MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 7 SOLUTION Exercise for home study O&W7.28 (a) Using the Fourier series coefficients of x(t), we can write swot k=-∞0 27 0.1se 20T rad/ The lowpass filter H(u) has a cutoff frequency wc=205T rad/ sec. Thus, c(t)is r(t) where all terms with frequency above we are removed by the lowpass filter. The terms which are kept have kwol 205T rad /sec k|< 10.25, so the output, ac(t),is r(t)= To obtain n, we sample c(t) every T=5 10-3 seconds with an impulse train The sampling frequency is 400T=2 x maximum frequency in c(t). Therefore, e can writ k=-10 ∑() where Wp=t0T=0.1丌rad Note that the complex discrete-time exponentials eJkuDn are all periodic with period N=2T/p= 20(although N is not the fundamental period for all of them). Hence n] must also be periodic with period N=20
� MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 7 Solution Exercise for home study O&W 7.28 (a) Using the Fourier series coefficients of x(t), we can write: �∞ � � k 1 | | x(t) = ejkw0t , 2 k=−∞ where 2π w0 = = 20π rad/sec. 0.1 sec The lowpass filter H(jw) has a cutoff frequency wc = 205π rad/sec. Thus, xc(t) is x(t) where all terms with frequency above wc are removed by the lowpass filter. The terms which are kept have |kw0| ≤ 205π rad/sec =⇒ |k| ≤ 10.25, so the output, xc(t), is � 10 � � k 1 | | xc(t) = ejkw0t 2 k=−10 To obtain x[n], we sample xc(t) every T = 5 × 10−3 seconds with an impulse train. The sampling frequency is 2 T π = 400π = 2 × maximum frequency in xc(t). Therefore, we can write, x[n] = xc(nT) 10 � � k 1 | | jkw0(nT) = e 2 k=−10 � 10 � � k 1 | | = ejkwDn, where wD = w0T = 0.1π rad. (1) 2 k=−10 Note that the complex discrete-time exponentials ejkwDn are all periodic with period N = 2π/wD = 20 (although N is not the fundamental period for all of them). Hence x[n] must also be periodic with period N = 20. 1
(b) To find the Fourier series representation for x[n], we rewrite Equation(1)in the form of a Fourier series synthesis equation: Note that there are 21 terms(k=-10.10 )in the left-hand side of Equation(2), but there are only 20 terms in the Fourier series on the right-hand side(as rn is periodic with N=20). Looking at the k= 10 and k=-10 terms k=10 (是)(e)=()(-1)n k (是) (-10)(0.1m)n We can add these two terms since they involve the same complex exponential, (-1)n Now we have the full set of Fourier series coefficients for an. The coefficients are periodic with period N= 20, and over the period -9 k 10 have values (是),因 2(是),k=10 O&W8.23 (a) The block diagram of modulation and demodulation for this problem is shown below () v() r(t) cos(oct) cos(@t) We want to show that the output of the lowpass filter in the demodulator is proporti to r(t)cos(Awt), where Aw=wd-wc. We show this algebraically as well as grap for the example in Figure P8.23. y(t) is just a(t)multiplied by cos(wct). Using the multiplication property, Y(w) consists of two shifted copies of X(ju), centered at Wc, with amplitude scaled by 1/ 2 Y(元u) [(w-wc))+X((w +we))I YOw) is sketched below
