)= q(t–nT )= δ(t–nT ) ∗ q(t)=p(t) ∗ q(t):
p(t)
pq (t)
q(t)
*
=
✻ ✻ ✻ ✻ ✻
• Then: F{x(t)pq (t)}= pq,k F{x(t)ej2πkt/T }= pq,k X(ω–2π Tk ).
P
P
P
P
–2T
0
✲
2T t
P
–2T
0
✲
2T t
–2T
0
• Sample: x(t)=cos(2π600t) with
• Get: Samples
Set
x(nT )=cos(2π 600n
1000 )=cos(1.2πn)=cos(0.8πn).
• Since: cos(1.2πn)=cos(1.2πn–2πn)=cos(−0.8πn)=cos(0.8πn).
F{x(t)p(t)}
✟❍
❍❍
✟✟
✟
❍ ✲
B
f
✏✏PPP
P
✏✏
–B–S
✟❍
❍❍
✟✟
✟
❍
B–S –B
✏✏PPP
P✲
✏✏
B S–B
f
S+B
UNDERSAMPLING AND ALIASING [2/4]
• Image spectrum impersonates actual spectrum! Overlap!
• Low-pass filtering this→400 Hz, not 600 Hz, sinusoid!
• Aliasing: image spectrum masquerades as actual spectrum!
n
t=nT = 1000
:
P
F{x(t)}
✲
MAX.
= S < 2B = 2 FREQ
=Nyquist frequency?
1
T = 1000
.
P
P
2T t
SAMPLE
• EX: 600 Hz sinusoid sampled at 1000 SECOND
. 1000<2(600).
sin( kπ∆
T ).
• S= T1 → F{x(t)p(t)}= pq,k X(2π(f – Tk ))= pq,k X(2π(f –kS)).
UNDERSAMPLING AND ALIASING [1/4]
SAMPLE
RATE
1
−j2πkt/T
dt= kπ
−∆/2 e
P
–B
• What if
R ∆/2
0.5
0.4
0.3
0.2
0.1
0
−1500
−1000
−500
0
500
1000
1500
UNDERSAMPLING AND ALIASING [3/4]
UNDERSAMPLING AND ALIASING [4/4]
SAMPLE
• 500-Hertz sinusoid sampled at 450 SECOND
. Samples: circles.
SAMPLE
• x(t)=cos(2π300t+1) sampled at 500 SECOND
. Reconstructed=?
• Did samples come from 450 Hertz or 50 Hertz sinusoid?
n
→ x(nT )=cos(2π 300
• Sample: t= 500
500 n+1)=cos(1.2πn+1).
• But: cos(1.2πn+1)=cos(1.2πn–2πn+1)=cos(−0.8πn+1)=cos(0.8πn–1).
1
• Reconstruct: n=500t → cos(0.8π(500t)–1)= cos(2π200t–1)
0.5
0
• 300 Hertz aliased down to 200 Hertz. Phase also changed.
−0.5
−1
0
0.005
0.01
0.015
0.02