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 -T<w< T is given by X(e) -Wo tWM wo t WM ac[n]=an] cos[won Using table 5.2 and taking Fourier transform of coswonI DTFTicoswonJ=T>8( 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