� � � (b) To find the Fourier series representation for x[n], we rewrite Equation (1) in the form of a Fourier series synthesis equation: 10 � � k 1 | | jkwDn jkwDn e = ake (2) 2 k=−10 k= Note that there are 21 terms (k = −10...10) in the left-hand side of Equation (2), but there are only 20 terms in the Fourier series on the right-hand side (as x[n] is periodic with N = 20). Looking at the k = 10 and k = −10 terms: k = 10 : � 1 �10 ej10(0.1π)n � 1 �10 (−1) = n � 1 �10 (ejπ) 2 2 n = 2 k = −10 : � 1 �10 ej(−10)(0.1π)n � 1 �10 = � 1 �10 (e−jπ)n = (−1)n 2 2 2 We can add these two terms since they involve the same complex exponential, (−1)n. Now we have the full set of Fourier series coefficients for x[n]. The coefficients are periodic with period N = 20, and over the period −9 ≤ k ≤ 10 have values � � k 1 | | , |k| ≤ 9 ak = � 2 1 �10 2 , k = 10. 2 O&W 8.23 (a) The block diagram of modulation and demodulation for this problem is shown below: -w co wco x(t) y(t) w(t) r(t) 2 ω cos(ωct) cos(ωdt) We want to show that the output of the lowpass filter in the demodulator is proportional to x(t) cos( ∆ωt), where ∆ω = ωd−ωc. We show this algebraically as well as graphically for the example in Figure P8.23. y(t) is just x(t) multiplied by cos(ωct). Using the multiplication property, Y (jω) consists of two shifted copies of X(jω), centered at ω = ωc and ω = −ωc, with amplitude scaled by 1/2: 1 Y (jω) = 2 [X(j(ω − ωc)) + X(j(ω + ωc))] Y (jω) is sketched below: 2
-0+ωm AtoM Similarly, w(u) will consist of two shifted and scaled copies of r(w) Wd)=2((-ca)+Y0(+) e))+XG(w-wd+wo))+X((w+ +xG(w+wd +we)) IOw d)+X(j(u-△u)+X(j(u+△u)+X(j( If we assume that the two copies centered at Aw and -Aw do not overlap, then w(o will look like the followin 1/4 △o-0m△o△oH+0 Finally, R(u)is wGw)passed through a lowpass filter with cutoff frequency between m+△uand2u+△-tm=ue+wd-ωim, This filtering process removes the high frequency content of w(w)so that only the two triangles at low frequencies remain Their amplitudes are scaled by 2, since the gain of the lowpass filter is 2. Therefore, R(w), the output of the demodulator, will be B()=2x((-△)+X((+△u which in the time domain gives us r It cos (b) From part(a), the spectrum of the output of the demodulator will be [X(j(u-△u)+X(j(u+△u) If the two copies of X(u) do not overlap, as shown in the figures above, this will look
Y(j ) ω −ωc−ωm −ωc −ωc+ωm ω c−ωm ω c ωc+ωm ω 1/2 1/2 Similarly, W(jω) will consist of two shifted and scaled copies of Y (jω): 1 W(jω) = [Y (j(ω − ωd)) + Y (j(ω + ωd))] 2 1 = [X(j(ω − ωd − ωc)) + X(j(ω − ωd + ωc)) + X(j(ω + ωd − ωc)) 4 +X(j(ω + ωd + ωc))] 1 = [X(j(ω − ωc − ωd)) + X(j(ω − ∆ω)) + X(j(ω + ∆ω)) + X(j(ω + ωc + ωd))] 4 If we assume that the two copies centered at ∆ω and −∆ω do not overlap, then W(jω) will look like the following: W(j ) ω −ω c −ω d ∆ω−ωm ∆ω+ωm ωc+ ωd 1/4 1/4 1/4 1/4 ω −∆ω ∆ω Finally, R(jω) is W(jω) passed through a lowpass filter with cutoff frequency between ωm + ∆ω and 2wc + ∆w − wm = ωc + ωd − ωm. This filtering process removes the high frequency content of W(jω) so that only the two triangles at low frequencies remain. Their amplitudes are scaled by 2, since the gain of the lowpass filter is 2. Therefore, R(jω), the output of the demodulator, will be 1 R(jω) = [X(j(ω − ∆ω)) + X(j(ω + ∆ω))] 2 which in the time domain gives us r(t) = x(t) cos( ∆ωt) (b) From part (a), the spectrum of the output of the demodulator will be 1 R(jω) = [X(j(ω − ∆ω)) + X(j(ω + ∆ω))] 2 If the two copies of X(jω) do not overlap, as shown in the figures above, this will look like: 3
RGO △①m△o△o+O1 This will happen if Aw WM. If this is not true, then the two copies will overlap, and the output will look like RGw)
∆ω−ωm ∆ω+ωm R(j ) ω 1/2 1/2 −∆ω ∆ω ω This will happen if ∆ω ≥ ωM . If this is not true, then the two copies will overlap, and the output will look like: R(jω) ✻ 1 2 ❅❅ ❅ ❅ ❅❅ ❅❅ ❅ ✲ ω −∆ω − ωm −∆ω ∆ω ∆ω + ωm 4
