Answer.AI
Find the inverse system for the given systems. If the system is in IIR form, also find the FIR form. Assume x n = 0 x_(n)=0 for n 0 n <= 0.
  1. y n = x n + x n 1 y_(n)=x_(n)+x_(n-1)
  2. y n = x n 1 2 x n 1 y_(n)=x_(n)-(1)/(2)x_(n-1)
  3. y n = x n x n 1 x n 2 y_(n)=x_(n)-x_(n-1)-x_(n-2)
  4. y n = i = n x n y_(n)=sum_(i=-oo)^(n)x_(n)
  5. y n = x n + y n 1 y_(n)=x_(n)+y_(n-1)
  6. y n = x n x n 1 + y n 1 y_(n)=x_(n)-x_(n-1)+y_(n-1)
See Answers
Get the Answer.AI App
Solve problem with AI
Best Answer
Step 1/6
  1. y n = x n + x n 1 y_(n)=x_(n)+x_(n-1)
To find the inverse system, we need to reverse the operations. In this case, we can subtract x n 1 x_(n-1) from y n y_(n) to get x n x_(n). Therefore, the inverse system is x n = y n x n 1 x_(n)=y_(n)-x_(n-1).
To convert this to FIR form, we need to express it as a finite sum. Since the system only depends on the current and previous input values, we can write it as x n = y n x n 1 = y n x n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-x_(n-1)=y_(n)-x_(n-1)+0x_(n-2)+0x_(n-3)+.... Therefore, the FIR form is x n = y n x n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-x_(n-1)+0x_(n-2)+0x_(n-3)+....
Step 2/6
2. y n = x n 1 2 x n 1 y_(n)=x_(n)-(1)/(2)x_(n-1)
Similarly, we can subtract 1 2 x n 1 (1)/(2)x_(n-1) from y n y_(n) to get x n x_(n). Therefore, the inverse system is x n = y n + 1 2 x n 1 x_(n)=y_(n)+(1)/(2)x_(n-1).
To convert this to FIR form, we can write it as x n = y n + 1 2 x n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)+(1)/(2)x_(n-1)+0x_(n-2)+0x_(n-3)+.... Therefore, the FIR form is x n = y n + 1 2 x n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)+(1)/(2)x_(n-1)+0x_(n-2)+0x_(n-3)+....
Step 3/6
3. y n = x n x n 1 x n 2 y_(n)=x_(n)-x_(n-1)-x_(n-2)
To find the inverse system, we can add x n 1 x_(n-1) and x n 2 x_(n-2) to y n y_(n) to get x n x_(n). Therefore, the inverse system is x n = y n + x n 1 + x n 2 x_(n)=y_(n)+x_(n-1)+x_(n-2).
To convert this to FIR form, we can write it as x n = y n + x n 1 + x n 2 + 0 x n 3 + 0 x n 4 + . . . x_(n)=y_(n)+x_(n-1)+x_(n-2)+0x_(n-3)+0x_(n-4)+.... Therefore, the FIR form is x n = y n + x n 1 + x n 2 + 0 x n 3 + 0 x n 4 + . . . x_(n)=y_(n)+x_(n-1)+x_(n-2)+0x_(n-3)+0x_(n-4)+....
Step 4/6
4. y n = i = n x n y_(n)=sum_(i=-oo)^(n)x_(n)
This system is a cumulative sum of the input values. To find the inverse system, we need to find the difference between consecutive output values. Therefore, the inverse system is x n = y n y n 1 x_(n)=y_(n)-y_(n-1).
To convert this to FIR form, we can write it as x n = y n y n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-y_(n-1)+0x_(n-2)+0x_(n-3)+.... Therefore, the FIR form is x n = y n y n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-y_(n-1)+0x_(n-2)+0x_(n-3)+....
Step 5/6
5. y n = x n + y n 1 y_(n)=x_(n)+y_(n-1)
To find the inverse system, we can subtract y n 1 y_(n-1) from y n y_(n) to get x n x_(n). Therefore, the inverse system is x n = y n y n 1 x_(n)=y_(n)-y_(n-1).
To convert this to FIR form, we can write it as x n = y n y n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-y_(n-1)+0x_(n-2)+0x_(n-3)+.... Therefore, the FIR form is x n = y n y n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-y_(n-1)+0x_(n-2)+0x_(n-3)+....
Answer
6. y n = x n x n 1 + y n 1 y_(n)=x_(n)-x_(n-1)+y_(n-1)
To find the inverse system, we can subtract y n 1 y_(n-1) from y n y_(n) and add x n 1 x_(n-1) to get x n x_(n). Therefore, the inverse system is x n = y n y n 1 + x n 1 x_(n)=y_(n)-y_(n-1)+x_(n-1).
To convert this to FIR form, we can write it as x n = y n y n 1 + x n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-y_(n-1)+x_(n-1)+0x_(n-2)+0x_(n-3)+.... Therefore, the FIR form is x n = y n y n 1 + x n 1 + 0 x n 2 + 0 x n 3 + . . . x_(n)=y_(n)-y_(n-1)+x_(n-1)+0x_(n-2)+0x_(n-3)+....
You might be interested in...
Explore more...
Get the Answer.AI App
Solve problem with AI