remco litjens - university of california, berkeleyee120/fa99/summary/summary.pdf · f ourier series...
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Continuous-time Signals and Systems
General / Miscellaneous
1. !o = 2�To
= 2�fo angular or fundamental frequency
2. f(�t) = f(t) even function (e.g. cos t)
3. f(�t) = �f(t) odd function (e.g. sin t)
4. sinc z = sin�z�z
sinc-function
5.R11
sin axbx
dx = �b
area under a sinc-function
6. e�t�(t) = e�0�(t) = �(t)
7.p!2 = j!j
8. Hlpf(!) = 1 � Hhpf(!)
9. arctan(!0 ) = �
2 sgn(!)
Trigonometry
1. sinx = cos(x� 12�)
2. sin2x+ cos2x = 1
3. cos2x = 2cos2x� 1 = 1� 2sin2x
4. sin2x = 2sinxcosx
5. cos(�+ �) = cos�cos� � sin�sin�
6. sin(�+ �) = sin�cos� + cos�sin�
7. 2 cos� cos� = cos(�+ �) + cos(�� �)
8. �2 sin� sin� = cos(�+ �) � cos(�� �)
9. dd!
sin! = cos!
10. dd!
cos! = �sin!11. d
d!arctan(!) = 1
1+!2
Complex Numbers
1. j =p�1 imaginary constant
2. z = a+ bj = Re[z] + jIm[z] complex number (rectangular or Cartesian form)
3. z� = a� bj = Re[z]� jIm[z] complex conjugate of z
4. z = jzjej 6 z complex number (polar form)
5. jzj =p
Re2[z] + Im2[z] magnitude of complex number z
6. 6 z = arctan( Im[z]Re[z] ) angle or phase of complex number z
7. ejx = cosx + jsinx Euler's formula
8. in IC : 1 = ej0; �1 = ej� ; j = e12 j�; �j = e�
12 j� basic points on unit circle in IC
9. ejn� = (�1)n10. cos !ot = 1
2 (ej!ot + e�j!ot)
11. sin !ot = 12j (e
j!ot � e�j!ot)
12. 1a+ bj
= a� bja2 + b2
13. j(a+ bj) � (c+ dj)j = j(a+ bj)j � j(c+ dj)j 18. 6 f(a+ bj) � (c+ dj)g = 6 fa+ bjg+ 6 fc+ djg14. z + z� = 2Re[z] 19. z � z� = 2jIm[z]
15. zz� = jzj2 20. 6 z�1 = �6 z16. 1
z�= ( 1
z)� 21. jzj = jz�j
17. 6 z� = �6 z 22. jz�1j = jzj�1
Singularity Signals
1. u(t) =R t�1
�(s)ds = 1 t � 0; and 0 elsewhere unit step function
2. r(t) =R t�1
u(s)ds = tu(t) unit ramp function
3. p(t) =R t�1
r(s)ds = 12 t
2u(t) unit parabola function
4. �(t) = u(t+ 12 ) � u(t� 1
2 ) = 1; jtj � 12 ; and 0 elsewhere unit pulse function
5. �(t) = r(t+ 1) � 2r(t) + r(t� 1) unit triangle function
6. �(t) = lim�!012��(
t2� ) unit impulse or delta function (a.k.a. the "spike")
7. _�(t) = ddt�(t) doublet
8. �(at) = 1jaj
�(t); for a 2 IRnf0g9. u(at) = u(t)
10. r(at) = a r(t)
11.R1�1
�(t)dt = 1
12. x(t)�(t� a) = x(a)�(t� a)
13.R bax(t)�(t)dt = x(0); if a < 0 < b; and 0; elsewhere
14.R1�1
x(s)�(t� s)ds = x(t) sifting or convolution property (integral)
15.R1�1
x(s)�(n)(t� s)ds = (�1)nx(n)(t)
Power and Energy
1. E = lim�!1
R ���jx(t)j2dt average energy (in joules)
2. P = lim�!112�
R ���jx(t)j2dt average power (in watts)
� if 0 � E <1 then P = 0 and the signal is called a energy signal.
� if 0 < P <1 then E = 1 and the signal is called a power signal.
� the energy of a periodic signal is in�nite and the power equals P = 1To
R to+Toto
jx(t)j2dt; with to arbitrary.
Convolution
� de�nition: h(t) x(t) =R1�1
h(t� �)x(�)d� =R1�1
h(�)x(t� �)d�
1. x1(t) x2(t) = x2(t) x1(t) commutativity property of convolution
2. x1(t) (x2(t) x3(t)) = (x1(t) x2(t)) x3(t)) associativity property of convolution
3. x1(t) (x2(t) + x3(t)) = x1(t) x2(t) + x1(t) x3(t) distributivity property of convolution
4. trick 1: x(t� �) = x(t) �(t� �) delay system
5. trick 2: x(t) = x(t) �(t); 8x(t)6. trick 3: convolving a signal x(t) with u(t) means taking the running time integral of x(t):
7. trick 4: convolving a signal x(t) with ddt�(t) means taking the derivative of x(t):
8. u(t) = u(t) �(t) =R t�1
�(s)ds
9. r(t) = u(t) u(t) =R t�1
u(s)ds
10. p(t) = u(t) r(t) =R t�1
r(s)ds
"
and conversely
#8�. �(t) = d
dt�(t) u(t) = d
dtu(t)
9�. u(t) = ddt�(t) r(t) = d
dtr(t)
10�. r(t) = ddt�(t) p(t) = d
dtp(t)
General Systems
1. y(t) = H[x(t)] system H[�]2. H[�1x1(t) + �2x2(t)] = �1H [x1(t)] + �2H[x2(t)] condition for linearity of H[�]3. H[D� [x(t)]] = D� [H[x(t)]] condition for time-invariance of H[�]4. y(to) = H[x(t0)] depends only on x(t) for t = to condition for memoryless property of H[�]5. y(to) = H[x(t0)] depends only on x(t) for t � to condition for causality of H[�]
� note that H[x(t)] = x(t) + 1 is non-linear.
� memoryless systems are causal.
Linear Systems
1. g(t; s) = H[�(t� s)]
2. y(t) =R1�1
g(t; �)x(�)d�
3. check for causality: input �(t� s) and check if for any s the output is zero 8 t > s:
4. H is time-invariant() g(t+ �; s+�) = g(t; s)
� if H[�] is linear and memoryless, then H[�] will be of the form H[�] = �(t)x(t):
Linear and Time-Invariant (LTI) Systems
1. H1[�]; H2[�] LTI , H1[H2[�]] LTI2. H1[�]; H2[�] LTI , (H1 +H2)[�] LTI
1. h(t) = H[�(t)] impulse response of H [�]2. ystep = H[u(t)] step response of H[�]3. yramp = H[r(t)] ramp response of H [�]4. yparabola = H[p[(t)] parabola response of H[�]5. H(!) =
R1�1
h(t) e�j!t dt = jH(!)jej 6 H(!) frequency response of H[�] (Fourier transform of h(t)
6. H(s) =R10�
h(t) e�st dt transfer function of H[�] (Laplace transform of h(t)
7. h(t) real) H(�!) = H�(!)
