Signals & Systems - Chapter 6

Signals & Systems - Chapter 6 S. A real-valued signal x( is knon to be uniquely determined by its samples hen the sampling frequeny is s = 0,000π. For hat values of is (j) guaranteed to be zero? From the Nyquist sampling theorem, it is kno that (j) = 0 for > s /. In other ord signal frequenies above s / are not reoverable. Therefore: anser is any frequeny suh that > 5,000π U. A real-valued signal x( is knon to be uniquely determined by its samples hen the sampling frequeny is f s = 5,000. For hat values of is (j) guaranteed to be zero? S. A ontinuous-time signal x( is obtained at the output of an ideal lopass filter ith ut off frequeny =,000π. If impulse-train sampling is performed on x(, hih of the folloing sampling periods ould guarantee that x( an be reovered from its sampled version using a appropriate lopass filter? a) T = 0.5 x 0-3 Se. b) T = x 0-3 Se. ) T = 0-4 Se. So Sampling period, T S, < (π/ m )= π/,000π = x 0-3 Seonds a and meet this ondition. U. A ontinuous-time signal x( is obtained at the output of an ideal lopass filter ith ut off frequeny =,500. If impulse-train sampling is performed on x(, hih of the folloing sampling periods ould guarantee that x( an be reovered from its sampled version using a appropriate lopass filter? a) T =.0 x 0-3 Se. b) T =0.5 x 0-3 Se. ) T = 0-4 Se. H lp (j) - 0 3S. The frequeny hih, under the sampling theorem, must be exeeded by the sampling frequeny is alled the Nyquist rate. Determine the Nyquist rate orresponding to eah of the folloing signals: a) x( = + os(,000π + sin(4,000π sin(4,000π b) x(= π t Signal ith maximum frequeny m =,000π pass through Sampling rate s > m =,000π.EngrCS.om, ik Signals and Systems page 64

sin(4,000π ) x(= π t Nyquist rate = x maximum signal frequeny Sampling Rate must exeed Nyquist rate in order to be able to fully reonstrut the signal. a) x( = + os(,000π + sin(4,000π The frequeny for eah term is a follos Term is DC = 0 Term =,000π Term 3 3 = 4,000π Maximum Signal Frequeny m = 4,000π Another ay of saying this is that (j) =0 for > 4,000π Sampling theorem says that s > m = 8,000π Therefore Nyquist rate is 8,000π sin(4,000π b) x( = π t Using Fourier Transform table, e have (j) = for < 4000π 0 for > 4000π Therefore Maximum Signal Frequeny m = 4,000π Sampling theorem says that s > m = 8,000π Therefore Nyquist rate is 8,000π sin(4,000π ) x( = π t sin(4,000π We an rerite the above funtion as x( = x ( x (( here x (( π t Using the Convolution property (j) = (/π) (j)* (j) We kno that onvolving a signal ith itself ill double the maximum frequeny therefore: Therefore Maximum Signal Frequeny m = 8,000π Sampling theorem says that s > m = 6,000π Therefore Nyquist rate is 6,000π 3U. The frequeny hih, under the sampling theorem, must be exeeded by the sampling frequeny is alled the Nyquist rate. Determine the Nyquist rate orresponding to eah of the folloing signals: a) x( = + os(3,000π + sin(6,500π.engrcs.om, ik Signals and Systems page 65

