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Understanding Energy, Power, and Signal Types in ECE313

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University of Texas**We aren't endorsed by this school
Course
ECE 313
Subject
Electrical Engineering
Date
Dec 11, 2024
Pages
14
Uploaded by DoctorRhinoceros2814
The University of Texas at AustinDepartment of Electrical and Computer EngineeringECE313: Linear Systems and Signals — Fall 2024Lecture 2Aryan MokhtariThursday, August 29, 2024Goal:In this lecture, we discuss energy and power of signals, Odd and Even Signals, and trans-formations on signals.1Energy and (time-averaged) Power of SignalsNext, we define the energy and power of continuous-time and discrete-time signals.1.1Continuous-time SignalsThe total energy of a continuous-time signalx(t) over the intervala≤t≤bis defined asZba|x(t)|2dt, where|x(t)|is the magnitude of the possibly complex numberx(t).The energy of the signalx(t) over an infinite time interval is given byE∞:= limT→∞ZT−T|x(t)|2dt=Z∞−∞|x(t)|2dtWe can define the time-averaged power of the signalx(t) over an infinite time interval asP∞:= limT→∞12TZT−T|x(t)|2dt1.2Discrete-time SignalsWe similarly define energy and power for discrete time signals as (replace integral with sum)E∞:=limN→∞NXn=−N|x[n]|2=∞Xn=−∞|x[n]|2,P∞:=limN→∞12N+ 1NXn=−N|x[n]|2where|x[n]|is the magnitude of the signalx[n].1.3Important remarks and empalesDefinition 1.An energy signal is a signal with finiteE∞. Indeed, for an energy signal we haveP∞= 0.1
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Definition 2.A power signal is a signal with finite and nonzeroP∞. Indeed, for a power signalwe haveE∞=∞.Examples:•Consider the signalx(t) =e−|t|/2. We show that it is an energy signal:E∞= limT→∞ZT−T(e−|t|/2)2dt= limT→∞2ZT0e−tdt= limT→∞2(−e−t)T0= limT→∞2(1−e−T) = 2P∞= limT→∞12TZT−Te−|t|dt= limT→∞212TZT0e−tdt= limT→∞1T(−e−t)T0= limT→∞1T(1−e−T) = 0•Consider the signalx[n] = 4. We show that it is a power signal:E∞=limN→∞NXn=−N42=limN→∞16(2N+ 1) =∞P∞=limN→∞12N+ 1NXn=−N42=limN→∞16(2N+ 1)2N+ 1= 16•Consider the signalx(t) = sin(t). It can be shown that it is a power signal. (Exercise!)•Consider the signalx[n] =n. It can be shown that for this signal both energy and power arenot finite. (Exercise!)2Odd and Even SignalsDefinition 3.A continuous-time signalx(t)is called anevensignal ifx(t) =x(−t).Similarly, A discrete-time signalx[n]is called anevensignal ifx[n] =x[−n].Examples:x(t) = cos(2πt) andx[n] = cos(πn/5) are even signals.-3-2-10123-1-0.8-0.6-0.4-0.200.20.40.60.81-1-0.8-0.6-0.4-0.200.20.40.60.81-20-15-10-505101520Proof:Ifx(t) = cos(2πt) then we havex(−t) = cos(−2πt) = cos(2πt) =x(t)Similar argument holds forx[n] = cos(πn/5). Hence they are both even signals.2