8. y(t) = h(t) x(t) =R1�1
h(t� �)x(�)d� =
=R1�1
h(�)x(t� �)d� = h(t) x(t) superposition or convolution integral
9. H[ej!t] = H(!) ej!t for periodic signals (scaling; eigenfunctions / -value)
10. Y (!) = X(!)H(!) Fourier transform of (8)
11. Y (s) = X(s)H(s) Laplace transform of (8)
12. h(t) = ddt�(t) ystep(t) = d
dtystep(t)
13. ystep = ddt�(t) yramp(t) = d
dtyramp(t)
14. yramp = ddt�(t) yparabola(t) = d
dtyparabola(t)
"and conversely
#12�. ystep = h(t) u(t) =
R t�1
h(s)ds
13�. yramp = h(t) r(t) = h(t) u(t) u(t) = ystep u(t) =R t�1
ystep(s)ds
14�. yparabola = h(t) p(t) = h(t) r(t) u(t) = yramp u(t) =R t�1
yramp(s)ds
Frequency, Phase and Magnitude Response
1. H(!) =R1�1
h(t) e�j!t dt frequency response is F[h(t)]2. from an LDE with CC: let x(t) = ej!t and y(t) = H(!)ej!t; solve for H(!)
3. from an LDE with CC: let X(!) = F[x(t)] and Y (!) = F[y(t)]; solve for H(!) = Y (!)X(!) ; using F[ _x(t)] = j!X(!)
4. if H(!) = K 1+aj!1+bj! ; then jH(!)j = jKj j1+aj!j
j1+bj!j = jKjp1+(a!)2p1+(b!)2
5. if H(!) = K 1+aj!1+bj! ; then
6 H(!) = 6 K + 6 (1 + aj!) � 6 (1 + bj!) = 6 K + arctan(a!) � arctan(b!)
BIBO-Stability
� de�nition: system H[�] is BIBO-stable i� every bounded input results in a bounded output
1. system H[�] is BIBO-stable i�R1�1jh(t)j < 1
2. if system H[�] is BIBO-stable then H(s) = N(s)D(s)
must be proper, i.e. degree N(s) � degree D(s)
3. if causal system H[�] is BIBO-stable then all the poles of H(s) must lie in the open left-half plane
� necessary conditions 2 and 3 for stability are su�cient if H(s) is rational (generally the case)
� for all stable causal systems, the frequency response exists
Distortion
� an LTI system is distortionless if the output y(t) = H[x(t)] = K x(t� �); a scaled and shifted replica of the input x(t) ()() H(!) = K e�j!� () jH(!)j is constant and 6 H(!) = �!� is linear in !:
� an LTI system shows amplitude distortion when jH(!)j varies with !:
� an LTI system shows phase distortion when 6 H(!) is non-linear in !; and then we de�ne �g(!) = � d6 H(!)d!
as the group delay.
Minimum Phase
A system H[�] is said to be minimum phase, if all poles and zeros of H(s) lie in the open left half plane, so that H�1(s) is also minimum
phase; hence a causal system that is minimum phase, is also causally invertible.
Feedback
Given a feedback system block diagram, we can write the equations:
� e(t) = x(t)� y(t) h2(t)L ! E(s) = X(s)� Y (s)H2(s) (error function);
� y(t) = e(t) h1(t)L ! Y (s) = E(s)H1(s);
� combine: y(t) = x(t) h1(t)� y(t) h2(t) h1(t);L ! Y (s) = X(s)H1(s)� Y (s)H1(s)H2(s) , Y (s)
X(s) =H1(s)
1�H1(s)H2(s):
� in control problems, where we try to optimize feedback, we often use the �nal value theorem for Laplace transforms;
Finding the Output of LTI Systems
1. y(t) = x(t) h(t) straight convolution
2. y(t) = F�1[X(!)H(!)] using Fourier transforms
2. y(t) = L�1[X(s)H(s)] using Laplace transforms
3. if x(t) = ej!ot; then y(t) = H(!o)ej!ot using Fourier eigenfunctions
4. if x(t) = est; then y(t) = H(s)est using Laplace eigenfunctions
5. if x(t) = cos(!ot + �) and h(t) is real, then y(t) = jH(!o)jcos(!ot + � + 6 H(!o))
steady-state sinusoidal response
6. if x(t) = sin(!ot + �) and h(t) is real, then y(t) = jH(!o)jsin(!ot + � + 6 H(!o))
steady-state sinusoidal response
7. if x(t) =P1
n=�1Xnejn!ot; then y(t) =
P1
n=�1XnH(n!o)ejn!ot =
P1
n=�1XnjH(n!o)jej(n!ot+ 6 H(n!o))
using Fourier series
� the transfer function and consequently the impulse response can be derived from any LDE by assuming zero initial conditions.
Remarks
� in LTI systems it makes no di�erence if we see h(t) as the input signal and x(t) as the impulse response
of the LTI system (follows directly from the commutativity property of convolution).
� both h(t) and H(!) completely characterize an LTI system.
� for an LTI system H[�] : h(t) = 0 for t < 0 , H[�] is causal.� a di�erentiator (h(t) = _�(t)) is causal.
� an ideal di�erentiator _�(t) is a causal system.
� H(!) real , h(�t) = h(t) ) H[�] is non-causal (unless h(t) = 0 8t 6= 0) =) �lters are non-causal
� an LTI system converts a periodic input signal into a periodic output signal.
� if H[�] is LTI and memoryless, then h(t) = 0 for t 6= 0 and H [�] will be of the form H[�] = �x(t):
Fourier Series
De�nition of Fourier Series
We can decomposing any periodic power signal (with period To) into discrete frequencies, each a multiple of !o; yielding the Fourier series
representation of that signal:
� x(t) =P1
n=�1Xnejn!ot synthesis equation
� Xn = 1To
RTo
x(t)e�jn!otdt analysis equation
Properties of Fourier Series (given x(t)FS ! Xn):
� real signals: if x(t) is real, then X�n = X�n
� real and even signals: if x(t) is real and even in t; then Xn is real and even in n
Ev[x(t)]FS ! Re[X(!)] and Re[x(t)]
FS ! Ev[X(!)]
� real and odd signals: if x(t) is real and odd in t; then Xn is imaginary and odd in n
Od[x(t)]FS ! Im[X(!)] and Im[x(t)]
FS ! Od[X(!)]