sin(,000π b) x(= π t sin(4,000π ) x(= π t 4S. Let x( be a signal ith Nyquist rate o. Determine the Nyquist rate for eah of the folloing signals: a) x( + x(t ) dx ( b) dt ) x ( d) x(os( o Nyquist rate = x maximum signal frequeny Sampling Rate must exeed Nyquist rate in order to be able to fully reonstrut the signal. a) y( = x( + x(t-) Fourier transform Y(j) = (j) + e -j (j) Sine the Maximum Frequeny for Y(j) is the same as (j) then y( Nyquist rate is also 0. dx ( b) y( = dt Fourier transform Y(j) =j(j) Sine the Maximum Frequeny for Y(j) is the same as (j) then y( Nyquist rate is also 0. ) y( = x ( We an rerite the above funtion as y( = x ( x(( Using the Convolution property Y(j) =(/π) (j)* (j) We kno that onvolving a signal ith itself ill double the maximum frequeny therefore: Therefore Y(j) =0 for > 0 in other ord Maximum Signal Frequeny m = 0 d) Therefore Nyquist rate is 0 FourierTra nsform ( j( 0 )) ( j( + y( = x( os( 0 Y ( j) = + Note: Use os Fourier transform and onvolution property to find Y(j) 0 )) We see that Y(j) = 0 hen > 0.+ 0 / sine (j)=0 hen > 0 / Therefore Nyquist rate = m = 3 0 4U. Let x( be a signal ith Nyquist rate o. Determine the Nyquist rate for eah of the folloing signals: a) x(- + x(t 3).EngrCS.om, ik Signals and Systems page 66

dx ( t 3) b) dt j0t ) x( e d) x(sin( o 5S. Let x( be a signal ith Nyquist rate o. Also, let y( -= x(p(t ) here p( = n= δ ( t nt) and π T < Speify the onstraints on the magnitude and phase of the frequeny response of a filter that gives x( as its output hen y( as the input. o Nyquist rate = x maximum signal frequeny Sampling Rate must exeed Nyquist rate in order to be able to fully reonstrut the signal. p( p( t ) FourierTransform Shifting Pr operty π T FourierTransform Sine y( = x(p(t-) k= π e T δ ( kπ / T ) j k = δ ( kπ / T ) = π T k = δ ( kπ / T ) e jk π / T jk π / T y( j) = ( )[ ( j) * FT{ p( t )} = ( j( kπ / T ) e π T k = Therefore Y(j) onsists of opies of (j) shifted by kπ/t and added together as shon belo (j) A - 0 / 0 / (A/T)e jπ/t Y(j) A/T (A/T)e -jπ/t.. -π/t - 0 / 0 / In order to reover x( from y(, e need to be able to isolate one opy of (j) from Y(j). From the figure e see that if e multiply Y(j) ith filter H(j): π/t H(j) = T for 0 for >..EngrCS.om, ik Signals and Systems page 67

Where ( 0 /) < < (π/t) ( 0 /) 5U. Let x( be a signal ith Nyquist rate o. Also, let y( -= x(p(t 3) here p( = n= δ ( t nt) and π T < Speify the onstraints on the magnitude and phase of the frequeny response of a filter that gives x( as its output hen y( us the input. o 6S. In the system shon belo, to funtions of time, x ( and x (, are multiplied together, and the produt ( is sampled by a periodi impulse train. x ( is band limited to, and x ( is band limited to ; that is (j) = 0 for (j) = 0 for Determine the maximum sampling interval T suh that ( is reoverable from p ( through the use of an ideal lopass filter. x ( P( = δ (t nt ) n= ( p ( x ( (j) (j) - - ( = x (x ( W(j) = (/π){ (j) * (j)} We have the folloing fats: ) (j)=0 for > ) (j)=0 for > Convolution to signal ill result a signal that is non-zero ith at least on of the signals is non-zero Therefore: W(j)=0 for > ( + ) Nyquist rate = M = ( + ) hih is also the minimum sampling frequeny for the signal to be reoverable. Maximum sampling period = π / ( minimum sampling frequeny) = π / ( + ) = π / ( + ) 6U. In the system to funtions of time, x ( and x (, are multiplied together, and the produt ( is sampled by a periodi impulse train here:.engrcs.om, ik Signals and Systems page 68

