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 t0 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 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 t0 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 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 (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, 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 (4) Q(jω) is shown below: Q(jω) ω −2π 0 2π 1 Thus, C(jω) can be drawn as shown below: 5
C(u) (d)To compute c(t), we multiply b(t) with q(t). BGw)is the sum of two ideal frequency shifted unity-gain filters. Filters in the frequency domain become sinc functions in the time domain. In addition, a frequency shift of wo corresponds to multiplying by e-Jwot in the time domain hence b(t) rt + -/art sin 2t e- art sin2丌t sin 2It cos 4t Therefore c(t)=b(t)q(t)=cos 4t sin- 2rt Problem 3 O&W 4.44. In addition to parts(a) and(b), answer the following (c) Find the differential equation relating the input and output of this system Solution We are given the following equation that relates the output y(t) of a causal LTI system to the input a (t dy(t) dt-+10v(t) r(Tz(t-r)dr-a(t) where z(t)=e-t(t)+36(1) We want to find the frequency response H(u)=Y(w)/X(w)of this system. To do this we take the transform of each term in Equation 5. Because of linearity, to find the Fourier transform of the entire equation, we take the Fourier transform of each term in the equation. For the first term, du(), we use the differentiation property to find that x{0}乙,jYGa)
C(jω) ω −8π −4π 0 4π 8π 1 (d) To compute c(t), we multiply b(t) with q(t). B(jω) is the sum of two ideal frequencyshifted unity-gain filters. Filters in the frequency domain become sinc functions in the time domain. In addition, a frequency shift of ωo corresponds to multiplying by e −jωot in the time domain. Hence, b(t) = 1 2 e −j4πt sin 2πt πt + 1 2 e j4πt sin 2πt πt = cos 4πt sin 2πt πt . Therefore c(t) = b(t)q(t) = cos 4πt sin2 2πt π 2 t 2 . Problem 3 O&W 4.44. In addition to parts (a) and (b), answer the following . (c) Find the differential equation relating the input and output of this system. Solution: (a) We are given the following equation that relates the output y(t) of a causal LTI system to the input x(t). dy(t) dt + 10y(t) = Z ∞ −∞ x(τ )z(t − τ )dτ − x(t) (5) where z(t) = e −tu(t) + 3δ(t). We want to find the frequency response H(jω) = Y (jω)/X(jω) of this system. To do this we take the transform of each term in Equation 5. Because of linearity, to find the Fourier transform of the entire equation, we take the Fourier transform of each term in the equation. For the first term, dy(t) dt , we use the differentiation property to find that F dy(t) dt F←→ jωY (jω). 6
For the second term, we have 10r(w). For the next term, the integral, we recognize this integral as being the convolution of x(t) and a(t). Thus, the convolution property tells us that Fc(t)*2(t))+ X(w)z(ju) Finally, .(t)becomes X(w). The new equation in the frequency domain is juY(ju)+10Y(ju)=X(j)Z(ju)-X(ju) Algebraic manipulations are used to separate out Y(u)on the left side of the equation nd separate out X(u) on the right side of the equation. We then find H(j)Y(0j)_2(u)-1 10+ju We insert the function Z(w). By linearity, F((t)= Fle-tu(t)+F38(t).The Fourier transform for each of the terms can be found ino w's Table 4.2 4+3j 1+u This can be plugged into Equation( 6)to give Y 3+2 H(0u)=x(0)=0+11+0) (b) The impulse response can be found by doing a partial fraction expansion of H(u)as found in Equation(7) HGw) 3+2 17/9 1/9 (10+ju)(1+j) +Jw 10+ The inverse Fourier transform of the last two terms can be determined from table 4.2 of o& w to give a(t) (c)To find the differential equation relating the input to the output we go back to Equa- tion 7 and rewrite it as 3+2 HGw) Xou) (ju)2+llu+ After cross-multiplying we get
