SOLUTIONS TO ECE 3084

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1 SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no held, a simple counerexample is sufficien as an explanaion. (a) The phase modulaor y ( ) = cos(384 + x( )). (i) YES MEMORYLESS: The oupu a any given insan of ime depends only on he inpu a ha same momen of ime; here is no dependence on eiher he pas or fuure inpus. (ii) YES CAUSAL: Since is memoryless (no dependence on pas or fuure), i is auomaically causal (no dependence on fuure). (iii) NOT INVERTIBLE: In order o be inverible, he inpu mus be recoverable from he oupu. For example, adding 2 o any inpu will no change he oupu. So looking a he oupu, we wouldn be able o deermine which signal was a he inpu. A specific example: Boh x ( ) = and x 2 ( ) = 2u( ) resul in he same oupu y( ) = cos(384). (iv) NOT LINEAR: I fails he zero-in zero-ou es. (v) NOT TIME-INVARIANT: : Le T = /542 be he period of he 542-Hz sinusoid cos(384). The oupu in response o x ( ) = 384(u( ) u( 4T)) will hus be:, when (, 4T ), y ( ) =. cos(384), when 6 (, 4T ) On he oher hand, if x 2 ( ) = x (.5T), hen he corresponding oupu will be:, when (.5T, 4.5T ), y 2 ( ) =. cos(384), when 6 (.5T, 4.5T ) The wo oupus y ( ) and y 2 ( ) are skeched a he op of he nex page. We see ha y 2 ( ) 6= x (.5T), and herefore he sysem is ime-varian. (The oupu depends no only on wha he inpu is, bu when i is applied.)

2 y ( ) 4T.5T y 2 ( ) 4.5T (vi) YES BIBO STABLE: Since cos(anyhing) is always bounded, he oupu of he sysem will always be bounded. So yes i is rue ha a bounded inpu will always resul in a bounded oupu. (b) The even-par exracor y 2 ( ) = (x( ) + x( ))/2. (i) NOT MEMORYLESS: The oupu a (for example) depends no only on he inpu a ime, bu also on he inpu a ime +. (ii) NOT CAUSAL: We can re-use he example from (i): The oupu a ime depends on he inpu a ime +. I looks ino he fuure. (iii) NOT INVERTIBLE: In order o be inverible, he inpu mus be recoverable from he oupu. Bu when we ake he even par of he signal, is odd par is los! A specific example: Boh x ( ) = 2 and x 2 ( ) = will resul in he same oupu (namely y( ) = 2 ). (iv) YES LINEAR: I passes he zero-in, zero-oupu es. I passes he double-in, double-ou es. In fac, i passes a full-blown / es, since he oupu in response o x( ) = x ( ) + x 2 ( ) is: y( ) = (x( ) + x( ))/2 = (x ( ) + x 2 ( ) + x ( ) + x 2 ( ))/2 = (x ( ) + x ( ))/2 + (x 2 ( ) + x 2 ( ))/2 = y ( ) + y 2 ( ). Therefore i is linear. (v) NOT TIME-INVARIANT:. The oupu in response o x ( ) = sin(2) is zero, since his is an odd funcion (is even par is zero). If he sysem were ime-invarian, he oupu would also be zero for any delayed version, say x 2 ( ) = x ( ). However, if we choose o be =.25 (one quarer of he sinusoid period), he delayed inpu would be x 2 ( ) = cos(2), whose even par is iself, no zero!

