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/ 30
Jan 15, Set, aya)
Define causal and non causal signal. Gh
(Refer Causal and Non-eansal Signals)
Ans:
Signal
i function which is used to deseribe a physical quantity ot Rees
information about the behavior or nature of the phenomenon. Mathemat
fanction nd itis represented by symbol x()
‘System c
A system is defined as the physical process by ami
Classification of Signals
Signals in electrical communication systems are categorized as,
i signal) Basically, a signal contains
a variable is a ie ed by atime dependent ordinary
a signal can
ahemtict model wai relates input signal to the oufput signal
ematical model
L Continuous-time and diserete-tinie signals
2. Analog and digital signals
2: Periodic and aperiodic signals
3. . “Energy and power signals
5. Deterministic and probabilistic sighals
& Causal and non-causal signals
7. Even aid odd signals,
Contnuous-ime signals as shown in figure (1),
x) .
Fi t
aie inure Conti
ee SIA sumtime Sony
Publishers sing oy ‘Stributors Pvt. Leg,
‘ Signals defined at discret-instant of ve are called a8
“anvortine uh Te dein sige isbn ns
Examples for these signals are 1
ls are quarterly Gross National
product (GNP), monthly sales ofa corporation and stock market
daily averages.
Continuous and discrete time signals specify the nature
she signal alongthe timeatis.
‘Analog and Digital Signals: Anal
Analog signal is a signal,
va take on any value fom the continous-inerva,
(-0,1, then the signal is compressed
a1, then'the signal is expanded. —_
This is described diagrammatically inthe below figure (4)
x(0) : x(W2)
@) )
Figure: (a) Original Signal (b) Expanded Signals (¢) Compressed Signal
Figure (4)
Explain amplitude scaling and amplitude shifting.
5
"Amplitude shitting
Amplitude shifting ofa signal can be obtained by adding the value of every signal sample by a constant ‘4
Amplitude shifting is mathematically expressed as, (0) = x() +4
Whee, x() is the original signal and A is shifting constant,
Fora signal x(
"he shifting constant A > 0, then the signal shift upwards,
"the shitting constant, A <0, then the signal shift downwards,
| SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING Staens . " w
iAing trshown in figure
representation of amplitude shi
exd=2
” rine
iow?
i
+ 7] nee
| y jorrnct
(0 igang oirwa>o
Figure
Amplitude Scaling jgnal sample by A. The mathematica)
“Amplinde scaling of signl ca Be obtained by mupying the vale f every SDs!
expression for amplitude scaling is, g
. HO =Ax(D, ;
‘Where, x(#) is original signal . .
Zis scaling constant. ‘ :
Fora signal (0),
‘IfA> 1, then the signal expands (doubles) é
IfA< 1, then the signal compress (becomes haif). .
‘The graphical representation of amplitude scaling is explained below.
Consider, ‘
0. :
O 1, then the signal expands.
IfA<1, then the signals compress. P :
‘The graphical representation of amplitude scaling is shown in figure (4).
x(n)
oe
y(n) =0.5x(n)
or
Given that,
Input signal, (1) =e u(0)
The energy ‘E” of the signal x(¢) is defined as, .
Jan.A5, Set, 210)
gu Jixor dt
flewuor di (s(t) = 0 fort <0) Find whether the given sequences are
ornot
© xy{)= sio( 2]
(er se 4}
{or)
Determine whether the sequence Is perlodic oF
not x,(n) = sin(n/8).
(Refer Only Tope @)
(or)
Determine whether the sequence Is portodic oF
not. x,(n) = sin(6xn/7).
‘Refer Only Topic Gi)
‘OctINov.18, Sot-4, 21(a)
Ans:
o xe sin]
Given sequence is,
i,
wom af
‘On comparing the given sequence with sin an we get,
—
ag
>
bem ohh
1
I Tee
In the above equation zis a real number. So, fis not the
ratio of two integers.
‘Therefore, the given sequence is not a periodic signal, »
x,(n) = si
Given sequence is,
(0) sin (6m)
a
On comparing the given sequence with sin wn, we get
The frequency fis the ratio of two integers, Therefore,
the given sequence is periodic.
ae
integer
yy —Nontzeto inleget
“ihe smallest vale of 37 istered as fundamental
a
period.
‘Then,
“deel
notde’ to satisfy the above equality, the following has
tobe true for some integer, F,
(Serva (SI
= aD.