Problem 1 (a) We are given x(t)=cos(10t). Here, wo= 10 rad/ sec. Taking the Fourier transform of (t), X Gu) 10 The sampling function,s()=∑杜、6(t-k), with T=需 S Taking the Fourier transform of s(t)(note that ws=T=90) (90) Using the multiplication property, z(t)=a(t)s(t) in frequency domain is Z(w) o(XGu)*SGu)), i.e. we need to convolve X(u) with the periodic impulse train in (w)and scale the amplitude by 2r(see section 7.1.1 in O&W) ∑x(n)(t-n 1 XGe)sGlw-e)de
� � � Problem 1 (a) We are given x(t) = cos( 10t). Here, ωo = 10 rad/sec. Taking the Fourier transform of x(t), X(jw) −10 10 (π) (π) w 2π The sampling function, s(t) = +∞ δ(t − kT), with T = 90 . k=−∞ s(t) −2T −T 0 T 2T (1) ··· ··· t Taking the Fourier transform of s(t) (note that ωs = 2 T π = 90), S(jw) (90) ··· ··· −180 −90 0 90 180 w Using the multiplication property, z(t) = x(t)s(t) in frequency domain is Z(jw) = 2 1 π (X(jw) ∗ S(jw)), i.e. we need to convolve X(jw) with the periodic impulse train in S(jw) and scale the amplitude by 2 1 π (see section 7.1.1 in O&W). +∞ z(t) = x(nT)δ(t − nT) n=−∞ 1 +∞ Z(jw) = X(jθ)S(j(w − θ))dθ 2π −∞ 5
Therefore, Z((w) is as follows (j) 190-170 100-80 10 80100 1701900 (b)y(t)is the output from the band-pass filter, HGw), with input z(t)as derived in part (a). We know, Gw)= hgu)zg Let us consider Yw)l and zY(w) separately. Y(a)I is the band-pass filtered version of ZG)l with frequency components between 90 to 180 and-180 to-90 rad/sec YGu) 170 100 100 170 ∠Y(j ∠H(1u)+∠Z(u aU 200 200 Combining the magnitude and angle, Y(w)=Y(jw)lejzrgu Consider Y(w)as the Fourier transform of the sum of two sinusoidal signals; one with Wo= 100 and another with Wo= 170. Using the time-shifting property of Fourier transform,a(t-to)+e-3uto X(u) cos(100(t 200+=cos(170(t 45 17 cos(100t 2)+zc(17ot (c)Now the sampling function s(t)is changed with T= 90
Therefore, Z(jw) is as follows: Z(jw) (45) ··· ··· −190−170 −100 −80 −10 10 80 100 170 190 w (b) y(t) is the output from the band-pass filter, H(jw), with input z(t) as derived in part (a). We know, Y (jw) = H(jw)Z(jw) Let us consider |Y (jw)| and ∠Y (jw) separately. |Y (jw)| is the band-pass filtered version of |Z(jw)| with frequency components between 90 to 180 and -180 to -90 rad/sec. |Y (jw)| (45) −170 −100 0 100 170 w ∠Y (jw) = ∠H(jw) + ∠Z(jw) πw πw = − + 0 = − 200 200 Combining the magnitude and angle, Y (jw) = |Y (jw)|ej∠Y (jw) . Consider Y (jw) as the Fourier transform of the sum of two sinusoidal signals; one with wo = 100 and another with wo = 170. Using the time-shifting property of Fourier ←→ e transform, x(t − to) −jwtoX(jw), FT 45 π 45 π y(t) = cos(100(t − )) + cos(170(t − )) π 200 π 200 45 π 45 17π = cos(100t − ) + cos(170t − ) π 2 π 20 (c) Now the sampling function s(t) is changed with T = 2π 90 , 6
s(t) 6(t-kn)-∑6( kT Taking the Fourier transform 2 6( k=-0 k=-0 2 TT k=-0 k=-0 2丌 6( k=-0 2丌 ∑( k=-0 Seperating the odd and even terms of k T 6(-k -even k=even +>(0-k)+>60-62F ∑6(m r(t)=cos(10t)as before. To find Z(w), we need to convolve X(o) with the impulse train in S(ju)and scale the result by 2T SGa) is as sketched below