� half-wave odd symmetry: if x(t� 12To) = �x(t); 8t (e.g. x(t) = sint), then Xn = 0 for n even (in particular, X0 = 0)
� time reversal: x(�t) FS ! X�n
� complex conjugate: x�(t)FS ! X��n and x�(�t) FS ! X�n
� delay: x(t� �)FS ! e�jn!o�Xn
� di�erentiation: _x(t)FS ! jn!o Xn
� integration: if Xo = 0 then u(t) x(t)FS ! Xn
jn!o; for n 6= 0
� linearity: �x(t)+ �y(t)FS ! �Xn + �Yn
� PARSEVAL's identity: P =P1
n=�1X�nXn =
P1
n=�1jXnj2
� power at fundamental frequency of x(t) =P1
n=�1Xnejn!ot equals jX�1j2 + jX1j2
Some Examples
� DC: x(t) = 1FS ! Xo = 1; Xn = 0 otherwise
� cosine: cos !ot = 12(ej!ot + e�j!ot)
� sine: sin !ot = 12j (e
j!ot � e�j!ot)
� even square wave: period To; x1(t) = 1 for jtj < T1 and x1(t) = 0 elsewhere in that period:
x1(t)FS ! Xn = 1
n�sin n!oT1
� odd square wave: x2(t) is shifted x1(t)
� even triangle wave: x3(t) is running time integral of x1(t)
x3(t)FS ! Xn = � 4
n2�2; for n odd, and Xn = 0; for n even
� pulse train: x4(t) is same as x1(t) but shifted by � and with added DC (this changes X0 only):
x4(t) =P1
k=�1�( t�kTo��
to)FS ! Xn = to
Toe�jn!o� sinc(nto
To)
� even impulse train: x5(t) =P1
k=�1�(at� kT )
FS ! Xn = 1jajT
Fourier Transform
De�nition and Interpretation
The Fourier transform X(!) is a decomposition of an (a)periodic signal x(t) into ej!t-type signals with ! real (x(t) and X(!) uniquely
de�ne each other); we write x(t)F ! X(!) :
� x(t) = F�1[X(!)] = 12�
R1�1
X(!)ej!td! synthesis equation
� X(!) = F[x(t)] =R1�1
x(t)e�j!tdt analysis equation
The Fourier transform X(!) exists when it's �nite for all !: Allowing �-functions, we can accept a broader class of signals: su�cient conditions
for the existence of the generalized Fourier transform of x(t) are (1) x(t) is bounded; and (2) x(t) is periodic (E =1) with �nite power.
The interpretation of the Fourier transform is that X(!) contains the frequency content of the signal x(t) for each !: For example, consider
x(t) = 1F ! �(!) (DC only), x(t) = ej!ot
F ! 2��(! � !o) (single frequency at !o), and �nally x(t) = cos !otF ! �[�(! + !o) + �(! � !o)]
(contains two frequencies: !o and �!o). Note: if the signal x(t) is time-limited, X(!) has in�nite bandwidth.
Properties of Fourier Transforms (given x(t)F ! X(!)):
� real signals: if x(t) is real, then X(!) = X�(�!) � check resultsRe[X(!)] = Re[X(�!)] and Im[X(!)] = �Im[X(�!)] � check results
� real and even signals: if x(t) is real and even in t; then X(!) is real and even in ! � check resultsEv[x(t)] F ! Re[X(!)] and Re[x(t)] F ! Ev[X(!)] � check results
� real and odd signals: if x(t) is real and odd in t; then X(!) is imaginary and odd in ! � check resultsOd[x(t)] F ! jIm[X(!)] and jIm[x(t)]
F ! Od[X(!)] � check results� time reversal: x(�t) F ! X(�!)� complex conjugate: x�(t)
F ! X�(�!) and x�(�t) F ! X�(!)
� linearity: �1x1(t) + �2x2(t)F ! �1X1(!) + �2X2(!)
� delay: x(t� �)F ! X(!)e�j!� � phase change (of �!�) only
� scale: x(at)F ! 1
jajX(!
a) � wide (narrow) in t , narrow (wide) in !
� duality: x(t)F ! X(!) , X(t)
F ! 2�x(�!)� frequency translation: x(t)ej!ot
F ! X(! � !o) � used for modulation
� di�erentiation: dn
dtnx(t)
F ! (j!)nX(!) � help solve LDEs
� integration: u(t) x(t)F ! 1
j!X(!) + �X(0)�(!) � �-function accounts for DC in x(t)
� convolution: x1(t) x2(t)F ! X1(!) �X2(!)
� multiplication: x1(t) � x2(t) F ! 12�X1(!)X2(!)
� t-multiplication: t x(t)F ! j d
d!X(!)
� modulation: x(t)cos!otF ! 1
2 [X(!� !o) +X(!+ !o)]
� PARSEVAL's identity: E =R1�1jx(t)j2dt = 1
2�
R1�1jX(!)j2d!
Some Examples and Facts
� X(0) =R1�1
x(t)dt is the DC level/power of signal x(t)
� x(0) = 12�
R1�1
X(!)d! is the total power in x(t)
� h(t) F ! H(!) frequency response
� 1�t
F ! �j sgn(!) = e�j�2 sgn(!) Hilbert transform; FT exists, even though x(t) is unbounded
� hHT x(t) phase shifts x(t) only
� HHT [cos t] = sin t and HHT [sin t] = cos t
� HHT [HHT [x(t)]] = �x(t)� sgn(t) F ! 2
j!
� �( tto)
F ! tosinc(!to2� )
� sinc( t�)
F ! ��(�!2� )
� for � > 0; �e��tu(t)F ! �
�+j!
� for � > 0; �e��jtjF ! 2��
�2+!2
� e��t2 F ! e��f2= e�
!2
4�
� sin !ot F ! �j[�(!� !o)� �(!+ !o)]
� cos !ot F ! �[�(! � !o) + �(! + !o)]
� 0 F ! 0
� 1 F ! 2��(!)
� �(t) F ! 1
� �(t� to)F ! e�j!to
� _�(t)F ! j!
� u(t) F ! 1j!
+ ��(!)
� �( tT) = 1
T�( t
T) �( t
T)
F ! Tsinc2(!T2� )
� ej!ot F ! 2��(! � !o)
�P1
n=�1�(t� nT )
F ! 2�T
P1
n=�1�(!� n!o)
� _�(t) + �(t) + 12 sgn(t)
F ! j! + 1+ 1j!
� the Fourier transform of a Gaussian e�12 (x���
)2 is another Gaussian of the same form
� the Fourier transform of a Fourier Series x(t) =P1
k=�1Xke
jk!otF ! 2�
P1
k=�1Xk�(!� k!o)
Laplace Transform
De�nition of Laplace Transform
The Laplace transform X(s) is a decomposition of an (a)periodic signal x(t) into signals of the type est = e�tej!t with s = � + j! complex.