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Definition 4.A continuous-time signalx(t)is called anoddsignal ifx(t) =−x(−t).Similarly, A discrete-time signalx[n]is called anoddsignal ifx[n] =−x[−n]Examples:x(t) = sin(2πt) andx[n] = sin(πn/5) are even signals.-3-2-10123-1-0.8-0.6-0.4-0.200.20.40.60.81X 3Y 0-1-0.8-0.6-0.4-0.200.20.40.60.81-20-15-10-505101520Proof:Ifx(t) = sin(2πt) then we havex(−t) = sin(−2πt) =−sin(2πt) =−x(t)Similar argument holds forx[n] = sin(πn/5). Hence. they are both odd signals.2.1Odd and even parts of a signalEach signal can be written as a sum of an even signal and an odd signal.Ev{x(t)}:=x(t) +x(−t)2,Od{x(t)}:=x(t)−x(−t)2Ev{x[n]}:=x[n] +x[−n]2,Od{x[n]}:=x[n]−x[−n]2Exercise:show that the even part is an even signal and the odd part is an odd signal.3Elementary Operations on Signals•Amplitude Scale:y(t) =ax(t) ory[n] =ax[n] whereais a real (or complex) constantExamples:x[n] = sin(πn/10) andy[n] = 0.5x[n] = 0.5 sin(πn/10)Examples:x(t) = sin(2πt/5) andy(t) = 2x(t) = 2 sin(2πt/5)-1-0.8-0.6-0.4-0.200.20.40.60.81-20-15-10-505101520-10-8-6-4-20246810-2-1.5-1-0.500.511.523
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•Amplitude Shift:y(t) =x(t) +bory[n] =x[n] +bwherebis a real (or complex) constantExamples:x[n] = sin(πn/10) andy[n] =x[n]−2 = sin(πn/10)−2Examples:x(t) = sin(2πt/5) andy(t) =x(t) + 1 = sin(2πt/5) + 1-3-2.5-2-1.5-1-0.500.51-20-15-10-505101520X -4Y -2.951-10-8-6-4-20246810-1-0.500.511.52•Addition:y(t) =x(t) +z(t) ory[n] =x[n] +z[n] (add corresponding components)Examples:x[n] = 3 andz[n] =nwe havey[n] =x[n] +z[n] =n+ 3Examples:x(t) = cos2(2πt/5) andz(t) = sin2(2πt)/5 we havey(t) =x(t) +z(t) = 1-20-15-10-50510152025-20-15-10-505101520-10-8-6-4-2024681000.10.20.30.40.50.60.70.80.91•Multiplication: :y(t) =x(t).z(t) ory[n] =x[n].z[n] (multiply corresponding components)Examples:x[n] = 3 andz[n] =nwe havey[n] =x[n].z[n] = 3nExamples:x(t) = sin(2πt/5) andz(t) =et/10we havey(t) =x(t).z(t) =et/10sin(2πt/5)-60-40-200204060-20-15-10-505101520-10-8-6-4-20246810-3-2-101234Elementary Operations on the Independent Variable•Time Shift:y(t) =x(t−t0) ory[n] =x[n−n0], wheret0is a real value andn0is an integer.Whent0for C.T. (andn0for D.T.) signals is positive the signal moves to right, and when itis negative the it moves to left. [Think of playing a recording earlier or later in time]4
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012345678910-20-15-10-505101520012345678910-20-15-10-505101520-10-8-6-4-2024681000.10.20.30.40.50.60.70.80.91•Time Reversal:y(t) =x(−t) ory[n] =x[−n]. This is equivalent to flipping the signal withrespect to the vertical axis. [Think of playing a recording backward]012345678910-20-15-10-505101520012345678910-20-15-10-505101520-10-8-6-4-2024681000.511.522.53•Time Scaling:y(t) =x(at) ory[n] =x[an], where a is a real constant.–Ifa >1, then the resulted signal is a time-compressed signal[Think of playing a recording twice faster (a= 2)]–If 0< a <1, then the resulted signal is time dilation[Think of playing a recording at half speed (a= 1/2)]5
