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BC:5.3 In the flow diagram shown in Fig. 3, the blocks with the "D" represent a one-step time delay. Find the difference equation in the form N M Σakyn -k] = bp x[n − p] k=0 p=0 where ao = 1. x[n] 40 -25 q[n] D -0.9 w[n] D ↑ 0.2 M y[n]

BC:5.3 In the flow diagram shown in Fig. 3, the blocks with the "D" represent a one-step time delay. Find the difference equation in the form N M Σakyn -k] = bp x[n − p] k=0 p=0 where ao = 1. x[n] 40 -25 q[n] D -0.9 w[n] D ↑ 0.2 M y[n]

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Problem BC:5.3** In the flow diagram shown in Figure 3, the blocks with the "D" represent a one-step time delay. Find the difference equation in the form: \[ \sum_{k=0}^{N} a_k y[n-k] = \sum_{p=0}^{M} b_p x[n-p] \] where \( a_0 = 1 \). **Flow Diagram Explanation:** The diagram consists of interconnected blocks and summing junctions as follows: 1. **Inputs and Outputs:** - Input: \( x[n] \) - Output: \( y[n] \) 2. **Blocks and Operations:** - The input \( x[n] \) is multiplied by 40, leading to the first summing junction. - The initial summation forms a signal \( q[n] \) which passes through a delay block labeled "D". - This delayed signal is then multiplied by \(-0.9\) and fed back into the first summing junction, completing the feedback loop for \( q[n] \). - Separately, another path multiplies \( x[n] \) by \(-25\), leading to a second summing junction, forming signal \( w[n] \). - This \( w[n] \) also passes through a delay block "D". - The delayed \( w[n] \) signal is multiplied by 0.2 and added as feedback into the second summing junction. - Both processed paths, \( q[n] \) and \( w[n] \), combine at the final summing junction resulting in the output \( y[n] \). **Figure Caption:** Figure 3: Flow diagram for Problem BC: 5.3 This diagram visually represents a linear, time-invariant digital filter system, focusing on how different components influence the system's response to the input signal.
Transcribed Image Text:**Problem BC:5.3** In the flow diagram shown in Figure 3, the blocks with the "D" represent a one-step time delay. Find the difference equation in the form: \[ \sum_{k=0}^{N} a_k y[n-k] = \sum_{p=0}^{M} b_p x[n-p] \] where \( a_0 = 1 \). **Flow Diagram Explanation:** The diagram consists of interconnected blocks and summing junctions as follows: 1. **Inputs and Outputs:** - Input: \( x[n] \) - Output: \( y[n] \) 2. **Blocks and Operations:** - The input \( x[n] \) is multiplied by 40, leading to the first summing junction. - The initial summation forms a signal \( q[n] \) which passes through a delay block labeled "D". - This delayed signal is then multiplied by \(-0.9\) and fed back into the first summing junction, completing the feedback loop for \( q[n] \). - Separately, another path multiplies \( x[n] \) by \(-25\), leading to a second summing junction, forming signal \( w[n] \). - This \( w[n] \) also passes through a delay block "D". - The delayed \( w[n] \) signal is multiplied by 0.2 and added as feedback into the second summing junction. - Both processed paths, \( q[n] \) and \( w[n] \), combine at the final summing junction resulting in the output \( y[n] \). **Figure Caption:** Figure 3: Flow diagram for Problem BC: 5.3 This diagram visually represents a linear, time-invariant digital filter system, focusing on how different components influence the system's response to the input signal.
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