For the second term, we have 10Y (jω). For the next term, the integral, we recognize this integral as being the convolution of x(t) and z(t). Thus, the convolution property tells us that F{x(t) ∗ z(t)} F←→ X(jω)Z(jω). Finally, x(t) becomes X(jω). The new equation in the frequency domain is jωY (jω) + 10Y (jω) = X(jω)Z(jω) − X(jω). Algebraic manipulations are used to separate out Y (jω) on the left side of the equation and separate out X(jω) on the right side of the equation. We then find H(jω) = Y (jω) X(jω) = Z(jω) − 1 10 + jω (6) We insert the function Z(jω). By linearity, F{z(t)} = F{e −tu(t)} + F{3δ(t)}. The Fourier transform for each of the terms can be found in O & W’s Table 4.2. Z(jω) = 1 1 + jω + 3. = 4 + 3jω 1 + jω This can be plugged into Equation( 6) to give H(jω) = Y (jω) X(jω) = 3 + 2jω (10 + jω)(1 + jω) (7) (b) The impulse response can be found by doing a partial fraction expansion of H(jω) as found in Equation( 7). H(jω) = 3 + 2jω (10 + jω)(1 + jω) = A 10 + jω + B 1 + jω = 17/9 10 + jω + 1/9 1 + jω . The inverse Fourier transform of the last two terms can be determined from Table 4.2 of O & W to give: h(t) = 1 9 e −t + 17 9 e −10t u(t). (c) To find the differential equation relating the input to the output we go back to Equation 7 and rewrite it as H(jω) = Y (jω) X(jω) = 3 + 2jω (jω) 2 + 11jω + 10 . After cross-multiplying we get 7
10Y(ju)+11jY(ju)+(ju)2Y(ju)=3X(ju)+2juX(ju) Using the differentiation property, we do an inverse Fourier transform to get the dif 10y(+1(),,2y(=3r()+2 dt Problem 4 O&W 521(c),(g) (c) We need to compute the Fourier transform of n]=5al-n-2. To do this we use the analysis equation as shown below X(eu 3 (g) We need to compute the Fourier transform of x/n= sin(n)+ cos(n). To do this we use Table 5.2 in o w to find the Fourier transform of each term and then sum the two transforms for a[) F{他m(2)=∑(60-2-20-6a+=2-2 cos(n T 2rl)+6(u+1-2r) )+6(+1-2l) Problem 5 The following are fourier transforms of discrete-time signals. Determine the signal corresponding to each transform (a) X(eu) e+6+8e-3-16e 1,0≤<票,<ll≤ 0.哥<同<吾 1+ X
10Y (jω) + 11jωY (jω) + (jω) 2Y (jω) = 3X(jω) + 2jωX(jω). Using the differentiation property, we do an inverse Fourier transform to get the differential equation 10y(t) + 11 dy(t) dt + 11 d 2 y(t) dt 2 = 3x(t) + 2 dx dt . Problem 4 O&W 5.21 (c), (g) Solution: (c) We need to compute the Fourier transform of x[n] = 1 3 |n| u[−n − 2]. To do this we use the analysis equation as shown below: X(e jω ) = X−2 n=−∞ 1 3 −n e −jωn = X∞ n=2 1 3 e jω n = 1 1 − 1 3 e jω − 1 − 1 3 e jω = 1 9 e j2ω 1 − 1 3 e jω (g) We need to compute the Fourier transform of x[n] = sin( π 2 n) + cos(n). To do this we use Table 5.2 in O & W to find the Fourier transform of each term and then sum the two transforms for x[n]. F n sin π 2 n o = π j X∞ l=−∞ (δ(ω − π 2 − 2πl) − δ(ω + π 2 − 2πl)) F{cos(n)} = π X∞ l=−∞ (δ(ω − 1 − 2πl) + δ(ω + 1 − 2πl)) X(e jω ) = π j X∞ l=−∞ (δ(ω− π 2 −2πl)−δ(ω+ π 2 −2πl))+π X∞ l=−∞ (δ(ω−1−2πl)+δ(ω+1−2πl)) Problem 5 The following are Fourier transforms of discrete-time signals. Determine the signal corresponding to each transform. (a) X(e jω ) = 4e j4ω − e jω + 6 + 8e −j3ω − 16e −j11ω (b) X(e jω ) = ( 1, 0 ≤ |ω| < π 4 , π 2 < |ω| ≤ π 0, π 4 < |ω| < π 2 (c) X(e jω ) = 1 + 3e −j3ω 1 + 1 4 e −jω 8