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-10-8-6-4-20246810-0.500.511.5-1.5-1-0.500.511.5-20-15-10-505101520-10-8-6-4-20246810-0.500.511.5-1.5-1-0.500.511.5-20-15-10-505101520-10-8-6-4-20246810-0.500.511.5-1.5-1-0.500.511.5-20-15-10-505101520Remark 1.In discrete-time signals whenais an integer different froma̸= 1ora̸=−1,loss of information in the signal occurs, unlike the continuous-time case.4.1Combination of Shift, Reverse, and ScaleWhen you have a combination of all these three operators the best order to apply them is thefollowing: 1-shift 2-reverse 3-scaleConsider the following example. If signalx(t) is defined as the following,x(t) =1if−1< t <0,−t+ 1if 0< t <1,0otherwise,then find signaly(t) =x(−32t+ 1).Analytical approach:think of−32t+ 1 as a new variablet′and exploit the conditions for signalx(t).•If−1< t′<0 we havex(t′) = 1. Considering the fact thatt′=−32t+ 1 we have−1< t′<0⇐⇒23< t <436
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Hence, we obtain that for23< t <43we havey(t) =x(−32t+ 1) =x(t′) = 1•If 0< t′<1 we havex(t′) =−t′+ 1. Considering the fact thatt′=−32t+ 1 we have0< t′<1⇐⇒0< t <23for which we havey(t) =x(−32t+ 1) =x(t′) =−t′+ 1 =−(−32t+ 1) + 1 =32t•Ift′>1 ort′<−1 we havex(t′) = 0. Considering the fact thatt′=−32t+ 1 we havet′>1 ort′<−1⇐⇒t <0 ort >43for which we havey(t) =x(−32t+ 1) =x(t′) = 0•Therefore, we havey(t) =x(−32t+ 1) =32tif 0< t <23,1if23< t <43,0otherwise.7
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Geometric approach:Shift, flip, scale!Note that signalx(t) can be plotted asFirst, shift by 1 to the left:Then flip since there is a negative sign beforetNow scale (in this case compress) by a factor of 2/3.8
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5Additional Examples1. (10 pts) Determine if the following signals are power signals or energy signals by computingtheir total energy and time-average power.(a)x(t) =−2 cos(2πt/5)Solution: Using the fact that cos2(x) =1+cos(2x)2, the energy of this signal is given byE∞= limT→∞ZT−T| −2 cos(2πt/5)|2dt= 4 limT→∞ZT−Tcos2(2πt/5)dt= 4 limT→∞ZT−T1 + cos(4πt/5)2dt=4t2+20 sin(4πt/5)8π∞−∞=∞The power of this signal is also given byP∞= limT→∞12TZT−T| −2 cos(2πt/5)|2dt= 4 limT→∞12TZT−Tcos2(2πt/5)dt= 4 limT→∞12TZT−T1 + cos(4πt/5)2dt= limT→∞12T4t2+20 sin(4πt/5)8πT−T!= limT→∞12T4T2+20 sin(4πT/5)8π−(−4T2+20 sin(−4πT/5)8π)= limT→∞12T(4T+ 5 sin(4πT/5)) = 2Hence,x(t) is a power-signal.9
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(b)x[n] =n+ 20if−20≤n≤0−n+ 20if 1≤n≤200otherwise.Solution: Considering the fact that∑Nn=1n2=N(N+1)(2N+1)6, the energy of this signalcan be computed asE∞=∞Xn=−∞|x[n]|2=0Xn=−20(n+ 20)2+20Xn=1(−n+ 20)2=20Xn=0n2+19Xn=0n2=20×21×416+19×20×396= 2870 + 2470= 5340and the power of this signal is given byP∞=limN→∞12N+ 1NXn=−N|x[n]|2=limN→∞12N+ 10Xn=−20(n+ 20)2+20Xn=1(−n+ 20)2!=limN→∞12N+ 120Xn=0n2+19Xn=0n2!=limN→∞53402N+ 1= 0Hence, this signal is an energy-signal.10
