Solution_manual_discrete_time_signal_pro

Solutions - Chapter 2\nDiscrete-Time Signals and Systems 2.1. (a) T{x[n]} = g[n]x[n]\n\n* Stable: Let |h[n]| ≤ M then |T{y[n]}| ≤ |g[n]|M. So, it is stable if |y[n]| is bounded.\n* Causal: y[n] = g[n]x[n] and y[n] = g[n]x[n-n0], so if x[n] = 2x[n] for all n < n0, then y[n] = y[n] for all n < n0, and the system is causal.\n* Linear:\n\nT{(a1x[n] + b2x[n])} = g[n](a1x[n] + b2x[n])\n = ag[n]x[n] + bg[n]x[n]\n = aT{x[n]} + bT{z[n]}\n\nSo this is linear.\n• Not time-invariant:\n\nT{(z[n - n0])} = g[n]z[n - n0]\n ≠ y[n - n0] = g[n - n0]x[n - n0]\n\nwhich is not T.I.\n* Memoryless: y[n] = T{x[n]} depends only on the n-th value of z, so it is memoryless.\n\n(b) T{x[n]} = Σ_n a[k]\n\n* Not Stable: |x[n]| ≤ M ⇒ |T{x[n]}| ≤ Σ_n |x[k]| ≤ n - n0|M. As n → ∞, T → ∞, so not stable.\n* Linear:\n\nT{(a1x[n] + b2x[n])} = Σ_k a1x[k] + b2x[n] = Σ_n x[n]\n = aT{x[n]} + bT{(z[n])}\n\nThe system is linear.\n* Not T.I:\n\nT{(z[n - n0])} = Σ_k z[k - n0]\n = Σ_k z[k]\n\n≠ y[n - n0] = Σ_k z[k]\n\nThe system is not T.I.\n* Not Memoryless: Values of y[n] depend on past values for n > n0, so this is not memoryless. * T{x[n]} = Σ_n a[k]\n\n• Stable: |T{x[n]}| ≤ Σ_n |x[n]| ≤ 2n + 1|M| ≤ M, so stable.\n• Causal: If n ≥ 0, this is causal, otherwise it is not causal.\n* Linear:\n\nT{(a1x[n] + b2x[n])} = a1x[n] + b2x[n - n0]\n = aT{x[n]} + bT{(z[n])}\n\nThis is linear.\n\n* T.I:\n\nT{(z[n - n0])} = z[n - n0] = y[n - n0]\n\nThis is T.I.\n\n* Not memoryless: The values of y[n] depend on the n0 other values of z, not memoryless.\n\n(d) T{x[n]} = z[n - n0]\n* Stable: |T{x[n]}| = |z[n - n0]| ≤ M if |x[n]| ≤ M, so stable.\n* Linear:\n\nT{(a1x[n] + b2x[n])} = a1x[n - n0] + b2x[n - n0]\n = aT{x[n]} + bT{(z[n])}\n\nThis is linear.\n* T.I:\n\nT{(z[n - n0])} = z[n - n0] = y[n - n0]. This is T.I.\n\n* Not memoryless: Unless n0 = 0, this is not memoryless.\n\n(e) T{x[n]} = e[n]\n\n• Stable: |x[n]| ≤ M, |T{x[n]}| ≤ e[n]| ≤ e[n]| ≤ e^M, this is stable.\n• Causal: It doesn’t use future values of x[n], so it is causal.\n• Not linear:\n\nT{(a1x[n] + b2x[n])} = e[n] = a1x[n] + b2x[n] = e[n].\n = a1z[n - n0] + b2z[n - n0]\n\n≠ aT{x[n]} + bT{(z[n])}\n\nThis is not linear. • TI: T[x[n - n0]] = ax[n - n0] + b = y[n - n0]. It is TI.\n• Memoryless: y[n] depends on the n th value of x[n] only, so it is memoryless.\n\n(g) T[x[n]] = x[-n]\n• Stable: |T[x[n]]| ≤ |x[n]| ≤ M, so it is stable.\n• Not causal: For n < 0, it depends on the future value of x[n], so it is not causal.\n• Linear:\n T(a1[x[n1]) + b2[x[n2]]) = aT[x[n1]] + bT[x[n2]]\n = a(x[-n1] + b2x[-n2])\n ≠ aT[x[n1]] + bT[x[n2]]\n\nThis is linear.\n• Not TI:\n T[x[n - n0]] = x[-n + n0]\n ≠ y[n - n0] = x[-n + n0]\n\nThis is not TI.\n• Not memoryless: For n ≠ 0, it depends on a value of x other than the n th value, so it is not memoryless.