= A
|
In the above equation, for N= 7, 14, 21
to be an integer. :
__,Herice, the given signal x(n] is periodic a fundament
period N=7 (least possible.)
(ii) The given signal is,
F x(n) = Jeet
Fot the given signal to
condition,
be periodic, if it is satisfy th
Wen = 3004) for ally?
N—Not-zero integer
‘And the smal
oT Test value of >i referred as fandom
‘Then, TT pit 1 gnwodoaipn) sss" << te
ve Torder to satigly the abOve equality, he following haa vg Ba ioe for ome imtemets he -+ ~ . bs
3 {nef Jeane 28{nowy ed] at MS .
aN
= * ‘0
tn the above equation, for N = 10, 20, 30, ... turns out to be an integer. iia
Hence, the given signal x(n) is periodic a fundamental period, N= 10 (least possible value),
- of power of
7 Check whether the given signi re energy signals or power signal and find the energy
the signal.
[eos(rn); -4ens4
0
) x-{ 3 Otherwise |
nye [Comte no
* 0 5 Otherwise
anst
(0) Given signal .
aon [es -déns4
© ; Otherwise.
‘he sketches of the above signal is Sbtained as shown in figure (1).
(0) = coxen)
Figure (1)
From the above graph shown in figure (1), it is clear that the amplitude of the signal afprroaches zero as time tends to
‘ehnity, Hence the given signal is energy signal. The energy of the signal is given by,
e- Dm - Sees Fee
ee
= (cos nny! = cos?(-4n) + cos*(-3n) + cos*(-2n) + cos*(-m) + cos? 0+ cosn + cos? 2x + cost3n + cos? 4
ne
adebededoded ed etal
=
A) =9 ’
Signal energy,
‘The given signal
costa nz0 “The sketches of above signal is obtained as shows if
figure (2).
x(n) = costn
Figure (2)
From the above plot shown in figure (2), under the
condition 2 0; the signal amplitude does not tends to zero
as time approaches to infinite. Hence the given signal is power
signal. The average power of the signal is given as,
x,
= Need 2
PeNS “le (n)
First of all calculate ‘N’, the given signal is periodic
signal with fundamental time period 7 then,
Time Period
aa
N
samples
Then the power of the signal is,
1S.
=p
Hwee (ny
= = (008704005? n)
£ Power of the signal is :
JF, Find th power and rms value of signal
cos 2nt. ee 2s
Ans:
Given that,
(0) = 20 cos 2nr
Then, power of sign
RMS value of signal
The expression for power of signal is given by,
SIA visi
(Model Paper, Q1() | Dec-16,Se2, a2)
A
oy AO
7 Orx2
7 ob
np 9) fare fovea]
rmearlh
a,
400| 7 ante)
Not |
Alon ( 4x Jr
100
he P|
(ys singn— sindne (— 7) = sindre? — (sind
u
1
J 400s 2(2nt) dt
es
0
"
Tt, .
[r-c+gebinger =sin4n(-T)]
= 2sin4nT = 2(0) = 0)
mu 100 100
F rl? GT) +01 = een
100 *
th FD
P= 200 watts
© Power of signal = 200 watts
Thetms value of si
sim eeu
4 cos) given by ee signal with amplitude A(
Then,
BMS value 020 60s ane is given as 22
sie =
ctiNov.-17, Sot, €2(0)
‘The given signal is shown in figure.
x)
Ans?
Figure
From the figure, the signal can be represented as,
x) =sint, O<¢< pe
Where, i
Amplitude =
Energy of a signal =?
‘The energy ofa signal is expressed as,
E= fora
is ee ON ee
DSF Test whether the signal x(n) = (%4)" u(n) energy
or power signal.
Ans: ee-5, Set, 2a)
Given signal,
© xGi) = (Ay ur)
The expression for energy of signal is given a,
‘Energy of signal,
B= Lixo
eer Sor uel?
Wim ay
“nt Spoor hot i
Wud
‘The expression for power of signal is given as,
Power of signal,
Spoor
hb
ena
++ Power of signal x(n)=0
Henee, energy of signal is finite and power of signal is
zer0.
So, given ee isa energy signal. . 4 pop
Tost whether the signal xi) =o ul energy!
