Übungsaufgaben zum Kapitel 4 von Signale und Systeme - Peter Klein
Kapitel 4 Diskrete Signale und Systeme
Aufgabe mit Kommentaren.
Aufgabe zum Abtasttheorem, Diskrete Signale, Differenzengleichungen, z-Transformationen, Rücktransformationen, Pol- und Nullstellenplan, bilineare Transformationen ...
Hinweis : Die Ableitung der Einheits
sprungfunktion Elt)
ergibt
die Dirac -
Funktion GH) → DEH) = GH)
°
)
Ulf) [ 10 EH) % (Hms 3) (t 3ms )
-
E
-
= - -
! ]
t
(Hms -
9) -
EH -9ms ) V
Ult) =
100ft) U t
[ -
I -
Ef -
3ms ) t
¥ <
EH-9ms)
] Ins
b)
alt) =
[aoelt) -
5E ( t -3ms) -
SE (t -
9ms) ]V
=
[ aodlt) -
56ft -3ms ) -
56ft -
9ms)
]V
dlt) hat Einheit
Bezug ! : die als
du
du
Hyatt in 4ms
^
Hyatt in 4ms
•
3 9 3
91 I I e >
Hans
!
- - . .
. . .
-
s - K U
, 2
K
Durch Abtastung entstehen die
Frequenzen :
fd =
Ifk ± mifs mit m :O , ta , IZ
,
. . .
$ ↳
Frequenzen des zeitkontinvirlichen
der beidseitigen
Frequenzen
Signals
Spektrums nach der
Abtastung
Bei den Frequenzen f- 20kHz und
=
fu = 10kHz , 30kHz
ist das Abtasttheorem erfüllt →
fa =
f =
10kHz, 20 kHz
,
30kHz
Bei den Frequenzen f- #= =
50kHz ,
70kHz und 110 kHz
ist das Abtasttheoreiu nicht erfüllt →
Aliasing -
Frequenzen
fa =
30kHz, 10kHz
,
30kHz
•
÷
, 3
→ N -
fslfmess = 3
-
d) Ts =
¥ =
0,05ms
III | 906/-4031010,03/0,0619
-
↳ vuertetabelk
4,33 -4,33 0 4,33 -4,33 0
U
,
-
Fläche =
4,33
-
Ts
a a a
⑨ d 1 O d d ⑨ a d ⑨ >
-0,1 0,1 0,2 Hms
v v v
b) Komplexe
Fourier Koeffizienten
~
e- Ünv
1 13
[n =
§
g U ( v) .
Io =
Eg % Uhr) -
1 =
{[ Ot 5. sin ( ZH} ) t 5. sinkt
-
§ )] V
=
0%
¥
JUZITB
[1 =
Ig Uhr) .
e-
Ig [
""
""
= Ot 5. sin (E) e- ts.sinfztt.az/.e-i-ziTBJV
j 2,56
-
= -
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