MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 Problem set SolutioN Exercise for home study o&W8.35 (a) From the system diagram, we see that a(t)=a(t)cos(wet) Using the multiplication property ZGu)= o(X(u)* FT(cos wct)) FT of cos wt is two impulses with area T at +wc. Therefore, Z(w)is the spectrum of X (jw)shifted to be centered at w and -wc and scaled by 2Xm=3. The real and imaginary parts of Z(w) are shown below: Relzloy 1/2 1/2 l/2 To find FT of p(t), let us first define a signal, q(t), such that q(t)
� MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 8 Solution Exercise for home study O&W 8.35 (a) From the system diagram, we see that z(t) = x(t) cos(ωct) Using the multiplication property 1 Z(jω) = 2π (X(jω) ∗FT {cos wct}) FT of cos wct is two impulses with area π at ±wc. of X(jω) shifted to be centered at ωc and −ωc and scaled by Therefore, Z(jw) is the spectrum 1 2 1 2π ×π = . The real and imaginary parts of Z(jω) are shown below: Re{ Z(j ω)} 1/2 1/2 −ωc ω ω −ω −ω c m −ω +ω c m −ω ω c +ω ω m c c m Im{ Z(jω)} − 1/2 c m c m c m c 1/2 −ω +ω −ω −ω −ω ω +ω ω ω m To find FT of p(t), let us first define a signal, q(t), such that: π 2, 2wc q(t) = 0, |w| ≤ π w > 2wc | | 1
Therefore, p()=(∑ Taking Fourier transform ∑ Following is the plot of P(ja). Sm(PGo)=0 PGo) Note that the impulse at the origin disappeared in the above graph because of the subtraction with 2(w) Now, using the multiplication property again, Y(u)=2(Z(u)* P(u) The real and imaginary part of Y(a) is shown below Ryujo) 2=m-2a2-2a2+om 20 20+00 Myo)/ 20c=
� � Therefore, � � � +∞ 2π p(t) = q(t) ∗ δ(t − n wc ) − 1 n=−∞ Taking Fourier transform, ∞ 4 sin( πω ) � P(jω) = 2ωc ωc δ(ω − kωc) − 2πδ(ω) ω k=−∞ Following is the plot of P(jw). m{P(jw)} = 0. P(jω) −2ω c −ωc −4ω c −3ωc −6ω c −5ωc ω c 2ω c 3ω c 4ω c 5ω c 6ω c ω 0 4 4 Note that the impulse at the origin disappeared in the above graph because of the subtraction with 2πδ(ω). Now, using the multiplication property again, Y (jω) = 2 1 π (Z(jω) ∗ P(jw)). The real and imaginary part of Y (jw) is shown below: Re{ Y(jω)} π −2ω 2ω 2 ω −2ω −ωm −2ωc+ωm −ωm ωm 2ωc−ωm 2ωc c c c +ωm 2 π Im{ } Y(j ) ω −2ω c−ωm −2ω c+ωm −ω m ωm 2ω c−ωm 2ω c+ωm ω 2
(b)To let r(t)=u(t), we must retain the frequency content between twm of Y (jw) and scale it properly. Therefore, Hw) is a low-pass filter with cutoff frequency at wm and gain =, as sketched belo HGO)
(b) To let x(t) = v(t), we must retain the frequency content between ±wm of Y (jω) and scale it properly. Therefore, H(jω) is a low-pass filter with cutoff frequency at ωm and gain π 2 , as sketched below: H(jω) π 2 ω −ω m ω m 3
Problem 1(o W734) X(eu) has characteristics as graphed below (where A is any real number) X(ejo) To let X(eju)occupy the entire range between and T, we need to scale the spectrum from-it to it by a factor of 3, and therefore we need to change the sampling rate by a factor of 3. Since we can only upsample and downsample by integer factors, we need to upsample by a factor of 3, and then downsample by a factor of 14 When we upsample by 3, we compress the spectrum of X(e?u) by a factor of 3, and then pass this through a low-pass filter with a gain of 3. The result is shown in the following gure 3A 14|74 Next, when we downsample the above spectrum by 14, we expand the spectrum by a factor of 14 and scale the height by 14, as graphed below Therefore,L=3 and M=14