x ( = d( e e t j500 t + 0e ) dt Cos(5000π Sin(000π t j000 π j 000t x ( = ) Determine the maximum sampling interval T suh that ( is reoverable from the samples. 7S. Determine hether eah of the folloing statement is true or false: a) The signal x( = u(t + T o ) u(t T o ) an undergo impulse-train sampling ithout aliasing, provided that the sampling period T < T o. b) The signal x( ith Fourier transform (j) = u( + o ) u( o ) an undergo impulse-train sampling ithout aliasing, provided that the sampling period T < π/ o. ) The signal x( ith Fourier transform (j) = u() u( - o ) an undergo impulse-train sampling ithout aliasing, provided that the sampling period T< π/ o. a) x( = u(t + T o ) u(t T o ) j) = e + πδ ( ) e j + jt jt0 ( + πδ ( j 0 Meaning that x( is not a band-limited signal ( M is not finite) therefore e an not sample it at a high enough rate so that it an be reonstruted. {Anser: False} b) (j) = u(+ 0 ) u(- 0 ) (j)=0 for > 0 x( is band limited ) Nyquist rate = M = 0 s > 0 for no aliasing (π/t s ) > 0 Therefore sampling period ithout aliasing is T s < (π/ 0 ) {Anser: True} ) First dra (j) and its onvolution ith Impulse train ith Sampling frequeny = π/t > 0 (j) 0 P(j)*(j) - 0 0 + 0 So if e Filter the x(p( through a lo pass filter ith the ut off frequeny of = 0 e an reover the signal. {Anser: True} 7U. Determine hether eah of the folloing statement is true or false: a) The signal x( = 7u(t + T o ) u(t T o ) an undergo impulse-train sampling ithout aliasing, provided that the sampling period T < 4T o. b) The signal x( ith Fourier transform (j) = u( + o ) u( o ) an undergo impulse-train.engrcs.om, ik Signals and Systems page 69

sampling ithout aliasing, provided that the sampling period T < π/ o. ) The signal x( ith Fourier transform (j) = 5u() u( - o /) an undergo impulse-train sampling ithout aliasing, provided that the sampling period T< π/ o. 8S. A signal x( ith Fourier transform (j) undergoes impulse-train sampling to generate n= x ( = x( nt ) δ ( t nt ) p here T=0-4. For eah of the folloing sets of onstraints on x( and/or (j). Does the sampling theorem guarantee that x( an be reovered exatly from x p (? a) (j) = 0 for > 5,000π b) (j) = 0 for > 5,000π ) {Real (j)} = 0 for > 5,000π d) x( is real and (j)=0 for > 5,000π e) x( is real and (j)=0 for < -5,000π f) (j)*(j)=0 for > 5,000π g) (j) =0 for > 5,000π For all the setion sampling frequeny is Ws = π/t = 0,000π. for signal to be reoverable x(max. Signal Frequeny, W M ) < Ws a) Maximum signal Frequeny = M = 5,000π W M = 0,000 π < Ws = 0,000π Therefore (j) is fully reoverable. b) Maximum signal Frequeny = M = 5,000π W M = 30,000 π > Ws = 0,000π Therefore (j) is not fully reoverable. ) Sine e do not have the imaginary portion of (j), e an determine Nyquist rate is indeterminate hih means e annot guarantee reovery. d) Maximum signal Frequeny = M = 5,000π W M = 0,000 π < Ws = 0,000π Therefore (j) is fully reoverable. e) Maximum signal Frequeny = M = 5,000π W M = 30,000 π > Ws = 0,000π Therefore (j) is not fully reoverable. f) Convolution property says that: (j) = 0 for > (j)*(j) = 0 for > Therefore in this problem: (j) = 0 for > 5,000π/ Maximum signal Frequeny = W M = 5,000π/ W M = 5,000 π < Ws = 0,000π Therefore (j) is fully reoverable. g) f s =0,000 s = 0,000π. Maximum signal Frequeny = M = 5,000π W M = 0,000 π < Ws = 0,000π Therefore (j) is fully reoverable. 8U. A signal x( ith Fourier transform (j) undergoes impulse-train sampling to generate n= x ( = x( nt ) δ ( t nt ) p.engrcs.om, ik Signals and Systems page 70

