MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 5 SOLUTIONS Issued: October 22. 2003 Exercise for home study O&W4.47 tea this problem examines the Fourier transform of a continuous-time LTI system with a real, causal impulse response, h(t) (a) To prove that H(w) is completely specified by eh(u) for a real and causal h(t) lore the even part of a function, he(t) By definition, he(t)=3h(t)+2h(-t). Since h(t)=0 for t< 0, 2he(t) for t>0 h(t)=he(t) for t=0 0 for t<0 Therefore, if we know He(u), then we can find both he(t) and h(t). If h(t) was not causal, we couldn't determine h(t)from he(t)alone. We would need ho(t), the odd part of h(t)also. What is He(jw )? If h(t)is real then He(w)=ReH(ju). This is shown below He(ju)= he(t)e -judt h(t)e- jut dt+/ 5h(-t)e-yutdt h(t)e-yudt+ 5h(t)e3udt h(t)cos( at dt= Reh(u)) (b)Given we know Relhgu)) is cos w, we need to find h(t) Relhgu)= helu)=cow=0.5e/+0.5e-y h(t)=f{cos(u)}=F-{0.5-}+{0.5e-}=hal(t)+h2(t) For hel(t) we use the time shift property that for any to, e -Juto X (jw)+a(t-to) Thus for hel(t), to=-1. We have 1}=0.56(t+1)
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 5 Solutions Issued: October 22, 2003 Exercise for home study: O&W 4.47 This problem examines the Fourier transform of a continuous-time LTI system with a real, causal impulse response, h(t). (a) To prove that H(jω) is completely specified by <e{H(jω)} for a real and causal h(t), we explore the even part of a function, he(t). By definition, he(t) = 1 2 h(t) + 1 2 h(−t). Since h(t) = 0 for t < 0, h(t) = 2he(t) for t > 0 he(t) for t = 0 0 for t < 0 (1) Therefore, if we know He(jω), then we can find both he(t) and h(t). If h(t) was not causal, we couldn’t determine h(t) from he(t) alone. We would need ho(t), the odd part of h(t) also. What is He(jω)? If h(t) is real then He(jω) = <e{H(jω)}. This is shown below: He(jω) = Z ∞ −∞ he(t)e −jωt dt = Z ∞ −∞ 1 2 h(t)e −jωt dt + Z ∞ −∞ 1 2 h(−t)e −jωt dt = Z ∞ −∞ 1 2 h(t)e −jωt dt + Z ∞ −∞ 1 2 h(t)e jωt dt = Z ∞ −∞ h(t) cos(ωt)dt = <e{H(jω)}. (b) Given we know <e{H(jω)} is cos ω, we need to find h(t). <e{H(jω)} = He(jω) = cos ω = 0.5e jω + 0.5e −jω . he(t) = F −1 {cos(ω)} = F −1 {0.5e jω } + F −1 {0.5e −jω } = he1(t) + he2(t) For he1(t) we use the time shift property that for any to, e −jωtoX(jω) F←→ x(t − to). Thus for he1(t), to = −1. We have he1(t) = F −1 {0.5e −jω·−1 · 1} = 0.5δ(t + 1). 1
We find he2(t) using the same method We combine the two signals to get he(t)=0.50(t+1)+0.58(t-1). Finally, we know because h(t)is causal, our final answer is h(t)=2he(t)=8(t-1) (c We need to show that a real and causal h(t) can be recovered from ho(t) everywhere except at t=0. By definition, ho(t)=3h(t)-5h(-t). Because h(t) is caus ho(t)=sh(t)for t>0 and ho(t)=5h(t) for t< 0. This means that if we know ho(t), we know h(t)=2ho(t)when t>0 and h(t)=0 whent <0. However, at t=0 we have a problem. ho(0)=0 no matter what h(O)is. For example, if h(t)=8(t)+d(t-1) then ho(t)=0(t-1)+8(t-1). The delta function at t=0 was lost when we looked at the odd part of h(t) If we do not have a singularity at t=0, but instead has some arbitrary finite value at t=0, then the imaginary part of H( w) can be used to specify H(w). If we have h(t)= 1+u(t)for t=0 u()fort≠ Then H(w)=_o u(te-jutdt. The finite value of 1 at t=0 has no area so it do show up under the integral Ho(w)=2oo(h(t)-h(t)e udt=20-o h(t)e- yu-_o h(t)eJutdt)=-j -oo h(t)sin wt This shows that HoGjw)= SmfH(w). This also shows that HoGw)=HGjw) H(ju). Thus, H(w) can be recovered from Ho(w). Also the imaginary part can be used to find ho(t) which can be used to find h(t)everywhere except at t=0 Problems to be turned in: Problem 1 Consider the signal a(t)with spectrum depicted in Figure p4. 