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MAE143A Signals & Systems Winter 2016 1
1.5 Discrete Signals
Discrete-time signals Generating discrete-time signals by sampling continuous-time signals will be a major subject of this course Some signals are inherently discrete-time, e.g. sunset time We consider periodic sampling
Fixed time between samples: Ts seconds Ts is the sampling period 1/Ts Hz is the sampling rate or sampling frequency
The interpretation of a discrete-time signal relies on knowing the sampling rate There other sampling strategies such as event-based or event-triggered sampling. We do not study these.
MAE143A Signals & Systems Winter 2016 2
1.5 Discrete Signals
Sampled exponential
time (s)0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97
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Continuous-time exponential: samping at 100Hz, 20Hz, 30 Hz
contiuous100Hz20Hz30Hz
x(t) = exp(1.5 t)
100Hz
20Hz
30Hz
MAE143A Signals & Systems Winter 2016 3
1.5 Discrete Signals time (s)
0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97
sign
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Continuous-time exponential: samping at 100Hz, 20Hz, 30 Hz
contiuous100Hz20Hz30Hz
Sampled exponential continued
Sampling period T (0.01s, 0.05s, 1/30s) Continuous signal (function) x(t) Discrete signal (vector) x[n] In electronics, this is done by a circuit - sample and hold and - analog to digital converter
x[n] = x(nT )
MAE143A Signals & Systems Winter 2016 4
1.5 Discrete Signals
Sampled signals … for now
We shall study sampling in greater detail later It is a nuanced field … and very important! A sampled signal is an ordered sequence, which is the same as a vector We can consider sequences with a (countably) infinite number of elements
x[n] = x(nT )
x(t) = exp(1.5 t)
T = 0.05s, n = [1 : 5]0, x[n] =
2
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1.07791.16181.25231.34991.4550
3
77775
MAE143A Signals & Systems Winter 2016 5
1.5 Discrete Signals
Discrete impulse and step signals
Discrete impulse Discrete step No craziness of functions Note:
�[n] =
(1, n = 0
0, else
1[n] =
(1, n � 0
0, else
1[n] =nX
k=�1�[k]
time (samples)-3 -2 -1 0 1 2 3 4 5
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1.2Discrete impulse function
time (samples)-3 -2 -1 0 1 2 3 4 5
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1.2Discrete step function
Not the sampled version of �(t)
MAE143A Signals & Systems Winter 2016 6
1.5 Discrete Signals
Sampling sinusoids
Jump to Matlab for demo [diary file on class website: jan14.txt]
MAE143A Signals & Systems Winter 2016 7
1.5 Discrete Signals
The Z-transform of a discrete signal
Discrete signal Z-transform of the signal Discrete-time alternate to the Laplace transform for continuous-time signals z is a complex variable in the complex z-plane Just like the Laplace transform, there are unilateral and bilateral versions We will be concerned only with the unilateral version
x[n], n = . . . ,�2,�1, 0, 1, 2, . . .
X (z) =1X
k=0
z
�kx[k]
k 2 [0,1)
k 2 (�1,1)
MAE143A Signals & Systems Winter 2016 8
1.5 Discrete Signals
Computing Z transforms
Sampled exponential example, sample rate 20Hz continuous-time signal sample period sample times for sample values Z transform
x(t) = e
1.5t
Ts =1
20s = 0.05 s
nTs = {0, 0.05, 0.10, 0.15, 0.20, . . . } sn = {0, 1, 2, 3, 4, . . . }
x[n] = {e0, e1.5⇥0.05, e
1.5⇥0.10, e
1.5⇥0.15, . . . }
X (z) =1X
k=0
z
�kx[k] =
1X
k=0
z
�ke
1.5⇥0.05⇥k =1X
k=0
�z
�1e
1.5⇥0.05�k
=1
1� z
�1e
1.5⇥0.05=
z
z � e
1.5⇥0.05
x[k] =
(e
1.5⇥0.05⇥k, k � 0,
0, else.
MAE143A Signals & Systems Winter 2016 9
1.5 Discrete Signals
Computing Z transforms 2 Geometric series formula for any value of a Infinite sums provided Our Z transform provided
(1� a)(1 + a+ a2 + a3 + · · ·+ aN ) = 1� aN+1
1 + a+ a2 + a3 + · · ·+ aN =1� aN+1
1� aNX
k=0
ak =1� aN+1
1� a1X
k=0
ak =1
1� a|a| < 1
X (z) =1X
k=0
�z�1e1.5⇥0.05
�k=
1
1� z�1e1.5⇥0.05=
z
z � e1.5⇥0.05
��z�1e1.5⇥0.05�� < 1 or |z| > e1.5⇥0.05
MAE143A Signals & Systems Winter 2016 10
1.5 Discrete Signals
Computing Z transforms 3
Higher-order poles 1
(1� z�1a)2=
✓1
1� z�1a
◆✓1
1� z�1a
◆
= (1 + z�1a+ z�2a2 + z�3a3 + . . . )(1 + z�1a+ z�2a2 + z�3a3 + . . . )
= 1 + z�12a+ z�23a2 + z�34a3 + . . .
= Z {(n+ 1)an}
1
(1� z�1a)3=
✓1
1� z�1a
◆✓1
(1� z�1a)2
◆
= (1 + z�1a+ z�2a2 + z�3a3 + . . . )(1 + z�12a+ z�23a2 + z�34a3 + . . . )
= 1 + z�13a+ z�26a2 + z�310a3 + . . .
= Z⇢n+ 2
2(n+ 1)an
�
MAE143A Signals & Systems Winter 2016 11
1.5 Discrete Signals
Z transforms of sampled exponential signals
Consider a continuous-time (complex) exponential signal with b a complex number Sample this at sampling period Ts Take the Z transform of this discrete-time signal Convergence provided
x(t) = e
bt
X (z) =1X
k=0
z�kekbTs =1X
k=0
�z�1ebTs
�k=
1
1� z�1ebTs=
z
z � ebTs
x[n] = e
nbTs
L�ebt
=
1
s� bZ�enbt
=
z
z � ebTs
pole at s=b convergence if Re(s)>Re(b)
pole at z=ebTs
convergence if
|z| >��ebTs
�� = eRe(b)Ts
|z| >��ebTs
�� |z| > eRe(b)Ts
MAE143A Signals & Systems Winter 2016 12
1.5 Discrete Signals
Foreshadowing MAE143C Digital Control
General transformation between s and z:
L�ebt
=
1
s� bZ�enbt
=
z
z � ebTs
pole at s=b convergence if Re(s)>Re(b)
pole at z=ebTs
convergence if |z| > ebTs
z = esTs
z = esTs = e�Ts+j!Ts = e�Tsej!Ts
Re(s) > Re(b) , |z| >��ebTs
��So
Also Re(s) < 0 , |z| < 1
The open left half s-plane corresponds to the inside of the unit disk in the z-plane
MAE143A Signals & Systems Winter 2016 13
1.5 Discrete Signals
Summary Continuous-time signals
t takes values in a real interval Signal x(t) is a real function Laplace transforms are used s is a complex variable Fourier series for periodic signals Fourier transform for bounded energy signals
Discrete-time signals n takes integer values Signal x[n] is a real sequence Z transforms are used z is a complex variable Discrete (Fast) Fourier Transform used to analyze a finite sequence
L{f(t)} = F (s) Z {x[n]} = X(z)
