100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Solution Manual - Digital Signal Processing: A Computer-Based Approach Fourth Edition ( Sanjit K. Mitra (Author) latest Editon $20.49   Add to cart

Exam (elaborations)

Solution Manual - Digital Signal Processing: A Computer-Based Approach Fourth Edition ( Sanjit K. Mitra (Author) latest Editon

 119 views  0 purchase
  • Course
  • Digital Signal Processing: A Computer-Based Approa
  • Institution
  • Digital Signal Processing: A Computer-Based Approa

Solution Manual Digital Signal Processing: A Computer-Based Approach Fourth Edition Sanjit K. Mitra (Author)

Preview 4 out of 593  pages

  • July 11, 2024
  • 593
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers
book image

Book Title:

Author(s):

  • Edition:
  • ISBN:
  • Edition:
  • Digital Signal Processing: A Computer-Based Approa
  • Digital Signal Processing: A Computer-Based Approa
avatar-seller
kylaexcell
Kylaexcell work 1 Solution Manual Digital Signal Processing: A Computer -Based Approach Fourth Edition Sanjit K. Mitra (Author) Kylaexcell work 2 0, 0,  0, 0, 0,  2 Chapter 2 2.1 (a) x1 1 = 22.85, x1 2 = 9.1396, x1  = 4.81, (b) x2 1 = 18.68, x = 7.1944, 2 x2  = 3.48. 2.2 [n] = 1,  n  0, n  0. Henceforth , [−n − 1] = 1,  n  0, n  0. n Thus, x[n] = [n] + [−n − 1]. 2.3 (a) Consider the sequence defined by x[n] =  [k]. k =− If n < 0, then k = 0 is not included in the sum and Henceforth , x[n] = 0 for n < 0. On the other hand, for n  0, k = 0 is included in the sum, and as a result, x[n] =1 for n  0. Therefore, x[n] = n [k] = 1, n  0, = [n].  k =− 0, n  0, (b) Since [n] = 1,  n  0, n  0, it follows that [n − 1] = 1,  n  1, n  1. Henc eforth , [n] − [n − 1] = 1,  n = 0, = [n]. n 0, 2.4 Recall [n] − [n − 1] = [n]. Henceforth , x[n] = [n] + 3[n − 1] − 2[n − 2] + 4[n − 3] = ([n] − [n − 1]) + 3([n − 1] − [n − 2]) − 2([n − 2] − [n − 3]) + 4([n − 3] − [n − 4]) = [n] + 2[n − 1] − 5[n − 2] + 6[n − 3] − 4[n − 4]. 2.5 (a) (b) c[n] = x[−n + 2] = {2 0 − 3  d[n] = y[−n − 3] = {− 2 7 8 − 2 1 0 − 1 5 − 4}, − 3 6 0 0},  (c) e[n] = w[−n] = {5 − 2 0 − 1 2 2 3 0 0},  (d) u[n] = x[n] + y[n − 2] = {− 4 5 1 − 2 3  − 3 1 0 8 7 − 2}, (e) v[n] = x[n]  w[n + 4] = {0 15 2 − 4 3 0  − 4 0}, (f) s[n] = y[n] − w[n + 4] = {− 3 4 − 5 0  0 10 2 − 2}, (g) r[n] = 3.5 y[n] = {21 − 10.5  − 3.5 0 2.8 24.5 − 7}. 