‘Ansi
Given that,
x= oF u(e)
“The expression fr energy of signal is given
Energy of signal = flx@OP a
- j et uP at
‘The expression for power signal is given as,
r
"ower 1
nal = fe e
Power of signal = It ar {ho at
its
at fee
= Maelle na
= Abr :
18 OF
1
fe; 50
AsToe, 79
Powerof signal =0
Hence, the enerayof seals 4 ond poe of signal
: is
So, the given signal isa energy signal, seis mine the power and RMS value of the following signals,
(@__¥(t)= 5 cos (Sot + ni3)
(i) y(t) = 10cosstcostot,
ans:
@
Dec.48; Set, G20)
Given that, ,
40) = Scos (so +3)
3
The expression for power of a signal is given as,
Power of signal = A
fhe hbo dt
xh. yf IScos| 50r a
hap JB +f) ae
mil ar Jo" [so(soet ya
cos* 0=
= zee a foo{oee3)]
nyt
sin2| 50047
3
eee
= t Awl, i
ane |S 250) .
afer ae ret
re (-T)+
=f =
Ti
andl Beans I ee irene 5
Toe
Power of signal yatts
{SPECTRUM ALLIN-ONE JOURNAL FOR ENGINEERING STUDENTS RMS value = t
BS 2
(os) RMS value otf GB
3
RMS value: z
(i) Giventha, '
A-B)
x = 10c0sS1 cos10? exis
sfomstons is femeact?
{f= Seost51+ c0s54]
a) = vale SeosSt
BR
coswor is expressed 45 “>”
‘The power of a signal fora simi signal A
‘From equation (1), the expression for power signal is given 35,
Power of signal (P)
= Power of signal = 25 watts
‘The expression for RMS value is given as,
rats vane « [444
= eae
RMS value = 5
(or)
RMS value = yPower = 4/75 <5
Q17. Find 1d odd com,
a Ponents of the signal x(
, (t) = cos(oogt + 1/3),
5 from equation (1))
Given signal is,
x) ~cos[04/+)
Even component =
Odd component
)
Papors, at(a) | Octinov-t7, SH Fee re poipenet el od component oft iignal is determined by the expressions,
oe FeO+xco
=)
20 F60- 20) 2
2-1 is obtained by replacing ‘? with (in equation (1) ie
(—1) =. a
11) =c05(o4(- os). cos foge- +)
waco ~eos{oge 4) (re0s(- 0) = 6090) REG
Even Component
On substituting equations (1) and (4) in equation (2).3,( is obtained
210 = jfc0s{age+ §)reos( t- 4
1
-3P os 0505] [is cos(d + 8) + c0s(d ~ B) = 2 cosd eosit]
= Cosengf. | 1
4d Component
‘On substituting (1) and (4) in equation (3), x,(0) is obtained as,
140~ Heoslaye+ 3) ~ cos{oy-*}]
= ies ogtsi
[+> cos(d +B) cos(d ~ B) =~2sind sinB]
zee (iE
x= sin oot
(peyFina the even and odd components of the signal x(t) = sin2t + sin2t.cos2t + cos2t.
‘Ans:
ctiMtow-17, Set2, €2fa)
Given that,
(0) = sin2t + sin21.cos2t + cos2t
a)
Even component =?
Odd component = ?
© The even component and odd component of the signal are determined by the expressions,
x= FO +a) *
X(0= Fb ACO) °
2(-1) is obtained by replacing ‘' with ‘fi equation (1), :
ie, x{c) = sin2(-1) + sin2(-1) c0s2(-1) + c0s2(-1)
in(—x) = |
cos(-x) = cosx
(ARAN srecTran ALLiN-ONe JOURNAL FOR ENGINEERING stavents ae
=~ sin2t— sin2¢ cos2t + cos2r ( Even Comiponent aol tain
‘On'substituting equations (1) and (4) in equation a Ca
240 = 3 [sin2e + sin2reos2t + cos2r -sin2¢— sit
1
z Reos2n)
= cos2r
28 = cos?
‘Odd Component yptained #5+
z i " (1) is 0
‘Gn substituting equations (1) and (4) in equation (3) 44(0 at + 60820)
= sin2t 008
ss(t= ina sina ost + coat sind sin
= Fresinars 2sin2¢ e082]
= sin2r + sin2r cos2t
in 21 + sin2¢ cos 21
3,0)
Gio, Find the even and odd components of the signal x(t
Ans:
Given that,
x = (0)
Even component = ?
(Odd component =? y
The even and. a components of the signa is determined by the expreSsions,
== i [x +x] .