Compared to the Fourier transform, the extra (real) exponential e�t allows us to treat a broader class of signals. The e�t part is used to model
growth or decay, while the ej!t part models the oscillatory behavior of x(t): Again, x(t) and X(s) uniquely de�ne each other and here we write
x(t)L ! X(s) :
� unilateral LT: X(s) = L[x(t)] =R10�
x(t)e�stdt =R10�
x(t)e��te�j!t
(� bilateral LT: X(s) = L[x(t)] =R1�1
x(t)e�stdt =R1�1
x(t)e��te�j!t)
Note that the ULT only works for (real world) "causal" signals while the BLT permits us to consider any signal; furthermore, if x(t) = 0 for t < 0
(causal) then the ULT and the BLT are the same. Both the ULT and the BLT have regions of convergence (ROC) over which the transform
exists. A ROC is the range of values of s over which the integral converges. A ROC cannot contain a pole, in fact for causal signals/systems, the
ROC contains all s to the right of the rightmost pole. If x(t) is of �nite duration, ROCx = IC:
Any signal x(t) that is of exponential order, i.e. jx(t)j does not increase faster than Aect with A and c real constants, can be transformed into
X(s): Examples of signals that are not of exponential order, consider x(t) = et2u(t) and x(t) = ttu(t) = et ln tu(t):
Properties of Unilateral Laplace Transforms (given x(t)L ! X(s) with ROCx):
� linearity: �1x1(t) + �2x2(t)L ! �1X1(s) + �2X2(s) at least ROCx1 \ROCx2
� delay (� > 0): x(t� �)u(t� �)L ! X(s)e�s� ROCx
� scale: for real a > 0 : x(at)L ! 1
aX( s
a) fs j s
a2 ROCxg
� frequency translation: x(t)e��tL ! X(s+ �) fs j (s+ �) 2 ROCxg
� di�erentiation: dn
dtnx(t)
L ! snX(s)� sn�1x(0�)� :::� sx(n�2)(0�)� x(n�1)(0�) at least ROCx
� integration: for t > 0 : u(t) x(t)L ! 1
sX(s) + 1
s
R 0��1
x(�)d� at least ROCx \ fs jRe(s) > 0g� convolution: x1(t) x2(t)
L ! X1(s) �X2(s) at least ROCx1 \ROCx2� t-multiplication: t � x(t) L ! � dX(s)
dsROCx
� initial value: lims!1 sX(s) = limt!0+ x(t); if this limit exists
� �nal value: lims!0 sX(s) = limt!1 x(t); if this limit exists
IMPORTANT: sin t u(t)L ! 1
s2+1; so lims!0 sX(s) = 0; but lim t!1 sin t u(t) clearly does not exist!!!
Some Examples
� h(t) L ! H(s) transfer function
� 0 L ! 0 � _�(t)L ! s all s
� �(t) L ! 1 all s � �(t� �)L ! e�s� all s
� u(t) L ! 1s
Re[s] > 0 � �u(�t) L ! 1s
Re[s] < 0
� r(t) L ! 1s2
Re[s] > 0 � �r(�t) L ! 1s
Re[s] < 0
� tn�1
(n�1)! u(t)L ! 1
snRe[s] > 0 � � tn�1
(n�1)! u(�t)L ! 1
snRe[s] < 0
� e��tu(t) L ! 1s+� Re[s] > �Re[�] � �e��tu(�t) L ! 1
s+� Re[s] < �Re[�]� tn�1
(n�1)! e��tu(t)
L ! 1(s+�)n Re[s] > �Re[�] � � tn�1
(n�1)! e��tu(�t) L ! 1
(s+�)n Re[s] < �Re[�]� (sin !ot)u(t) L ! !o
s2+!2oRe[s] > 0 � (e��t sin !ot)u(t) L ! !o
(s+�)2+!2oRe[s] > �Re[�]
� (cos !ot)u(t) L ! ss2+!2o
Re[s] > 0 � (e��t cos !ot)u(t) L ! s+�(s+�)2+!2o
Re[s] > �Re[�]
Linear Di�erential Equations
Solving LDE w/CC using Fourier transforms
One can solve LDE w/CC using Fourier transforms or Laplace transforms. Below we show how to do it with Laplace transform. For Fourier
transforms, the procedure is identical: �nd h(t) = F�1[H(!)] with H(!) = Y (!)X(!) and X(!) = F[�(t)] = 1: Y (!) follows from using the
di�erentiation property for Fourier transforms and some algebraic manipulations.
Example: if �y(t) � _y(t)� 30y(t) = x(t); then for x(t) = �(t) we �nd (j!)2H(!) � (j!)H(!)� 30H(!) = 1 () H(!) = 1(j!)2�(j!)�30
; so
h(t) = F�1[ 1(j!+5)(j!�6) ] = F�1[
�1=11j!+5 ] + F�1[
1=11j!�6 ] = � 1
11 e�5tu(t) + 1
11 e6tu(t): Note that this system is BIBO unstable.
Solving LDE w/CC using Laplace transforms
We show how the unilateral Laplace transform can be used to solve linear di�erential equations with constant coe�cients. In particular, the
integration property of the unilateral Laplace transform is utilized. The procedures are not shown for the general case, but rather for an example.
Example: Assume we are given the following LDE w/CC that we are trying to solve:
�y(t) + 2 _y(t) + y(t) = 2 _x(t)
with initial conditions _y(0�) = 2; y(0�) = 2 and x(0�) = 0:
First, take the unilateral Laplace transform of both sides, using the di�erentiation property:
[s2Y (s)� sy(0�)� _y(0�)] + a[sY (s)� y(0�)] + Y (s) = c[sX(s)� x(0�)] () Y (s) =s(2 + 2X(s))+ 2 + 4
s2 + 2s+ 1
For input signal x(t) = cost u(t)L ! s
s2+1:
Y (s) =2s+ 6 + 2s � s
s2+1
s2 + 2s+ 1=
2s3 + 8s2 + 2s+ 6
(s+ 1)2(s2 + 1)=
1
s+ 1+
5
(s+ 1)2+
s
s2 + 1() y(t) = L�1[Y (s)] = (e�t + 5te�t + cost)u(t)
ZIR and ZSR
The zero-input response yzir(t) is y(t) when the non-zero intial conditions are assumed to contain all information about what happened in the
past, and for t � to; there is no input to the system: x(t) = 0; for t � to: So yzir(t) is the system response to initial conditions only. In the example
above, yzir(t) = (2e�t + 4te�t)u(t):
The zero-state response yzsr(t) is y(t) when zero initial conditions are assumed and for t � to; the input x(t) is known. In the example,
yzsr(t) = (�e�t + te�t + cost)u(t):
Note that y(t) = yzir(t) + yzsr(t) always holds. In the example: yzir(t) + yzsr(t) = (2e�t + 4te�t)u(t) + (�e�t + te�t + cost)u(t) =
(e�t + 5te�t + cost)u(t) = y(t):
Partial Fraction Expansions
Sorry, I never found the time to do this section: see your notes!