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2. Compute the power of the following continuous time signal:x(t) = 3 cos57t+π4+jsin34t−π8Solution:Note that since this signal is periodic, we only need to compute its power over oneperiod of that. Hence,P=156πZ56π0|x(t)|2dt=156πZ56π09 cos257t+π4+ sin234t−π8dtNow first we simplify the first term in the integral (cos2(x) =1+cos(2x)2):156πZ56π09 cos257t+π4dt=956πZ56π01 + cos(107t+π2)2dt=956πZ56π012+cos(107t+π2)2dt=956π×56π×12+"956π×27 sin(107t+π2)10|56π0#=92+ 0 = 4.5Similarly we can simplify the second integral as (sin2(x) =1−cos(2x)2)156πZ56π0sin234t−π8dt=156πZ56π01−cos(32t+π4)2dt=156π×56π×12−"156π×22 sin(32t+π2)3|56π0#=12+ 0 = 0.5Hence, the overall power of the signalx(t) isP= 4.5 + 0.5 = 5.11
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3. For the following signals, Either prove that the signal is an even signal or prove that it is notan even signal.(a)x[n] =n2−cos(45πn)Solution:Note that in this case we havex[−n] = (−n)2−cos−45πn=n2−cos45πn=x[n]Hence, the above signal is even.(b)x[n] =n4−cos(25πn+π3)Solution:Note that in this case we havex[−n] = (−n)4−cos−25πn+π3=n4−cos−25πn+π3̸=n4−cos25πn+π3=x[n]Hence, we can’t argue that the signal is even. Now using the above analysis we can find acounter example. For instance, atn= 2 we havex[2] = 24−cos4π5+π3= 16−cos17π15= 15.08whilex[−2] = 24−cos−4π5+π3= 16−cos−7π15= 15.89therefore,x[2]̸=x[−2] and hence the signal is not even.12
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4. (10 pts) Consider the signalx(t) given byx(t) =−2t−8if−4< t <−23if−2< t <1t+ 1if 1< t <30otherwise.Find the following signaly(t) =x(−2t+ 4) usingboth analytical and geometric ap-proaches:Analytical approach:think of−2t+ 4 as a new variablet′and exploit the conditions forsignalx(t).(a) If−4< t′<−2 we havex(t′) =−2t′−8. Considering the fact thatt′=−2t+ 4 wehave−4< t′<−2⇐⇒ −4<−2t+ 4<−2⇐⇒ −8<−2t <−6⇐⇒3< t <4Hence, we obtain that for 3< t <4 we havey(t) =x(−2t+ 4) =x(t′) =−2t′−8 =−2(−2t+ 4)−8 = 4t−16(b) If−2< t′<1 we havex(t′) = 3. Considering the fact thatt′=−2t+ 4 we have−2< t′<1⇐⇒ −2<−2t+ 4<1⇐⇒ −6<−2t <−3⇐⇒1.5< t <3for which we havey(t) =x(−2t+ 4) =x(t′) = 3(c) If 1< t′<3 we havex(t′) =t′+ 1. Considering the fact thatt′=−2t+ 4 we have1< t′<3⇐⇒1<−2t+ 4<3⇐⇒ −3<−2t <−1⇐⇒0.5< t <1.5for which we havey(t) =x(−t−3) =x(t′) =t′+ 1 = (−2t+ 4) + 1 =−2t+ 5(d) Ift′>3 ort′<−4 we havex(t′) = 0. Considering the fact thatt′=−2t+ 4 we havet′>3 ort′<−4⇐⇒ −2t+ 4>3 or−2t+ 4<−4⇐⇒t <0.5 ort >4for which we havey(t) =x(−2t+ 4) =x(t′) = 0(e) Therefore, we havey(t) =x(−2t+ 4) =4t−16if 3< t <4,3if 1.5< t <3,−2t+ 5if 0.5< t <1.5,0otherwise.13
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Geometric approach:Shift, flip, scale! Note thatx(t) can be plotted asFirst, shift by 4 to the left:Then flip since there is a negative sign beforetThen scale by half since there is a 2 coefficient beforet14
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