1
zk -x09)
(1) is obtained by replacing ‘r’ with ‘7’ in equation (1).
) = tu(t)-
rl
ie, x-f) = 1-1)
Even Component :
On substituting equations (1) and (4) in equation (2), x,(0)s obtained as,
240 ~ Firm)
= Flee -u-o
1
= ZF sent) [" u(t) ul-1) = sen(9)]
‘Odd Component
‘On substi
tng equations (1) and (4) in equation (3, x. (1 ig obtained ag
20 = 5 [tu(l) ~~ tu(-1)))”
1
= Flue) + u-D))
1
3 () + u(-1)= 1]
0-5
Oct Nov.-17, Set, 2 even and odd components of the signal
x(9 = (1+ 8+) co:
ee SctiNov.17, Sets, 02a)
Given that
x0 = (142 +8) 0082 10;
ep eat
on (Ms
Odd component =
seein De ret 6 organs ofthe
20 = FbO +H) a
10 70-9) @)
x) is obtained by replacing * with *-? in equation (1).
ie, = (1+ C0? + 09) cost10(-1)
=(1+#-8)cost1or [+ cos(-6) = c0s6]
(4)
ven Component
on substituting equations (D and (4) in equation (2),
s{pisoblained as,
xf) = Fu +240) cos? 104 (1+2-A) cos*10q
= Fie +212) cos? 107)
= Fea +1) cos? 101)
=(1+P) cos*10r
7-x,(0 = (1+ P) cos*10r
044 Component
On substituting equations (1) and (4) in equation (3),
2{isobtained as,
af) = Fa ++ 2) cos? 10r—(1 + P= P) cos*10¢]
Fee cos? 10/]
val
Signal x(t) = u(t).
Ans: Jan 15, S0t4, G2(e)
The given signal is, ,
+ x) = u(t)
Power of the signal, P=?
RMS of the signal =? ,
‘SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
1,23:
“The expression Tor power of the signal x(0) is Bi
r
1 e
a ek fier a
oo bea 0
ut Afnor a
Pe aT
- el WP at
=
p
2
= be a” at
= AROS
Le-o
igo
ut
2
The-RMS value of a signal is the square root of
power (P).
ie, RMS = VP
“Tals
= 0.707
RMS value = 0.707
we Examit following systems with respectto -
the above properties,
(y(t) = cosx(ey
(y(t) = Log, Ixt8.
(Mode! Papertt, Q1(b) | Jan.-15, Set-3, Q4(b))
Ans:
© yO =coslt0]
‘The given system equation is,
20) 008 [x()]
Causal or Non-causal
From the given system,
For t= 0, (0) = cos [x(0)]
For f= 1, (1) = cos [x(1)] - te Sable oF Unstable ;
From the giva system, »(#) og), [x9]
Stability j4Pltes that for any boutided in
should also be *0unded,
AS,
put, the output
S
SPECTRUM ALLIN-ONE JOURNAL FOR ENGINEERING STUDENTS |
= (@l)
= agny(n= ng) +238") ‘The linear combination of individual outputs iS
© ayyln) + ayn) = ayny(n =n) + 4,6,00~ "0
‘Equations (1) and (2) are equal.
Hence, the system is linea.
‘Causality
From the given system,
a(n) = x(n)
Forn=0, (0) =x(—%)
Form=1, 91) =a(1 =M%)
Forn=2, 2—m),
Forn=~1, (-1) = (-1- 1)
MD) = HD =)
From the above analysis, itis clear thatthe output of the
@
‘system depends on ‘n," value, which may result in dependency
‘on past and future inputs also.
Hence, the system is non-causal.
‘Time-variant
From the given system is,
3{nt) = x(n ng) and
Kn) = Hs(n)] = x(— m9)
If the input is delayed by k units in time, then,
20, B= Hat B)}
=x(n-k=m)
2 (ay B) = x(n kn)
If the output is delayed by & units in time, then,
Ho-K) = Z*H Gn)
=n k—n)
2 n-h =n ~k-m)
From equations (3) and (4),
Wn, k) =n -h),
Hence, the system is time-invariant,
= Q)
=)
i)’ The given system equation is,
2472) = c05(n0>4) x)
Linearity
Lety,(n) and y,(r) be the outputs of the
x,(n) respectively,
‘Then, y,(7) = eos(nag)x (n) = Hx(n)]
Jx(N) = cos(rtog x(n) = Hfx(n))
For inputs x(n) nd, the linear combing
44% \(1) + azxfn)
inputs (0) and
ion is,
pees so" a o
(n+ ana
coro individual Bouts is,
i eog)apeilt) ¥ C8)
12,10) + a)
‘the lin
ay
n="
= c0s("io)
re equal: %
Equations (5) 82 Oe
the system iS
Hences
causal
From the Bi ‘
= cos(ntg)"
Fe a 40) = €08(0)#(0) = 00) \
oe=0,
y= cos(@p)x(1)
forn=2; 92)= cos(20g)*(2)_
Forn=-1, xl cos(wo)e(-!)