Discrete-time Signals and Systems
General Remarks
Since in discrete time, when observing some continuous-time signal at times t = nT; complex exponentials ej!1nT and ej!2nT with
!2 = !1 +2�T; look exactly identical, we introduce the concept of normalized frequency: � !T (note that we basically only
rescaled the time axis). All of this implies that DISCRETE-TIME SIGNALS HAVE PERIODIC DTFTs, with period 2�T
in unnormalized frequency and period 2� in normalized frequency.
In discrete-time, high frequencies lie around odd multiples of �; while low frequencies lie around even multiples of �: Consequently,
an ideal low-pass �lter with cut-o� frequency 1 < � has impulse response hlpf[n] =1�sinc n1
�and a pulse train as frequency response.
The frequency response of a high-pass, bandpass or notch �lter is simply a shifted and possibly duplicated version of the frequency
response of a low-pass �lter.
Singularity Signals
1. �[n] = u[n] � u[n � 1] = 1; n = 0; and 0 elsewhere delta function
2. u[n] =Pn
k=�1�[k] =
P1
k=0�[n� k] = 1 n � 0; and 0 elsewhere unit step function
3. x[n] =P1
k=�1x[k] �[n] sifting property (sum)
Convolution
� de�nition: h[n] x[n] =P1
k=�1x[k]h[n� k] =
P1
k=�1x[n� k]h[k] often very easy to do!
1. trick 1: x[n� no] = x[n] �[n� no] delay system
2. trick 2: x[n] = x[n] �[n]; 8x[n]3. trick 3:
Pn
k=�1x[n] = u[n] x[n] convolving with u[n] means taking the running sum
� example: u[n] u[n] =P1
k=�1u[k]u[n � k] =
Pn
k=01 = (n+ 1)u[n]
BIBO-Stability
� de�nition: system H[�] is BIBO-stable i� every bounded input results in a bounded output
1. system H[�] is BIBO-stable i�P1
k=�1jh[k]j <1
2. causal system H[�] is BIBO-stable i� all the poles of H(z) are strictly inside the unit circle
� for all stable causal systems, the frequency response exists
Finding the Output of LTI Systems
1. h[n] = H[�[n]] unit sample response
2. y[n] = x[n] h[n] straight convolution
3. y[n] = F�1[X(ej)H(ej)] using Discrete-time Fourier transforms
4. y[n] = L�1[X(z)H(z)] using Z-transforms
5. if x[n] = ejon; then y[n] = H(ejo)ejon using Fourier eigenfunctions
6. if x[n] = zn; then y[n] = H(z)zn using Z-eigenfunctions
7. if x[n] = cos(on + �) and h[n] is real, then y[n] = jH(ejo)jcos(on + � + 6 H(ejo))
steady-state sinusoidal response
8. if x[n] = sin(on + �) and h[n] is real, then y[n] = jH(ejo)jsin(on + � + 6 H(ejo))
steady-state sinusoidal response
� the transfer function and consequently the impulse response can be derived from any LDE by assuming
zero initial conditions.
Discrete-time Fourier Transform
De�nition of Discrete-time Fourier Transform:
The Discrete-time Fourier transform X() is a decomposition of an (a)periodic signal x[n] into ejn-type signals with real (x[n] and
X() uniquely de�ne each other). X() is periodic in with period 2�: We write x[n]F ! X() :
� x[n] = F�1[X()] = 12�
R o+2�o
X()ejnd synthesis equation
� X() = F[x(t)] =P1
n=�1x[n]e�jn analysis equation
The discrete-time Fourier transform X() exists when it's �nite for all : The interpretation of the discrete-time Fourier transform is that X(!)
contains the frequency content of the signal x[n]: The DTFT is always periodic. Often, the DTFT of a discrete-time signal can be found
more easily, by writing x[n] as a sampled continuous-time signal x[n] = x(nT ) =P1
n=�1x(nT )�(t� nT ); which has Fourier transform
P1
n=�1x(nT )e�j!nT ; the change of variables !T = gives the DTFT of x[n] : X() =
P1
n=�1x[n]e�jn (EXTREMELY HELPFUL!).
Properties of Discrete-time Fourier Transforms (given x[n]F ! X()):
� real signals: if x[n] is real, then X() = X�(�) � check resultsRe[X()] = Re[X(�)] and Im[X()] = �Im[X(�)] � check results
� real and even signals: if x[n] is real and even in n; then X() is real and even in � check resultsEv[x[n]] F ! Re[X()] and Re[x[n]] F ! Ev[X()] � check results
� real and odd signals: if x[n] is real and odd in n; then X() is imaginary and odd in � check resultsOd[x[n]] F ! jIm[X()] and jIm[x[n]]
F ! Od[X()] � check results� time reversal: x[�n] F ! X(�)� complex conjugate: x�[n]
F ! X�(�)� linearity: �1x1[n] + �2x2[n]
F ! �1X1() + �2X2()
� delay: x[n� no]F ! X()e�jno � phase change (of �no) only
� frequency translation: x[n]ejonF ! X(� o) � used for modulation
� �rst di�erence: x[n]� x[n� 1]F ! X()(1� e�j)
� running sum: u[n] x[n]F ! 1
1�e�jX() + �X(0)
P1
k=�1�(� 2�k) � �-functions account for DC in x[n]
� convolution: x1[n] x2[n]F ! X1() �X2()
� multiplication: x1[n] � x2[n] F ! 12�
R o+2�o
X1(�)X2(� �)d� � periodic convolution� n-multiplication: nx[n]
F ! j ddX()
� upsampling: x(k)[n]F ! X(k) � x(k)[n] = x[n
k]; if n is a multiple of k
and 0 otherwise
� PARSEVAL's identity: E =P1
n=�1jx[n]j2 = 1
2�
R o+2�o
jX()j2d
Some Examples
� h[n] F ! H() frequency response
� sinon F ! �j
P1
k=�1[�(� o � 2�k)� �( + o � 2�k)]
� coson F ! �P1
k=�1[�(� o � 2�k) + �( + o � 2�k)]
� 0 F ! 0 � 1 F ! 2�P1
k=�1�(� 2�k)
� �[n] F ! 1 � �[n� no]F ! e�jno
� u[n] F ! 11�ej
+ �P1
k=�1�(� 2�k) � ejon F ! 2�
P1
k=�infty�(� o � 2�k)
� 1�sinc(1n
�)
F !P1
k=�1�(�2�k21
)
Z-Transform or DT Laplace Transform
De�nition of Z-Transform:
The Z-transform X(z) is a decomposition of an (a)periodic signal x(t) into signals of the type z�n = r�ne�jn with z = rej complex.