Problem 1 (O & W 7.34) X(ejω) has characteristics as graphed below (where A is any real number) : X(e jω) 3π 14 − 3π 14 ω A −π 0 π 2π To let X(ejω) occupy the entire range between 3 , and therefore we need to change the sampling rate by a −π and π, we need to scale the spectrum from − to 3π 3π by a factor of 14 14 14 factor of 14 . Since we can only upsample and downsample by integer factors, we need to 3 upsample by a factor of 3, and then downsample by a factor of 14. When we upsample by 3, we compress the spectrum of X(ejω) by a factor of 3, and then pass this through a low-pass filter with a gain of 3. The result is shown in the following figure: j ω X (e ) p − 14 π 14 π ω 3A −π π 2π Next, when we downsample the above spectrum by 14, we expand the spectrum by a factor of 14 and scale the height by 14 , as graphed below: 1 Xd(e j ω) 3A 14 ω −π 0 π 2π Therefore, L = 3 and M = 14. 4
Problem 2 (a)a[n is a real-valued DT signal whose DTFT for -T8( 2rl)+6(+ DT) 0 Using the multiplication property from table 5.1, Zc(eu) is the periodic convolution of X(e) and DTFT(cos[won) over period 2m and then scaled by 2. We take one period, from to of DT won) and do regular convolution with X(e) Centered at w=0, we get the superposition of two X(eu) scaled by 3. Zc(eu)is shown below for the interval-丌to丌 Ze(e) 12
� Problem 2 (a) x[n] is a real-valued DT signal whose DTFT for −π < ω < π is given by X(ejω) 2 ω0 = 3π,ωM = π ✟✟✟✟ 4 ✟✟✟✟ ❍❍ 1 ✟✟✟ ❍❍❍❍❍❍❍❍❍ ω −π 0 ω0 π −ω0 − ωM −ω0 −ω0 + ωM ω0 − ωM ω0 + ωM Let zc[n] = x[n] cos[won] Using table 5.2 and taking Fourier transform of cos[won], +∞ DT FT {cos[won]} = π {δ(w − wo − 2πl) + δ(w + wo − 2πl)} l=−∞ DT FT {cos[won]} −π −ω0 0 ω0 π π π ω Using the multiplication property from table 5.1, Zc(ejw) is the periodic convolution on]} over period 2π and then scaled by 1 . We take one 2π of X(ejw) and DT FT {cos[w period, from −π to π, of DT FT {cos[won]} and do regular convolution with X(ejw). 1 Centered at w = 0, we get the superposition of two X(ejw) scaled by 2 . Zc(ejw) is shown below for the interval −π to π. Zc(ejw) 1 = 1 π × 2π 2 11π 5π 5π 11π −π− 12 − 12 −ωM 0 ωM 12 12 π ω 5
Z(e) is then passed through a low-pass filter with cut-off frequency wm and gain of 1. DTFT of Ten is shown below X(ea x)=丌-M uM丌(2x-M)2ru n=an sinon Using table 5.2 and taking Fourier transform of sinon D{nn}=∑{6(m-m-2)-b(+-2m)} We find Zs(ej) using the periodic convolution as before. The superposition terms centered at w=0 from X(e)(in dashed lines)are shown below. Adding the super- position terms, resulting Zs(eu) is shown for interval -T to T 12 Zs(eu) goes through the low-pass filter with cut-off frequency wM and gain of 1, we find DTFT of s[n] as shown below
� Zc(ejw) is then passed through a low-pass filter with cut-off frequency wM and gain of 1. DTFT of xc[n] is shown below. Xc(ejw) 1 2 −2π −(2π − ωM) −π −ωM 0 ωM π (2π − ωM ) 2π ω Let zs[n] = x[n] sin[won] Using table 5.2 and taking Fourier transform of sin[won], +∞ π [won]} = j DT FT {sin {δ(w − wo − 2πl) − δ(w + wo − 2πl)} l=−∞ We find Zs(ejw) using the periodic convolution as before. The superposition terms centered at w = 0 from X(ejw) (in dashed lines) are shown below. Adding the superposition terms, resulting Zs(ejw) is shown for interval −π to π. Zs(ejw) 5π 12 11π 12 −5π − 12 11π −π 12 −ωM π ωM 1 2j − 1 2j ω Zs(ejw) goes through the low-pass filter with cut-off frequency wM and gain of 1, we find DTFT of xs[n] as shown below. 6