Forn=-2, 4-2) = cos(2a9)x(-2)
+ Brom the above analysis, it is clear that the output oftie
system depends only on. present input.
Hence, the system is causal.
‘Time-variant
From the given system,
y(n) = cos(n,)x(n) and
40) = Hfx(n)] = cos(n,) (7),
If the,input is delayed by k units in time, then,
01 #) = Hien — B)]
= eos(rtrg) x(n k)
+ 3404 B) = cos(nen,) x1 — b)
iven system
Forn= 15
0
, Ifthe output is delayed by k units in time, then,
Yn—K) = rt (ny]
= cos[(n — k)wg).x(n — k)
a Ke B= cosl(n— kdl x(n— k)
ene and (8), (n, &) # (n= 2)
Tee vem is time-vatian,
Biven system equation
20) = any,
~@
(ily
is,
Linearity
Li
ct,
(0) respec
Pectively,
dy,
: A(n) be the Sutputs of the inputs 1, (0)
en y(n) =
atx,
OR + ap
HOV steonip ge
Nuts, = ala,x\(n) + a,x, +b
‘The linear combination of individual outputs is,
ayi(e) + ayn) = a,lalx, mp + 6) + alate, P+ 5)
= aa, [x\(n)? + a,b +aa,[x,(n)P + ayb
Equations (9) and (10) are not equal.
Hence, the system is non-linear,
Causality
From then given system,
Qn) = alx(nyP +d
Forn=0, 3(0)=afx(O)}? +
Forn=1, (1) =alx()P +5
Forn=2, »(2)=alx(2)2 +5
Forn==1, -h)=afx(-1)P +5
Forn=~2, y(-2)=alx(-2)P +b
From the above analysis, itis clear that the output of the system depends only on present input.
Hence, the system is causal.
‘Time-variant
Ans:
@
From the given system,
3) = alx(npP +b
and y(n) = H[x(o)] = afx(ny)? +
If the input is delayed by & units in time, then,
af) = Hix - 1)
(nH +b
2 yn, =alx(n WP +b
If the output of delayed by & units in time, then,
n—B) =2* Hx)
[x(n HP +b
2 yn —h) = afx(n— WP +b
From equations (11) and (12), y(n, &) = (i)
Hence, the system is time-variant.
(1
(2
Determine whether the systems described th l/p o/p equations are linear, time invariant, dynamic an
‘stable.
yext=3) (8-0
ast)
() yatt) = ae
(i) yale) = natn] + Be? fo.
The given system is,
yO =xt=3)+ 1)
- Linearity
‘The system which satisfies the superposition theorem is called a linear system i,
ay) + by g(t) = Max (0) +2x,00] = ail (01 + HO]
1), (0 is the output ofa system for an applied input x,(), then the equation (1) becomes as,
y,O=x¢-3)4+G-9
"SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
Cet Nov.-19, Set-t, O2/
asl UNIt=1~ UntToaUetion)*
Q28. Detormi
time Invariant or not, Ch lowing systine
tems are causal or not, Whether th
y(t) = tx(t)
y(n) =x(2n)
Ans:
‘Time Invariant
(Given that,
WO =a
If the input of | is
Patient ctiaqecruaey
46 fe) = SOC ~ fe)
OctiNovt, Sat, az)
o)
layed by fie. x(¢~%),
as,
Whe .Q)
. ere-)Xf, fp) is the response of th
delayed input. ie system due to
If the output is delayed by “then replace f by tf
Then, equation (1) becomes,
=) = (=) =F)
~@)
From equations (2) and (3),
We, te) #0 —t)
Hence, the system is time variant.