Compared to the discrete-time Fourier transform, the extra (real) exponential r�n allows us to treat a broader class of signals. The r�n part
is used to model growth or decay, while the e�jn! part models the oscillatory behavior of x(t): Again, x[n] and X(z) uniquely de�ne each
other and here we write x[n]Z ! X(z) :
� unilateral ZT: X(z) = Z[x[n]] =P1
k=0x[n]z�n
(� bilateral ZT: X(z) = Z[x[n]] =P1
k=�1x[n]z�n)
Note that the UZT only works for (real world) "causal" signals while the BZT permits us to consider any signal; furthermore, if x[n] = 0 for
n < 0 (causal) then the UZT and the BZT are the same. Both the UZT and the BZT have regions of convergence (ROC) over which the
transform exists. A ROC is the range of values of z over which the integral converges. A ROC cannot contain a pole, in fact for causal signals/
systems, the ROC contains all z on the outside of the circle that crosses the outermost pole. If x[n] is of �nite duration, ROCx = IC; except
possibly for z = 0 and z =1:
Any signal x[n] that is of exponential order, i.e. jx[n]j does not increase faster than Acn with A and c real constants, can be transformed
into X(z): As an example of a signal that is not of exponential order, consider for instance x[n] = 2n2u[n]
Properties of Unilateral Z-Transforms (given rightsided x[n]Z ! X(z) with ROCx):
� linearity: �1x1[n] + �2x2[n]Z ! �1X1(z) + �2X2(z) at least ROCx1 \ROCx2
� time reversal: x[�n] Z ! X(z�1) fz j z�1 2 ROCxg� delay (M > 0): x[n�M ]
Z ! z�MX(z) ROCx
� advance (M > 0): x[n+M ]Z ! zM fX(z)� x[M � 1]z�(M�1) � � � � � x[0]g ROCx
� frequency translation: x[n]p�nZ ! X(pz) fz j pz 2 ROCxg
� running sum: u[n] x[n] =Pn
k=�1x[k]
Z ! 11�z�1
X(z) at least ROCx \ fz j jzj > 1g� convolution: x1[n] x2[n]
Z ! X1(z) �X2(z) at least ROCx1 \ROCx2� n-multiplication: n � x[n] Z ! �z dX(z)
dzROCx
� initial value: limz!1X(z) = x[0]; which always exists
� �nal value: limz!1 (1� z�1)X(z) = limn!1 x[n]; if this limit exists
FOR A TWOSIDED x[n]Z ! X(z) (useful for INITIAL CONDITIONS in LDEs):
� delay (M > 0): x[n�M ]Z ! z�MX(z) + fx[�M ] + x[�M + 1]z�1 + � � �+ x[�1]z�M+1g
Some Examples
� h[n] Z ! H(z) transfer function
� 0 Z ! 0
� �[n] Z ! 1 all z � �[n�M ]Z ! z�M all s except 0 (1) if m > 0 (m < 0)
� u[n] Z ! 11�z�1
jzj > 1 � u[�n � 1]Z ! 1
1�z�1jzj < 1
� �nu[n] Z ! 11��z�1
jzj > j�j � ��nu[�n � 1]Z ! 1
1��z�1jzj < j�j
� n�nu[n] Z ! �z�1
(1��z�1)2jzj > j�j � �n�nu[�n � 1]
Z ! �z�1
(1��z�1)2jzj < j�j
� [coson]u[n] Z ! 1�[coso]z�1
1�[2coso]z�1+z�2jzj > 1 � [sinon]u[n] Z ! [sino]z
�1
1�[2coso]z�1+z�2jzj > 1
� [rncoson]u[n] Z ! 1�[rcoso]z�1
1�[2rcoso]z�1+r2z�2jzj > jrj � [rnsinon]u[n] Z ! [rsino]z
�1
1�[2rcoso]z�1+r2z�2jzj > jrj
Linear Di�erence Equations
Solving LDE w/CC using Z-transforms
We show how the unilateral Z-transform can be used to solve linear di�erential equations with constant coe�cients. In particular, the delay
property of the unilateral Z-transform is utilized. The procedures are not shown for the general case, but rather for an example.
Example: Assume we are given the following LDE w/CC that we are trying to solve:
y[n] � 2y[n� 1] = x[n] + x[n� 1]
with initial conditions y[�1] = 1 and x[�1] = 0:
First, take the unilateral Z-transform of both sides, using the delay property:
Y (z)� [2z�1Y (z) + 2y[�1]] = X(z) + [z�1X(z) + x[�1]] ()
() Y (z) =1 + z�1
(1� z�1)(1� 2z�1)+
x[�1] + 2y[�1]1� 2z�1
;
wherein inverse Z-transform of the �rst part equals the zero-state-response:
yzsr [n] = Z[ 1 + z�1
(1� z�1)(1� 2z�1)] = Z[ 1
(1� z�1)(1� 2z�1)] + Z[ z�1
(1� z�1)(1� 2z�1)] = (�[n] + �[n� 1]) Z[ 1
(1� z�1)(1� 2z�1)] =
= (�[n] + �[n� 1]) Z [ �11� z�1
+2
1� 2z�1] = (�[n] + �[n� 1]) (�u[n] + 2 � 2nu[n]) = �u[n]� u[n� 1] + 2n+1u[n] + 2nu[n� 1]:
Secondly, we have the zero-input-response:
yzir [n] = Z[ 2
(1� 2z�1)] = 2 � 2nu[n] = 2n+1u[n]:
The actual output y[n] is the sum of the two:
y[n] = yzir [n] + yzir[n] = �u[n] � u[n � 1] + 2n+1u[n] + 2nu[n � 1] + 2n+1u[n] = �u[n] � u[n � 1] + 2n+2u[n] + 2nu[n� 1]:
Partial Fraction Expansions
Suppose we are given Y (z) = 1(1�z�1)(1�2z�1)
(see also above) that we wish to expand into partial fractions. The procedure for doing this, is to
�rst factor and expand into partial fractionsY (z)z
; expressed in terms of z :
Y (z)
z=
z
z� 1
(1� z�1)(1� 2z�1)=
z
(z � 1)(z� 2)=�1z � 1
+2
z � 2;
whereafter we multiply both sides again by z and write the partial fractions in terms of z�1; so that we can use the Z-transform tables in order to
�nd y[n] :
Y (z) =�1
1� z�1+
2
1� 2z�1=) y[n] = �u[n] + 2 � 2nu[n] = �u[n] + 2n+1u[n]:
Comparison of FS and FT
Decomposing a periodic signal into discrete frequencies, each a multiple of some basic frequency !o; gives us the Fourier series representation
of that signal. With Fourier transforms, we can decompose aperiodic signals into its component frequencies, which, however, are not multiples
of some frequency, but rather range from ! = �1 to ! = 1: For periodic signals, x(t); F[x(t) and the FS represenation all carry the same
information.
How to get from FS to FT? If a periodic signal x(t) is written as a Fourier Series, x(t) =P1
n=�1Xnejn!ot; then its Fourier Transform is
X(!) = F[x(t)] = F[1X
n=�1
Xnejn!ot] =
1X
n=�1
XnF[ejn!ot] = 2�
1X
n=�1
Xn�(!� n!o):
How to get from FT to FS (if the signal is periodic)? If a periodic signal x(t) is written as a Fourier Series,
Xn =1
ToX(!) j !=n!o :
THIS SECTION IS FINISHED NOR CLEAR TO MYSELF; APOLOGIES.