3 (vi) YES BIBO STABLE: If he inpu is bounded, so ha x( ) <B for all ime, hen he magniude of he oupu is: y( ) = x( ) + x( ) /2 x( ) /2 + x( ) /2 (his is he riangle inequaliy) < B/2 + B/2 < B. Therefore, a bounded inpu yields a bounded oupu. In oher words, he even par of a bounded signal is bounded. (c) The ampliude modulaor y 3 ( ) = x( )cos(2). (i) YES MEMORYLESS: The oupu a any given insan depends only on he inpu a ha same momen of ime; here is no dependence on eiher he pas or fuure inpus. (ii) YES CAUSAL: Since is memoryless (no dependence on pas or fuure), i is auomaically causal (no dependence on fuure). (iii) ESSENTIALLY INVERTIBLE: The sinusoid cos(2) has zero crossings a imes =.25 + k.5 for all inegers k.we can recover he inpu a any oher momen of ime simply by dividing he oupu by ha ime cos(2), namely x( ) = y( )/cos(2). The only inpu values ha are no recoverable are is values a he zero crossings. If he inpu were known o be coninuous, for example, his would no be a problem, we could easily fill in he missing values. If he inpu included a Dirac impulse ha happened o fall a one of he zero crossings, however, ha componen of he inpu would no be recoverable. (iv) YES LINEAR: I passes he zero-in zero-oupu es. I passes he double-in, double-ou es. In fac, he response o x( ) = x ( ) + x 2 ( ) is: y( ) = cos(2)x( ) = cos(2)(x ( ) + x 2 ( )) = y ( ) + y 2 ( ). Therefore, since i passes he full-blown / es, i is linear. (v) NOT TIME-INVARIANT: For example, he oupu in response o ( ) is ( ), while he oupu in response o (.25) is zero. (vi) YES BIBO STABLE: Muliplying a bounded inpu by cos(2) (which is also bounded o he range ±) will resul in a bounded oupu.

4 (d) x( ), when x( ) > The half-wave recifier y 4 ( ) = max{x( ), } =, when x( ) < _. A picure makes i easy o see wha his sysem does: I replaces any negaive values by zero, as illusraed in he following example: y( ) x( ) (i) YES MEMORYLESS: The oupu a any given insan depends only on he inpu a ha same momen of ime; here is no dependence on eiher he pas or fuure inpus. (ii) YES CAUSAL: Since is memoryless (no dependence on pas or fuure), i is auomaically causal (no dependence on fuure). (iii) NOT INVERTIBLE: In order o be inverible, he inpu mus be recoverable from he oupu. Bu his sysem discards he inpu whenever he inpu is negaive, and here is hus no way o recover hese negaive inpus from he oupu. For example, here are wo disinc inpus ha resul in he same oupu: x ( ) = u( ), and x 2 ( ) = 5u( 384). (iv) NOT LINEAR: Ineresingly, his is an example of a sysem ha passes he zero-in zero-ou limus es and he doubling he inpu doubles he oupu limus es, and ye is nonlinear. A specific couner example: The response of his sysem o he uni sep is he uni sep; bu he response of his sysem o he uni sep scaled by a = is zero! While x( ) = u( ) resuls in y( ) = u( ), x( ) = u( ) resuls in y( ) =. (If he sysem were linear, scaling he inpu by would produce he same oupu, only scaled by he same consan. The fac ha we ge a compleely differen oupu is wha makes his sysem nonlinear.) (v) YES TIME-INVARIANT: The oupu in response o x( ) = x ( ) is: y( ) = max{x ( ), } = y ( ). (vi) YES BIBO STABLE: The oupu is he same as he inpu, excep wih negaive values replaced by zero. Therefore, if he inpu is bounded, he oupu is bounded.

5 PROBLEM 2.2. A sysem is a mapping from an inpu signal x( ) o an oupu signal y( ). To help make precise saemens, le us denoe his ransformaion by he funcional noaion y( ) = T(x( )). A sysem T(. ) is said o be linear when i saisfies wo properies (for all possible inpus x ( ) and x 2 ( ), and for all possible scaling consans a): addiiviy: T(x ( ) + x 2 ( )) = T(x ( )) + T(x 2 ( )). homogeneiy: T (ax ( )) = at(x ( )). This problem explores he quesion of wheher we need boh properies, or wheher one implies he oher x 2 ( ), when x( ) 6= (a) Consider he sysem y( ) = x( )., when x( ) = Only one of he wo properies (addiiviy or homogeneiy) is saisfied. Which one? Presen your answer in wo pars: (i) explain why one of he properies is held; (ii) explain why he oher is no. To es for homogeneiy, we need o deermine he oupu in response o he scaled version of he inpu. In paricular, if he inpu is x( ) = ax ( ) for some nonzero a, he oupu will be: a 2 x 2 ( ) , when ax ( ) 6= ax y( ) = ( ) = a x 2 ( ) x ( ) = ay ( )., when ax ( ) =, when x( ) 6=, when x ( ) = (i) The sysem saisfies he homogeneiy propery! x ( ) (ii) Bu he sysem is no linear, because i does no saisfy he addiiviy propery. 2 y ( ) = T (x ( )) For example, consider he signals shown on he righ: 2 The response o he op signal x ( ) is he second signal y ( ). x 2 ( ) Similarly, he response o he hird signal x 2 ( ) is he fourh signal y 2 ( ). If he sysem were addiive, he response o he sum x ( )+ x 2 ( ) would be he sum of he responses o each, namely he black curve labeled T (x ( )) + T (x 2 ( )). Bu insead he response o he sum is he green curve shown a he boom, which is differen. The sysem is hus no addiive. 2 4 y 2 ( ) = T (x 2 ( )) 3 4 y ( ) + y 2 ( ) = T (x ( )) + T (x 2 ( )) T (x ( ) + x 2 ( )) 4