\n\n(b) T[x[n]] = x[n] + u[n + 1]\n• Stable: |T[x[n]]| ≤ M + 3 for n > -1 and |T[x[n]]| ≤ M for n < -1, so it is stable.\n• Causal: Since it doesn't use future values of x[n], it is causal.\n• Not linear:\n T(a1[x[n1]] + b2[x[n2]]) = aT[x[n1]] + 3u[n + 1]\n ≠ aT[x[n1]] + bT[x[n2]]\n\nThis is not linear.\n• Not TI:\n T[x[n - n0]] = x[n - n0] + 3u[n + 1]\n = y[n - n0]\n = x[n - n0] + 3u[n - n0 + 1]\n\nThis is not TI.\n• Memoryless: y[n] depends on the n th value of x only, so this is memoryless. 2.2. For an LTI system, the output is obtained from the convolution of the input with the impulse response of the system:\n\n y[n] = ∑ h[k]x[n - k] k = -∞ to ∞ \n\n(a) Since h[k] ≠ 0, for (N0 ≤ n ≤ N1), \n y[n] = ∑ h[k]x[n - k] n1≤ n ≤ n2 \n\nThe input, x[n] ≠ 0, for (N2 ≤ n ≤ N3), so \n z[n] = k h[n - k] ≠ 0, for N2 ≤ (n - k) ≤ N3. Note that the minimum value of (n - k) is N2. Thus, the lower bound on n, which occurs for \n k = N0 is\n N4 = N0 + N2.\n N5 = N1 + N2.\n\nTherefore, the output is nonzero for\n (N0 + N2) ≤ n ≤ (N1 + N2).\n\n(b) If x[n] ≠ 0, for some n0 ≤ n ≤ (n0 + N1 - 1), and h[n] ≠ 0, for some n1 ≤ n ≤ (n1 + M - 1), the results of part (a) imply that the output is nonzero for:\n (n0 + n1) ≤ n ≤ (n0 + n1 + M + N - 2)\n\nSo the output sequence is M + N - 1 samples long. This is an important quality of the convolution for finite length sequences as we shall see in Chapter 8.\n\n2.3. We desire the step response to a system whose impulse response is\n h[n] = a*u[-n], for 0 < a < 1.\n\nThe convolution sum:\n y[n] = ∑ h[k]x[n - k], k = -∞ to ∞ \n\nThe step response results when the input is the unit step:\n z[n] = u[n] = { 1, for n ≥ 0 0, for n < 0 }\n\nSubstitution into the convolution sum yields:\nFor n ≤ 0:\n y[n] = ∑ a*u[-k]\n = ∑ a^{k}\n = a^{n}*1/(1 - a)\n\nFor n > 0:\n y[n] = ∑ a^{k}\n = ∑ a^{k}\n = k=0 to ∞ 1/(1 - a) 8\n(c) Let z[n] = u[n] (unit step), then\nX(z) = 1\n1 - z^{-1}\n\nand\nY(z) = X(z) . H(z)\n= 2z^{-1}\n(1 - z^{-1})(1 - 2z^{-1})(1 - 3z^{-1}) .\nPartial fraction expansion yields\n\nY(z) = 1\n1 - z^{-1} - 4\n1 - 2z^{-1} + 3\n1 - 3z^{-1}.\nThe inverse transform yields:\n\ny[n] = u[n] - 4(2)n u[n] + 3(3)n u[n].\n2.6. (a) The difference equation:\n\ny[n] - 1/2 y[n - 1] = z[n] + 2z[n - 1] + z[n - 2]\n\nTaking the Fourier transform of both sides,\nY(e^{j ω }) (1 - 1/2 e^{-j ω }) = X(e^{j ω}) (1 + 2 e^{-j ω } + e^{-2j ω }).\nHence, the frequency response is\n\nH(e^{j ω }) = Y(e^{j ω }) / X(e^{j ω })\n= 1 + 2 e^{-j ω } + e^{-2j ω }\n1 - 1/2 e^{-j ω }\n\n(b) A system with frequency response:\nH(e^{j ω }) = 1 - 1/2 e^{-j ω } + e^{-j ω }\n1 + e^{-j ω } + 1 - e^{-2j ω } = Y(e^{j ω }) / X(e^{j ω })\n\ncross multiplying,\nY(e^{j ω }) (1 + 1/2 e^{-j ω } + 3 e^{-2j ω }) = X(e^{j ω }) (1 - 1/2 e^{-j ω } + e^{-j ω}),\nand the inverse transform gives\n\ny[n] + 1/2 y[n - 1] + 3/4 y[n - 2] = z[n] - 1/2 z[n - 1] + z[n - 3].