Comparison of FT and ULT
The Laplace transform is a more general transform than the Fourier transform, meaning that it can be used for all signals that can be Fourier
transformed, andmore. From the de�nitons, the relationshipbetween the Laplace and Fourier transform seems to follow relatively straightforwardly:
X(s) = F[x(t) e��t] and F[x(t)] = X(j!)
However, we should distinguish between three cases:
Case (a): when ROCx includes the j!-axis then X(s) j s=j! = X(j!) = F[x(t)] and F�1[X(j!)] = x(t):
Case (b): when ROCx is just bounded by the j!-axis then F[x(t)] can sometimes be computed, but if so, then F[x(t)] 6= X(j!) and
F�1[X(j!)] 6= x(t): For example, u(t)L ! 1
swith ROC = fs jRe[s] > 0g; but u(t) F ! 1
j!+ ��(!) 6= 1
j!= X(j!): Furthermore,
F�1[ 1j!] = 1
2 sgn(t):
Case (c): when ROCx doesn't include the j!-axis then F[x(t)] cannot be computed. Although X(j!) may be de�ned, F�1[X(j!)] 6= x(t):
For example, etu(t)L ! 1
s�1 with ROC = fs jRe[s] > 1g; F[etu(t)] is unde�ned and F�1[ 1j!�1 ] = �e�tu(�t):
� F�1 [ 1j!��
] = �e�tu(�t) for complex �
� F�1 [ 11�!2
] = F�1[ 1(j!+j)(j!�j) ] = F�1[
j2
(j!+j) ] � F�1[j2
(j!�j) ] =j4 sgn(t)(e
�jt + ejt) = 12 sgn(t)sin(t)
Comparison of DTFT and UZT
The Z-transform is a more general transform than the discrete-time Fourier transform, meaning that it can be used for all signals that can
be Fourier transformed, and more. From the de�nitons, the relationship between the Laplace and Fourier transform seems to follow relatively
straightforwardly:
X(z) = F[x[n] r�n] and F[x[n]] = X(ej)
However, as above, we should distinguish between the cases where the ROC contains the unit circle (given by jzj = 1 , z = ej), is just bounded
by it, or does not contain it at all. These three cases lead to perfectly similar conclusion about the existence of the Fourier transform as in the
continuous-time case above.
Modulation
NOTE: � modulators are generally time-variant systems;
� some modulators are linear, some non-linear;
� asynchronous demodulators can only demodulate DSB-AM-LC;
� synchronous demodulators can demodulate all linear schemes discussed;
� (with synchronous we mean that the received signal is multiplied with a copy of the carrier;)
�WB-FM is much more noise immune than AM, since noise mainly e�ects the amplitude
of a signal and not so much the (instantaneous) frequency; furthermore, the amount of
used bandwidth is much larger for WB-FM, leading to better noise immunity.
Motivation for Modulation
� baseband signals will not propagate, but higher frequency signals will;� in order to simultaneously send multiple signals, one must use (distinct) carrier frequencies to prevent (limit) interference.
Assumptions and Terminology
� In H[m(t)] = x(t); H[�] is the modulation system, m(t) the message or modulating signal and x(t) the transmittedmodulated signal;
� the message signal, m(t)F ! M(!); is bandlimited to j!j � 2�BM rad=s; furthermore, we assume jm(t)j � 1; 8t;
� the modulated signal, x(t)F ! X(!); is a narrowband bandpass signal, centered at !c and has bandwidth 2�BX � !c rad=s;
Underlying Fourier Transform Properties
� ej!ot F ! 2��(! � !o)
� x(t)ej!ot F ! X(!� !o) � frequency translation
Overview of Analog Modulation Systems
Amplitude Modulation (linear): � Double-Sideband Amplitude Modulation (DSB-AM)
� Double-Sideband Amplitude Modulation with Large Carrier (DSB-AM-LC; a.k.a. broadcast AM)
� Single-Sideband Amplitude Modulation (SSB-AM)
� Quadrature Amplitude Modulation (QAM; see notes 15)
Angle Modulation (non-linear): � Narrowband Frequence Modulation (NB-FM)
� Narrowband Phase Modulation (NB-PM)
� Wideband Frequency Modulation (WB-FM)
� Wideband Phase Modulation (WB-PM)
Double-Sideband Amplitude Modulation (DSB-AM)
� modulation: H[m(t)] = x(t) = m(t) cos !ctF ! X(!) = 1
2�M(!) �[�(! + !c) + �(! � !c)] =12 [M(!+ !c) +M(!� !c)];
� demodulation: y(t) = x(t) cos !ct = m(t) cos2 !ct =12m(t) + 1
2m(t)cos 2!ctF ! Y (!) = 1
2M(!) + 14 [M(!+ 2!c) +M(!� 2!c)];
subsequently, an ideal LPF with Hlpf (!) = 1(0); j!j � (>)2�BM passes 12m(t) only;
� remarks: � demodulator must have the cos !ct carrier in phase lock to carrier at modulator, which requires a phase locked loop;
� you can demodulate SSB-AM using imperfect carrier cos(!ct+ �) : received signal is 12 cos(�)m(t) : �ne as long as cos(�) 6= 0;
� you can demodulate SSB-AM using imperfect carrier cos([!c +�!]t);
� the modulated signal x(t) takes up twice the bandwidth 2�BM of the message signal m(t) : BX = 2BM ;
� DSB-AM: power-e�cient, wastes bandwidth, requires synchronized reception.
Double-Sideband Amplitude Modulation w/ Large Carrier (DSB-AM-LC)
� modulation: H[m(t)] = x(t) = [1 + �m(t)] cos !ctF ! X(!) = �[�(! + !c) + �(!� !c)] +
�2[M(!� !c) +M(!+ !c)];
� demodulation: (sync) y(t) = x(t) cos !ct = [1 + �m(t)] cos2 !ct =F ! Y (!) = �YDSB-AM(!) +
�4 [�(! � 2!c) + 2�(!) + �(!+ 2!c)];
subsequently, an ideal LPF with Hlpf (!) = 1(0); j!j � (>)2�BM passes 14 +
�2m(t) only;
(async) x(t) �! recti�er �! y(t) = jx(t) j �! LPF �! z(t) = 1 + �m(t) �! capacitor �! �m(t);
� remarks: � demodulation of DSB-AM-LC can be done asynchronously, i.e. there is no need for synchronization of
demodulator to modulator;
� � is themodulation index;
� if the amplitude of �m(t) = (>) 1 we have full modulation (overmodulation �! distorted recovery);
� power e�ciency is poor (=12�
2PM12 (1+�
2PM )< 50%); because of large carrier;
� the modulated signal x(t) takes up twice the bandwidth 2�BM of the message signal m(t) : BX = 2BM ;
� DSB-AM-LC: power-ine�cient, wastes bandwidth, does not require synchronized reception.