6 (b) I urns ou ha addiiviy implies homogeneiy (and hus implies lineariy) for he special case when he sysem inpu and oupu are resriced o be real-valued signals. (This is a helpful fac, if you ever find yourself working only wih real signals you can use addiiviy as he sole es for lineariy!) Bu addiiviy does no imply homogeneiy when compex signals are involved. For example, consider he conjugaion sysem y( ) = x*( ), whose oupu is he complex conjugae of he inpu. Only one of he wo properies (addiiviy or homogeneiy) is saisfied. Which one? As in par (a), presen your answer in wo pars: (i) explain why one of he properies is held; (ii) explain why he oher is no. (Noe ha he scaling consan a in he homogeneiy es can be a complex number here.) (i) The conjugaion sysem saisfies addiiviy, because he conjugae of he sum is he sum of he conjugaes: T (x ( ) + x 2 ( )) = (x ( ) + x 2 ( ))* = x *( ) + x 2 *( ) = T (x ( )) + T (x 2 ( )). (ii) The conjugaion sysem does no saisfy homogeneiy, because he scaling consan ges conjugaed oo: T (ax ( )) = (ax ( ))* = a*x *( ) 6= ax *( ) = at (x ( )).

7 PROBLEM 2.3. Consider a linear and ime-invarian (LTI) sysem whose impulse response is recangular, as shown below: x( ) h( ) y( ) For each inpu signal lised below, provide boh (i) an equaion and (ii) a skech of he corresponding oupu signal y( ): (a) x ( ) = u( ). Subsiuing x ( ) = u( ) ino he convoluion inegral proceeding mechanically: y( ) = h( )x ( )d = h( )u( )d = h( )d From here we can break he soluion down ino hree cases: Case : When <, he inegral reduces o h( )d ()d= Case 2: When (, ), he inegral reduces o h( )d= ( )d = Case 3: When >, he inegral reduces o h( )d= ( )d = Overall, he answer is hus:, y( ) =, for < for < < as skeched below:, for >, y( )...

8 (b) x 2 ( ) = u( ) u( 5). For he sake of variey (and o show you he differen sraegies), we will solve his problem in a more algebraic manner. The oupu is deermined by he convoluion: y( ) = x 2 ( ) h( ) = (u( ) u( 5)) (u( ) u( )) = u( ) u( ) u( 5) u( ) u( ) u( ) + u( 5) u( ) = r( ) r( ) r( 5) + r( 6) = r( ) r( ) r( 5) + r( 6), where we have inroduced r( ) = r( ) o denoe he uni ramp. Along he way, we exploied he fac ha he convoluion of he uni sep wih iself is he uni ramp, u( ) u( ) = r( ), and he convoluion of a delayed uni sep wih he uni sep is a delayed ramp, u( ) u( ) = r( ). Here is a skech: y( ) 5 6 Saniy checks: he convoluion of wo recangles having differen widhs is a rapezoid. he convoluion of wo signals ha boh sar a ime is a signal ha sars a ime. he convoluion of a duraion-5 signal wih a duraion- signal has duraion 5 + = 6. (c) x 3 ( ) = sin()( u( ) u( )). Subsiuing x 3 ( ) = sin()( u( ) u( )) ino he convoluion inegral yields: y( ) = x 3 ( )h( )d = ( u( ) u( ))sin()h( )d = sin()h( )d