(i) Given that,
Yn) = 2n - w= 4)
If the input of the system is delayed by samples, then
the output of the system is given as,
We, Mg) =f x(n = rg)
= x[2(1—n9)]
TF the output is delayed by ng then replace n by 1"~ ry
Then, equation (1) becomes,
w= M9) = f=]
=x12(n—n0)]
(8)
S 6)
‘From equations (2) and (3),
W(, Mg) = {= Mo)
Hence, the system is time invariant.
Causal
) Thegiven system is,
For 1= 0, (0) =0. (0) =0
For r= 1, (1) = 1. x(1) = 1x01)
For 1= 2, (2) = 2.x(2) = 2202)
For ?= 1, x1) =CD.2CD = CD
For t= 2, (2) = (2)-x(-2) = 2-2)
‘SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS
the system depends only on present int
élear
put, Hence, the system
From the above analysis,
is causal:
(ii) Thé given system
an) =x2n)
For n =0, (0) = x(2(0)) = x(0)
For n= 1,y(1)=(2(1) = x2) ”
For n = 2,9(2) = *(2Q)) =*(4)
For n= 1, (1) = (2-1) = #2)
For n= 2, 2) = x2(-2)) = 1-4)
From th: above analysis, itis clear that the output of the
system dependson future and past inputs along with the present
input.
Hence, thesystem is non-causal.
41.4. SINGULARITY FUNCTIONS AND
RELATED FUNCTIONS: IMPULSE FUNCTION,
STEP FUNCTION, SIGNUM FUNCTION AND
RAMP FUNCTION i
ox Define and sketch the following elementary
continuous time signals,
(iy Unit step function
(ii) Unit impulse signal
(ii) Signum function.
(Model Papers, Qt(a) OctiNow-12,Set-4, Q2a) | Jan-18, Set Q2())
(Refer Excluding Relation berween Step andl Sigiim
Functions)
(or)
Define and sketch. the following elementary
signals,
(i): ‘Unit impulse signal
(ii) Unit step signal
(ii) Signum function. |
andl Signum
(Refer Excluding Relation bebween Step
Functions) ety
(or)
Define and sketch the unit step function and
signum function. Bring out the relation between
these two functions.
‘OctyNov.:18, Set, 02(b)
(Refer Only Excluding Unit Iripulse Signal) © Unit Impiitse Signal ‘
signal.
For answer refer Unit, Q31 Topi: Unit ns Si
(MH) Unit Step Signal |
| uy
For answer refer Unit-1, Q31, Topic: Unit Ste oh
(ii) Signum Function
Signum function sani) is defined as,
1fort>0 i
sony = 4 Ofort=0
[=I fore<0
shown in figute-
jgntm function is a8
‘The graphical representation of signtm so)
Ieee:
dab eet
yy
Figure: Signum Funetion
Relation Between Step and Signum Function
‘The graphical representation of signum function determines the relation between step and signum function. Signum function
‘can also be represented as shown in figure below.
sentt) (0) 0)
Figure
From figure, the mathematical description of signaly
=u) 1 AM10'5 given by
=)
Thus, the expression for slgnum funtion interns ot yey
sen()= u(t) —u(-2) OO igen by, ‘ wo
Define the fotlowi
ing basic —__
representation, ‘Slanale with graphical
(Ramp signal
+ (i), Sinusoidal signal, -
Ramp Si
The ramp function is the inte
“ ‘eral
xo) with respect to time and its sone ean fin
continuous-time unit ramp signal 1) ig lsat a
aft rz0
ra it <0
Treampltude of unit-ramp signa increases linearly with
reference to its time ‘P and itis graph
iphically
as shown in figure (1), pinoy
r(t)
ny) t
Figure (1}: Continuous-Time Unit Ramp Signal
(can also be expressed in terms of u(t) as,
ri) =tu(y)
Where,
1a <0
Sinusoidal Signal
u(t
The sinusoidal signal is defined as,
Asint, O<1<0
*O= 19 | elsewhere
The graphical representation of the sinusoidal signal is
shown in figure (2).
x)
1.33,
Define continuous and discrete time unit step
and unit imputse signals.
(or)
Define continuous time unit step and unit
impulse. %
Dee, St2, Qa)”
(RefeP Only Continunis Ont Tapa Sigal ar Con
tious Unit Step Signal)
(or)
Give the mathematical and graphical represen-
tation of continuous time and Discrete time im-
pulse function.
(Refer Onby Ont Impu5e Sigal),
Oct iNov.-19, Set, QA(a)
Ans:
Continuous.Unit Impulse Signal
Unit impulse signal 8(0) is defined as,
I fort=0
B= ort
0 fort #0
‘The sketch of unit impulse signal is as shown in figure (1).