� � � � s(t) 0 (1) (−1) −T 2 T 2 −T T −3T 2 3T 2 −2T 2T ··· ··· t � � T s(t) = δ(t − kT) − δ(t − kT − ) ∞ ∞ 2 k=−∞ k=−∞ Taking the Fourier transform, ∞ ∞ 2π 2π 2π 2π e−jw T S(jw) = δ(w − k ) − 2 T T δ(w − k ) T T k=−∞ k=−∞ 2π �∞ 2π 2π ∞ T 2π 2 δ(w − k π e− T jk 2 T δ(w − k ) − T T = ) T k=−∞ k=−∞ � 2π 2π ∞ = 2π ∞ δ(w − k ) − (e−jπ) kδ(w − k 2π) T T T T k=−∞ k=−∞ � 2π 2π ∞ 2π ∞ � 2π = δ(w − k ) − (−1)kδ(w − k ) T T T T k=−∞ k=−∞ Seperating the odd and even terms of k, 2π � 2π 2π � 2π S(jw) = δ(w − k ) − δ(w − k ) T T T T k=even k=even 2π � 2π 2π � 2π + δ(w − k ) + δ(w − k ) T T T T k=odd k=odd 4π � 2π = δ(w − k ) T T k=odd x(t) = cos( 10t) as before. To find Z(jw), we need to convolve X(jw) with the impulse train in S(jw) and scale the result by 2 1 π . S(jw) is as sketched below, 7
(180) The convolution will place two scaled impulses(from X(a)) centered at each impulse in the impulse train of S(w). Finally, HGw) will only pass impulses that exist between 90 to 180 and -180 to-90 radians. We plot Y(w)l(output from H(w)) as follows 100 As derived in part(b),∠Y(1)=∠H(j)=-2 From the plot of Y(w)l and the zy(u), we can view y(t) as a time-shifted cos functions. Therefore y() 9os(100t-200 s(100t
S(jw) 0 (180) −270 −90 90 270 ··· ··· w The convolution will place two scaled impulses (from X(jw)) centered at each impulse in the impulse train of S(jw). Finally, H(jw) will only pass impulses that exist between 90 to 180 and −180 to −90 radians. We plot |Y (jw)| (output from H(jw)) as follows: |Y (jw)| (90) −100 0 100 w As derived in part (b), ∠Y (jw) = ∠H(jw) = −πw . 200 From the plot of |Y (jw)| and the ∠Y (jw), we can view y(t) as a time-shifted cos functions. Therefore, 90 π y(t) = cos(100(t − )) π 200 90 π = cos(100t − ) π 2 8
Problem 2(O&W 7.30 except let rc(t)=d(t-2)) (a) We are given wc(t) T X(0u) We take the Fourier transform of the system s differential equation and find the fre- quency response, HGw), of the system ye(t) jwC(o)+Y(u)= X gw) B()=0) Xlw) 1+jw Now we can write Y(0)=X(j)(il)=c-m31 (b)y[n= ye(nr)where ye(t)is as defined in part(a). Therefore, ye(nr)will pick-up values from ye(t) at nT time values with n=0, 1, 2, yn=ye(nT) -nT+2u (e2)(e-)( Using the time-shifting property of DTFT and basic DTFT table Y(e Now we choose H(e) such that y[n]*hn]= wn]=8[n H(eru) Taking the inverse FT hn=ein+l-ei8nl
Problem 2 (O&W 7.30 except let xc(t) = δ(t − T 2 )) (a) We are given xc(t) T xc(t) = δ(t − ) 2 Xc(jw)=e−jw 2 T We take the Fourier transform of the system’s differential equation and find the frequency response, H(jw), of the system. dyc(t) + yc(t) = xc(t) dt jwYc(jw) + Yc(jw) = Xc(jw) H(jw) = Yc(jw) Xc(jw) = 1 1 + jw Now, we can write, Yc(jw) = Xc(jw)H(jw)=e−jw 2 T 1 1 + jw yc(t) = e−(t− 2 T ) u(t − T ) 2 (b) y[n] = yc(nT) where yc(t) is as defined in part (a). Therefore, yc(nT) will pick-up values from yc(t) at nT time values with n = 0, 1, 2, ... 2 T e u[ −nT + y[n] = yc(nT) = n − 1] 2 T )(e−T )(e−T ) n−1 = (e u[n − 1] Using the time-shifting property of DTFT and basic DTFT table, jw) = e− 2 T e−jw 1 e−jw 1 − e−T Y (e Now we choose H(ejw) such that: y[n] ∗ h[n] = w[n] = δ[n] jw Y (e )H(ejw)=1 1 H(ejw) = e−jw (1 − e ) −T 2 T e−jw e− T T H(e 2 jw) = jw − e 2 − e e Taking the inverse FT, T T − e− h[n] = e δ[n + 1] δ[n] 9 2 2
Problem 3 First, we need to find frequency response of the DT filter, y[n]=iy[n-2+a[n]+a[n-1] When a/n]=8n, yIn]=h/n]. Therefore h/nl hn-2]+6{m]+76{n Hleny-3e-j2 H(en)+114e2 H(e 9|
� � Problem 3 First, we need to find frequency response of the DT filter, y[n] = 3 4 y[n−2]+x[n]+1x[n−1] 4 When x[n] = δ[n], y[n] = h[n]. Therefore, 3 1 h[n] = h[n − 2] + δ[n] + δ[n − 1] 4 4 3 1 e−j2ΩH(ejΩ H(ejΩ) = )+1+ e−jΩ 4 4 1 + 1 e−jΩ H 4 (ejΩ) = 1 − 3 e−j2Ω , |Ω| π T 1 4 −jωT 1+ e ωs |w| ≤ π T = π −j2ωT 3 , 4 H 2 c(jω) = 1− e 0, |w| > T 10