2.6 (a) x[n] = −4[n + 3] + 5[n + 2] + [n + 1] − 2[n] − 3[n − 1] + 2[n − 3], y[n] = 6[n + 1] − 3[n] − [n − 1] + 8[n − 3] + 7[n − 4] − 2[n − 5], w[n] = 3[n − 2] + 2[n − 3] + 2[n − 4] − [n − 5] − 2[n − 7] + 5[n − 8], (b) Recall [n] = [n] − [n − 1]. Henceforth , x[n] = −4([n + 3] − [n + 2]) + 5([n + 2] − [n + 1]) + ([n + 1] − [n]) − 2([n] − [n − 1]) − 3([n − 1] − [n − 2]) + 2([n − 3] − [n − 4]) Kylaexcell work 3 h[0] + w[n] + z x[n _ 1] _ 1 11 z + w[n _ 1] _ 1 12 + z _ 1 z _ 1 x[n _ 2] 21 w[n 2] _ 22 = −4[n + 3] + 9[n + 2] − 4[n + 1] − 3[n] − [n − 1] + 3[n − 2] + 2[n − 3] − 2[n − 4], 2.7 (a) x[n] FROM the above figure it follows that y[n] y[n] = h[0]x[n] + h[1]x[n − 1] + h[2]x[n − 2]. (b) x[n] y[n] FROM the above figure we get w[n] = h[0](x[n] + 11x[n − 1] + 21x[n − 2]) and y[n] = w[n] + 12 w[n − 1] + 22w[n − 2]. we arrive at Making use of the first equation in the second y[n] = h[0](x[n] + 11x[n − 1] + 21x[n − 2]) + 12h[0](x[n − 1] + 11x[n − 2] + 21x[n − 3]) + 22h[0](x[n − 2] + 11x[n − 3] + 21x[n − 4]) = h[0](x[n] + (11 + 12 )x[n − 1] + (21 + 1211 + 22 )x[n − 2] + (1221 + 2211 )x[n − 3] + 2221x[n − 4]). (c) Figure P2.1(c) is a cascade of a first -order section and a second -order section. The input -output relation remains unchanged if the ordering of the two sections is interchanged as shown below. x[n] y[n] z _ 1 x[n-1] z _ 1 x[n-2] h[0] h[1] h[2] + + + w[n] 0.6 + u[n] + y[n+1] _ 1 _ 0.8 z z _ 1 0.3 0.4 + + w[n _ 1] z _ 1 _ 0.5 0.2 w[n _ 2] Kylaexcell work 4 The second -order section can be redrawn as shown below without changing its input - output relation. x[n] y[n] The second -order section can be seen to be cascade of two sections. Interchanging their ordering we finally arrive at the structure shown below: x[n] 0.6 _ 1 s[n] + + u[n] + _ 1 y[n+1] _ 1 z x[n _ 1] _ 1 0.3 + _ 0.8 + z u[n _ 1] _ 1 z 0.4 y[n] z x[n _ 2] 0.2 _ 0.5 z u[n _ 2] Analyzing the above structure we arrive at s[n] = 0.6x[n] + 0.3x[n − 1] + 0.2x[n − 2], u[n] = s[n] − 0.8u[n − 1] − 0.5u[n − 2], y[n + 1] = u[n] + 0.4 y[n]. FROM u[n] = y[n + 1] − 0.4 y[n]. Substituting this in the second equation we get after some algebra y[n + 1] = s[n] − 0.4 y[n] − 0.18y[n − 1] + 0.8y[n − 2]. Making use of the first equation in this equation we finally arrive at the desired input -output relation y[n] + 0.4 y[n − 1] + 0.18y[n − 2] − 0.2y[n − 3] = 0.6 x[n − 1] + 0.3x[n − 2] + 0.2 x[n − 3]. (d) Figure P2.19(d) is a parallel connection of a first -order section and a second -order section. The second -order section can be redrawn as a cascade of two sections as indicated below: x[n] y 2[n] + w[n] 0.6 + u[n] + y[n+1] _ 1 z _ 1 z _ 1 _ 0.8 z 0.3 0.4 + w[n _ 1] + z _ 1 z _ 1 _ 0.5 w[n _2] 0.2 + w[n] _ 1 _ 1 _ 0.8 z z 0.3 + w[n _ 1] + z _ 1 z _ 1 _ 0.5 w[n _2] 0.2

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller kylaexcell. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $20.49. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

62890 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$20.49
  • (0)
  Add to cart