&(t)|
1
4327012345
>t
Figuré (1): Unit Impulse Signal
Discrete Time Impulse Function
AA discrete-time unit impulse function {7} also known
as unit sample and defined as,
0, n 40
Ci f n=0
Graphical representation of discrete-time unit impulse
is as shown in figure (1).
3)
t
eg ot O23 tee
Figure (1): Diserote-Time Unit Impulse
‘The amplitude of unit impulse signal is one when time
interval is zero..It is also called Dirac Delta function.
Continous Unit Step Sigi
Unit step signal u(t) is defined as,
Figure (2)
= , RUM ALLIN-ONE JOURNAL FOR ENGINEERING STUDENTS |
6 1fort20
WO” Vo fore frosoa - Proomoa
= FO) Oa
But, focoa =1
2 J7@8EOdr= £0)
Property-3
Bar) = ia 5,
Proof
Considering the integral,
Jr@8candt, for(a> 0)
“Let, at =y
4)
mane
a
2 fre@beanar- I s010980) #
=+ frolasore
a
-+%0
a
Using property-2, (0) can be written as,
Jr@scana = 4 f(), for (a<0)
mm” (a<0)
yen
= ast S(HBdt :
* Jroxana - froyso"
LHS = R.H.S when,
Bat) = me
* SPECTROM ALL-IN-ONE JOURNAL FOR, ENGINEERING STUDENTS )/ 7
Jro%e-tott = 100) .
a
tion (2) can be written as,
Consider LHS. i
Jroe-1) nome
= at f= lo
If £0) is continuous function at ¢= fy the the valve ee
Frommer =p0y Jato AOA
= LHS =RHS
2 [rose -t)ar= Ft)
G36. Define an impulse function and plot a(t
Ans:
Impulse Function
For answer refer Unit-l, Q31, Topie: Unit Impulse Signal.
Given function is,
= 8(¢+ 2)- 8-3) 2
‘The graphical representation of given function is as shown in figure.
Be+2) = 3-3)
+2)= t= 8) OctNov-A7, Ss
a(t +2)-8(t=3)
57. Define a unit step function and plot uta,
Ans:
Paes soar Gs ‘OctJNov.-17, Set!
‘or answer refer Unit-I, Q31, Topic: Unit is
pene opie: Unit Step Signal,
a= u(t ~2)
‘The signal ¢~ 2) is obtained by right shifting the signal i) on time ax
cm t=2) “sis by 2 units as shown in figure. Ans:
© — ftae+3)-28 ua
31
‘The given integral is,
*
4 1
Jue+3)-28 ua = Jace ante — f28eu(nae
4
4 ‘
‘ etfs nbd 1; 130
uta) = seautnulls
[ iB lsewhere S449 {i aaa
4 4
= far—o- fowoa = tf, -2facar
She Beg a
=(8-C1]-201) [ wo={h cal
1
2
The given integral is,
3
Joon
1
2
Put k= 3¢
ee east
1
=< dk
= Aes
Upper Limit
oy spECTRUM ALLAN-ONE JOURNA
‘ « *
Jucesayde [28enuryde— f26(Quat [:4
3 >
IL FOR ENGINEERING STUDENTS
ana 32
Then, f 8(32)ar can be written as,
ia
y 1
Teoje
2
J sana.
h a
1g
1
=z J5eak
34
2) ;
egal pe
J 3 0G0a = 50) [sw {i tt]
i) u(t) + L(t) - uft—1) + uft-+ 4) Suit 2),
Ans:
5 fe
® al J+ne-0
The given signal is,
s0= {Zn
oa (l)
‘In general, the function n(¢) can be represented as,
| Eat elt
| wedded
PN ft) tt
tr) }ta4)
rah)
A
@)
pe)
(c+ From equation Qy)
gga)
sions (3) and (4) im equation (
qu :
{3
1 3
1) -ufo-3
goy=no-4o-*(0 2) | j
3
ee cor(-4)-«(3)
=(0)-" 3 5
ro eet
porto
>
)
=1-0+1-0 i
Aly=2
Fort=2
= me) fapoo fey
=u() ~ u(0) + (})
=l-1e1ey
> f2)=0
Forts.1
S flys fst
2
bauer tof a]
GH) .4¢ (introductiony’y)—7 2