Single-Sideband Amplitude Modulation (SSB-AM)
Upper-Sideband Amplitude Modulation (USB-AM)
� modulation: H[m(t)] = xu(t) = m(t) cos !ct � m̂(t) sin !ctF ! Xu(!) = M(!+ !c)u(�! � !c) +M(!� !c)u(! � !c);
� demodulation: y(t) = xu(t) cos !ct = � � � = 12m(t) + 1
2m(t) cos 2!ct � 12 m̂(t) sin 2!ct �! LPF �! z(t) = 1
2m(t);
Lower-Sideband Amplitude Modulation (LSB-AM)
� modulation: H[m(t)] = xl(t) = m(t) cos !ct + m̂(t) sin !ctF ! Xl(!) = M(! + !c)u(! + !c) +M(!� !c)u(�! + !c);
� demodulation: y(t) = xl(t) cos !ct = � � � = 12m(t) + 1
2m(t) cos 2!ct +12 m̂(t) sin 2!ct �! LPF �! z(t) = 1
2m(t);
Remarks on SSB-AM
� Hilbert transformer is a system with hHT (t) =1�t
and HHT (!) = �j sgn(!) = e�j�2 sgn(!) (phase shift); x̂(t) = x(t) hHT ;
� power e�cient: no carrier transmitted;
� bandwidth e�cient: now the modulated signal x(t) takes up the same amount of bandwidth as the message signal: BX = BM :
� you can demodulate SSB-AM using imperfect carrier cos(!ct+ �) : received signal is 12 cos(�)m(t) : �ne as long as cos(�) 6= 0 (see notes 15);
� you can demodulate SSB-AM using imperfect carrier cos([!c +�!]t) (see notes 15);
� SSB-AM: power-e�cient, no waste of bandwidth, requires synchronized reception.
Design of AM Receivers (DSB-AM-LC)
� total received signal is xtot =PN
n=1[1 + �mn(t)] cos !cnt; the sum of N DSB-AM-LC signals; we need message signal mn(t) only;
� solution A : use a tunable BPF; however it is hard to make a (multi-stage) �lter with su�ciently shapr cuto�;
� solution B : heterodyne receiver;
� solution C : superheterodyne receiver;
� SEE NOTES - SEE NOTES - SEE NOTES - SEE NOTES - SEE NOTES
Angle Modulation (FM and PM)
� A: PM: x(t) = Acos[!ct+ �(t)]; with �(t) = ��m(t); with j��j � � the phase deviation constant;
�(t) = ��m(t) is the instantaneous phase;
� B: FM: x(t) = Acos[!ct+ �(t)]; with �(t) = !�R t�1
m(s)ds; with !� the frequency deviation constant;
!(t) = ddt[!ct+ �(t)] = !c + !�m(t is the instantaneous frequency;
� we speak of NARROWBAND angle modulation when j�(t)j � 1; hence ej�(t) � 1 + j�(t); so x(t) � cos !ct � �(t) sin !ct;
� note that in narrowband angle modulation a LARGE CARRIER cos !ct is transmitted (ine�cient!)
NARROWBAND PHASE MODULATION (NB-PM)
� modulation: � = ��m(t); so H[m(t)] = x(t) = cos !ct � ��m(t) sin !ctF ! X(!) = �[�(! � !c) + �(!+ !c)]+
+ j2��[M(!� !c)�M(!+ !c)] (similar to DSB-AM-LC);
NARROWBAND FREQUENCY MODULATION (NB-FM)
� modulation: � = !�R t�1
m(s)ds; with j�(t)j � 1; hence m(t) cannot have a DC component, henceM(!)j!=0 = 0;
H[m(t)] = x(t) = cos !ct � �(t) sin !ctF ! X(!) = �[�(! � !c) + �(!+ !c)] +
j2 [�(! � !c)��(! + !c)];
with �(!) = !�M(!)j!
; so X(!) = �[�(! � !c) + �(!+ !c)] +!�2 [M(!�!c)
!�!c� M(!+!c)
!+!c];
Examples of Systems and their Properties
time BIBO
system linear invariant causal memoryless stable
(CT-1) y(t) = x(t) sin(t)u(t) yes yes yes no
(CT-2) y(t) = x(t) � cos(t)u(t) yes no yes yes
(CT-3) y(t) =R t�1
px(s)ds no yes yes no
(CT-4) y(t) = x(2t)+ 1 no no no no
(CT-5) y(t) = x(t) etu(t+ 1) yes yes no no
(CT-6) y(t) = x(�t) yes no no no
(CT-7) y(t) =R t�1�1
x(s)ds yes yes yes no
(CT-8) y(t) =R t�1
e�(t�s)x2(s)ds no yes yes no
(CT-9) y(t) = jx(t)j no yes yes yes
(CT-10) y(t) =R1�1
e�jt�sjx(s)ds yes yes no no
(DT-1) y[n] = 0 yes yes yes yes
(DT-2) y[n] = cos(n)2x[n] no no yes yes
(DT-3) y[n] + y[n� 1] = x[n] yes yes yes no
(DT-4) y[n] =Pn
k=�1x[k] yes yes yes no
Electronics
1. v(t) = i(t)R RESISTOR
2. R1 and R2 in series equals R1 + R2 I same; V = V1 + V2 (splits)
3. R1 and R2 in parallel equals (R�11 + R�12 )�1 V same; I = I1 + I2 (splits)
4. i(t) = Cdv(t)dt
CAPACITOR
5. C1 and C2 in series equals (C�11 + C�12 )�1 q (charge) same; V = qC�1
6. C1 and C2 in parallel equals C1 + C2 V same; q = CV
7. v(t) = Ldi(t)dt
INDUCTOR
8. L1 and L2 in series equals
9. L1 and L2 in parallel equals
10. KCL Kircho�'s Current Law (KCL)
11. KVL Kircho�'s Voltage Law (KVL)
� inductors and capacitors (amount of charge) are components with memory.
� resistors and opamps are memoryless components.
� a circuit with an L and a C in parallel is of order 2; a circuit with two L's or C's in parallel is of order 1:
LDE
����
h(t)
���� H(!)
����
jH(!)j6 H(!)
�
�
�
�
����
����
@@@@
@@@@
Fourier transform
substitute x(t) = ej!t
and y(t) = H(!)ej!t;
solve for H(!)
H(!) = jH(!)jej6 H(!)
Figure 0.1: Relations between System Descriptors.
LDE
����
h(t)
���� H(s)
����
Poles
Zeroes
�
�
�
�
����
����
@@@@
@@@@
Laplace transform
substitute x(t)L ! X(s)
and y(t)L ! Y (s);
solve for H(s) =Y (s)X(s)
write H(s) =N(s)D(s)
and
solve N(s) = 0 �! Zeroes
solve D(s) = 0 �! Poles
Figure 0.2: Relations between System Descriptors.