here T=x0-5. For eah of the folloing sets of onstraints on x( and/or (j), does the sampling theorem guarantee that x( an be reovered exatly from x p (? a) (j) = 0 for > 5,000π b) (j) = 0 for > 5,000π ) (j)*(j)=0 for > 30,000π d) (j) =0 for > 49,000π 9S. Using the folloing system in hih sampling signal is an impulse train ith alternating sign. x( p( x p ( H(j) y( p( - (j) - M M W H(j) -3π/ -π/ π/ 3π/ a) For < π/( M ), sketh the Fourier transform of x p ( and y( b) For < π/( M ), determine a system that ill reover x( from x p (. ) For < π/( M ), determine a system that ill reover x( from y(. d) hat is the maximum value of in relations to M for hih x( an be reovered from either x p ( or y(? a) We an rite p( = p ( p (t- ) here p ( = δ ( t k ) P ( j) = δ ( π / ) k = k = using the above information and the time shifting property e an rite P(j) as: π P(j) = P (j) e -j P (j) π πk πk j j jk P( j) = { δ ( ) δ ( ) e } here e = e = ( ) k = k = FT xp( = x( p( p( j) = [ ( j) * P( j) ] π k here =πk/.engrcs.om, ik Signals and Systems page 7

P(j) π/ -3π/ -π/ -π/ 0 π/ π/ 3π/ p(j) / -3π/ -π/ -π/ 0 π/ π/ 3π/ H(j) -3π/ -π/ -π/ 0 π/ π/ 3π/ Y(j)= p (j)h(j) / -3π/ -π/ -π/ 0 π/ π/ 3π/ b) reovering x( from x p ( Let s use: ) FT{os( 0 } = π[δ(- 0 ) + δ(- 0 ) ) Convolutions FT{x p (os(πt/ )} = (/π) p (j)*{ π[δ(-π/ ) + δ(-π/ )} os(πt/ ) H(j) x p ( x( - M M ) reovering x( from y( use similar proess as b..engrcs.om, ik Signals and Systems page 7

os(πt/ ) H(j) y( - M M x( d) As an be seen from figure in setion a, e an avoid aliasing by having M < π/ sine sampling rate is S = π/ and it has to be larger the M for guaranteed reover. 9U. Using the folloing system in hih sampling signal is an impulse train ith alternating sign. x( p( x p ( H(j) y( p( - (j) 4 W - M M H(j) -4π/ -π/ π/ 4π/ What is the maximum value of in relations to M for hih x( an be reovered from either x p ( or y(? 0S. The sampling theorem states that a signal x( must be sampled at a rate greater than its bandidth (or equivalently, a rate greater than tie its highest frequeny). This implies that if x( has a spetrum as indiated in figure (a) then x( must be samples at a rate greater than. Hoever, sine the signal has most of its energy onentrated in a narro band, it ould seem reasonable to expet that a sampling rate loer than tie the highest frequeny ould be used. A signal hose energy is onentrated in a frequeny band is often referred to as a bandpass signal. There are a variety of tehniques for sampling suh signals, generally referred to as bandpasssampling tehniques. To examine the possibility of sampling a bandpass signal at a rate less than the total bandidth, onsider the system shon in figure (b). Assuming the >, find the maximum value of T and the values of the onstant A, a and b suh that x r ( = x(..engrcs.om, ik Signals and Systems page 73

Figure (a) (j) Figure (b) p(= + δ (t nt ) k = x( x p ( H(j) x f ( - - p( T H(j) t A - b - a a b π We have that P ( j) = δ ( kπ / T ) T k = Sine x p ( = x(p( p ( j) = [ ( j) * P( j)] = δ ( kπ / T ) π T P(j) π/t k = -4π/T -π/t 0 π/t 4π/T p(j) /T -π/t 0 π/t π/t - Consider that aliasing does not ours hen (π/t ) > 0 Sampling period, T < π/ hih means T Max = π/ and result in folloing P (j):.engrcs.om, ik Signals and Systems page 74

p(j) /T -π/t 0 π/t π/t - In order to have x r ( = x( A= T, a = π/t & a = b - 0U. Consider the system shon in figure (b). Assuming the >, find the maximum value of T and the values of the onstant A, a and b suh that x r ( = x(. (a) (j) (b) p(= + δ ( t 4nT ) k = x( x p ( H(j) x f ( - - p( 4T H(j) t A - b - a a b S. The system shon belo onsists of a ontinuous-time LTI system folloed by a sampler, onversion to a sequene, and an LTI disrete-time system. The ontinuous-time LTI system is ausal and satisfies the linear, onstant-oeffiient differential equation dy ( + y dt The input x ( is a unit impulse δ( ( = x ( a) Determine y (. b) determine the frequeny response H(e j ) and the impulse response h[n] suh that [n] = δ[n]..engrcs.om, ik Signals and Systems page 75