9 From here we can break he soluion down ino four cases: Case h( ) sin() Case : When <, he facor h( ) is zero hroughou he inegraion window (he wo facors don overlap ), so ha he inegral reduces o zero. Case 2: When (, ), he facor h( ) effecively changes he upper inegraion limi o (he wo facors overlap only for < < ), so ha he inegral reduces o: y( ) = sin()h( )d sin()d= -- cos() Case 3: When (, 2), he facor h( ) effecively changes he lower inegraion limi o (he wo facors overlap only for < < ), so ha he inegral reduces o: y( ) = sin()h( )d sin()d Case 2 Case 3 h( ) Case 4 = -- cos(( )) + = -- cos() Case 4: When > 2, he facor h( ) is zero hroughou he inegraion window (no overlap), so ha he inegral reduces o zero. -- cos(), for < < 2 Overall, he answer is: y( ) =., for 6 (, 2) 2 (Using rig ideniy, anoher way o wrie he answer is y( ) = -- sin 2 (.5 ) for (, 2).) Here is he skech: y( ) h( ) sin() sin() sin() h( ) 2

10 PROBLEM 2.4. If he sep response of an LTI sysem is s( ) = e 5 u( ), wha is is impulse response h( )? Provide boh (i) an equaion and (ii) a skech of h( ). The impulse response is he response of he sysem o a Dirac impulse, so ha when ( ) goes in, we mus have h( ) coming ou: ( ) LTI SYSTEM h( ) Since he Dirac impulse is he derivaive of he uni sep, we can generae ( ) in he above diagram by feeding he uni sep ino a differenaor: u( ) d ( ) ---- LTI h( ). d SYSTEM Since convoluion is commuaive, we can swap he order of he wo LTI sysems wihou changing he oupu: u( ) LTI ---- d h( ) SYSTEM. d From his picure we conclude ha he impulse response can be found by differeniaing he oupu of he LTI sysem in response o he uni sep, i.e., he impulse response is he derivaive of he sep response: d d h( ) = ---- s( ) = ---- e 5 u( ). d d By he chain rule for differeniaing a produc (see Prob..3(b) for similar mah): h( ) = e 5 d ---- u( ) + d ---- e 5 u( ) d d = e 5 ( ) 5e 5 u( ) = ( ) 5e 5 u( ). Here is a skech: h( ) 5

11 PROBLEM 2.5. Suppose an inpu signal of he form x( ) = m u( ) is applied o an LTI sysem whose impulse response is h( ) = 2 u( ), as shown below: his skech is for he special case of m = 5, he shape of he inpu signal will be differen for oher values of m! x( ) h( ) y( ) The inpu signal depends on he parameer m ha is known o be a nonnegaive ineger bu is oherwise unspecified. The picure shows wha he inpu would look like only when m = 5, bu i could be any value in he se m {,, 2, 3,...}. (The firs wo cases have special names: When m =, he inpu reduces o he uni sep. When m =, he inpu reduces o he uni ramp.) Wha does he oupu signal look like? To help you ou I will give you he form of he oupu signal: The oupu signal will have he same form as he inpu signal, only wih a differen exponen parameer, and wih an exra scaling consan oo. In paricular, he oupu signal can be wrien as: y( ) = A B u( ). Find values for he consans A and B, expressed as a funcion of he parameer m. (Hin: As a saniy check, you migh separaely compue he sep response of his LTI sysem, and hen verify ha i agrees wih your answer for he special case when m =.) The convoluion inegral is: y( ) = x( )h( )d In he firs case, when <, here is no overlap, and he above inegral is zero. ( ) 2 case m hese skeches are for he special case of m = 5, he shape of he blue signal will be differen for oher values of m! In he second case, when >, boh facors of he inegrand are nonzero only over he window from < <, so ha he limis of he convoluion inegral can be changed: case 2 ( ) 2 m y( ) = m ( ) 2 d = m ( )d = 2 m d 2 m+ d + m+2 d m + 3 = + = m+3 m m + 3 m + 3 m + 2 m m + m m + 3 = m+3 2 A =, B =. m + m + 2m m + m + 2m + 3 m + 3