Solutions: (a)When the Fourier transform is a sum of exponentials, often the easiest way to find zn is to use the analysis equation to match each term to what the particular value of n is. For example, in this problem we can see that the first term, 4ej4w can only result from the n=-4 term. Thus, we can rewrite the equation as X(e)=x[-4le-(-4)+x-1]e-(-1)+x0e-10+x图3e-1(3)+l1le-1(1) We match terms to find r-4]=4,x{-1]=-1,x(0]=6,x(3=8,x[1]1=-16 n=0 for all other n (b) X(e) for this problem looks like: X(eu) The Fourier transform for this signal as the sum of 3 ideal low pass filters with two of them frequency-shifted to w=% and w=-. Table 5. 1 in o W shows that a frequency-shift corresponds to multiplication by an exponential in the time domain X(ej(wwwo))++ejuwonc[nI Thus. in the time domain we can write this as the sum of three sinc functions each with W=4 and with two of them multiplied by the appropriate exponential. This Sin -n sin -n +e乎nSin Sinl=n 2 cos(-n)+1
Solutions: (a) When the Fourier transform is a sum of exponentials, often the easiest way to find x[n] is to use the analysis equation X(e jω ) = X∞ n=−∞ x[n]e −jωn to match each term to what the particular value of n is. For example, in this problem we can see that the first term, 4e j4ω can only result from the n = −4 term. Thus, we can rewrite the equation as X(e jω ) = x[−4]e −jω(−4) + x[−1]e −jω(−1) + x[0]e −jω0 + x[3]e −jω(3) + x[11]e −jω(11) . We match terms to find x[−4] = 4, x[−1] = −1, x[0] = 6, x[3] = 8, x[11] = −16. x[n] = 0 for all other n. (b) X(e jω ) for this problem looks like: X(e jω ) ω −π − π π 2 π − 2 π 4 π 0 4 1 The Fourier transform for this signal as the sum of 3 ideal low pass filters with two of them frequency-shifted to ω = 3π 4 and ω = − 3π 4 . Table 5.1 in O & W shows that a frequency-shift corresponds to multiplication by an exponential in the time domain: X(e j(ω−ωo) ) F←→ e jωonx[n]. Thus, in the time domain, we can write this as the sum of three sinc functions each with W = π 4 and with two of them multiplied by the appropriate exponential. This gives x[n] = e −j 3π 4 n sin π 4 n πn + sin π 4 n πn + e j 3π 4 n sin π 4 n πn = 2 cos(3π 4 n) + 1 sin(π 4 n) πn . 9
(c)This Fourier transform looks similar to the Fourier transform for a"u[n] in Table 5.2 of o& w. We will manipulate it to look more like that. First, we separate it into two terms so that 3e-j3 1+ (e)+X2(e) +=e By linearity we can solve for the time signal for each transform and then a[n= 1n]+r2n]. X1(eu)matches to the transform pair mentioned above in Table 5.2 with a=-. Thus, X2(eu )matches to the transform pair in Table 5.2 also except for the numerator term of 3 This numerator term corresponds to a scaling of 3 and a time shift of 3 Thus r2]=3 n Summing the two terms, we get +3 Problem 6 Let X(e) denote the Fourier transform of the signal n] depicted below n (a)Find X(1)=X(e0) (b)Find a such that eauX(eu)is real (c) Evaluate∫x(ey)d (d)Find X(em)
(c) This Fourier transform looks similar to the Fourier transform for a nu[n] in Table 5.2 of O & W. We will manipulate it to look more like that. First, we separate it into two terms so that X(e jω ) = 1 1 + 1 4 e −jω + 3e −j3ω 1 + 1 4 e −jω = X1(e jω ) + X2(e jω ). By linearity we can solve for the time signal for each transform and then x[n] = x1[n]+x2[n]. X1(e jω ) matches to the transform pair mentioned above in Table 5.2 with a = − 1 4 . Thus, x1[n] = − 1 4 n u[n]. X2(e jω ) matches to the transform pair in Table 5.2 also except for the numerator term of 3e −j3ω . This numerator term corresponds to a scaling of 3 and a time shift of 3. Thus, x2[n] = 3 − 1 4 n−3 u[n − 3]. Summing the two terms, we get x[n] = − 1 4 n u[n] + 3 − 1 4 n−3 u[n − 3]. Problem 6 Let X(e jω ) denote the Fourier transform of the signal x[n] depicted below. x[n] n −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 −2 −1 2 2 −1 −2 (a) Find X(1) = X(e j0 ). (b) Find α such that e jαωX(e jω ) is real. (c) Evaluate R π −π X(e jω )dω. (d) Find X(e jπ ). 10