SIGNAL PROCESSING
LTI
Systems
Linear Time Invariant
Systems are: Stability
· Additive: (x,(t)
J + x(z(+)7= J-(x,(+)) +
F(xe(+)1 · A
system is stable if it's
· Scaler: 5(ax,(+F aF(x,(+1Y =
&
tends
inepulse response
Time Invariant:
y(t -x) 590,(t-4)7
=
·
Zero in the time domai
Can derive the output y(t) if the impulse response
↓ It is known. Output is the convolution of impul
·
If all poles are on the h
and the impulse Response < x(+) *hC+
y(t
=
Analogue Frequacy Re
Laplace Transforms
Continuous Fourie Transform
u(t) =
b eat it
a referred to as the
spectrum
response *
cos(at) cal an
Efte
sin (at) -just
in #
(j) : dt
297'CHY: sF(s) -
f(0) aSf "CHF=s"FCs)-sf(0) -f'co
Convolution is to
Ignoring the real compo
in the time domain equal
ks
multiplication in the
Laplace Domain
H(w)
=
-
Time Domain Output:
1. Transform (cCt) and hCt) into heplace Domain
2. Find Product Y(s): x(s)+(s)
·
Magnitude:
3. Take
haplace YCs) for
yCH) n w -z
(H (jwll 1k
inverse of
=
Regular system responses: r x/j p =
-
1. ie.(c(H=f2t) Phase:
If the input is a delta function ·
then X (s):) Y (s) M(s)
11(jw (
=
and
2. If the input is step function ie.x(t-u(t<H(jw) =
21)
-
-
a
tha x(s) =
> and Y(s) H(s) =
J
I
k <O add T
Analogue Poles and zeros
Ideal Responses
H(s) kk.S-
=
z;5Zees
K =
overal transfer |+(ju))
a
Mi. Poles function gain 1
1.
s-pi
& -
Real Poles:
Step Response Example: ia
Low Pass
Y(s) - H(s)
eeopolynomia...
to
= =
|+(ju))
a
= constant
ousian
&onstant
po
1 -
-> -
, SIGNAL PROCESSING
Butterworth Filters
Due the
to
superpositio
·
maximally flat the
passband.
·
in
system produces an ou
Transfer Norder Butterworth
E,r(jlewd X ye
How
·
function of a
Pass filler: Where
y(+) =
k=
co
-
a
He(s): pn= Mope
wc half power cut off
&)
=
requency igital and
Analogu
Glinear -
↓ Signal Processing work
Half power vz
out offrequency:
Cr 1. Low Pass filly applie
· Poles are located at
EdB=-3dB frequencies.
[E(2n -1)
2.
Ancelogue to
digita
quanlises the continu
Pn jw,e
=
connect it to a disc
=-wisin (Itn-1) +
jw,cos (tn-1)
3. The digital
signal p
the spection
requi
how Pass Butterworth filter
signal
frequency Response: 4. A to
analog
digital
In (jwll: roc
p
operation to recons
In (wll =
from y[n].
5. An output low pa
normalised
cut-off comparat
requency
frequency wc = 1
operation to t
give
Butteworth Filter:
Designing
·
a
x(t) Low Pass
-> ADC Eb
-> Filter
1. Translate band and stop band requirements
↑
pass
Cia Gp. wp, Gs, ws) to suitable order N.
2. Determine out
of frequency wa Sampling Analogue S
3.
Scaling the normalized
frequency w) = 1
-
Minimum order for N (low-pass) =
sumpling prod:
·Sat..so
Undersampling causes
·
the lower
frequency is c
wjoo.1) En · Discrete-time
signals h
Ws= 2πfs there are in
each
separated by U
Periodic Analogue Functions
|5(jw))
Euler's Formula:e
cossetjsinc
A
LTI
Systems
Linear Time Invariant
Systems are: Stability
· Additive: (x,(t)
J + x(z(+)7= J-(x,(+)) +
F(xe(+)1 · A
system is stable if it's
· Scaler: 5(ax,(+F aF(x,(+1Y =
&
tends
inepulse response
Time Invariant:
y(t -x) 590,(t-4)7
=
·
Zero in the time domai
Can derive the output y(t) if the impulse response
↓ It is known. Output is the convolution of impul
·
If all poles are on the h
and the impulse Response < x(+) *hC+
y(t
=
Analogue Frequacy Re
Laplace Transforms
Continuous Fourie Transform
u(t) =
b eat it
a referred to as the
spectrum
response *
cos(at) cal an
Efte
sin (at) -just
in #
(j) : dt
297'CHY: sF(s) -
f(0) aSf "CHF=s"FCs)-sf(0) -f'co
Convolution is to
Ignoring the real compo
in the time domain equal
ks
multiplication in the
Laplace Domain
H(w)
=
-
Time Domain Output:
1. Transform (cCt) and hCt) into heplace Domain
2. Find Product Y(s): x(s)+(s)
·
Magnitude:
3. Take
haplace YCs) for
yCH) n w -z
(H (jwll 1k
inverse of
=
Regular system responses: r x/j p =
-
1. ie.(c(H=f2t) Phase:
If the input is a delta function ·
then X (s):) Y (s) M(s)
11(jw (
=
and
2. If the input is step function ie.x(t-u(t<H(jw) =
21)
-
-
a
tha x(s) =
> and Y(s) H(s) =
J
I
k <O add T
Analogue Poles and zeros
Ideal Responses
H(s) kk.S-
=
z;5Zees
K =
overal transfer |+(ju))
a
Mi. Poles function gain 1
1.
s-pi
& -
Real Poles:
Step Response Example: ia
Low Pass
Y(s) - H(s)
eeopolynomia...
to
= =
|+(ju))
a
= constant
ousian
&onstant
po
1 -
-> -
, SIGNAL PROCESSING
Butterworth Filters
Due the
to
superpositio
·
maximally flat the
passband.
·
in
system produces an ou
Transfer Norder Butterworth
E,r(jlewd X ye
How
·
function of a
Pass filler: Where
y(+) =
k=
co
-
a
He(s): pn= Mope
wc half power cut off
&)
=
requency igital and
Analogu
Glinear -
↓ Signal Processing work
Half power vz
out offrequency:
Cr 1. Low Pass filly applie
· Poles are located at
EdB=-3dB frequencies.
[E(2n -1)
2.
Ancelogue to
digita
quanlises the continu
Pn jw,e
=
connect it to a disc
=-wisin (Itn-1) +
jw,cos (tn-1)
3. The digital
signal p
the spection
requi
how Pass Butterworth filter
signal
frequency Response: 4. A to
analog
digital
In (jwll: roc
p
operation to recons
In (wll =
from y[n].
5. An output low pa
normalised
cut-off comparat
requency
frequency wc = 1
operation to t
give
Butteworth Filter:
Designing
·
a
x(t) Low Pass
-> ADC Eb
-> Filter
1. Translate band and stop band requirements
↑
pass
Cia Gp. wp, Gs, ws) to suitable order N.
2. Determine out
of frequency wa Sampling Analogue S
3.
Scaling the normalized
frequency w) = 1
-
Minimum order for N (low-pass) =
sumpling prod:
·Sat..so
Undersampling causes
·
the lower
frequency is c
wjoo.1) En · Discrete-time
signals h
Ws= 2πfs there are in
each
separated by U
Periodic Analogue Functions
|5(jw))
Euler's Formula:e
cossetjsinc
A