28(a)of O&w Sketch the spectrum of y(t)=(t)[cos(t/2)+cos(3/2)] Solution To draw Y(u), the spectrum of y(t), we use the linearity and the multiplication property. Thus, Y(jw)=X(u)* Fcos 50+X(u)* Ffcos 3) {s}=x[6(u-0.5)+6(u+0.5) F{cosa}=丌(u-1.5)+6(+1.5) Thu Y(j)=X(j)*丌(u-0.5)1 2 X(ju)*丌6(u+0.5)
We find he2(t) using the same method: he2(t) = F −1 {0.5e −jω·1 · 1} = 0.5δ(t − 1). We combine the two signals to get he(t) = 0.5δ(t + 1) + 0.5δ(t − 1). Finally, we know because h(t) is causal, our final answer is h(t) = 2he(t) = δ(t − 1). (c) We need to show that a real and causal h(t) can be recovered from ho(t) everywhere except at t = 0. By definition, ho(t) = 1 2 h(t) − 1 2 h(−t). Because h(t) is causal, ho(t) = 1 2 h(t) for t > 0 and ho(t) = − 1 2 h(−t) for t < 0. This means that if we know ho(t), we know h(t) = 2ho(t) when t > 0 and h(t) = 0 when t < 0. However, at t = 0 we have a problem. ho(0) = 0 no matter what h(0) is. For example, if h(t) = δ(t)+δ(t−1), then ho(t) = 1 2 δ(t − 1) + 1 2 δ(−t − 1). The delta function at t = 0 was lost when we looked at the odd part of h(t). If we do not have a singularity at t = 0, but instead has some arbitrary finite value at t = 0, then the imaginary part of H(jω) can be used to specify H(jω). If we have h(t) = 1 + u(t) for t = 0 u(t) for t 6= 0 (2) Then H(jω) = R ∞ −∞ u(t)e −jωtdt. The finite value of 1 at t = 0 has no area so it doesn’t show up under the integral. Ho(jω) = 1 2 R ∞ −∞(h(t)−h(−t))e −jωtdt = 1 2 ( R ∞ −∞ h(t)e −jωtdt− R ∞ −∞ h(t)e jωtdt) = −j R ∞ −∞ h(t)sin ωt. This shows that Ho(jω) = =m{H(jω)}. This also shows that Ho(jω) = 1 2H(jω) − 1 2H(−jω). Thus, H(jω) can be recovered from Ho(jω). Also the imaginary part can be used to find ho(t) which can be used to find h(t) everywhere except at t = 0. Problems to be turned in: Problem 1 Consider the signal x(t) with spectrum depicted in Figure p4.28 (a) of O&W. Sketch the spectrum of y(t) = x(t)[cos(t/2) + cos(3t/2)] . Solution: To draw Y (jω), the spectrum of y(t), we use the linearity and the multiplication property. Thus, Y (jω) = 1 2π [X(jω) ∗ F{cos t 2 }] + 1 2π [X(jω) ∗ F{cos 3t 2 }]. F{cos t 2 } = π[δ(ω − 0.5) + δ(ω + 0.5)]. F{cos 3t 2 } = π[δ(ω − 1.5) + δ(ω + 1.5)]. Thus, Y (jω) = 1 2π X(jω) ∗ πδ(ω − 0.5) + 1 2π X(jω) ∗ πδ(ω + 0.5) 2
X(ju)*r6(u-1.5)+X(j元u)*丌6(u+1.5) 2(x((u-05)+X(+0.5)+X0u-1.5)+X((u+15) o, X(w)is convolved with 4 shifted impulse functions. Convolving a signal with a shifted impulse function causes the signal to be shifted and replicated about the location on the whe we replicate X Gw). We also need to scale these 4 replications by a factor of i due to the multiplication of the constants, o. T. This can be seen in the figure below 0.5 -2.5-1.5 Problem 2 Consider the system depicted below: a() b(t) .c(t) H(w) c(t) p() gt) where r(t)rt,p(t)=cos 2nt, q(t)=sIn 2t sln4丌 and the frequency response of H (u)is 丌t gl