x ( LTI System y ( Conversion of y[n] LTI impulse train to H(e j ) a sequene [n] p(= + δ (t nt ) k = y[n]=y (nt) a) x ( = δ( dy ( + y ( = δ ( dt Take F. T. of both side jy ( j) + Y Y ( j) = ( j) = IFT y j + ( = e t u( b) y ( = e y[ n] = y t u( ( nt ) = e j Y ( e ) = T j e e j W ( e given : H ( e ) = Y (( e FT given : [ n] = δ[ n] W ( e j H ( e ) = T e e Therefore h[ n] = δ[ n] e T nt j u[ n] j j δ[ n ] ) ; ) = e T j e ) = j U. A ontinuous-time LTI system is ausal and satisfies the folloing linear, onstant-oeffiient differential equation: dy( 3 + 5y( = x( dt x( System y( Determine y( and h( if input x( is a unit impulse δ(..engrcs.om, ik Signals and Systems page 76

S. A signal, x (, undergoes sampling using the folloing pulse train: T p ( = δ ( t ) 0,000 n= Explain if x ( an be fully reovered from the sampled signal here (j) =0 for > 8,000 rad/se. Ts = 0.000 Ws = π/t = 0,000 π Sampling Frequey Mag. F{x(x(}= (j)*(jw) =0 for >6,000 M = 6,000 Sine Ws < W M signal is NOT reoverable from the sampled data 3S. x( has a Nyquist rate of 0. Determine the Nyquist rate of the folloing signal: Solution y( = x(os( o Nyquist rate = x maximum signal frequeny Sampling Rate must exeed Nyquist rate in order to be able to fully reonstrut the signal. FourierTra nsform ( j( 0 )) ( j( + y( = x( os( 0 Y ( j) = + Note: Use os Fourier transform and onvolution property to find Y(j) We see that Y(j) = 0 hen > 0.+ 0 / sine Y(j)=0 hen > 5 0 / 0 )) Therefore Nyquist rate = m = 5 0 4S. What is the maximum alloable sampling period suh that the folloing signal an be reovered from the sampled signal? Solutions: x( = Cos (58π * sin(774π Note: * indiates onvolution (j) = π[δ(-58) + δ(+58)]+ π/j[δ(-774) + δ(+774)] = 0 for all T Wrong Approah This approah ould be orret if the onvolution as in time domain. = 58πt = 774πt M = (58π + 774π)=0000π S > M must be true for the signal to be reoverable π/t s > 0000π T s < /0000 se or 00 use. 5S. x p ( is a sampled signal from x( ith Fourier transform (j) as shon belo: m= x [ t] = x( nt ) δ ( t nt ) p.engrcs.om, ik Signals and Systems page 77

Determine the limits of sampling period that guarantees x( is reoverable ompletely from the signal x p ( hen (j)*(j)=0 for > 500π. Solutions: Based on Convolution property e kno that If (j)=0 for > then (j)*(j)=0 for > Therefore, e an onlude that x(j)=0 for >500π/ = 750π In order to avoid aliasing and be able to reover x( from the x p ( Sampling Frequeny = s > m = 500π π/t s > 500π T s < /750 Seonds 6S. Let x( be a signal ith Nyquist rate 000 rad/se. Determine the Nyquist rate for the folloing signal: x(os(3000 FourierTransform ( j( 3000)) y( = x( os(3000 Y ( j) = + Nyquist rate is M therefore x(j) = 0 hen > 000 rad/se. ( j( + 3000)) Therefore Y(j) = 0 hen > 000+ 3000 = 4000 rad/se. Therefore Nyquist rate = m = * 4000 = 8,000 rad/se..engrcs.om, ik Signals and Systems page 78