Xs(eu) -WM (b) Maximum possible downsampling is achieved once the non-zero portion of one period of the discrete-time spectrum has expanded to fill the entire band from to T Therefore (c) Following is the system diagram to recover n co[n] ys 2 sin(won) After upsampling by m, we get back cn] and as n] from ye[n] and ys [n] respectively Note that upsampling by m has zero-insertion block(up-arrow m) and a low-pass filter for time-domain interpolation. DTFT of acn] and sn are derived in part a According to the system diagram on]=cn Using the multiplication property and doing periodic convolution, we get Xco(e/)as shown below
Xs(ejw) 1 2π −2π −π π −(2π − ωM) (2π − ωM ) −ωM ωM 2j − 1 2j ω (b) Maximum possible downsampling is achieved once the non-zero portion of one period of the discrete-time spectrum has expanded to fill the entire band from −π to π. Therefore, π π m = = = 4 w π M 4 (c) Following is the system diagram to recover x[n]. 2 cos(ω0n) yc[n] ↑ m × xco[n] ys[n] ↑ m × xso[n] + x[n] 2 sin(ω0n) After upsampling by m, we get back xc[n] and xs[n] from yc[n] and ys[n] respectively. Note that upsampling by m has zero-insertion block (up-arrow m) and a low-pass filter for time-domain interpolation. DTFT of xc[n] and xs[n] are derived in part a. According to the system diagram, xco[n] = xc[n] × 2 cos[won] Using the multiplication property and doing periodic convolution, we get Xco(ejw) as shown below. 7
×27×= Xso(e/u) Wo -uM Similarly, aso[]=Ts[n x 2 sin(won), and we get Xso(eu) as shown in the figure Adding Xco(e) and Xso(e/), we get back the spectrum of X(e). Thus, we recover
Xco(ejw) π × 2π × = 1 2 1 2 1 2 π −wo 0 wo π ω −wo − wM wo − wM Xso(ejw) π −wo wo π −wo − wM wo − wM 1 2 −1 2 ω Similarly, xso[n] = xs[n] × 2 sin[won], and we get Xso(ejw) as shown in the figure. Adding Xco(ejw) and Xso(ejw), we get back the spectrum of X(ejw). Thus, we recover x[n]. 8
Problem 3 a(t) u(-t)+2e-2a(t) Using Laplace transforms of elementary functions(table 9. 2), we find 1 R1=Re{s}<-1 R2= ReIsh s+2 Using the linearity property, ROC containing R1∩f2 Resh The pole-zero plot is shown below p-z map of X(s
Problem 3 (a) −2t x(t) = e−t u(−t)+2e u(t) Using Laplace transforms of elementary functions (table 9.2), we find, −t e u(−t) −1 ←→ , R1 = Re{s} −2 s + 2, Using the linearity property, 2 � X(s) = −1 + , ROC containing R1 R2 s + 1 s + 2 s = (s + 1)(s + 2) ROC = −2 < Re{s} < −1 The pole-zero plot is shown below: × −2 × −1 e m p-z map of X(s) 9
(b) (t)=(e cos t)u(-t)+u(-t) u(-t)+e(1-0(-t)+u(-t) Using Laplace transforms of elementary functions(table 9.2), we find 14+yu(-t) 1 (1+j) R1=Re{s}<1 1-)2(-t)→ R2=Re{s}<1 R3=Re{s}<0 USing the linearity property, 1/2 X(s) s-(+)8-(1-+s, ROC containing R1∩B∩R3 (1/2)s2+(1/2)s(1-j)-(1/2)s2+(1/2)s(1+j)-s2+2s s(s2-2s+2) ROC= Res)<o 1(2s2-3 8(-2+2)2BO=Rt<0 Solving the denominator, we see that poles are located at Solving the numerator. we see that zeros are located at 3 4 The pole-zero plot is shown below
� � � � (b) x(t)=(et cost)u(−t) + u(−t) t = e ( 1 ejt + 1 e−jt) u(−t) + u(−t) 2 2 1 1 (1+j)t u(−t) + e(1−j)t = e u(−t) + u(−t) 2 2 Using Laplace transforms of elementary functions (table 9.2), we find, (1+j)t −1 e u(−t) ←→ R1 = Re{s} < 1 s − (1+ j) , (1−j)t −1 e u(−t) ←→ R2 = Re{s} < 1 s − (1 − j) , u(−t) −1 ←→ R3 = Re{s} < 0 s Using the linearity property, −1/2 X(s) = −1/2 + + −1 , ROC containing R1 R2 R3 s − (1+ j) s − (1 − j) s −(1/2)s2 +(1/2)s(1 − j) − (1/2)s2 +(1/2)s(1+ j) − s2 + 2s − 2 = ROC= Re{s} < 0 s(s2 − 2s + 2) −1(2s2 − 3s + 2) = ROC= Re{s} < 0 s(s2 − 2s + 2) Solving the denominator, we see that poles are located at, s = 1 ± j s = 0 Solving the numerator, we see that zeros are located at, 3 √7 s = 4 ± j 4 The pole-zero plot is shown below: 10
