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 $n\le 0$n <= 0.
1. ${y}_{n}={x}_{n}+{x}_{n-1}$y_(n)=x_(n)+x_(n-1)
2. ${y}_{n}={x}_{n}-\frac{1}{2}{x}_{n-1}$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. ${y}_{n}=\sum _{i=-\mathrm{\infty }}^{n}{x}_{n}$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)
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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. ${y}_{n}={x}_{n}-\frac{1}{2}{x}_{n-1}$y_(n)=x_(n)-(1)/(2)x_(n-1)
Similarly, we can subtract $\frac{1}{2}{x}_{n-1}$(1)/(2)x_(n-1) from ${y}_{n}$y_(n) to get ${x}_{n}$x_(n). Therefore, the inverse system is ${x}_{n}={y}_{n}+\frac{1}{2}{x}_{n-1}$x_(n)=y_(n)+(1)/(2)x_(n-1).
To convert this to FIR form, we can write it as ${x}_{n}={y}_{n}+\frac{1}{2}{x}_{n-1}+0{x}_{n-2}+0{x}_{n-3}+...$x_(n)=y_(n)+(1)/(2)x_(n-1)+0x_(n-2)+0x_(n-3)+.... Therefore, the FIR form is ${x}_{n}={y}_{n}+\frac{1}{2}{x}_{n-1}+0{x}_{n-2}+0{x}_{n-3}+...$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. ${y}_{n}=\sum _{i=-\mathrm{\infty }}^{n}{x}_{n}$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)+....
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)+....