\n2.7. z[n] is periodic with period N if z[n] = z[n + M] for some integer N.\n(a) z[n] is periodic with period 12:\nz[n + 12] = z[n + M] = z(n + 2k)\n==> 2πk = 2π/N, for integers k, N.\nMaking k = 1 and N = 12 shows that z[n] has period 12. 9\n(b) z[n] is periodic with period 8:\ne^{j (2πN)} = e^{j (2π(k + N))} = e^{j (2π(n + 2k))}\n==> 2πk = 2π/N, for integers k, N\n==> N = 3k / 4,\nfor integers k, N.\nThe smallest k for which both k and N are integers are is 3, resulting in the period N being 8.\n(c) z[n] = [sin(πn/5)/(πn)] is not periodic because the denominator term is linear in n.\n(d) We will show that z[n] is not periodic. Suppose that z[n] is periodic for some period N:\ne^{j (2πN)} = e^{j (2π(k + N))} = e^{j (2π(n + 2k))}\n==> 2πk = π/N, for integers k, N\n==> N = 2√2k, for some integers k, N\nThere is no integer k for which N is an integer. Hence z[n] is not periodic. 10\n2.8. We take the Fourier transform of both h[n] and z[n], and then use the fact that convolution in the time domain is the same as multiplication in the frequency domain.\n\nH(e^{j ω }) = 5\n1 + j e^{j ω }\nY(e^{j ω }) = H(e^{j ω}) * X(e^{j ω})\n= 5\n1 + j e^{-j ω} - 1 - j e^{-2j ω}\n= 3\n2\n\ny[n] = 2(3)n u[n] + 3(-2)^{n} u[n].\n2.9. (a) First the frequency response:\nY(e^{j ω }) - 5/6 - y Y(e^{j ω }) + 1/6 - 2y Y(e^{j ω }) = 1/3 e^{-j ω} X(e^{j ω })\n\nH(e^{j ω }) = Y(e^{j ω }) / X(e^{j ω })\n= 1 - 1/3 e^{-j ω} + 2 Y(e^{j ω}) =Y(e^{j ω })/X(e^{j ω })\nNow we take the inverse Fourier transform to find the impulse response:\nH(e^{j ω }) = -2\n1 - j e^{-j ω} + 2 j e^{-j ω}\nh[n] = -2(3)i u[n] + 2(2)i^2 u[n].\n2.10. (e)\ny[n] = h[n] * e^{j n t}\n= ∑ a^{k} u[n - k] - 1 u[n - k]\n= ∑ a^{k}, n ≤ -1, n > -1\n= { -1/{a}, n ≤ -1\n 1/a, n > -1\n = { 1 -1/{a}\n 1/(1 - 1/a) . (b) First, let us define y[n] = 2^n u[n - 1]. Then, from part (a), we know that \nu[n] = u[n] * y[n] = { 2^n, n ≤ -1 1, n > -1 Now, y[n] = y[n - 4] * y[n] = u[n - 4] = { 2^n - 4, n ≤ 3 1, n > 3 (c) Given the same definitions for y[n] and y[n] from part(b), we use the fact that h[n] = 2^n * u[n - (n - 1)] = y[n] – 1 to reduce our work: y[n] = z[n] * h[n] = z[n] = y[n - 1] = u[n - 1] = { 2^n, n ≤ 0 1, n > 0 (d) Again, we use y[n] and y[n] to help us. y[n] = z[n] * h[n] = (u[n] = u[n - 10]) * y[n] = u[n] = u[n - 10] = (2^(n - 1) * (n - 1) + y[n]) - (2^n * y[n - 9]) + u[n - 10] { 2^(n - 1) - 2^(n - 9), n ≤ 2 1 - 2^(n - 9), -1 ≤ n ≤ 8 0, n ≥ 9 2.11. First we re-write z[n] as a sum of complex exponentials: z[n] = sin(nπ/4) = e^(jnπ/4) - e^(-jnπ/4)/2j Since complex exponentials are eigenfunctions of LTI systems, y[n] = H(e^(jω)n)e^jnπ/4 - H(e^(-jω))e^(-jnπ/4)/2j Evaluating the frequency response at ω = π/4: H(e^(jπ/4)) = 1 - e^(-jπ/4) 1 + 1/e^(jπ/4) = 2√2e^(-jπ/4) H(e^(-jπ/4)) = 1 - e^(jπ/4) 1 + 1/e^(-jπ/4) = 2√2e^(jπ/4) We get: y[n] = √2e^(-jπ/4)e^jnπ/4 - √2e^(jπ/4)e^(-jnπ/4)/2j = √2/2 sin(nπ/4 - π/4).