+ 1 2π X(jω) ∗ πδ(ω − 1.5) + 1 2π X(jω) ∗ πδ(ω + 1.5). = 1 2 (X(j(ω − 0.5)) + X(j(ω + 0.5)) + X(j(ω − 1.5)) + X(j(ω + 1.5))) So, X(jω) is convolved with 4 shifted impulse functions. Convolving a signal with a shifted impulse function causes the signal to be shifted and replicated about the location on the x-axis where the impulse function is located. Thus, centered at t = −1.5, −0.5, 0.5,and 1.5, we replicate X(jω). We also need to scale these 4 replications by a factor of 1 2 due to the multiplication of the constants, 1 2π · π. This can be seen in the figure below: X(jω) ω −2.5 −1.5 1.5 2.5 0.5 Problem 2 Consider the system depicted below: x(t) × p(t) a(t) H(jω) b(t) × q(t) c(t) where x(t) = sin 4πt πt , p(t) = cos 2πt, q(t) = sin 2πt πt , and the frequency response of H(jω) is given by H(jω) ω 1 −2π 2π 3
(a)Let A(u)be the Fourier transform of a(t). Sketch and clearly label Agw) (b) Let B(ju) be the Fourier transform of b(t). Sketch and clearly label B(jw) (c) Let C(ju) be the Fourier transform of c(t). Sketch and clearly label C(w) (d) Compute the output c(t) Solution (a) To find A(u), we use the multiplication property. Since a(t)=a(t)x p(t), then A(w)=2X(w)* P(jw)]. We need to find X(u)and P(w). To find X(w)from (t), we recognize a(t)as being in o& Ws Table 4.2 Basic Fourier Transform Pairs It is a sinc function with w= 4T. Thus XGw) 1 for wl 0 for Jwl XGw 0 4丌 Because p(t)=cos 2mt, PGu)=T[S(w-2T+8(w+2T). Since P(u) is two impulse functions, the convolution of X(w) with P(w)results in the superposition of two copies of X(w), one centered at w= 2T and the other centered at -2丌.The resulting A Gjw) is shown below 0.5
(a) Let A(jω) be the Fourier transform of a(t). Sketch and clearly label A(jω). (b) Let B(jω) be the Fourier transform of b(t). Sketch and clearly label B(jω). (c) Let C(jω) be the Fourier transform of c(t). Sketch and clearly label C(jω). (d) Compute the output c(t). Solution: (a) To find A(jω), we use the multiplication property. Since a(t) = x(t) × p(t), then A(jω) = 1 2π [X(jω) ∗ P(jω)]. We need to find X(jω) and P(jω). To find X(jω) from x(t), we recognize x(t) as being in O & W’s Table 4.2 Basic Fourier Transform Pairs. It is a sinc function with W = 4π. Thus, X(jω) = 1 for |ω| < 4π 0 for |ω| > 4π (3) X(jω) ω −4π 0 4π 1 Because p(t) = cos 2πt, P(jω) = π[δ(ω − 2π) + δ(ω + 2π)]. Since P(jω) is two impulse functions, the convolution of X(jω) with P(jω) results in the superposition of two copies of X(jω), one centered at ω = 2π and the other centered at ω = −2π. The resulting A(jω) is shown below: A(jω) ω −6π −2π 0 2π 6π 1 0.5 4
(b) To find B(w), we use the convolution property. Thus, B(w)=A(w)H(u). Ag is a low pass filter and H (w)is a high pass filter. Multiplying the two together cre- ates a bandpass filter. A(jw) cuts off all frequencies for wl (ju) cuts off all frequencies for w< 2. The resulting signal, B(u)is shown below BOw) 1.0 (c) To find C(w), we need to convolve B(u)with Q(w) 0 for w> 2, Q(w) is shown below 0 2丌 Thus, C(u) can be drawn as shown below
(b) To find B(jω), we use the convolution property. Thus, B(jω) = A(jω)H(jω). A(jω) is a low pass filter and H(jω) is a high pass filter. Multiplying the two together creates a bandpass filter. A(jω) cuts off all frequencies for |ω| > 6π. H(jω) cuts off all frequencies for |ω| < 2π. The resulting signal, B(jω) is shown below: B(jω) ω 0.5 1.0 −6π −2π 0 2π 6π (c) To find C(jω), we need to convolve B(jω) with Q(jω). Q(jω) = 1 for |ω| < 2π 0 for |ω| > 2π (4) Q(jω) is shown below: Q(jω) ω −2π 0 2π 1 Thus, C(jω) can be drawn as shown below: 5